aqa core 3 revision booklet - douis. · pdf filecore 3 specifications candidates will be...

AQA ‐ Core 3 ‐ Revision booklet

Name: ……...............................................................................

Tutor group: .............................

AQA - Core 3 1

Key dates

Core 3 exam: 10th June 2014 am

Term dates:

Term 1: Monday 2 September 2013 – Friday 25

October 2013

Term 4: Monday 24 February 2014 ‐ Friday 4

April 2014

Term 2: Monday 4 November 2013 ‐ Friday 20

December 2013

Term 5: Tuesday 22 April 2014 ‐ Friday 23 May

2014

Term 3: Monday 6 January 2014 ‐ Friday 14

February 2014

Term 6: Monday 2 June 2014 ‐ Tuesday 22 July

2014

AQA - Core 3 2

Scheme of Assessment Mathematics Advanced Subsidiary (AS) Advanced Level (AS + A2)

The Scheme of Assessment has a modular structure. The A Level award comprises four compulsory Core units, one optional Applied unit from the AS scheme of assessment, and one optional Applied unit either from the AS scheme of assessment or from the A2 scheme of assessment. For the written papers, each candidate will require a copy of the AQA Booklet of formulae and statistical tables issued for this specification. All the units count for 331/3% of the total AS marks 162/3% of the total A level marks Written Paper 1hour 30 minutes 75 marks

Grading System The AS qualifications will be graded on a five-point scale: A, B, C, D and E. The full A level qualifications will be graded on a six-point scale: A*, A, B, C, D and E. To be awarded an A* in Mathematics, candidates will need to achieve grade A on the full A level qualification and 90% of the maximum uniform mark on the aggregate of the best three of the A2 units which contributed towards Mathematics. For all qualifications, candidates who fail to reach the minimum standard for grade E will be recorded as U (unclassified) and will not receive a qualification certificate.

CORE 3 subject content Algebra and Functions

Trigonometry Exponentials and Logarithms

Differentiation Integration

Numerical Methods

AQA - Core 3 3

Core 3 specifications

Candidates will be required to demonstrate:

a) construction and presentation of mathematical arguments through appropriate use oflogical deduction and precise statements involving correct use of symbols and appropriate connecting language;

b) correct understanding and use of mathematical language and grammar in respect ofterms such as "equals", "identically equals", "therefore", "because", "implies", "is implied by", "necessary", "sufficient" and notation such as , , and .

c) methods of proof, including proof by contradiction and disproof by counter‐example.

Candidates may use relevant formulae included in the formulae booklet without proof.

Candidates should learn the following formulae, which are not included in the formulae booklet, but which may be required to answer questions.

Trigonometry 2 2

2 2

sec 1 tan

cosec 1 cot

A A

A A

Differentiation

Volumes

Integration

AQA - Core 3 4

Algebra and functions Definition of a function. Domain and range of a function.

Notation such as f (x) = x2 − 4may be used. Domain may be expressed as x >1 for example and range may be expressed as f (x) > −3 for example.

Composition of functions. fg (x) = f (g (x))

Inverse functions and their graphs.

The notation f−1 will be used for the inverse of f. To include reflection in y = x.

The modulus function. To include related graphs and the solution from them of inequalities such as |x + 2|< 3|x| using solutions of |x + 2|=|3 x|.

Combinations of the transformations on the graph of y = f (x) as represented by y = af(x), y =f(x) + a, y = f(x + a), y = f(ax).

For example the transformations of: ex leading to e2x −1 ; Ln(x) leading to 2ln (x −1) ; sec x leading to 3sec2x Transformations on the graphs of functions included in modules Core 1 and Core 2.

Trigonometry Knowledge of sin‐1, cos‐1 and tan‐1 functions. Understanding of their domains and graphs.

Knowledge that –π/2 ≤ sin‐1x ≤π/2 ; 0≤ cos‐1x≤ π ; ‐π/2<tan‐1x< π/2. The graphs of these functions as reflections of the relevant parts of trigonometric graphs in y = x are included. The addition formulae for inverse functions are not required.

Knowledge of secant, cosecant and cotangent. Their relationships to cosine, sine and tangent functions. Understanding of their domains and graphs. Knowledge and use of 1+ tan2 x = sec2 x 1+ cot2 x = cosec2x .

Use in simple identities. Solution of trigonometric equations in a given interval, using these identities.

Exponentials and logarithms The function ex and its graph.

The function ln x and its graph; ln x as the inverse function of ex .

Differentiation Differentiation of ex , ln x, sin x, cos x, tan x , and linear combinations of these functions.

Differentiation using the product rule, the quotient rule, the chain rule and by the use of

1dxdydy

dx

22 3

2

1 2 1. . ln ; sin ; ,

1 3 2

xx

x

e xE g x x e x

e x

2. .A curve has equation 4 1. 1dy

E g x y y Find when ydx

AQA - Core 3 5

Integration Integration of ex ,1/x , sin x, cos x .

Simple cases of integration: by inspection or substitution;

3 2. . ; sin 4 ; 1xE g e dx x dx x x dx

by substitution; 6. . (2 ) ; 2 3E g x x dx x x dx

and integration by parts. 2. . ; 3 ; lnxE g xe dx xSin x dx x x dx

These methods as the reverse processes of the chain and product rules respectively.

Including the use of by inspection

or substitutio

'( )ln ( )

(

.

)

n

f xdx f x c

f x

Evaluation of a volume of revolution.

The axes of revolution will be restricted to the x − and y − axis .

Numerical methodsLocation of roots of f (x) = 0 by considering changes of sign of f (x) in an interval of x in which f (x) is continuous.

Approximate solutions of equations using simple iterative methods, including recurrence relations of the

form 1 ( )n nx f x .

Rearrangement of equations to the form x = g(x). Staircase and cobweb diagrams to illustrate the iteration and their use in considerations of convergence.

Numerical integration of functions using the mid‐ordinate rule and Simpson’s Rule.

To include improvement of an estimate by increasing the number of steps.

AQA - Core 3 6

The formulae booklet

AQA - Core 3 7

AQA - Core 3 8

AQA - Core 3 9

AQA - Core 3 10

Content

Algebra and functions .................................................................................................. 12

Functions – definitions and vocabulary .................................................................................................................. 12

The modulus function ............................................................................................................................................. 13

Transformations of graphs ..................................................................................................................................... 14

Functions – Test yourself ........................................................................................................................................ 15

Trigonometry ............................................................................................................... 17

Inverse trigonometric functions ............................................................................................................................. 17

Other trigonometric functions ............................................................................................................................... 18

Trigonometry ‐ Test yourself .................................................................................................................................. 19

Exponentials and logarithms ........................................................................................ 20

Definition and properties ....................................................................................................................................... 20

Calculus with ”ln” and “exp” .................................................................................................................................. 21

Differentiation ............................................................................................................. 22

Usual functions and rules of differentiation ........................................................................................................... 22

Differentiation – Test yourself ................................................................................................................................ 23

Integration ................................................................................................................... 24

Techniques of integration ....................................................................................................................................... 24

Integration – Test yourself...................................................................................................................................... 25

Using integration .................................................................................................................................................... 27

Volumes of revolution ‐ exercises .......................................................................................................................... 28

Numerical methods ..................................................................................................... 29

Solving equations using an iterative method ......................................................................................................... 29

Iterative methods ‐ exam questions ....................................................................................................................... 30

Iterative methods ‐ exam questions ‐ MS............................................................................................................... 32

Integration using numerical methods .................................................................................................................... 33

Integration by numerical methods – Exam questions ............................................................................................ 34

Past papers .................................................................................................................. 37

AQA - Core 3 11

Algebraandfunctions

Functions–definitionsandvocabulary

AQA - Core 3 12

Themodulusfunction

Solving equations involving the modulus function

Solving inequalities involving the modulus function Specifications:

Once you have solved the equation, use the graph to solve the inequality.

AQA - Core 3 13

Transformationsofgraphs

AQA - Core 3 14

Functions–Testyourself

AQA - Core 3 15

AQA - Core 3 16

Trigonometry

Inversetrigonometricfunctions

The graphs of the inverse trig functions

AQA - Core 3 17

Othertrigonometricfunctions

The graphs of these trig functions

AQA - Core 3 18

Trigonometry‐Testyourself

Answers:

AQA - Core 3 19

Exponentialsandlogarithms

Definitionandproperties

Properties of the “ln” function

Properties of the “ln” function

AQA - Core 3 20

Calculuswith”ln”and“exp”

AQA - Core 3 21

Differentiation

Usualfunctionsandrulesofdifferentiation Logarithm and exponential

AQA - Core 3 22

Differentiation–Testyourself

AQA - Core 3 23

Integration

Techniquesofintegration

Integration by parts

AQA - Core 3 24

Integration–Testyourself

AQA - Core 3 25

AQA - Core 3 26

Usingintegration

AQA - Core 3 27

Volumesofrevolution‐exercises

Answers:

AQA - Core 3 28

Numericalmethods

Solvingequationsusinganiterativemethod

AQA - Core 3 29

Iterativemethods‐examquestionsQuestion 1: Jan 2007 – Q8

Question 2: Jun 2007 – Q4

AQA - Core 3 30

Question 5: Jan 2009 – Q3

AQA - Core 3 31

Iterativemethods‐examquestions‐MS Question 1: Jan 2007 – Q8



AQA - Core 3 32

Integrationusingnumericalmethods

AQA - Core 3 33

Integrationbynumericalmethods–ExamquestionsQuestion 1: Jan 2006 – Q2






AQA - Core 3 34

Exam questions ‐ MS Question 1: Jan 2006 – Q2






AQA - Core 3 35

AQA - Core 3 36

Pastpapers

AQA - Core 3 37

GeneralCertificate

ofEducation

January

2008

AdvancedLevelExamination

MATHEMATICS

MPC3

UnitPure

Core

3

Thursday17January

2008

1.30pm

to3.00pm

Forthis

paperyoumusthave:

*an8-pageanswerbook

*theblueAQAbookletofform

ulaeandstatisticaltables.

Youmayuseagraphicscalculator.

Tim

eallowed:1hour30minutes

Instructions

*Use

blueorblack

inkorball-pointpen.Pencilshould

only

beusedfordrawing.

*Write

theinform

ationrequired

onthefrontofyouransw

erbook.TheExaminingBodyforthis

paper

isAQA.ThePaperReference

isMPC3.

*Answ

erallquestions.

*Show

allnecessary

working;otherwisemarksformethodmay

belost.

Inform

ation

*Themaxim

um

markforthispaper

is75.

*Themarksforquestionsareshownin

brackets.

Advice

*Unless

stated

otherwise,

youmay

quote

form

ulae,

withoutproof,from

thebooklet.

Answ

erallquestions.

1(a)

Finddy

dxwhen:

(i)

y¼

ð2x2�5xþ1Þ20

;(2

marks)

(ii)

y¼

xcosx.

(2marks)

(b)

Given

that

y¼

x3

x�2

show

that

dy

dx¼

kx2ðx

�3Þ

ðx�2Þ2

wherekis

apositiveinteger.

(3marks)

2(a)

Solvetheequationcotx

¼2,givingallvalues

ofxin

theinterval

04

x4

2p

in

radiansto

twodecim

alplaces.

(2marks)

(b)

Show

that

theequationcosec2

x¼

3cotx

þ4

2canbewritten

as

2cot2x�3cotx

�2¼

0(2

marks)

(c)

Solvetheequationcosec2

x¼

3cotx

þ4

2,givingallvalues

ofxin

theinterval

04

x4

2p

inradiansto

twodecim

alplaces.

(4marks)

AQA - Core 3 38

3Theequation

xþð1

þ3xÞ1 4

¼0

has

asingle

root,a.

(a)

Show

that

alies

between�0

.33and�0

.32.

(2marks)

(b)

Show

that

theequationxþð1

þ3xÞ1 4

¼0canberearranged

into

theform

x¼

1 3ðx

4�1Þ

(2marks)

(c)

Use

theiterationx n

þ1¼

ðx4n

�1Þ

3withx 1

¼�0

:3to

findx 4

,givingyouransw

erto

threesignificantfigures.

(3marks)

4Thefunctionsfandgaredefined

withtheirrespectivedomainsby

fðxÞ¼

x3,

forallreal

values

ofx

gðxÞ

¼1

x�3,

forreal

values

ofx,

x6¼

3

(a)

State

therangeoff.

(1mark)

(b)

(i)

FindfgðxÞ

.(1

mark)

(ii)

SolvetheequationfgðxÞ

¼64.

(3marks)

(c)

(i)

Theinverse

ofgis

g�1

.Findg�1

ðxÞ.

(3marks)

(ii)

State

therangeofg�1

.(1

mark)

5(a)

(i)

Given

that

y¼

2x2�8xþ3,finddy

dx.

(1mark)

(ii)

Hence,orotherwise,

find ð 6 4

x�2

2x2�8xþ3dx

givingyouransw

erin

theform

kln3,wherekis

arational

number.

(4marks)

(b)

Use

thesubstitutionu¼

3x�1to

find

ð xffiffiffiffiffiffi

ffiffiffiffiffiffiffi

3x�1

pdx,givingyouransw

erin

term

s

ofx.

(4marks)

6(a)

Sketch

thecurvewithequationy¼

cosecxfor0<x<p.

(2marks)

(b)

Use

themid-ordinaterule

withfourstripsto

findan

estimatefor

ð 0:5

0:1cosecxdx,giving

youransw

erto


(4marks)

7(a)

Describeasequence

oftw

ogeometricaltransform

ationsthat

mapsthegraphofy¼

x2

onto

thegraphofy¼

4x2�5.

(4marks)

(b)

Sketch

thegraphofy¼

j4x2�5j,

indicatingthecoordinates

ofthepointwherethe

curvecrosses

they-axis.

(3marks)

(c)

(i)

Solvetheequationj4x2�5j¼

4.

(3marks)

(ii)

Hence,orotherwise,

solvetheinequalityj4x2�5j5

4.

(2marks)

8(a)

Given

that

e�2x¼

3,findtheexactvalueofx.

(2marks)

(b)

Use

integrationbyparts

tofind

ð xe�2

xdx.

(4marks)

(c)

Acurvehas

equationy¼

e�2xþ6x.

(i)

Findtheexactvalues

ofthecoordinates

ofthestationarypointofthecurve.

(4marks)

(ii)

Determinethenature

ofthestationarypoint.

(2marks)

(iii)

TheregionRis

bounded

bythecurvey¼

e�2xþ6x,thex-axis

andthe

lines

x¼

0andx¼

1.

Findthevolumeofthesolidform

edwhen

Ris

rotatedthrough2pradiansabout

thex-axis,givingyouransw

erto


(5marks)

END

OF

QUESTIO

NS

AQA - Core 3 39

AQA – Core 3 – Jan 2008 – Answers

Question 1: Exam report

2 19 1

3 2 3

2

3 2 3 3 2

2

( )) ) : '

) 1 ( ) ( ) (Pro

20(4 5)(2 5 1)

( ) duct rule)

3 ( 2) 1) Using the quotient rule:

2 ( 2)

3 6 2 6

( 2) (

(

2

)

nndy d u

a i Chain rule n u udx dxdy

ii Cos x x Sin xdx

x dy x x xb

x x x

C

yx dx x

dy x x x x x

dx

os x

x

xS n x

x

i

2

2 2

2 ( 3)

2)2

) (

x x

xk

Part (a)(i) was well answered by the majority of candidates. Many fully correct responses were seen, and, if there were errors, it was usually through further incorrect work or by the omission of brackets. Very few incorrect responses were seen to part (a)(ii). Most candidates appeared to be able to use the product rule successfully. Some errors with signs were made, leading to the loss of the accuracy mark. Part (b) was very well answered.


1 1

2

22

2 2

2

2

1) ( ) 2 ( )

2

tan (0.5) tan (0.5)

3cot( ) 4)cosec ( ) 2cosec ( ) 3cot( ) 4

2

2cosec ( ) 3cot( ) 4 0

Now use the identity :

2(cot 2cot (

0.46 3

) 3

.61

cosec ( ) cot 1

1) 3cot 4 0

a Cot

x or x

x x

x

x Tan x

so x or x

xb x x x

x x

x x

2

1

) We now factorise this quadratic equation in "cot "

2cot ( ) 3cot( ) 2 0 (2cot 1)(cot 2) 0

1 1cot cot 2 tan 2 tan

2 2

tan ( 2) 1.107 1.107 1.107 22.03

co

from Qa) we a

t( ) 2

.1

0

5 8

c x

x x x x

x or x x or x

so x or

an

x

d

0.ls 46o hav 6e 3. 1x or x

Part (a) was well answered by the majority of candidates. Tan x = 0.5 and x = 0.46, 3.60 was a common error. The second angle was often incorrect. Few cases of angles in degrees were seen. In part (b), most candidates used the correct identity and were successful in answering this part of the question. The main error was cosec²x = cot² x – 1 followed by fudging of the rest of the solution. In part (c), most candidates attempted to factorise the quadratic expression, although some used the quadratic formula. Those who factorised were usually correct, although solutions of ½ and –2 were not uncommon. Those who used the formula often made the error of cot²x = 2 or cot²x = –0.5. Candidates with 3.60 as a solution in (a) were able to recover here but often failed to obtain both marks; 5.17 was a common error.

Question 3: Exam report 1

4

1 1

4 4

change of sign ru

) ' ( ) (1 3 )

( 0.33) 0.014 0

( 0.32) 0.127 0

According the the ,

we know that there is a root so tha

le

0.33 0.3t

) (1 3 ) 0 (1 3 )

by elevating bo h

2

t

a Let s call f x x x

then f

and f

b x x x x

1 2

4

4

4

4

42

3 4

3

0.3, 0.331, 0.329,

sides to the power of 4, w have

1 3 ( )

3 1

)

( : " 0.3 " "( 1

0.32

) 3 "

1( 1

( )

"

9

)

)

3

" .

x x x x

x x x

x x and

c

Type then Ans for x

then again for x etc

x x

Part (a) of this question was reasonably well answered with most candidates obtaining the method mark for substitution of the two given values. Some candidates then lost the accuracy mark through inaccurate evaluation, although the main reason for the loss of the final mark was that many candidates just stated “change of sign therefore root” without stipulating that –0.33 < α < –0.32. Part (b) was answered very badly. x4 + (1+3x) = 0 was a common error as was (1+3x)¼ = –x, which then became (1+3x) = –x4 . This part was omitted on many scripts. Most candidates were able to score full marks on part (c), with the correct answer often seen. One error was to write the final answer to two decimal places as –0.33. The other main error was calculating –0.334 and not (–0.33)4 for xn4 in the numerator.

AQA - Core 3 40


1

3

3

3

3

1

3

1

4

1( ) ( )

3)The range of

1 1) ) ( ) ( )

3 ( 3)

1 1 1) 64 ( 3) 3

( 3) 64 4

1 1 1) ) ( ) 3 3

3

)The range of g is the dom

3 3.25

1( ) 3

f x x and g xx

a f is

b i fg x f g xx x

ii x xx

c

all real val

i g x y so x an

ues

x

g xx

d xx y y

ii

1 al

ain

l t

of g:

he reaThe range of l values exc g is ept 3

Part (a) was not very well answered with many candidates putting x = R. “x > 0” was also common. Part (b)(i) was very well answered. Part (b)(ii) was answered well by the majority of candidates. Errors were made by those candidates who produced further working in part (b)(i), and hence tried to work with expressions such as

3

1

27x Other common incorrect responses were seen, such as

x = 7 through mishandling of the 3. Part (c)(i) was very well answered by the majority of candidates,

although 1

3x was a common incorrect response, which lost

the accuracy mark. Part (c)(ii) was quite well answered with many correct solutions seen. Common incorrect responses were f(x) ≠ 0 and f(x) ≠ –3.


2

6 6

2 24 4

6 622 44

) ) 2 8 3

4 8

2 1 4 8)

2 8 3 4 2 8 3'

Using the result: ln | | c

1 4 8 1ln | 2 8 3 |

4 2 8 3 41 1 27 1 1

ln 27 ln 3 ln ln 9 2ln 34 4 3 4 4

1 1

1ln 3

2

) 3 13 3

a i y x x

dyx

dxx x

ii dx dxx x x x

ff

f

xdx x x

x x

b u x so x u dand x

3 1

2 2

5 3

2

3

2 2

5

2

1 1 1 13 1

3 3 3 9

1 2 2(3 1

1

3

) (3 1)45 79 2

2 2

5 3

x x dx u u du u u

x x

d

du

u u c

u

c

Part (a)(i) was well answered. Candidates who saw the connection with part (a)(i), usually made a very good attempt at part (a)(ii), often with complete success. Limits did cause a problem where candidates had not bracketed 2x2– 8x +3, and some candidates did not leave the answer in the required form

but left their answer as 1

ln 94

.

Part (b) was well answered by candidates who were able to write the integral in terms of u successfully. Many candidates lost a mark through omission of either du or c. A disturbing number of candidates thought that

1 1 1

3 3 6 . Some clearly able candidates lost the last

mark by spoiling an otherwise acceptable answer.


0.5

0.15 0.25 0.35 0.450.1

Set your c

1.

alculator to RAD

)

)

cos ec ( ) 0.1

0.1 6.691732 4.041972 2.916321 2.2

I

59

A

99033

3 .

NS

.

a Sketch

b

x dx y y y y

to sig fig

Many candidates lost marks on this question through careless work or a failure to write answers to the correct degree of accuracy. In part (a), most candidates produced the correct shaped response, but many lost the accuracy mark by failing to indicate the coordinates of the minimum point. Part (b) was generally well answered. The majority of candidates attempted the mid‐ordinate rule, and many fully correct responses were seen. Errors occurred in working with three significant figures and in writing the final answer to an inappropriate degree of accuracy. An answer of 1.60 was a very common error.

AQA - Core 3 41


2

2

Stretch scale factor 4 in the y-direction

0fo

) the graph of the curve with equation 4 5c

llowed by a translation of vecto

an be

obtained fro

r

m after a

)The roots

5

5

o 4 55

2f

2

a y x

y x

b y x are and

2 2 2

2 2

2

0, | 5 | 5

) )| 4 5 | 4 4 5 4 4 5 4

9 1

4 4

)| 4 5 | 4 when the curve is "above" the line y = 4

(0,5)

3 3 1 1

2 2 2 2

3 1 1 3

2 2 2 2.

for x y

c i x means x or x

x or x

ii x

x or x

happ

or x or x

x o

ens

i e r x or x

Part (a) was not very well answered, with few candidates gaining all 4 marks. The most common problem was an inability to give the appropriate scale factor when stretching in the x‐axis. Part (b) was well answered. Marks were often lost due to the quality of the sketch. In part (c)(i), there were many fully correct responses, but marks were frequently lost because candidates only gave the positive values of ½ or 1.5. Part (c)(ii) was not very well answered; –½ ≤ x ≤ ½ was the most common way of getting 1 mark. Many incorrect solutions involving the two positive solutions were seen.


2

2 2

2

2

2

2

22

1) 3 2 ln 3 ln 3

21 1

)2 2

) 6

1) 2 6 0 3 ln 3 ( . )

21 1

ln 3, 3 6 ln 3 3 3ln 32 2

The stationary point has coordinates

1 1

2 4

1ln 3

2

x

x

x x x

x

x x

x

a e x x

b xe dx xe e dx

c y e x

dyi e when e x Q a

dx

When x

xe e c

y

2 22

2 2

1 1 122 2 4 2 2

0 0 0

1

4 2 2 3

0

,3 3ln 3

minimum

1) 4 ln 3, 4 3 12 0

2The stationary point is a .

) 2 6 12 36

1 1 112 12

4 2 4

1

4

x

x x x

x x x

d y d yii e and for x

dx dx

iii V y dx e x dx e xe x dx

V e xe e x

V e

4 2 2 4 214

1 16 3 12 3 15 9

4 4

44.1 3 . .V to sig fig

e e e e

Part (a) was well answered by the majority of candidates. A common error was

3ln

2x .

The integration by parts in part (b) was well done, with many candidates achieving full marks. There were a few candidates who used u = e–2x and dv = x and hence produced a more complicated expression, although there were fewer this session than in the past. In part (c)(i), many candidates obtained the correct derivative and went on to obtain the correct x‐coordinate. Unfortunately they then lost marks by not finding the appropriate y‐value. Candidates who made earlier errors were usually able to obtain some credit from a correct method. In part (c)(ii), most candidates at least obtained the method mark, and many using the correct exact value of x successfully completed the question. In part (c)(iii), most candidates achieved the method mark and many went on to get the mark for the expansion, although

e4x and 24 xe were common errors for the

first term. The integration was poorly done with most candidates failing to realise that they could use their earlier result from part (b). Some fully correct solutions were seen.

GRADE BOUNDARIES

Component title Max mark A B C D E

Core 3 – Unit PC3 75 60 53 46 39 32

AQA - Core 3 42

Key

to m

ark

sche

me

and

abbr

evia

tions

use

d in

mar

king

M

mar

k is

for m

etho

d m

or d

M

mar

k is

dep

ende

nt o

n on

e or

mor

e M

mar

ks a

nd is

for m

etho

d A

m

ark

is d

epen

dent

on

M o

r m m

arks

and

is fo

r acc

urac

y B

m

ark

is in

depe

nden

t of M

or m

mar

ks a

nd is

for m

etho

d an

d ac

cura

cy

E m

ark

is fo

r exp

lana

tion

or ft

or F

fo

llow

thro

ugh

from

pre

viou

s in

corr

ect r

esul

t M

C

mis

-cop

y C

AO

co

rrec

t ans

wer

onl

y M

R

mis

-rea

d C

SO

corr

ect s

olut

ion

only

R

A

requ

ired

accu

racy

A

WFW

an

ythi

ng w

hich

falls

with

in

FW

furth

er w

ork

AW

RT

anyt

hing

whi

ch ro

unds

to

ISW

ig

nore

subs

eque

nt w

ork

AC

F an

y co

rrec

t for

m

FIW

fr

om in

corr

ect w

ork

AG

an

swer

giv

en

BO

D

give

n be

nefit

of d

oubt

SC

sp

ecia

l cas

e W

R

wor

k re

plac

ed b

y ca

ndid

ate

OE

or e

quiv

alen

t FB

fo

rmul

ae b

ook

A2,

1 2

or 1

(or 0

) acc

urac

y m

arks

N

OS

not o

n sc

hem

e –x

EE

dedu

ct x

mar

ks fo

r eac

h er

ror

G

grap

h N

MS

no m

etho

d sh

own

c ca

ndid

ate

PI

poss

ibly

impl

ied

sf

sign

ifica

nt fi

gure

(s)

SCA

su

bsta

ntia

lly c

orre

ct a

ppro

ach

dp

deci

mal

pla

ce(s

)

No

Met

hod

Show

n

Whe

re t

he q

uest

ion

spec

ifica

lly r

equi

res

a pa

rticu

lar

met

hod

to b

e us

ed, w

e m

ust

usua

lly s

ee e

vide

nce

of u

se o

f th

is

met

hod

for a

ny m

arks

to b

e aw

arde

d. H

owev

er, t

here

are

situ

atio

ns in

som

e un

its w

here

par

t mar

ks w

ould

be

appr

opria

te,

parti

cula

rly w

hen

sim

ilar

tech

niqu

es a

re in

volv

ed.

You

r Pr

inci

pal E

xam

iner

will

ale

rt yo

u to

thes

e an

d de

tails

will

be

prov

ided

on

the

mar

k sc

hem

e.

Whe

re th

e an

swer

can

be

reas

onab

ly o

btai

ned

with

out s

how

ing

wor

king

and

it is

ver

y un

likel

y th

at th

e co

rrec

t ans

wer

can

be

obt

aine

d by

usi

ng a

n in

corr

ect

met

hod,

we

mus

t aw

ard

full

mar

ks.

How

ever

, th

e ob

viou

s pe

nalty

to

cand

idat

es

show

ing

no w

orki

ng is

that

inco

rrec

t ans

wer

s, ho

wev

er c

lose

, ear

n no

mar

ks.

Whe

re a

que

stio

n as

ks th

e ca

ndid

ate

to st

ate

or w

rite

dow

n a

resu

lt, n

o m

etho

d ne

ed b

e sh

own

for f

ull m

arks

.

Whe

re th

e pe

rmitt

ed c

alcu

lato

r ha

s fu

nctio

ns w

hich

rea

sona

bly

allo

w th

e so

lutio

n of

the

ques

tion

dire

ctly

, the

cor

rect

an

swer

with

out w

orki

ng e

arns

ful

l mar

ks, u

nles

s it

is g

iven

to le

ss th

an th

e de

gree

of

accu

racy

acc

epte

d in

the

mar

k sc

hem

e, w

hen

it ga

ins n

o m

arks

.

Oth

erw

ise

we

requ

ire

evid

ence

of a

cor

rect

met

hod

for

any

mar

ks to

be

awar

ded.

QSo

lutio

nM

arks

T

otal

Com

men

ts

1(a)

(i)

()20

22

–5

1y

xx

=+

()

()

192

d20

2–

51

4–

5dy

xx

xx

=+

OE

M1

A1

2 ch

ain

rule

(

)(

)19

20f

xw

ith n

o fu

rther

inco

rrec

t wor

king

(ii)

cos

yx

x=

d

–si

nco

sdy

xx

xx

=+

M

1 A

1 2

prod

uct r

ule

sin

cos

xx

x±

±

CSO

(b)

3

2x

yx

=− (

) ()

23

2

–2

31

d d–

2

xx

xy x

x

−×

=M

1

A1

quot

ient

rule

(

)2

''

–2

vuuv

x±

±

cond

one

mis

sing

bra

cket

s 3

23

2

36

(2)

xx

xx

−−

=−

2

2

2(

3)(

2)x

xx

−=

−A

1 3

CSO

Tot

al

7 2(

a)

cot

2ta

n0.

5x

x=

=

M1

0.

46,3

.61

x=

A

1 2

AW

RT;

no

othe

rs w

ithin

rang

e

(b)

23c

ot4

cose

c2

xx

+=

()

22

1co

t3c

ot4

xx

+=

+M

1 C

orre

ct u

se o

f 2

2co

sec

1co

tx

x=

+

()

22c

ot3c

ot2

40

xx

−+

−=

22c

ot2

03c

ot–

xx

=−

A

1 2

AG

; cor

rect

with

no

slip

s fro

m li

ne

with

no

frac

tions

(c

) (

)()(

)2c

ot1

cot

20

xx

+−

=M

1 A

ttem

pt to

solv

e

1co

t,

22

x=

−A

1

tan

2,0.

5x

=−

0.

46,3

.61,

2.03

,5.1

8x

=

B1

B1

4 2

corr

ect

Allo

w 3

.6(0

) 4

corr

ect (

with

no

extra

s in

rang

e) A

WR

T D

egre

es

26.5

7, 2

06.5

7 B1

for 2

cor

rect

116.

57,

296.

57

SC

Tot

al8

Cor

e 3

- Jan

2008

- M

ark

sche

me

AQA - Core 3 43

QSo

lutio

nM

arks

Tot

alC

omm

ents

3(a)

1 4

(13

)0

xx

++

=

()

f–0

.32

0.1

=

()

f–0

.33

–0.0

1=

M1

AW

RT;

allo

w +

ve,

–ve

Cha

nge

of si

gn–0

.33<

<–0

.32

x∴

A

1 2

(b)

1 4(1

3)

xx

=−

+

41

3x

x=

+

M1

Atte

mpt

to is

olat

e 4 x

4

3–1x

x=

A

1 2

AG

(c)

1–

0.3

x=

(

)2

–0.

331

x=

A

WR

T

M1

()

3–

0.32

9x

=

AW

RT

A1

4–

0.32

9x

=

A1

3 T

otal

7 4(

a)

all

()

real

valu

es

B1

1 N

o x

in a

nsw

er, u

nles

s f (x

)

(b)(

i) (

)3

1fg

–3

xx

=B

1 1

ISW

(ii)

31

64–

3x

=

14

–3

x=

M

1 3

1–

34

x=

M1

Inve

rt

13

4x

=A

1 3

(c)(

i) 1 –

3y

x=

1 –3

xy

=M

1 Sw

ap

and

xy

()

–3

1x

y=

–3

1xy

x=

M

1 at

tem

pt to

isol

ate

()

–11

3g

xy

xx+

==

or

13

x+

A1

3

(ii)

()

real

val

ues

()

()

–1 g3

x≠

B1

1 T

otal

9

Q

Solu

tion

Mar

ksT

otal

C

omm

ents

5(

a)(i)

2

28

3y

xx

=−

+

d4

–8

dyx

x=

B1

1

(ii)

6

24

–2

d2

–8

3x

xx

x+

62

4

1ln

2–

83

4x

x=

+M

1A1

M1

for k

ln (

)2

–2

83

; allo

w k

lnx

xu

+

[]

1ln

27–

ln3

4=

m1

Cor

rect

subs

titut

ion

into

k

ln (

)2

2–

83

xx

+or

3, 2

7 in

to k

ln u

1ln

94

=

1ln

32

=A

1 4

(b)

(31)

dx

xx

−

31

d3d

ux

ux

=−

=

B1

OE

()

31

22

1d

9u

uu

=+

M1

2 te

rms i

n w

ith ra

tiona

l ind

ices

u

()

53

22

15

39

22

uu

c=

++

A

1F

Mus

t be

2 te

rms w

ith c

orre

ct in

dice

s –1

only

ft fo

r =

3ux

53

22

22

(31)

(31)

4527

xx

=−

+−

+c

A1

4 C

SO

OE

Tot

al9

6(a)

M1

A1

2

Cor

rect

shap

e

Ver

tex

(b)

x y

0.15

6.

692

0.25

4.

042

0.35

2.

916

0.45

2.

299

M1

B1

Cor

rect

x v

alue

s 3

cor

rect

y v

alue

s

()

0.1

15.9

49y

y×

=B

1 co

rrec

t h u

sed

corr

ectly

= 1

.59

A1

4 T

otal

6

AQA - Core 3 44

QSo

lutio

nM

arks

Tot

alC

omm

ents

7(a)

St

retc

h (I

)

Scal

e fa

ctor

1 2 (

II)

para

llel t

o x-

axis

(III

)

(Or s

cale

fact

or 4

par

alle

l to

y-ax

is)

M1

A1

I + (I

I or I

II)

All

corr

ect

Tran

slat

ion

M1

0 5−

OE

A1

4

Alte

rnat

ives

0 5–

4 01

–52

trans

late

, s

tretc

h sf

4

-axi

s

trans

late

, s

tretc

hsf

-a

xis

y x

Mar

k tra

nsla

tion

first

. M

ark

stre

tch

as

abov

e, b

ut re

lativ

e to

thei

r tra

nsla

tion.

