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SERANGOON JUNIOR COLLEGE

2011 JC2 MID YEAR EXAMINATION

MATHEMATICS

Higher 2 9740/1

Wednesday 18 May 2011

Additional materials: Writing paper

List of Formulae (MF15)

TIME : 3 hours

READ THESE INSTRUCTIONS FIRST Write your name and class on the cover page and on all the work you hand in. Write in dark or black pen on both sides of the paper. You may use a soft pencil for any diagrams or graphs. Do not use staples, paper clips, highlighters, glue or correction fluid. Answer all the questions. Give non-exact numerical answers correct to 3 significant figures, or 1 decimal place in the case of angles in degrees, unless a different level of accuracy is specified in the question. You are expected to use a graphic calculator. Unsupported answers from a graphic calculator are allowed unless a question specifically states otherwise. Where unsupported answers from a graphic calculator are not allowed in a question, you are required to present the mathematical steps using mathematical notations and not calculator commands. You are reminded of the need for clear presentation in your answers. The number of marks is given in brackets [ ] at the end of each question or part question. At the end of the examination, fasten all your work securely together.

Total marks for this paper is 100 marks.

This question paper consists of 6 printed pages and no blank pages.

Answer all questions [100 marks].

1. Without the use of the calculator, solve the inequality , 5

3x ≠ − .

[3] Hence, solve . [2] Solution:

M1

– + – +

4

M1

or A1

M1

From earlier part, we have

(rejected 0x ≥ ) or

( 0x ≥ )

A1

2. In a certain culture of bacteria, the rate of increase of the population is proportional to the number, x of bacteria present at any time t hours. (i) Write down a differential equation connecting x and t. [1] (ii) The initial number of bacteria is 0x and the number of bacteria doubles after 4 hours. Find , in terms of 0x , the number of bacteria at the end of

12 hours. [5]

Comment [YUN1]: Misconception Some students performed cross multiplication which is incorrect as one will not know whether the expression 3x−5 is always positive or negative.

Comment [YUN2]: Careless Mistake Some students forgot to realise that it should not have the “=” sign as we can’t have division of zero.

Comment [YUN3]: Misconception Some students either missed out this word or would have used the incorrect word “AND” (“AND” cannot be used as we can’t have a number that satisfy both inequalities at the same time.

Comment [YUN4]: Presentation of Answers Some students did not state the reason for rejecting the answer.

Comment [YUN5]: Presentation of Answers Some students did not indicate why the above inequality is reduced to this form.

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Solution: Let x be the number of bacteria present at t hours. d(i)

dx kxt=

B1 1(ii) d dx k t

x=∫ ∫

M1 ln x kt C= + kt Cx e += C ktx e e=± ktx Ae= A1 When 0 00,t x x A x= = ⇒ = B1

0ktx x e=

When t = 4, 4 40 02 2k kx x e e= ⇒ =

ln 24

k = B1

When t = 12, ( )3

ln 212 ln 212 3ln 240 0 0 0 08kx x e x e x e x e x

= = = = = B1

3. The diagram below shows the graph of f ( )y x= . There is a minimum point at ( 3,0)− and the equations of the asymptotes are 0x= and 4y = . Sketch, on separate clearly labelled diagrams, the graphs of (i) f ( )y x=− [2]

(ii) ( )1

fy

x=

[2] (iii) ( )f 'y x= [2] (iv) f (1 2 )y x= − [2]

x

y

o

4

( )0,3−

f ( )y x=

Comment [YUN6]: Presentation of working:

Many failed to include ce± after the elimination of

modulus sign

Comment [YUN7]: Accuracy:

Many failed to realised that 3ln 2

0 08e x x=

Solution:

(i) f ( )y x=−

Asymptote – L1 (x = 0 and y = -2) Correct Graph – G1

(ii)

( )1

fy

x=

Asymptote – L1 Correct Graph – G1

(iii) ( )f 'y x=

x

y

O

14

-3

x

y

O

-2

( )0,3−

Comment [YUN8]: Presentation: Many failed to label the fixed point and the equations of the asymptotes

Comment [YUN9]: Presentation: Many failed to label the fixed point and the equations of the asymptotes

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One part correct – P1 Correct Graph – G1

(iv) f (1 2 )y x= −

Correct Asymptote – L1 Correct Point (2,0) – P1 Correct Graph – G1

x

y

O

4

( )0,2 12

x

y

O -3

Comment [YUN10]: Theory: Most of the students have no basic concept on sketching the graph of y=f’(x) Many students never attempt this part

Comment [YUN11]: Misconception: Many have forgotten the basic theory on transformation and they have completely messed up the order of transformation.

