Solutions to O Level Add Math paper 2 2011 · 2011. 11. 14. · Solutions to O Level Add Math paper...

Solutions to O Level Add Math paper 2 2011

By KL Ang, Nov 2011 Page 1

1. The equation of a curve is kkxxy 62 2 , where k is a constant.

(i) Find the range of values of k for which the curve lies completely above the x-axis.

[4]

(ii) In the case where 2k , find the values of m for which the line 4mxy is a

tangent to the curve. [4]

Solution :

(i) The curve is to hang completely above x-axis.

That is kkxx 620 2 has no real root.

It is necessary and sufficient that the discriminant

of the equation kkxx 620 2 is negative.

06242 kk

04882 kk

0412 kk

From the sketch,

412 k

(ii) When 2k , kkxxy 62 2 has no real root.

4

62 2

mxy

kkxxy has repeat roots.

4224 2 xxmx

8220 2 xmx

The discriminant,

82420 2 m

22 820 m 6100 mm

10m or 6m

[Analysis]

This question is on nature of roots, in relation to quadratic equation. A sketch of its graph

would have made the problem a lot clearer. Need to know how to answer quadratic

inequality. Part (ii) is about a curve tangent to a line, or a repeated root where discriminant is

zero.

x

y

O

y

k O

x

y

O 4

4 mxy



Alternative solution:

(i) Given that the quadratic curve is above x-axis, the minimum value of y must be positive.

ymin is located on the line of symmetry.

Line of symmetry 422

kkx

kk

kk

y

6

442

2

min

68

2

min kk

y

Since ymin > 0,

068

2

kk

04882 kk

0412 kk

From the sketch,

412 k

(ii) Given that the line 4 mxy is tangent to the curve kkxxy 62 2 , when 2k ,

the gradient at the point of tangent must be the same.

24d

d x

x

y , must be the same as the gradient of the line, m.

24 xm

422424 2 xxxx

042 x

2x or 2x

When 2x , 6224 m When 2x , 10224 m

In Summary:

The topic on Quadratics remains the most important content to be mastered.

k O

x

y

O

ymin



2. The function f is given by bxax tan:f , where a and b are positive integers and

22

x .

(i) Given that 0f x when 2

x , find the smallest possible value of b. [1]

(ii) Using the value of b found in part (i) and given that the gradient of the graph of xy f

is 12 at the point where 8

x , find the value of a. [3]

(iii) Sketch the graph of xy f . [3]

Solution :

(i) Refer to the sketch, the period of bxay tan is 2

.

2

b

2b

(ii) When 8

x , 12

d

d

x

y

xax

y2sec2

d

d 2

82sec212 2

a

4sec6 2

a

4cos6 2

a

2

1

4cos

3a

(iii) as shown above.

[Analysis]

Tangent curve might not be familiar to many students. A sketch of the curve would be useful.

Part (i), the period of tan x is 180 . Part (ii) is about differentiation.

4

O

y

x

4

2

2



3. The expression baxx 32 , where a and b are constants, has a factor of 2x and leaves a

remainder of 35 when divided by 3x .

(i) Find the value of a and of b. [4]

(ii) Using the values of a and b found in part (i), solve the equation 02 3 baxx ,

expressing non-integer roots in the form 2

dc , where c and d are integers. [4]

Solution :

(i) Let baxxx 32p By Factor Theorem, 02p ,

ba 2222p 3 12160 ba By Remainder Theorem, 353p ,

ba 3323p 3

ba 35435

23190 ba 12 ,

7a

Sub. 7a into (1),

2b

272p 3 xxx

(ii) 0272 3 xx

By long division,

1422272 23 xxxxx 01422 2 xxx

2x or

222

22

124164

x

[Analysis]

This is a typical Factor and Remainder Theorems question. Part (ii) simply asks for

application of quadratic formula or completing square after factorised x+2.



4. The roots of the quadratic equation 0542 2 xx are and .

(i) Show that 5

222

. [4]

(ii) Given that the roots of 02 baxx are 2

and 2

, find the value of a and

of b, where a and b are constants. [4]

Solution :

(i) Given that 0542 2 xx ,

22

4

2

5

12

5242

222

5

2

2

5

122

(ii)

5

184

5

2422

22

5

212524122

22

05

21

5

182 xx

[Analysis]

Recognize that this is a Sum-of-roots, Product-of-roots question, a

b and

a

c of

a quadratic equation 02 cbxax .

