january 2006 mathematics general proficiency … maths january 2006.pdfjanuary 2006 mathematics...

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JANUARY 2006 MATHEMATICS GENERAL PROFICIENCY (PAPER 2)

Section I

1. a. (i) Required To Calculate:

Calculation: Numerator

Denominator

Hence,

(ii) Required To Calculate:

Calculation:

b. Data: Amount of loan = $12 000 Rate = 14% per annum (i) Required To Calculate: Interest on the loan at the end of the first year. Calculation:

Interest on loan at the end of 1st year

21

53

54

412

-

´

59

54

4954

412

=

´=

´

101

105621

53

=-

-

10159

21

53

54

412

=-

´

form)exact(in18110

59

=

´=

( )211.275.18 -

( )

figurestsignifican3to3.147992.14

4521.475.1811.275.18 2

==

-=-

1200010014

´=

1680$=

Copyright © 2019. Some Rights Reserved. www.faspassmaths.com Pg 1 of 33

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(ii) Required To Calculate: Total amount owing at the end of 1st year. Calculation: Total amount owing at the end of 1st year = Original amount borrowed + Interest after 1 year. = $12 000 + $1 680 = $13 680

Data: Repayment at start of 2nd year = $7 800

(iii) Required To Calculate: Amount outstanding at the start of 2nd year. Calculation: Amount owed at start of second year = $13 680 - $7 800 = $5 880

(iv) Required To Calculate: Interest on the outstanding amount at the end of

2nd year. Calculation: Interest on the outstanding amount at the end of 2nd year

2. a. Data: and Required To Calculate: Calculation:

b. Data: ,

Required To Calculate: The value of x and of y Calculation:

Let …(1) and …(2) Equation (2) …(3) Equation (1) + (3)

20.823$

588010014

=

´=

2-=m 4=n( )( )nmnm -+ 22

( )( ) ( )( ) ( )( )( )( )

080

444442242222

=-´=

--+-=--+-=-+ nmnm

3765 =+ yx 432 =- yx

3765 =+ yx 432 =- yx2´

864 =- yx

459

)3(864)1(3765

=

=-=+

x

yxyx


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When Substitute in equation (1)

Hence and OR Let …(1) and …(2) From (2)

Substituting in (1)

Substitute in

Hence, and OR

5945

=

=

x

x

5=x

( )

2126

25376376553765

==

-==+=+

yyyyyx

5=x 2=y

3765 =+ yx 432 =- yx

243432

432

+=

+==-

yx

yxyx

( ) ( ) ( )

254277412201537262435

3762435

===++=++

=+÷øö

çèæ +

yyyyyy

yy

2=y243 +

=yx

( )

52102423

=

=

+=x

5=x 2=y


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Obtaining 2 points on the straight line

x y -1 7 11 -3

Obtaining 2 points on the straight line

x y -1 -2 8 4

Plotting both straight lines on the same axes.

Point of intersection is (5, 2), therefore, and . OR and and

…matrix equation

Let

det

3765 =+ yx

432 =- yx

5=x 2=y

3765 =+ yx 432 =- yx

÷÷ø

öççè

æ=÷÷

ø

öççè

æ÷÷ø

öççè

æ- 4

373265

yx

÷÷ø

öççè

æ-

=3265

A

( ) ( )2635 ´--´=A

271215

-=--=


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Matrix equation

Equating corresponding entries and c. Required To Factorise: (i) , (ii) , (iii) Factorising: (i)

(ii)

( )( )

