Solutions to O Level Add Math paper 1 2014 - korlinang · PDF fileSolutions to O Level Add...

Solutions to O Level Add Math paper 1 2014

By KL Ang, Feb 2014, source of questions ©UCLES & MOE 2012

1. Given that the line cxy 2 is a tangent to the curve kxy 2 , where c and k are positive

constants, prove that k is a multiple of c. [4]

Solution :

Given cxy 2 and kxy 2 ,

kxcx 2

2

kxccxx 22 44

044 22 cxkcx

The discriminant 22444 ckc

222 16816 ckckc

082 ckk

08 ckk

Since 0k , 0k , we have

08 ck ck 8

Alternative solutions:

Let mck , 0m

mcxcx 2

2

mcxccxx 22 44

044 22 ccxmx

The discriminant 222 444 cmc

16816 22 mmc

0822 mmc

Since 0c , 0c and 0m we have

08 mm 8m

ck 8

[Analysis]

To prove that mck . A curve is tangent to a line. Discriminant is zero.



Alternative solution:

Given cxy 2 ,

2d

d

x

y

Given kxy 2 ,

y

x

k

y

d

d2

y

k

x

y

2d

d

y

k

22

4

ky

kxk

2

4

16

kx

ckk

162

4

ck

8

ck 8



2. Find the coordinates of the stationary point of the curve x

xy4

for 0x , and determine

the nature of this stationary point. [5]

Solution :

Given x

xy4

,

2

41

d

d

xx

y

When 0d

d

x

y,

2

410

x

14

2

x

42x

2x

2x , 0x

42

42 y

the stationary point is 4,2

32

2 8

d

d

xx

y

When 2x , 012

8

d

d32

2

x

y, the turning point is a minimum.

[Analysis]

Finding turning points of a curve by differentiation, 0d

d

x

y.

2

2

d

d

x

y is to determine min/max

points.




Given x

xy4

, 0x , 04

xxy

42 xyx

042 yxx

Taking discriminant,

0442

y

162 y

4y or 4y (rejected as 0y )

y has a minimum value of 4.

When 4y ,

0442 xx

022x

2x

the stationary point 4,2 is a minimum point.



3. It is given that 7d

3

1

2 xax , where a is a constant.

(i) Find the value of xax d

6

2

2

. [3]

(ii) Express xbax d

3

1

2

in terms of the constant b . [2]

Solution :

(i) Given 7d

3

1

2 xax ,

73

26

33

27

1

3

3d

33

1

2

aaaaxxax

56783

268

3

208

3

8

3

216

2

6

3d

36

2

2

aaaaaxxax

(ii) bxbxbxaxxbax 271

37ddd

3

1

3

1

2

3

1

2


(i) Given 7d

3

1

2 xax ,

73

26

33

27

1

3

3d

33

1

2

aaaaxxax

73

26

a

26

21a

[Analysis]

Some technical aspect of definite integration.



56873

208

26

21

3

8

3

216

26

21

2

6

326

21d

36

2

2

xxax



4. It is given that xx 112 32 .

(i) Find the exact value of x12 . [3]

(ii) Hence find the value of x corrected to 2 decimal places. [2]

Solution:

(i) Given that xxxx 323212 22 ,

xx 112 32

xx

12

32

2

xx 12 322

xxx 33212 1

3212 x

612 x

(ii) From (i), we have 612 x .

6lg12lg x

6lg12lg x

72.012lg

6lgx (2 d.p.)

[Analysis]

Part (i) is to find the value of x12 , not x. Then, find the value of x in Part (ii).




Given that xx 112 32 ,

xx 112 3lg2lg ,

3lg12lg12 xx

3lg3lg2lg2lg2 xx

2lg3lg3lg2lg2 xx

6lg12lg x --------- (a)

6lg12lg x

612 x

(ii) From (a),

72.012lg

6lgx (2 d.p.)



5.

The diagram shows points A, B and C lying on a circle. The point T is such that the lines TA

and TC are tangents to the circle. Given that angle ABC = angle ATC, prove that triangle

ACT is equilateral. [4]

Solution:

TCTA (property of tangents from an external point)

ATC is an isosceles

TACTCA

TACCBATCA (tangent chord theorem)

Given that ATCABC ,

TACATCTCA

ATC is an equilateral.

[Analysis]

From the “tangents from an external point”, we know TC = TA, any one of the 3 angles is 60°

will have made an equilateral.

A

B

T C




Given that ATCABC ,

Let the centre of the circle be O.

2AOC (angle at the centre is twice the angle at the circumference)

90TAOTCO (radius tangent)

180AOCATC (sum of angles in quadrilateral OATC)

1802

60

TCTA (property of tangents from an external point)

ATC is an isosceles

602

60180TACTCA

TACCBATCA

ATC is an equilateral.

A

B

T C

2 O



6.

In the diagram, M is the midpoint of the line joining the points 4,1A and 6,7B .