(b)

M1

A1

A1

3

Mod

ulus

gra

ph sy

mm

etric

al a

bout

y-a

xis

left

of

5 2−

and

right

of

5 2

(0, 5

), cu

sps d

raw

n an

d no

stra

ight

line

s be

twee

n cu

sps

(c)(

i) 2

45

4x

−=

2

49

x=

3 2

x=

±

O

E

B1

24

54

x−

=−

M

1 4

216

–40

90

xx

+=

2

41

x= 1 2

x=

±

A1

3

(ii)

33

, 22

xx

≤−

≥

B1F

2

corr

ect s

tate

men

ts

11

,2

2x

x−

≤≤

B

1F

2 4

corr

ect s

tate

men

ts

SC

c(ii

)

1 m

ark

pena

lty fo

r stri

ct in

equa

litie

s T

otal

12

QSo

lutio

nM

arks

T

otal

Com

men

ts8(

a)

e3

x−2

=

2ln

3x

−=

M

11

ln3

2x

=−

A1

2 O

E IS

W

(b)

2e

dx

xx

−

2d

ed

xv

ux

x−

==

2d

11

ed

2x

uv

x−

==

−M

1 di

ffer

entia

ting

and

inte

grat

ing

()

22

11

ee

d2

2x

xx

x−

−=

−+

m1

A1

corr

ect s

ubs o

f the

ir va

lues

into

par

ts

form

ula

22

11

ee

24

xx

xc

−−

=−

−+

A

1 4

No

furth

er in

corr

ect w

orki

ng

(c)(

i) 2

e6

xy

x−

=+

2

d2e

+6

dx

y x−

=−

= 0

M

1 –2 e

60

xk

+=

()

2d

02

e–

30

dx

y x−

=−

=

1ln

32

x=

−A

1 O

E

13

6ln

32

y=

+−

M1

Cor

rect

subs

titut

e of

thei

r val

id x

33l

n3=

−

A1

4 O

E I

SW

(ii)

22

2

d4e

12d

0

xy x

−=

= >

M1

Oth

er m

etho

ds n

eed

just

ifica

tion

Allo

w e

rror

in

2

2

d dy x

or x

-val

ue, b

ut n

ot

both

m

inim

um∴

A

1 2

(iii)

()

()

()

()

()

()

11

22

–2

00

de

6d

xV

yx

xx

+=

=M

1 Ei

ther

()

()

()

() 1

42

2

0

e12

e36

dx

xx

xx

−−

=+

+B

1 C

orre

ct e

xpan

sion

()

()

()13

0

42

21

e6

e3e

124

xx

xx

x−

−−

=−

−−

+A

1

A1

3 co

rrec

t ter

ms;

‘–6

’,‘–3

’ cor

rect

or

12 ×

thei

r (b)

A

ll co

rrec

t 4

21

1e

9e12

34

4−

−=

−−

+−

−−

24

11

159e

e4

4−

−=

−−

44.1

=

B1

5 A

WR

T T

otal

17

T

OT

AL

75

AQA - Core 3 45

GeneralCertificate

ofEducation

June2008


MATHEMATICS

MPC3

UnitPure

Core

3

Friday23May2008

9.00am

to10.30am

Forthis

paperyoumusthave:

*an8-pageanswerbook




Tim


Instructions

*Use

black

inkorblack

ball-pointpen.Pencilshould

only

beusedfordrawing.

*Write

theinform

ationrequired



paper


isMPC3.

*Answ

erallquestions.

*Show

allnecessary


belost.

Inform

ation

*Themaxim

um

markforthispaper

is75.


brackets.

Advice

*Unless

stated

otherwise,

youmay

quote

form

ulae,

withoutproof,from

thebooklet.

Answ

erallquestions.

1Finddy

dxwhen:

(a)

y¼

ð3xþ1Þ5;

(2marks)

(b)

y¼

lnð3xþ1Þ;

(2marks)

(c)

y¼

ð3xþ1Þ5

lnð3xþ1Þ.

(3marks)

2(a)

Solvetheequationsecx¼

3,givingthevalues

ofxin

radiansto

twodecim

alplacesin

theinterval

04

x<2p.

(3marks)

(b)

Show

that

theequationtan2x¼

2secxþ2canbewritten

assec2

x�2secx�3¼

0.

(2marks)

(c)

Solvetheequationtan2x¼

2secxþ2,givingthevalues

ofxin

radiansto

two

decim

alplacesin

theinterval

04

x<2p.

(4marks)

AQA - Core 3 46

3A

curveis

defined

for04

x4

p 4bytheequationy¼

xcos2x,andis

sketched

below.

(a)

Finddy

dx.

(2marks)

(b)

ThepointA,wherex¼

a,onthecurveis

astationarypoint.

(i)

Show

that

1�2atan2a¼

0.

(2marks)

(ii)

Show

that

0:4<a<0:5.

(2marks)

(iii)

Show

that

theequation1�2xtan2x¼

0canberearranged

to

becomex¼

1 2tan�1

1 2x

�� .

(1mark)

(iv)

Use

theiterationx n

þ1¼

1 2tan�1

1 2x n

��

withx 1

¼0:4

tofindx 3

,givingyour

answ

erto

twosignificantfigures.

(2marks)

(c)

Use

integrationbyparts

tofind

ð 0:5

0xcos2xdx,givingyouransw

erto

threesignificant

figures.

(5marks)

y

A

Oa

xp 4



fðxÞ¼

x2,

forallreal

values

ofx

gðxÞ

¼1

2x�3,

forreal

values

ofx,

x6¼

3 2

(a)

State

therangeoff.

(1mark)

(b)

(i)

Theinverse

ofgis

g�1

.Findg�1

ðxÞ.

(3marks)

(ii)

State

therangeofg�1

.(1

mark)

(c)


¼9.

(3marks)

5(a)

Thediagram

showspartofthecurvewithequationy¼

fðxÞ.

Thecurvecrosses

the

x-axis

atthepointða,0Þandthey-axis

atthepointð0,�b

Þ.

Onseparatediagrams,sketch

thecurves

withthefollowingequations.

Oneach

diagram,indicate,

interm

sofaorb,thecoordinates

ofthepoints

wherethecurve

crosses

thecoordinateaxes.

(i)

y¼

jfðxÞ

j.(2

marks)

(ii)

y¼

2fðx

Þ.(2

marks)

(b)

(i)

Describeasequence

ofgeometricaltransform

ationsthat

mapsthegraphof

y¼

lnxonto

thegraphofy¼

4lnðx

þ1Þ�

2.

(6marks)

(ii)

Findtheexactvalues

ofthecoordinates

ofthepoints

wherethegraphof

y¼

4lnðx

þ1Þ�

2crosses

thecoordinateaxes.

(4marks)

y Ox

ð0,�b

Þða,0Þ

AQA - Core 3 47

AQA – Core 3 – Jun 2008 – Answers


4

1

4

54

15(3 1)

3

3 1

3(3 1)15(3 1) ln(

) 5 3 3 1

( ): '

(ln ) ') :

3 1)1

)3

nn

x

x

dya x

dx

d uChain rule n u u

dxdy d u u

b Chain ruledx dx u

dyc

x

dxx x

x

Part (a) was well answered by the majority of candidates. Many fully correct responses were seen and if there were errors it was usually for failing to multiply by the derivative of (3x + 1) so answers of 5(3x+1)4 were not uncommon. Part (b) was not as well answered as part (a) but many correct responses were seen. The main error was again to miss the factor of 3 and the very common incorrect answer was 1/(3x+1). In part (c) most candidates used the product rule successfully but a number of candidates believed that the differential of a product was the product of the differentials. Candidates who made errors in part (b) were able to recover and earn 2 marks. A very common incorrect response was to simply multiply the answers to parts (a) and (b) resulting in 45(3x+1)3 Another common error was to ‘simplify’ ln(3x+1)(3x+1)4 to obtain ln(3x+1)5


2

1 1

2

2

2

2

2

1)sec 3 cos

31 1

cos 2 cos3 3

) tan 2sec 2

Using the identity ,

sec 1 2sec 2 0

)sec 2sec 3 0

(sec 3)(sec 1) 0

sec 3 sec 1

1cos

3

tan s

1.23 5.05

sec 2

e

3 0

c

ec

1

s

x or x

x x

a x x

so x or x

b x x

we have

x x

c x x

x x

x

x o

o

x

r x

x

1.23 5.0

c 1

5

os

x or x o

r x

r x

In general this question was done well by candidates of all abilities. Part (a) was very well answered by the majority of candidates, although tan x = 1/3 and sin x =1/3 were also seen. 1.23 was usually seen but the second result was often incorrect. There were very few cases of results given in degrees. Most candidates used the correct identity part (b) in and were successful in answering this part of the question. The main error was candidates using tan²x = sec²x + 1 and then fudging the rest of the solution. In part (c) most candidates attempted to factorise the result from part (b), although some used the quadratic formula. Those who factorised were usually correct, although solutions of ‐1/3 and 1 were not uncommon. For the final 3 answers many totally correct solutions were seen. Candidates with an incorrect solution in part (a) were able to recover here from follow through marks, but often candidates failed to obtain both marks since 0 or 6.18 often accompanied 3.14.

AQA - Core 3 48


) cos 2 2sin 2

) ) is such that ( ) 0

cos 2 2 sin 2 0 dividing both sides by cos2 ,

) ' ( ) 1 2 tan 2 ,

(0.4) 0.18 0

(0.5) 0.56 0

According to the

cos 2 2 sin 2

1 2 tan 2 0

chan

dya x x x

dxdy

b i xdx

so and

ii let s c

x

all x

f

f

x

f x x

x

1

1 2

0.50

00

3

.5

1

, we know that there is a root

so that 0.4 0.5

1 1)1 2 tan 2 0 tan 2 2

2 2

ge of sign

) 0.4 , 0.44

r

8 , 2 . .

1 1c) cos 2 sin 2

2

ule

0.

2

1

2 2

42

1

iii x x x x Tanx x

iv x x to dec pl

x x

x Tanx

x

dx x x

0.5

0

0.5 0.5

0 0

sin 2

1 1 1 1 1sin 2 cos 2 s 0in1 cos1 0

2 4 4 4.0954

4

x dx

x x x

Part (a) of this question was reasonably answered with most candidates obtaining the method mark for the product rule. Some candidates then lost the accuracy mark through incorrect evaluation of the constant associated with the derivative of cos2x; 2 and ½ were frequently seen. Answers to part (b)(i), although frequently correct, were often badly set out with the function being equated to zero or x being changed to α at various points in the solution. Division of cos2x by cos2x often resulted in zero before going on to fudge the correct answer. Part (b)(ii) was usually well answered with correct evaluations of f(0.4) and f(0.5). Some candidates then lost marks by just saying α was ‘between these two values’ without stipulating 0.4 and 0.5. Part (b)(iii) was usually well done but there were many cases when the tan became removed from its 2x and ½ x tan‐1 = x was often seen. Part (b)(iv) was generally well answered but marks were lost from the use of degrees and for not writing the answer to the required degree of accuracy. Many fully correct answers and many answers which only lost the final accuracy mark were seen in part (c). Other answers which had the wrong coefficient associated with the sin 2x often got the method marks. There were candidates who started with u = x and v = cos 2x or

started with dv

xdx

and obtained a more

difficult integral.


1

1

2

1

2

2

) for all , 0

:

1) ) ( )

2 31 1 3 3

2 32 2 2

)The range of g is t

(

he domain of g

1) ( ) 9 9

2 3

1 1(2 3) 2 3

9 31 1 1 1

3

)

32 3 2 3

0

3( )

2

3g ( )

2

5

3

a x x

range

b i g x yx

yx x

y y y

ii

c f

f x

xg x

x

g xx

x x

x

x or x

4

3

Part(a) was fairly well answered but, for many candidates, putting x = IR+ and x ≥0 was also common. Part (b)(i) was very well answered. Most candidates at least obtained the 2 method marks but several lost the accuracy mark for an incorrect sign in the numerator. A few candidates tried using a flow chart but these were generally unsuccessful. Part (b)(ii) was answered well by the majority of candidates. Common errors were responses of 0 or 2/3. Part (c) was not very well answered by the majority of candidates, although most gained part marks. Many candidates only gave the result from the positive square root of 9 or 1/9. Those candidates who formed a quadratic and solved it were far more successful in obtaining both results for the final accuracy mark.

AQA - Core 3 49


ln

) i) ( )

) 2 ( )

) ) 4 ln( 1) 2

1 ln( 1) 4 ln( 1) 4 ln( 1) 2

vector ,

f

1Translation

0

stretch scale factor 4 in the y-dir

0translation vect

ollowed by a

followed or 2

by a

) 0

a y f x

ii y f x

b i y x

x x x x x

ii for x

1

1 1

2 2

2

, 4 ln( 1) 2 4ln(1) 2 2

The curve

14ln( 1) 2

crosses the y-axis at (0, 2)

crosses the x-axis at (

0 ln( 1) 1 12

The curv ,e e 1 0)

x

x x x e x e

Part (a)(i) was probably the worst answered question on the paper. Most scored B0B1because the curvature to the left of (a, 0) was wrong. Part (a)(ii) was similar to part (a)(i) but this time it was usually B1B0 with the shape being correct but the coordinates of the intercepts on the axes were often incorrect. In part (b) many fully correct answers were seen and most candidates earned partial credit. Mistakes usually occurred where candidates started with the translation and gave the combined vector. Where candidates started with the stretch they were far more successful. Part (c) was very well answered with many candidates earning full marks.


13 2

1 13ln 2 3ln 2 ln8 ln82 2

1

2

12 3 2

0.25 0.75 1.25 1.

2

0

3 3

75

113 1

2

The gradient is 4

(e 1)

ln(2)

3 3( 1) 1

2 23

8 9 42

) ( 1) 0.5

0.5 1.7655 3.2385 6.59705 13.84

12.7

7

3

0 4

x x

x

x

y

dy

dxfor x

dye e e e

dxdy

dx

b e dx y y y

e e

to

y

22 22 3 3

0 00

6 6

1) ( 1)

3

1 12

3

. .

533

x xc V y dx e dx e x

V e

sig fig

e

Most candidates produced the correct derivative in part (a). The main error was the omission of the 3. Substitution of x = ln 2 was usually correct although there were several cases of e3ln2 = 6. Part (b) was generally very well answered. The majority of candidates attempted the mid‐ordinate rule with many fully correct responses seen. Errors occurred with working with 3 s.f. and writing the final answer to an inappropriate degree of accuracy. Part (c) was done very well up until the last step where a very large number of candidates thought that 2 – 1/3 = ‐ 5/3 and failed to gain the final mark.

AQA - Core 3 50


2 2

2 2

2

22 2

2

3 3 32 2 22 2 2

2

) ,

Cos

)1

)

1 1

(1

1

) (1 ) ( )

1

Cos

1

Sin dy Cos Cos Sin Sina y

Cos dx Cos

dySec

dx Cosx Sin Sin Sin

b TanCosx

c x

Co

Sin then dx Cos d

Cosdx Cos d d

x Sin Cos

d

s Si

n Co

n

Si s

21

xTan

xc c

Part (a) was very well answered with most candidates gaining all 3 marks but it was not uncommon to see an incorrect quotient formula used (usually the wrong order in the numerator, but sometimes the wrong denominator) Part (b) was reasonably well answered. However a number of candidates tried to tackle this question by squaring, generally without success. There were very few successful attempts seen in part (c).

Most candidates failed to provide dx

d at all. Many of those

who did make an attempt chose to try a different substitution of their own, either immediately or subsequently.

GRADE BOUNDARIES


Core 3 – Unit PC3 75 60 53 46 40 34

AQA - Core 3 51

QSo

lutio

nM

arks

Tot

alC

omm

ents

1(a)

(

)4d

53

1dy

xx

=+

×3

M

1

(

)43

1k

x+

()4

153

1x

=+

A1

2 w

ith n

o fu

rther

err

ors (

w.n

.f.e)

(b)

d3

d3

1y x

x=

+M

1

31

k x+A

1 2

w.n

.f.e

(c)

d dy x=

()

()

()

54

33

1ln

31

153

+13

1x

xx

x+

×+

+×

+M

1 A

1 A

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ule

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′+

(fro

m (a

) and

(b))

ei

ther

term

cor

rect

C

SO w

ith n

o fu

rther

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ors

()

()

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ln3

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31

xx

xx

=+

++

=+

++

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1,A

1 3

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RT

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for e

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rang

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SC 7

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2x

x−

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M

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xx

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2se

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0x

x−

−=

A

1 2

AG

; CSO

(c)

2se

c2s

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(

)()

sec

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−o.

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1 T

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no fu

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xx

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=

–11

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x=

B

1 1

AG

; C

SO

(iv)

10.

4x

=

20.

4480

...x

=

M1

225

.7x

=

30.

4200

...x

=

=

0.4

2 A

1 2

(c)

cos2

yx

x=

d1 sin2

dco

s22

ux

ux

vx

v

==

==

M1

diffe

rent

iate

one

term

inte

grat

eon

ete

rmm

ust b

e si

n2k

x

m1

corr

ect s

ubst

itutio

n of

thei

r val

ues i

nto

parts

form

ula

usin

g u

= x

sin2

sin2

–(d

)2

2x

xx

x=

(0.5

)

(0)

sin

2co

s22

4x

xx

=+

A1

sin1

cos1

cos0

–4

44

=+

m1

corr

ectly

subs

titut

ing

valu

es fr

om

prev

ious

2 m

etho

d m

arks

=

0.0

954

A1

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WR

T T

otal

14

Cor

e3 -

June

200

8 - M

ark

sche

me

AQA - Core 3 52

Q

Solu

tion

Mar

ksT

otal

C

omm

ents

4(

a)

()

f0

xB

1 1

allo

w f

0,0,

0y

(b)(

i) 1

2–

3y

x=

12

–3

xy

=M

1 sw

ap x

and

y

()

2–

31

xy

=2

–3

1xy

x=

2

13

xyx

=+

M

1 at

tem

pt to

isol

ate

()

–11

3g

2x

yx

x+

==

o.e

. A

1 3

w.n

.f.e

(ii)

()

()

13

g2

x−

≠B

1 1

(c)

21

92

–3

x=

B

1

12

–3

3x

=±

M1

squa

re ro

ot a

nd in

vert

(con

done

mis

sing

±

) al

tern

ativ

e: a

ttem

pt to

solv

e a

quad

ratic

that

com

es fr

om

21

412

99

x−

+=

o.e

.

54 , 33

x=

o.

e.

A1

3

Tot

al

8

Alte

rnat

ive

4(b)

(i)

2

3di

vide

into

1x

y→

×→

−→

→

13

23

divi

dein

to1

22

13

yy

y

+←

÷←

+←

←

+

M1

Q

Solu

tion

Mar

ksT

otal

C

omm

ents

5(a)

(i)

B1

B1

2

shap

e

coor

dina

tes

(ii)

B1

B1

2

shap

e

coor

dina

tes

(b)(

i) Tr

ansl

atio

n

M1

OR

I s

tretc

h M

1

I +

(II o

r III

) –1 0

A

1 II

SF

4 II

I ⁄⁄

y-ax

is A

1

(I

+ I

I + II

I)

Stre

tch

I

M1

()

I+II

or II

ITr

ansl

atio

n M

1 SF

4

I

I ⁄⁄

y-ax

is

III

A

1 I +

II +

III

1 2− −A

1 B

1

Tran

slat

ion

0 –2

B1

both

All

corr

ect a

nd n

o m

ista

kes o

n or

der e

tc

A1

6 A

ll co

rrec

t A1

Alte

rnat

ive:

()

()

14l

n1

24

ln1

2y

xx

=+

−=

+−

(B1)

Tran

slat

ion

(M

1)

–1 1 2−

(A

1)

Stre

tch

I

(M1)

(

)I+

II or

III

SF 4

II

⁄⁄y-

axis

I

II

(A1)

I +

II +

III

All

corr

ect a

nd n

o m

ista

kes o

n or

der e

tc

(A1)

(6

)

AQA - Core 3 53

Q

Solu

tion

Mar

ksT

otal

C

omm

ents

5(

b)(ii

) (

)4l

n1

–2

yx

=+

0–2

xy

==

B

1 0

y=

()

4ln

12

x+=

()

1ln

12

x+=

M

1

isol

ate

()

ln1

x+=

or

4(

1)x+

1 21

ex+

=

A1

1ek

x+=

1 2 e

–1x

=

o.e.

A

1 4

CSO

is

w

Tot

al14

6(a)

(

)13

2e

1x

y=

+

(

)1 –3

32

d1

e1

ed

2x

xy x

=+

×3

M

1

()1 –

23

1e

12

x+

A1

3 ex

A1

3 2 (

allo

w 1

32

×)

w.n

.f.e

ln2

x=

: ()1

–2

ln8

ln8

d3

e1

ed

2y x

=+

×M

1 co

rrec

t sub

stitu

tion

into

thei

r d dy x

(mus

t use

ln8

or ln

23)

3

12

3=

××

8

4=

A

1 5

(b)

x y

0.25

1.

765(

5)

0.75

3.

238(

5)

1.25

6.

597(

1)

1.75

13

.84(

1)

B1

B1

corr

ect x

val

ues

3 or

4 c

orre

ct y

val

ues

4 s.

f. or

bet

ter

0.

5y

=×

P.I

M1

=

12.

7 A

1 4

sc 1

2.7

with

no

wor

king

24

(c)

2 dv

yx

=

()

()

3 e1

(d)

xx

=+

M1

()

(2)

3

(0)

1 e 3x

x=

+A

1 3 e

xk

x+

()

60

11

e2

–e

03

3=

++

m

1 co

rrec

t sub

stitu

tion

into

f 3

()x

e

61

5e 3

3=

+A

1 4

CSO

6(

5)3

e=

+

Tot

al13

QSo

lutio

nM

arks

Tot

alC

omm

ents

7(a)

si

nco

sy

θ θ=

()

2co

sco

s–

sin

–sin

d dco

sy

θθ

θθ

θθ

=

M1

A1

22

2co

ssi

nco

sθθ

θ±

±

21

cos

θ=

o

.e.

2(1

tan

)θ+

2se

cθ

=

A1

3 A

G; C

SO

(b)

sin

xθ

=

OR

LH

S =

2

2si

nx

θ=

2

sin

1si

nθθ

−

2

2co

s1–

xθ

=

sin

cosθ θ

=

M1

us

e of

2

2co

s1

xθ

+=

sin

tan

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θθ

=

2

1–x

x=

tanθ

=

A

G

A1

2 A

G; C

SO

(c)

()3

22

1d

1x

x−

si

nx

θ=

d

cos

dx

θθ

=

o.e.

M

1 dx

cos

dθ

θ=

±

()3

22

cos

(d)

1–si

nθθ θ

=

m1

all i

n te

rms o

f θ

()3

22

cos

(d)

cos

θθ

θ=

A1

2

sec

(d)

θθ

=

A1

tanθ

=

2

()

1–

xc

x=

+

A1

5 C

SO in

clud

ing

d'sθ

Tot

al10

T

OT

AL

75

Alte

rnat

ive

7(a)

ta

ny

1θ=

2 2

d1s

ec0

d1

yθ

θ−

=

M1

A1

=2

sec

θ

A1

AQA - Core 3 54

GeneralCertificate

ofEducation

January

2009


MATHEMATICS

MPC3

UnitPure

Core

3

Monday19January

2009

1.30pm

to3.00pm

Forthis

paperyoumusthave:

*an8-pageanswerbook


ulaeandstatisticaltables

*aninsertforusein

Question3(enclosed).


Tim


Instructions

*Use

black

inkorblack


only

beusedfordrawing.

*Write

theinform

ationrequired



paper


isMPC3.

*Answ

erallquestions.

*Show

allnecessary


belost.

*Fillin

theboxes

atthetopoftheinsert.

Inform

ation

*Themaxim

um

markforthispaper

is75.


brackets.

Advice

*Unless

stated

otherwise,

youmay

quote

form

ulae,

withoutproof,from

thebooklet.

Answ

erallquestions.

1Use

Sim

pson’s

rule

with5ordinates

(4strips)

tofindan

approxim

ationto

ð 9 1

1

1þ

ffiffiffi xpdx,

givingyouransw

erto


(4marks)

2Thediagram

showsthecurvewithequationy¼

ffiffiffiffiffiffiffiffiffiffiffiffi

ffiffiffiffiffiðx

�2Þ5

qforx5

2.

Theshaded

regionRis

bounded

bythecurvey¼


ffiffiffiffiffiðx

�2Þ5

q,thex-axis

andthelines

x¼

3

andx¼

4.

Findtheexactvalueofthevolumeofthesolidform

edwhen

theregionRis

rotatedthrough

360�aboutthex-axis.

(4marks)

y O2

34

x

R

AQA - Core 3 55

3[Figure

1,printedontheinsert,is

provided

foruse

inthis

question.]

Thecurvewithequationy¼

x3þ5x�4intersects

thex-axis

atthepointA,wherex¼

a.

(a)

Show

that

alies

between0.5

and1.

(2marks)

(b)

Show

that

theequationx3þ5x�4¼

0canberearranged

into

theform

x¼

1 5ð4

�x3Þ

(1mark)

(c)

Use

theiterationx n

þ1¼

1 5ð4

�x n

3Þwithx 1

¼0:5

tofindx 3

,givingyouransw

erto

threedecim

alplaces.

(2marks)

(d)

Thesketch

onFigure

1showsparts

ofthegraphsofy¼

1 5ð4

�x3Þand

y¼

x,and

thepositionofx 1

.

OnFigure

1,draw

acobweb

orstaircasediagram

toshow

how

convergence

takes

place,indicatingthepositionsofx 2

andx 3

onthex-axis.

(2marks)

4(a)


3 2,givingallvalues

ofxto

thenearest

degreein

theinterval

0�<

x<360�.

(2marks)

(b)

Byusingasuitable

trigonometricalidentity,solvetheequation

2tan2x¼

10�5secx

givingallvalues

ofxto

thenearest

degreein

theinterval

0�<

x<360�.

(6marks)



fðxÞ¼

2�x4

forallreal

values

ofx

gðxÞ

¼1

x�4

forreal

values

ofx,

x6¼

4

(a)

State

therangeoff.

(2marks)

(b)

Explain

whythefunctionfdoes

nothavean

inverse.

(1mark)

(c)

(i)

Write

downan

expressionforfgðxÞ

.(1

mark)

(ii)


¼�1

4.

(3marks)

6A

curvehas

equationy¼

e2xðx

2�4x�2Þ.

(a)

Findthevalueofthex-coordinateofeach

ofthestationarypoints

ofthecurve.

(6marks)

(b)

(i)

Find

d2y

dx2.

(2marks)

(ii)

Determinethenature

ofeach


ofthecurve.

(2marks)

7(a)

Given

that

3ex

¼4,findtheexactvalueofx.

(2marks)

(b)

(i)

Bysubstitutingy¼

ex,show

that

theequation3ex

þ20e�

x¼

19canbewritten

as3y2�19yþ20¼

0.

(1mark)

(ii)

Hence

solvetheequation3ex

þ20e�

x¼

19,givingyouransw

ersas

exactvalues.

(3marks)

AQA - Core 3 56

8Thesketch

showsthegraphofy¼

cos�

1x.

(a)

Write

downthecoordinates

ofPandQ,theendpoints

ofthegraph.

(2marks)

(b)

Describeasequence

oftwogeometricaltransform

ationsthat

mapsthegraphof

y¼

cos�

1xonto

thegraphofy¼

2cos�

1ðx

�1Þ.

(4marks)

(c)

Sketch

thegraphofy¼

2cos�

1ðx

�1Þ.

(2marks)

(d)

(i)

Write

theequationy¼

2cos�

1ðx

�1Þin

theform

x¼

fðyÞ.

(2marks)

(ii)

Hence

findthevalueofdx

dywhen

y¼

2.

(3marks)

9(a)

Given

that

y¼

4x

4x�3,use

thequotientrule

toshow

that

dy

dx¼

k

ð4x�3Þ2,wherekis

aninteger.

(2marks)

(b)

(i)

Given

that

y¼

xlnð4x�3Þ,

finddy

dx.

(3marks)

(ii)

Findan

equationofthetangentto

thecurvey¼

xlnð4x�3Þat

thepointwhere

x¼

1.

(3marks)

(c)

(i)

Use

thesubstitutionu¼

4x�3to

find

ð4x

4x�3dx,givingyouransw

erin

term

s

ofx.

(4marks)

(ii)

Byusingintegrationbyparts,orotherwise,

find

ð lnð4x�3Þd

x.

(4marks)

END

OF

QUESTIO

NS

Py O

Qx

Figure

1(foruse

inQuestion3)

y Ox 1

¼0:5

x

y¼

x y¼

1 5ð4

�x3Þ

AQA - Core 3 57



9

0 4 1 3 21

1 14( ) 2

311

2 (1) (9) 4( (3) (7)) 2 (5)32

0.5 0.25 4(0.3660 0.2743)

2.62 3 . .

2 0.30903

dx h y y y y yx

y y y y

to sig fig

y

This was well answered by the majority of candidates giving them a good start to the paper, although many candidates did not apply brackets accurately. Common errors were to write the answer 2.619 or 2.6195 or to

incorrectly evaluate 1 12

3 6as leading to a final

answer of 0.655. The 2 and the 4 inside the bracket were sometimes reversed.


44

4 5 6

3

6

2

33

6

1( 2) ( 2)

6

63 21 12 1

1

6 6 6 2

V x dx x

V

y dx

This question was very well answered by the majority of candidates with many achieving full marks. Some candidates lost the final mark for several reasons: the introduction of (+ c), the omission of π or by not giving the exact answer as required. Some candidates gained no credit as they were unable to handle the squaring of

5( 2)x and other candidates lost 3 marks by differentiating

rather than integrating (x – 2)5.


3

3

5 4 intersect the x-axis so

when , 0

We have to solve the equation 5 4 0

) (0.5

the

) 1.375 0 (1) 2 0

According to change of sign rul , we know

that there is a root so that 0.5 1

e

) 5

y x x

x y

x x

a y and y

b x

3

1 2

3

3

2 3

3

4 0 5 4

) 0.5 , 0.775

14

50.707, 3 .

(Type :"0.5 " "(1/ 5)(4 ) "( ) " " )

)

x x x

c x x to dec places

x

then Ans for x then again for x

d

x

x

Part (a) of the question was reasonably answered with most candidates obtaining the method mark for substitution of the two given values. Some candidates then lost the accuracy mark through inaccurate evaluation, although the main reason for the loss of the final mark was because many candidates just put ‘change of sign therefore root between’ without stipulating 0.5 < root < 1. Part (b) was answered very well with most candidates obtaining the mark. Most candidates were able to do part (c), with the correct answers often seen. In part (d), the cobweb diagram was very well done, with the majority of candidates obtaining both marks. Some candidates lost the final mark through inaccurate positioning of x2 and x3 on the actual graph and not on the horizontal axis.

AQA - Core 3 58


2

1

2

2

1

2

2

tan sec

3 2)sec cos

2 32 2

cos 360 cos3 3

) 2 tan 10 5sec

Using the identity

2(sec 1) 10 5sec 0

2sec 5sec 12 0

(2sec 3)(sec 4) 0

3sec sec 4

49 312

22 1

49

cos c

1

os3 4

o o

o

a x x

so x or x

b x x

x x

x x

x x

x or x

x or

x o x

x

x

x

x

r

312 104 256o o oor x or x or x

Part (a) was well answered by the majority of candidates. Sec x being changed to sin x was a common error. The first angle was usually found correctly but the second angle was often incorrect, with 318 being a common error. There were very few cases of angles in radians being found. In part (b), most candidates used the correct identity and were successful in answering this part of the question. The main error was the use of tan2x = sec2x + 1, but 5sec x = 5(tan x + 1) was also seen. For those who successfully reached the correct quadratic, the factorisation, or use of the quadratic formula, was usually well done, though there were some sign errors. For the final four answers, many totally correct solutions were seen. A common error at this stage was 284 degrees, from 180 +104, and not the correct answer of 256. There were a number of candidates who failed to correctly round their answers to the nearest degree, with 104.47 = 104.5 = 105 seen.


4 4

4

4

2

4

4

) , 0

0 2 2

The

) does not have an inverse because it is not one-to-one.

1) ) ( ) ( ( )) 2

4

1 1) ( ) 14 2 14 16

( 4) ( 4)

range is ( ) 2

12

(

1 1( 4) 4

4)

16 2

a for all x x

x and x

b f

c i fg x f g xx

ii

f

fg xx x

x s

x

o x

x

1 12 2

14

24 3x x

or x

or

Part (a) was reasonably well answered although many candidates again lost a mark through poor notation. In part (b), candidates usually gave a correct response that f(x) was a many–one relationship or that it was not a one‐to‐one relationship. Few numerical examples were seen and there were many responses which simply stated it was because it was x4. Part (c)(i) was usually correctly answered, with the majority of candidates evaluating fg(x) in the correct order. Surprisingly few totally correct responses to part (c)(ii) were seen, with the majority of incorrect results coming from candidates not realising that (x − 4)4 = ... had both a positive and a negative solution. Consequently, those candidates who were able to invert correctly often only obtained one solution.


