Solutions to O Level Add Math paper 1 2014 - korlinang · Solutions to O Level Add Math paper 1 2014 By KL Ang, Jan 2015 Page 11 7. x The diagram shows a trapezium OABC in which OA

Solutions to O Level Add Math paper 1 2014

By KL Ang, Jan 2015 Page 1

1. Find the value of k for which the coefficient of x3 in the expansion of 6532 xkx

is 860. [5]

Solution :

Given that 6532 xkx , we have

rr

rkx

rkx

5

0

552

52

rr

r

rxk

r

5

0

52

5

When 3r , the coefficient of the x3 term,

332335402

!3

3452

3

5kkk

rr

rx

rx

6

0

663

63

When 3r , the coefficient of the x3 term,

5403!3

4563

3

6336

Therefore,

86040540 3 k

83 k

2k

[Analysis]

The question is looking for the x3 term in the expansion of 52 kx and 63 x .

We should use the term expression, rr

rkx

rkx

5

0

552

52 and r

r

rx

rx

6

0

663

63 .



Alternative solution:

Let 32

210

6586032 xxaxaaxkxy

2

21

54860323625

d

dxxaaxkxk

x

y

xaxkxkx

y86062330220

d

d2

432

2

2

86063120260d

d 323

3

3

xkxkx

y

Let 0x ,

86063120260323 k

433233 k

274323 k

83 k

2k

In Summary:

Binomial Theorem is a simple topic that will continue to feature in the future

examination. Each year, there will be one question on this.



2. The acute angles A and B are such that 8tan BA and 3tan B . Without using a

calculator, find the exact value of Asin . [5]

Solution :

Given that 8tan BA ,

8tantan1

tantan

BA

BA

8tan31

3tan

A

A

For 3

1tan A ,

AA tan2483tan

5

1tan A

Therefore, 26

26

26

1sin A


From the ratios of tan A and tan (A+B) , we know that 90,,0 BABA

1013 22 PT

6518 22 PR

TSRPQR ~ ,

565

1 ST

65

5ST

26

26

10

65

5

sin PT

STA

[Analysis]

90,0 BA and 1800 BA .

To find the exact value of trigo ratio of Asin , we know that A is an acute angle, in quadrant I , its

value must be positive. Since 8tan BA , we can deduce that angle BA is in quadrant I

too. Therefore 900 BA .

5

1

A

5

1

A 3 B

P Q

R

S

T



In Summary:

Exact value refers to non-rounding of the answer. Students should be well

aware of doing mathematics without calculator.



3. A particle moves along the curve 2

12

xy in such a way that the y-coordinate of the

particle is increasing at a constant rate of 0.03 units per second. Find the y-coordinate of

the particle at the instant that the x-coordinate of the particle is increasing at 0.12 units per

second. [5]

Solution :

Given that 2

12

xy

32d

d xx

y ------ (1)

12.0d

d03.0

x

y

4

1

d

d

x

y ------ (2)

(1) = (2),

4

12 3 x

33 2 x

2x

4

31

2

12

2y

[Analysis]

Rate of change t

x

x

y

t

y

d

d

d

d

d

d . Given that 03.0

d

d

t

y, when 12.0

d

d

t

x, we have

12.0d

d03.0

x

y , the connected rate of change. Take note that the question is asking for

the y-coordinate .

In Summary:

Typically question on rate of change is straight forward. The only issue is the

finding of y-coordinate, instead of the usual value of x.



4. Express 2

22

2

xx

x as the sum of 3 partial fractions. [6]

Solution :

Let 22

222

2

x

C

x

B

x

A

xx

x.

22222 CxxBxAxx

When 2x ,

2224 C

4C

When 0x ,

222

B

2B

When 1x ,

CBA 1132

3A

2

423

2

222

2

xxxxx

x


Let 22

222

2

x

C

x

B

x

A

xx

x.

22222 CxxBxAxx

BxABxCAxx 2244 22

2B

3A

4C

2

423

2

222

2

xxxxx

x

[Analysis]

The numerator is of degree 2, the denominator is of degree 3. This is a proper fraction.

In Summary:

Make sure the given fraction is a proper fraction. Must also check the

correctness of the decompositions with any value of x.



5. An experiment in Physics to find the focal length, f m, of a lens, requires the student to

place an object at a distance, u m, from the lens and to record the distance, v m, at which

the image is seen on the other side of the lens. The table below shows some results.

