Q&A STPM MATHEMATICS T ( PENGGAL 1.2.3 )

CONTENTS

STPM Scheme of Assessment ............................................................................................ ii

Term 1 Functions 1

1Chapter

2Chapter Sequences and Series 64

3Chapter Matrices 88

4Chapter Complex Numbers 125

5Chapter Analytic Geometry 144

6Chapter Vectors 163

STPM Model Paper 954/1................................................................................................. 191

Term 2 Limits and Continuity 194

1Chapter

2Chapter Differentiation 212

3Chapter Integration 241

4Chapter Differential Equations 272

5Chapter Maclaurin Series 301

6Chapter Numerical Methods 318

STPM Model Paper 954/2................................................................................................. 344

Term 3 Data Description 347

1Chapter

2Chapter Probability 382

3Chapter Probability Distributions 405

4Chapter Sampling and Estimation 429

5Chapter Hypothesis Testing 460

6Chapter Chi-Squared Tests 496

STPM Model Paper 954/3................................................................................................. 530
Answers................................................................................................................................... 533

iv

Chapter Matrices

3

Term Question 1

Matrix A over the set of integers is given as follows:

1 –2 3
1 21
0 4 –1
–5 3 2

(a) What is the order of A?
(b) What is the element in row 1, column 2?
(c) What is the second column of A?
(d) What are the elements in the leading diagonal of A?

Answer:

(a) The order of A is 3 × 3.
(b) The element in row 1, column 2 is –2.
–2
1 2(c) The second column of A is4.
3
(d) The elements in the leading diagonal of A are 1, 4, 2.

Question 2

For each of the following matrices, identify which is null, identity, diagonal,
triangular, symmetric matrix or none of them.

(b) 10 01 (c) 001 600 –003
11 2 1 2 1 2(a) 0
0 1

1 2 1 2 1 2(d) 25 6
0 7 –9 (e) 00 (f ) –15 –35
0 0 1

Answer: (b) none of them
(a) identity matrix (d) triangular matrix
(c) diagonal matrix (f) symmetric matrix
(e) null matrix

88

Mathematics T Term 1 STPM Chapter 3 Matrices

Question 3

1 3c2 a2 2b2 .
5b
Given a symmetric matrix E = –2a – 1 a–b c
2b c+1

Find the values of a, b and c.

Answer: Term

If E is a symmetric matrix, then ET = E. 1

1 2 13c2 22b
–2a – 1
2b a–b
a2 b2 3c2 –2a – 1 c+1
5b c = a2 5b
a–b c+1 b2 c

a2 = –2a – 1
a2 + 2a + 1 = 0
(a + 1)2 = 0
a = –1

b2 = 2b
b(b – 2) = 0
b = 0 or 2

c = a – b
= –1 – 0 or –1 – 2
= –1 or –3

Question 4 2a + b + 1

13 a–3 b – 2c + 6 is a triangular matrix, find the values of a, b and c.
–4
If A = 1 5
2 –1

Answer:

1If A = 3 a–3 2a + b + 1 is a triangular matrix, then
1 5
2 –1 b – 2c + 6
–4

a – 3 = 0
a = 3

89

Term Mathematics T Term 1 STPM Chapter 3 Matrices

a + b + 1 = 0
3 + b + 1 = 0
b = –4
b – 2c + 6 = 0
–4 – 2c + 6 = 0
c = 1

Question 5

2
1 21 a–5 a+b
If M = 3c –6 a + b + c is a diagonal matrix, find the values of a,
b+5 c 7

b, and c.

Answer: a–5 2a + b
–6
12 c a + b + c is a diagonal matrix, then
7
If M = 3c
b+5

a – 5 = 0
a = 5
b + 5 = 0
b = –5

c = 0

Question 6

1 2 1 2 1 2Given A =x 1 ,B= 2 1 ,C= x 1 0 .
3 5 3 5 3 5 0

(a) Find x, if A = B.
(b) Is A = C? Is B = C?