2 2

2 2 2

2 2

2 2

2 2

2 2

2

( 4 2)

) 2 ( 4 2) 2 4

2 8 4 2 4

To find the x-coordinateof the stationary points,

let's solve 2 6 8 0

, 0, 0 2 6 8 0

3 4

2 6 8

0

x

x x

x

x

x

x

y e x x

dya e x x e x

dx

e x x x

dye x x

dxdy

Because for all x e when x xdx

e x x

x x

2

2 2 22

28

2

22

2

2

2

( 4)( 1) 0

The x-coordinates of the stationary points are

)

4 1

(

4 This is a minimu

4 8 22)) 2 2 6 8 4 6

) m

1 This i

, 10 0

, 10 0 s a maxi mum

x x x

x x

d yb i e x x e x

dx

d yii for e

dx

d yfor e

d

x or x

e x x

x

xx

In part (a), most candidates used the product rule correctly, but at this stage e2x ≠ 0 was rarely seen, although many candidates obtained the correct quadratic equation. Factorisation, or use of the formula, was usually done well and many candidates obtained both of the correct solutions. Where candidates seemed unsure of what to do with the e2x, many expressions involving the use of ln e2x = 2x were seen together with the formation of a cubic expression in x. A third value of x = 0 was also seen by candidates who thought this was the solution of e2x = 0. Part (b) was generally well answered by those candidates able to do the product rule in part (a). Some errors with the coefficients meant candidates lost accuracy marks. In part (c), although many correct solutions were seen some

candidates lost marks through incorrect evaluation of 2

2

d y

dx

AQA - Core 3 59


22

2

4)3 4

3

1) )3 20 19

203 19 0 and multiplying by y bo

4ln

3

th sides:

3 20 19 0

)3 19 20 0

4(3 4)( 5) 0 5

3 19 20

3

53

n3

0

4

4l

x x

x x x x

x x

a e e

b i e e with e y and therfore ey

yy

y y o

x

r

ii y y

y y so y or y

e or e

x

y

o

y

r

ln(5)

Part (a) was well answered, with most candidates gaining both marks. The most common error was 3ln ex = ln 4,

resulting in ln 4

3x .

Part (b) was generally well done. Part (c) was not very well answered. Most candidates obtained 1 mark for the y‐values but many candidates stopped at this stage and therefore lost the final 2 marks. The error in part (a) was often repeated.


1

1

1

)The point P( 1, ) and Q(1,0)

1) A translation of vector

0

a stretch scale factor 2 in the y-dir

) 2cos ( 1)

) ) 2cos ( 1)

cos ( 1) cos 12 2

1)

1 cos2

0

a

b

AND

c Sketch of y x

d i y x

y yx so

yx

dxii

dy

x

1

sin2 2

1si

sin2 2

n )2, (12

yy

dxand when y

dy

Candidates were generally successful in part (a), sometimes aided by the allowances made for (–1,180). Part (b) was well answered by many candidates. Common errors were the use of ‘shift’ and ‘transformation’ to describe the word ‘translation’, and the stretch was often given a scale factor of ½ in the x‐axis. In part (c), the shape of the graph was usually correct, but many candidates lost this mark by going into other quadrants. The mark for the labelling of the axis was not so readily earned, as these values were often missing or incorrect. In part (d)(i), many candidates gave the correct response. The major error was to state that 2y =cos–1(x – 1) for the first step resulting in no marks being awarded. In part (d)(ii), most candidates at least obtained the method mark. Where candidates correctly differentiated, they went on to achieve full marks.

AQA - Core 3 60


2 2

4 4(4 3) 4 4 12)

4 3 (4 3) (4 3)

4) ) ln(4 3) so ln(4 3)

4 3

4) When 1, ln(4 3) 4

4 3When 1, (1) 1ln(4 3) 0

The equation of the ta

12

4ln(4

ngent at t

3)4 3

h

x dy x xa y so

x dx x x

dyb i y x x x

k

dy xx

xdx x

dyii x m

d

x

y

dx

xx

e point where 1 is

0 4( 1)

) ) 4 3 4 3 4

4 3 1 1 3 11 3ln

4 3 4 4 4

in terms of :

4 4

1 3(4 3 3l

3( )

4

) ln(4 3) 1 ln(4

n(4 3)) ln(4 3)4

3) ln(

4

x

y x

c i u x so x u and du dx

x udx du du u u c

x u u

y x

x x c x xx

with C c

ii x dx x dx x

C

4

4 3)4 3

4ln(4 3)

3ln(4 3) ln(4 3

4)

43

x x dxx

xx x dx

xx x x x C

Very few candidates achieved completely correct solutions to this question. Part (a) was well answered by the majority of candidates. The main error was the two terms in the numerator being interchanged, resulting in k = +12. Other errors came from the expansion of the brackets, with various numerators including 8x, and with –3 not being multiplied by 4, giving a final constant of –3. In part (b)(i), the expression ln(4x – 3) was rarely differentiated completely correctly. A common error was to get

ln(4 3),4 3

4ln(4 3)

4 3

xx

xx

rather than xx

,

which earned two method marks.

In part (b)(ii),dy

dx was frequently calculated

correctly, following through from part (b)(i), but then this value was used as the gradient of the normal, with some candidates explicitly stating that they were finding the normal. In part (c)(i), most of the candidates earned a method mark for ‘du = 4dx’. Many of these candidates went on to write the integral completely in terms of u. Some candidates lost marks at this stage through either obtaining an extra factor of 4 or getting

34 3

4

ux when rearranging u x

.

Although many fully correct responses were seen, some candidates were unable to proceed beyond this point: part (c)(ii) proved to be beyond many candidates, as the correct application of integration by parts was not common. Most candidates started the

question with u = ln and 4 3dv

xdx

. Those

candidates who correctly started this part of the question often failed to get the first accuracy mark because of their derivative of ln(4x – 3), losing the factor of 4 in the numerator. However, many were able to get the next method mark because they correctly used their value from part (c)(i). A few candidates spotted that the result obtained in part (b)(i) could provide an alternative method of solution.

GRADE BOUNDARIES


Core 3 – Unit PC3 75 59 51 43 36 29

AQA - Core 3 61

QSo

lutio

nM

arks

Tot

alC

omm

ents

1 x

y 1

0.5

3 0.

366(

0)

5 0.

309(

0)

7 0.

274(

3)

9 0.

25

B1

B1

x va

lues

and

no

extra

val

ues

4+ c

orre

ct y

val

ues

1or

et

c1+

3

()

()

()

0.5

0.25

12

34

0.36

600.

2743

20.

3090

=

++

××

++

M

1 C

orre

ct a

pplic

atio

n of

Sim

pson

’s ru

le

for t

heir

x va

lues

(x o

dd)

= 2

.62

A1

4 C

SO

mus

t be

3sf

Tot

al4

2 (

)2

dV

yx

= ()

()5

2d

xx

=−

M1

()

()

4 3

6–

2 6x

=A

1 lim

its n

ot re

quire

d

()

6 21

–6

6=

m

1co

rrec

t sub

stitu

tion

into

()

()6

–2

kx

10.5

=A

1 4

allo

w e

quiv

alen

t fra

ctio

n 63

etc

6

(AW

RT

10.

5 or

10.

5π

m1,

A0)

T

otal

4

Q

Solu

tion

Mar

ksT

otal

C

omm

ents

3(

a)

()

3f

5–

4x

xx

=+

()

f0.

5–1

.375

=M

1 C

ondo

ne f

(0.5

) rou

ndin

g to

– 1

.4

()f1

2=

Cha

nge

of si

gn

0.5

1α

∴<

<

A1

2 B

oth

stat

emen

ts n

eede

d

(b)

35

–4

0x

x+

=

35

4–

xx

=

Mus

t be

seen

()

31

4–

5x

x=

B1

1 A

G

(c)

10.

5x

=

() (

)2

310.

775

40x

==

M1

For

()

23

or2

sfx

x=

30.

707

x=

A

1 2

(d)

M1

A1

2

From

0.5

ver

tical

to c

urve

th

en h

oriz

onta

l to

line

CA

O

Tot

al

7

Cor

e 3

- Jan

200

9 - M

ark

sche

me

AQA - Core 3 62

Q

Solu

tion

Mar

ksT

otal

C

omm

ents

4(a)

3

sec

2x

=

2co

s3

x=

x =

48, 3

12

B1

1 co

rrec

t

(Con

done

ans

wer

s rou

ndin

g to

) B

1 2

2 co

rrec

t and

no

extra

s in

inte

rval

(b

) 2

2tan

10–

5sec

xx

=

()

22

sec

–110

–5s

ecx

x=

M1

Use

of t

rig id

entit

y co

rrec

tly

()

22s

ec5s

ec–1

20

xx

+=

A1

()(

)()

2sec

–3

sec

40

xx

+=

m1

Atte

mpt

to so

lve

or fa

ctor

ise

1 sl

ip u

sing

form

ula

3

sec

,–4

2ei

ther

of t

hese

2

1co

s,–

34

x x= =A

1

x =

48, 3

12, 1

04, 2

56

B1

AW

RT

3 c

orre

ct

con

done

105

or 2

55

B1

6 A

ll co

rrec

t and

no

extra

s in

inte

rval

2 2 22

22

2

Alte

rnat

ive:

2sin

510

–co

sco

s2s

in10

cos

–5c

os2

–2

cos

10co

s–

5cos

12co

s–

5co

s–

20

xx

x xx

xx

xx

xx

= ==

=th

en re

st o

f sch

eme

as a

bove

(M1)

(A1)

Tot

al

8

5(a)

(

)f

2,f

2,2

xy

≤≤

≤B

2 2

()

2,f

2,2

2,f

2x

xy≤

<≤

<<

B1

(b)

()

fx

is n

ot o

ne to

one

E1

1

Allo

w m

any

to o

ne o

r num

eric

al e

xam

ple

(c)(

i) (

)4

1fg

2–

4x

x=

−B

1 1

(ii)

41

214

–4

x−

=−

41

16–

4x

=

()4

14

16 1–

42

x x

−=

=±

M1

M1

Cor

rect

han

dlin

g of

four

th ro

ot

Mus

t hav

e

±

Cor

rect

han

dlin

g of

reci

proc

al

11

4,3

22

x=

A1

3

Tot

al7

Q

Solu

tion

Mar

ksT

otal

C

omm

ents

6(

a)

()

22

e–

4–

2x

yx

x=

()

2d

e2

–4

dx

yx

x=

M1

Prod

uct r

ule;

allo

w 1

slip

()

22

–4

–2

2ex

xx

+A

1

()

22

de

2–

42

–8

–4

dx

yx

xx

x=

+M

1 Fa

ctor

isin

g (

)2

2e

a6

0x

xx

++

()

22

e2

–6

–8

xx

xA

1 or

2

–3

–4

0x

x=

2 e0

x≠

(

)()

–4

10

xx+

=m

1 So

lvin

g 3

term

qua

drat

ic

Dep

ende

nt o

n bo

th M

mar

ks

4,1

x=

−

A1

6 A

nd n

o ex

tras e

g x

= 0

(b)(

i) (

)2

22

2

de

.22

–4

2ed

xx

yx

x=

+

()

()

22

2–

42

4e2e

2–

4x

xx

xx

+−

+

M1

A1

Prod

uct r

ule

from

thei

rd dy xin

form

2 ex

(qua

drat

ic)

()

22 e

4–

8–

22x

xx

2 O

r

()

()

22

22

2d

e4

–6

2–

6–

82e

dx

xy

xx

xx

=+

M1

A1

(ii)

4x

=:

()

8"

e10

0M

INy

=>

∴M

1 Th

eir 2

x’s

in th

eir

2

2

d dy x

only

of f

orm

2 ex

(qua

drat

ic)

–1x=

:(

)–2

"e

–10

0M

AX

y=

<∴

A1

2 C

SO

Bot

h co

rrec

t A

llow

val

ues e

ither

side

of y

or

y′

Tot

al10

7(

a)

3e4

x=

4

e3

x=

M1

4ln

3x

=A

1 2

(b)(

i) –

3e20

e19

xx

+=

2

203

19or

3e20

19e

xx

yy

+=

+=

2

3–1

920

0y

y+=

B

1 1

AG

(ii

) (

)()

3–

4–

50

yy

=4 ,5 3

y=B

1

4ln

,ln5

3x

∴=

M1

A1

3 ln

(the

ir +

ve y

’s)

Tot

al6

AQA - Core 3 63

Q

Solu

tion

Mar

ksT

otal

C

omm

ents

8(

a)

()

–1,

PB

1 C

ondo

ne (

)–1

,180°

()

1,0

QB

1 2

(b)

Tran

slat

e

E1

1 0

B1

or e

quiv

alen

t in

wor

ds

Stre

tch

SF 2

⁄⁄

y-ax

isM

1 St

retc

h +

one

othe

r cor

rect

A

1 4

all c

orre

ct

(c)

B1

B1

2

Cor

rect

shap

e in

1st

qua

dran

t

2an

d 2

mar

ked

corre

ctly

(d)(

i) (

)1

cos

12y

x−

=−

M1

cos

12y

x=

−

cos

12y

x=

+

A1

2

(ii)

1si

n2

2y−

M

1

A1

()

sin

...k d

corre

ctdx y

d1

At

2,si

n1d

2x

yy

==

−A

1 3

Con

done

AW

RT

–0.4

2

Tot

al13

Q

Solu

tion

Mar

ksT

otal

C

omm

ents

9(a)

4

43

xy

x=

− ()

()

()2

43

.44

4d d

43

xx

y xx

−−

=−

M1

Mus

t use

quo

tient

rule

C

ondo

ne o

ne sl

ip

()2

124

3x−

=−

A1

2 12

k=−

(b)(

i) (

)ln

4–

3y

xx

=

()

d.4

ln4

–3

d4

–3

yx

xx

x=

+M

1

m1

A1

3

()

()

f 4–

3x

gx

x+

‘f(x

)’ m

ay b

e co

nsta

nt

()

ln4

–3

43

kxx

x+

−

(ii)

10

xy

==

B

1 d

4dy x

=M

1 Su

b x

= 1

into

thei

r d dy x

()

4–1

yx

∴=

an

y co

rrec

t for

m

A1

3 C

SO

Mus

t hav

e fu

ll m

arks

in (b

)(i)

(c)(

i) 4

–3

ux

=

d4d

ux

=

M1

43

dd

4–

34

xu

ux

xu+

=A

1 4

3O

rd

1d

43

43

xx

xx

x=

+−

−

()

13

1(d

)4

uu

=+

m

1 3

1du

u=

+ e

tc

()

13l

n4

uu

=+ (

)(

)(

)1

4–

33l

n4

–3

4x

xc

=+

+A

1 4

CSO

Con

done

mis

sing

du

(ii)

()

ln4

–3

dx

x

()

dln

4–

31

dvu

xx

==

M

1 In

cor

rect

dire

ctio

n

d4

d4

–3

uv

xx

x=

=

()

4ln

4–

3–

d4

–3xx

xx

x=

A1

()

()

()

1ln

4–

3–

4–

33l

n4

–3

4(

)

xx

xx

c

=+

+

m1

A1

4 (

)ln

4–

3–

thei

r(c)

(i)x

x

Tot

al16

T

OT

AL

75

AQA - Core 3 64

GeneralCertificate

ofEducation

June2009


MATHEMATICS

MPC3

UnitPure

Core

3

Friday5June2009

1.30pm

to3.00pm

Forthis

paperyoumusthave:

*an8-pageanswerbook




Tim


Instructions

*Use

black

inkorblack


only

beusedfordrawing.

*Write

theinform

ationrequired



paper


isMPC3.

*Answ

erallquestions.

*Show

allnecessary


belost.

Inform

ation

*Themaxim

um

markforthispaper

is75.


brackets.

Advice

*Unless

stated

otherwise,

youmay

quote

form

ulae,

withoutproof,from

thebooklet.

Answ

erallquestions.

1(a)

Thecurvewithequation

y¼

cosx

2xþ1,

x>�1 2

intersects

theliney¼

1 2at

thepointwherex¼

a.

(i)

Show

that

alies

between0andp 2.

(2marks)

(ii)

Show

that

theequation

cosx

2xþ1¼

1 2canberearranged

into

theform

x¼

cosx�1 2

(1mark)

(iii)

Use

theiterationx n

þ1¼

cosx n

�1 2withx 1

¼0to

findx 3

,givingyouransw

er

tothreedecim

alplaces.

(2marks)

(b)

(i)

Given

that

y¼

cosx

2xþ1,use

thequotientrule

tofindan

expressionfordy

dx. (3marks)

(ii)

Hence

findthegradientofthenorm

alto

thecurvey¼

cosx

2xþ1at

thepointonthe

curvewherex¼

0.

(2marks)

AQA - Core 3 65



fðxÞ¼


ffi2xþ5

p,

forreal

values

ofx,

x5

�2:5

gðxÞ

¼1

4xþ1,

forreal

values

ofx,

x6¼

�0:25

(a)

Findtherangeoff.

(2marks)

(b)

Theinverse

offis

f�1

.

(i)

Findf�1

ðxÞ.

(3marks)

(ii)

State

thedomainoff�1

.(1

mark)

(c)

Thecomposite

functionfg

isdenotedbyh.

(i)

Findan

expressionforhðxÞ

.(1

mark)

(ii)

SolvetheequationhðxÞ

¼3.

(3marks)

3(a)

Solvetheequationtanx¼

�1 3,givingallthevalues

ofxin

theinterval

0<x<2pin

radiansto

twodecim

alplaces.

(3marks)

(b)

Show

that

theequation

3sec2

x¼

5ðta

nxþ1Þ

canbewritten

intheform

3tan2x�5tanx�2¼

0.

(1mark)

(c)

Hence,orotherwise,

solvetheequation

3sec2

x¼

5ðta

nxþ1Þ

givingallthevalues

ofxin

theinterval

0<x<2p

inradiansto

twodecim

alplaces.

(4marks)

4(a)

Sketch

thegraphofy¼

j50�x2j,

indicatingthecoordinates

ofthepointwherethe

graphcrosses

they-axis.

(3marks)

(b)

Solvetheequationj50�x2j¼

14.

(3marks)

(c)

Hence,orotherwise,

solvetheinequalityj50�x2j>

14.

(2marks)

(d)

Describeasequence


ationsthat

mapsthegraphofy¼

x2

onto

thegraphofy¼

50�x2.

(4marks)

5(a)

Given

that

2lnx¼


(1mark)

(b)

Solvetheequation

2lnxþ

15

lnx¼

11

givingyouransw

ersas

exactvalues

ofx.

(5marks)

AQA - Core 3 66

6Thediagram



ffiffiffiffiffiffiffiffiffi

100�4x2

p,wherex5

0.

(a)

Calculate

thevolumeofthesolidgenerated

when

theregionbounded

bythecurve

shownaboveandthecoordinateaxes

isrotatedthrough360�aboutthey-axis,giving

youransw

erin

term

sofp.

(5marks)

(b)

Use

themid-ordinaterule

withfivestripsofequal

width

tofindan

estimatefor

ð 5 0



100�4x2

pdx,givingyouransw

erto


(4marks)

(c)

ThepointPonthecurvehas

coordinates

ð3,8Þ.

(i)

Findthegradientofthecurvey¼



100�4x2

pat

thepointP.

(3marks)

(ii)

Hence

show

that

theequationofthetangentto

thecurveat

thepointPcanbe

written

as2yþ3x¼

25.

(2marks)

(d)

Theshaded

regionsonthediagram

below

arebounded

bythecurve,

thetangentat

P

andthecoordinateaxes.

Use

youransw

ersto

part(b)andpart(c)(ii)to

findan

approxim

atevalueforthetotal

area

oftheshaded

regions.

Giveyouransw

erto


(5marks)

y

10 O

5x

Pð3,8Þ

x5

y

10 O

7(a)

Use

integrationbyparts

tofind

ð ðt�1Þln

tdt.

(4marks)

(b)

Use

thesubstitutiont¼

2xþ1to

show

that

ð 4xlnð2xþ1Þd

xcanbewritten

asð ðt

�1Þln

tdt.

(3marks)

(c)

Hence

findtheexactvalueof

ð 1 04xlnð2xþ1Þd

x.

(3marks)

END

OF

QUESTIO

NS

AQA - Core 3 67



cos 1 1) intersect.

2 1 2 2cos 1

is solution of the equation 2 1 2

cos 1' ( )

2 1 2cos 0 1

) (0) 0.5 01 20 1

( ) 0.5 02 1 2

According to ,

we know that t

the change of sign rule

here

xa y for x and y

xx

xx

Let s call f xx

i f

f

1

2

2

2 3

is a root so that 0 .2

cos 1) 2cos 2 1 2 2cos 1

2 1 2

)

cos sin (2 1) cos 2) )

2 1 (2 1)

2co

1cos

20 , 0.5 , 0.3

s (2 1)sin(2 )

7

(2 1)

2 1 1 0) ,

8

0

xii x x x x

x

iii

x dy x x xb i y so

x dx x

dy x x x

dx x

dyi

x

i When xd

x

x

x

x x

2

1gradient of the normal

21

The2

is

Part (a)(i) was not very well answered by the majority of candidates. Fully correct responses were seen but it was usually from candidates who successfully rearranged the equation into the form f(x) = 0, which was seen in a number of acceptable forms. Where candidates simply substituted in the two values given into the LHS they obtained 1 and 0 but still indicated that this was a change of sign and therefore there was a root. Most candidates achieved at least one mark but lost the second by simply stating change of sign therefore a root. Very few incorrect responses were seen in part (a)(ii). Part (a)(iii) was very well answered. The main error was with the candidates who used degrees rather than radians Part (b)(i) was very well answered, with most candidates successfully using the quotient rule. Where errors occurred it was usually through missing brackets, although many candidates were able to recover at some stage in the working. As part (b)(ii) followed on from part (i) most candidates were able to obtain the mark for substitution; even those who had incorrectly simplified their work. Many fully correct responses were seen. Several candidates also stopped after the substitution of x = 0 and left their answer as ‐2.


2

2

2

1

1

2

1

( ) 2 5 , 2.5

1( ) , 0.25

4 1

) 2.5, 2 5 0 2 5 0

) ) 2 5 2 5

55 2

2

)The domain of is the range of

The doma

( ) 0

5

in of

1) ) ( ) ( ( )) 2

4 1

( )2

0

f x x x

g x xx

a for all x x and x

b i y x so y x

yy x an

f

d x

ii f f

f is

c i fg x f

x

xf x

xx

x

g

5

2) ( ) 3 5 3

4 12 2 4 1 1

5 9 44

25

4 1

1

8

1 4 1 2 4

4(4 1) 2 16 4 2

ii h xx

xso

x x

x x

x

x

Part (a) was not very well answered by the majority of candidates. Errors occurred due to poor notation. Part (b)(i) was very well answered, with most candidates achieving full marks. Part (b)(ii), like part (a), was not very well answered, with poor notation. Part (c)(i) was well answered, although candidates often spoiled their work with incorrect subsequent working which was then penalised in the next part. Many totally correct responses were seen in part (c)(ii) and those candidates who worked with an incorrect h(x) often achieved the method mark for squaring. A common error was to substitute x = 3 into h(x).

AQA - Core 3 68


1

2

2

2 2

2

2

1 1) tan . tan ( ) 0.32

3 30 2 , 0.32 2 0.32

)3sec 5(tan 1)

Using the identity ,

3(tan 1) 5 tan 5

3tan 5 tan 2 0

)3tan 5 tan 2 0

(3tan 1

2.82

tan

)(tan 2) 0

tan

5 9

1 ec

. 6

s

a x rad

because x x or x

b x x

x x

x x

x x

c x x

x x

x

1

1tan 2

3

tan 2

1.11

2.82 5.96 1.11

4.25

or x

or x orx r x

x

o

Most candidates obtained the method mark in part (a) by obtaining ‐0.32. Many candidates went on to complete this part correctly, although incorrectly writing their answer to 2dp was a common error. Answers in degrees were not common, but they were seen. Part (b) was answered very well by the majority of candidates with the correct trigonometric identity being used. In part (c) most candidates were able to successfully factorise the quadratic expression, and many went on to complete the question correctly. Marks were lost by ‘extra’ values being given or poor accuracy of writing their answers to 2dp.


2 2 2

2 2

2

) 50

0, 50 50

The graph crosses the y-axis at

b) 50 14 50 14 50 14

36 64

)The

(0,50

curve is "above" the line when

) 50 can be obta

)

6 6 8 8

8 6 6

in

8

a y x

for x y

x x or x

x or x

c

d y

x or x or x or x

x

x or x or x

2from

a followed byre

a

flection in the x-axis

0translation of vector

50

y x

through

Candidates scored well on this question, with many correct graphs seen in part (a). Where candidates only scored 2 marks it was

generally the curvature beyond 50 50

which was at fault. In part (b) x² =36 was the most common answer to earn the method mark. Many candidates also just gave the two positive solutions of 8 and 6 earning one accuracy mark. Quite a few candidates did come up with all four solutions. Part (c) was not very well answered. The inequalities for x < –8 and x > 8 were often seen but ‐6 < x < 6 was not often encountered. In part (d) most candidates knew they needed a reflection and a translation, gaining two of the marks, but often had the translation first and hence the corresponding line of reflection was incorrect.


2

53

2

2

2

5) 2ln 5 ln

215

) 2 ln 11 (multiplying by ln )ln

2(ln ) 15 11ln

2(ln ) 11ln 15 0

(2 ln 5)(ln 3) 0

2ln 5 0 ln 3

a x x

b x xx

x

x e

x

x x

x

x e

x

x

or x e

or x

This question proved to be the downfall for many of the candidates. It was the poorest answered question on the paper. Many candidates answered part (a) correctly with the correct answer often seen. A common error was not to give the answer in an exact form, but this only affected candidates who showed no working. The solution x = ln(5/2) was also common. In part (b) the only candidates who were really successful here were the ones who used a substitution such as y = ln x, who then were able to formulate and solve the quadratic very easily. Those candidates who attempted to work in ln x usually (despite condoning poor notation) failed to obtain a quadratic and hence scored zero. Many candidates just substituted 2ln x =5 and verified that it fitted the equation.

AQA - Core 3 69

Question 6: Exam report 10 2

0

2 2 2 2 2 2

1010 2 3

00

5 20.5 1.5 20

)

We need to rearrange the equation, making the subject

1100 4 4 100

500

3

(100 )4

1 1(100 ) 100

4 4 3

11000 1000

4 3

) 100 4 1

a V x dx

x

y x x y x y

V y dy y y

V

b x dx y y y

.5 3.5 4.5

12 2

2

9.949874 9.539392 8.660254 7.141428 4.358899

1 4) ) 8 (100 4 )

2 100 412 12

When 3,8100 36

)The equation of the tangent at P is :

39

38 ( 3

.6 3 . .

3

22

2

)

y y

dy xc

t

i x xdx x

dyx

dx

ii

y x

o sig fig

m

16 3 9

25)The tangent crosses the y-axis at (0, )

225

and the x-axis at ( ,0)3

The area of the made with the two axes and the tangent

1 25 25 is

2 3 2

triangle

625

1An approximation of the area

2 3 25

2

y x

d

y x

underneath the curve is 39.6

625Therefore, an approximation of the shaded area is 39.6

1.

212 5

Many candidates lost marks on this question from careless work. Some candidates scored very well on this question and full marks were not uncommon amongst the more able candidates. Many candidates obtained full marks in part (a). Where candidates scored 4 out of 5 it was generally because at no stage did they indicate ‘dy’ and hence lost the B mark. A very common error was in ∫(100‐y²)dy , where the result was often (100x – y³/3), again with candidates very unsure as to whether they were integrating w.r.t. x or y. Many special cases were seen and, where attempted, candidates often achieved 2 marks. Several candidates found it difficult to isolate x². Although many candidates answered part (b) correctly they often lost the final mark for not writing the answer to 3sf. A common error was also to miss out one of the x values, usually the 0.5. Candidates still seem not to understand that if they require an answer to 3sf then they should be showing either exact values of y or working to 4 sf. Part (c)(i) was not particularly well answered, although many candidates did have ½(100‐4x²)‐½ and earned the method mark. Various errors occurred with the 8 and the ½ and many candidates lost the negative sign. In part (c)(ii) those candidates who made an error in part (i) lost marks for this part. Although candidates who followed through with their incorrect gradient gained the method mark, many fudged solutions were seen as ‐3/2 suddenly appeared from nowhere. Part (d) was quite well answered by most candidates and full marks were often seen. The method of finding the intercepts on the axes was used most frequently, although use of integration was not uncommon. Candidates who integrated often made errors on the limits to be used

AQA - Core 3 70


2 2

2

1

0

2 2

1 1 1) ( 1) ln ( ) ln ( )

2 21 1

ln 12 2

1 1 1) 2 1

2 2 21 1 1

4 ln(2 1) 4( ) ln2 2 2

) When 0, 1 1, 3

4

1

( 1) l

1ln

2 4

ln(2

n

1)

a t t dt t t t t t dtt

t t t t dt

b t x so x

t t

t and dx dt

x x dx t t dt

c x t and x t

x

t t d

x x

t

t t c

d

t

33 2 2

11

1 1( 1) ln ln

2 4

9 9 1 1 3 3 1( 3) ln 3 3 ( 1) ln1 1 ln 3 12 4 2 4 2 4

3ln 3

2

4

t t dt t t t t t

Most candidates lost marks on this question. Full marks were not common although they were seen from the more able candidates. Not many candidates obtained full marks for part (a). Most candidates scored 2 out of 4 since they started correctly by differentiating ln t and integrating (t –1). Obvious errors were differentiating both terms or starting by trying to integrate ln t. Several candidates managed to obtain the second accuracy (A) mark for simplification to (t²/2 – t) ln t ‐ ∫(t/2 – 1 ) dt but lost the final accuracy mark by ending up with a final term of –t, not + t. Part (b) was done well by most candidates, although poor manipulation often cost the loss of the final accuracy mark. Some candidates also confused the issue by trying to introduce terms in u and du. Part (c) was not very well answered, with candidates failing to change the limits as a common error. Many candidates did not appreciate that they should be using their answer to part (a) and tried to start again, often obtaining different answers to those they had found in part (a). After correctly approaching the choice of u and dv/dt, most were then defeated by the required manipulation of the subsequent algebra.

GRADE BOUNDARIES


Core 3 – Unit PC3 75 61 53 46 39 32

AQA - Core 3 71

Q

Solu

tion

Mar

ks

Tot

al

Com

men

ts

1(a)

(i)

()

cos

1f

–2

12

xx

x=

+

OE

()

1f

02

=;

1f

–2

2=

M1

0LH

S1,

LHS

02

xx

==

==

Cha

nge

of si

gn

02

α<

<

A1

2 Ei

ther

side

of

1,

02

2α

<<

∴

(ii)

cos

12

12

xx

=+

2cos

21

1or

, co

s2c

os–1

22

xx

xx

xx

=+

=+

=

Eith

er li

ne

1co

s2

xx

=−

B1

1 A

G; o

r 1

cos

2x

x−

=

All

corr

ect w

ith n

o er

rors

(iii)

10

x=

20.

5x

=

30.

378

x=

M1

A1

2

Atte

mpt

at i

tera

tion

()

23

allo

w0.

5,0.

38,0

.4x

x=

−=

CA

O

(b)(

i) (

)()

()2

21

–si

n–

cos

2d d

21

xx

xy x

x

+×

=+

M1

Atte

mpt

at q

uotie

nt ru

le:

() (

)2

21

sin

2cos

21

xx

xx

±+

±

+A

1 Ei

ther

term

cor

rect

A

1 3

All

corr

ect

I

SW

(ii)

0x

=d

–2

dy x=

m1

Cor

rect

ly su

bst.

0x

= in

to th

eir

d dy x1

Gra

dien

t of n

orm

al =

2∴

A1

2 C

SO

Tot

al

10

QSo

lutio

nM

arks

T

otal

Com

men

ts2(

a)

()

f0

xM

1

()

For

0, f

0x

>

A1

2 C

orre

ct; a

llow

0,

f0

y�

(b)(

i) 2

5y

x=

+2

5x

y=

+

M1

xy

⇔

22

5x

y=

+

M1

Atte

mpt

to is

olat

e, sq

uarin

g fir

st

()

21

–5

f2

xx

−=

A1

3 co

ndon

e (

) y

=

(ii)

0x

B1F

1

ft th

eir (

a), b

ut m

ust b

e x

2(c)

(i)

()

()

hfg

xx

=

12

54

1x

=+

+

B1

1

(ii)

12

53

41

x+

=+

12

59

41

x+

=+

M

1 on

e co

rrec

t ste

p fr

om (c

)(i),

squa

ring

12

41

x=

+1

41

or16

42

2x

x+

=+

=

A1

1–

or e

quiv

8x

=A

1 3

CSO

T

otal

10

3(a)

1

1ta

n0.

323

−−

=−

M

1 Si

ght o

f 0.

32or

18.4

3±

2.82

,5.9

6x

=

A1

A1

3 a

corr

ect a

nsw

er

AW

RT

–1

for a

ny e

xtra

in ra

nge,

igno

re e

xtra

an

swer

s not

in ra

nge.

[S

C 1

61.5

7,34

1.57

AW

RT

M1A

1

(max

2/3

)]

(b)

()

23

tan

15t

an5

xx

+=

+2

3ta

n–

5tan

–2

0x

x=

B

1 1

AG

3(c)

(

)()

3tan

1ta

n–

20

xx

+=

M1

Atte

mpt

at f

acto

risat

ion/

form

ula

1ta

n2,

–3

x=

A1

1.11

,4.2

5,2.

82,5

.96

x=

A

WR

T B

1 3

corr

ect

[SC

x

= 1

.11,

4.2

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thei

r tw

o a

nsw

ers f

rom

(a)]

B

1 4

4 co

rrec

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ext

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161

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341.

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8

eith

er

Cor

e 3

- Jun

2009

- M

ark

sche

me

AQA - Core 3 72

QSo

lutio

nM

arks

T

otal

Com

men

ts4(

a)

M1

A1

A1

3

Mod

ulus

gra

ph, 3

sect

ion,

con

done

shap

e in

side

+ o

utsi

de

50±

Cus

ps +

cur

vatu

re o

utsi

de

50±

V

alue

of

()

and

shap

e in

side

50

y±

(b)

250

–14

x=

2

2

22

50–

1436

50–

–14

64x

xx

x=

==

=M

1

Eith

er

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x=

±±

A

1 2

corr

ect,

from

cor

rect

wor

king

A

1 3

All

4 co

rrec

t, fr

om c

orre

ct w

orki

ng

(c)

–66

x<

<B

18,

8x

x>

<−

B1

2

(d)

Ref

lect

in x

-axi

s

Tran

slat

e 0 50

M1,

A1

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m1

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ula

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52

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=

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1,A

1 5

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for s

ubst

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100

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may

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scor

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4)

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(i)

()

()

12

2d

110

0–

4–8

d2

yx

xx

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M1

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in ru

le (

)(

)1 2

fx

−×

; allo

w f(

x) =

k

()

()

1f

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xx

==

()1

–2

d3

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100

–36

dyx

x=

=A

1

3–

oreq

uiva

lent

2=

A1

3 C

SO

(ii)

()

38

––

32

yx

−=

M1

()

d8

thei

r3

dyy

xx

−=

−

dor

thei

r dyy

xc

x=

+an

d su

bst.