÷÷÷÷

ø

ö

çççç

è

æ

-=

÷÷ø

öççè

æ-

---=\ -

275

272

276

273

5263

2711A

1-´ A

÷÷ø

öççè

æ=

÷÷÷÷

ø

ö

çççç

è

æ

=

÷÷÷÷

ø

ö

çççç

è

æ

÷øö

çèæ ´-+÷

øö

çèæ ´

÷øö

çèæ ´+÷

øö

çèæ ´

=

÷÷ø

öççè

æ

÷÷÷÷

ø

ö

çççç

è

æ

-=÷÷

ø

öççè

æ

÷÷ø

öççè

æ=÷÷

ø

öççè

æ´

÷÷ø

öççè

æ=÷÷

ø

öççè

æ´´

-

--

25275427135

427537

272

427637

273

437

275

272

276

273

437

437

1

11

yx

Ayx

I

Ayx

AA

5=x 2=y

254 2 -x qsqpsp 6496 -+- 443 2 -+ xx

254 2 -x( ) ( )

( )( )5252squares2ofDifference

52 22

+-=

-=

xx

x

qsqpsp 6496 -+-( ) ( )

( )( )qpssqsp

2332322323

+-=-+-=


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(iii)

3. a. Data:

Required To Express: u in terms of v, s and t. Solution:

b. Data: Venn diagram illustrating the customers’ purchases at a bakery. (i) Required To Complete: The Venn diagram to represent the data given.

Solution:

(ii) Required To Find: An expression in terms of x to represent the total number of customers to visit the bakery that day. Solution: Total no. of customers who visited the bakery

443 2 -+ xx( )( )223 +- xx

( )tvus +=21

( )

( )

( )

utvts

utvtsvtutstvus

tvus

tvus

=-

=-+=+=

+=

+=

22

22

21

21

vts

tvtsu -

-=\

2or2

80702 +++= xx1503 += x


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(iii) Required To Calculate: The number or customers who bought bread only. Calculation:

(data)

No. of customers who bought bread only

4. a. Data: Line, l, with equation (i) Required To Find: The gradient of and line parallel to l.

Solution:

is of the form , where is the gradient.

(ii) Required To Find: The equation of the line parallel to l that passes

through (2, -6) Solution:

The gradient of the required line = 4 (Parallel lines have the same gradient). Hence equation of the required line is

3001503 =+x

501503

==\xx

\ x2=( )100502

==

54 += xy

54 += xy cmxy += 4=m


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b. Data: Map showing the position of three cities A, B and C with a scale of

1 : 20 000 000. (i) Required To Find: The length of line segment BC. Solution:

cm (by measurement)

(ii) Required To Calculate: The actual shortest distance from B to C. Calculation: The shortest distance between B and C cm

km

(iii) Required To Find: The bearing of B from A. Solution:

By measurement, the bearing of B from A = 050°

( )

( )

144846246

426

-=-=+-=+

=---

xyxyxy

xy

7.6=BC

000000207.6 ´=

1001000000000207.6

´´

=

km1340=


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5. a. Data: Diagram illustrating a solid glass paper weight consisting of a hemisphere

mounted on a cylinder.

(i) Required To Calculate: The curved surface area of the cylinder

Calculation: The area of the curved surface of the cylinder is

(ii) Required To Calculate: the surface area of the hemisphere.

Calculation:

Surface area of the hemisphere

(iii) Required To Calculate: total surface area of the solid paper weight

Calculation: Total surface area of the paper weight = Area of curved surface of cylinder + Area of curved surface of hemisphere + Area of the circular base.

( )( )( )2

2

cm72.150cm8314.322

=

=rhp

( )2421 rp=

( )( )2

2

cm52.56

314.3421

=

´´=

( )2

2

cm50.235314.352.5672.150

=

++=


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b. Required to Construct: Parallelogram KLMN with KL = 8 cm, KN = 6 cm and

Solution:

6. Data: Diagram illustrating PQRS and its image under a rotation.

a. Required To Find: the coordinates of and Solution:

°= 60NK̂L

SRQP ¢¢¢¢

R¢ S ¢

( )4,2-=¢R


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(from the given diagram) b. Required To Describe: the rotation completely.

Solution: Two object vertices – P and R Two image vertices - and Broken line segments join P to and R to Perpendicular bisectors of these line segments were constructed and found to intersect at the origin (0, 0).

