PYQ SM015

MATHEMATICS SM015

14. Given a polynomial P(x)  2x3  ax2  bx  30 has factors (x  2) and (x  5) .

(a) Find the value of the constants a and b.

(b) Factorize P(x) completely.

(c) Obtain the solution set for P(x)  0 ( 2009/10 )

15. A polynomial f (x)  px3  ( p  q)x2  ( p  2q)x  1 has a factor x  1.

(a) Express q in terms of p.
(b) Write f (x) in terms of p and x. Determine the quotient when f (x) is divided by

x 1.
(c) Hence, find the value of p if x  3 is one of the roots for f (x)  0 . Using the value

of p, factorize f (x) completely. ( 2010/11 )

16. Dividing M (x)  x2  ax  b by (x 1) and (x 1) give a remainder of -12 and -16.

Determine the values of a and b . ( 2010/11 )

17. Express 6x 13 in the form of partial fractions. ( 2011/12 )
(3x  4)2

18. The polynomial P(x)  x3  2x2  ax  b , where a and b are constants, has a factor

(x  2) and leaves a remainder of a3 when it is divided by (x  a) .

(a) Find the values of a and b .
(b) Factorise P(x) completely by using the values of a and b obtained from part (a).

Hence, find the real roots of P(x) , where a and b are not equal to zero. ( 2011/12 )

2x3  7x2 17x 19 ( 2012/13 )
19. Express 2x2  7x  6 in the form of partial fractions.

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MATHEMATICS SM015

20. The polynomial P(x)  2x3  ax2  bx  24 has a factor (x  2) and a remainder 15 when

divided by (x  3) .

(a) Find the values of a and b

(b) Factorise P(x) completely and find all zeroes of P(x) ( 2012/13 )

21. Express x2 in partial fractions form. ( 2013/14 )

x2  3x  2

22. A cubic polynomial P(x) has a remainder 3 and 1 when divided by (x 1) and

(x  2) respectively.

(a) Let Q(x) be a linear factor such that P(x)  x 1(x  2)Q(x)  x   , where  and

 are constants. Find the remainder when P(x) is divided by (x 1)(x  2) .

(b) Use the values of  and  from part (a) to determine Q(x) if the coefficient of x3 for

P(x) is 1 and P(3)  7 . Hence, solve for x if P(x)  7  3x . ( 2013/14 )

23. Given that (x  2) is a factor of a polynomial f (x)  ax3 10x2  bx  2 where a and

b are real numbers. If f (x) is divided by (x 1) the remainder is -24, find the values of

a and b . Hence, find the remainder when f (x) is divided by (2x 1) . ( 2014/15 )

24. Expand (x  a)(x  b)2 , a and b are real numbers with b  0 . Hence, find the values of

a and b if (x  a)(x  b)2  x3  3x  2 . Express x4  4x2  5x 1
in the form of partial
x3  3x  2

fractions. ( 2014/15 )

25. Express 5x2  4x  4 in the form of partial fractions. ( 2015/16 )
(x2  4)(x  2)

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MATHEMATICS SM015

26. A polynomial P(x)  2x4  ax3  bx2 17x  c where a , b and c are constants, has
factors (x  2) and (x 1) . When P(x) is divided by (x 1) , the remainder is 8. Find the
values of a , b and c . Hence, factorise P(x) completely and state its zeroes. ( 2015/16 )

27. Express x2 in partial fractions form. ( 2016/17 )
x2  2x  3

28. (a) Polynomial P(x) has a remainder of 3 when divided by x  3 . Find the remainder
of P(x)  2 when divided by x  3 .

(b) Polynomial P1(x)  x3  ax2  5bx  7 has a factor x 1 and remainder R1 when

divided by x 1, while a polynomial P2 (x)  x3  ax2  bx  6 has a remainder
R2 when divided by x 1. Find the value of the constants a and b if R1  R2  5 .

Hence, obtain the zeroes for P1(x) ( 2016/17 )

29. Express 3x2  5 in partial fractions. ( 2017/18 )
(x  3)(x2  2)

30. Given the polynomial P(x)  x2  4 and Q(x)  x4  x3  2x2  x  28.

(a) Find all zeroes of P(x)

(b) When Q(x) is divided by P(x) , the remainder is 14x  52 . Use the remainder

theorem to find the value of  and  .

(c) Using the values of  and  obtained from part (b), find the remainder when

2Q(x)  x is divided by P(x) . ( 2017/18 )

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MATHEMATICS SM015

SUGGESTED ANSWERS

1. 3  2  5
x  2 x  1 (x  1)2

2. a  12 , b  26 .. P(x)  2(x 1)2(x  2)(2x  3)

3. P(x)  (x  3)2(1  2x) , 13x  18  1 3 2
P(x) x3 (x  3)2 1 2x

4. p  3 , q  2 P(x)  (x  2)(2x 1)(x 1)
5. a  5 , b  6
6. a  2 , b  5 , third factor (x  3) , 1  1  2

x 1 x 2 x 3
7. A  2 , B  3 , C  10 , D  0

33
8. (a) P(x)  (x  3)(x  2)(x 1)

(b) a  4, b  12, c  9

9.  1  x  4
4(x  2) 4(x2  2x  4)

10. (a) Q(x)  x3  6x2  3x 10 (b) a  1,b  3, P(x)  (x  2)(x  3)2

11. 4  1  5
1 x 1 x (1 x)2

12. m  3,n  5, P(x)  (x  3)(3x  2)(x 1) x  3, x  1, x  2 , x  2
3

13. 1   x  4
2(x  2) 2(x 2  2x  2)

14. (a) a  3 b  29
(b) P(x)  (x  2)(x  5)(2x  3)
(c) x  (,2)  ( 3 ,5)
2

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MATHEMATICS SM015

15. (a) q 1 p

(b) f (x)  px3  x2  2  px 1 , quotient is px2  1 px 1
(c) p   2 ; f (x)   1 x 12x 1x  3