In Summary:

A routine question, Be competent in handling the algebraic indentity.



5. (a) Solve the equation xx 273 log4log .

(b) The curve naxy , where a and n are constants, passes through the points (2, 40),

(3, 135) and (4, k). Find the values of n, a and k.

Solution:

(a) Given

xx 273 log4log

27log

log4log

3

33

xx

3

3

333

3log

log3log4log

xx

3log3

log3loglog

3

34

33

xx

3

log3loglog 3433

xx

433

3 3log3

loglog

xx

433 3loglog3

2x

433 3log2

3log x

23

4

33 3loglog

x

633 3loglog x

63x

729x

[Analysis]

Part (a) requires base change formula. Part (b) simply find the values of the unknowns.



(b)

When 2x , 40y , na240

When 3x , 135y , na3135

n

n

3

2

135

40

n

3

2

27

8

n

3

2

3

23

3

n

3

2

3

23

3n

therefore 3240 a

5a

When 4x , ky , 345k 320k

In Summary:

Application of Log and Indices laws are key elements in this paper.



6.

A garden is being designed to include a semicircular pond and a lawn. The radius of the pond

is r m and the length of the lawn, which is in the shape of a rectangle with the semicircle

removed, is l m as shown in the diagram above.

(i) Given that the area of the lawn is 400 m2, express l in terms of r. [2]

(ii) Given that the perimeter of the lawn is P m, show that r

rP400

22

3

. [2]

(iii) Given that r and l can vary, find the value of r for which P has a stationary value and

determine whether this value of P is a maximum or a minimum. [5]

Solution:

(i) Area of the lawn = 400 m2,

4002

12 2 rrl

rr

l 4

1

2

400

2r m

l m

r m

POND

LAWN

[Analysis]

Part (i) is to form an equation with the area. Part (ii) is to form an equation with the perimeter,

and then replace l from part (i) to obtain the desired expression. Part (iii) is to differentiate P,

then set it to zero to locate the turning points.



(ii)

rlrP 22

rrr

rP

42

40022

r

rP400

22

3

(iii)

2

4002

2

3

d

d

rr

P

When 0d

d

r

P,

2

4002

2

30

r

2

434002

r

43

8002

r

43

800

r take 0r only.

72.7r m (3 s.f.)

0800

d

d32

2

rr

P the turning point is a minimum.



7.

The diagram shows two intersecting circles C1 and C2, with centres P and Q respectively.

The point R lies on both circles and the line PR is a tangent to C2. A line L passes through Q.

The point E lies on C2. The line PE is perpendicular to L and intersects C2 at A, C1 at B and

L at D.

(i) By considering triangles QAD and QED show that AD = ED. [4]

(ii) Show that PEPAADPD 22 . [3]

(iii) Hence show that 222 PBADPD . [2]

Solution:

(i) Consider QDA and QDE

EQAQ (radius of circle C2)

QDQD (common side)

90QDEQDA

Therefore, QDA QDE (RHS)

Therefore, AD = ED

A

D

L E

B

C1

R

P

Q

C2

[Analysis]

Part (i) requires proof of congruent triangles. Part (ii) applies algebraic identity and finally part

(iii) uses tangent-secant theorem.



(ii)

22 ADPD

ADPDADPD PADEPD DEAD , from part (i) PAPE

(iii)

By Tangent-Secant Theorem,

2PRPAPE

222 PRADPD PEPAADPD 22

222 PBADPD PBPR , radius of circle C1

In Summary:

This geometrical proof is the easiest since 2009.



8. (i) Express 22

12

138

xx

xx in partial fractions. [5]

(ii) Hence find

x

xx

xxd

12

1382

2

. [5]

Solution:

(i) Let

222

121212

138

x

C

x

B

x

A

xx

xx

CxxxBxAxx 1212138 22

When 0x ,

A1

CxxxBxxx 1212138 22

CxxxBxxx 1212138 22

CxxxBxx 124 2

When 1x ,

)1(33 CB

When 1x ,

)2(5 CB

(1)+(2),

B28

B4

9C

222

12

9

12

41

12

138

xxxxx

xx

[Analysis]

First recongise that the numerator is of degree 2 and the denominator is of degree 3. Therefore

this is a proper fraction. Again, observe that 2x + 1 is a repeated factor. Apply

'fln f

f

xdx x c

x

in part (ii)



(ii)

x

xx

xxd

12

1382

2

xxxx

d12

9

12

412

cx

xx

12

1

2

912ln2ln

In Summary:

This s a typical question that combining partial fraction with integration.