The centre of rotation is (0, 0).

Line segments are drawn from an object vertex and the image vertex to the centre of rotation, example OP and . The angle between both line segments is the angle of rotation.

The angle of rotation is 90°.

( )1,4-=¢S

P¢ R¢P¢ R¢

\

PO ¢

\


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Quadrilateral PQRS is mapped onto under a rotation of 90° anti-

clockwise about O. This may be represented by the matrix .

c. Data: is a reflection of the image in the x – axis.

Required To Draw: Solution:

d. Required To Describe: the transformation that maps PQRS onto

Solution:

The simple transformation that maps PQRS onto is

. This is equivalent to a reflection in the line .

SRQP ¢¢¢¢

÷÷ø

öççè

æ -0110

SRQP ¢¢¢¢¢¢¢¢ SRQP ¢¢¢¢SRQP ¢¢¢¢¢¢¢¢

SRQP ¢¢¢¢¢¢¢¢

SRQPPQRS

SRQPSRQPPQRS

SRQPSRQPPQRS axisxinflectionOaboutclockwiseanti

ofRotation

¢¢¢¢¢¢¢¢¾¾¾¾¾¾¾ ®¾

¢¢¢¢¢¢¢¢¾¾¾ ®¾¢¢¢¢¾¾¾ ®¾

¢¢¢¢¢¢¢¢¾¾¾¾¾ ®¾¢¢¢¢¾¾¾¾ ®¾

÷÷ø

öççè

æ-

-=÷÷ø

öççè

æ -÷÷ø

öççè

æ-

÷÷ø

öççè

æ-÷÷

ø

öççè

æ -

--

°

0110

0110

1001

1001

0110

Re

90

\ SRQP ¢¢¢¢¢¢¢¢

÷÷ø

öççè

æ-

-0110

xy -=


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7. Data: The lengths of the right foot of 25 students. a. Required to complete: the grouped frequency table for the given data. Solution:

Modifying and completing the table of values for the continuous variable. Length of right foot,

x cm L.C.B U.C.B. Frequency Mid-class Interval

14 – 16 4

17 – 19 6

20 – 22 8

23 – 25 5

26 – 28 2

Frequency for class interval 17 – 19

b. Required To Find: The lower class boundary of the class interval 14 – 16 Solution: The lower class boundary of the class interval 14 – 16 is 13.5 (as shown in the above table).

c. Required To Find: The width of class interval 20 – 22.

Solution: The width of class interval 20 – 22 is

d. Required To Find: The class interval in which a measurement of 16.8 cm would

lie. Solution: A measurement of 16.8 cm lies in the class interval 17 – 19.

e. Required To Calculate: The probability that a randomly chosen student has a right foot measuring greater than or equal to 20 cm.

2U.C.B.L.C.B +

5.165.13 <£ x15

25.165.13=

+

5.195.16 <£ x18

25.195.16=

+

5.225.19 <£ x21

25.225.19=

+

5.255.22 <£ x24

25.255.22=

+

5.285.25 <£ x27

25.285.25=

+

25=å f( )258425 +++-=

6=

35.195.22L.C.B.U.C.B

=-=-


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Calculation:

f. Required to find: The modal length of a student’s right foot.

Solution: Modal class is 20 – 22 since this class corresponds to the greatest frequency.

Modal length of student’s foot

g. Required To Estimate: the mean length of a student’s foot using the mid-class

intervals for the data given. Solution: Mean length of student’s foot where

f = frequency x = mid-class interval of class

8. Data: Path of a ball follows the equation , h = height in m and t = time in seconds.

a. Required To Complete: the table of values of t and corresponding values of h. Solution:

t 0 0.5 1 1.5 2 2.5 3 3.5 4 h 0.0 8.8 15 18.8 (20) 18.8 (15) 8.8 0

When When

( )

532515

classtheinstudentsofnumberTotalcm20footwithstudentsofNo.cm20footsstudent'ofLength

=

=

³=³P

\2L.C.B.U.C.B +

=

cm212

5.225.19

=

+=

x=

åå=ffx

x

( ) ( ) ( ) ( ) ( )

cm4.2025

272245218186154

=

´+´+´+´+´=x

2520 tth -=

2=t ( ) ( )225220 -=h20=

3=t ( ) ( )235320 -=h15=


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b. Data: Scale of 2 cm to represent 0.5 seconds on t – axis for . Scale of 1 cm to represent 1 metre on the h – axis for .