33

16. a  2,b  15

17. 6x 13 = 2  5
(3x  4)2 3x  4 (3x  4)2

18. (a) a  0;  2, b  0; 4
(b) For a  b  0 ; P(x)  x(x2  2)
For a  2,b  4 ; P(x)  (x  2)(x2  2)
x   2, 2,2

19. 2x3  7x2 17x 19 = x  5  3
2x2  7x 6 2x  
3 x 2

20. (a) a  7,b  10
(b) (x  2)(2x  3)(x  4); 2,- 3 ,4
2

21. x2 1 1  4
x2  3x  2 x 1 x  2

22. (a) R(x)  2x  5
(b) Q(x)  x 1; x  2, x  0

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MATHEMATICS SM015

23. a  3, b  9; - 75
8

24. x3  (2b  a) x 2  (2ab  b2 )x  ab2 ; a  2, b  1; x  x 1 2  2  x 3
 x 1
 12

25. 5x2  4x  4 = 2  3  4
(x2  4)(x  2) x  2 x  2 (x  2)2

26. a  13, b  12, c  10; P(x)  (x  2)(x 1)(x  5)(2x 1);  5,2, 1 ,1
2

27. 1 9  1
4(x  3) 4(x 1)

28. (a) 5 (b) a  11, b  1 , zeroes of P1(x) :  6  29 ,1
29. 2  x  3

x 3 x2  2

30. (a) 2,  2
(b)  1,   10
(c) 29x 104

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MATHEMATICS SM015

TOPIC 7 : TRIGONOMETRY

1. If tan x   t , find sin x and cos x in terms of t. Hence, solve cos x  7sin x  5 for
2

0 x ( 2004/05 )

2. Show that tanA  B  tan A  tan B ( 2005/06 )

1 tan A tan B

3. If tan x   t , express sin x and cos x in terms of t. Hence, find all values of t which
2

satisfy 3cos x  4sin x  5 ( 2005/06 )

4. Show that cos 6x  cos 2x(4cos2 2x  3) ( 2007/08 )

5. (a) Let P(x, y) be a point on a unit circle with center O at the origin, such that OP make

an acute angle  with the positive x-axis. Prove that sin 2   cos 2  1and hence show

that sec2  1  tan2  . ( 2007/08 ).
(b) Show that the equation cos x(sin x  cos x) 1  0 can be reduced to

tan x(1  tan x)  0 . Hence, solve for x on the interval 0,2 .

6. Find A and B if sin 2x cos 3x  Asin 5x  Bsin x ( 2008/09 )
( 2009/10 )
7. Prove that for  n , where n is an integer,
sin  tan sec4 .

cos (1  2sin 2   sin 4  )

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MATHEMATICS SM015

8. (a) Prove that cos 3x  4cos3 x  3cos x. Hence, show that

cos3 2x  1 cos 6x  3cos 2x

4

(b) Use the above identity to find all the solutions in the interval 1800  x 1800 of the

equation 2cos 3x  cos 2x 1  0. ( 2010/11 )

9. (a) Given that tan p  3 and tan p  1.
34

Express tan 7 p in the form a  b where a and b are integers.
12

Hence, show that tan 7 p  1 .
6b

(b) Find R and  such that the expression 9sin x 12cos x can be expressed in the form
of Rsin(x  ) , where R  0,00    900.

Hence, if 9sin x 12cos x  5solve for x in the interval 00  x  3600. ( 2011/12 )

10. Prove that 1 tan 2x tan x  sec 2x . ( 2012/13 )

11. Given f ( )  3sin  2cos ( 2012/13 )
(a) Express f ( ) in the form of Rsin(  a) , where R  0,00  a  900.
Hence, find the maximum and minimum values of f ( ) .

(b) Solve f ( )  13 for 00    3600.
2

12. State the values of R and  such that 3sin  6cos  Rsin(  ) where R  0 and

00    900.Hence, solve 3sin  6cos  5 for 00    1800. ( 2013/14 )

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MATHEMATICS SM015

13. (a) Show that sin   sin   cot    .
cos  cos   2

(b) Use trigonometric identities to verify that

2 tan  1  tan2 

(i) sin   2 (ii) cos  2
tan2  
1 1 t an 2
2 2

Hence, solve the equation 3sin  cos  2 for 00    1800. Give your answer

correct to three decimal places. ( 2013/14 )

14. Solve the equation 2cos2 x 1  sin x , for 0  x  2. Give your answer in terms of  .
( 2014/15 )

15. (a) Express sin 6x  sin 2x in a product form. Hence, show that
sin 6x  sin 2x  sin 4x  4cos3xsin 2x cos x .

(b) Use the result in (a) to solve sin 6x  sin 2x  sin 4x  sin 2x cos x for 00  x  1800.
( 2014/15 )

16. Given cos ec2 x  cot x  3, show that cot 2 x  cot x  2  0 . Hence, solve the equation

cos ec2 x  cot x  3for 0  x  . ( 2015/16 )

17. (a) Determine the values of R and a , where R  0 , and 00    900 so ( 2015/16 )
tha 3sin  4cos  Rsin( ) .

(b) Hence, solve the equation 3sin  4cos  2 for 00    3600 .

18. Show that sin 2 x  1  cos x . Hence, solve sin 2 x  cos 2x for 0  x  360
1 cos x 1 cos x

( 2016/17 )

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MATHEMATICS SM015

19. Consider a function f (x)  3 cos 2x  2sin 2x

(a) Express f in the form of R cos(2x ) for R  0 , 0    90 and  to the nearest

minute. State the maximum and minimum values of f .