It is known that u, v and f are related by the equation fvu

111 . It is believed that an

error was made in recording one of the values of v.

(i) Plot v

1against

u

1 and hence determine which value of v, in the table above, is the

incorrect recording. [2]

(ii) Draw the straight line graph and use it to estimate a value of v to replace the incorrect

recording of v found in part (i). [2]

(iii) Estimate the value of f. [2]

Solution:

(i) From the graph, it shows that v = 0.263

is probably the incorrect recording.

(ii) From the graph, the estimate value of 30.41

v. v is 0.233

(iii) The v

1-axis intercept is 8.2.

u 0.150 0.200 0.250 0.300

v 0.603 0.299 0.263 0.201

u

1 6.67 5.00 4.00 3.33

u 0.150 0.200 0.250 0.300

v 0.603 0.299 0.263 0.201

v

1 1.66 3.34 3.80 4.98

[Analysis]

We were told that one of the data of v is wrong. By comparing the u and v, we suspect that the

first point, 0.603 might be the one. Then find, graphically the suitable value of v.



ufv

111

Therefore, 2.81

f. Hence f = 0.122

In Summary:

Graph plotting is a very time consuming activity. It is usually best to be done as

the last question of the paper.





6. (i) Prove that

tantanseccosec1

1

. [4]

(ii) Find, in radians, the acute angle for which

cot3tanseccosec1

1

. [2]

Solution:

(i)

tansectanseccosec1

tansec

tanseccosec1

1

22 tanseccosec1

tansec

cosec1

tansec

sin

11

cos

sin

cos

1

sin

sin1cos

sin1

tan

cos

sin

(ii) Given that

cot3tanseccosec1

1

.

tan

3tan

3tan2

3tan

Since 2

0

, 3tan 3

[Analysis]

Part (i) has numerous ways to do. Part (ii) is for 2

0



7.

The diagram shows a trapezium OABC in which OA is parallel to CB and O is the origin.

The side AB is parallel to the x-axis and the diagonal AC is parallel to the y-axis. The side

OA has equation xy 2 and the side OC has equation xy2

1 . The x-coordinate of A is h.

(i) Express the coordinates of A, B and C in terms of h. [5]

(ii) In the case where h = 4, find the area of the trapezium OABC. [2]

Solution:

(i) hhA 2, ,

2,h

hC ,

Equation of BC, hxh

y

2

2

When hy 2 ,

hxh

h

2

22

xh 22

7

hx4

7

[Analysis]

The first challenge is the find the x-coordinate of C. There are numerous ways to find the area.

Without knowing (i), we can still solve for area.

A

O

xy 2

x

y B

C xy2

1



hhB 2,

4

7

(ii)

When 4h ,

8,4A , 8,7B , 2,4C

the area of OABC

02880

04740

2

1

325614322

1

88462

1

21

the area of OABC is 21 square units

In Summary:

This question is very straight forward. Finding area of a quadrilateral is a

regular affair in this exam.



8. It is given that f(x) is such that f (x) xx 2cos4sin . Given also that 02

f

, show

that 14cos3f4''f xxx . [7]

Solution:

Let xy f , given that xxx

y2cos4sin

d

d and when

2

x , 0y .

xxxxx

yy d2cos4sind

d

d

Cxx

y 2

2sin

4

4cos where C is a constant.

When 2

x , 0y .

C2

sin

4

2cos0

C

4

10

4

1C

2

2sin

4

4cos

4

1 xxy

xxx

y2sin24cos4

d

d2

2

2

2sin

4

4cos

4

142sin24cos44

d

d2

2 xxxxy

x

y

xxxxyx

y2sin24cos12sin24cos44

d

d2

2

14cos34d

d2

2

xyx

y

[Analysis]

This is the only non-routine question in this paper. Firstly, xx

fd

d'f . Secondly

xx

fd

d''f

2

2

. Let xy f , xxx

y2cos4sin

d

d . 14cos34

d

d2

2

xyx

y.

Without knowing (i), we can still solve for area.



In Summary:

The best part of this question is that if you make any mistake along the way, you

will know it in the end and let you work backward to correct the error.