The perpendicular bisector of AB intersects the line l at the point P. Given that line l is

parallel to the line 32 xy , find the coordinates of P. [6]

Solution:

Given that 4,1A and 6,7B ,

5,42

64,

2

71MM

Gradient of AB, 3

1

6

2

71

64

ABm

Gradient of PM, 31

AB

PMm

m , AB PM

Equation of PM,

435 xy

173 xy ---------- (1)

Equation of AP,

124 xy , AP // 32 xy

22 xy ---------- (2)

[Analysis]

Create equation for PA and PM, then solve for the intersection of these two lines.

A 4,1

O

M

x

y l

P B 6,7



Therefore, solve AP and PM,

173 xy and 22 xy

22173 xx

155 x

3x

8232 y

8,3P


Gradient of AB, 3

1

6

2

71

64

ABm

Gradient of PM, v

v

mm

AB

PM

31

,

( AB PM )

Gradient of AP, u

umAP

22

From the diagram,

3 vu vu 3 vu 262

132 vu

1326 vv

55 v

1v

213 u

8,3224,21 PP

A 4,1

O

M

x

y l

P

B 6,7

1

3

u

u2 v

v3



7.

The diagram above shows part of the graph of xy 4 . In each of the following cases

determine the number of intersections of the line cmxy with xy 4 , justifying your

answer.

(i) 1m and 2c . [2]

(ii) 2

1m and 0c . [2]

(iii) 2

1m and 2c . [2]

Solution:

From xy 4 , we get

When 0x ,

44 y

When 0y ,

x 40

4x

[Analysis]

Testing on the graph of modulus function and a straight line. Good graph sketching is

necessary.

O x

y

O x

y

4

4



(i) 1m and 2c . cxy

cxy intersects xy 4 ,

// 4 xy

One point of intersection.

(ii) 2

1m and 0c . xy

2

1

From the sketch,

there are 2 points of intersections.

(iii) 2

1m and 2c . cxy

2

1

When 0y ,

cx 2

10

42 cx

From the sketch, there is

no intersection.

O x

y

4

4

2

2 xy

4 xy

xy 4

O x

y

4

4

xy2

1

4 xy

xy 4

O x

y

4

4

2

22

1 xy

4 xy

xy 4




(i) 1m and 2c . cxy

When 4x , 44 xxy

4 xcx

4c , no solution

When 4x , xxy 44

xcx 4

42

4

cx

c 4

Since 2c , there is a solution.

There is 1 intersection.

(ii) 2

1m and 0c . xy

2

1

When 4x ,

42

1 xx

8x

When 4x ,

xx 42

1

3

8x

There are 2 points of intersections.

(iii) 2

1m and 2c . cxy

2

1

When 4x ,

42

1 xcx

43

28

cx

2c , but 2c , no solution.

When 4x ,

xcx 42

1



428 cx

2c , but 2c , no solution.

There is no intersections.



8. The roots of the quadratic equation 0532 xx are and .

(i) Express 22 in terms of and . [1]

(ii) Find a quadratic equation whose roots are 3 and 3 . [6]

Solution:

Given that 0532 xx , 3 , 5 ,

(i) 32222222

(ii) 322233 ,

185333 233

12553333

0125182 xx


(i) Let ba 222

When 0 , 22 a 1a

When 1 , 1 , b2

2111 , 3b

3222

(ii) Given that 0532 xx , 3 , 5 ,

0532 xx

532 xx

xxx 53 23

[Analysis]

A question on SOR and POR, where (i) is a preparation for (ii).



xxx 55333

1543 xx

1543

1543

30433

18303433

12553333

0125182 xx



9. (i) Express the equation 1cosec5cot2 2 as a quadratic equation in cosec . [2]

(ii) Hence solve the equation 1cosec5cot2 2 for 3600 . [4]

(iii) State the number of solutions of the equation 1cosec5cot2 2 in the range

720720 . [1]

Solution:

(i) Given that 1cosec5cot2 2 , knowing 22 coseccot1 , we get

1cosec51cosec2 2

03cosec5cosec2 2

(ii) 03cosec5cosec2 2 , 3600

03cosec12cosec

012cosec or 03cosec

2

1cosec 3cosec

(Rejected as 1cosec ) 3sin

1

3

1sin

Therefore

3

1sin 1

5.19 or 5.1605.19180

(iii) For 720720 , there are 4 cycles. 824 solutions.

[Analysis]

(i) Apply 22 coseccot1 .

(ii) Solve quadratic equation .

(iii) Extend the range of solution.



10. A particle travels in a straight line, so that t seconds after passing through a fixed point O ,

its acceleration, a m/s2, is given by

22

8

ta . The particle comes to rest when 2t .

Find

(i) an expression for the velocity of the particle in terms of t , [3]

(ii) the distance from O at which the particle comes to rest. [4]

Solution:

(i) Given that 22

8

ta ,

C

tt

tt

tV

2

8d

2

18d

2

822

When 2t , 0V ,

C

22

80

2C

2

82

tV

(ii) From 2

82

tV ,

20

2

0

2

0

2ln82d2

182d

2

82

ttt

tt

tS

2ln84ln84 S

2ln162ln84 S

55.1S (3s.f.)