(b) Hence, solve 3 cos 2x  2sin 2x   2 for 0  x  180 . Give your answer to the

nearest minute. ( 2016/17 )

20. Solve the equation cos  cos 5  2cos 3 for 0     . Give your answer in terms of
 . ( 2017/18 )

21. Express cos  2 sin in the form Rsin   , where R  0 and  is an acute angle.

Hence,

(a) Solve the equation cos  2 sin  3 by giving all solutions between 0 and
2

360 .

(b) Show the greatest value of 1  5  3 ( 2017/18 )
cos  2 sin  5 22

SUGGESTED ANSWERS

1. sin x  2t , cos x  1  t 2 , x  0.205 , 0.705
1 t2 1 t2

2. Shown

3. t   1
2

4. Shown

5. (a) Shown (b) 0, ,  , 5 ,2
44

6. A  1 , B   1
22

7. Shown

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MATHEMATICS SM015

8. Shown (b) –90, 90, –41.4, 41.4 , 180

9. (a) tan 7 p  2  3, tan 7 p  1 (b) 15sin(x  53.130 ) , x  107.40 ,326.340
12 6 3

10. Shown

11. (a) 13 sin(  33.690 ) , min f ( )   13 ,max f ( )  13

(b)   78.690 ,168.690

12. R  45,  63.430 ,  97.10

13. (b)   20.7410 ,122.3200

14. x   , 5 , 3 (b) x  00 ,25.170 ,900 ,94.830 ,145.170 ,1800
66 2

15. (a) 2cos 4xsin 2x

16. x  0.148rad, 3 rad (b)   76.70 ,209.50
4

17. (a) R  5,  53.10

18. x  141.35 , 218.65

19. (a) 3 cos 2x  2sin 2x  7 cos(2x  496 ') , max = 7 , min =  7

(b) x  8534', 14324'

20.   0,  ,  , 5 ,
62 6

 21. cos  2 sin  3 sin   35.3

(a)   114.7 , 354.7

(b) Shown.

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MATHEMATICS SM015

TOPIC 8 : LIMIT AND CONTINUITY

1. Find the following limit, if they exists :

(a) lim x3 1 (b) lim x  3 ( 2003/04 )
x1 x 1 x x  9

2. Find a value of k so that the function

f (x)  kx2 if x  2
is continuous. Hence, by using definition of
2x  k if x  2

f '(a)  lim f (x)  f (a) , determine whether f '(2) exists or not. ( 2003/04 )
xa x  a

3. Let f (x)  2x for x  0
x2 1 for x  0

(a) Find lim f (x) ( 2003/04 )

x0

(b) Is the function f is continuous at x  0 ? Give your answer.

4. Find the following limit, if they exists :

(a) lim e3x 1 (b) lim x2  2 ( 2004/05 )
x0 ex 1 x 3x  6
Page 62
5. (a) State the condition for the function f to be continuous at x  c .

(b) Given that q(x)  x2  4, x  2
x2

3, x  2

(i) Sketch the graph of q.
(ii) Discuss the continuity of q at x  2

BKS 2018/19

MATHEMATICS SM015

(c) Determine the values of A and B such that

f (x)  x, x 1
Ax  B,
 2x, 1  x  4 is continuous on the interval  ,. ( 2004/05 )

x4

6. A function f is defined by f (x)  x
x2 9

(a) State the domain of f.

(b) Find the vertical asymptotes.

(c) Determine lim f (x) and lim f (x) . Hence, state the horizontal asymptotes
x x  

( 2004/05 )

7. Find the following limits (b) lim  x 1 x  1  ( 2005/06 )
(a) lim x2  4 1 x
x2 x  2 x0

8. If function g is defined as

Ax  2, 3 x 1

g(x)  x2  Bx  A, 1  x  2

1 1, 2x5
x

Find the values of A and B such that g is continuous in the interval  3, 5 ( 2005/06 )

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MATHEMATICS SM015

9. Evaluate each of the following limits, if it exists.

(a) lim x  4 (b) lim 3x4  x ( 2006/07 )
x4 x  2 x2 6
x  

x, x  1

10. Let g(x)  ax  b, 1  x  4 . Find the values of a and b so that g is continuous on

 2x, x  4

the interval  , ( 2006/07 )

5x2  m, x2

11. Given f (x)  k, x  2 . Find the value of m such that lim f (x) exists.
mx3 1, x2
x2

Hence, find the value of k such that f is continuous at x  2 . ( 2006/07 )

12. Given f x  x 2x  3 3 . Find

1x 

(a) the domain of f,

(b) the x-intercept and y-intercept of f,

(c) the vertical asymptote(s) of f,

(d) lim f x and lim f x. Hence, state the horizontal asymptote of f.
x x

( 2007/08 )

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MATHEMATICS SM015

13. The function f is defined as

 x2  x  12 , x3

 x3
A
f x  , x3

2x  B , 3 x4

C , x  4

(a) Find lim f x and lim f x.
x3 x3

(b) Use the definition of continuity to determine the values of A and B if f is

continuous at x  3 .

(c) For what values of C is f discontinuous at x  4 ? ( 2007/08 )

14. Given f x  x x x 1 2

1x 

x , x 1
(x  , x 1
(a) Show that f x is equivalent to g( x)   2)
 (x x
 2)

(b) Determine the asymptotes and the points of discontinuity of g

(c) Find the points of intersection of g(x) with the straight line y  x  2 ( 2008/09 )

15. A function f is defined by

 34 , x  4

 0 , x2
f (x)   17, , x4

 x4  3x2  4 , x  4, x  3, x  2, x  4

 x2  x  6

Evaluate lim f (x). ( 2009/10 )

x2 Page 65

BKS 2018/19

MATHEMATICS SM015

16. (a) A function f is given as

x 1 , x0
 , x0
f (x)   2 , x0

e 2x

Find lim f (x), lim f (x) and lim f (x) .
x0 x0 x0

Hence, determine whether f is continuous at x  0. Give a reason to your answer.