9. The equation of a curve is 552 axaxy where a is a constant.

(i) In the case where a = 2, find the set of values of x for which the curve lies completely

above the line y = 9. [3]

(ii) In the case where a = 4, show that the line 2 xy is a tangent to the curve. [2]

(iii) Find the other value of a for which the line 2 xy is tangent to the curve. [3]

Solution:

(i) Given that 552 axaxy , when 2a , 352 2 xxy

For 9y , we get

9352 2 xx

01252 2 xx

0432 xx

From the graph, 4x or 2

3x

2

3or4: xxx

(ii) When 4a , the curve is 154 2 xxy . The intersection of the line and the curve is

2154 2 xxx

0144 2 xx

the discriminant of the quadratic equation,

014442

Hence the line 2 xy is tangent to the curve.

[Analysis]

(i) solve 9352 2 xx . (ii) To show 154 2 xxy and 2 xy has a repeated root.

(iii) To find a for 552 axaxy tangent to 2 xy .

4 O x

2

3

| |



(iii) Let 552 2 axaxx

0342 axax

Set the discriminant to zero,

03442

aa

041216 2 aa

034 2 aa

014 aa

4a or 1a

1a is the other value.

In Summary:

(i) needs answer in Set notation. (ii) is to show discriminant is zero. (iii) is to let

discriminant to be zero.



10.

A gardener uses 130 m of fencing to enclose a plot of the shape shown above. The shape

consists of two equilateral triangles of side x m and a rectangle with sides x m and l m.

(i) Show that the area of the plot is 2

34130 2xx m2. [3]

(ii) Given that x can vary, find the value of x for which the area of the plot is stationary.

[4]

(iii) Explain why this value of x gives the gardener the largest area possible. [1]

Solution:

(i) The perimeter, 13042 xL

xL 265 ---------- (3)

The area, xxLxA2

3

2

2

3xLxA ---------- (4)

Substitute (3) into (4),

2

2

3265 xxxA

2

34130 2xxA

[Analysis]

(i) This is about surd conjugates, simplifying surd. (ii) is to find the turning point and (iii) is

to determine the nature of this turning point.

L m

x m

60°

x m

60°



(ii)

2

342130

d

d x

x

A

xx

A3465

d

d

Let 0d

d

x

A,

x34650

34

65

x

3434

3465

x

22 34

3465

x

13

3465 x

345 x

(iii) Since 034 , the coefficient of the x2 term is negative, hence equation

2

34130 2xxA

has a maximum turning point at 345 x .

In Summary:

Simplifying surds and Properties of quadratic curve.



11. The point P lies on the curve xxy ln . The tangent to the curve at P is parallel to the line

32 xy .

(i) Find the coordinates of P. [5]

The normal to the curve xxy ln at P meets the line 32 xy at point Q.

(ii) Show that the x-coordinate of Q is 2ek , where k is a constant to be found. [3]

Solution:

(i) The gradient of point P is 2.

xxy ln

The gradient of the curve,

xxx

x

y 1ln

d

d

xx

yln1

d

d

Therefore,

xln12

1ln x

ex

ey

e,eP

(ii) The gradient of the normal at P is 2

1 .

The equation of the normal at P is,

e2

1e xy

e2

3

2

1 xy

Let e2

3

2

132 xx

e2

33

2

5x

e365

1x

[Analysis]

Gradient of the curve at the point of tangent is the same as the gradient of the line. Forming

equation of the normal to find the intersection.



e25

3x

In Summary:

A simple routine question on coordinate geometry. The handling of e may give

problem to some students.



12. A curve has the equation 4322 xy .

(i) Explain why the lowest point on the curve has coordinates

4,

2

3. [1]

(ii) Find the coordinates of the points at which the curve intersects the x-axis. [2]

(iii) Sketch the graph of 4322 xy . [3]

(iv) Using your graph, state the number of solutions to each of the following equations.

(a) 74322

x [1]

(b) 34322

x [1]

(c) 024322

x [1]

Solution:

(i) Given that 4322 xy ,

0322x

44322

x

4y , y has a minimum value of 4 .

When 2

3x , 4y .

(ii) When 0y ,

43202 x

4322x

232 x

322 x

2

5x or

2

1x

[Analysis]

Sketching Quadratic curve and modulus quadratic curve.



(iii) the graph of 4322 xy

(iv)

(a) 74322

x , 2 solutions.

(b) 34322

x , 4 solutions.

(c) 24322

x , no solution.

In Summary:

Questions on Curve sketching will continue to appear in the future papers.

4

O x

2

3

| |

5

|

2

1

2

5

y

7y

3y

2y

Documents

Solutions to O Level Add Math paper 1 2014 - korlinang · Solutions to O Level Add Math paper 1 2014 By KL Ang, Jan 2015 Page 11 7. x The diagram shows a trapezium OABC in which OA