(3,8

) to

find

c (

)2

–16

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9y

x=

+

23

25y

x+

=A

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AG

; all

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AQA - Core 3 73

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125

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162

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n an

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3−

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(

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=A

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1

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into

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L75

AQA - Core 3 74

GeneralCertificate

ofEducation


January

2010

Mathematics

MPC3

UnitPure

Core

3

Friday15January

2010

1.30pm

to3.00pm

Forthis

paperyoumusthave:

*an8-pageanswerbook


ulaeandstatisticaltables

*aninsertforusein

Question2(enclosed).


Tim

eallowed

*1hour30minutes

Instructions

*Useblackinkorblackball-pointpen.Pencilshould

only

beusedfordrawing.

*Write

theinform

ationrequiredonthefrontofyouranswerbook.TheExaminingBodyfor

this

paperis

AQA.ThePaperReferenceis

MPC3.

*Answerallquestions.

*Show

allnecessary

working;otherw

isemarksformethodmaybelost.

*Fillin

theboxesatthetopoftheinsert.

Inform

ation

*Themarksforquestionsare

shownin

brackets.

*Themaxim

um

mark

forthis

paperis

75.

Advice

*Unlessstatedotherw

ise,youmayquote

form

ulae,withoutproof,from

thebooklet.

Answ

erallquestions.

1A

curvehas

equationy¼

e�4xðx

2þ2x�2Þ.

(a)

Show

that

dy

dx¼

2e�

4xð5

�3x�2x2Þ.

(3marks)

(b)

Findtheexactvalues

ofthecoordinates


ofthecurve.

(5marks)

2[Figure

1,printedontheinsert,is

provided

foruse

inthis

question.]

(a)

(i)

Sketch

thegraphofy¼

sin�1

x,whereyis

inradians.

State

thecoordinates

of

theendpoints

ofthegraph.

(3marks)

(ii)

Bydrawingasuitable

straightlineonyoursketch,show

that

theequation

sin�1

x¼

1 4xþ1

has

only

onesolution.

(2marks)

(b)

Therootoftheequationsin�1

x¼

1 4xþ1is

a.Show

that

0:5<a<1.

(2marks)

(c)

Theequationsin�1

x¼

1 4xþ1canberewritten

asx¼

sin� 1 4

xþ1� .

(i)

Use

theiterationx n

þ1¼

sin� 1 4

x nþ1� w

ithx 1

¼0:5

tofindthevalues

of

x 2andx 3

,givingyouransw

ersto

threedecim

alplaces.

(2marks)

(ii)

Thesketch

onFigure

1showsparts

ofthegraphsofy¼

sin� 1 4

xþ1� an

d

y¼

x,andthepositionofx 1

.

OnFigure

1,draw

acobweb

orstaircasediagram

toshow

how

convergence

takes


andx 3

onthex-axis.

(2marks)

AQA - Core 3 75

3(a)

Solvetheequation

cosecx¼

3

givingallvalues

ofxin

radiansto

twodecim

alplaces,in

theinterval

04

x4

2p.

(2marks)

(b)

Byusingasuitable

trigonometricidentity,solvetheequation

cot2x¼

11�cosecx

givingallvalues

ofxin

radiansto

twodecim

alplaces,in

theinterval

04

x4

2p.

(6marks)

4(a)

Sketch

thegraphofy¼

j8�2xj.

(2marks)

(b)

Solvetheequationj8

�2xj¼

4.

(2marks)

(c)

Solvetheinequality

j8�2xj>

4.

(2marks)

5(a)

Use

themid-ordinaterule

withfourstripsto

findan

estimatefor

ð 12 0lnðx

2þ5Þd

x,

givingyouransw

erto


(4marks)

(b)

Acurvehas

equationy¼

lnðx

2þ5Þ.

(i)

Show

that

this

equationcanberewritten

asx2¼

ey�5.

(1mark)

(ii)

Theregionbounded

bythecurve,

thelines

y¼

5andy¼

10andthey-axis

is

rotatedthrough360�aboutthey-axis.Findtheexactvalueofthevolumeofthe

solidgenerated.

(4marks)

(c)

Thegraphwithequationy¼

lnðx

2þ5Þis

stretched

withscalefactor4parallelto

the

x-axis,andthen

translated

through

0 3�� to

givethegraphwithequationy¼

fðxÞ.

Write

downan

expressionforfðx

Þ.(3

marks)



fðxÞ¼

e2x�3,

forallreal

values

ofx

gðxÞ

¼1

3xþ4,

forreal

values

ofx,

x6¼

�4 3

(a)

Findtherangeoff.

(2marks)

(b)

Theinverse

offis

f�1.

(i)

Findf�

1ðxÞ

.(3

marks)

(ii)

Solvetheequationf�

1ðxÞ

¼0.

(2marks)

(c)

(i)

Findan

expressionforgfðx

Þ.(1

mark)

(ii)

Solvetheequationgfðx

Þ¼1,givingyouransw

erin

anexactform

.(3

marks)

7Itis

given

that

y¼

tan4x.

(a)

Bywritingtan4xas

sin4x

cos4x,use

thequotientrule

toshow

that

dy

dx¼

pð1

þtan24xÞ,

wherepis

anumber

tobedetermined.

(3marks)

(b)

Show

that

d2y

dx2¼

qyð1

þy2Þ,

whereqis

anumber

tobedetermined.

(5marks)

8(a)

Usingintegrationbyparts,find

ð xsinð2x�1Þd

x.

(5marks)

(b)

Use

thesubstitutionu¼

2x�1to

find

ðx2

2x�1dx,givingyouransw

erin

term

sofx.

(6marks)

END

OF

QUESTIO

NS

AQA - Core 3 76



4 2

4 2 4

4 2

4 2 4 2

4 2 2

4

( 2 2)

) 4 ( 2 2) (2 2)

4 8 8 2 2

10 6 4 2 (5 3 2 )

)To find the stationary points, solve 0

2 (5 3 2 ) 0 5 3 2 0

( , 0)

5 3

x

x x

x

x x

x

x

y e x x

dya e x x e x

dx

e x x x

e x x e x x

dyb

dx

e x x when x x

because for all x e

x

4

2 2

4

10

10

2 0 2 3 5 0

5(2 5)( 1) 0 1

2

1,

5 3,

5 3The stationary point

2

s

4

are 1, ,2 4

x x x

x x x or x

for x

e and e

y e

for x y e

Part (a) was well answered by the majority of candidates. Many fully correct responses were seen and, if there were errors, it was usually the final accuracy mark that was lost through the omission of brackets at some stage in the solution. Part (b) was not answered as well as part (a). Although many candidates were successful in factorising the required quadratic, there were also many solutions which were accompanied by terms in e–4x. Where marks were lost, it was mainly due to incorrect signs, though some candidates did manage to obtain follow through marks. A few candidates attempted to substitute their values of x back into the derivative rather than y. Many candidates stopped when they had found the two values of x, presumably thinking they had finished the question.


1

) ) ( )

The end points have coordinates

1) Plot the line with equation

-1,- 1,2 2

This line crosses the cu

1.4

1therefore t

rve only once

he equation sin ( ) 1has only one sol4

a i y Sin x

ii y

and

x

x x

1

2 31

change of sign rule

ution.

1b) Let's call ( ) 1

4(0.5) 0.60 0 (1) 2.32 0

according to the , we know

that there is a root so that 0.5 1

) ) 0. 0.902 , 0.945 ,

)

1x x

f x Sin x x

f and f

c i x

ii

In part (a)(i), even though the sketch was often poorly drawn, many candidates obtained full marks, and, where they did not, it was often because sketches went beyond the correct end points. Some correct graphs had the wrong end points marked, reversed coordinates being the most common error. There were however many candidates who had no idea what the graph looked like. Not all candidates attempted part (a)(ii). Where lines were drawn, they were often not accompanied by sufficient explanation to award the accuracy mark. There were also many instances of lines with a positive gradient intersecting in the third quadrant. In part (b), where candidates clearly defined which function they were using many achieved full marks. There are still candidates who lose marks by saying “change of sign so the root lies between these two values” without stipulating 0.5 < x < 1. Part (c)(i) was usually well answered, with answers given to the required degree of accuracy. Part (c)(ii) was very well answered with most candidates obtaining both marks.

AQA - Core 3 77


2

1

2

2

2

2

1)cosec 3 sin

31

sin ( ) 0.343

)cot 11 cosec

Using the identity

cosec 1 cos ec 11 0

cosec cos ec 12 0

(cosec 3)(cosec

cot cosec 1

4) 0

cosec 3 cosec 4

1 1sin sin

0.34 2.80

3 4

a x x

x or x

b x x

x x

x x

x x

x or x

or

x

x

x

x

1

1

1

sin ( 0.25) 0.253 0

2 sin ( 0.25)

s

0.34 2.80 6.03

3.in ( 0.25) 39

orx or x

or x

x

Part (a) was reasonably well answered, with most candidates obtaining both values. Some candidates lost the accuracy mark through inaccurate evaluation of the second angle, with 5.94 being

a common incorrect answer. A few cases of 1

cosec( )cos( )

xx

were seen. Part (b) was answered very well, with most candidates who obtained full marks in part (a) also obtaining full marks in part (b). The majority of candidates earned the first 4 marks but some then lost the final mark(s) through inaccurate values. There were a few candidates who started with the wrong identity and hence scored zero.


) 8 2

) 8 2 4 8 2 4 8 2 4

2 4 2 12

) 4

8 2 4 means that the graph of 8 2

is "above" the line 4

This happens

2

when

6

2 6

a y x

b x x or x

x or x

c Plot the line y

x y

x or x

x or x

x

y

Part (a) was well answered by the majority of candidates. The main errors were not crossing the y‐axis into the second quadrant or failing to give the points of contact with the axes. In part (b), most candidates obtained the required two values. In most cases where candidates lost marks, it was for only giving one value, usually 2. Four values of +2, –2, +6, and –6 was not uncommon. Part (c) was not as well answered with many candidates trying to write down a single inequality for an answer.


12 21.5 4.5 7.5 10.50

2 2

1

2

0 10 102

55 5

1 50 5

2

10

) ln( 5) 3

3 1.981001 3.228826 4.114964 4.747104

) ) ln 5 5

) 5 5

50 25

42.2 3 .

) ln( 5)

.

5

25

y

y y

y

a x dx y y y y

b i y x e x

ii V x dy e dy e y

V e e

c y

to sig fi

x bec

g

x e

e e

2

ln 5 34

omx

es y

Part (a) was very well answered by the majority of candidates. Some candidates lost a mark through not giving the answer to the required degree of accuracy. Candidates losing the final mark had sometimes lost the previous mark by showing their working to insufficient accuracy (usually 3sf). Few attempts at anything other than the mid‐ordinate rule were seen. In part (a)(i), most candidates finished with the required expression though not all had derived it through valid means. Most candidates obtained some marks on part (a)(ii), and many fully correct responses were seen. The final A mark was often lost because dy had been

omitted. Other errors occurred during the integration, with 5ye x being

very common and 5ye

yy also being popular.

In part (c), there were very few fully correct expressions seen. Most candidates earned the mark for +3, but –3 was also quite common. The main error was dealing with the stretch scale factor 4 parallel to the x axis. Although

this was mostly seen correctly as a 1

4, it was often outside the brackets:

21ln( 5)

4x . The translation was handled much better, although expressions

such as 21ln( 5) 3

4x were seen.

AQA - Core 3 78


2

2

2

2 2

1

2

1

( ) 3

1 4( ) ,

3 4 3

) for all , 0

3 3

The range is

) ) 3 3

12 ln( 3) ln( 3)

2

1) ( ) 0 ln( 3) 0

23 1

1) ) ( ) (

( ) 3

1

)3(

( ) ln( 3)2

2

x

x

x

x x

f x e for all x

g x for all x xx

a x e

e

b i y e e y

x y x y

ii f x x

x

c i gf x g f x

f x

f x

e

x

x

2

2

2

3) 4

) ( ) 1 when 3e 5 1

2

1

3 5

1ln 2

2

2 ln 2

x

x

x

xe

x

ii gf x

e x

Considerably less than half the candidates gained 2 marks in part (a); f(x) > –2 was common, as was f(x) > 3. Part (b)(i) was generally well answered with many fully correct responses seen. The majority of candidates earned the mark for swapping x and y. Marks were lost in the attempt to isolate x or y because many candidates could not cope with changing e2x = y + 3 into 2x = ln (y + 3), the most common error being ln y + ln 3. In part (b)(ii), the majority of candidates who had been successful in part (b)(i) and knew that e0 =1 went on to earn both marks. There did, however, seem to be a significant number of candidates who did not know that e0 =1. Part (c)(i) was well answered by most candidates. Part (c)(ii) was reasonably well answered by the majority of candidates, with many earning full marks. Candidates who had trouble with e2x in part (b)(i) also had the same problems in this part.


2

2 22

2

2 2

2

2

2

2

22

2

4(

4) tan 4

44 4 4 4 4 4

4

4 4 4 4 )4 4 4

4

) 4(1 4 ) 4

1 4 )

32 (

4 4

0 4 2 (4 4 4 ) 4

32 4 (1 4 ) 1

Tan x

Sin xa y x

Cos xdy Cos x Cos x Sin x Sin x

dx Cos x

dy Cos x Sin xTan x

dx Cos xdy

dxdy

b Tan x Tan xdx

d yTan x Tan x

dx

d yTan x Tan x

dxy y

2 )

Part (a) was well answered with many candidates gaining full marks. Where part marks were earned, the most common error was to lose the factor of 4 in the numerator by incorrectly writing the derivatives of sin 4x and cos 4x as cos 4x and –sin 4x. Many candidates did not attempt part (b). Although fully correct responses were seen, this was the question part where most candidates scored very few if any marks. Where candidates split the term 4tan24x up into (4tan 4x)(tan 4x) or similar expressions they seemed to have more success. In general, the use of the chain rule from whatever starting point was not coped with satisfactorily.

AQA - Core 3 79


2 2 2

2 2

1 1(2 1)

1 1) (2 1

(2 1)2 4

) (2 1) (2 1)2 2

1 1 1 1) 2 1 ( 1)

2 2 2 2

1 ( 1) 1 1 2 1

2 1 4 2 8

1 1 12 2 ln

2 1 8 8 2

In terms

a xSin x dx x Cos x Cos x dx

b u x so x u u and dx du

x u u udx du du

xCos x S

x u u

x udx u du u u c

in x c

x u

221 1 1

(2 1) (2 1) ln(2 1) c2 1 16 4

o

8

f ,

xdx x x

x

xx

Part (a) was well answered by many of the candidates. The majority of candidates differentiated the x and integrated the sine function. Those candidates who made an error in the integration were able to gain the method marks. Those candidates who gained the first accuracy mark usually went on to give fully correct solutions. Where candidates gained no marks, it was usually due to setting up the parts incorrectly: it was common to see u = xsin and v = 2x – 1. In part (b), most candidates made the appropriate start,

with 2du

dx . Although fully correct responses were

seen, this question part was not very well answered by many candidates. It was essential to substitute for dx, (2x – 1) and x2 in the integral to earn the second method mark and some omitted the first of these or the integral sign. In putting x2 in terms of u there were several hurdles: some only substituted for x, many forgot to square the 2 on the denominator, many had (u – 1) instead of (u + 1) and many only had two terms when they attempted to square. Work correct up to this point often did not go any further.

GRADE BOUNDARIES


Core 3 – Unit PC3 75 57 49 41 34 27

AQA - Core 3 80

QSo

lutio

nM

arks

Tot

alC

omm

ents

1(a)

(

)4

e2

2x

yx

−′ =

+

()

42

4e2

2x

xx

−−

+−

M1

()

()

2–4

–4e

e2

–2

xx

yA

axb

Bx

x′ =

+±

+

whe

re A

and

B a

re n

on-z

ero

cons

tant

s A

1 A

ll c

orre

ct

()

42

e2

24

88

xx

xx

−=

+−

−+

or

2-4

-4-4

-4e

-6e

+10e

xx

xx

x

()

42

2e5

32

xx

x−

=−

−A

1 3

AG

; all

corr

ect w

ith n

o er

rors

, 2nd

line

(OE)

mus

t be

seen

C

ondo

ne in

corr

ect o

rder

on

final

line

or

2

–4

–4

–4

e2

e–

2ex

xx

yx

x=

+

2–

4–

4–

4

–4–4

4e

2e

2.–

4e

2e8e

xx

x

xx

yx

xx

′ =−

++

++

(M1)

(A

1)

2–

4–

4–

4–4

– 4

ee

e e

ex

xx

xx

AxBx

Cx

DE

++

++

All

corr

ect

2–

4–

4–

4–

4e

–6

e10

ex

xx

xx

=+

()

42

2e5

32

xx

x−

=−

−(A

1)A

G; a

ll co

rrec

t with

no

erro

rs,

3rd li

ne (O

E) m

ust b

e se

en

(b)

()(

)2

51

(0)

xx

−+

−=

M1

OE

A

ttem

pt a

t fac

toris

atio

n

(

25)

(1)

xx

±±

±±

or

form

ula

with

at m

ost o

ne e

rror

5 ,1

2x

−=

A1

Bot

h co

rrec

t and

no

erro

rs

SC x

= 1

onl

y sc

ores

M1A

0 4

1,e

xy

−=

=

m1

For

eby

a=

atte

mpt

ed

A1F

Eith

er c

orre

ct, f

ollo

w th

r oug

h on

ly fr

om

inco

rrec

t sig

n fo

r x

105

3,

e2

4x

y=

−=

−A

1 5

CSO

2

solu

t ions

onl

y

Not

e: w

ithho

ld fi

nal m

ark

for e

xtra

so

lutio

ns

Not

e: a

ppro

xim

ate

valu

es o

nly

for y

can

sc

ore

m1

only

T

otal

8

QSo

lutio

nM

arks

Tot

alC

omm

ents

2(a)

(i)

B1

corr

ect s

hape

pas

sing

thro

ugh

orig

in

and

stop

ping

atA

and

B

1,2

AB

1

1,2

B−

−B

1 3

SC

A(1,

90)

and

B(

)–1

,–90

scor

es B

1

(ii)

line

inte

rsec

ting

thei

r cur

ve (p

ositi

ve

grad

ient

, pos

itive

y in

terc

ept)

M1

Cor

rect

stat

emen

t A

1 2

one

solu

tion

only

, sta

ted

or in

dica

ted

on

sket

ch -

mus

t be

in th

e fir

st q

uadr

ant

(ie c

urve

inte

rsec

ts li

ne o

nce)

M

ust h

ave

scor

ed B

1 fo

r gra

ph in

(a)(

i) (b

) (

)(

)()

()LH

S 0.

50.

5R

HS

0.5

1.1

LHS

11.

6R

HS

11.

3

==

==

M1

At

0.5

LHS

RH

S,

At1

LHS

RH

S<

>0.

51

α∴

<<

A

1 2

CSO

or

()

()

11

fsi

n1

4x

xx

−=

−−

f()x

mus

t be

defin

ed

()

()f0.

50.

6A

WR

Tf

10.

3=−

=

(M

1)

Allo

w

()

()f

0.5

0f

10

<>

Cha

nge

of si

gn

0.5

1α

<<

(A

1)

or

()

1f

sin

1–

4x

xx

=+

f()x

mus

t be

defin

ed

()

()f0.

50.

4A

ttem

ptf

1–

0.1

=

=

(M1)

Cha

nge

of si

gn

0.5

1α

<<

(A

1)

or (

)–1

f4s

in–

–4

xx

x=

f()x

mus

t be

defin

ed

()

()f0.

5–

2.4

atte

mpt

f1

1.3

= =

(M1)

Cha

nge

of si

gn

0.5

1α

<<

(A

1)

A

B

Cor

e 3

- Jan

2010

- M

ark

sche

me

AQA - Core 3 81

QSo

lutio

nM

arks

Tot

alC

omm

ents

2(c)

(i)

20.

902

x=

M

1Si

ght o

f AW

RT

0.90

2 or

AW

RT

0.94

1

30.

941

x=

A

1 2

Thes

e va

lues

onl

y

(ii)

M1

Stai

rcas

e, (v

ertic

al li

ne) f

rom

x1 t

o cu

rve,

ho

rizon

tal t

o lin

e, v

ertic

al to

cur

ve

A1

2 2

3,

xx

appr

ox c

orre

ct p

ositi

on o

n x-

axis

Tot

al11

O

1x

2

3x

x

QSo

lutio

nM

arks

Tot

alC

omm

ents

3(a)

1

sin

, 3or

sigh

t of

0.34

,0.

11 o

r19

.47

x=

±±

±

(o

r bet

ter)

M

1

()

0.34

,2.8

0x=

AW

RT

A1

2 Pe

nalis

e if

inco

rrec

t ans

wer

s in

rang

e;

igno

re a

nsw

ers o

utsi

de ra

nge

(b)

2co

sec

111

cose

cx

x−

=−

M

1C

orre

ct u

se o

f 2

2co

tco

sec

–1x

x=

(

)2

cose

cco

sec

120

xx

+−

=A

1

()(

)()

cose

c4

cose

c3

0x

x+

−=

m1

Atte

mpt

at F

acto

rs

Giv

es c

osec

x o

r – 1

2 w

hen

expa

nded

Fo

rmul

a on

e er

ror c

ondo

ned

cose

c4,

31

1si

n,

43

x

x

=−

=−

A1

Eith

er L

ine

1si

n4

x=− 3.

39,6

.03

x=

A

WR

T B

1F3

corr

ect o

r th

eir t

wo

answ

ers f

rom

(a)

and

3.39

, 6.0

3 (

)0.

34,2

.80

A

WR

T B

1 6

4 co

rrec

t and

no

extra

s in

rang

e ig

nore

ans

wer

s out

side

rang

e

SC 1

9.47

, 160

.53,

194

.48,

345

.52

B

1

A

ltern

ativ

e

2 2

cos

111

–si

nsi

nx

xx

=

22

cos

11si

n–

sin

xx

x=

(M

1)

Cor

rect

use

of t

rig ra

tios a

nd m

ultip

lyin

g by

2

sin

x 2

21

–si

n11

sin

–si

nx

xx

=

20

12si

n–

sin

–1x

x=

(A

1)

()(

)0

4sin

13s

in–1

xx

=+

(m1)

A

ttem

pt a

t fac

tors

as a

bove

1

1si

n–

, 43

x=

(A1)

(B1F

) A

s abo

ve

(B1)

T

otal

8

AQA - Core 3 82

QSo

lutio

nM

arks

Tot

alC

omm

ents

4(a)

M1

Mod

ulus

gra

ph V

shap

e in

1st q

uad

goin

g in

to 2

nd q

uad,

touc

hing

x-a

xis.

Mus

t cro

ss

y-ax

is

Con

done

not

rule

d A

1 2

4 an

d 8

labe

lled

(b)

2x

=

B1

One

cor

rect

ans

wer

6

x=

B

1 2

Seco

nd c

orre

ct a

nsw

er a

nd n

o ex

tras

Con

done

ans

wer

s sho

wn

on th

e gr

aph,

if

clea

rly in

dica

ted

(c)

6x>

2x<

B

1 B

1 2

One

cor

rect

ans

wer

Se

cond

cor

rect

ans

wer

and

no

extra

s and

no

furth

er in

corr

ect s

tate

men

t eg

62

or2

6x

x<

<<

>

SC

6x≥

, 2

x≤sc

ores

B1

Tot

al6

5(a)

x

y1.

5 1.

9810

0 4.

5 3.

2288

3 7.

54.

1149

610

.5

4.74

710

B1

M1

x va

lues

cor

rect

PI

3+ y

val

ues c

orre

ct to

2sf

or b

ette

r or

exac

t val

ues

A1

()

1.98

1,3.

228

/9,4

.114

/5,4

.747

for y

or b

ette

r3

y=

×

42.2

=

A1

4 (N

ote:

42.

2w

ith e

vide

nce

of m

id-o

rdin

ate

rule

with

four

strip

s sco

res 4

/4)

(b)(

i) (

)2

ln5

yx

=+

2e

5y

x=

+

OE

2e

–5

yx

=

B1

1 A

G M

ust s

ee m

iddl

e lin

e, a

nd n

o er

rors

(ii)

()

()(

)e

5d

yy

−M

1C

ondo

ne o

mis

sion

of b

rack

ets a

roun

d f (

y) th

roug

hout

()

()

()

10 5e

5y

y=

−A

1

()

()

()

105

e50

e25

=−

−−

m1

F(

)(

)10

–F

5

105

ee

25V

=−

−

A1

4 C

SO

incl

udin

g co

rrec

t not

atio

n –

mus

t see

dy

ISW

if e

valu

ated

(c)

2

()l

n5

34x

y =+

+

M1

4x se

en, c

ondo

ne l

n2

......

....

4x+

B1

… +

3

A1

3 C

SO m

ark

final

ans

wer

(no

ISW

) T

otal

12

8

4 x

y

QSo

lutio

nM

arks

Tot

alC

omm

ents

6(a)

(

)f

3x

>−

M1

‘3

>−

’, ‘

3x>

−’ o

r ‘

()

f3

x≥

−’

A1

2 A

llow

–

3y

>

(b)(

i) 2 e

3x

y=−

2

3e

xy+

=

()

ln3

2y

x+

=M

1

swap

x a

nd y

M1

atte

mp t

to is

olat

e:

()

lny

ABx

±=

or

reve

rse

()

()

()

11

fln

32

xx

−=

+A

1 3

OE

with

no

furth

er in

corr

ect w

orki

ng

Con

done

y =

…..

A

ltern

ativ

e x

× 2

e

-3

÷

2 ←

ln ←

+ 3

← x

(M

1)

(M

1)

ln(

3)2x

y+

=

(A1)

(ii)

31

x+=

M

1fo

r put

ting

thei

r (

)p

1x

= fr

om

()

()

lnp

kx

in th

eir p

art (

b)(i)

2

x=−

A

1 2

CSO

SC

: B

2 x

= -2

with

no

wor

king

, if f

ull

mar

ks g

aine

d in

par

t (b)

(i)

(c)(

i) (

)(

)2 2

1(g

f)

3e

34

eith

er1

(=)

3e5

x x

x=

−+

−

OE

B1

1 su

bstit

utin

g f i

nto

g IS

W

(ii)

211

3e5

x=

−

21

3e5

x=

−

OE

M1

Cor

rect

rem

oval

of t

heir

frac

tion

2 e2

x=

2ln

2x=

m

1 C

orre

ct u

se o

f log

s lea

ding

to

lna

kxb

=

1ln

22

x=

OE

A

1 3

CSO

N

o IS

W e

xcep

t for

num

eric

al

eval

uatio

n T

otal

11

AQA - Core 3 83

QSo

lutio

nM

arks

Tot

alC

omm

ents

7(a)

2

dco

s44c

os4

sin4

4sin

4d

cos

4.

.y

xx

xx

xx

−−

=M

1 2

2

2

cos

4si

n4

cos

4A

xB

xx

±±

22

24c

os4

4sin

4co

s4

xx

x+

=

or b

ette

r

A1

Bot

h te

rms c

orre

ct

()

24

1ta

n4x

=+

CSO

A

1 3

All

corr

ect

or

2

cos4

4cos

4si

n4

4sin

4d d

cos

4.

.x

xx

xy x

x−

−=

(M1)

22

2

cos

4si

n4

cos

4A

xB

xx

±±

= 4c

os4

cos4

4sin

4si

n4

cos4

cos4

cos4

cos4

xx

xx

xx

xx

+

or b

ette

r (A

1)

()

24

1ta

n4x

=+

C

SO

(A1)

All

corr

ect

(b)

2

2d

42t

an4

dy

xx

=×

× …

…

M1

tan

4A

x× f(

4x)

24s

ec4x

m

1f(

4x) =

2

sec

4B

x

232

tan4

sec

4x

x=

A

1Fft

8th

eir p

×fr

om p

art (

a)

()

232

tan4

1ta

n4

xx

=+

m1

Prev

ious

two

met

hod

mar

ks m

ust h

ave

been

ear

ned

()

232

1y

y=

+A

1 5

CSO

Alte

rnat

ive

Solu

tions

2

22

sin

44

4tan

44

4co

s4x

yx

x′ =

+=

+

22

4

4

cos

42s

in4

4cos

4si

n4

2cos

44s

in4

cos

4

y

xx

xx

xx

x

′′ =×

+(M

1)

(m1)

33

4

cos

4si

n4

cos

4A

xB

xx

± w

here

A a

nd B

are

cons

tant

s or t

rig fu

nctio

ns.

Whe

re A

is m

sin4

x an

d B

is n

cos4

x 2

2

4

48s

in4

cos4

cos

4si

n4

cos

4

xx

xx

x

×+

=(A

1F)

ft 8

thei

r p×

from

par

t (a)

2

32ta

n4

sec

4x

x=

(m

1)

2ta

n4

sec

4k

xx

()

232

1y

y=

+(A

1)C

SO

or

2

d4s

ec4

dyx

x=

2

2

d4

2se

c4.4

sec4

tan

4d

yx

xx

x=

×

(M1)

(m

1)

sec4

Ax

× f(

4x)

f(4x

) =

sec4

tan

4B

xx

2

32se

c4

tan

4x

x=

(A

1F)

ft 8

thei

r p×

from

par

t (a)

()

232

1ta

n4

tan

4x

x=

+(m

1)Pr

evio

us tw

o m

etho

d m

arks

mus

t hav

e be

en e

arne

d (

)2

321

yy

=+

(A1)

CSO

QSo

lutio

nM

arks

Tot

alC

omm

ents

7(b)

or

2

2

d4(

1ta

n4

)d

dta

n4

4

4d

yx

xy

ux

ux

=+

==

+

2

2

dd

(8)

dd

yu

ux

x=

(M1)

22

d4

4ta

n4

44

dux

ux

=+

=+

(m1)

2

22

d8

(44

)d

yu

ux

=+

(A

1)

2

32(1

)u

u=

+

(m1)

232

(1)

yy

=+

(A

1)

Tot

al8

8(a)

(

)si

n2

1d

xx

x−

()

dsi

n2

1dv

ux

xx

==

−M

1 (

)(

)d

sinf

,d

xx

x a

ttem

pted

()

d1

1co

s2

1d

2u

vx

x=

=−

−A

1 A

ll co

rrec

t – c

ondo

ne o

mis

sion

of b

rack

ets

()

()

cos

21

2xx

=−

−m

1 co

rrec

t sub

stitu

tion

of th

eir t

erm

s int

o pa

rts

()

1co

s2

1(d

)2

xx

−−

−

()

()

1co

s2

1co

s2

1(d

)2

2x

xx

x=

−−

+−

A

1 A

ll co

rrec

t – c

ondo

ne o

mis

sion

of b

rack

ets

()

()

1co

s2

1si

n2

12

4x

xx

c=

−−

+−

+

A1

5 C

SO c

ondo

ne m

issi

ng +

c a

nd d

x C

ondo

ne m

issi

ng b

rack

ets a

roun

d 2x

– 1

if

reco

vere

d in

fina

l lin

e IS

W

(b)

21

ux

=−

'd

2d'

ux

=

M1

OE

()2

21

dd

21

42

ux

ux

xu+

=−

m1

A1

All

in te

rms o

f u

All

corr

ect

PI fr

om la

ter w

orki

ng

21

21

d8

uu

uu+

+=

11

2d

8u

uu

=+

+

A1

21

2ln

82u

uu

=+

+

B1

or

()2

21

ln8

2u

u+

+

()

()

()

22

11

22

1ln

21

82x

xx

c−

=+

−+

−+

A1

6 or

()

()

22

11

ln2

18

2xx

c+

=+

−+

CSO

con

done

mis

sing

+ c

onl

y

IS

W

Tot

al11

T

OT

AL

75

AQA - Core 3 84

GeneralCertificate

ofEducation


June2010

Mathematics

MPC3

UnitPure

Core

3

Friday11June2010

9.00am

to10.30am

Forthis

paperyoumusthave:




Tim

eallowed

*1hour30minutes

Instructions


only

beusedfor

drawing.

*Fillin

theboxesatthetopofthis

page.


*Write

thequestionpartreference(eg(a),(b)(i)etc)in

theleft-hand

margin.

*Youmustanswerthequestionsin

thespacesprovided.Donotwrite

outsidetheboxaroundeachpage.

*Show

allnecessary

working;otherw

isemarksformethodmaybe

lost.

*Doallroughwork

inthis

book.Crossthroughanywork

thatyoudo

notwantto

bemarked.

Inform

ation


shownin

brackets.

*Themaxim

um

mark

forthis

paperis

75.

Advice

*Unlessstatedotherw

ise,youmayquote

form

ulae,withoutproof,

from

thebooklet.

1Thecurvey¼

3xintersects

thecurvey¼

10�x3at

thepointwherex¼

a.

(a)

Show

that

alies

between1and2.

(2marks)

(b)(i)

Show

that

theequation3x¼

10�x3canberearranged

into

theform

x¼


ffiffiffiffi10�3x

3p.

(1mark)

(ii)

Use

theiterationx n

þ1¼


ffiffiffiffiffi10�3x

3pnwithx 1

¼1to

findthevalues

ofx 2

andx 3

,

givingyouransw

ersto

threedecim

alplaces.

(2marks)

2(a)

Thediagram

showsthegraphofy¼

secxfor0�4

x4

360�.

(i)

ThepointAonthecurveis

wherex¼

0.State

they-coordinateofA.

(1mark)

(ii)

Sketch,ontheaxes

given

onpage3,thegraphofy¼

jsec2xjf

or0�4

x4

360�.

(3marks)

(b)


2,givingallvalues

ofxin

degrees

intheinterval

0�4

x4

360�.

(2marks)

(c)

Solvetheequationjse

cð2x�10�Þj

¼2,givingallvalues

ofxin

degrees

inthe

interval

0�4

x4

180�.

(4marks)

y A O90�

180�

270�

360�

x

AQA - Core 3 85

3(a)

Finddy

dxwhen:

(i)

y¼

lnð5x�2Þ;

(2marks)

(ii)

y¼

sin2x.

