Solutions to O Level Add Math paper 2 2014 · · 2015-01-22Solutions to O Level Add Math paper 2...

Solutions to O Level Add Math paper 2 2014

By KL Ang, Jan 2015

1. Baby food is heated in a microwave to a temperature of 80°C. It subsequently cools in such

a way that its temperature, T °C, t minutes after removal from the microwave, is given by

ktAT e20 , where A and k are constants.

(i) Explain why A = 60. [1]

When t = 1 the temperature of the food is 65°C.

(ii) Find the value of k correct to 3 significant figures. [2]

A baby should only be given this food when the temperature of the food is less than 40°C.

(iii) Determine, with working, whether it is safe to give the food 4 minutes after removal

from the microwave. [2]

Solution :

(i) When 0t , 80T ,

0e2080 A

60A

(ii) When 1t , 65T ,

k e602065

k e60

45

k4

3ln

3

4lnk

288.0k

(iii) For 40T ,

40e6020 kt

3

1e kt

[Analysis]

Understand the modelling of cooling with exponential function.


By KL Ang, Jan 2015

3

1ln kt

k

t3ln

3ln4ln

3ln

t

82.3t (3 s.f)

Therefore, when 4t , 40T . It is safe to feed the baby.

In Summary:

Modelling question similar to this will continue to be in the future paper.


By KL Ang, Jan 2015

2. Given that 61132f 23 xxxx ,

(i) find the remainder when xf is divided by 2x , [2]

(ii) show that 2x is a factor of xf and hence solve the equation 0f x . [4]

Solution :

(i) Given that 61132f 23 xxxx ,

1262212162f

(ii) 062212162f , therefore 2x is a factor of xf .

Let 32261132 223 kxxxxxx

When 1x ,

32361132 k

536 k

7k

Therefore, 372261132 223 xxxxxx

061132 23 xxx

03722 2 xxx

03122 xxx

2x or 2

1x or 3x

In Summary:

A very typical question on R/F theorems.

[Analysis]

Applying Remainder and Factor Theorems.


By KL Ang, Jan 2015

3. The area of a quadrilateral is 4813 cm2.

(i) In the case where the quadrilateral is a rectangle with width 33 cm, find, without

using a calculator, the length of the rectangle in the form 3ba cm. [4]

(ii) In the case where the quadrilateral is a square with side c32 cm, find, without

using a calculator, the value of the constant c. [3]

Solution :

(i) 33

4813

3333

3331613

22 33

333413

6

1231231339

6

327

36

1

2

9

(ii) 4813322

c

34133432 22

cc

34133412 2 cc

1c

[Analysis]

Simplifying surd.

In Summary:

Surd will still be in the future paper.


By KL Ang, Jan 2015

4. The quadratic equation 0452 2 xx has roots α and β.

(i) Find the value of α2 + β2. [3]

(ii) Find the quadratic equation whose roots are α3 and β3. [5]

Solution :

(i) Given that 0452 2 xx , 2

5 2

2222 .

2222

222

52

44

25

4

12

(ii) SOR: 2233

33

2

523

2

53

8

515

8

125

POR: 8333 .

The quadratic equation is 088

52 xx or 06458 2 xx

[Analysis]

SOR: 2

5 ; POR: 2 .

In Summary:

As expected for the new cubic identity.


By KL Ang, Jan 2015

5. (a) Express 34loglog2 22 xx as a quadratic equation in x and explain why there are

no real solutions. [5]

(b) Given that

08log

log2

x

y

y

x , express y as a power of x. [4]

Solution:

(a) Given that 34loglog2 22 xx , 4x

4log3log 2

2

2 xx

4log2log3log 22

2

2 xx

4log2loglog 2

3

2

2

2 xx

42loglog 3

2

2

2 xx

3282 xx

03282 xx

Taking discriminant, 06412864321482

Hence, 03282 xx has no real root.

(b) Given that

08log

log2

x

y

y

x , 0log xy , 1, yx , 0, yx

0log8log2

xy yx

0log8log

log2

x

x

yy

y

y

0log8log

12

xx

y

y

0log813 xy

8

1log

3xy

[Analysis]

Log rules, discriminant. base changing.


By KL Ang, Jan 2015

2

1log xy

2

1

yx

2

1

xy

In Summary:

The last part in (ii) is a little non-routine.


By KL Ang, Jan 2015

6.

