solutions to o level add math paper 2 2012 - korlinang · pdf filesolutions to o level add...

Solutions to O Level Add Math paper 2 2012

By KL Ang, Nov 2011

1. The equation of a curve is xxy cos3sin2 for x0 .

(i) Write down expression for x

y

d

d and

2

2

d

d

x

y. [3]

(ii) Find the value of x for which the curve has a stationary point. [2]

(iii) Determine the nature of this stationary point. [2]

Solution :

(i) Given that xxy cos3sin2 ,

xxx

ysin3cos2

d

d

xxx

ycos3sin2

d

d2

2

(ii) When 0d

d

x

y,

xx sin3cos20

3

2tan x

55.2x rad or 70.5x rad (rejected)

(iii) When 55.2x rad,

55.2cos355.2sin2d

d2

2

x

y < 0, hence a maximum point.

[Analysis]

This question is on differentiation and nature of turning points.


By KL Ang, Nov 2011

Alternative solution:

(i) Given that xxy cos3sin2 ,

xxy cos

13

3sin

13

213

sincoscossin13 xxy where 2

3tan

xy sin13 983.0 rad

xx

ycos13

d

d

xx

ysin13

d

d2

2

(ii) When 0d

d

x

y,

xcos130

2

x or

2

3 x

55.2x rad 70.5x rad (rejected)

(iii) When 55.2x rad,

2sin13

d

d2

2

x

y < 0, hence a maximum point.

In Summary:

Be very careful in handling the radian. The actual value of 2nd

derivative is not

as important, as we just need to know the sign of it only.


By KL Ang, Nov 2011

2. (i) Given that xBxAxx coscoscos3cos , state the value of A and of B. [1]

(ii) Hence, or otherwise, solve the equation 0cos23cos xx for 1800 x . [5]

Solution :

(i)

2

3cos

2

3cos2cos3cos

xxxxxx

xxxx cos2cos2cos3cos

Hence, 2A , 2B

(ii) Given that 0cos23cos xx for 1800 x , so 36020 x

0coscos2cos2 xxx

012cos2cos xx

12cos2 x or 0cos x

2

12cos x 90x

principle angle, 1202x

240,1202x

120,60x

Alternative solution:

(i) xxxxxxx cossin2sincos2coscos3cos

xxxxx coscossin2cos2cos 2

xxx 2sin212coscos

xxcos2cos2

[Analysis]

Part (i) apply the sum of angle identity. Part (ii) is about solving a trigo equation.

In Summary:

Be very careful in handling equation in part (ii). Take note of the double angle

solution, because the range of 2x is larger.


By KL Ang, Nov 2011

3. (i) Express 383

202 xx

in partial fractions. [4]

(ii) Find x

xxd

383

202

and hence evaluate xxx

d383

207

22

. [4]

Solution :

(i) Since 313383 2 xxxx ,

let 313383

12

x

B

x

A

xx

when 3x , 10

1

10

1

B ,by cover-up method

when 3

1x ,

10

3

33

1

1

A

3

10

1

13

10

3

383

12

xxxx

3

2

13

6

383

202

xxxx

(ii) x

xxd

383

202

x

xxd

3

2

13

6

x

xx

xd

3

12d

13

32

cxx 3ln213ln2 where c is an integrating constant

[Analysis]

Part (i) is a simple 2 parts decomposition. Part (ii) will need integration into ln function.


By KL Ang, Nov 2011

xxx

d383

207

22

273ln213ln2 cxx

cc 32ln216ln237ln2121ln2

cc 5ln25ln210ln220ln2

2ln2

39.1 (3 s.f.)

In Summary:

Nothing complicated in this question, except the integration into ln function.

Students are reminded that exponential and log functions are VERY important.


By KL Ang, Nov 2011

4.

Two circles C1 and C2, intersect at P and Q as shown in the diagram. The tangent to C1 at P

meets C2 at R and the tangent to C2 at P meets C1 at S. Prove that

(i) triangle PQR and SQP are similar, [3]

(ii) Show that 2QPQRQS . [2]

Points A and B lie on the circumferences of C1 and C2 respectively,

(iii) Prove that angle SAP = angle PBR. [3]

Solution :

(i) Consider PQR and SQP

PSQRPQ (Alternate Segment Theorem)

QPSQRP (Alternate Segment Theorem)

Therefore, PQR SQP (AA)

(ii) Since PQR SQP ,

PS

RP

QP

QR

SQ

PQ .

QP

QR

SQ

PQ

QRSQQPPQ 2QPQRQS

[Analysis]

Part (i) needs to find 2 congruent angles in both triangles. Part (ii) can be arranged into length ratio.

Part (iii) may need properties of cyclic quadrilaterals.

S

A

B

C1

R

P

Q

C2


By KL Ang, Nov 2011

(iii)

SAPQ and PBRQ are cyclic quadrilaterals.