(2marks)

(b)

Thefunctionsfandgaredefined


fðxÞ¼

lnð5x�2Þ,

forreal

values

ofxsuch

that

x5

1 2

gðxÞ

¼sin2x,

forreal

values

ofxin

theinterval

�p 44

x4

p 4

(i)

Findtherangeoff.

(2marks)

(ii)

Findan

expressionforgfðx

Þ.(1

mark)

(iii)Solvetheequationgfðx

Þ¼0.

(3marks)

(iv)Theinverse

ofgis

g�1

.Findg�1

ðxÞ.

(2marks)

y O90�

180�

270�

360�

x

4(a)

Use

Sim

pson’s

rule

with7ordinates

(6strips)

tofindan

approxim

ation

to

ð 2 0:5

x

1þx3dx,givingyouransw

erto


(4marks)

(b)

Findtheexactvalueof

ð 1 0

x2

1þx3dx.

(4marks)

5(a)

Show

that

theequation

10cosec2

x¼

16�11cotx

canbewritten

intheform 10cot2xþ11cotx

�6¼

0(1

mark)

(b)

Hence,given

that

10cosec2

x¼

16�11cotx

,findthepossible

values

oftanx.

(4marks)

6Thediagram

showsthecurve

y¼

lnx

x.

Thecurvecrosses

thex-axis

atAandhas

astationarypointat

B.

(a)

State

thecoordinates

ofA.

(1mark)

(b)

Findthecoordinates

ofthestationarypoint,B,ofthecurve,

givingyouransw

erin

an

exactform

.(5

marks)

(c)

Findtheexactvalueofthegradientofthenorm

alto

thecurveat

thepoint

wherex¼

e3.

(3marks)

y

B

Ax

O

AQA - Core 3 86

7(a)

Use

integrationbyparts

tofind:

(i)

ð xcos4xdx;

(4marks)

(ii)

ð x2sin4xdx.

(4marks)

(b)

Theregionbounded

bythecurvey¼

8x


ffiffiffiðsi

n4xÞ

pandthelines

x¼

0andx¼

0:2

isrotatedthrough2pradiansaboutthex-axis.Findthevalueofthevolumeofthe

solidgenerated,givingyouransw

erto


(3marks)

8Thediagram

showsthecurves

y¼

e2x�1andy¼

4e�

2xþ2.

Thecurvey¼

4e�

2xþ2crosses

they-axis

atthepointAandthecurves

intersectat

thepointB.

(a)

Describeasequence


ationsthat

mapsthegraphof

y¼

exonto

thegraphofy¼

e2x�1.

(4marks)

(b)

Write

downthecoordinates

ofthepointA.

(1mark)

(c)(i)

Show

that

thex-coordinateofthepointBsatisfiestheequation

ðe2xÞ2

�3e2

x�4¼

0(2

marks)

(ii)

Hence

findtheexactvalueofthex-coordinateofthepointB.

(3marks)

(d)

Findtheexactvalueofthearea

oftheshaded

regionbounded

bythecurves

y¼

e2x�1andy¼

4e�

2xþ2andthey-axis.

(5marks)

END

OF

QUESTIO

NS

y

A Ox

B

y¼

4e�

2xþ2

y¼

e2x�1

AQA - Core 3 87



3 3

3

3

The number is solution of the equation

3 10 3 10 0

) ' ( ) 3 10 0

(1) 3 1 10 6

the ch

0

(2) 9 8 10 7 0

According to ,

there is a root so that 1 2

ange of sign ru

) )3 1 0

l

0

e

x x

x

x

x or x

a Let s call f x x

f

f

b i x

2 3

3

1

3 10 3

1.913 , 1.2

10

) , 21

3

1

x

x

x

ii x x x

x

Part (a) This was well answered by the majority of candidates. Many fully correct responses were seen. Most candidates used f(x) = 3x –10 + x3 and evaluated f(1) and f(2) correctly. There are still many candidates who still then write ‘change of sign’ therefore a root without clarification of where the root lies. Those candidates who used the alternative LHS RHS method were less successful, as they appeared to be unable to then make a correct statement, with many just putting ‘change of sign’ therefore a root. Part (b)(i) This part was very well answered, the only real error being 3x becoming 3x during rearrangements. Part (b)(ii) Again this part was very well answered. Many fully correct responses were seen. The main error was with candidates who wrote answers to 3 significant figures rather than 3 decimal places.


0

) ) sec

sec(0) 1

) sec 2

1)sec 2 cos

2

360 60

) sec(2 10) 2

sec(2 10) 2 sec(2 10) 2

2 10 60 2 10 300

2 10 2

(0,1

10

2 70 2 310 2 130 2 25

)

60 300

3

1 2

0

40

5

20

o

o

a i y x

so

ii Graph of y x

b x x

or x

c x

x or x

A

x

x

x or x

or x or x

x or x or x or x

or x

155 65 125o o oor x or x

Part (a)(i) This part was very well answered by most candidates. Part (a)(ii) The majority of candidates drew a modulus graph and therefore earned the method mark. Many correct responses to the number of sections were seen although some candidates appeared to try to sketch y = |sec½x|. Most candidates who earned the first two marks went on to earn the final A mark although poor sketching did lose the A mark for variable heights in some cases. Part (b) This was very well answered, with most candidates obtaining both marks. Where marks were lost it was usually through only offering the answer of 60°. A few candidates thought sec = 1/tan or 1/sin and hence obtained no marks. Part (c) Although many fully correct responses were seen many found this part difficult and scored few marks. Where candidates earned the first method mark many were able to go on to earn the A mark for 60° and 300°. As many did not consider (cos2x – 10)=‐1/2 they gained no further marks, since they only obtained two final answers. Candidates who did use (cos2x – 10)=‐1/2 usually obtained full marks. Handling of 2x – 10 = 60 proved difficult for some with 60/2 + 10 being seen.


(ln( )) ') )

)

1) ( ) ln(5 2)

21 5 1

5 5 22 2 2

1ln 5 2 ln

2

The range is

) ( ) ( ) 2 ln(5 2)

) ( ) 0

5

5 2

2 (2 )

( ) ln(0

2

.5

ln( )

5

)

5 2 0

d u ua i chain rule

dx u

ii

b f x x for x

if x then x and x

s

dy

dx xdy

Cos

o x

ii gf x g f x Sin x

iii gf x

xdx

f x

when x

x

1

1 1

1

2 1 5 3

) sin(2 ) 2 sin

1s

3

5

1( ) si

2

n2

in

x

i

x

v y x so x y

x

x x

y

g

Part (a) (i) The first part of this question was reasonably answered with many candidates obtaining both marks. Where candidates obtained 1 mark it was because many ended up with the answer 1/(5x – 2). Part (a)(ii) Again this part was very well answered with candidates obtaining both marks. The majority of candidates arrived at Kcos 2x but K = –2 was a common error. Part (b)(i) This part was not answered very well. Many candidates lost a

mark through using –0.693 instead of 1ln

2

and f(x) ≥ 0 was a common

response. Part (b)(ii) Most candidates were able to do this part with the correct answers often seen. The main error was the omission of brackets around 5x – 2 obtaining sin2(ln5x – 2). The expression for fg(x) was also often seen. Part (b)(iii) For those candidates with a correct starting expression many went on to get full marks . Those candidates who used sin ln (5x‐2)² often lost an accuracy mark for not rejecting one of their answers. Most candidates obtained the first method mark for making the correct initial step for their expression. Part (b)(iv) This was usually well done but there were common errors of dividing by sin2 obtaining y/sin2 = x or even y/sin = 2x and y/2 = sinx

AQA - Core 3 88


2

30.5

2 21 1 13

3 3 00 0

11 1.5

(0.5) (2) 4 (0.75) y(1.25) y(1.75) 2 (1) y(1.5)3 61

0.2222 0.4444 4 0.5275 0.4233 0.2752 2 0.5 0.342912

1 3 1 1) ln(1 ) ln(2) ln(1)

1 3 1 3 3

0.605

1ln(2

3

xdx

x

y y y y

x xb dx dx x

x x

(This integral is of the for lm

)

')n

ff c

f

Part (a) This was well answered by the majority of candidates with many obtaining full marks. Some candidates lost the final accuracy mark through premature rounding. Most candidates obtained the first B mark and the second B mark although substitution into 1/(1 + x³) was seen quite often. The main error within the brackets for Simpson’s rule was to have the 2 and the 4 reversed, although this was not common. Part (b) Most candidates who realised the numerator was related to the derivative of the denominator and hence required a ln function were successful in obtaining all 4 marks. Three marks were obtained by candidates who failed to evaluate ln1 as 0. The major error was the use of an incorrect k in kln(1 + x³) with k = ½ being the most common. These candidates usually obtained both method marks. Many candidates scored zero on this part as they obtained answers by attempts at integrating

2 3 1(1 )x x dx obtaining expressions involving

x³/3.


2

2

2

2

2 2

10cot 1

)10c

5 2tan tan

2 3

1co

osec 16 11cot

Using the identity 1 cot cosec

10 1 cot 16 11cot 0

)10cot 11cot 6 0

(5cot 2)(2c

t 6 0

ot 3) 0

2 3cot cot

5 2

a x x

x x

x x

b x x

x x

x or

x

x

x

or x

x

Part (a) Very well answered by the majority of candidates. Most candidates used the correct identity and were successful in answering this part of the question. Part (b) Most candidates were successful in answering this part of the question. When answered correctly the factorisation of the resulting quadratic or use of the quadratic formula was usually well done but there were some sign errors. Many totally correct solutions were seen.


2 2

1

1

3

3

3

1

3 2 6 6

ln) 0 ln 0 , 1

1ln 1 1 ln

)

0 ln 1

ln 1,

)The gradient of the curve at

1 ln 1 3 2( )

( )

(

The gradient of the norm

1,0)

( , )

A

B

xa y when x x

x

x xdy xxbdx x xdy

when x x e edx

eFor x e y e so

e e

c x e is

ed

e

ye

dx e e e

e

6

al at B s 2

ie

Many candidates lost marks on this question from careless work when handling the number of e2x terms involved. Part (a) Most candidates obtained the correct coordinates although (lnx, 0) was a common error. Part (b) This was generally well answered by those candidates able to use the quotient rule. Some errors with the numerator being reversed to (ln x – 1) meant candidates lost accuracy marks later in the question. Many candidates obtained 3 marks for reaching ln x – 1 = 0 and many went on to find x = e. However many stopped here without finding the value of y. Part (c) Although many correct solutions were seen some

candidates lost marks through leaving 23xe rather than

simplifying it. There was also a large number of candidates who did not answer part (b) but started again here, used the quotient rule correctly and went on to then answer the remainder of the question correctly. They seemed to realise that they needed the derivative to find a gradient but not for finding stationary points

AQA - Core 3 89


2 2

2

2

2

1 1) cos(4 ) sin(4 ) sin(4 )

4 4

1 1) sin(4 ) cos(4 ) 2 cos(4 )

4 41 1

cos(4 ) cos(4 )4 21 1 1 1

cos(4

1 1sin(4 ) cos(4 )

4

) sin(4 ) cos(4 )4 2 4

16

1 1cos(4 ) si

16

n4 8

i x x dx x x x dx

ii x x dx x x x x dx

x x x x dx

x x x x x

x c

x

c

x x

x x

0.2 0.2 0.22 2 2

0 0 0

0.22

0

2

b) 64 sin(4 ) 64 sin(4 )

16 cos(4 ) 8 s

1(4

in(4 ) 2cos(4 )

16 0.2 cos(0.8) 8 0.2sin(0.8) 2cos(0.8)

) cos(4 ) c32

0.299

2

V y dx x x dx x x dx

V x x x x

x x

x

V

V

Part (a)(i) Very few candidates did not attempt this part and it was well answered with most candidates gaining at least part marks. The most common error was to get v = 4sin x rather than ¼ (sin4x). This still gave the possibility of earning both method marks. Part (a)(ii) The first method mark was gained by most candidates for setting up the parts correctly. Many candidates who had trouble integrating cos 4x in part (a)(i) repeated their error here with v = 4cosx rather than – ¼ (cos4x) and therefore lost the A mark. Good candidates at this stage realised they could use their answer to part (a)(i) to write down the required solution but I am afraid most chose to do integration by parts twice in order to find a solution. Part (b) This was not very well answered since the relationship to part (a)(ii) often went unnoticed. Most candidates obtained the first method mark for ∫x²sin4x dx . The majority of candidates started again, some of them with some success, but for most their expressions usually were integrated to Kx³sin4x and no further credit was available.

AQA - Core 3 90


2

0

) A and

a

)The curve with equation

1stretch scale factor in

4 2 crosses

the y-axis when 0: 4 2 6

) )The point B is the intersection

the x-direction2

0translation of vector

1

(0,6)

x

A

a

b y e

x y e

c i

2 2 2

2 2

2 2 2

2

2 2

2 2

of the two curves

the x-coordiante of B is solution of the equation

e 1 4 2

3 4 0 (multiplying by e )

)This is a quadratic equation in , wefact

( ) 3 4

orise

4 1 0

0

4

x x

x x

x x x

x

x x

x x

e

e e

ii e

e e

e or

e

e

e

2

ln 2 ln 22 2

0 0

ln 2ln 22 2

00

2ln 2 2ln 2

2

1

2 ln 4 0

)The shaded area is A= 4 2 1

12 2

2

1 12 2ln 2 2 0 ln 2

2 2

(2

1ln 4

ln 2 ln 2 ln 4 2l

ln 22

x

x x

x x

x or no solution as e for all x

d e dx e dx

A e x e x

A e e

x

and

2 1n 2 ln 2 ln )

41 1 1

2 2ln 2 2 4 ln 24 2

3ln 22

A

Part (a) This was well answered by the majority of candidates with many fully correct responses. The main error with the stretch was using the y‐direction with a scale factor of 2 and with the translation the vector often contained +1. Part (b) This part was well answered by many candidates. Part (c)(i) This was not very well answered by weaker candidates with many just rearranging the expression and then writing down the answer given without any convincing step being shown. Part (c)(ii) This was answered better than part c(i) with many correct factorisations seen. The main error was with the rejection of the solution e2x = –1. Part (d) Some candidates obtained the correct answer and earned full marks. Most candidates made a reasonable attempt at the integrals with reasonable success and most showed enough to earn the B mark for subtraction of their areas. Incorrect limits of 6 and 0 often spoiled what would have been very good attempts.

GRADE BOUNDARIES

Component title Max mark A* A B C D E

Core 3 – Unit PC3 75 67 62 54 46 39 32

AQA - Core 3 91

Key

to m

ark

sche

me

and

abbr

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tions

use

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mar

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M

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for m

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f1

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Mus

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< R

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This

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AQA - Core 3 92

QSo

lutio

nM

arks

Tot

alC

omm

ents

2(a)

(i)

()1

y=B

1 1

Con

done

1 m

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A,

A =

1 et

c

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ot

1co

s0, s

ec 0

(ii)

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Mod

ulus

gra

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0y>

A1

13

22

+×

sect

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roug

hly

as sh

own,

cond

one

sect

ions

touc

hing

, va

riabl

e m

inim

um h

eigh

ts

A1

3 C

orre

ct g

raph

with

cor

rect

beh

avio

ur a

t 4

asym

ptot

es b

ut n

eed

not s

how

bro

ken

lines

; and

roug

hly

sam

e m

inim

a

(b)

1co

s2

x=

or

11

cos

2−

seen

M

1 or

sigh

t of

60or

,1.

05(A

WR

T)3

±±

±

60,3

00x

=A

1 2

Con

done

ext

ra v

alue

s out

side

0

360

x°<

<°,

but

no

extra

s in

inte

rval

(c)

sec(

210

)2

,se

c(2

10)

2x

x−

=−

=−

1co

s(2

10)

2x−

= o

r 1

cos

(210

)2

x−=

−M

1 Ei

ther

of t

hese

, PI b

y fu

rther

wor

king

210

60,3

00

or2

1012

0,

240

x

x

−=

−=

°(ig

nore

val

ues o

utsi

de 0

360

x°<

<°)

A1

Bot

h co

rrec

t val

ues f

rom

one

equ

atio

n

or 2

cor

rect

val

ues a

nd n

o w

rong

val

ues

from

bot

h eq

uatio

ns,

but m

ust h

ave

“210

x−=

”

PI b

y 2

70,1

30,2

50,3

10x=

35,6

5,1

25,1

55x=

B1

3 co

rrec

t (an

d no

t mor

e th

an 1

ext

ra v

alue

in

018

0x

°<<

°)

B1

4 A

ll 4

corr

ect (

and

no e

xtra

s in

inte

rval

) T

otal

10

90°

180°

27

0°

360°

O

y

x

Q

Solu

tion

Mar

ksT

otal

Com

men

ts3(

a)(i)

ln

(52)

yx

=−

M1

52

k x−d

5d

52

y xx

=−

A1

2 N

o IS

W, e

g 5

15

22

xx

=−

−(M

1A0)

(ii)

sin2

yx

=

M1

co

s2k

x d

2cos

2dy

xx

=

A1

2

(b)(

i) M

1(

)f

ln0.

5x

or

()

fln

2x

−�A

1 2

(ii)

()

()

()

gfsi

n2l

n5

2x

x=

−

or

()

()

()2

gfsin

ln5

2x

x=

−B

1 1

Con

done

()

sin2

ln5

2x−

or

()

()

sin2

ln5

2x−

but n

ot s

in2(

ln5

2)x−

or

sin

2ln5

2x

−

(iii)

()

gf0

x=

()

sin

2ln

52

0x−

=

()

2ln

52

0x−

=M

1 C

orre

ct fi

rst s

tep

from

thei

r (b)

(ii)

52

1x−

=

m1

Thei

r (

)(

)(

)f

1fro

m

lnf

0x

kx

==

3 5x=

A

1 3

With

hold

if c

lear

err

or se

en o

ther

than

om

issi

on o

f bra

cket

s

(iv)

sin2

xy

=

1si

n2

xy

−=

()

1or

sin

2y

x−

=M

1

Cor

rect

equ

atio

n in

volv

ing

1si

n−

()

11

1g

()

sin

2x

x−

−=

A

1 2

Tot

al12

AQA - Core 3 93

QSo

lutio

nM

arks

T

otal

Com

men

ts4(

a)

x y

0.5

40.

49

=

0.75

48

0.52

7591

=

1 1

0.5

2=

1.25

80

0.42

3318

9=

1.5

120.

3429

35=

1.75

11

20.

2752

407

=

2 2

0.2

9=

B1

B1

x va

lues

cor

rect

PI

At l

east

5 y

val

ues t

hat w

ould

be

corr

ect

to 2

sf o

r bet

ter,

or e

xact

val

ues.

May

be

seen

with

in w

orki

ng.

42

4880

112

112

42

99

9118

940

72

35+

++

++

+

M1

Cle

ar a

ttem

pt to

use

‘the

ir’ y

val

ues

with

in S

imps

on’s

rule

[]1

0.25

3=

×

0.60

5=

A

1 4

Ans

wer

mus

t be

0.60

5 w

ith n

o ex

tra sf

(N

ote

0.60

5 w

ith n

o ev

iden

ce o

f Si

mps

on’s

rule

scor

es 0

/4)

(b)

21

30

d1

xx

x+

()

31 ln

13

x=

+M

1 (

)3

ln1

kx

+ c

ondo

ne m

issi

ng b

rack

ets

A1

Cor

rect

. A

1 m

ay b

e re

cove

red

for

mis

sing

bra

cket

s if i

mpl

ied

late

r 1

1ln

(11)

ln1

33

=+

−

m1

F(1)

(–

F(0

))

1 ln2

3=

A1

4 ln

1 m

ust n

ot b

e le

ft in

fina

l ans

wer

A

ltern

ativ

e 3

lu

x=

+

2

d3

du

xx

=

d 3u u=

(M1)

d du x c

orre

ct a

nd in

tegr

al o

f for

m

duk

u

[]

1ln

3u

=(A

1)

11

ln2

ln1

33

=−

(m

1)C

orre

ct su

bstit

utio

n of

cor

rect

u li

mits

or

conv

ersi

on b

ack

to x

and

F(1

) (–

F(0

))

1 ln2

3=

(A1)

ln

1 m

ust n

ot b

e le

ft in

fina

l ans

wer

Tot

al8

QSo

lutio

nM

arks

T

otal

Com

men

ts5(

a)

210

cose

c16

11co

tx

x=

−

210

(1co

t)

1611

cot

xx

+=

−

210

cot

11co

t6

0x

x+

−=

B

1 1

AG

Mus

t see

evi

denc

e of

cor

rect

iden

tity

and

no e

rror

s.

(b)

Atte

mpt

at f

acto

rs, g

ivin

g 2

10co

t6

x±

±w

hen

expa

nded

. M

1 U

se o

f for

mul

a: c

ondo

ne o

ne e

rror

()(

)(

)5c

ot2

2cot

30

xx

−+

=A

1 C

orre

ct fa

ctor

s 2

3co

t,

52

x=

−

52

tan

, 23

x=

−A

1,A

1 4

1st A

1 m

ust b

e ea

rned

C

ondo

ne A

WR

T –0

.67

ISW

if x

val

ues a

ttem

pted

A

ltern

ativ

e 1

210

cot

11co

t6

0x

x+

−=

2 2

cos

cos

1011

60

sin

sin

xx

xx

+−

=

22

10co

s11

cos

sin

6sin

0x

xx

x+

−=

(

)()

()

5cos

2sin

2cos

3sin

0x

xx

x−

+=

(M

1)

(A1)

Atte

mpt

at f

acto

rs, g

ives

2

210

cos

6sin

xx

±±

whe

n ex

plai

ned

A

s abo

ve

(5co

s2s

in2c

os3s

inx

xx

x=

=−

)

52

tan

tan

23

xx

=−

=(A

1),

(A1)

1st A

1 m

ust b

e ea

rned

C

ondo

ne A

WR

T –0

.67

ISW

if x

val

ues a

ttem

pted

A

ltern

ativ

e 2

210

11ta

n6t

an0

xx

+−

=

()(

)(

)5

2tan

23t

an0

xx

−+

=(M

1)

(A1)

A

ttem

p t a

t fac

tors

giv

es2

106t

anx

±±

52

tan

, 23

x=

−(A

1),

(A1)

1st A

1 m

ust b

e ea

rned

C

ondo

ne A

WR

T –0

.67

ISW

if x

val

ues a

ttem

pted

T

otal

5

AQA - Core 3 94

QSo

lutio

nM

arks

T

otal

Com

men

ts

6(a)

ln

xy

x=

(whe

n)0

1y

x=

=

or

(1

, 0)

B1

1 B

oth

coor

dina

tes m

ust b

e st

ated

, not

1

sim

ply

show

n on

dia

gram

(b)

2

1ln

d d

xx

yx

xx

×−

=M

1Q

uotie

nt/p

rodu

ct ru

le

2lnx x

xx

±±

21

lnx

x−=

o

r

22

lnx

xx

−−

−

A1

OE

mus

t sim

plify

x x

21

lnA

t ,

0x

Bx−

=m

1Pu

tting

thei

r d

0dy x

= o

r num

erat

or =

0

ex

=

A1

CSO

con

done

1 e

x=

11

ore

ey

−=

A1

5 C

SO m

ust s

impl

ify ln

e

(c)

Gra

dien

t at

3 ex

=

3

32

1ln

e(e

)−

=M

1Su

bstit

utin

g 3

de

into

thei

r dy

xx

= (c

ondo

ne

1 sl

ip) b

ut m

ust h

ave

scor

ed M

1 in

(b)

62 e−=

or

62e

−−

A

1PI

Gra

dien

t of n

orm

al

61

e2

=A

1 3

CSO

sim

plifi

ed to

this

Tot

al9

QSo

lutio

nM

arks

Tot

alC

omm

ents

7(a)

(i)

dco

s4d

cos4

dvx

xx

ux

xx

==

M

1 (

)d

cos4

,d

xx

x a

ttem

pted

dsi

n41

4du

xv

x=

=A

1 A

ll co

rrec

t

sin4

sin4

d4

4x

xx

x=

−m

1 C

orre

ct su

bstit

utio

n of

thei

r ter

ms i

nto

parts

form

ula

()

sin4

cos4

416

xx

xc

=+

+A

1 4

OE

with

frac

tions

uns

impl

ified

(ii)

22

dsi

n4d

sin4

dd

cos4

24

d

vx

xx

ux

xx

ux

xv

x==

==

−M

1 (

)2

dsi

n4,

dx

xx

atte

mpt

ed

2co

s42

cos4

d4

4x

xx

xx

−−

=−

A

1

2co

s41

cos4

d4

2x

xx

xx

−=

+

[]2

1co

s44

2x

x−

=+

si

n4co

s44

16x

xx

+

m1

Cle

ar a

ttem

pt to

repl

ace

inte

gral

usi

ng

thei

r ans

wer

from

par

t (a)

(i)

()

2co

s4si

n4

cos4

48

32x

xx

xx

c−

=+

++

A1

4 O

E w

ith fr

actio

ns u

nsim

plifi

ed

(b)

()

()

()

()(

)0.

22

064

sin

4d

Vx

xx

=M

1

()

2co

s4si

n4co

s464

48

32x

xx

xx

−=

×+

+

Mus

t see

evi

denc

e of

thei

r (a)

(ii) r

esul

t or

star

ting

agai

n ob

tain

ing

3 te

rms o

f the

fo

rm

2co

s4si

n4co

s4Ax

xBx

xC

x±

±±

[

]2.

0952

92

=−

m1

AN

D F

(0.2

) – F

(0) a

ttem

pted

0.29

9=

A

WR

T A

1 3

Acc

ept A

WR

T 0.

0953

T

otal

11

AQA - Core 3 95

QSo

lutio

nM

arks

Tot

alC

omm

ents

8(a)

2

ee

1x

xy

=→

−

Stre

tch

(I)

scal

e fa

ctor

1 2 (I

I)

M1

I + (I

I or I

II)

in x

-dire

ctio

n

(III

) A

1 I +

II +

III

Tran

slat

ion

E1

A

llow

“tra

nsla

te”

0 1 −B

1 4

OE

“1

unit

dow

n” e

tc

(b)

06

xy

==

or

(0, 6

) B

1 1

Bot

h co

ordi

nate

s mus

t be

stat

ed, n

ot

sim

ply

6 m

arke

d on

dia

gram

(c)(

i) 2

2e

14e

2x

x−

−=

+

42

2e

e4

2ex

xx

−=

+

or (

)22

22

ee

42e

xx

x−

=+

M1

Mul

tiply

ing

both

side

s by

2 ex

22

2(e

)3e

40

xx

−−

=

A1

2 A

G W

ith n

o er

rors

seen

(ii)

()(

)2

2e

4e

1x

x−

+M

1 (

)()

22

e4

e1

xx

±±

ln2

x=

or

1ln

42

A1

Rej

ect

2 e1

x=

−

O

E A

1 3

eg

2 e0

x>

, 2 e

1x

≠−

, im

poss

ible

etc

(d)

()

24e

2d

xx

−+

(I

) ln

22

0

4e2

2x

x−

=+

−M

1 I o

r II a

ttem

p ted

and

2

2e

ore

xx

− in

tegr

ated

co

rrec

tly

2ln

24e

42l

n2

02

2

−

=+

−+

−−

m

1 F[

‘thei

r ln

2’ fr

om (c

)(ii)

] –

F[0]

13

2ln2

22l

n22

2=

−+

++

=+

()

2 e1

dx

x−

(II

) ln

22

0

e 2x

x=

−A

1 B

oth

I and

II c

orre

ctly

inte

grat

ed

2ln

2e

1ln

20

22

=−

−−

13

2ln

2ln

22

2=

−−

=−

33

=2l

n2

ln2

22

A+

−−

B1

Atte

mpt

to fi

nd d

iffer

ence

of

‘thei

r I –

thei

r II’

3l

n2

= o

r ln

8 o

r 3

ln4

2 O

E A

1 5

CSO

m

ust b

e ex

act

QSo

lutio

nM

arks

Tot

alC

omm

ents

8(d)

A

ltern

ativ

e (

)(

)2

24e

2d

e1

dx

xA

xx

−=

+−

−(B

1)

Con

done

func

tions

reve

rsed

()

()(

)ln

22

2

04e

e3

dx

xx

−=

−+

ln2

22

0

4ee

32

2

xx

x−

=−

+−

(M1)

(A

1)

22

eor

ex

x−

corr

ectly

inte

grat

ed

2ln

22l

n2

11

2ee

3ln2

22

2−

=−

−+

−−

−(m

1)

Cor

rect

subs

titut

ion

of th

eir

ln2

from

(c

)(ii)

into

thei

r int

egra

ted

expr

essi

on

3ln

2=

or

ln 8

or

3ln

42

OE

(A1)

C

SO

mus

t be

exac

t

Tot

al15

TO

TA

L75

AQA - Core 3 96

GeneralCertificate

ofEducation


January

2011

Mathematics

MPC3

UnitPure

Core

3

Wednesday19January

2011

1.30pm

to3.00pm

Forthis

paperyoumusthave:




Tim

eallowed

*1hour30minutes

Instructions


only

beusedfor

drawing.

*Fillin


page.


*Write


theleft-hand

margin.




*Show

allnecessary

working;otherw


lost.

*Doallroughwork

inthis


thatyoudo

notwantto

bemarked.

Inform

ation


shownin

brackets.

*Themaxim

um

mark

forthis

paperis

75.

Advice

*Unlessstatedotherw

ise,youmayquote

form

ulae,withoutproof,

from

thebooklet.

1(a)

Finddy

dxwhen

y¼

ðx3�1Þ6.

(2marks)

(b)

Acurvehas

equationy¼

xlnx.

(i)

Finddy

dx.

(2marks)

(ii)

Findan


thecurvey¼

xlnxat

thepointonthecurve

wherex¼

e.

(3marks)

2A

curveis

defined

bytheequationy¼

ðx2�4Þln

ðxþ2Þforx5

3.

Thecurveintersects

theliney¼

15at

asingle

point,wherex¼

a.

(a)

Show

that

alies

between3.5

and3.6.

(2marks)

(b)

Show

that

theequationðx

2�4Þln

ðxþ2Þ¼

15canbearranged

into

theform

x¼

�ffiffiffiffiffiffi



4þ

15

lnðx

þ2Þ

s(2

marks)

(c)

Use

theiteration

x nþ1

¼ffiffiffiffiffiffi


ffiffiffiffiffiffiffiffiffiffiffi

4þ

15

lnðx n

þ2Þ

s

withx 1

¼3:5

tofindthevalues

ofx 2

andx 3,givingyouransw

ersto

threedecim

al

places.

(2marks)

3(a)

Given

that

x¼

tanð3yþ1Þ:

(i)

finddx

dyin

term

sofy;

(2marks)

(ii)

findthevalueofdy

dxwhen

y¼

�1 3.

(2marks)

(b)

Sketch

thegraphofy¼

tan�1

x.

(2marks)

AQA - Core 3 97



fðxÞ¼

3cos1 2x,

for04

x4

2p

gðxÞ

¼jx

j,forallreal

values

ofx

(a)

Findtherangeoff.

(2marks)

(b)

Theinverse

offis

f�1.

(i)

Findf�

1ðxÞ

.(3

marks)

(ii)


1ðxÞ

¼1,givingyouransw

erin

anexactform

.(2

marks)

(c)(i)

Write

downan

expressionforgfðx

Þ.(1

mark)

(ii)

Sketch

thegraphofy¼

gfðx

Þfor04

x4

2p.

(3marks)

(d)

Describeasequence


ationsthat

mapsthegraphof

y¼

cosxonto

thegraphofy¼

3cos1 2x.

(3marks)

5(a)

Find

ð1

3þ2xdx.

(2marks)

(b)

Byusingintegrationbyparts,find

ð xsinx 2dx.

(4marks)

6(a)

Use

themid-ordinaterule

withfourstripsto

findan

estimatefor

ð 0:4

0cos


ffi3xþ1

pdx,

givingyouransw

erto


(4marks)

(b)

Use

thesubstitutionu¼

3xþ1to

findtheexactvalueof

ð 1 0x


ffi3xþ1

pdx.

(6marks)

7(a)


�5,givingallvalues

ofxin

radiansto

twodecim

al

placesin

theinterval

0<

x<

2p.

(3marks)

(b)

Show

that

theequation

cosecx

1þcosecx�

cosecx

1�cosecx¼

50

canbewritten

intheform

sec2

x¼

25

(4marks)

(c)

Hence,orotherwise,

solvetheequation

cosecx

1þcosecx�

cosecx

1�cosecx¼

50

givingallvalues

ofxin

radiansto

twodecim

alplacesin

theinterval

0<

x<

2p.

(3marks)

AQA - Core 3 98

8(a)

Given

that

e�2x¼


(2marks)

(b)

Thediagram

showsthecurvey¼

4e�

2x�e�

4x.

Thecurvecrosses

they-axis

atthepointA,thex-axis

atthepointB,andhas

a

stationarypointat

M.

(i)

State

they-coordinateofA.

(1mark)

(ii)

Findthex-coordinateofB,givingyouransw

erin

anexactform

.(3

marks)

(iii)Findthex-coordinateofthestationarypoint,M,givingyouransw

erin

anexact

form

.(3

marks)

(iv)Theshaded

regionRis

bounded

bythecurvey¼

4e�

2x�e�

4x,thelines

x¼

0

andx¼

ln2andthex-axis.

Findthevolumeofthesolidgenerated

when

theregionRis

rotatedthrough360�

aboutthex-axis,givingyouransw

erin

theform

p qp,wherepandqareintegers.

(7marks)

B

My

A

R

Oln2

x

AQA - Core 3 99



62 3 2 3 5) 6 3 2

1) ) ln( ) 1 ln

)The gradient of the tan

ln 1 2

18 (

gent at

, ln

The equation of the tangent at

2)

ln

1

2

2( )

dya x x

dxdy

b i y x x so x xdx x

ii x e is

when x e y e e e

x e is

y e x

x x

x

y

e

x

e

e

m

Part (a) This part was well answered by the majority of candidates with many fully correct responses seen. Most candidates used substitution of u =(x3 – 1) to give y = u6. Although (x3 – 1)5 was usually seen, errors did occur with the associated function of x. Part (b)(i) This was very well answered, with a majority of candidates earning both marks. The major error was from candidates who did not treat this as a product and simply differentiated x to get 1 and lnx to get

1

x.