4. The diagram below shows the graph of 5 2xy x= − . The two roots of the

equation 5 2 0xx − = are denoted by α and β , where α β< .

(i) Find the values of α and β , each correct to 3 decimal places. [2]

A sequence of real numbers 1 2 3, , ,...x x x satisfies the recurrence relation

( )11 25n

nxx + = for 1n ≥ .

(ii) Prove algebraically that, if the sequence converges, then it converges to either α

or β . [2]

(iii) Use a graphic calculator to determine the behaviour of the sequence for each of the cases 1 4x = and 1 5x = . [2]

(iv) By considering 1n nx x+ − , show that

1n nx x+ < if nxα β< <

1n nx x+ > if or > n nx xα β< [2] Solution (i) From the G.C, 0.235 and 4.488α β= = B2 (ii) Suppose the sequence converges to L. As n∴ →∞ 1 and n nx L x L+⇒ → → M1

So ( )1 25

LL =

5 2 0LL − = From (i), 0.235 or 4.488L L= = A1 Hence the sequence converges to either α

or β .

(iii) Using a G.C, When 1 4x = , the sequence is decreasing and converges to 0.235 B1

x

y

α β

5 2xy x= −

Comment [YUN12]: G.C Skills Some students are not familiar how to use the graphic calculator to find the axial intercept. It is incorrect to just move the cursor to some region near the x-intercept to obtain the answer.

Comment [YUN13]: Accuracy Many failed to follow the instructions to leave the answer in 3 decimal places.

Comment [YUN14]: Presentation of Answers Many students did not indicate what is the meaning of L when they introduce the symbol in the working.

Comment [YUN15]: Presentation of Answers Some students failed to point out that both

1 and n nx x+tends to the same value (NOT EQUALS

TO L) as n tends to infinity.

Comment [YUN16]: Presentation of Answers It is important to indicate how the value of L is obtained using the above results.

Comment [YUN17]: G.C Skills Some scripts shows lack of understanding how to use the sequence mode and the table option to obtain the answer. Presentation of Answers Some words in which students can used to describe the behaviour of the sequence are (i) Increasing (ii) Decreasing (iii) Constant (iv) Converging (v) Diverging (vi) Alternating

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When 1 5x = , the sequence is increasing and diverges B1

(iv) ( )11 25n n n

nxx x x+ − = −

( )1 5 25 n

nxx= − −

For xα β< < , from the given sketch of 5 2xy x= − , y > 0

5 2 0 for n nnxx xα β⇒ − > < < B1

( )1 5 2 05 n

nxx∴− − <

Hence 1 0n nx x+ − < 1n nx x+⇒ < For or x xα β< > , from the given sketch of 5 2xy x= − , y < 0

5 2 0 for or n n nnxx x xα β⇒ − < < > B1

( )1 5 2 05 n

nxx∴− − >

Hence 1 0n nx x+ − > 1n nx x+⇒ >

5.

The diagram shows the region R bounded by the curve 2 6x y y= − , the x-axis and the line 9x=− . (a) Find, without the use of Graphic Calculator, the exact area of R. [3] (b) Find the volume of the solid formed when R is rotated completely about (i) the x-axis, [2] (ii) the y-axis, [3] giving your answer correct to 1 decimal place.

x

y

O

9x=−

2 6x y y= −

R

Comment [YUN18]: Presentation of Answers Students should not use the phrase does not converge to describe the behaviour as does not converge can mean alternating or mean it diverges.

Comment [YUN19]: Careless Mistakes

( )1 2 2 55 n n

n nx xx x− ≠ −

Comment [YUN20]: Presentation of Answers Students needs to be very clear in the working especially when it is a “proving” question. So in this case, students need to show clearly the link between the given graph and how it is related to the inequality that was found.