[Analysis]

Part (i) is when the time 2t , the particle has zero velocity. Part (ii) is to find the distance between

the particle when it is zero velocity from point O.



11. Variables x and y are connected by the equation naxy , where a and n are constants.

Using experimental values of x and y, a graph was drawn in which yln was plotted on the

vertical axis against xln on the horizontal axis. The straight line which was obtained passed

through the points 5.2,6.1 and 1.1,3.2 .

Estimate

(i) the value of n and , to 2 significant figures, the value of a, [4]

(ii) the coordinates of the point on the line at which xy 2 . [4]

Solution:

(i) Given the straight line passed through the points 5.2,6.1 and 1.1,3.2 ,

At 5.2,6.1 , 6.1ln5.2 na ---------- (1)

At 1.1,3.2 , 3.2ln1.1 na ---------- (2)

(1) – (2) , n7.04.1

2n

6.12ln5.2 a

7.5ln a

7.5ea

300a (2 s.f.)

2300 xy

(ii) xy 2 xy ln2lnln .

Let yY ln and xX ln

XY 2ln

and XY 27.5

XX 27.52ln

0.52ln7.53 X

67.1X (3 s.f.) and 36.2669.127.5 Y (3 s.f.)

The coordinates of the point on the line at which xy 2 is 36.2,67.1 .

[Analysis]

(i) the equation of the straight line is xnay lnlnln .

(ii) xy 2 , xy ln2lnln intersects xnay lnlnln . To find yx ln,ln .




Given that 2300 xy and xy 2 ,

23002 xx

1503 x

313.5x and 626.10313.52 y

67.1ln x and 36.2ln y

The coordinates of the point on the line at which xy 2 is 36.2,67.1 .



12.

[The volume of a cone of height H and base radius R is given by HR2

3

1 ]

The diagram shows a hollow conical tank of height 30 cm and radius 15 cm. The tank is

held fixed with its circular rim horizontal. Water is then poured into the empty tank at a

constant rate of 20 cm3/s. After t seconds the depth of water is h cm.

(i) Show that the volume of water in the tank, V cm3, at time t is given by

12

3hV

. [2]

(ii) Find the rate of change of the depth when 5h . [4]

(iii) State, with a reason, whether this rate will increase or decrease as t increases. [2]

[Analysis]

(i) is about formulating an equation, the equation has no t and r.

(ii) applies chain rule to find t

h

d

d.

(iii) need to deduce the change in t

h

d

d, i.e.

2

2

d

d

t

h .

30 cm

h cm

15 cm



Solution:

(i) Given tV 20 ,

Let 30

15

h

r

2

hr

hrV 2

3

1

hh

V

2

23

1

12

3hV

(ii) Let 20d

d

t

V and

4d

d 2h

h

V ,

t

h

h

V

t

V

d

d

d

d

d

d

5h , t

h

d

d

4

520

2

5

16

d

d

t

h

02.1d

d

t

h (3 s.f.)

(iii) t

hh

d

d

420

2

2

80

d

d

ht

h

32

2 160

d

d

ht

h

When 5h ,

05

160

d

d32

2

t

h

t

h

d

d is at a decreasing rate.



13.

The height above ground level, h m, of a capsule on the Singapore Flyer is modelled by the

equation, kth cos180 , where k is a constant and t is the time in minutes after starting

the ride at ground level. The total time to complete one revolution is 30 minutes.

(i) Explain why this model suggests that the height of the Singapore Flyer is 160 m. [1]

(ii) Show that the value of k is 15

radian per minute. [2]

It is possible for a person riding in a capsule to see a certain landmark, provided the capsule

is at least 100 m above ground level.

(iii) Find the length of time for which the landmark will be in view during one revolution.

[5]

[Analysis]

Need to relate the formula kth cos180 to the flyer. Consider cos180 h where

being the angle of rotation of the flyer. Then kt , k is the speed of rotation. The flyer

rotates 3602 in 30 minute.



Solution:

Make a sketch of kth cos180 .

(i) when 1cos kt , flyer moved by half

of the revolution,

1601180 h

(ii) The flyer takes 30 min to complete one

Revolution, 2 radians.

230 k

15

k radian/min

(iii) when 100h ,

100cos180 kt ktcos4

1

Consider 4

1cos kt

4

1cos 1kt

8235.1kt rad or 4597.4kt rad

70.88235.1 kt or 29.214597.4 kt

The duration of above 100 m is 6.1270.829.21 minutes


(iii) Consider 100h ,

100cos180 kt 4

1cos kt

4

1cos 1kt

8235.1kt rad

70.88235.1 kt , it takes 15 minutes to reach the top. So, 6.127.8152 minutes

30

h

t (min) 0

100

80

160

15

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