Part (b)(ii) Again, this part was very well answered. Many fully correct responses were seen. The main error was with candidates who wrote the gradient as 1+ ln(e) which earned the method mark. Then this was often evaluated as 1 + ln(e) = 1 or left unsimplified. Several candidates who

obtained 2dy

dx then went on to use a gradient of – ½ .


2

2

2

A curve has equation ( 4) ln( 2) 3

A line has equation 15

)They intersect at means that is solution of

the equation ( 4) ln( 2) 15

( 4) ln( 2) 15 0

Let's call ( ) ( 4) ln( 2) 15

(3.

y x x for x

y

a x

x x

x x

f x x x

f

2

2

2 3

2

1

5) 0.936 0 (3.6) 0.436 0

According to , we know

that has a root so that 3.5 3.6

) ( 4) ln( 2) 15

15 154 4

ln( 2)

the change of sign rule

154

l

ln( 2)

) 3.5

n( 2)

3.578 , 3.568, (

and f

f

b x x

x xx x

c x

xx

x x

3 . )dec places

Part (a) This was well answered by the majority of candidates. Candidates should be encouraged to write down their relevant function before they begin to substitute – too many fudged their approach. Most candidates used f(x) = (x² ‐ 4) ln (x + 2) ‐ 15 and evaluated f(3.5) and f(3.6) correctly. There are still many candidates who still then write ‘change of sign’ therefore a root without clarification of where the root lies. Those candidates who used the alternative f(x) = (x² ‐ 4) ln (x + 2) and then compared it with y = 15 were less successful, as they appeared to be unable to then make a correct statement. Part (b) This was very well answered, the only real errors being ± missing in the final answer and some candidates losing a bracket in their working. Part (c) This part was very well answered with full marks often obtained. The main error was stating x3 = 3.567 through using premature approximation of x2.


2 2

2 2

1

2

3

) ta

(3

n(3 1)

) 3(1 (3 1)) 3 (3 1)

Cos (3 1) 1 (0)) for

3 3

1)

3

) ta

1

3

n

a x y

dxi Tan y Sec y

dy

dy y dy Cosii and y is

dx

Cos y

dx

b Sketch of y x

Part (a)(i) It was clear that dx/dy confused many, they simply wrote down dy/dx = . Many solutions were correct, but some candidates missed the multiplier and some were clearly trying to use the product rule or split 3y and +1, which was an indication of poor understanding of work on trigonometric functions. Part (a)(ii) This part was not very well answered by many candidates. Those who had managed to get the method mark in part (a)(i) often gained the method mark for obtaining

23sec (0)dx

dy . However, many incorrectly rearranged the

function to 2

2

1 cos (3 1)

3cos (3 1) 3

dy ynot

dx y

, so substituting

into an incorrect function. Part (b) A large number of candidates had no idea of the shape of the inverse tangent graph. A few incorporated a turning point and many failed to put the correct values on the y ‐axis. For the correct shapes that were seen, the y ‐axis was often not labelled, so only part marks were awarded. The main error seen was a reflection of y = tan x in the y‐axis and not in y = x.

AQA - Core 3 100


1

1

1 1

1

3 ( ) 3

1) When 0 2 0

21

(0) cos( ) cos2

1 11 1 3 3 3

2 2The range is :

1 1) ) 3

2 3 21

23 2 3

The inverse of ( )

)

23

) 1 1( (

a x then x

and Cos x

Cos x so Cos x

yb i y Cos x Cos x

y yCos x so x Cos

f is

ii f x means

f x

xf x Cos

x fthat

1) 3

2

Stretch in the x-dir scale factor 2

and a stretch in

1

the y-dir

)

scale fa

) ( ) ( ) 32

) ( ) 0 2

)

ctor 3

c i gf x g f x Cos x

ii Sketch y gf x for x

d

Cos

This question was generally answered very well and full marks were often seen. Part (a) Many correct answers with correct notation were seen but there were many cases of 3 ( ) 3f x also seen. Where candidates lost a mark it was

usually for poor notation. Part (b)(i) Most candidates earned the method mark for swapping x and y but the ½ confused many when trying to obtain

1 1 12cos , cos cos

3 3 6

x x xwith and being common errors.

Part (b)(ii) Having the correct inverse function generally led to the correct answer here; unfortunately repeating incorrect algebra for part (b)(i) sometimes gave the ‘right’ answer, but obviously without reward. Part (c)(i) This was very well answered. Part (c)(ii) Most candidates achieved the method mark by drawing at least two continuous parts. Unfortunately many lost the final mark, as drawing multiple curves was a common error, as was labelling the x‐axis incorrectly.

Part (d) This was well done; a few candidates had 1 1

3 2or as the scale factor

and a handful introduced another wrong transformation.


1

'

ln(3 2 )2

2 4

1 1 2)

3 2

ln(

2 3 2

remember:

) 2 1 22 2

2

)

2

2

ff c

a dx dxx x

x x xb x Sin dx x

f

x c

x xxCos S

Cos Cos

in c

dx

Part (a) This was very well answered, with the majority of candidates obtaining the correct function. The main error made by candidates who obtained a ln function was with the multiplying constant, which was often 2 or was completely missing. Omission of brackets was also a problem. Part (b) Not many of the candidates answered this part fully correctly. Only a few integrated by parts ‘the wrong way round’ and some

weaker candidates put u = x and v = sin2

x. However most knew that

they should integrate the sin(x/2) term, but did it wrongly, getting ± cos(x/2) or more commonly ± ½ cos(x/2). If the first method mark was earned, the second was generally scored as well.


0.4

0

1

0

) Let's call ( ) 3 1

3 1 0.1 (0.05) (0.15) (0.25) (0.35)

0.1 0.4780 0.3585 0.2454 0.1386

1 1 1) 3 1

3

0.122

3 30, 1 1, 4

1 1 13

3 .

3

.

13 3

a y x Cos x

Cos x dx y y y y

b u x so x u an

to si

d dx du

when x u and x u

x x d

f

d

g g

x u

i

u

3 1

4 42 2

1 1

45 31

2

0

1

1

2

0

1

9

1 2 2 1 2 2 1 2 23 1 32 8

9 5 3 9 5

1

3 9 5 3

163 1

135

u u u du

x x dx

x x dx

u u

Part (a) Those candidates who used radians usually went on to earn full marks. Many candidates, however, used degrees without showing unsimplified correct expressions for y, and hence only obtained the B mark for 4 correct x‐values. Part (b) Almost everyone earned the first mark for

3du

dx ; this was a distinct improvement from

previous years. Many candidates earned the first three method marks. The first accuracy mark was sometimes then lost through having an incorrect value for 1/9 with 1/3and1/6 and being quite common errors. Incorrect limits were seen but candidates usually managed to use the appropriate values either for u or for x.

AQA - Core 3 101


1

22

2

1)sec 5

51

( ) 2 1.775

cosec cosec ) 501 cosec 1 cosec

cosec (1 cosec ) cosec (1

1.77 4.5

cosec )50

(1 cosec )(1 cosec ) (1 cosec )(1 cosec )

2cosec50 2cosec 50

1 c

1

osec

a x Cos x

x Cos or x

x xb

x x

x x x x

x x x x

xx

x

2

2 2 2

2 2

2

1

2

50cosec

50 4848cosec 50 cosec

48 5048 48 2 1

1 150 50 50 25

1 1 1) cos cos

25 5 51

cos 2 1.

25

1.37 4.91 1.77 1375

4.5

Se

x

x x Sin x

Cos x Cos x

c Cos x x or x

x or x or x or x

c x

Part (a) Most candidates were able to obtain the first angle of 1.77 radians and hence obtain 2 marks. Many candidates also obtained the correct second solution, although 4.91 was a common error. Part (b) This was the worst answered question part on the paper, with very few candidates having any idea of how to go about combining the two fractions. Very few completely correct solutions were seen. Those who managed to

reduce the LHS to 2

2

2cosec

cot

x

x

could often not

complete the question. It was sad to see some very poor algebraic techniques in evidence here. There were many cases where candidates did not attempt the question. Part (c) Far too many candidates failed to recognise both the positive and the negative roots for sec x and in consequence they did not score any marks, since they only found the solutions for cos x = 0.2.

AQA - Core 3 102


2

2 4

0 0

2 4

4 2

2

2 4

4

) 4 2 ln 4

) 4

) 4 4 1 3

) is solution of 4e 0

( ) : 4 1 0

1 12 ln

4 4

) Let's solve 0

8 4 0

1( ) 2

1ln 4

2

(0,3)

1 1 1ln ln

2 2

4

4

x

x x

A

x xB

x x

x

x x

x

a e x

b y e e

i y e e

ii x e

multiplying by e e

e x

dyiii

dx

x

dye e

dx

e e

A

x

4

2

2

ln 2 ln 22 2 4 2

0 0

ln 2 4 6 8

0

ln 24 6 8

0

4ln 2 6ln 2 8ln 2

1ln4ln 2 ln(2 ) 16

1 0

1 12 ln

2 2

) (4 )

16 8

4 14

3 8

4 1 4 14 4

1 1ln

2

3 3 8

2

8

1

x

x

x x

x x x

x x x

e x

iv V y dx e e dx

V e e e dx

V e e e

V e e e

e e e

x

16

1 4 1 1 1

5247

204

4 1

8

4 416 3 64 8 256 3 8

V

V

Part (a) This was quite well answered by the majority of candidates. Part (b)(i) Again, this was well answered. Part (b)(ii) This was very poorly answered, with many candidates being unable to handle the negative indices. Some used logs wrongly and some assumed the value 4 for e−2x from part (a). There was also evidence of incorrect factorisation, but some used substitution effectively. Part (b)(iii) Although most candidates were able to differentiate the function, once the derivative had been found candidates were then unable to solve the equation for the same reason given in part (b)(ii). Part (b)(iv) Most candidates lost the first B mark, usually for missing the ‘dx’. Attempts at squaring the function often went awry; losing the middle term, losing the negative index, getting an x2 power were all seen. After such errors only a few more marks were available. Treating the function as a square and thinking that they could integrate to a cube was very disappointing to see on an A2 paper.

GRADE BOUNDARIES


Core 3 – Unit PC3 75 66 59 52 45 38 31

AQA - Core 3 103

Q

Solu

tion

Mar

ksT

otal

C

omm

ents

1(a)

(

)53

d–1

dyk

xx

=M

1W

here

k is

an

inte

ger o

r fun

ctio

n of

x

()5

23

63

–1x

x=

×

(

ISW

) A

1 2

But

note (

)53

2d

–1p

dyk

xx

x=+

M

0

Or (

)(

)3

6–1

ux

yu

==

5d

6dy

uu

= a

nd

2d

3du

xx

= M

1

(

)53

26

–13

xx

=×

A

1

Not

e

(

)52

3d

63

–1dy

xx

cx

=×

+ s

core

s M

1 A

0

(pen

alis

e +

c in

diff

eren

tial o

nce

only

in

pape

r)

(b)(

i) d

1ln

dyx

xx

x=

±×

±

M1

Prod

uct r

ule

atte

mpt

ed a

nd

diff

eren

tial o

f ln

x 1

lnx

=+

(IS

W)

A1

2

(ii)

(x =

e)

y

= e

PI

B1

Mus

t hav

e re

plac

ed ln

e b

y 1

Con

done

y =

2.7

2 (A

WR

T)

()

d1

lne

=2

dy x=

+M

1C

orre

ct su

bstit

utio

n in

to t h

eir

d dy x

But

mus

t hav

e sc

ored

M1

in (b

)(i)

()

e2

ey

x−

=−

or

2e

yx

=−

O

E, IS

W

A1

3 M

ust h

ave

repl

aced

ln e

by

1 T

otal

7

Q

Solu

tion

Mar

ksT

otal

C

omm

ents

2(

a)

()

()

()

2f

–4

ln2

–15

xx

x=

+O

r re

vers

e

()

()

f3.

5–

0.9

f3.

60.

4

= =M

1(

)(

)f

3.5

0.9

M1

f3.

60.

4

= =−

Atte

mpt

at e

valu

atin

g bo

th f

(3.5

) and

f (

3.6)

B

ut m

ust s

ee

()

()

()

2f

15–

–4

ln2

xx

x=

+

befo

re A

1 m

ay b

e ea

rned

Con

done

()

()

f3.

5 0

f3.

6 0

< >

Or

()

()

3.5

14.1

15

3.6

15.4

15

xy

xy

==

<

==

>

M1

Cha

nge

of si

gn,

3.5

3.6

α∴

<<

OE

A1

2 Ei

ther

side

of 1

5,

3.5

3.6

α∴

<<

OE

A1

(b)

()

()

2–

4ln

215

xx+

=

()

215

–4

ln2

xx

=+

M1

()

215

4ln

2x

x=

++ (

)15

4+ln

2x

x=

±+

A

G

A1

2 M

ust h

ave

both

mid

dle

lines

and

no

erro

rs se

en

(c)

()

13

5x

=⋅

23.

578

x=

C

AO

B

13

3.56

8x

=

CA

O

B1

2 Si

ght o

f AW

RT

3.58

or 3

.57

scor

es B

1 B

0

O

r ±

3.5

78 o

r ±

3.5

68 sc

ores

B1

B0

12

3.57

8,3.

568

xx

==

scor

es B

1B0

Tot

al6

Onl

y if

()

fde

fined

x

M

1

Eith

er o

f the

se li

nes c

orre

ct

Con

done

poo

r use

of b

rack

ets

for M

1 on

ly

Cor

e3 -

Jan2

011

- Mar

k sc

hem

e

AQA - Core 3 104

Q

Solu

tion

Mar

ksT

otal

C

omm

ents

3(a)

(i)

()

2d

sec

31

dxk

yy

=+

M1

W

here

k is

an

inte

ger

Con

done

om

issi

on o

f d dx y

But

()

2d

sec

31

dyk

yx

=+

sc

ores

M

1 A

0

()

23s

ec3

1y

=+

ISW

A

1 2

Alte

rnat

ive

met

hods

()

–11

tan

–13

yx

=

()

2d

1dx

kx

y=

+ M

1

()

()

23

1ta

n3

1y

=+

+ A

1

Or

()

()

sin

31

cos

31

yx

y+=

+

(

)(

)(

)2

2

2

cos

31

sin

31

d dco

s3

1k

yk

yx y

y±

+±

+=

+M

1

(

)2

3co

s3

1y

=+

A1

(ii)

2d

13s

ec3

1d

3x y

=×

−+

M1

Subs

titut

ion

of

1–

3y

=in

to th

eir

= 3

sec2 0

dd

dd

xy

yx

or B

UT

mus

t hav

e sc

ored

M1

in

(a)(

i) d

1d

3y x

=

CSO

A1

2 C

ondo

ne 0

.333

or b

ette

r

Or

2

d1

d3s

ec(3

1)y x

y=

+

2

13s

ec0

=A

s abo

ve

1 3=

3(b)

M1

A1

2

App

rox

corr

ect s

hape

with

no

tur n

ing

poin

ts, t

hrou

gh (0

,0) a

nd o

nly

1 cu

rve

Asy

mpt

otic

at b

oth

2±

and

bot

h va

lues

show

n C

ondo

ne ±

90

(deg

rees

) C

ondo

ne

tan

yx

= a

lso

draw

n bu

t cle

arly

id

entif

ied,

oth

erw

ise

M0

Tot

al6

Q

Solu

tion

Mar

ksT

otal

C

omm

ents

4(

a)

()

–3f

3x

M1

()

–33,

–3

f3

xx

<<

–3f

3,–

33

y<

<<

<–3

f<3,

–3

f3

<A

1 2

Allo

w–

33,

–3

f3

y

(b)(

i) 1

3cos

2y

x=

1co

s3

2y

x= –1

1co

s3

2y

x=

M

1O

r –1

cos

3x=

–12c

os3y

x=

Eith

er o

rder

–12c

os3x

y=M

1Sw

ap x

and

y

()

–1–1

f2c

os3x

x=

A

1 3

(ii)

1co

s3

2x

=M

1If

inc

orre

ct in

(b)(

i) B

UT

ans

wer

in

form

(

)–1

cos

p

qx(c

ondo

ne p

, q =

1)

13c

os2

x=

ISW

A

1 2

Then

1

cos

qxp

= M

1 o

r

f(1)

M1

x=

13c

osA

12

x=

(c)(

i) (

)1

gf3c

os2

xx

=B

1 1

(ii)

Mod

ulus

gra

ph in

1st q

uadr

ant,

star

ting

from

a +

ve y

-inte

rcep

t, at

leas

t 2

M1

cont

inuo

us p

arts

, firs

t des

cend

ing,

then

se

cond

incr

easi

ngIG

NO

RE

CU

RV

E O

UTS

IDE

RA

NG

E A

1 C

orre

ct c

urva

ture

, cur

ves r

each

ing

x-ax

is,

cond

one

mul

tiple

cur

ves (

no tu

rnin

g po

ints

at a

xis)

A

1 3

App

roxi

mat

ely

sym

met

rical

gra

ph w

ith

3,

, 2

indi

cate

d (m

ust h

ave

scor

ed

prev

ious

2 m

arks

)

C

ondo

ne

13c

os2

yx

= a

lso

draw

n bu

t

clea

rly id

entif

ied,

oth

erw

ise

M0

(d)

STR

ETC

H +

dire

ctio

n M

1 Ei

ther

in x

-dire

ctio

n or

y-d

irect

ion

s.f

. 3, p

aral

lel t

o y-

axis

A1

Ei

ther

ord

er

s.f

. 2, p

aral

lel t

o x-

axis

A1

3 T

otal

14

3

π 2π

AQA - Core 3 105

Q

Solu

tion

Mar

ksT

otal

C

omm

ents

5(a)

(i)

1d

32

xx

+

()

ln3

2k

x=

+M

1W

here

k is

a ra

tiona

l num

ber

()

1ln

32

2x

c=

++

A

1 2

Or

if su

bstit

utio

n 3

2,d

2du

xu

x=

+=

1

dln

2uk

uu

==

M1

(

)1

ln3

22

xc

=+

+

A1

(b)

dsi

n2x

ux

v=

=

M1

()

()

dsi

nd

cos

,1

22

dx

xx

kx

x=

=

whe

re k

is a

con

stan

t

d1

–2c

os2x

uv

==

A

1A

ll co

rrec

t

()

–2

cos

2cos

d2

2x

xx

x−

−=

m1

Cor

rect

subs

titut

ion

of th

eir t

erm

s int

o pa

rts fo

rmul

a (w

atch

sign

s car

eful

ly)

–2

cos

4sin

22

xx

xc

=+

+A

1 4

CA

O

Tot

al6

Q

Solu

tion

Mar

ksT

otal

C

omm

ents

6(a)

x

y

0.05

co

s1.

15

= 0.

4780

0.

15

cos

1.45

=

0.35

85

0.25

co

s1.

75

= 0.

2454

0.

35

cos

2.05

=

0.13

86

B1

M1

Usi

ng 4

cor

rect

x-v

alue

s, PI

At l

east

3 c

orre

ct y

-val

ues,

(con

done

uns

impl

ified

cor

rect

ex

pres

sion

s),

Or c

orre

ct v

alue

s rou

nded

to 2

s.f.

or

trunc

ated

to 2

s.f.

0.1

y×

Σ

m1

Use

d an

d m

ust b

e w

orki

ng in

radi

ans

= 0.

122

C

AO

A

1 4

Mus

t be

3 s.f

.

(b)

d3

du x=

M1

d3d

ux

= O

E

1d

3uu

ku

±=

×m

1A

ll in

term

s of u

, with

k =

3 o

r 1 3

Con

done

om

issi

on o

f du

()

31

22

1d

9u

uu

=±

m1

()

31

22

dp

uu

u±

(mus

t hav

e sc

ored

firs

t 2 m

arks

) 5

32

2

53

22

1–

9u

u=

A

1O

E

53

22

12

22

24

–4

––

95

35

3=

××

m

1

Mus

t hav

e ea

rned

all

prev

ious

met

hod

mar

ks a

nd th

en c

orre

ct su

bstit

utio

n, in

to

thei

r int

egra

l, of

1, 4

for u

or

0, 1

for x

an

d su

btra

ctin

g 11

613

5=

I

SW

A1

6 O

r equ

ival

ent f

ract

ion

Tot

al10

AQA - Core 3 106

Q

Solu

tion

Mar

ksT

otal

C

omm

ents

7(

a)

cos

–0.2

x=

M1

Or

()

tan

24x

=±

1.77

,4.

51x=

AW

RT

A1

O

ne c

orre

ct v

alue

A1

3 Se

cond

cor

rect

val

ue a

nd n

o ex

tra v

alue

s in

inte

rval

0 to

6.2

8…

Igno

re a

nsw

ers o

utsi

de in

terv

al

SC 1.

8,4.

5w

ithor

with

out w

orki

ngx=

M

1 A

1 A

0

SC (u

sing

deg

rees

)

101.

54,

281.

54

M

1 A

1 A

0

101.

5,28

1.5

M

1 A

0 A

0

SC No

wor

king

show

n 2

corr

ect a

nsw

ers 3

/3

1 co

rrec

t ans

wer

2/3

(b)

LHS

()

()

()(

)co

sec

1 –

cose

c –

cose

c 1

cose

c 1

cose

c 1–

cose

c x

xx

xx

x+=

+M

1C

orre

ctly

com

bini

ng fr

actio

ns b

ut

cond

one

poor

use

, or o

mis

sion

, of

brac

kets

2

2

2

cose

c –

cose

c–

cose

c –

cose

c1–

cose

cx

xx

xx

=A

1 A

llow

reco

very

from

inco

rrec

t bra

cket

s

2

2

2cos

ecco

tx

x−

=−

or

()

2

2

21

cot

cot

xx

−+

−m

1C

orre

ct u

se o

f rel

evan

t trig

iden

tity

eg c

osec

2 x =

1 +

cot

2 x

22s

ec50

x=

2se

c25

x=

A

GA

1 4

All

corr

ect w

ith n

o er

rors

seen

IN

CLU

DIN

G c

orre

ct b

rack

ets o

n 1st

line

O

r cose

c co

sec

501

cose

c 1

cose

c x

xx

x−

=+

−

()

()

cose

c 1

cose

c co

sec

1co

sec

xx

xx

−−

+

()(

)50

1co

sec

1co

sec

xx

=+

−(M

1)C

orre

ctly

elim

inat

ing

frac

tions

but

co

ndon

e po

or u

se, o

r om

issi

on, o

f br

acke

ts

22

cose

cco

sec

cose

cco

sec

xx

xx

−−

−

()

250

1co

sec

x=

−(A

1)

Allo

w re

cove

ry fr

om in

corr

ect b

rack

ets

248

cose

c50

x=

2

224

1si

nco

s25

25x

x=

=(m

1)C

orre

ct u

se o

f rel

evan

t trig

iden

tity

eg si

n2 x =

1 –

cos

2 x

2se

c25

x=

AG

(A

1)A

ll co

rrec

t with

no

erro

rs se

en

INC

LUD

ING

cor

rect

bra

cket

s on

1st li

ne

Q

Solu

tion

Mar

ksT

otal

C

omm

ents

7(

c)

sec

5x

=±

M

1O

r co

s0.

2x

=±

Or

tan

24x

=±

1.77

,4.

51,1

.37,

4.91

x=

(A

WR

T)

A1

3 co

rrec

t A

1 3

4 co

rrec

t and

no

othe

r ans

wer

s in

inte

rval

Ig

nore

ans

wer

s out

side

inte

rval

SC

1.8,

4.5,

1.4,

4.9

With

or w

ithou

t wor

king

M

1 A

1

SC thei

r 2 a

nsw

ers f

rom

(a)

+1.3

7, 4

.91

(AW

RT)

2

/3

SC F

or th

is p

art,

if in

deg

rees

m

ax m

ark

is

M1

A0

SC No

wor

king

show

n 4

corr

ect a

nsw

ers

3/3

3

corr

ect a

nsw

ers

2/3

0,

1, 2

cor

rect

ans

wer

s 0/3

T

otal

10

AQA - Core 3 107

Q

Solu

tion

Mar

ksT

otal

C

omm

ents

8(

a)

–2

e4

x=

–2

ln4

x=M

11

ln4

2x

=−

ISW

A

1 2

OE,

eg

1

ln4

ln,–

ln2,

22−

(b)(

i) (

)3y=

B1

1 C

ondo

ne (

)(

)0,

3bu

tnot

3,0

(ii)

0y=

–2

–44e

–e

0x

x=

2

4e1

0x

−=

M

12

e=

0x

ab

±±

22

1e

or e

4

4x

x−

==

A1

1ln

2x=

ISW

A

1 3

OE,

eg

11

1–

ln4,

–ln

2,

ln2

24

and

no e

xtra

solu

tions

Or

24

4ee

xx

−−

=

ln4

24

xx

−=

−

(M1)

2

ln4

x=−

(A

1)

O

E 1

ln4

2x

=−

(A1)

O

E

(iii)

()

–2–4

–8e

4ex

xy′

=+

B1

–4–2

4e8e

xx

=

22

22

2e1

0or

e2

01

ore

ore

22

orln

44

ln8

2

xx

xx

xx

− −

−=

−=

==

−=

−

M1

Equa

ting

d0

dy x=

and

get

ting

2e

=0

xa

b±

± fr

om

–2–4

de

ed

xx

yp

qx

=+

11

ln2

2x

=

IS

W

A1

3 O

E, e

g (

)1

ln4

ln8

2−

and

no e

xtra

solu

tions

Q

Solu

tion

Mar

ksT

otal

C

omm

ents

8(

b)(iv

)

()

ln2

2–

2–

4

04e

–e

dx

xV

x=

B1

Mus

t be

com

plet

ely

corr

ect i

nclu

ding

dx

seen

on

this

line

or n

ext l

ine

Lim

its, b

rack

ets a

nd

PI f

rom

late

r w

orki

ng

()

()

–4

–8–6

16e

e–8

ed

xx

xx

=+

B1

Cor

rect

exp

ansi

on, P

I fro

m la

ter w

orki

ng

()

()

()

ln2

–6–4

–8

0

14e

–4e

–e

83

xx

x=

+B

14

16e

4x

−

− O

E

B1

–81

–e

8x

OE

B1

–68

e6

x− −

OE

may

be

two

sepa

rate

term

s

()

–4ln

2–8

ln2

–6ln

21

4–

4e–

ee

83

=+

M

1

Cor

rect

subs

titut

ion

of

ln2

and

0x=

into

th

eir i

nteg

rate

d ex

pres

sion

(

)4

68

mus

t be

of fo

rm

ee

ex

xx

ab

c−

−−

++

00

01

4–

–4e

–e

e8

3+

and

subt

ract

ing.

PI

52

4720

48=

A1

7 O

E ex

act f

ract

ion

eg

2518

5698

304

Tot

al16

T

OT

AL

75

AQA - Core 3 108

GeneralCertificate

ofEducation


June2011

Mathematics

MPC3

UnitPure

Core

3

Monday13June2011

9.00am

to10.30am

Forthis

paperyoumusthave:




Tim

eallowed

*1hour30minutes

Instructions


only

beusedfor

drawing.

*Fillin


page.


*Write


theleft-hand

margin.




*Show

allnecessary

working;otherw


lost.

*Doallroughwork

inthis


thatyoudo

notwantto

bemarked.

Inform

ation


shownin

brackets.

*Themaxim

um

mark

forthis

paperis

75.

Advice

*Unlessstatedotherw

ise,youmayquote

form

ulae,withoutproof,

from

thebooklet.

1Thediagram


lnð6xÞ.

(a)

State

thex-coordinateofthepointofintersectionofthecurvewiththex-axis. (1

mark)

(b)

Finddy

dx.

(2marks)

(c)

Use

Sim

pson’s

rule

with6strips(7

ordinates)to

findan

estimatefor

ð 7 1lnð6xÞdx,

givingyouransw

erto


(4marks)

2(a)(i)

Finddy

dxwhen

y¼

xe2x.

(3marks)

(ii)

Findan


thecurvey¼

xe2xat

thepointð1,e2Þ.

(2marks)

(b)

Given

that

y¼

2sin3x

1þcos3x,use

thequotientrule

toshow

that

dy

dx¼

k

1þcos3x

wherekis

aninteger.

(4marks)

y Ox

AQA - Core 3 109

3Thecurvey¼

cos�

1ð2x�1Þintersects

thecurvey¼

exat

asingle

point

wherex¼

a.

(a)

Show

that

alies

between0.4

and0.5.

(2marks)

(b)

Show

that

theequationcos�

1ð2x�1Þ¼

excanbewritten

asx¼

1 2þ

1 2cosðe

xÞ.

(1mark)

(c)

Use

theiterationx n

þ1¼

1 2þ

1 2cosðe

x nÞwithx 1

¼0:4

tofindthevalues

ofx 2

and

x 3,givingyouransw

ersto

threedecim

alplaces.

(2marks)

4(a)(i)

Solvetheequationcosecy¼

�4for0�<

y<

360�,

givingyouransw

ersto

the

nearest

0.1�.

(2marks)

(ii)

Solvetheequation 2cot2ð2xþ30�Þ

¼2�7cosecð2

xþ30�Þ

for0�<

x<

180�,

givingyouransw

ersto

thenearest

0.1�.

(6marks)

(b)

Describeasequence


ationsthat

mapsthegraphof

y¼

cosecxonto

thegraphofy¼

cosecð2

xþ30�Þ.

(4marks)



fðxÞ¼

x2

forallreal

values

ofx

gðxÞ

¼1

2xþ1

forreal

values

ofx,

x6¼

�0:5

(a)

Explain

whyfdoes

nothavean

inverse.

(1mark)

(b)

Theinverse

ofgis

g�1

.Findg�1

ðxÞ.

(3marks)

(c)

State

therangeofg�1

.(1

mark)

(d)


¼gðxÞ

.(3

marks)

6(a)

Given

that

3lnx¼


(1mark)

(b)

Byform

ingaquadraticequationin

lnx,solve3lnxþ

20

lnx¼

19,givingyour

answ

ersforxin

anexactform

.(5

marks)

7(a)

Onseparatediagrams:

(i)

sketch


j3xþ3j;

(2marks)

(ii)

sketch


jx2�1j.

(3marks)

(b)(i)

Solvetheequationj3xþ3j¼

jx2�1j.

(5marks)

(ii)

Hence

solvetheinequalityj3xþ3j<

jx2�1j.

(2marks)

8Use

thesubstitutionu¼

1þ2tanxto

find

ð1

ð1þ2tanxÞ2

cos2xdx

(5marks)

AQA - Core 3 110

9(a)

Use

integrationbyparts

tofind

ð xlnxdx.

(3marks)

(b)

Given

that

y¼

ðlnxÞ2

,finddy

dx.

(2marks)

(c)

Thediagram


ffiffiffi xplnx.

Theshaded

regionRis

bounded

bythecurvey¼

ffiffiffi xplnx,thelinex¼

eandthe

x-axis

from

x¼

1to

x¼

e.


when

theregionRis



erin

anexactform

.(6

marks)

END

OF

QUESTIO

NS

y O1

ex

R

AQA - Core 3 111



7

1

) let's solve ln(6 ) 0

6 1

1) 6

61

) ln(6 ) 1 (1) (7) 4 (2) (4) (6) 2 (3) (5)3

11.7918 3.7377 4 2.4

1

61

18

84

.4 3 . .

9 3.1781 3.5835 2 2.8904 3.40123

a y x

x

dyb

dx x

c x

x

x

to sig f

dx y y y y y

ig

y y

Part (a) This was generally correct 0

6

e was

not good enough and neither was 0.16. Wrong answers included 1 and ln6. Part (b) Many solutions were correct, often

in an unsimplified form, but 1 6

6and

x x and

were common. Those who included dx or +c in their solution were penalised. Part (c) The majority of the candidates got this fully correct, although a few failed to correct to 3 significant figures as required. Those who chose to keep exact values in terms of ln were among the most successful. A few made a slip in writing down digits from their calculator, and a few, after writing the correct expression failed to multiply by their 1/3. The most common error in the application of Simpson’s rule was to confuse “odd” with “even”.


2

2

2 2

2 2

2

2

2

2

) ) 1 2

) for 1,

and an equation of the tangent at 1

is 3 ( 1)

2 3)

1 36 3 (1 3 ) 2 3 ( 3 3

(2 1)

3 2

)

(1 3 )

6 3 6

3

(

xx xdya i e x e

dx

ii x

x

y e e x

Sin xb y so

Cos xdy Cos x Cos x Sin x Sin x

dx Cos x

dy Cos x C

x

os x

dx

dye

x

e

y e x e

d

2

2

2 2

)

(1 3 )

6 3 6 6(1 3 )

(1 3 ) (1 3 )

66

1 3

Sin x

Cos x

dy Cos x Cos x

dx Cos x Cos x

dy

dx Cos xk

Part (a)(i) The majority of candidates were able to differentiate e2x and apply the product rule correctly but a few made an error in one or other of these processes. Occasionally an attempt to simplify was incorrect. Part (a)(ii) When finding the equation of the tangent (a line) it is essential to find the gradient (a constant) at the requisite point first. Most did this but it is disconcerting, at this level, to find candidates who give a non‐linear equation as their answer by failing to do the first step. A few added e²+2e² wrongly or lost exactness by evaluating 3e², and a few found the equation of the normal. Part (b) Although the quotient rule was known by almost everyone, and almost all earned the first mark, some appalling algebra and trigonometry abounded thereafter. A few failed even to get the first mark as they put a term in the wrong place on the numerator, had sinx or cosx instead of sin3x or cos3x in one of their differentials or else their denominator was 1+cos²3x instead of (1+cos3x)². A few failed to use the chain rule when differentiating sin3x or cos3x. Those who missed the brackets around (1+cos3x) in the numerator sometimes recovered but far too many falsely cancelled cos3x at this stage. If the numerator was correctly expanded another mark was available, although cos3x² and cos3²x instead of cos²3x for (cos3x)² was common; this was recoverable, but cos9x² was not. After that more incorrect cancellation was seen. Taking out a common factor of 6 was the safest route forward, as those who tried to deal with 6cos²3x + 6sin²3x often put it equal to 1 or 18 instead of 6. Even being correct to that stage did not ensure completion as, again there was false cancellation. Those who split the terms, then used (1‐cos²3x) for sin²3x and factorised to (1‐cos3x)(1+cos3x) generally completed successfully. It was quite extraordinary how many methods with completely wrong working landed up at

6

1 cos3x.