9.

The diagram shows part of the curve 2

sin3x

y . The line 3

4x meets the curve at P and

the x-axis at Q. The tangent to the curve at P meets the x-axis at R. Find

(i) the length of QR.

(ii) the area of the shaded region.

Solution:

(i)

3

2sin3,

3

4 P for

2

3

3sin

3

2sin

2

33,

3

4P

x

P

Q R O

y 3

4x

2sin3

xy

[Analysis]

For part (i), PQ can be found by simply substituting 3

4x into

2sin3

xy . Then by

finding the gradient with 2

cos2

3

d

d x

x

y , then QR is found with similar triangles. Part (ii) is to

find the area under the curve from 0x to 3

4x by integration, then combines with the

area of triangle PQR.



2cos

2

3

d

d x

x

y

When 3

4x ,

3

2cos

2

3

d

d

x

y

4

3

d

d

x

y for

2

1

3

2cos

Therefore,

4

3

QR

PQ

PQQR3

4

32QR units for 2

33PQ

(ii) the area of the shaded region = area under the curve from 0x to 3

4x + area of triangle

PQR.

the area of the shaded region

QRPQxx

2

1d

2sin3

3

4

0

323

4

2

1d

2sin

2

16

3

4

0

xx

3

34

2cos6

3

4

0

x

3

340cos

3

2cos6

3

341

2

16

3.163

349

unit

2 (3 s.f.)



10. Without using a calculator

(i) find the exact value of 75cos15cos , [3]

(ii) find the exact value of 75cos15cos , [2]

(iii) show that 2

375cos15cos 22 , [2]

(iv) state the acute angle such that 75cossin , [1]

(v) use the results of parts (iii) and (iv) to find the exact value of 15cos2 . [3]

Solution:

(i)

Given that 75cos15cos

30sin45sin2 30sin45sin2 for xx sinsin

2

1

2

12

2

1

2

175cos15cos

(ii)

Given that 75cos15cos

30cos45cos2 30cos45cos2 for xx coscos

2

3

2

12

[Analysis]

This question examines trigo identities and ratios of special angles.



2

3

2

3

(iii)

Given that 75cos15cos 22

75cos15cos75cos15cos

2

1

2

3

2

3

(iv)

Given that 75cossin

15sinsin

15 or 165 (rejected)

(v)

75cos2

315cos 22

275cos2

3

215sin2

3

2

315sin15cos 22

2

3115cos2 2 for xx 22 cos1sin

2

1

4

315cos2

4

2315cos2



11. A circle, centre C, has a diameter AB where A is the point (1, 1) and B is the point (7, 9).

(i) Find the coordinates of C and the radius of the circle. [4]

(ii) Find the equation of the circle. [2]

(iii) Show that the equation of the tangent to the circle at B is 5734 xy . [3]

The lowest point on the circle is D.

(iv) Explain why D lies on the x-axis. [1]

(v) Find the coordinates of the point at which the tangents to the circle at B and D intersect.

[1]

Solution:

(i)

5,42

91,

2

71CC

coordinates of 5,4C

(ii)

1064361917 22 AB

2554 22 yx

(iii)

Gradient of AB, 3

4

17

19

ABm

Gradient perpendicular of AB, 4

3ABm

[Analysis]

Part (i) makes use of midpoint formula; part (ii) first find the length AB; part (iii) is to first find

the gradient AB then forming the equation of tangent; Part (iv) is to related the centre of the

circle, its radius and x-axis; Part (v) is to set 0y .



The equation of the tangent at B,

74

39 xy

5734 xy

(iv)

For point 5,4C , and radius of 5 units, the circle touches x-axis. So, the lowest point, D on

the circle is also tangent to the x-axis.

(v)

The tangent of the circle at D is the x-axis ,i.e. 0y .

When 0y ,

57304 x

19x

coordinates of the point of intersection 0,19

Solutions to O Level Add Math paper 2 2011 · 2011. 11. 14. · Solutions to O Level Add Math paper...

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