Answer:
(a) If A = B, then x = 2.
(b) A is 2 × 2, C is 2 × 3, therefore A ≠ C
B is 2 × 2, C is 2 × 3, therefore B ≠ C

90

Mathematics T Term 1 STPM Chapter 3 Matrices

Question 7

1 2 1 2Given that A =–13 7 5 –2 4
2 6 10 and B = 1 3 9 , find

(a) A – B Term
(b) A + (–1) B
Prove that A – B = A + (–1)B.

Answer: 1

1 2 1 2(a) A – B =–13 7 – 5 –2 4
2 6 10 1 3 9

–1 – 5 3 – (–2) 7–4
2–1 6–3 10 – 9
1 2 =

1 2 = –6 5 3
1 3 1

1 2 1 2(b) A + (–1)B =–13 7 5 –2 4
2 6 10 + (–1) 1 3 9

1 2 1 2=–13 7 + –5 2 –4
2 6 10 –1 –3 –9

–1 + (–5) 3+2 7 + (–4)
2 + (–1) 6 + (–3) 10 + (–9)
1 2=

1 2=–6 5 3
1 3 1

Thus, A – B = A + (–1)B

Question 8

1 2 1If A =3–2 3 2–6
6 5 and B = 9 , show that
4

(a) (3 + 5)A = 3A + 5A
(b) 2(A + B) = 2A + 2B

91

Mathematics T Term 1 STPM Chapter 3 Matrices

Answer:

1 2 1 2(a) (3 + 5)A = 8A = 8 3 –2 = 24 –16
6 5 48 40

1 2 1 2 3A + 5A = 3 3 –2 +5 3 –2
6 5 6 5

Term 1 2 1 2 = 9 –6 + 15 –10
18 15 30 25

1 21 = 24 –16
48 40

∴ (3 + 5)A = 3A + 5A

31 2 1 24 1 2 1 2(b) 2(A + B) = 23–2+ 3 –6 =2 6 –8 = 12 –16
6 5 9 4 15 9 30 18

1 2 1 2 2A + 2B = 2 3 –2 +2 3 –6
6 5 9 4

1 2 1 2 = 6 –4 + 6 –12
12 10 18 8

1 2 = 12 –16
30 18

∴ 2(A + B) = 2A + 2B

Question 9

Evaluate.

3 –1
(a) 5 2

1 2 1 2 1 2 1 2–2 0
1 (b) 23 –61 –53 4
6 0

1 21 21 3 4 5

(c) 2 5 –1 2
0 –2 6 1

Answer: 3×1–1×6 –3
5×1+2×6 17
13 –2 × 1 + 0 × 6 –2
21 2 1 2 1 2–11= =
(a) 5 6
–2 2
0

92

Mathematics T Term 1 STPM Chapter 3 Matrices

2 –1 –3 4 2 × (–3) – 1 × 5 2×4–1×0 –11 8
3 6 5 0 3 × (–3) + 6 × 5 3×4+6×0 21 12
1 21 2 1 2 1 2(b) = =

5 1 × 5 + 3 × 2 + 4 × 1 15
13 4
(c) 2 5 –1

1 2 1 2 1 2 1 20 –2 6
2 = 2 × 5 + 5 × 2 – 1 × 1 = 19
1 0×5–2×2+6×1 2

Term

Question 10 1

5 –1
1 2 1 2(a) 5CA
If A = (2  –2  4), B = (0  1  2), C = 0 and D = –1 , find
12

(b) 5AC – (2A – B)D

Answer:

1 25

(a) 5CA = 5 0 (2 –2 4)
1

10 –10 20
=5 0 0 0

2 –2 4
1 2

50 –50 100
00 0
10 –10 20
1 2 =

5 –1
(b) 5AC = 5(2  –2  4) 0 = 5(14) = (70) –1
2
1 21
1 2
1 2
(2A – B)D = [2(2 –2 4) – (0 1 2)]