AQA - Core 3 112


1

1

1

(2 1) intersects

this means that is solution of the

equation Cos (2 1)

(2 1) 0

) Let's call ( ) (2 1)

the chang

(0.4) 0.28 0 (0.

According

5

et

) 0.

o

0

08

x

x

x

x

y Cos x y e

where x

x e

Cos x e

a f x C

f an

s e

d

x

f

o

1

2 31

, we know

that there is a root of , so that 0.4 0.5

)Cos (2

of sign rule

1 1( )

2 20.53

1)

2 1 (e )

9 , 0) 0.4 .42, (3 . )8

x

x

xx Cos e

f

b x e

so x Cos

c x dec cx pla esx

Part (a) It was good to see that a much greater proportion of the candidates defined f(x) first before they substituted values. As it is essential to get the correct numerical outcomes here it is well worth checking before moving on (eg quite a few carelessly rounded 0.0779 to 0.8). Those who used degrees could make no progress. Many lost the second mark by failing to state their conclusion and that α was in the required interval. Those who found the four numerical values needed to be precise in their statements to earn the accuracy mark. Part (b) was well done. Part (c) was well done with almost everyone giving both answers to the required accuracy, but those whose calculator was in degree mode earned zero.


2

1

2

2

2

2

1) ) cosec 4 0 360

41

( ) 14.54

180 14.5 14.48 3

co

60

) 2cot (2 30) 2 7cosec(2 30)

Using the identity

2cosec (2 30) 2 2 7co

t cos

194.5 3

sec(2 30)

2cosec (2

5

ec

3

4 .5

1

o

o o

o

a i Sin for

Sin

so x or x

ii x x

A

x

A

x

x

0) 7cosec(2 30) 4 0

(2cosec (2 30) 1)(cosec (2 30) 4) 0

1cosec(2 30) cosec (2 30) 4

21

sin(2 30) 2 sin(2

82.3 157.8

1stretch s

30)4

2 30 194.5 2 30 345.5

) A cale fact in the xr2

-o

x

x x

x or x

x or x

no so

x or x

lution x or x

b

30translation o

direction follo

f vector

we

0

d

bya

In part (a)(i), although the majority of candidates scored both marks here some rounded to the nearest degree, some truncated, some rounded 14.477 to 14.8, some only considered +4, and some erroneously added ‐14.5 or 14.5 to 90. A handful took cosec to be 1/tan or 1/cos Part (a)(ii) There were some excellent full solutions here but also some disturbing misunderstanding of notation. Many candidates found it disconcerting to deal with the functions of (2x + 30), for example 2cosec²x − 1 + (2x + 30) = 2 – 7cosec(2x + 30) was seen. They would have been well advised to use a substitution such as Y for (2x + 30). A few missed the brackets around (1 − cosec²Y) thus getting a wrong equation and no further marks; some had this correct but failed to deal with the constants correctly and again could not progress. Sadly the incorrect substitution of cotY + 1 for cosecY was also seen on several occasions. Once the correct factors were obtained a few stopped there (particularly those who unfortunately substituted x for 2x + 30). Quite a few handled the −4 solu on well but not the as they went on to assume sin(2x + 30) was ½ and not 2. Part (b) Most candidates recognised that the required transformations were a stretch and a translation. The majority started with the stretch and then, wrongly took the translation to be [‐30,0] instead of [‐15,0]. A few missed the correct term “stretch” and many failed to use the right terminology of “scale factor” ½. A few had the direction of the stretch in the y‐axis or the SF as 2 and a few had their translation in the y‐direction.

AQA - Core 3 113


1

1

1

1

) ( )

has no inverse because it is

1 1 1) 2 1 2 1

2 1

1 1

2 2

)The range of g is the domain

not one-to-one

1 1

of g

The range

1( )

2 2

of g

) ( ) (

2

1

1

)

1

(

)

2

2

xg x

a f x x for all x

f

b y x xx y y

xy

c

is

d

x x

g x

fg x g x

x

2

2

1 1 1

2 1 (2 1) 2 1

( (2 1))

11 2 1 1

2 10

x x x

multiplying by x

so xx

x

Part (a) Candidates should be advised to answer this question as “not 1 to 1”. For those who chose this phrasing, it was not always clear whether they were referring to f or its inverse when they said that it was “many to one” or perhaps “one to many”. There were many imprecise statements about square roots. Part (b) This part was generally well done with only the odd algebraic error. However those who failed to swap x and y at the end were penalised, and those who did this as a first step tended to score better. Part (c) Too often the negative sign was omitted here. Part (d) It was good to see fg(x) almost always correct. Many correct approaches also gave x = ‐1/2 as an answer. Equating and then trying to square root both sides, or taking both terms to the same side seldom proved fruitful. A few took (2x + 1)² as 4x² + 1 which was disappointing at this level, and 2x² + 4x + 1 was also quite common.


3

453

2

2

4)3ln 4 ln

320

)3ln 19 0 ( ln )ln

3(ln ) 20 19ln 0

3(ln ) 19ln 20 0

(3ln 4)(ln 5) 0

4ln ln 5

3

a x x

b x xx

x x

x x

x x

x or

x e

x e or x e

x

Part (a) This was well done. Part (b) The majority recognised what was required and obtained the relevant quadratic even if their notation was sometimes a bit suspect. A few used a substitution for lnx which helped, but those who used x instead of, say Y, occasionally forgot to change back to earn the final marks. The occasional candidate changed 3lnx to ln(x³) which was not helpful, and after multiplication by lnx there were a few (3lnx)² terms seen.


2

2

2 2

2 2

2

) ) 3 3

) 1

) ) 3 3 1

3 3 1 3 3 1

3 4 0 3 2 0

( 4)( 1) 0 ( 2)( 1) 0

) with support of the graph, we have

4 1

3 3 1

2

2 4

a i y x

ii y x

b i x x

x x or x x

x x or x x

x x or x x

ii

x x wh

x or x or x

x or xen

A small number of candidates clearly did not recognise the modulus function at all. Part (a)(i) Most graphs were in the correct place and looked linear. However a few failed to extend into the first quadrant and many were lacking the values of either one or both intercepts and a few had the vertex labelled ‐1/3 or −3. Part (a)(ii) Although almost all graphs had the three requisite sections, quite a number had the outer parts linear or curving convexly instead of concavely. Many did not have all three intercepts correctly labelled. Part (b)(i) The more able candidates were able to form two different quadratic equations and solve them, but many got only one pair of roots. However many formed wrong equations such as 3x + 3 = x² + 1 as well and got extra roots. Having found correct roots some assumed that their negatives would also be roots, misunderstanding the modulus. Those who tried to square both sides usually made algebraic errors, and only a handful were able to establish correct roots from their quartic equation. Another misunderstanding about the modulus function caused some to discard the negative values that they had found. Part (b)(ii) It was very common for the answers here to be between their values rather than outside them. Only more able students produced the correct intervals.

AQA - Core 3 114


2 2

2 2 2

1 2

2 1

21 1 1

(1 2 )

1

2(1 2

1

)

2

1

2

u Tan x

du dxso du

dx Cos x Cos x

dx duTanx Cos x u

cu

cTanx

Almost everyone earned the mark for differentiating 1 + 2tanx correctly. It was good to see some excellent fully correct solutions, but many candidates made no further progress. Some had problems dealing with the 2, either at the substitution stage or when trying to simplify in order to integrate. Many failed to recognise that cos²x × sec²x equals 1. It was disconcerting to see these terms jumping from one side of the integral sign to the other on some scripts. Some tried to write these expressions in terms of u and did not progress further. A few failed to substitute back into an expression in x at the end.


2 2

2

2 2

1

2 2 2

11

22

11

22 2

2 2

1 1 1) ln ln

2 2

1) ln 2 ln

) (ln )

1 1 2ln(ln )

2

1 1ln

2 42l

2

(ln ) ln2

1(ln ) ln

2 2

n

e

ee

ee

a x x dx x x x dxx

dyb y x so x

dx x

c V y dx x x dx

xV x x x dx

x

xV x x x dx

xV x

x x x c

x

x

x x

2

1

2 2 2

2

2

1

4

1 1 1 1 1

2 2 4 4 4 4

14

e

x

eV e e

e

e

V

Part (a) In applying integration by parts, it was essential to start off in the correct direction and many failed at the first fence. A few fell by integrating x to x² and losing the ½. It was disappointing how many candidates failed to simplify the second integral properly and integrate it to get x2/4. Part (b) This differentiation was generally well done, mainly using the chain rule but also the product rule. Part (c) It should be clear from previous reports that the initial mark here requires a fully correct integral, simplified in terms of x, including dx, with the limits and in this case also π. The function needed to be squared and this proved to be the downfall of many candidates as they needed to use correct notation. Able candidates then recognised how to split the integral, apply integration by parts and use part (b) and then part (a) [or vice versa] to complete. A significant number had a sign error in their final expression as, again, they failed to use brackets correctly.

GRADE BOUNDARIES


Core 3 – Unit PC3 75 68 59 52 46 40 34

AQA - Core 3 115

QSo

lutio

nM

arks

T

otal

Com

men

ts

1 (a

) 1 6

or

1 ,0 6

B1

1 co

ndon

e 0.

167

AW

RT

(b)

d1

dy xx

=

M1

k x w

here

1

1,6

or6

k=

A1

2k

= 1

(c)

x

y

1 ln

61.

7918

=

2 ln

122.

4849

=

3 ln

182.

8904

=

4 ln

243.

1781

=

5 ln

303.

4012

=

6 ln

363.

5835

=

7 ln

423.

7377

=

M1

A1

5+ y

-val

ues c

orre

ct, e

ither

exa

ct o

r cor

rect

to

3SF

(rou

nded

or t

runc

ated

) or b

ette

r

all 7

y-v

alue

s cor

rect

(and

onl

y th

ese

7 va

lues

), ei

ther

exa

ct o

r cor

rect

to 3

SF

(rou

nded

or t

runc

ated

) or b

ette

r

()

1A

1

1.79

183.

7377

3=

×+

()

42.

4849

3.17

813.

5835

++

+

(

)2

2.89

043.

4012

++

M

1co

rrec

t use

of S

imps

on’s

rule

on

thei

r 7

y-va

lues

, con

done

mis

sing

squa

re b

rack

ets

18

.4=

A1

4 C

AO

this

val

ue o

nly

Tot

al

7

QSo

lutio

nM

arks

Tot

al

Com

men

ts

2(a)

(i)

2 ex

yx

=

22

d2

ee

dx

xy

xx

=+

M1

A1

A1

ISW

3

22

ee

xx

kxl

+ w

here

k a

nd l

are

1s o

r 2s

2k

=

1l=

(2 e

(21)

xx

=+

)

(ii)

1x

=2

d3e

dy x=

M1

corr

ect s

ubst

itutio

n of

x =

1 in

to th

eird dy x

but m

ust h

ave

earn

ed M

1 in

par

t (i)

tang

ent:

()

22

e3e

1y

x−

=−

O

E A

1 2

CSO

(no

ISW

), m

ust h

ave

scor

ed fi

rst 4

m

arks

co

mm

on c

orre

ct a

nsw

er:

22

3e2e

yx

=−

(b)

2sin

31

cos3x

yx

=+

2

(1co

s3)6

cos3

2sin

3(

3sin

3)

(1co

s3)

d dx

xx

xx

y x+

−−

+=

M1

2

(1co

s3)c

os3

sin3

(sin

3)

(1co

s3)

px

xq

xx

x±

+±

+w

here

p a

nd q

are

ratio

nal n

umbe

rs

cond

one

poor

use

/om

issi

on o

f bra

cket

s PI

by

furth

er w

orki

ng

22

2

6cos

36c

os3

6sin

3(1

cos3

)x

xx

x+

+=

+A

1

this

line

mus

t be

seen

in th

is fo

rm (i

e in

te

rms o

f cos

2 3x

and

sin2

3x),

but a

llow

2

sin

3x re

plac

ed b

y 2

1co

s3x

−

cond

one

deno

min

ator

cor

rect

ly e

xpan

ded

m1

corr

ect u

se o

f 2

2si

n3

cos

3k

xk

xk

+=

or

()

22

sin

31

cos

3k

xk

x=

−

2

6cos

36

(1co

s3)

xx+

=+

61

cos3

x=

+A

1 4

CSO

Tot

al

9

Inde

pend

ent o

f eac

h ot

her

Cor

e3 -

Jun2

011

- Mar

k sc

hem

e

AQA - Core 3 116

QSo

lutio

nM

arks

Tot

al

Com

men

ts

note

: if d

egre

es u

sed

then

no

mar

ks in

(a)

and

(c)

3(a)

(

)(

)–1

fco

s2

–1–

exx

x=

or re

vers

e

()

()

f0.

40.

3

f0.

5–

0.1

= =M

1si

ght o

f ±0.

3 (A

WR

T) A

ND

0.

1 (A

WR

T)

chan

ge o

f sig

n 0.

40.

5α

∴<

<

A1

2 C

SO, n

ote

f (x)

mus

t be

defin

ed, c

ondo

ne

0.4

0.5

α≤

≤

(M1)

(A1)

alte

rnat

ive

met

hod

e0.4 =

1.5

, cos

–1 (2

×0.4

– 1

) = 1

.8

e0.5 =

1.6

5, c

os–1

(2×0

.5 –

1) =

1.5

7

at 0

.4 e

x < c

os–1

(2x

– 1)

at

0.5

ex >

cos

–1 (2

x –

1)

0.4

0.5

α∴

<<

(b)

()

–1co

s2

–1ex

x=

2–1

cos(

e)x

x= (

)1

11

cos(

e)

1co

s(e

)2

22

xx

x=

+=

+B

1 1

AG

m

ust s

ee m

iddl

e lin

e, a

nd n

o er

rors

seen

, bu

t con

done

cos

ex

(c)

10.

4x

=

20.

539

x=

B

1C

AO

3

0.42

8x

=B

1 2

CA

O

Tot

al

5

QSo

lutio

nM

arks

T

otal

Com

men

ts

4(a)

(i)

()

1si

n0.

2514

.5−

±=

±M

1PI

by

sigh

t of 1

94.5

etc

co

ndon

e ±

14.4

19

4.5,

345.

5θ

=

(AW

RT)

A

1 2

no e

xtra

s in

inte

rval

, ign

ore

answ

ers

outs

ide

inte

rval

(ii)

22c

ot(2

30)

27c

osec

(230

)x

x+

=−

+

cond

one

repl

acin

g 2x

+ 3

0 by

Y

22(

cose

c(2

30)

1)2

7cos

ec(2

30)

xx

+−

=−

+

M1

corr

ect u

se o

f 2

2co

sec

1co

tY

Y=

+(

)2

2cos

ec(2

30)

7cos

ec(2

30)

40

xx

++

+−

=A

1 m

ust b

e in

this

form

(

)(2

cose

c(2

30)

1)(c

osec

(230

)4)

0x

x+

±+

±=

m1

atte

mpt

at f

acto

risat

ion

1co

sec(

230

)or

42

x+=

−A

1 m

ust b

e th

is li

ne u

sing

f (2

x +

30)

230

194.

5,34

5.5

x+=

82.2

,157

.8x

=

(AW

RT)

B

1 on

e co

rrec

t ans

wer

, allo

w 8

2.3,

igno

re

extra

solu

tions

B1

6 C

AO

bot

h an

swer

s cor

rect

and

no

extra

s in

inte

rval

, ign

ore

answ

ers o

utsi

de in

terv

al

(b)

stre

tch

(I)

scal

e fa

ctor

1 2 (

II)

para

llel t

o x-

axis

(II

I)

M1

I and

eith

er II

or I

II

A1

I+II

+III

tra

nsla

teE1

15 0−

B1

4 co

ndon

e ‘1

5 to

left’

or ‘

–15

in x

(d

irect

ion)

’ al

tern

ativ

e m

etho

d tra

nsla

te(E

1)30 0−

(B

1)

stre

tch

(M1)

as

abov

e

scal

e fa

ctor

1 2pa

ralle

l to

x-ax

is(A

1)

as

abov

e

Tot

al

12

AQA - Core 3 117

QSo

lutio

nM

arks

T

otal

Com

men

ts

5(a)

(

)f

not

1–

1x

E1

1 O

E

(b)

12

1y

x=

+ 12

1x

y=

+1

21

yx

+=

M1

M1

swap

x a

nd y

eith

er o

rder

a

corr

ect n

ext l

ine

()

11

11

2g

xx

−=

−O

EA

1 3

[]1

11

2y

x=

−

(c)

1g

()

0.5

x−

≠−

B1

1 si

ght o

f 0.

5≠

− O

E

(d)

21

12

12

1x

x=

++

B1

sigh

t of

21

21

x+ o

r (

)2

12

1x+

()

()2

21

21

xx

+=

+or

2

21

44

1x

xx

+=

++

or

11

21

x=

+

or 2

x +

1 =

1

M1

one

corr

ect s

tep,

mus

t be

one

of th

ese

four

lin

es

0x

=

A1

3 C

SO

Tot

al8

6(a)

3l

n4

x=

4

ln3

x= 4 3 e

x=

B1

1 IS

W.

Con

done

34 e

(b)

203l

n19

lnx

x+

=

23(

ln)

2019

lnx

x+

=

M1

corr

ectly

mul

tiply

ing

by ln

x.

()

23(

ln)

19ln

200

xx

−+

=

A1

()

(3ln

4)(ln

5)0

xx

±±

=m

1us

e of

form

ula,

or c

ompl

etin

g th

e sq

uare

m

ust b

e co

rrec

t 4

ln,5 3

x=

A1

45

3 e,e

x=

A

1 5

cond

one

34 e

Tot

al6

QSo

lutio

nM

arks

T

otal

Com

men

ts

7(a)

(i)

M1

mod

ulus

gra

ph, a

ppro

xim

ate

V sh

ape,

to

uchi

ng n

egat

ive

x-ax

is a

nd c

ross

ing

y-ax

is

A1

2 1,

3−

mar

ked,

gra

ph sy

mm

etric

al, s

traig

ht

lines

(ii)

M1

mod

ulus

gra

ph in

3 se

ctio

ns, t

ouch

ing

x-

axis

and

cro

ssin

g po

sitiv

e y-

axis

A1

co

rrec

t cur

vatu

re

t hei

r 1,

thei

r 1

xx

><

−

A1

3 co

rrec

t cur

ve –

1 x

1

an

d x

= ±1

, y =

1 m

arke

d

(b)(

i)2

33

–1

xx

+=

()

23

3–

1x

x+

=

()

20

–3–

4x

x=

—A

M

1 ei

ther

A o

r B se

en, a

ll te

rms o

n on

e si

de

x =

4, –

1 A

1,A

1 (

)2

33

1–

xx

+=

()

23

20

xx

++

= —

B

x =

–1,

–2

A1,

A1

5

∴x

= –2

, –1,

4

SC N

MS

or p

artia

l met

hod

1 co

rrec

t val

ue 1

/5

2 co

rrec

t val

ues 2

/5

ind

epen

dent

of

3 co

rrec

t val

ues 5

/5

met

hod

mar

k m

ore

than

3 d

istin

ct v

alue

s max

2/5

(ii)

4,–

2x

x>

<

M1,

A1

2 x

> th

eir l

arge

st, x

< th

eir s

mal

lest

; C

AO

Tot

al12

–1

3

–1

1

1 in

depe

nden

t

AQA - Core 3 118

QSo

lutio

nM

arks

T

otal

Com

men

ts

82

2

1d

cos

(12t

an)

xx

x+

12t

anu

x=

+

2d

2sec

dux

x=

OE

M1

cond

one

2d

sec

dua

xx

=w

here

a is

a

cons

tant

2

d 2u u

=m

12

(d)

ku

u, w

here

k is

a c

onst

ant

A1

corr

ect,

or

()

21

d2

uu

−

11 2

1u−

=−

A1F

corr

ect i

nteg

ral o

f the

ir ex

pres

sion

but

m

ust h

ave

scor

ed M

1 m

1 1 2u

=−

()

12(

12t

an)

cx

=−

++

A1

5 C

SO, n

o IS

W

Tot

al5

QSo

lutio

nM

arks

T

otal

Com

men

ts

9 (a

) ln

dx

xx

lnu

x=

()

d dvx

x=

()

dl

du xx

=2 2x

v=

M1

corr

ect d

irect

ion

and

sigh

t of

1 x,

2 2x

()

22

1ln

d2

2x

xx

xx

=−

×A

1

()

22

ln2

4x

xx

c=

−+

A1

3

(b)

2(ln

)y

x=

d

12l

ndy

xx

x=

×M

1 ln

kx

x w

here

1

,1or

22

k=

A1

22

k=

(c)

lny

xx

=

()

()

e2

1ln

dV

xx

x=

B1

all c

orre

ct, i

ncl b

rack

ets,

, lim

its a

nd d

x (b

ut d

x m

ay b

e se

en B

EFO

RE

this

line

)

()

()

2d

lndv

ux

xx

==

()

2d

12l

nd

2u

xx

vx

x=

=M

1co

rrec

t dire

ctio

n w

ith (

)d

lndu

kx

xx

= w

here

1,1

or2

2k

=an

d si

ght o

f 2 2x

()

22

22

ln–

lnd

22

()

xx

xx

xx

=×

m

1co

rrec

t sub

stitu

tion

of th

eir t

erm

s int

o th

e pa

rts fo

rmul

a

()

22

(ln

–n

d2

)l

xx

xx

x=

A1

inte

gral

nee

ds to

be

sim

plifi

ed to

ln x

x

()

()

22

21

ln–

2ln

–12

4x

xx

x=

OE

()

()

22

2e 1

1(

)ln

–2l

n–1

24

xV

xx

x=

22

e1

1(

)–

e–

02

44

=+

m1

corr

ect s

ubst

itutio

n of

1 a

nd e

into

thei

r ex

pres

sion

s of t

he fo

rm

22

22

(ln)

lnpx

xqx

xrx

++

whe

re p

, q a

nd

r are

non

-zer

o ra

tiona

l num

bers

, and

an

inte

ntio

n to

subt

ract

D

o no

t con

done

F(1

) – F

(e)

2 e–1

4=

OE

A

1 6

2 e1

–4

4 e

tc

Tot

al

11

TO

TA

L

75

AQA - Core 3 119

GeneralCertificate

ofEducation


January

2012

Mathematics

MPC3

UnitPure

Core

3

Friday20January

2012

1.30pm

to3.00pm

Forthis

paperyoumusthave:




Tim

eallowed

*1hour30minutes

Instructions


only

beusedfor

drawing.

*Fillin


page.


*Write


theleft-hand

margin.




*Show

allnecessary

working;otherw


lost.

*Doallroughwork

inthis


thatyoudo

notwantto

bemarked.

Inform

ation


shownin

brackets.

*Themaxim

um

mark

forthis

paperis

75.

Advice

*Unlessstatedotherw

ise,youmayquote

form

ulae,withoutproof,

from

thebooklet.

*Youdonotnecessarily

needto

useallthespaceprovided.

1(a)

Use

Sim

pson’s

rule

with7ordinates

(6strips)

tofindan

estimatefor

ð 3 04xdx.

(4marks)

(b)

Acurveis

defined

bytheequationy¼

4x.Thecurveintersects

theliney¼

8�2x

atasingle

pointwherex¼

a.

(i)

Show

that

alies

between1.2

and1.3.

(2marks)

(ii)

Theequation4x¼

8�2xcanberearranged

into

theform

x¼

lnð8

�2xÞ

ln4

.

Use

theiterativeform

ula

x nþ1

¼lnð8

�2x nÞ

ln4

withx 1

¼1:2

tofindthevalues

of

x 2andx 3,givingyouransw

ersto

threedecim

alplaces.

(2marks)

2Thecurvewithequationy¼

63

4x�1is

sketched

below

for14

x4

16.

Thefunctionfis

defined

byfðx

Þ¼63

4x�1for14

x4

16.

(a)

Findtherangeoff.

(2marks)

(b)

Theinverse

offis

f�1.

(i)

Findf�

1ðxÞ

.(3

marks)

(ii)


1ðxÞ

¼1.

(2marks)

(c)

Thefunctiongisdefined

bygðxÞ

¼x2for�4

4x4

�1.

(i)

Write

downan

expressionforfgðxÞ

.(1

mark)

(ii)


¼1.

(3marks)

y Ox

116

AQA - Core 3 120

3(a)

Given

that

y¼

4x3�6xþ1,finddy

dx.

(1mark)

(b)

Hence

find

ð 3 2

2x2�1

4x3�6xþ1dx,givingyouransw

erin

theform

plnq,where

pandqarerational

numbers.

(5marks)

4(a)

Byusingasuitable

trigonometricalidentity,solvetheequation

tan2y¼

3ð3

�secyÞ

givingallsolutionsto

thenearest

0.1�in

theinterval

0�<

y<

360�.

(6marks)

(b)

Hence

solvetheequation

tan2ð4x�10�Þ

¼3½3�secð4

x�10�Þ�

givingallsolutionsto

thenearest

0.1�in

theinterval

0�<

x<

90�.

(3marks)

5(a)

Describeasequence


ationsthat

mapsthegraphof

y¼

lnxonto

thegraphofy¼

4lnðx

�eÞ.

(4marks)

(b)

Sketch,ontheaxes

given

below,thegraphofy¼

j4lnðx

�eÞj

,indicatingtheexact

valueofthex-coordinatewherethecurvemeets

thex-axis.

(3marks)

(c)(i)

Solvetheequationj4lnðx

�eÞj

¼4.

(3marks)

(ii)

Hence,orotherwise,

solvetheinequalityj4lnðx

�eÞj

54.

(3marks)

y

xO

�ee

6(a)

Given

that

x¼

1

siny,use

thequotientrule

toshow

that

dx

dy¼

�cosecycoty

.

(3marks)

(b)

Use

thesubstitutionx¼

cosecy

tofind

ð 2 ffiffi 2p1

x2


ffix2�1

pdx,givingyouransw

erto


(9marks)

7(a)

Acurvehas

equationy¼

x2e�

x 4.

Show

that

thecurvehas

exactlytwostationarypoints

andfindtheexactvalues

of

theircoordinates.

(7marks)

(b)(i)

Use

integrationbyparts

twiceto

findtheexactvalueof

ð 4 0x2e�

x 4dx.

(7marks)

(ii)

Theregionbounded

bythecurvey¼

3xe

�x 8,thex-axis

from

0to

4andtheline

x¼

4is

rotatedthrough360�aboutthex-axis

toform

asolid.

Use

youransw

erto

part(b)(i)to

findtheexactvalueofthevolumeofthesolid

generated.

(2marks)

AQA - Core 3 121



3

0

1) 4 0.5 (0) (3) 4 (0.5) (1.5) y(2.5) 2 (1) y(2)

31

0.5 1 64 4(2 8 32) 2(4 16)3

) is solution of the equation 4 8 2

' ( ) 4 2 8

4

(1.

Acco

5.

2) 0.322 0 (1.3) 0.663

5

rding to

0

the

x

x

x

a dx y y y y y

b x

Let s cal

f

l f x

and f

x

1 2 3

, we know

that there is a root of , so that 1.2 1.3

) 4 8 2 ln 4 ln(8 2 )

ln 4 ln(8 2

change of sign rule

1.243 , 1.

)

1.

ln(8 2 )

232,ln 4

2

x x

xx

f

ii x

x x x

x

x x

In part (a), Simpson’s Rule was generally well done; a few omitted x = 0, thus invalidating the formula; a few reversed the 4 and 2 multipliers; and a few made calculator errors, with (1 + 64) = 64 and 2(4 + 16) = 20 being the common ones. Part (b)(i) was well answered by the majority of candidates. Many fully correct responses were seen. Most candidates used f(x) = 4x + 2x – 8 or g(x) = 8 – 2x – 4x and evaluated either g or f(1.2) and f(1.3) correctly. There are still many candidates who then write “change of sign therefore a root” without clarification of where the root lies. Those candidates who used the alternative LHS/RHS method were less successful as they appeared to be unable to then make a correct statement, with many still just putting “change of sign therefore a root”. Part (b)(ii) was very well answered. The main error was with candidates who wrote answers to three significant figures, 1.24 and 1.23, rather than three decimal places. A few candidates gave answers to more than three decimal places.


1

2

1

2

2

)1 16 4 4 64

1 1 13 4 1 63

63 4 1 363

1 214 1

The range of

63 1 4 1) )

4 1 63

63

1 ( ) 21

63 1

63 14 1

4 4

) (

( )4 4

(1) 21

63( )

) 1

) ) ( )

63) ( ) 1 1

4 1

4

4 1

1

a x x

xx

xf is

xb i y

x y

x xy y

ii f x

f x

f xx

x f

f g xc i fg x

iix

x

fg x

x

263 1

4 4

6

x or x

x

In part (a), there were many fully correct answers, but often the accuracy mark was lost for two separate sets not connected. Some candidates gave their answers as strict inequalities, and some gave their answers as x instead of f(x) and a few gave the range to be 21 – 1 = 20. However many weaker candidates did not know how to tackle this part at all. In part (b)(i), it was good to see that most got a correct

expression, although a few candidates wrote 63 1 252 1

4 4x as

x

.

Some candidates made a sign error, getting –1 instead of +1. It was good to see that hardly anyone left their expression in terms

of y or took the function to be 4 1

63

x .

Part (b)(ii) was very well answered with most candidates obtaining both marks. Those candidates who had made an error in part (b)(i) were usually able to obtain the method mark for a correct step. Part (c)(i) was, again, mostly well done with just a few having

2 2

63 631

(4 ) (4 1)or

x x

In part (c)(ii), very few candidates, even the most able, gained full marks. Most candidates gained the first two marks by equating to 1 and obtaining an equivalent expression to x² = 16, but most then offered the two solutions +4 and –4 and failed to state that the only possible solution was –4, thus taking no account of the domain for g.

AQA - Core 3 122


3

2 23 3

3 32 2

33

2

2

) 4 6 1

2 1 1 12 6)

4 6 1 6 4 6 1

This integral is n'

ln( )ow of the form

1 1ln 4 6 1 l

12 6

1 13ln

n 91 ln 21

6 3

6 61 91

ln6 21

f

a y x x

so

x xb dx dx

d

x x x x

x

y

f cf

xd

x

x

In part (a), almost everyone earned the mark. Part (b) was answered very well. Most candidates who realised the numerator was related to the derivative of the denominator and hence required a ln function were successful in obtaining full marks. There were two very common errors: • candidates losing the final accuracy by writing

1 1 1ln91 ln 21 ln 70

6 6 6

• candidates losing 2 marks by obtaining an incorrect value of k; the most common value was 6 although ½ and 1/3 were also seen. The major errors seen were expressions such as

2

32

2 1ln 4 6 1

12 6

xx x

x

. Some candidates were able to

recover by simplifying to the correct expression

31ln 4 6 1

6x x before substitution of the limits; however

some candidates substituted for 3 and 2 through the whole unsimplified expression and therefore gained no credit.


2 2

2

2

1

) tan 3(3 sec )

Using the identity tan sec 1

sec 1 9 3sec

sec 3sec 10 0

(sec 5)(sec 2) 0

sec 5 sec 2

1 1cos cos

5 21

101.5 258.5

60 30

cos ( ) 360 101.55

)This is the sa

0

o o

o o

a

A A

or

or

or

or or

b

me equation with 4 10

The solutions are given by solving

4 10 1 27.9

67.

01.5

4 10 258.5

4 10 60

4 1

1

17.5

0 77.0 530

o

o

o

o

x

x

x

x

x

x

x

x

x

A few candidates made no attempt at this question at all. Part (a) was very well answered by the majority of candidates, with many obtaining full marks. Most candidates used the correct substitution for tan²θ and went on to produce the correct quadratic function. Factorisation was usually handled correctly with the common error of secθ = –2 or 5 occasionally seen. The major error noticed was with the answers produced: 60°, 300°, 101.5° were usually correct but the fourth value was often given as 281.5° (from 101.5° + 180°), rather than the correct solution of 180° + 78.5° = 258.5° . In part (b), although most candidates recognised the connection with part (a), not all realised what ‘hence’ implied. Many gave fully correct answers, but it was disappointing, at this level, how many candidates got the algebra wrong by dividing by 4 first and then adding 10. However, a significant number of candidates chose to start again and most were unable to handle the new expression and gained no further marks.

AQA - Core 3 123


translation of vector 0

stretch scale factor 4 in the y-directi

) ln 4ln( )

a

) :ln( )

4 ln( ) 0

1

) ) 4ln( ) 4

4ln( ) 4

o

4ln( )

l

n

4

n( )

1

a y x and y x e

and

b Note x e exists only for

y x e

when x e

c i x e

x e or x

x e

x e

e

e

e

x

1

1

1

1 ln( ) 1

) Plot the line 4

and state the values of for which

the grap

2

h 4ln( ) " "

2

or x e

x e e or x e e

ii y

x

y x e is above

x e or x e e

e x e

the line

e or x e

Very few candidates scored full marks for this question. Part (a) was well done with only a handful failing to recognise a stretch and most giving the correct direction and scale factor; most who were wrong in one of these were wrong in both. It was essential to use the correct term ‘translate’ for the other, so ‘shift’ did not earn marks and ‘transformation’ was inadequate. A few made an error in the vector but most were correct. Most candidates were unsuccessful in answering part (b) of the question as their curve was either below the x‐axis or continued into the second quadrant. Many placed the curve through e instead of e + 1, and the curvature, for x < e + 1, was often wrong. Correct answers in part (c)(i) were sometimes spoiled by poor algebra. It was worrying to see the misunderstanding of the log function, with ln(x – e) expanded as ln x – ln e and incorrect subsequent work once a correct answer had been found: e–1 + e = 0 was a common such response. In part (c)(ii), although the first mark for x ≥ 2e was often earned (though x ≤ 2e was common), the second mark was only gained by a very few candidates, as the part of the inequality concerning e was generally omitted completely.