Required To Draw: the graph of for Solution:

40 ££ t0.210.0 ££ h

2520 tth -= 40 ££ t


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c.

(i) Required To Find: the greatest height above the ground reached by the

ball. Solution: The greatest height reached by the ball is 20 m.

(ii) Required To Find: the duration for which the ball was more than 12 m

above the ground. Solution:

m from the ground to

Time

That is, m for 2.52 seconds.

12>h74.0=t 26.3=t

74.026.3 -=seconds52.2=12>h


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(iii) Required To Find: the time interval during which the ball was moving upwards. Solution: The ball is moving upwards from to , that is for 2 – 0 = 2 seconds.

Section II 9. a. Data: and

Required To Calculate: x and y Calculation: Let …(1) and …(2) From (1)

Substitute in (2)

When

When

Hence, and or and

b. Required To Express: in the form

Solution:

Half the coefficient of x is

0=t 2=t

72 =+ yx 62 =- xyx

72 =+ yx 62 =- xyx

xyyx

2772-=

=+

( )

( )( ) 0323067306270627

2

22

2

=-+=--

=-+-

=---

xxxxxxxxxx

3or32

-=x

32

-=x ÷øö

çèæ--=3227y

318

347

=

+=

3=x ( )327 -=y1=

32

-=x318=y 3=x 1=y

3124 2 -- xx ( ) khxa ++ 2

( ) 3343124 22 --=-- xxxx

( )233

21

-=-


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is of the form where

OR

Equating coefficient of Equating coefficient of x

Equating constants

12?3

1291244934

234

?2343124

2

22

22

-=-

-+-

÷øö

çèæ +-=÷

øö

çèæ -

+÷øö

çèæ -=--

and

xxand

xxx

xxx

122343124

22 -÷

øö

çèæ -º--\ xxx ( ) khxa ++ 2

ÂÎ-=

ÂÎ-=

ÂÎ=

12234

k

h

a

( )( )

kahahxaxkhhxxa

khxax

+++=

+++=

++=--

22

22

22

22

3124

2xÂÎ= 4a

( )

ÂÎ-=

-=

=-

238124212

h

h

122343124

1293

2343

22

2

-÷øö

çèæ -º--\

ÂÎ-=+=-

+÷øö

çèæ-=-

xxx

kk

k


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c. (i) Required To Find: the minimum value of Solution:

Minimum value of

(ii) Required To Find: the value of x at which the minimum point occurs. Solution: Minimum value occurs at

OR Let . The axis of symmetry passes through the minimum point and occurs at

Therefore,

Therefore,

3124 2 -- xx

xx

xxx

"³÷øö

çèæ -

-÷øö

çèæ -º--

0234

122343124

2

22

\ 3124 2 -- xx 120 -=12-=

23

023

023

0234

2

2

=

=-

=÷øö

çèæ -

=÷øö

çèæ -

x

x

x

x

3124 2 --= xxy

( )( )

238124212

=

=

--=x

12

32312

234

2

min

-=

-÷øö

çèæ-÷

øö

çèæ=y

23at12min =-= xy


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(iii) Required To Solve:

Solution:

to 3 significant figures. OR

to 3 significant figures

10. a. Data: No. of Sonix radios = x No. of Zent radios = y Space on shelf for up to 20 radios. (i) Required To Find: The inequality to represent the data given.