Solution: (i) 2 6x y y= − ( )23 9x y= − − ( )23 9x y= − − 3 9y x= ± + When x = − 9, y = 3. Since 3y≤ , 3 9y x∴ = − + B1 Area of R = ( )

102

93 9 dx x

−

− + ∫

M1

= ( )03

2

9

23 93

x x−

− +

= 9 A1 Or Area of R = ( )3 2

09 3 6 dy y y× − −∫ M1

A1

=33

2

0

9 3 33y y

× − −

= 9 A1 (ii) Required volume = ( )20

93 9 dx xπ

−− +∫ M1

= 42.4 A1 (iii) Required volume = ( ) ( ) ( )

232 2

09 3 6 dy y yπ π− −∫

M1 A1

= 356.3 A1

6. (a) The length of each side of a cube is increasing at a constant rate of10.1ms− . Find the rate of increase of the volume of the cube when the length

of each side is 2 m. [2]

(b) [3]

y y z

x

z

z z

24 cm

9 cm

Comment [YUN21]: Accuracy: Many failed to indicate the correct equation of the required graph

Comment [YUN22]: Accuracy: Many failed to realised that

( )3 2

06 d 0y y y− <∫

and hence forgot to add in the modulus sign

Comment [YUN23]: Accuracy: Many mixed up the formulae used for rotation about the x-axis and the y-axis. Wrong choice for the equation of the curve. GC skills: Many students did not realise that they can use GC to find the answer without performing any integration.

Comment [YUN24]: Accuracy: Many mixed up the formulae used for rotation about the x-axis and the y-axis. GC skills: Many students did not realise that they can use GC to find the answer without performing any integration.

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The diagram above shows a 24 cm by 9 cm aluminium sheet that can be folded into a box with length x cm, width y cm and height z cm. (i) Show that the volume of the box, 3 22 39 180= − + −V y y y .

(ii) Hence, find the maximum volume of the box. [4] Solution

(a)

Let x = length of a side of the cube

So volume, 3xV = ⇒ txx

tV

dd3

dd 2=

When 2=x & 1.0dd

=tx

,

( ) ( ) 132 sm2.11.023dd −==

tV

M1

A1

(b) (i) 2 2 24+ =y z ,------(1) 2 9+ =x z ------(2) Solving the above simultaneous equation, 12= −z y and 2 15= −x y (2 15)( )(12 )= = − −V xyz y y y

3 22 39 180= − + −y y y

M1

M1

A1

(ii)

d 0dVy=

26 78 180 0− + − =y y

2 13 30 0− + =y y

( )( )3 10 0− − =y y 3 or 10y y= =

2

2

d 12 78d

V yy

= − +

When 3=y , 2

2

d 42 0d

Vy

= > , When 10=y , 2

2

d 42 0d

Vy

= − <

Hence, when 10=y , maximum volume, V = (5)(10)(2)=100 cm3

M1

M1

M1

A1

7.

(a) Prove by mathematical induction, that for all

positive integers, n.

[4] Hence, or otherwise, find the sum to infinity of the series, [2]

Comment [YUN25]: Improper Algebraic Manipulation (1) Failed to differentiate volume correctly. (2) Failed to use change rule dv/dt = dv/dx X dx/dt

Comment [YUN26]: Improper Algebraic Manipulation

(1) Failed to derive the two equations from the diagram given. (2) Failed to multiply correctly to show V as requested in the question.

Comment [YUN27]: Improper Algebraic Manipulation

(1) Failed to differentiate V with respect to y.

(2) Failed to solve values for y.

Comment [YUN28]: Comprehension of question

(1) Failed to check for max value of V using second derivative test or first derivative test.

(2)Failed to find max V.

.

(b) Show that =

( ) ( )3 5 2

1 ! ! 1 !r r r− +

− +.

Hence find . [4]

(a) Solution:

Let Pn be the statement “ for all n +∈

When , LHS =

( )2

14 1 1

=−

and RHS = ( )

12 1 1

=+

.

Thus P1 is true.

Assume, when , Pk is true for some k +∈ , i.e.

M1

Need to show Pk+1 is true , i.e. .

( )

1

2 2 21 1

1 1 1L.H.S4 1 4 1 4 1 1

k k

r rr r k

+

= =

= = +− − + −∑ ∑

M1

M1

Pk is true Pk+1 is true Hence, since P1 is true, and Pk is true Pk+1 is true, by Mathematical

Induction, Pn is true for all positive integers n. M1

Comment [YUN29]: Presentation of Answers Some students wrote the R.H.S as

2 1r

r +.