2 2

2

2 2 22

1)

sin0 sin 1 cos cos

sin sincos 1

sin sin

) cosec cosec cot

12, cosec 2 sin 30

2

c

12,cosec 2 sin 45

21 cosec cot

1 cos

osec cot

ec

o

o

a x

dx

ddx

ddx

b x sod

when x so and

when x so and

dxx x

30

245

2 30 30

22 22 45 45

2 30

452 2

2

2

2

cosec 1

using the identity ,

1 cosec cotsin

cosec cot1

1cos cos30 cos 4

3 20.

cot co

51

1592 2

sec 1

d

dx d dx x

dxx x

In part (a), it was essential to use the quotient rule, as requested — it is specifically mentioned in the specification — and to show sufficient steps in the proof of the given answer. It was surprising that quite a number of candidates were unable to go

from 2

Cos

Sin

to the final expression.

In part (b), it was good to see some completely correct solutions though some used 30º and 45º, which was incorrect. A substantial number of candidates who correctly started the question lost a minus sign in their work. Many candidates fell at the first hurdle by not replacing dx by an expression in and d . There

were B marks available for a correct dx

d, for

2cosec 1 converted to 2cot and for the

limits in radians, but even these marks were often not earned. Unfortunately, many confused dx d

withd dx

with and of course made no headway

after that. Candidates who produced the correct numerical answer from their calculators without correct working did not gain marks.

AQA - Core 3 124


2 4

1 124 4

124

124

124

102 4

2

)

12

4

(2 )4

To find the stationary points, let's solve

0 (2 ) 04

2 0 04

(2 ) 04

0 8

for 0, 0 0

for 8, 8

x

x x

x

x

x

a y x e

dyx e x e

dx

dy xx e

dx

dy xx e

dx

xx because e for all x

xx

x or x

x y e

x y

2

18 24

41 1 14 42 24 4 4

0 00

141 4

0

41 141 4 4

00

11 1 4

64

) ) 4 2 4

64 8

64 8 4 4

6

The stationary points are (0,0) and (8,64

128

e

4 32

)

x x x

x

x x

e e

b i I x e dx x e x e dx

I e xe dx

I e xe e dx

I e e e

41

4 1 4

00

1 1

1 14 4 42 2 24 4

0 0 0

1

1

320 12

192 128

1 8

9 (

92 128 12

320 128)

8

) 9 9

x x

x x

dx e e

I e e

ii V y dx x e dx x e

e

d

V

x

e

In part (a), most candidates scored the first two marks, but some missed out the minus sign when differentiating. The main problem was candidates’ inability to deal with

1

4x

e

, with 1

4

4

xxe being the most common error.

Where correct initial solutions were given, the next problem was forming a quadratic and then dealing with

1

4x

e

≠ 0. Very few candidates obtained the E mark, and

many lost the m mark for not showing a viable quadratic. Another very common error was when x = 0 to give the coordinates as (0, 1). A significant number of candidates failed to find the y values after an otherwise‐correct solution. In part (b)(i), the concept of integration by parts seemed to have been met by the majority of candidates and most

went in the right direction; those who chose dv

dx to be x2

could not recover any marks, and the same was true of

values of v such as 1

44 x

v ex

, which were often seen.

Those candidates who correctly obtained 1

44x

v e

usually obtained the first 4 marks, but there were often sign errors in the final expression, with

1 1 12 4 4 44 32 128

x x xx e xe e

being the most common

error. 144exx−−14ex− 14ex− 14ex− 14ex− Candidates who lost marks in part (b)(i) often obtained both the marks in part (b)(ii) by following through correctly with their solution. The main error with this part was using a coefficient of 3 rather than 9 after squaring the expression for y.

GRADE BOUNDARIES


Core 3 – Unit PC3 75 64 57 50 43 37 31

AQA - Core 3 125

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AQA - Core 3 126

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AQA - Core 3 127

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lutio

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arks

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sd d

sin

x

M1

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quot

ient

rule

2

sin

1co

s

sink

whe

re k

= 0

or

1

mus

t see

the

‘0’

eith

er in

the

quot

ient

or

in

eg d

0d

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c

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cos

–

or

si

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AQA - Core 3 128

GeneralCertificate

ofEducation


June2012

Mathematics

MPC3

UnitPure

Core

3

Thursday31May2012

9.00am

to10.30am

Forthis

paperyoumusthave:




Tim

eallowed

*1hour30minutes

Instructions


only

beusedfor

drawing.

*Fillin


page.


*Write


theleft-hand

margin.

*Youmustanswereachquestionin

thespaceprovidedforthat

question.Ifyourequireextraspace,useanAQA

supplementary

answerbook;donotusethespaceprovidedforadifferentquestion.

*Donotwrite


*Show

allnecessary

working;otherw


lost.

*Doallroughwork

inthis


thatyoudo

notwantto

bemarked.

Inform

ation


shownin

brackets.

*Themaxim

um

mark

forthis

paperis

75.

Advice

*Unlessstatedotherw

ise,youmayquote

form

ulae,withoutproof,

from

thebooklet.

*Youdonotnecessarily

needto

useallthespaceprovided.

1Use

themid-ordinaterule

withfourstripsto

findan

estimatefor

ð 1:2

0:4cotðx

2Þd

x,

givingyouransw

erto

threedecim

alplaces.

(4marks)

2For0<

x4

2,thecurves

withequationsy¼

4lnxandy¼

ffiffiffi xpintersectat

a

single

pointwherex¼

a.

(a)

Show

that

alies

between0.5

and1.5.

(2marks)

(b)

Show

that

theequation4lnx¼

ffiffiffi xpcanberearranged

into

theform

x¼

e

ffiffi xp 4��

(1mark)

(c)

Use

theiterativeform

ula

x nþ1

¼e

ffiffiffi x np 4��

withx 1

¼0:5

tofindthevalues

ofx 2

andx 3

,givingyouransw

ersto

threedecim

al

places.

(2marks)

(d)

Figure

1,onthepage3,showsasketch

ofparts

ofthegraphsofy¼

e

ffiffi xp 4�� an

d

y¼

x,andthepositionofx 1

.

OnFigure

1,draw

acobweb

orstaircasediagram

toshow

how

convergence

takes


andx 3

onthex-axis.

(2marks)

AQA - Core 3 129

Figure

1

3A

curvehas

equationy¼

x3lnx.

(a)

Finddy

dx.

(2marks)

(b)(i)

Findan


thecurvey¼

x3lnxat

thepointonthecurve

wherex¼

e.

(3marks)

(ii)

This

tangentintersects

thex-axis

atthepointA.Findtheexactvalueofthe

x-coordinateofthepointA.

(2marks)

y

xx 1

y¼

x

y¼

e

ffiffi xp 4��

O

4(a)

Byusingintegrationbyparts,find

ð xe6

xdx.

(4marks)

(b)

Thediagram


ffiffiffi xpe3

x.

Theshaded

regionRis

bounded

bythecurvey¼

ffiffiffi xpe3

x,thelinex¼

1andthe

x-axis

from

x¼

0to

x¼

1.


when

theregionRis



erin

theform

pðpe6

þqÞ,

wherepandqare

rational

numbers.

(3marks)



fðxÞ¼


ffi2x�5

p,

forx5

2:5

gðxÞ

¼10 x,

forreal

values

ofx,

x6¼

0

(a)

State

therangeoff.

(2marks)

(b)(i)

FindfgðxÞ

.(1

mark)

(ii)


¼5.

(2marks)

(c)

Theinverse

offis

f�1.

(i)

Findf�

1ðxÞ

.(3

marks)

(ii)


1ðxÞ

¼7.

(2marks)

y

x1

O

R

AQA - Core 3 130

6Use

thesubstitutionu¼

x4þ2to

findthevalueof

ð 1 0

x7

ðx4þ2Þ2

dx,givingyour

answ

erin

theform

plnqþr,wherep,qandrarerational

numbers.

(6marks)

7Thesketch


fðxÞ.

(a)

OnFigure

2onpage6,sketch


jfðxÞ

j.(3

marks)

(b)

OnFigure

3onpage6,sketch


fðjxjÞ.

(2marks)

(c)

Describeasequence


ationsthat

mapsthegraphof

y¼

fðxÞonto

thegraphofy¼

1 2fðx

þ1Þ.

(4marks)

(d)

Themaxim

um

pointofthecurvewithequationy¼

fðxÞhas

coordinates

ð�1,10Þ.

Findthecoordinates

ofthemaxim

um

pointofthecurvewithequationy¼

1 2fðx

þ1Þ.

(2marks)

y

xO

�4�3

�2�1

12

34

(a)

Figure

2

(b)

Figure

3

y

xO

�4�3

�2�1

12

34

y

xO

12

34

�4�3

�2�1

AQA - Core 3 131

8(a)

Show

that

theequation

1

1þcosyþ

1

1�cosy¼

32

canbewritten

intheform

cosec2

y¼

16

(4marks)

(b)

Hence,orotherwise,

solvetheequation

1

1þcosð2

x�0:6Þþ

1

1�cosð2

x�0:6Þ¼

32

givingallvalues

ofxin

radiansto

twodecim

alplacesin

theinterval

0<

x<

p.

(5marks)

9(a)

Given

that

x¼

siny

cosy,use

thequotientrule

toshow

that

dx

dy¼

sec2

y(3

marks)

(b)

Given

that

tany¼

x�1,use

atrigonometricalidentity

toshow

that

sec2

y¼

x2�2xþ2

(2marks)

(c)

Show

that,if

y¼

tan�1

ðx�1Þ,

then

dy

dx¼

1

x2�2xþ2

(1mark)

(d)

Acurvehas

equationy¼

tan�1

ðx�1Þ�

lnx.

(i)

Findthevalueofthex-coordinateofeach


ofthecurve.

(4marks)

(ii)

Find

d2y

dx2.

(2marks)

(iii)Hence

show

that

thecurvehas

aminim

um

pointwhichlies

onthex-axis.

(2marks)

AQA - Core 3 132



1.2 2

0.4cot( ) 0.2 (0.5) (0.7) (0.9) (1.1)

0.2 3.9163 1.8748 0.9520 0.3773

1.424 3 .

x dx y

t

y y

o dec p ce

y

la s

This was well answered by the majority of students. Many fully correct responses were seen. The major error was from students who worked in degrees, but they usually worked to the correct degree of accuracy and hence earned the initial B mark for their mid‐ordinates and the special case mark for their final answer. Students who worked to two decimal places (rounded or truncated) were liable to lose the last accuracy mark, but often lost more through not showing enough working.


4ln intersects

which means that is solution

of the equation 4 ln( ) 4 ln 0

Let's call ( ) 4 ln

)

According to

(0.5) 3.48 0 (1.5) 0.40

, we know

that there is

0

the change o

a ro

f sig rule

o

n

y x y x

x x x x

f x x x

a f and f

1

4

2 3

t so that 0.5 1.5

) 4 ln ln4

) 0.5 1.193 , 1., 314

)

x

x e

x

xb x x x

c xx

d

This was well done with many students obtaining full marks.

In part (a), the majority of students used 4ln 0x x and

evaluated it correctly for x = 0.5 and x = 1.5. There are still many students who then write ‘change of sign therefore a root’ without clarification of where the root lies and who therefore lose the A mark. Less successful were the students who tried the LHS‐RHS

approach on 4ln x x . Although correct numerical values

were seen earning the method mark, the final accuracy mark was often not earned. Parts (b) and (c) were well answered by the majority of students. When students failed to score full marks, it was usually due to poor notation. Part (d) was well answered by the majority of students. The main error was the incorrect labelling of the axes.


3

2 3

2 2

3 3

3

2 2

2

2

2 3

3

2

ln

1) 3 ln

) )The gradient of the tangent

4

where

is 3 ln

when , ln

The equation of the tangent where

4

)Solve 4

l

3

3 n

4 3

y x x

dya x x x

dx xb i x e

m e e e

A

dyx x x

dx

y

ND x e y e e e

x eis

y e e x e

ii y e x

e x e

e

e

2 3

0

4 33

4ex xe e

This question was in general a good source of marks for the majority of students. The main cause of loss of marks was a common desire to use an approximate numerical value for e.

AQA - Core 3 133


6

6 6 6

1 12 6

0 0

1 6 66 6

0

6

6

1

6 36

5 1

36 3

1 1) 1

6 6

)

1 10

6 36 6 36 36

6

x x x

x

x x

x x

a xe dx x e e dx

b V y dx xe dx

xe e c

V e

V e

x e ee

Part (a) was answered very well by many of the students, the main error being v = 6e6x instead of 1/6e6x. The other error was

in the final answer, where 1 1 1

6 6 12 was a common sight.

Again many correct responses were seen in part (b), but this was the first question where many students started to lose marks. Many could not set up the initial formula for the volume, with the major error being the omission of dx. A significant number of students also failed to see the connection with part (a) and started again often getting different results to their previous response. The other main error was the substitution of the limits with xe6x often being evaluated as 1/6 when x=0 was substituted.


2 2

1 2

1

) ( ) 2 5 2.5

10( ) 0

) 2.5 2 5 0

2 5 0

The range is :

) ) ( )

20) ( ) 5 5 5

20 205 25 30

1

20 30

) ) 2 5

1 52 5

2 21 5

(

( ) 0

20( ) 5

20 2

2

30 3

)2

) (

a f x x for x

g x for xx

a x so x

and x

b i fg x

ii fg xx

sox xx

c i y x

y x

f x

f

x y

f x x

x

ii f

g x

x

(7)7 14 5 3) x fx

Although this question was well answered very few students scored full marks. The majority of students scored full marks in parts (a) and (b), although there was a common mistake in part (b)

of 2025 5

x . Even the better students failed to obtain both

marks in part (c)(ii), with few scripts giving a justification of why there was only one solution and rejecting x = –3.


4 3 3

7 4 31 1

4 2 4 20 0

4 3

33 3

2 22 22

12 4

4 0, 2 1, 3

( 2) ( 2)

1We substitute 2

4

1 2 1 1 2 1 2ln

4 4 4

1 2ln 3 ln 2 1

4 3

1 3 1ln

4 2 12

duu x so x du x dx

dxwhen x u and when x u

x x xI dx dx

x x

x byu and x dx by du

uI du du u

u u u u

I

In this question, students tended to do very well or very badly. It was good to see that many students were able to complete this integration correctly. However there were many blank scripts, and the majority of students obtained only the B mark for a correct answer for du/dx. Some students were able to make slight progress, and where M1 was obtained, many did manage to produce some form of ln

function, although many made a sign error and obtained 2ln u

u

. There were also many fully correct solutions seen.

AQA - Core 3 134


) ( )

) ( )

)Translation vector 0

a stretch scale factor in the -dir.

)The maximum of has coordinates 1,10

1The maximum of ( 1) has coordinates

1

2

1

21 , 10

1

12

2,5

a y f x

b y f x

c

and y

d f

f x

There were many high marks on this question, although only a minority of students were able to answer part (b) correct. Students answered part (a) well, with the majority giving a correct modulus graph. The main error was incorrect curvature in the outside branches x > 3 and x< –2. In part (b), some students gave an identical response to the modulus graph in part (a), but the most common result was a correct plot for x � 0 but then the LHS of the curve was reflected in the y‐axis to produce a sketch in the first and fourth quadrants. There were many fully correct solutions to parts (c) and (d). The main errors were the use of the word transformation rather than translation and also the scale factor for the stretch stated as 2 in the x‐axis rather than ½ in the y‐axis. The common errors in part (d) were (0,5) and (–1,20)


2

2 2

2

22

1

1

2

1 1) 321 cos 1 cos

1 cos 1 cos32

(1 cos )(1 cos ) (1 cos )(1 cos )

2 132 16

1 cos sin

:

1 1) 16 sin

sin 161 1

sin sin4 4

1sin (

cosec 16

0.25) 0.

cos sin 1

2.8254

n

9

si

a

Identity

b

or

or

or

11 1( ) sin ( )

4 4The equation in part b) is the same as in part a) with

2 0.6

We need to solve

2 0.6 0.25

2 0.6 2.89

2 0.

0.25 3.39

0.43

1.75

0.18

2.00

All the values

6 0.25

2 0.

f

6 3.39

x

or

x

x

x

x

x

x

x

x

ound are between 0 and .

This question was poorly answered. In part (a), although many students achieved full marks, some found the combining of two fractions beyond them. For many students, an initial step was replacing 1+cosθ with sinθ or attempting to multiply through by 1+ cosθ , thus obtaining

expressions such as 1 cos1 32

1 cos

.

Despite the majority of students finding part (b) of this question difficult, full marks were occasionally seen. Students who were unsuccessful in part (a) were able to attempt this part using the result given. Most students earned the first method mark, but for many this was all they achieved since they only worked with sinθ = +¼ and then only worked to 2dp which resulted in just 2 solutions. For the final two B marks, the solution x= 0.17 was often missing, and 2.00 was often written as 1.99.

AQA - Core 3 135


2

2 2

2 2

2

2

2

2

2

2

1

2

2

2 2

2

2

sin)

cos

cos cos sin sin

cos

cos sin 1

cos cos

) 1 ( 1)

sec 2 2

) sec

) tan

sec 1 tan

sec

1 1

se

( 1) ln

1 1 ( 2 2))

2

2

2 (

c 2

y

dy

d

y

ya x

y

dx y y y

x y x

y

dy y

dx y y

d

y

y y y

b x

y x x

dxc y so

dy

d y x x

x

dy x x xi

dx x x x x x

2 2 2

2

2

2

2

12 1

22 2 2

2

2

2

2

2 2)

0 3 2 0

( 2)( 1) 0

)To wor

2 2

k out , we use

2 2

1 (2 2)( 2 2)

) for

1

( 2 2)

3

1, 0

2

( 2 2)

2 1

x

dywhen x x

dxx x

d y dyii writtenas

dx d

dy x x

dx x x x

x or x

xdy

x x xdx

d yx x x x

dx

d yiii x

x

dx

x

x x

1 1

1 1 0

Now let's show that this minimum belongs to the x-axis:

for 1, tan (1 1) ln1 tan 0 ln1 0

The minimum is on the x-axis.

Minimum

x y

Part (a) proved to be a good source of marks, although a significant number of students lost a mark through poor notation, such as cos y² or no dx/dy. Although many earned full marks in part (b), many other proofs were inadequate. Students were expected to use the identity tan²y + 1 = sec²y and to show the expansion of (x – 1)² clearly. Too many fudged their working, wrote tan y + 1 = sec y or only proved the known trig identity. The result in part (c) was easily derived from the previous two parts, but there were many blanks here and many confused statements about inverse functions which were often taken as 1/tan y. Students who used the formula book appropriately generally earned the mark. There were many fully correct solutions to part (d)(i). However it was distressing at this stage to see some horrendous algebra, with terms inverted in ways that were invalid and worthless. Only the more able students were able to make progress on the last two parts of the question. Many students differentiated their quadratic instead of the expression for dy/dx. Some students earned the method mark in the final part, but few completed the whole answer, as their final explanation often had an element missing.

GRADE BOUNDARIES


Core 3 – Unit PC3 75 67 61 55 49 43 38

AQA - Core 3 136

Key

to m

ark

sche

me

abbr

evia

tions

M

mar

k is

for

met

hod

m o

r dM

m

ark

is d

epen

dent

on

one

or m

ore

M m

arks

and

is f

or m

etho

d A

m

ark

is d

epen

dent

on

M o

r m

mar

ks a

nd is

for

acc

urac

y B

m

ark

is in

depe

nden

t of

M o

r m

mar

ks a

nd is

for

met

hod

and

accu

racy

E

m

ark

is f

or e

xpla

nati

on

or f

t or

F fo

llow

thro

ugh

from

pre

viou

s in

corr

ect r

esul

t C

AO

co

rrec

t ans

wer

onl

y C

SO

co

rrec

t sol

utio

n on

ly

AW

FW

an

ythi

ng w

hich

fal

ls w

ithi

n A

WR

T

anyt

hing

whi

ch r

ound

s to

A

CF

an

y co

rrec

t for

m

AG

answ

er g

iven

SC

spec

ial c

ase

OE

or e

quiv

alen

tA

2,1

2 or

1 (

or 0

) ac

cura

cy m

arks

–x

EE

de

duct

x m

arks

for

eac

h er

ror

NM

S

no m

etho

d sh

own

PI

poss

ibly

impl

ied

SC

Asu

bsta

ntia

lly

corr

ect a

ppro

ach

cca

ndid

ate

sfsi

gnif

ican

t fig

ure(

s)dp

deci

mal

pla

ce(s

)

No

Met

hod

Show

n

Whe

re th

e qu

esti

on s

peci

fica

lly

requ

ires

a p

arti

cula

r m

etho

d to

be

used

, we

mus

t usu

ally

see

evi

denc

e of

use

of

this

met

hod

for

any

mar

ks to

be

awar

ded.

Whe

re t

he a

nsw

er c

an b

e re

ason

ably

obt

aine

d w

ithou

t sh

owin

g w

orki

ng a

nd i

t is

ver

y un

likel

y th

at t

he

corr

ect

answ

er c

an b

e ob

tain

ed b

y us

ing

an i

ncor

rect

met

hod,

we

mus

t aw

ard

full

mar

ks.

How

ever

, th

e ob

viou

s pe

nalty

to c

andi

date

s sh

owin

g no

wor

king

is th

at in

corr

ect a

nsw

ers,

how

ever

clo

se, e

arn

no m

arks

.

Whe

re a

que

stio

n as

ks th

e ca

ndid

ate

to s

tate

or

wri

te d

own

a re

sult

, no

met

hod

need

be

show

n fo

r fu

ll m

arks

.

Whe

re t

he p

erm

itte

d ca

lcul

ator

has

fun

ctio

ns w

hich

rea

sona

bly

allo

w t

he s

olut

ion

of t

he q

uest

ion

dire

ctly

, th

e co

rrec

t an

swer

wit

hout

wor

king

ear

ns fu

ll m

arks

, un

less

it

is g

iven

to

less

tha

n th

e de

gree

of

accu

racy

ac

cept

ed in

the

mar

k sc

hem

e, w

hen

it g

ains

no

mar

ks.

Oth

erw

ise

we

requ

ire

evid

ence

of a

cor

rect

met

hod

for

any

mar

ks to

be

awar

ded .

QSo

lutio

nM

arks

Tot

alC

omm

ents

1x

y 0.

5 3.

9163

0.

7 1.

8748

0.

9 0.

9520

1.

1 0.

3773

B1

M1

All

4 co

rrec

t x v

alue

s (an

d no

ext

ras u

sed)

3+ y

dec

imal

val

ues r

ound

ed o

r tru

ncat

ed

to 2

dp

or b

ette

r (in

tabl

e or

in fo

rmul

a)

(PI b

y co

rrec

t ans

wer

)

0.2

ym

1 C

orre

ct su

bstit

utio

n of

thei

r 4 y

valu

es (o

f w

hich

3 a

re c

orre

ct),

eith

er li

sted

or

tota

lled

(

= 0.

2 ×

7.12

…)

=

1.42

4

A1

4 C

AO

Tot

al4

Cor

e3 -

Jun2

012

- Mar

k sc

hem

e

AQA - Core 3 137

QSo

lutio

nM

arks

Tot

alC

omm

ents

2(a)

f

4ln

–x

xx

O

r rev

erse

f0.

5–3

.5m

usth

aveb

oth

valu

esco

rrect

f1.

50.

4M

1

Allo

w f

0.5

0an

df

1.5

0on

ly if

f(

x) d

efin

ed

Cha

nge

of si

gn

0.5

15

A1

2 f(

x) m

ust b

e de

fined

and

all

wor

king

co

rrec

t, in

clud

ing

both

stat

emen

t and

in

terv

al (e

ither

may

be

writ

ten

in w

ords

or

sym

bols

)

OR

com

parin

g 2

side

s:

4l

n0.5

2.8

0.5

0.7

(M1)

4ln1

.51.

61.

51.

2

At 0

.5, L

HS

< R

HS;

at 1

.5, L

HS

> R

HS

0.5

1.5

(A

1)

(b)

4ln

ore

4x

xx

x

Mus

t be

seen

4e

x

xB

1 1

AG

; no

erro

rs se

en

(c)

21.

193

x

B1

31.

314

x

B1

2 If

B0B

0 sc

ored

but

eith

er v

alue

seen

co

rrec

t to

2 or

4 d

p, sc

ore

SC1

(d)

M1

Ver

tical

line

from

1xto

cur

ve (c

ondo

ne

omis

sion

from

x-a

xis t

o y

= x)

and

then

ho

rizon

tal t

o y

= x

A1

2 2nd

ver

tical

and

hor

izon

tal l

ines

, and

2x

,3x (n

ot th

e va

lues

) mus

t be

labe

lled

on x

-axi

s

Tot

al

7

x 1

x 2 x

3

QSo

lutio

nM

arks

Tot

alC

omm

ents

3(a)

3

2d

13

lndy

xx

xx

x

M1

32

1ln

pxqx

xx

whe

re p

and

q a

re in

tege

rs

A1

2 1,

3p

q

(b)(

i)2

2d

e3e

lne

dy x2

4eM

1 Su

bstit

utin

g e

for x

in th

eir

d dy x, b

ut m

ust

have

scor

ed M

1 in

(a)

33

eln

ee

yB

1

32

–e

4e–

ey

xA

1 3

OE

but m

ust h

ave

eval

uate

d ln

e(tw

ice)

fo

r thi

s mar

k (m

ust b

e in

exa

ct fo

rm, b

ut

cond

one

num

eric

al e

valu

atio

n af

ter

corr

ect e

quat

ion)

(ii)

32

23

–e

4e–

eor

4e3e

OE

xx

M1

Cor

rect

ly su

bstit

utin

g y

= 0

into

a c

orre

ct

tang

ent e

quat

ion

in (b

)(i)

3e

4x

A1

2 C

SO;

igno

re su

bseq

uent

dec

imal

eva

luat

ion

Tot

al7

4(a)

6 e

dx

xx

6

6

de

(d)

d1

e(d

)

x

x

vu

xx

uv

kx

M1

All

4 te

rms i

n th

is fo

rm,

1,1

or6

6k

A1

1 6k

66

11

e–

ed

66

xx

xx

A1F

Cor

rect

subs

titut

ion

of th

eir t

erm

s int

o pa

rts fo

rmul

a

66

11

e–

eO

E6

36x

xx

cA

1 4

No

ISW

for i

ncor

rect

sim

plifi

catio

n

(b)

16

0

ed

xV

xx

B1

Mus

t inc

lude

, l

imits

and

dx

66

11

1e

–e

––

636

36

M1

Cor

rect

subs

titut

ion

of 0

and

1 in

to th

eir

answ

er in

(a),

mus

t be

of th

e fo

rm

66

ee

xx

axb

, whe

re a

> 0

, b >

0

and

F(1

) – F

(0) s

een

65

1e

3636

A1

3 C

AO

; ISW

Tot

al7

AQA - Core 3 138

QSo

lutio

nM

arks

Tot

alC

omm

ents

5(a)

f

0x

M1

f0,

f0,

0,0,

rang

e0

xx

y

A1

2 C

ondo

ne

0y

(b)(

i)10

fg2

–5

xx

B1

1 N

o IS

W

20–

5O

Ex

(ii)

20–

55

x

220

55

x

M1

Cor

rect

ly sq

uarin

g th

eir f

g(x)

and

co

rrec

tly is

olat

ing

thei

r x te

rm

2 3x

A

1 2

No

ISW

(c)(

i)2

–5

yx

M1

M1

Swap

a

nd

eith

eror

der

Cor

rect

ly sq

uarin

gx

y

21

5f

()

2x

xA

1 3

(ii)

29

x or if

9

or 3

seen

M

1

Can

dida

te m

ust h

ave

scor

ed fu

ll m

arks

in

(c)(

i) (ie

no

follo

w th

roug

h)

x =

3

and

x =

3

reje

cted

A

1 2

Mus

t see

bot

h

Tot

al10

QSo

lutio

nM

arks

Tot

alC

omm

ents

64

2u

x

3d

4du

xx

B1

or

3d

4d

ux

x

7

42

7 4

22

3 4

d(

2)

d(

2)(

2)d

or(

2)

xx

x

uk

uk

uu

uu

u

M1

Eith

er e

xpre

ssio

n al

l in

ter m

s of u

incl

udin

g re

plac

ing

dx, b

ut c

ondo

ne

omis

sion

of

du

2

11

2–

d4

uu

um

1–1

–2d

kau

buu

, whe

re k

, a, b

are

cons

tant

s 1

2ln

4u

uA

1M

ust h

ave

seen

du

on

an e

arlie

r lin

e w

here

eve

ry te

rm is

a te

rm in

u

3 2

12

ln4

uu

14

4

0

21

ln2

24

xx

12

ln3

–ln

21

34

m1

Dep

ende

nt o

n pr

evio

us A

1

Cor

rect

cha

nge

of li

mits

, cor

rect

su

bstit

utio

n an

d F(

3) –

F(2

) or

co

rrec

t rep

lace

men

t of u

, cor

rect

su

bstit

utio

n an

d F(

1) –

F(0

) 1

31

ln–

42

12A

1 6

OE

in e

xact

form

Tot

al6

AQA - Core 3 139

QSo

lutio

nM

arks

Tot

alC

omm

ents

7(a)

M

1 M

odul

us g

raph

, 4 se

ctio

ns to

uchi

ng x

-axi

s at

2,

1,3

A1

C

orre

ct

3,–2

xx

A1

3 C

orre

ct

23

x w

ith m

axim

um a

t 2

low

er th

an m

axim

um a

t 1 a

nd c

orre

ct

cusp

s at x

=

2, x

= 1

and

x =

3

The

max

imum

s nee

d to

be

at x

=

1 an

d 2

(app

rox)

(b)

M1

Sy

mm

etric

al a

bout

y-a

xis,

from

orig

inal

cu

rve

for 0

< x

< 1

and

x >

3

A1

2 C

orre

ct g

raph

incl

udin

g cu

sp a

t x =

0

(c)

Tra

nsla

te–1 0

eith

eror

der

Stre

tch

(I)

1sf

(II)

2//

-axi

s (I

II)

y

E1

B1

M1

A1

4

I an

d (e

ither

II o

r III

)

I +

II +

III

(d)

–2

xB

1 5

y

B1

2 Ea

ch v

alue

may

be

stat

ed o

r sho

wn

as

coor

dina

tes

Tot

al11

–2

1 3

3 1

–1

–3

QSo

lutio

nM

arks

Tot

alC

omm

ents

8(a)

LH

S1–

cos

1co

s1

cos

1–co

sM

1C

o mbi

ning

frac

tions

2

21–

cos

A1

Cor

rect

ly si

mpl

ified

22si

n

m1

U

se o

f 2

2si

nco

s1

22c

osec

322

cose

c16

A1

4 A

G; n

o er

rors

seen

OR

1

cos

1co

s32

1co

s1

cos

(M

1)2

232

1co

s (A

1)

22

32si

n (m

1)

2co

sec

16 (A

1)

(b)

cose

c(

)16

yor

bet

ter (

PI b

y fu

rther

w

orki

ng)

M1

or

1si

n(

)16

yor

bet

ter

(y =

) 0.

253,

(2.8

89,)

(3.3

94,)

(6.0

31,)

(–0.

253)

B1

Sigh

t of a

ny o

f the

se c

orre

ct to

3dp

or b

ette

r

(y =

)

0.25

, 2.8

9, 3

.39

(o

r bet

ter)

A

1 M

ust s

ee th

ese

3 an

swer

s, w

ith o

r with

out

eith

er/b

oth

of

0.25

or 6

.03

Igno

re a

nsw

ers o

utsi

de in

terv

al

0.25

to

6.03

but

ext

ras i

n th

is in

terv

al sc

ores

A0

x =

0.43

, 1.7

4, 2

(.00)

, 0.1

7 B

1 B

1 5

3 co

rrec

t (m

ust b

e 2

dp)

All

4 co

rrec

t (m

ust b

e 2

dp) a

nd n

o ex

tras i

n in

terv

al (i

gnor

e an

swer

s out

side

inte

rval

)

Tot

al9

AQA - Core 3 140

QSo

lutio

nM

arks

Tot

alC

omm

ents

9(a)

2

dco

sco

s–

sin

sin

dco

sx

yy

yy

yy

M1

Con

done

inco

rrec

t sig

ns, p

oor n

otat

ion,

omis

sion

of d dx y

or u

sing

d dy x

22

2

cos

sin

cos

yy

yA

1

RH

S co

rrec

t with

term

s squ

ared

, in

clud

ing

corr

ect n

otat

ion

Mus

t see

this

line

221

or1

tan

cos

yy

2d

sec

dxy

y

A1

CSO

3

Mus

t see

one

of t

hese

AG

; al

l cor

rect

incl

udin

g co

rrec

t use

of

d dx yth

roug

hout

(b)

22

sec

1–1

yx

M

1 C

orre

ct u

se o

f 2

2se

c1

tan

yy

and

in

term

s of x

21

–2

1x

x

OE

2

–2

2x

x

A1

2 A

G;

mus

t see

“2

sec

y”,

2

(1)

x

expa

nded

and

no

erro

rs se

en

(c)

22

dd

1–

22

ord

dse

cx

yx

xy

xy

Mus

t be

seen

2

d1

d–

22

y xx

xB

1 1

AG

and

no

erro

rs se

en

QSo

lutio

nM

arks

Tot

alC

omm

ents

9 co

nt

(d)(

i)–1

tan

–1–

lny

xx

2

d1

1d

22

y xx

xx

M1

Mus

t be

corr

ect

d0

dy x 2(

0)x

bxc

m1

Expr

essi

on in

this

form

(gen

erou

s), w

here

b

and

c 0

23

20

xx

A1

Mus

t see

cor

rect

equ

atio

n =

0

1,2

xA

1 4

Bot

h an

swer

s mus

t be

seen

Th

e tw

o A

mar

ks a

re in

depe

nden

t

(ii)

M1

–22

–2–

22

2–

2y

px

xx

qx

whe

re p

and

q a

re c

onst

ants

–2

2–2

–2

22

–2

yx

xx

xA

1 2

p =

–1, q

= 1

incl

udin

g co

rrec

t bra

cket

s

(iii)

1,x

1y

M1

Mus

t hav

e sc

ored

full

mar

ks in

(d)(

i) an

d (ii

)

At

1,0

xy

min

Mus

t see

0

yor

in w

ords

W

hen

x =

1, y

= 0

hen

ce o

n x-

axis

A

1 2

Bot

h st

atem

ents

fully

cor

rect

Tot

al14

TO

TA

L75

AQA - Core 3 141

AQA - Core 3 142

aqa core 3 revision booklet - douis. · pdf filecore 3 specifications candidates will be...

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