Solution: Therefore, total must be less than or equal to 20. Hence, …(1)

03124 2 =-- xx

( ) ( ) ( )( )( )

0232.0or223.38192128

4814412

423441212

031242

2

-=

±=

+±=

---±--=

=--

x

xx

232.0or23.3 -=

0232.0or223.3732.15.1

323

323

323

412

23

012234

03124

2

2

2

2

-=±=

±=

±=-

=÷øö

çèæ -

=÷øö

çèæ -

=-÷øö

çèæ -

=--

x

x

x

x

x

xx

232.0or23.3 -=

20£+ yx


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(ii) Data: Cost of Sonix radio = $150 Cost of Zent radio = $ 300 Shop’s owner has $4500 Required To Find: The inequality to represent the data given Solution: Cost of x Sonix radios at $150 each and y Zent radios at $300 each Total available to spent is $4500.

(iii) Data: Owner decides to stock at least 6 Sonix and at least 6 Zent radios. Required To Find: The two inequalities to represent the data given. Solution: Stock is at least 6 Sonix and at least 6 Zent radios.

b. Data: Scale of 2 cm to represent 5 Sonix radios and 2 cm to represent 5 Zent radios.

Horizontal axis - Vertical axis - Required To: Draw the boundary lines for all four above inequalities, shade the region that satisfies all four inequalities and state the vertices of the shaded region. Solution: The line is a straight vertical line. The region is represented as

The line is a straight horizontal line. The region is represented as

( ) ( )300150 yx +=

302150

4500300150

£+÷

£+\

yx

yx

6and6 ³³\ yx

300 ££ x250 ££ y

6=x6³x

6=y6³y


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Obtaining two points on the line . When The line passes through the point (0, 20).

When The line passes through the point (20, 0).

The region with the smaller angle satisfies the region. The region is

Obtaining two points on the line When

The line passes through the point (0, 15).

20=+ yx0=x 200 =+ y

20=y20=+ yx

0=y 200 =+x20=x

20=+ yx

£20£+ yx

302 =+ yx0=x 3020 =+ y

15230302

=

=

=

y

y

302 =+ yx


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When The line passes through the point (30, 0).

The region with the smaller angle satisfies the region. The region is

The region which satisfies all 4 inequalities is the area where all shaded regions overlap.

0=y ( ) 3002 =+x30=x

302 =+ yx

£302 £+ yx


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(iv) The vertices of the shaded region are (6, 6), (14, 6), (10, 10) and (6, 12).


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c. Data: The owner of the shop sells to make a profit of $80 on each Sonix radio and $100 on each Zent radio. (i) Required To Express: The total profit in terms of x and y Solution: Profit, P on x Sonix radios at $80 each and y Zent radios at $100 each is

(ii) Required To Calculate: The maximum profit Calculation: Test Test

Test

Maximum profit = $1800 when the owner sells 10 Sonix radios and 10

Zent radios.

11. Data: Circle, centre O with and . and are chords. a. (i) Required To Calculate: Solution:

(Angle subtended by chord (AB) at centre of circle is twice the angle that the chord subtends at the circumference, standing on the same arc.)

( ) ( )( )yx

yxP10080$10080

+=´+´=

126 == yx 1010 == yx( ) ( )1680$

12100680=

+=P ( ) ( )1800$

101001080=

´+´=PP

614 == xx( ) ( )1720$

61001480=

´+´=P

\

°= 130ˆBOA °= 30ˆCAD AEC BEDACBÐ

( )

°=

°=

65

13021ˆBCA


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(ii) Required To Find: Solution:

(Chord CD subtends equal angles at the circumference, standing on the same arc).

(iii) Required To Find: Solution:

(Sum of angles in a triangle = 180°). (Vertically opposite angles).

b. Required To Show: is similar to Solution:

(Sum of angles in a triangle = 180°).

are equi-angular OR similar.