(b) Evaluate lim e2x  e2x . ( 2009/10 )
x0 e x  ex

17. (a) State the conditions of continuity of a function at a point x  c .
(b) A function f defined by

f (x)  x2
-5 x  2

x 2  3x 10
A 2x3

Ax  B x3

is continuous at x  2 and x  3 . ( 2010/11 )
(i) Find lim f (x) ( 2010/11 )

x2

(ii) Determine the values of the constants A and B.

18. (a) Evaluate (ii) lim 2  x 2  5
(i) lim 4x 2  2x 1
x3 x  3
x x 1

(b) If lim f (x)  5 1 , find lim f (x)
x4 x  2 x4

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MATHEMATICS SM015

19. Evaluate (b) lim x 1  x ( 2011/12 )
(a) lim x4  16
x x
x2 x  2

20. (a) Given that f (x)   x3  64 , x4
 x  4
 40,
x4

(i) Find lim f (x)

x4

(ii) Is f continuous at x  4 ? Give your reason.

(b) Determine the value of A and B such that the function

 Ax  B x  1
h(x)  2x2  3Ax  B 1 x 1

 4 x 1

is continuous for all values of x. ( 2011/12 )

1 e x x 1
 1
21. Given that f ( x)   x 1 Find lim f (x) and lim f (x) . Does the
x 1 x1 x1
 2  x

lim f (x) exist? State your reason. ( 2012/13 )

x1

22. Find the following limits:

(a) lim 2x2  x  4 (b) lim 3  x 7 ( 2012/13 )
x2 4
x 1 x2 x2

BKS 2018/19 Page 67

MATHEMATICS SM015

| x2  x  2 | x  0,2
 x2
23. Given that f ( x)   x2  2x
0

Find the lim f (x) . Is f (x) continuous at x  2 ? ( 2012/13 )

x2

24. (a) Find the value of m if lim mx  3x2  3
x0 4x  8x2

(b) Evaluate lim 3  x  3 ( 2013/14 )
x0 x

25. (a) State the definition of the continuity of a function at a point. Hence,

find the value of d such that

f ( x)  e3x  d x  0 is continuous at x  0 .
3x  x0
5

(b) A function f is defined by

f (x)   x2 1 x 1

k(x 1) x  1

Determine the value(s) 0f k if f is:

(i) Continuous for all x  R . ( 2013/14 )
(ii) Differentiable for all x  R .

26. Find the limit of the following, if it exist

(a) lim x  3 (b) lim 2x 1
x3 x3  27 x x2  9

(c) lim x2  3x  4 ( 2014/15 )
x4 x  2

BKS 2018/19 Page 68

MATHEMATICS SM015

f ( x)  1xe6x x0
27. Given that  0  x  4 where C is a constant.
 3  x
 C x4

(a) Determine whether f (x) is continuous at x  0 .

(b) Given that f (x) is discontinuous at x  4 , determine the value of C.

(c) Find the vertical asymptote of f (x) . ( 2014/15 )

28. Evaluate the following (if exists) (b) lim 1  x ( 2015/16 )
(a) lim x2  4x 12 x1 1  x
x2 x  2

29. (a) Evaluate lim 2x2  3x
x 5x  1


 5  px
  2  x  1
(b) Given f (x)  x 2  px  q 1 x  2

 x2  4 x2

 x2

(i) Find the value of p and q if f (x) is continuous for all real values of x.

(ii) Sketch the graph of f (x) using the values p and q obtained in part (i).

( 2015/16 )

30. Evaluate the following (if exists)

(a) lim x2 (b) lim (2  x)(x 1) ( 2016/17 )
x4 16 x (x  3)2
x2

BKS 2018/19 Page 69

MATHEMATICS SM015

2x lim f (x) and lim f (x) . Is f
31. (a) Given f (x)   5x . Compute
x0 x0
x
continuous at x  0 ?Give your reason.

(b) The continuous function g is defined by

g(x)  5 x, x  a
3x 1, x  a

Find the value of a. ( 2016/17 )

32. Evaluate the following (if exists)

(a) lim x3  8 (b) lim 5x  7 ( 2017/18 )
x2 x2  2x x 6x  5

33. State the conditions for continuity of f (x) at x  a .

(a) By using the conditions for continuity of f (x) at x  a . find the values of m and n

such that

f (x)  n  2 cos x, x0
2  mx2, 0 x2

m  x, x2

is continuous on the interval  ,.

(b) If m  2 and n  4 , determine whether f (x) is differentiable at x  2 or not.
( 2017/18 )

BKS 2018/19 Page 70

MATHEMATICS SM015

SUGGESTED ANSWERS

1. (a) 3 (b) 0

2. k  4 , f '(2) does not exist.
3

3. (a) lim f (x)  1 , f(x) is continuous at x  0 because lim f (x)  f (0)  1
x0 x0

4. (a) 3 (b) 1
3

5. (a) (i) f (c) is defined

(ii) lim f (x)  lim f (x)  lim f (x) exits
xc xc xc

(iii) lim f (x)  f (c)
xc

(b)

(c) A  3, B  4

6. (a) D f   ,3 3,

(b) V.A x  3 and x  3

(c) lim f (x)  1, lim f (x)  1 , H.A y  1 and y  1
x x  

7. (a)  4 (b)  1
2

8. A   11, B  1
2

9. (a) 2 (b) 3

10. a  3,b  4

11. m  3 , k  23

BKS 2018/19 Page 71

MATHEMATICS SM015

12. (a) Df   \  3,1 (b) x  intercept  3 , y  intercept  1 (c) x  3, x  1
2

(d) y  0

13. (a)  7 , 6  B (b) A  7 , B 13 (c) C  5 or C  ,5   5, 

14. (b) VA: x  2 , HA: y  1, x  2,1 (c) (4,2) and (1,1)