The diagram shows a triangle ABC whose vertices lie on the circumference of a circle. The

triangle DEF is formed by tangents drawn to the circle at the points A, B and C.

(i) Prove that angle DEF = 2 angle ABC. [4]

(ii) Make a similar deduction about angle DFE. [1]

(iii) Prove that 2 angle BAC = 180° + angle EDF. [3]

Solution:

(i)

ACEABC (Alternate Segment Theorem)

EAEC (Tangents from an external point)

Hence, AEC is an isosceles triangle.

Therefore, ABCACECAE

Therefore, AECABC 2180

DEFABC 1802180

Therefore, DEFABC 2

(ii)

Similarly, ACBDFE 2

[Analysis]

Properties of tangents to a circle are test in this question.

A

B D

C

F

E


By KL Ang, Jan 2015

(iii)

ACBABCBAC 180 (Sum of angles in a triangle)

22

180DFEDEF

BAC

DFEDEFBAC 3602

EDFBAC 1803602 (Sum of angles in a triangle)

EDFBAC 1802

In Summary:

Geometrical Proof is a weak link common to many students. Kids, do more

practice, it is not at all that difficult.


By KL Ang, Jan 2015

7. A curve has the equation 432 xy . The point ( p, q ) is the stationary point on the

curve.

(i) Determine the value of p and of q. [4]

(ii) Determine whether y is increasing or decreasing

(a) for values of x less than p, [1]

(b) for values of x greater than p. [1]

(iii) What do the results of part (ii) imply about the stationary point? [1]

(iv) What is the value of 2

2

d

d

x

y at the stationary point? [2]

Solution:

(i) Given that 432 xy ,

134d

d 3 x

x

y

334d

dx

x

y

Let 0d

d

x

y,

3340 x

x 30

3x

23324y

Therefore, 3p , 2q

(ii)

(a) When 3x ,

034d

d 3 x

x

y, y is increasing.

(b) When 3x ,

[Analysis]

Question on differentiation with graph.


By KL Ang, Jan 2015

034d

d 3 x

x

y, y is decreasing.

(iii) The turning point is a maximum point.

(iv)

1312d

d 2

2

2

xx

y

When 3x ,

013312d

d 2

2

2

x

y

In Summary:

Part (ii) may have been challenging for some students.


By KL Ang, Jan 2015

8. A particle travelling in a straight line passes through a fixed point O with a speed of 8 m/s.

The acceleration, a m/s2, of the particle, t s after passing through O, is given by ta 1.0e .

The particle comes to instantaneous rest at the point P.

(i) Show that the particle reaches P when 5ln10t . [6]

(ii) Calculate the distance OP. [4]

(iii) Show that the particle is again at O at some instant during the fiftieth second after first

passing through O. [2]

Solution:

(i) Given that ta 1.0e ,

The velocity, tv t de 1.0

Cvt

1.0

e 1.0

where C is an integration constant.

When 0t , 8v .

C1.0

e8

0

2C

21.0

e 1.0

t

v

When 0v ,

21.0

e0

1.0

t

t1.0e2.0

t1.05

1ln

t1.05ln

5ln10t

[Analysis]

Kinematics with integration.


By KL Ang, Jan 2015

(ii) Since 21.0

e 1.0

t

v ,

Let the distance, tst

d21.0

e 1.0

Dtst

21.0

e2

1.0

where D is an integration constant.

When 0t , 0s

D2

0

1.0

e0

100D

10021.0

e2

1.0

tst

When 5ln10t , OPs

1005ln1021.0

e2

5ln101.0

OP

1005ln1021.0

5

1

2OP

8.471005ln2020 OP m (3 s.f.)

(iii) When 50t ,

1005021.0

e2

501.0

s

5e

100s

674.0s (3 s.f.)

Since, when 5ln10t , 8.47s > 0 and when 50t , 674.0s < 0, the particle must have

return to point O, at least once.

In Summary:

Part (iii) may have given problem to some students.


By KL Ang, Jan 2015

9. (i) Solve the equation 02sin2cos3 AA for 3600 A . [4]

(ii) On the same axes sketch, for 1200 x , the graphs of

16cos2

3 xy and xy 3sin

2

12 . [6]

(iii) Explain how the solutions of the equation in part (i) could be used to find

the x-coordinates of the points of intersection of the graphs of part (ii). [2]

Solution:

(i) Given that 02sin2cos3 AA for 3600 A

02sinsincos3 22 AAA

02sinsinsin13 22 AAA

01sinsin6 2 AA

01sinsin6 2 AA

01sin21sin3 AA

3

1sin A or

2

1sin A

3

1sin 1A

2

1sin 1A

Principal angles, 5.19A 30A

5.199A or 5.340A 30A or 150A

[Analysis]

Trigo equation and its related graph.