180SQPSAP 180PQRPBR

PQRPBRSQPSAP

Since PQRSQP , from part (i) SQP PQR , therefore

PBRSAP

In Summary:

Need a keen sense of observation of the given diagram. When part (i) is solved,

the rest of the question becomes straight forward.


By KL Ang, Nov 2011

5. The equation of a curve is xx BAy ee2 , where A and B are constants. The point P(0, 4)

lies on the curve and the gradient of the tangent to the curve at P is 1 .

(i) Find the value of A and of B. [4]

(ii) Using your values of A and B from part (i), obtain an expression for xy d and hence

evaluate xy d1

0 corrected to one decimal place. [4]

Solution:

(i) Given xx BAy ee2 ,

At P(0, 4),

BA4 --------- (1)

xx BAx

y ee2d

d 2

At P(0, 4), 1d

d

x

y

BA 21 --------- (2)

(1) + (2),

1A

3B

(ii)

xxy e3e2

xy d

xxx de3e2

xx xx de3de2

[Analysis]

Part (i) requires differentiation. Part (ii) simply find the integration and the value of the definite

integration.


By KL Ang, Nov 2011

xx xx de3d2e

2

1 2

Cxx e3e2

1 2 where C is an arbitrary integrating constant

therefore xy d1

0

CC 0012 e3e

2

1e3e

2

1

CC 0012 e3e

2

1e3e

2

1

32

1e3e

2

1 12

1.5 (1 d.p.)

In Summary:

Application of differentiation and integration of exponential functions.


By KL Ang, Nov 2011

6. (a) (i) Given that ux 4

3

8 log4log , express u in terms of x. [3]

(ii) Find the value of x for which 4log

1log5log

3

3

8

2

4 xxx . [3]

(b) Solve the equation 152ee yy . [3]

Solution:

(a)

(i) Given that ux 4

3

8 log4log ,

2

2

2

3

2

3

2

2log

log4

2log

log ux

2

log4log 2

2

ux

ux 222 log2log8log2

ux 2

8

2

2

2 log2loglog

ux

2

2

2 log256

log

256

2xu

(ii) Given that 4log

1log5log

3

3

8

2

4 xxx where 0x and ( 0x or 5x )

4log

3log

2log

log

2log

5log

2

2

3

2

3

2

2

2

2

2 xxx

2

3loglog

2

5log 22

2

2

xxx

3loglog5log 2

2

2

2

2 xxx

3log5

log 22

2

2

x

xx

35

2

2

x

xx

22 35 xxx

[Analysis]

(a) Part (i) requires base change formula. Part (ii) simply find the values of the log equation.

(b) to solve a routine exponential equation.


By KL Ang, Nov 2011

052 2 xx

0x or 5.2x

(rejected)

(b) Given that 152ee yy ,

015e2e2 yy

03e5e yy

3e y or 5e y

(rejected) 61.15ln y (3 s.f.)

In Summary:

Part (a) requires application of log rules, part (b) is just another exercise.


By KL Ang, Nov 2011

7. (i) Find the value of a and of b for which 232 2 xx is a factor of

bxxaxx 234 32 . [6]

(ii) Using the value of a and b found in part (i), solve the equation

032 234 bxxaxx . [3]

Solution:

(i) Given that 212232 2 xxxx , let bxxaxxx 234 32f .

When 2x , 02f .

ba 2223222f234

ba 280 --------- (1)

When 2

1x , 0

2

1f

.

ba

2

1

2

1

2

13

2

12

2

1f

234

ba 4

3

2

10 --------- (2)

(1) – (2),

a4

5

2

150

a 60

6a

b 6280

4b

[Analysis]

212232 2 xxxx , apply factor theorem to find a and b. Part (ii) factorizes completely to

find the roots.


By KL Ang, Nov 2011

(ii) Given that 046632 234 xxxx

223246632 22234 kxxxxxxxx

When 1x ,

2123246632 k

133 k

0k

223246632 22234 xxxxxxx

02232 22 xxx

2x or 2

1x or 22 x

or 2x or 2x

In Summary:

Without fail, factor theorem and remainder theorem are regulars in this paper.


By KL Ang, Nov 2011

8. (i) Find the set of values of m for which the curve 22 xxy and the line 1 mxy do

not intersect. [4]

(ii) Sketch the curve 22 xxy , giving the coordinates of the maximum point and of the

points where the curve meets the x-axis. [3]

(iii) Find the number of solutions of the equation 12 2 mxxx when

(a) 1m , (b) 2

1m [4]

Solution:

(i) Given 22 xxy and 1 mxy ,

221 xxmx

0122 xmx

Discriminant of the equation is to be negative,

0422

m

422m

222 m

40 m

The set of values 40: mm

(ii) Given 22 xxy , when 0y ,

220 xx

xx 20

0x or 2x

Line of symmetry, 12

02

x

The locally maximum value, 112 y

The coordinate of the maximum point is (1, 1)

[Analysis]

(i) Solving simultaneous equations with negative discriminant.

(ii) Sketch the graph by flipping up the negative part to the upper half.