= (4 –5 6) –1
= (13) –1
2

5AC – (2A – B)D = (70) – (13) = (57)

93

Mathematics T Term 1 STPM Chapter 3 Matrices

Question 11

1 2Find A2 where A = –3 2 .
–2 2

Find the values of m and n such that A2 + mA + nI = 0, where I is the identity
2 × 2 matrix and 0 is the null 2 × 2 matrix.
Deduce that A4 = –5A + 6I and find A4.
Term
Answer:

1 21 2 1 21A2 =–3 2 –3 2 = 5 –2 Common error
–2 2 –2 2 2 0
Students often find
A2 + mA + nI = 0
1A2 = (–3)2 222
5 –2 +m –3 2 +n 1 0 = 0 0 (–2)2­
0 –2 2 0 1 0 0 22
1 2 1 2 1 2 1 22
instead of

5 – 3m + n –2 + 2m 0 0 A2 = AA
2 – 2m 2m + n 0 0
1 2 1 2 1 21 2 =
= –3 2 –3 2
–3 2 –3 2

2 – 2m = 0
m = 1
2m + n = 0
n = –2

A2 + A – 2I = 0  ⇒ A2 = 2I – A …… 1

AA2 + AA – 2IA = 0

A3 + A2 – 2A = 0

From 1, A3 + 2I – 3A = 0

AA3 + 2IA – 3AA = 0

A4 + 2A – 3A2 = 0

A4 + 2A – 6I + 3A = 0

A4 = –5A + 6I

A4 = –5A + 6I

1 2 1 2 = –5 –3 2 +6 1 0
–2 2 0 1

1 2 1 2 = 15 –10 + 6 0
10 –10 0 6

1 2 = 21 –10
10 –4

94

Mathematics T Term 1 STPM Chapter 3 Matrices

Question 12

1 2 1 2 1 2Let A =1 –1 7 –1
1 ,B= 2 and M = –2 8 .

Find MA and MB.
Hence express MnA and MnB in terms of A and B respectively, where n ∈ Z+.

Answer: Term

1 21 2 1 2 1 2MA =7–11 = 6 =6 1 = 6A 1
–2 8 1 6 1

1 21 2 1 2 1 2MB =7–1–1 = –9 =9 –1 = 9B
–2 8 2 18 2

M2A = M(MA)
= M(6A)
= 6MA
= 6(6A)
= 62 A
  

Similarly, MnA = 6nA
M2B = M(MB)
= M(9B)
= 9 MB
= 9(9B)
= 92 B
M3B = M(M2B)
= M(92B)
= 92 (MB)
= 93 B
  

Similarly, MnB = 9nB

Question 13

1 2 1 2Given P =1 –1 3 2 . Find the transpose of P and Q.
2 and Q = 0 4 –2
7

95

Mathematics T Term 1 STPM Chapter 3 Matrices

Answer: 1 2–1 0
PT = (1  2  7)
QT = 3 4
2 –2

Term Question 14

1 2 1 21 0 1
Find the values of u and v such that A2 + uA + vI = O where A = 3 –2 ,

I= 1 0 and O is 2 × 2 zero matrix.
0 1

Answer:
A2 + uA + vI = O

1 21 2 1 2 1 2 1 010 1 +u 0 1 +v 1 0 = 0 0 2
3 –2 3 –2 3 –2 0 1 0 0

3 –2 0 0 0 0 2
–6 7 3u v 0 0
1 2 1 2 1 2 1 + u + v =
–2u 0

3+v u–2 0 0
3u – 6 7 – 2u + v 0 0
1 2 1 = 2

3 + v = 0    , 3u – 6 = 0
v = –3 u=2

Question 15 20

11 2 2 , show that M3 = M. Hence, find M20.
–1
If M = –1 0
11

Answer:

M2 = MM 120
–1 0 2
1 1 –1
12 0
= –1 0 2

1 2 1 21 1 –1

1 2–1 2 4

= 1 0 –2  ……………… 1
–1 1 3

96

M3 = M2M 2 Mathematics T Term 1 STPM Chapter 3 Matrices Term
0
1 –1 1 2 1 24 1 2 0
2
= 1 0 –2 –1 0 2
–1 1 3 1 1 –1