By KL Ang, Jan 2015

(ii)

(iii)

Given that 16cos2

3 xy and xy 3sin

2

12 , the intersections

16cos2

33sin

2

12 xx

013sin2

16cos

2

3 xx

023sin6cos3 xx

Let Ax 3 , 36030 x , 3600 A

02sin2cos3 AA

From (i), the solutions of the equation are 5.199A , 5.340A , 30A , 150A

Therefore, 5.1993x , 5.3403x , 303x , 1503x

5.66x , 5.113x , 10x , 50x

1

O x

y

30

2

3

60 90 120

1

xy 3sin2

12

16cos2

3 xy


By KL Ang, Jan 2015

10. A circle, C1, has equation x2 + y2 + 4x – 6y = 36.

(i) Find the radius and the coordinates of the centre of C1. [3]

A second circle, C2, has a diameter AB. The point A has coordinates (5, 5) and the equation

of the tangent to C2 at B is 1543 xy .

(ii) Find the equation of the diameter AB and hence the coordinates of B. [4]

(iii) Find the radius and the coordinates of the centre of C2. [3]

(iv) Explain why the point (4, 6) lies within only one of the circles C1 and C2. [2]

Solution:

(i) Given that x2 + y2 + 4x – 6y = 36 ,

3664 22 yyxx

36996444 22 yyxx

94363222

yx

222732 yx

Centre 3,2 , radius 7 units

A second circle, C2, has a diameter AB. The point A has coordinates (5, 5) and the equation

of the tangent to C2 at B is 1543 xy .

(ii) Find the equation of the diameter AB and hence the coordinates of B. [4]

1543 xy

53

4 xy

Therefore, gradient of AB , 4

3

equation of the diameter AB, 54

35 xy

54

15

4

3 xy

4

5

4

3 xy


By KL Ang, Jan 2015

Let 4

5

4

35

3

4 xx

54

5

4

3

3

4 xx

4

25

12

25x

3x

1533

4y

1,3 B

(iii) radius of C2

52

515322

centre of C2

2

51,

2

53

2,12 C

(iv)

The distance between point (4, 6) and 3,21 C 4945362422

The point (4, 6) is inside circle 1C .

The distance between point (4, 6) and 2,12 C 2541261422

The point (4, 6) is outside circle 2C .


By KL Ang, Jan 2015

11. (a) Show that 312

1

12d

d

x

x

x

x

x. [3]

(b)

The diagram shows the line x = 5 and part of the curve

312

18

x

xy . The curve

intersects the x-axis at the point A. The line through A with gradient 1 intersects the

curve again at the point B.

(i) Verify that the y-coordinate of B is 2

3. [5]

(ii) Determine the area of the shaded region bounded by the curve, the line 5x , the

x-axis and the line AB. [4]

Solution:

(a) Let 12

x

xu .

2

2

1

2

1

2

1

12

2122

112

d

d

x

xxx

x

u

12

12

122

12

1

x

x

xx

12

12

12

2

1

x

x

xx

32

3

12

1

12

1

x

x

x

x

A

O

5x

x

y

B

312

18

x

xy


By KL Ang, Jan 2015

(b)

(i) When 0y ,

312

180

x

x where 2

1x

10 x

1x

0,1A

Equation of the line AB,

1 xy

Let

312

181

x

xx

0

12

181

3

x

xx

012

811

3

xx

01x or

012

81

3

x

112

83

x

8123x

64123x

412 x

2

5x

2

31

2

5y

2

3,

2

5B

(ii) area of the shaded region

x

x

xd

12

18

2

31

2

5

2

15

2

53

x

x

xd

12

18

8

95

2

53


By KL Ang, Jan 2015

2

55

128

8

9

x

x

12

52

2

5

152

58

8

9

4

5

3

58

8

9

12

140

8

9

3

10

8

9

24

107

46.4 square units (3 s.f.)

In Summary:

A simple routine question on integration. Part (a) is to facilitate the integration.

Solutions to O Level Add Math paper 2 2014 · · 2015-01-22Solutions to O Level Add Math paper 2...

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