15. 4

16. (a) (i) lim f (x)  1, lim f (x)  1 lim f (x)  1 (b) 2
x0 x0 x0

(ii) f is not continuous at x  0 because f (0)  lim f (x)
x0

17. (a) (i) f (c) is defined

(ii) lim f (x)  lim f (x)  lim f (x) exits
xc xc xc

(iii) lim f (x)  f (c)
xc

(b) (i)  1 (ii) A   1 , B  2

7 77

18. (a) (i) 2 , 3 (b) 7

(ii)

2

19. (a) 32 (b) 0

20. (a)(i) lim f (x)  48 (ii) lim f (x)  f (4) (b) A  3 , B   1
x4 x4
44

21. lim f (x) does not exists
x1

22. (a) -2 (b)  1

24

23. f (x) is not continuous at x  2

24. (a) 12 (b)  1

23

25. (a) ln 5 (b)(i) k is real number (ii) 2

26. (a) 1 (b) -2 (c) 20
27

27. (a) f (x) is continuous at x  0 (b) C  R \{10} (c) x  3

28. (a) lim f (x) does not exists (b) 1
x2 2

BKS 2018/19 Page 72

29. (a)  2 MATHEMATICS SM015

5 (b)(i) p  1, q  2 (ii)

-2 1 2

30. (a) 1 (b) 1
32

31. (a) lim f (x)  2 , lim f (x)  2 , f is discontinuous at x  0 because lim f (x)
x0 x0
x0

does not exist.

(b) a  1

32. (a) 6 (b) 5
6

33. (a) (i) f (a) is defined

(ii) lim f (x)  lim f (x)  lim f (x) exits
xa xa xa

(iii) lim f (x)  f (a)

xa

(b) m   4 , n  4
3

(c) f (x) is non-differentiable at x  2

BKS 2018/19 Page 73

MATHEMATICS SM015

TOPIC 9 : DIFFERENTIATION

1. If y  xex ( 2003/04 )
(a) find dy and d 2 y
dx dx2
(b) show that d 2 y  2 dy  y  0
dx2 dx

2. Given that x  t  xt and 2ty  y2  3, find dx and dy . Hence, find the values of dy
dt dt dx

when x  2 . ( 2003/04 )

3. Parametric equations of a curve is given by x  2t and y  3t 2  1 . Find
t2 1 t2 1

(a) dy in terms of t.
dx

(b) d2y when t  1 ( 2003/04 )
dx 2

4. If xy  2(x  y)2 , find the following values at the point 1, 2

(a) dy (b) d 2 y ( 2004/05 )
dx dx 2

5. (a) Given y  2x2 , find dy
dx

(b) If y  ex ln(1 x) , show that x  12  d2y  dy   xe x ( 2004/05 )
 dx 2 dx 

BKS 2018/19 Page 74

MATHEMATICS SM015

6. Given x  2t  1 and y  t  4 , where t is non-zero parameter.
tt

(a) Show that dy  1 1  9 
dx 2 2
2t 1

(b) Hence, deduce that dy  1 for all t.
dx 2

(c) Find d 2 y when t  1 ( 2004/05 )
dx 2

7. By taking logarithm on both sides of the equation y  (x2 ) x , show that

y ' 1 (ln x  2)(x2 ) x ( 2005/06 )
x

8. Given y  Ax  B , where A and B are constants and x  0 .
x2

(a) Find dy and d 2 y . Hence, show that x2 d 2 y  2x dy  2 y  0 .
dx dx2 dx2 dx

(b) Find the values of A and B if y  3 and y ' 6 when x  1 ( 2005/06 )

9. Consider the parametric equations, x  t 2 and y  t3  3t ( 2005/06 )

(a) Evaluate dy at the point 3, 0

dx

(b) Find the point x, ywhere dy is not defined.

dx
(c) Determine the interval of t such that d 2 y  0

dx 2

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MATHEMATICS SM015

10. Given that x 1 and y  1  t 2 , where t is a non-zero parameter. Show that
1 t2 t

dy  1 1  t 2 3 . Hence find its value at the point  1 , 0 . ( 2006/07 )
dx 2  t  2 

11. If y  ex ln x , show that x2  d2y  dy   e x (1  x)  0 ( 2006/07 )
 dx 2 dx 
( 2006/07 )
12. Let x2 y2  2xy  4y  4 . ( 2007/08 )
( 2007/08 )
(a) Find the values of A, B and C if dy  Ay(1 xy)
dx x(xy  B)  C ( 2007/08 )
( 2008/09 )
(b) Determine the value of d 2 y at the point 2, 2
Page 76
dx 2
13. If y3  ln(x3 y2 ) for x  0 , y  0 , then find dy when y  1 .

dx
14. Let y  x(ln x)2 , x  0. Show that

x2 d 2 y  x dy  y  2x.
dx2 dx

15. (a) Find dy when x  0
dx

y  e2x (2x3 1) .
x 1

(b) Given x  3t  2 , y  2t  3 ,t  0. Show that
tt

dy  2  1  13 2 
dx 3 3 3t 2  

Hence, find d2y.
dx 2

16. Given ln y  exy, find dy .
dx

BKS 2018/19

MATHEMATICS SM015 ( 2008/09 )

17. If y  2x2  5x  3, determine the domain of dy and find the respective
dx

intervals in which dy  0 and dy  0.
dx dx

18. Given that y  et  et and x  et

(a) Find the point (x, y) on the curve where dy  0.
dx

(b) Solve for t if

 d2y 2  dy 1  0. ( 2008/09 )
dx 2 dx

19. If y  x  ex , show that d 2 x  ex  0. ( 2009/10 )
dy 2 (1 ex )3 ( 2009/10 )
( 2010/11 )
20. A parametric curve is given by x  t  1, y  t  1, t  0.
tt