Comment [YUN30]: Presentation of Answers Some students failed to indicate the condition for n. It is also incorrect to write the condition as

n +∈ because n cannot take non –integer values.

Comment [YUN31]: Presentation of Answers Some students did not explicitly write down how the value of 1/3 is obtained.

Comment [YUN32]: Presentation of Answers Some students did not emphasize the word “SOME”. It is incorrect to use the phrase “FOR ALL”

Comment [YUN33]: Misconception and Presentation of Answers

Incorrect to write it as 1 1k kP P P+ = +

(because Pk and Pk+1 are symbolic representation of

a statement) or 1 1k kS S S+ = + (because sum

of first k+1 terms is not the same as the first term add to the sum of the first n terms.

Comment [YUN34]: Presentation of Answers Some students simplify the expression in an incorrect manner. Advise to combine by identifying the LCM of the denominator. Make sure working needs to be clear as it is a “PROVING” type of question.

Comment [YUN35]: Presentation of Answers This statement is often missing in most scripts. It is important to make a conclusion based on what was performed above.

Comment [YUN36]: Presentation of Answers Some students are not able to write it in the correct manner. It is incorrect to have “since Pk+1 is true => Pk is true”.

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M1

Thus A1

(b) L.H.S gives

( ) ( )( ) ( )

( )

( )

( )

2

2

3 1 5 1 23 5 21 ! ! 1 ! 1 !

3 3 5 5 21 !

3 2 31 !

R.H.S

r r rr r r r

r r rr

r rr

+ − + +− + =

− + +

+ − − +=

+

− −=

+

=

B1

( ) ( ) ( )2

1 1

3 2 3 3 5 21 ! 1 ! ! 1 !

n n

r r

r rr r r r

= =

− −= − + + − + ∑ ∑

M1

+…

M1

A1

Comment [YUN37]: Presentation of Answers Some students failed to provide reasons why the final answer is ½.

Comment [YUN38]: Presentation of Answers and Misconception Many students shows a lack of understanding on factorial and thus are not able to simplify the expression. Many students also performed partial fraction which is incorrect.

Comment [YUN39]: Presentation of Answers Some students failed to write down the summation sign.

Comment [YUN40]: Presentation of Answers For MOD on 3 terms, students need to show at least first 3 to 4 rows and the last 3 rows to identify clearly the cancellation.

8. The variable complex numbers w and z are such that 3 i 2w− − = and ( )arg 5 4iz .π− − =

Sketch, on a single Argand diagram, the loci representing the complex numbers w and z. [3]

(i) Find the minimum value of w z− ; [1] (ii) Show that the maximum value of ( )arg 2w+ is 0.600.

[3]

(iii) Hence, find the coordinates of the point P, which gives the maximum ( )arg 2w+ in part (ii). [3]

Solution

( )arg 5 4z i .π− − =

G1

G2

(i) minimum value of w z− = 1 B1

(ii) Dist bet (−2,0) and (3,1) = ( ) ( )2 22 3 0 1 26− − + − =

1 2sin26

α −= (or 0.40306) ,

1 1tan5

β −= (or 0.19740).

Maximum ( )arg 2w+ = α + β = 0.600 (to 3 s.f.)

M1

M1

B1

(iii) Dist bet (−2,0) and point P = ( ) ( )2 226 2 22− =

y-coordinate of P = 22 sin(0.600) =2.65

M1 M1

26

Im

–2

Re 0 5 0

(3,1)

1

3 2w i− − =

x

2

P

(5,4)

Ans (i) Comment [YUN41]: Accuracy (1) Failed to indicate exclusion of the

point (5,4) (2) Failed to label clearly the position of the point (5,4) (3) Indicating locus with a dotted line instead of a full line.

Comment [YUN42]: Accuracy (1) Failed to label clearly the centre and/or radius of the circle (2) Failed to adhere accurately to the scales defined for both axes

Comment [YUN43]: Misconception Wrongly conclude that

(1) w z w z− = −

(2) min 13 2w z− = −

Comment [YUN44]: Comprehension of question: Did not make use of the angle 0.600.