CBDÐ

°= 30ˆDBC

AEDÐ

( )°=

°+°-°=85

3560180ˆBEC

°= 85ˆDEA

BCED ADED

°= 65ˆADE

DC

EAngleAB

ˆˆcommonis

ˆˆ

=

=

ADEBCE DD\ and


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c. Data: CE = 6 cm, EA = 9.1 cm and DE = 5 cm (i) Required To Calculate: length of EB.

Calculation: If , then the ratio of their corresponding sides are the same. That is,

(ii) Required To Calculate: the area of

Calculation:

Area of

12. This question is not done since it involves latitude and longitude (Earth Geometry) which has been removed from the syllabus.

13. Data: A (1, 2), B (5, 2), C (6, 4) and D (2, 4) are the vertices of a quadrilateral ABCD.

a. (i)Required To Express: The position vectors of in the

form .

Solution:

If then is of the form where and .

ADEBCE DD ~

cm92.1056.5451.96

1.956

=

=

´=

=

==

EB

EBAEBE

DECE

ADBC

AEDD

( )( ) °=D 85sin1.9521AED

placedecimal1tocm7.22cm66.222

2

=

=

ODOCOBOA and,,

÷÷ø

öççè

æyx

( )2,1=A ÷÷ø

öççè

æ=21

OA ÷÷ø

öççè

æyx

1=x 2=y


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(ii) Required To Find: and Solution:

b. Required To Find: and the unit vector in the direction of

Solution:

( )2,5=B ÷÷ø

öççè

æ=25

OB ÷÷ø

öççè

æyx

5=x 2=y

( )4,6=C ÷÷ø

öççè

æ=46

OC ÷÷ø

öççè

æyx

6=x 4=y

( )4,2=D ÷÷ø

öççè

æ=42

OD ÷÷ø

öççè

æyx

2=x 4=y

AB DC

÷÷ø

öççè

æ=

÷÷ø

öççè

æ+÷÷ø

öççè

æ-=

+=

04

25

21OBAOAB

÷÷ø

öççè

æ=

÷÷ø

öççè

æ+÷÷ø

öççè

æ-=

+=

04

46

42OCDODC

AB AB


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Any vector in the direction of , where is a scalar.

If the vector is unit then the magnitude is 1 unit.

The unit vector in the direction of

.

c. (i)Required To Find: Two geometrical relationships between the line segments AB and DC. Solution:

Hence, and is parallel to .

(ii) Required To Explain: Why ABCD is a parallelogram. Solution:

If one pair of opposite sides of a quadrilateral are both parallel and equal

then the quadrilateral is a parallelogram. Hence, ABCD is a parallelogram.

( ) ( )units4

04 22

=

+=AB

÷÷ø

öççè

æ=

04

aAB a

( ) ( )

41104

104

22

=

=+

=\

a

a

a

÷÷ø

öççè

æ=

04

41AB

÷÷ø

öççè

æ=01

÷÷ø

öççè

æ=

÷÷ø

öççè

æ=

÷÷ø

öççè

æ=

04

1

04

04

DC

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c. Required To Find: The position vector of G, the midpoint of line AC and the coordinates of the point of intersection of the diagonals AC and BD. Solution:

Position vector of G is

The diagonals of a parallelogram bisect each other. AC and BD intersect at G.

Intersection of AC and BD is G = .

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14. a. Data:

(i) Required To Calculate: The determinant of L. Calculation: Det (ii) Required To Calculate: The values of x given that L is singular. Calculation: If L is singular, then det L = 0 When

b. Data:

(i) Required To Calculate: Calculation: det

(ii) Data:

Required To Calculate: The value of x and of y Calculation:

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Equating corresponding entries. and

c. Data:

(i) Required To Calculate: , the image of (3, -1) under N. Calculation:

When

The image of (3, -1) is (11, -4).

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(ii) Required To Calculate: when is (7, 4) Calculation:

Equation corresponding entries

and

The point which is mapped onto (7, 4) is (1, 3).

( )yx, ( )yx ¢¢,

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january 2006 mathematics general proficiency … maths january 2006.pdfjanuary 2006 mathematics...

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