(a) Find dy in terms of t and evaluate it at t  2.
dx

(b) Find the value of d2y at t 1, and evaluate d2y
lim .
dx 2 dxt 2

21. Find dy for each of the following :
dx

(a) y  ln x5

(b) xy2  yex  3

22. If f is a function with f '(1)  2 , find lim f (x)  f (1) . ( 2010/11 )
x1 x  1

BKS 2018/19 Page 77

MATHEMATICS SM015

23. Consider the parametric equations
x  2t  t 1 , y  2t t 1 , t  1

(a) Show that dy  2t 2 1
dx 2t 2 1

(b) Evaluate dy at the point 1,3
dx

(c) Find d2y in term of t. Hence show that d2y  8 ( 2010/11 )
dx 2 dx 2 y3 ( 2011/12 )

24. Find dy for the following equations:
dx

(a) y  32x1
(b) e xy  y  5x

25. Given that x  1 and y  1  t 2 , where t is a non zero parameter.
1 t2 t

(a) Show that dy  1  t 2 .
dx t3

(b) Find d 2 y when t=1. ( 2011/12 )
dx2

26. (a) If y  sin(x2  1) , show that x d 2 y  dy  4x3 y  0 . ( 2011/12 )
dx2 dx

(b) Find the gradient of a curve xexy  e2x  e3y at (0,0).

BKS 2018/19 Page 78

MATHEMATICS SM015 ( 2012/13 )
( 2012/13 )
27. (a) Given that y  1
2x 1

(i) By using the first principle of derivative, find dy .
dx

(ii) Find d 2 y
dx2

(b) Find dy of the following:
dx

(i) y  e2x tan x
(ii) y  xsec x

28. (a) Given 3y2  xy  x2  3. By using implicit differentiation,

(i) Find the value of dy at x  1.
dx

(ii) Show that 6 y  x d2y  6 dy 2  2 dy 20.
dx2  dx  dx

(b) Consider the parametric equations

x  3t  2 , y  3t  2 where t  0.
tt

(i) Show that dy  1 4 .
dx 3t 2  2

(ii) Find d 2 y when t  1.

dx2

BKS 2018/19 Page 79

MATHEMATICS SM015

29. (a) Find dy if y  cosec{sin[ln(x 1)]}
dx

(b) Obtain the second derivative of y  cos3x and express your answer in the simplest
e2x

form. ( 2013/14 )

30. (a) Find the derivative of f (x)  1 by using the first principle.
x 1

(b) Use implicit differentiation to find:

(i) dy if y ln x  e xy
dx

(ii) the value of dy if 1  1  3 when x  1 . ( 2013/14 )
dx y x 2

31. A curve is defined by parametric equations x  ln(1  t), y  et2 fort  1

(a) Find dy and d 2 y in terms of t.
dx dx2

(b) Show that the curve has only one relative extremum at (0,1) and determine the nature

of the point. ( 2013/14 )

32. Consider the parametric equations for the curve x  cos3  and y  sin 3 , 0   2

(a) Find dy and express your answer in terms of 
dx

(b) Find the value of dy if x  2 .

dx 4

(c) Show that d 2 y  1 d2y at    . ( 2014/15 )
. Hencecalculate
dx2 3cos4  sin  dx2 3

BKS 2018/19 Page 80

MATHEMATICS SM015

33. (a) Use the first principle to find the derivative of g(x)  x 1

(b) Given that e y  xy  ln(1  2x) 1, x  0 .

Show that (e y  x) d 2 y  e y ( dy)2  2 dy  4  0 . Hence find the value of d 2 y at
dx2 dx dx (1  2x)2 dx2

the point (0,0). ( 2014/15 )

34. Find the derivative of the following functions: ( 2015/16 )
(a) f (x)  cot 4x2 1
(b) f (x)  e2x ln(3x  4)

35. A curve is given by parametric equations x  t  1 , y  t  1
tt

(a) Find the dy d 2 y in term of t.
and
dx dx2

(b) Obtain the coordinates of the stationary points of the curve and determine the nature

of the points. ( 2015/16 )

36. If y 2  2y (1  x2  x2  0 , show that dy  x . ( 2015/16 )

dx (1  x 2 )

37. (a) Find the value of k if the slope of the curve x3  kx2 y  2y2  0 at the point (-1,1) is -3.

(b) Given y  sin x
1  cos x

(i) Find dy d 2 y in terms of x.
and
dx dx2

(ii) Hence, show that d 3 y  y d 2 y  ( dy)2  0 .
dx3 dx2 dx

( 2015/16 )

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MATHEMATICS SM015

38. The parametric equations of a curve is given by x  e2t 1 , y  e(2t 1)

(a) Find dy and d 2 y when t  1
dx dx2

(b) Given z  x2  xy . Express z in terms of t and find dz . Hence, deduce the set value
dt

of t such that dz is positive. ( 2016/ 17 )
dt

39. By writing tan x in terms of sin x and cos x , show that d (tan x)  sec2 x .
dx

(a) If y  tan x , find d 2 y in terms of y. Hence, determine the range of value of x such that
dx 2

d 2 y  0 for 0  x  .
dx 2

(b) If y  tan(x  y) , find dy in terms of x and y. Hence, show that dy   cos ec 2 2
dx dx

when x  y   . ( 2016/17 )

40. Given y  e2x sin 3x . Find dy and d 2 y . Hence, show that d 2 y  4 dy 13y  0.
dx dx2 dx2 dx

( 2017/18 )

41. A curve with equation x2  3y2  ae y2x  by  6 , where a and b are constants, passes

through the point 1, 2 .