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x-coordinate of P = 22 cos(0.600) 2 =1.87− Coordinates of point P is (1.87, 2.65)

A1

9. The line whose vector equation is r = 5i + 2j − 3k +α (7i − 2j − 3k) passes

through the point A with position vector 5i + 2j − 3k. The plane π whose vector equation is r = −5i +2j + 2k + β ( 7i − 2j − 3k) + γ ( 9i − j + k ) contains the point B with position vector −5i +2j + 2k.

(i) Find the position vector of the point P on the line segment AB such that AP:PB = 4:1. [2]

(ii) The plane 1π contains the line and the point P.

Write down a vector equation of the line of intersection of π and 1π . [2] (iii) Find the vector equation of 1π . [2] (iv) Calculate the angle between π and 1π [4] (v) Hence find the ratio of the perpendicular distances from P to the line and

from P to the plane π . [2] Solution

(i) By ratio theorem,

5 51 4 2 25

2 3OP

− = +

−

M1

3

21

− =

A1 (ii) 1π contains AP, hence it contains B.

Both π and 1π // to vector 723

− −

.

M1 Hence the line of intersection of π and 1π is

5 7

2 2 , 2 3

λ λ− = + − ∈ −

r

A1

(iii) 5 5 2

2 2 5 02 3 1

AB− − = − = −

M1

∴Eq. of 1π is 5 7 22 2 0 , ,3 3 1

s t s t−

= + − + ∈ − −

r

A1

A P B

4 1

Comment [YUN45]: Accuracy Answer not given to 3 significant figures

Comment [YUN46]: Incorrect application of formula Some students calculated

5 51 2 4 25

2 3OP

− = + −

instead .

Comment [YUN47]: Comprehension of question Some students did not understand the vector equation and provided the scalar form of the equation instead.

(iv) normal of π is 723

= − −

n x9 51 34

1 11

− − = −

B1

normal of 1π is 723

= − −

m x2 2

0 11 4

− − = − −

M1

cos

5 234 1

11 4

1302 21θ

− − − − − =

= 10 34 44 01302 21+ −

= M1

∴ between π and 1π is 90o. A1 (v) Let M be the foot of perpendicular of P to the line l and N be the foot of

perpendicular from P to the plane π . MP : PN = AP:PB M1 = 4:1 A1 10. (a) A convergent geometric progression has first term a and common ratio r,

where a and r are non-zero. The first three terms of this geometric progression are the fifteenth, seventh and the third term of an arithmetic progression.

(i) Show that 22 3 1 0r r− + = . [3] (ii) Find the least value of n for which the sum of the first n terms of the

arithmetic progression exceeds the sum to infinity of the geometric progression. [4]

(b) Jack took a housing loan of $250 000 in January 2011 from the bank to buy a 4-room flat in Punggol. The bank charges an annual interest rate of 5% on the outstanding loan at the end of each year. Jack pays $1500 at the beginning of each month until he finishes paying his loan. If Jack started paying back the loan in January 2011, show that the amount owe

by Jack at the end of nth year is ( )378000 128000 1.05 n− .

Find the minimum number of years he needs to clear his debts. [3] [2]

Solution (ai) Let b and d be the first term and common difference of the A.P

2

14 (1)6 (2)2 (3)

a b dar b dar b d

= += +

= +

M1 (1) (2) gives−

Comment [YUN48]: Comprehension of Question Some students misinterpreted as a =15, ar = 7 and ar2 = 3.

Comment [YUN49]: Presentation of Answers Many students failed to define any symbols that were introduced in the working.

Comment [YUN50]: Presentation of Answers Many students has the underlying assumption that the first terms of both A.P and G.P are the same, which is commonly written as a.

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8 (4)a ar d− = (2) (3) gives− M1 2 4 (5)ar ar d− = (4) and (5) gives ( )22 ar ar a ar− = − A1 22 3 1 0r r∴ − + = (ii) 22 3 1 0r r− + =

( 1)(2 1) 0r r− − = 1 or 1 (rejected convergent)

2r∴ =

A1 (2) (1) gives÷ 6 1

14 2b drb d+

= =+

2b d= and so subst into (1) gives 16 8a d b= = M1 For ( ) ( ). .n A P G P

S S∞>

( )2 1 12 2 12

n b ab n + − > −

M1 ( )1

164

bn nbn b

−+ >

2 3 64 0n n+ − > Using G.C 9.64 (rejected) or 6.64n n< − > Hence the least number of n is 7. A1 (b)