(iii) Add lines to the sketch in part (ii) to determine the number of solutions.


By KL Ang, Nov 2011

(a) 1m , 1 xy (b) 2

1m , 1

2

1 xy

number of solutions = 2 number of solutions = 3

In Summary:

This is a typical curve sketching question. Remember to label your curve and

lines clearly.

x

(1, 1)

2 O

y 1x

22 xxy

–1

1 xy

12

1 xy


By KL Ang, Nov 2011

9.

In the diagram ABC is a structure consisting of a rod AB of length 10 cm attached at B to a

rod BC of length 24 cm so that angle ABC = 90°. Small rings at A and B enable A to move

along a vertical wire Oy and B to move along a horizontal wire Ox. Angle OAB = Ɵ and can

vary. The horizontal distance of C from the vertical wire Oy is L cm.

(i) Explain clearly why cos24sin10 L . [2]

(ii) Express L in the form cosR , where R > 0 and 900 . [4]

(iii) Find the greatest possible value of L and the value of at which this occurs. [3]

(iv) Find the value of for which L = 20. [2]

Solution:

(i) sin10OB cm CBx , cos24xBC

cos24sin10 xBCOBL

(ii) sinsincoscoscos RRR

sinsincoscoscos24sin10 RRL

24cos R and 10sin R

12

5

24

10tan

6.22

12

5tan 1

[Analysis]

Part (i) is to formulate the projected horizontal length. Part (ii) is to express it into cosR .

Part (iii) is to find the maximum value of L. Part (iv) is simply solving the trigo equation.

O x

y

B

A

C

10 cm

24 cm

O x

y

B

A

C

10 cm

24 cm

Cx


By KL Ang, Nov 2011

22222410sincos RR

26R

cos26L where

12

5tan 1

(iii) When 1cos , L is at its maximum.

The largest 26L when 1cos

0

6.22

(iv) When 20L ,

cos2620

13

10cos

7.39

13

10cos 1

3.626.227.39

In Summary:

Question on R formula is a regular in this paper.


By KL Ang, Nov 2011

10.

The diagram shows a trapezium ABCD in which AD is parallel to BC and angle BCD = 90°.

The vertices of the trapezium are at the points A(8, k), B(3, 6), C(1, 2) and D. The line with

equation 253 xy passes through M, the mid-point of AB, and through N, which lies on

AD.

(i) Show that k = 11. [3]

(ii) Find the coordinates of N and of D. [6]

(iii) Find the area of the triangle AMN. [2]

Solution:

(i)

2

6,

2

11

2

6,

2

38 kkM

Since M lies on the line 253 xy ,

252

113

2

6

k

63350 k

11k

A(8, k)

O

253 xy

y

x

C

(1, 2)

(3, 6) B

M

N

D

[Analysis]

Part (i), mid-point M, is on the line. Part (ii) needs to find the equation of AD using gradient BC.

And also need to find the equation of CD. Part (iii) begins with finding the area of BNA.


By KL Ang, Nov 2011

(ii)

11,8A AD // BC, gradient of BC = gradient of AD 213

26

Equation of AD ,

8211 xy

52 xy

At point N, 52 xy intersects 253 xy ,

25352 xx

305 x

6x

7512 y

7,6N

CD BC, gradient of CD 2

1

Equation of CD ,

12

12 xy

5.22

1 xy

At point D, 52 xy intersects 5.22

1 xy ,

5.22

152 xx

5.75.2 x

3x

156 y

1,3D

(iii)

length CD 5211322 units

length AN 201178622 units

Area of triangle BNA 52052

1 square units

Area of triangle AMN = 2

1 of Area of triangle BNA 5.25

2

1 square units


By KL Ang, Nov 2011

11.

The diagram shows part of the curve xy 45 , meeting the x-axis at the point A and the

line 1x at the point B. The normal to the curve at B meets the x-axis at the point C. Find

(i) the coordinates of C , [6]

(ii) the area of the shaded region. [6]

Solution:

(i) When 1x , 345 y , coordinates of 3,1B

xxx

y

45

2

45

4

2

1

d

d

When 1x ,

3

2

45

2

d

d

x

y

The gradient of the tangent at B is 3

2, therefore the normal at B is

2

3 .

Equation of BC ,

12

33 xy

5.42

3 xy

x

B

A C O

y 1x

xy 45

[Analysis]

Part (i) requires differentiation to find gradient then the normal equation. Part (ii) breaks the shaded

area into two regions.


By KL Ang, Nov 2011

When 0y ,

5.42

30 x

3x

coordinates of 0,3C

(ii) When 0y ,

x450

4

5x

coordinates of

0,

4

5A

area of the shaded region

3132

145

1

4

5

dxx

34542

3

6

11

4

5

dxx

3456

11

4

52

3

x

30456

12

3

2

3

3276

1

35.4

5.7 square units

solutions to o level add math paper 2 2012 - korlinang · pdf filesolutions to o level add...

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