(a) Given dy  1 at 1, 2 , determine the values of a and b.

dx

(b) Evaluate d 2 y at 1, 2 ( 2017/18 )

dx 2

BKS 2018/19 Page 82

MATHEMATICS SM015

42. A curve is defined by parametric equations x  3t  1 and y  t  3 , where t  0 .
tt

(a) Show that dy  t 2  3 . Hence, find d 2 y .
dx 3t 2  1 dx 2

(b) Show that dy  1  10 . Hence, deduce that 3 dy 1 ( 2017/18 )
dx 3 3(3t 2  1) dx 3

SUGGESTED ANSWERS

1. (a) dy  e x (1  x) , d2y  e x (x  2) (b) Shown
dx dx 2

2. dx  x 1 , dy   y , x  2,t  2, y  1, dy  1, x  2,t  2, y  3, dy  3
dt 1  t dt t  y dx dx

3. (a) dy  2t (b) d 2 y does not exist.
dx 1  t 2 dx 2

4. (a) dy  2 (b) d2y 0
dx dx 2

5. (a) dy  2x2 1x ln 2 (b) shown

dx

6. (a) shown (b) shown (c) d 2 y  2
dx2 3

7. Shown

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MATHEMATICS SM015

8. (a) dy  A  2B , d 2 y  6B (b) A  0, B  3
dx x3 dx2 x4

9. (a) dy  3,  3 (b) 0, 0 (c)  , 0

dx

10. dy  4,  4
dx

11. Shown

12. (a) A  1, B  1,C  2 (b) d 2 y  37
dx2 54

1

13. 3e 3

14. Shown

15. (a) x  0, dy  3 (b) d2y  26 3t t 2 3
dx 2 dx 2  2 

16. dy  y 2e xy
dx 1 xye xy

17. (-∞, -3)  ( 1 , ∞), dy  0 in  1 ,  dy  0 in (,  3)
2 dx 2  dx

18. (x, y)  (1,2),t  0.3466

19. Shown Page 84

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MATHEMATICS SM015

20. (a) dy  t 2 1 ; 3 (b) d2y 1 ; d2y 0
dx t 2 1 5 lim
dx2 2 dxt0 2

 (b)
21. (a) dy  5 ln x4 dy  y ex  y
dx 2xy  ex
dx x

22. 4

23. (a) shown (b) dy  1
dx 3

d2y 8t 3
dx 2 2t 2  1 3
 (c) 

24. dy  32x1 (2) ln 3 (b) dy  5  yexy
dx dx xe xy  1

25. (a) Shown (b) 8 2

26. (a) Shown (b) 1
3
27. (a) (i) dy   1
dx 3 d 2 y 5
(2x  1) 2
(ii) dx2  3(2x  1) 2
(b) (i) dy  e2x (2 tan x  sec2 x)
dx (ii) dy  x sec x  sec x  (ln x)(sec x t an x)
dx  x

28. (a) (i) dy   1 , 8 (ii) Shown
dx 5 15 (ii) d 2 y  24

(b) (i) Shown dx2 125

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MATHEMATICS SM015 \
29. (a) dy   cosecsin(lnx  1)cotsin(ln[x  1])cos(ln[x  1])

dx x 1
(b) d 2 y  e2x (12(sin 3x)  5(cos3x))

dx2

30. (a) dy   1 (b) i) dy  xexy  y ii) dy  4
dx (x  1)2 dx 25
dx x(ln x  e xy )

31. (a) dy  2tet2 (1  t) , d 2 y  2et2 (2t 2  1  2t 3  2t)(1  t) (b) relative minimum
dx dx2

32. (a) dy   tan (b) -1,1 (c) 32 3
dx
9
33. (a) g'(x)  1
2 x 1 (b) d 2 y  4
dx2

34. (a) f '(x)   4x cosec2 4x2  1 (b) f '(x)  e2x  3  2 ln(3x  4)
4x2 1  3x  4 

35. (a) dy  t 2  1 , d 2 y  4t 3
dx t 2 1 dx2 (t 2  1)3

(b) minimum point (0,2) ; maximum point (0,-2)

36. Shown

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MATHEMATICS SM015

37. (a) k = 3

(b) (i) dy  1 , d 2 y  sin x , d 3 y  2  cos x
dx 1  cos x dx2 (1  cos x)2 dx3 (1  cos x)2

38. (a) dy  1 , d2y   2 (b) dz  4e4t2 , t 
dx e4 dx2 e7 dt

39. (a) d2y  2 y(1  y2) , 0x  (b) dy   cot2(x  y)  1
dx2 2 dx

40. dy  e2x (3cos 3x  2sin 3x) d 2 y  e2x (5sin 3x  12cos 3x)
dx dx2

41. (a) a  5,b  5 (b) d 2 y   3
42. (a) d 2 y  20t3 dx2 4

dx2 (3t2  1)3 (b) Shown

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MATHEMATICS SM015

TOPIC 10 : APPLICATION OF DIFFERENTIATION

1. Given the function f (x)  2x3  6x2 14x  2 ( 2003/04 )
(a) Determine the points on the graph of f (x) where the slope is 4.
(b) Find the normal line equation at the point obtained in (a) for x  0 .

2. A ladder 5 meter long is leaning against a vertical wall and is sliding down the wall.

When the foot at the ladder is 3 meters away from the wall, the top of the ladder slides

down at the rate of 0.5m / s . At what rate is the foot of the ladder moving away from the

wall at this instant. ( 2003/04 )

3. Milk is being poured into hemispherical bowl of radius 4 cm at the rate of 3π cm3/sec. If

the depth of the milk in the bowl is h cm, its volume V is given by

V   4h2  h3 cm3 . At the instant the milk is 3 cm deep, find
3 2

(a) The rate of change of h. ( 2006/07 )
(b) The rate of change of the radius of the milk’s surface

4. The total cost of manufacturing k boxes of chocolates ( a function of time, t ) is given

by Ck  2k 2  k  900, where kt  t 2 100t. Compute the rate of change of the

total cost with respect to time when t  1. ( 2007/08 )

5. Given that f x  x3  3x2  9x 11.

(a) If f intersects the x-axis at x  1, x  p and x  q , find p and q.
(b) Use the second derivative test to find the coordinates of the local extremum.