11 20

= –1 2
1 –1

= M

M3 = M 1

(M3)6 = M6

M18 = (M3)2

= M2

M18 M2 = M2 M2

M20 = M3 M

= MM

= M2

1 –1 2 24
0
= 1 1 –2 from 1
–1 3

Question 16

1 2 1 2Given that M =2 –3 5 9
8 6 and N = –1 4 , verify the commutative property

for addition by finding M + N and N + M.

Answer:

2 –3 5 9 2+5 –3 + 9 7 6
8 6 –1 4 8 + (–1) 6+4 7 10
1 2 1 2 1 2 1 2M + N = + = =

5 9 2 –3 5+2 9 + (–3) 7 6
–1 4 8 6 –1 + 8 4+6 7 10
1 2 1 2 1 2 1 2N + M = + = =

M + N = N + M ⇒ addition of matrices is commutative.

97

Answers

STPM Model Paper 954/1

1. y = | 1 – 2 x| = 1– 2x  if  1 – 2x > 0 0
–(1 – 2x)  if  1 – 2x ,

= 1 – 2x,  x < 1
2
1
2x – 1,  x . 2

x 0 —12 1
y101

y

y = |1 – 2x|

1 A y = –x1

–2 –1 0 12 x
–1

From the graph, A is the point (1, 1).
The set of values of x is {x | x , 0, x . 1}

11

2. (x + y) 2 – (x – y) 2

11

2 – x2 1 –
1 2 1 211+ y y 1
x x 2
= x2

3= 1 1 + y + 1 1– 1 2  1 y 22 + 1 1– 1 21– 3 2  1 y 23 + …4
2x 2 2 x 2 2 2 x
x2 2! 3!

3 1 2 1 2 1 21 2 1 2 4 –1 1 – 1 2+ 1 – 1 – 3
1 – y + 2 2  – y 2 2 2  – y 3+…
x2 2x 2! x 3! x

11 + y – y2 + y3 + … – 1 1 – y – y2 – y3 –…
2x 8x2 16x3 2x 8x2 16x3
1 2 1 2= x2 x2

1 21 y + y3
x 8x2
= x2

533

Mathematics T STPM Answers

Putting x = 4, y = 2,

1 11 2 + 8
4 8(43)
3 4(4 + 2) 2 – (4 – 2) 2 = 4 2

1 2 1 1
6 – 2 = 2 2 + 64

= 33
32
11

(x + y) 2 + (x – y) 2

1 1 + y – y2 + y3 + … 1 1 – y – y2 – y3 – …
2x 8x2 16x3 2x 8x2 16x3
1 2 1 2= x2 + x2

1 2≈ 1 2 – y2
4x2
x2

Putting x = 4, y = 2 1
1 1
42 4
2) 2 2) 2 = 4(42)
3 4(4+ + (4 – 2 –

6 + 2 = 31
8

3. First augment the matrix with the identity, forming (M | I), an n × 2n
omfatthriex.auLgemt Ren1,teRd2 manadtriRx3., represent row 1, row 2, and row 3 respectively

1 22 1 3 1 0 0
–1 0 4 0 1 0
310001

–R2 ↔ R1 11 20 –4 0 –1 0
2
3 13100
11 10001
R2 → R2 – 32RR11 0
R3 → R3 – 0 20 –4 0 –1 0

R3 → R3 – R2 11 1 11 1 2 0
0 1 12 0 3 1
0
1 20 –4 0 –1 0
1R1 → R1 + 141RR3 3 0
→ R2 – 0 1 11 1 2 0
R2 0 1 –1 1 1

21 0 –4 3 4

0 0 12 –9 –11
1 1 –1 1 1

1 2–4 3 4

Thus, M–1 = 12 –9 –11
–1 1 1

534