No. of years End of the year 1 ($250 000 − $18 00 )1.05 2 [($250 000 − $18 000)1.05 – $18 000]1.05

= ($250 000) 1.052 – ($18 000)1.052 – ($18 000)1.05 3 [($250 000) 1.052 – ($18 000)1.052 – ($18 000)1.05 –

($18000)](1.05) = ($250 000) 1.053 – ($18 000)1.053 – ($18 000)1.052 – ($18000)1.05

… n ($250 000) 1.05n – ($18 000)1.05n – ($18 000)1.05n−1 – …

– ($18000)1.05

At the end of n years, Amount owe =

( ) ( ) ( ) ( )250000 1.05 18000 1.05 18000 1.05 ... 18000 1.05n n n− − − − M1

Comment [YUN51]: Careless Mistakes Some wrote it as ar – a = 8d

Comment [YUN52]: Presentation of Answers Many failed to reject the answer 1 with the correct reason.

Comment [YUN53]: Presentation of Answers Many students has the underlying assumption that the first terms of both A.P and G.P are the same, which is commonly written as a.

Comment [YUN54]: Presentation of Answers Some students failed to present their solution neatly and in a very orderly manner.

Comment [YUN55]: Careless Mistake and Presentation of Answers Some students failed to write down the pattern for the nth year and even if they have done so, the values indicated is incorrect especially the last value.

( ) ( ) ( ) ( )250000 1.05 18000 1.05 18000 1.05 ... 18000 1.05n n n = − + + +

( )

( )1.05 1.05 1250000 1.05 18000

1.05 1

n

n − = − −

M1 ( ) ( )250000 1.05 378000 1.05 1n n= − − A1 ( )378000 128000 1.05 n= − To clear the debt, ( )378000 128000 1.05 0n− ≤ M1 ( )189 1.05

64n≤

( )

189ln64

ln 1.05n

≥

22.2n ≥ A1 Hence the least number of years is 23. 11.

The curve C has equation 2 2 1, .x axy a

a x− +

= ∈−

(i) Find the equations of the asymptotes of C. [3] (ii) Find the range of values of a such that C has no stationary points. [3] (iii) For a = 2, sketch C. Show clearly on your diagram the equations of the

asymptotes and the coordinates of the points of intersection of C with the axes and stationary points. [3]

(iv) By sketching a suitable graph, deduce the number of roots of the equation

22 4 1 1

2x x x

x − +

= − − .

[3] Solution

(i) 2 22 1 1x ax ay x a

a x a x− + −

= = − + +− −

M1

Oblique asymptote: y x a= − + . A1 Vertical asymptote : x a= A1 (ii)

( )( )

2

2

2

2 2

1

d 11 0d

1

ay x aa x

y ax a x

a x a

−= − + +

−−

= − + =−

⇒ − = −

M1

Comment [YUN56]: Careless Mistake Some students failed to realize that the first term of the G.P is 18000(1.05).

Comment [YUN57]: Careless Mistake Some students missed out the equal sign. Also, some students forgot about the fact that the given expression is on the amount of money that Jack still owes.

Comment [YUN58]: Presentation of Answers The inequality must be shown before the final conclusion of the answers is given.

Comment [YUN59]: Accuracy Many students cannot apply long division or juggling method correctly to obtain the equation of the oblique asymptote.

Comment [YUN60]: Misconception Many students did not differentiate the expression and concluded that the discriminant of the original expression is negative and proceeded to find the range of values of a.

[Turn Over

If no stationary points, 21 0a− < A1

1 or 1.a a⇒ > < −

A1

(iii)

G1 – shape G1 – correct asymptotes G1 - correct intercepts

(iv)

G1- shape G1 – correct x-intercept

From the graph, there are two intersection points between C and D. A1

x

y

-7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7

-3

-2

-1

0

1

2

3

2x = 2y x= − +

3.73 0.268

2 4 12

x xyx

− +=

−

0.5

2 4 12

x xyx

− +=

−

2 1y x= −

Comment [YUN61]: Misconception Some students sketched the graph of

1y x= − instead of 2 1y x= − . Other students sketched both graphs of

22 4 12

x xyx

− += − and

1y x= − which

is not allowed in the question.

Therefore there are two real roots for the equation

22 4 1 1.2

x x xx

− += − −