( 2007/08 )

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MATHEMATICS SM015

6. Water is leaking from the bottom of a conical tank with radius 1.5 meter and height

2 meter at a rate of 0.25 cubic meter per minute. The tank was initially full. If the

height of the water is 1 meter then find the rate of change of

(a) the water level,

( b) the radius of the water surface. ( 2008/09 )

7. The function f is defined by f x  x2  x  2 .
x2  x  4

(a) Find the y-intercept and determine the horizontal asymptote of f .

(b) Find the critical points of f . ( 2008/09 )

8. Given that f x  x3  8 . ( 2009/10 )

x
(a) State the asymptote of f .
(b) Find the critical point of f .

9. (a) The position of a particle moving along a straight line at any time t  0 is given

by st  tt 1t  2, where s is the distance of the particle from the origin. Find the

velocity of the particle at the instant when the acceleration becomes zero.

(b) A closed right circular cylindrical container of radius r and height h is to be

constructed with volume 4,000 cm3 . The cost for the construction is RM 1.00 per cm2

for the curved surface while RM 2.00 per cm2 for the top and bottom surfaces. State h

in terms of r and hence, find the radius of the cylinder so that the cost of the

construction is minimum. ( 2009/10 )

10. The surface area of a balloon in the shape of a sphere is decreasing at the rate of

2cm2 / min . Find the rate at which the volume is decreasing when the radius of the

balloon is 5cm. ( 2011/12 )

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MATHEMATICS SM015

11. The function f (x)  x3  6x2  9x  3 is defined on the interval [0,5]. Find the critical

points of f(x) on this interval and determine whether the critical points are local minimum

or maximum. ( 2011/12 )

12. (a) A conical tank is of height 12m and surface diameter 8m. Water is pumped into the

tank at the rate of 50m3 / min . How fast is the water level increasing when the depth

of the water is 6m.

(b) A cylindrical container of radius r and height h has a constant volume V. The cost of

the materials for the surface both of its ends is twice the cost of its sides. State h in

terms of r and V. Hence, find h and r in terms of V such that the cost is minimum.

( 2012/13 )

13. (a) A cylindrical container of volume 128πm3 is to be constructed with the same material

for the top, bottom and lateral side. Find the dimensions of the container that will

minimize the amount of the material needed.

(b) Gravel is poured onto a flat ground at the rate of 3 m3 per minute to form a conical

20

shape pile with vertex angle 600 as shown in the diagram below.

600

h

Compute the rate of change of the height of the conical pile at the instant t = 10

minutes. ( 2013/14 )

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MATHEMATICS SM015 ( 2014/15 )

14 . Find the relative extremum of the curve y  x3  4x2  4x .

15. Car X is travelling east at a speed of 80 km/h and car Y is travelling north at 100km/h as
shown in the diagram below. Obtain an equation that describes the rate of change of the
distance between two cars.

Hence, evaluate the rate of change of the distance between two cars when car X is 0.15

km and car Y is 0.08km from P. P
Car X

Car Y
( 2014/15 )

16. Water is running at a steady rate of 36cm3s1 into a right inverted circular cone with a

semi-vertical angle of 450.
(a) Find the rate of increasing in water depth when the water level is 3 cm.
(b) Find the time taken when the depth of the water is 18cm.

( 2015/16 )

17. (a) Use the derivative to find the maximum area of a rectangle that can be inscribed in a
semicircle of radius of 10 cm.

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MATHEMATICS SM015

(b) A cone-shaped tank as shown below.

60

Water flows through a hole A at rate of 6 cm3 per second. Find the rate of change in
height of the water when the volume of water in the cone is 24 cm3 ( 2016/17 )

18. The function f is defined by f (x)  ln(x 1) for x  1
x 1

(a) By considering the first and second derivative of f (x) , show that there is only one

maximum point on the graph y  f (x) .

(b) Use the result obtained in part (a) to state the exact coordinates of the maximum
point.

(c) Find the x-coordinate of the function f when d2y  0. ( 2017/18 )
dx 2

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MATHEMATICS SM015

SUGGESTED ANSWERS

1. (a) 3,  40 and  1, 8 (b) 4y  x  157  0

2. 2 m / s (b) 0.246 cm / s
3

3. (a) 4 cm / s
13

4. 41310

5. (a) 4.46,2.46

(b) 1,16 is a maximum point, 3,16 is a minimum point

6. (a) 0.1415 m/min (b) 0.1061 m/min

7. (a) y-intercept = 1 , horizontal asymptote: y  1
2

(b)  1, 1 ,3, 7 
 3  5

8. (a) vertical asymptote: x  0 (b) 1.59,7.56

9. (a) v  1 (b) h  4000 , r  6.8
r 2

10. dV  5cm3 / minute
dt

11. maximum point : (1,1) , minimum point: (3,-3)
12. (a) dh  25 m / minute

dt 2

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MATHEMATICS SM015

11

(b) h V The cost is minimum when h  4  V 3 , r   V 3
r2 4   4 

13. (a) r  4;h  8 (b) 0.0542m/ min

14. 2,0 is a minimum point;  2 , 32  is a maximum point
 3 27 

15. h dh  x dx  y dy ; 117.6km / h
dt dt dt (b) 54s

16. (a) 4 cms-1

17. (a) max area  100 unit 2 (b) dh  0.159 cm/s
dt

18. (a) Shown (b) e  1, 1  3
 e
(c) x  e 2  1  5.48

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