Section A [ 45 marks ]

Answer all questions in this sectionType equation here.

1. The function f and g is defined with their respective domains by

𝑓 : 𝑥 → √2𝑥 − 5, 𝑥  2.5
10
𝑔:𝑥 → 𝑥
, 𝑥  𝑅, 𝑥 ≠ 0

(a) State the range of f. [1 mark]

(b) Find fg. Hence, solve the equation fg(x)=5. [2 mark]
(c) Find 𝑓 −1 and sketch the curve of f and 𝑓 −1 on the same axes. [3 mark]

1 𝑛
2. (a) The nth term of a geometric series is 𝑈𝑛 , where 𝑈𝑛 = 48 (4) .

(i) Find the common ratio of the series. [2 marks]

(ii) Find the sum to infinity. Hence, evaluate ∑∞
𝑛=4 𝑈𝑛 . [2 marks]

(b) Given 𝑓(𝑟) = 𝑟(𝑟 + 1)(𝑟 + 2), find 𝑓(𝑟 + 1) − 𝑓(𝑟). [1 mark]
Hence, find ∑𝑛1 3(𝑟 2 + 3𝑟 + 2). [3 marks]
𝑛
Deduce that ∑𝑛1 𝑟 2 = 6 (𝑛 + 1)(2𝑛 + 1). [3 marks]

3 2 1
3. The 3×3 matrix M is given by 𝐌 = (1 −2 −1) .
1 0 3
−1
By using elementary row operations, find 𝐌 .
Hence solve the following system of equations. [7 marks]
3𝑥 + 2𝑦 + 𝑧 = 7
𝑥 − 2𝑦 − 𝑧 = 1
𝑥 + 3𝑧 = 11

4. (a) It is given that 𝑧 = 𝑥 + 𝑦𝑖, where x and y are real numbers.

Find, in term of x and y, the real and imaginary part of 𝑖(𝑧 + 7) + 3(𝑧 ∗ − 𝑖). [2 marks]
Hence, find the complex number z such that 𝑖(𝑧 + 7) + 3(𝑧 ∗ − 𝑖) = 0. [2 marks]
√5+𝑎𝑖
(b) Determine the value of a such that 1+√5𝑖 is a real number. Hence, find this real number. [4 marks]

5. A circle with centre C has a equation 𝑥 2 + 𝑦 2 + 2𝑥 − 6𝑦 − 40 = 0.

Express this equation in the form (𝑥 − 𝑎)2 + (𝑦 − 𝑏)2 = 𝑑. Hence state the centre C and the radius
of circle. [3 marks]
The point P with coordinates ( 4, k ) lies on the circle. Find the possible values of k. [2 marks]
The points Q and R also lie on the circle, and the length of chord QR is 2. Calculate the shortest
distance from C to the chord QR. [2 marks]
6. The points U, V, and W have position vectors u, v and w respectively relative to an origin O where
1 3 𝑎
𝐮 = [2] , 𝐯 = [−4] and 𝐰 = [ 7 ]
2 2 −2
(a) Find (𝐮 × 𝐯) ∙ 𝐰 in terms of a. [2 marks]
(b) Given u, v and w are linear dependent.
(i) Find the value of a. [1 mark]
(ii) Express u as a linear combination of v and w. [3 marks]

Section B [ 15 marks ]
Answer any one question in this section
7. (a) A teacher borrows RM400,000 from the government to buy a house. The interest charged on the
loan is 4% per annum and it is calculated based on the balance outstanding each year. He is required
to repay the loan by monthly instalments for a period of 30 years. If the interest charged is
unchanged, find the amount to be paid each month. [9 marks]

𝑎 𝑎
(b) If 𝑏 and 𝑐 are small, by ignoring the powers higher than square for products of these quantities, show
𝑏+𝑎 1 1 1 𝑏
that √𝑐+𝑎 = [1 + 2 𝑎 (𝑏 − 𝑐 )] √ 𝑐 [4 marks]

101 𝑚
Hence, find an approximation of √ in the form , where m and n are integers. [2 marks]
401 𝑛

8. (a) Three position vectors p, q, and r of point P, Q and R respectively are given by
𝐩 = 5i + 10j
𝐪 = 3i + 4j
𝐫 = –i + 6j
Point X is a midpoint of line PQ and point Y is a midpoint of line QR.
(i) Determine whether vector ⃗⃗⃗⃗⃗
𝑃𝑄 and 𝑄𝑅⃗⃗⃗⃗⃗ are perpendicular to each others. [4 marks]
(ii) Find the position vectors of points X and Y. [3 marks]
(iii) Find the value of 𝑃𝑄𝑅. [4 marks]

(b) Given 𝐴(4,3, −2), 𝐵(8,7, −8) and 𝐶(16,15, −20)

Show that the points A, B, and C are collinear. [4 marks]

The Paper End

Success is no accident.
It is hard work, perseverance, learning, studying, sacrifice
and most of all,
love of what you are doing or learning to do.

☺☺☺☺☺☺☺☺☺ All the best ☺☺☺☺☺☺☺

 Marking Scheme for TRIAL EXAM SEM 1 2022
Partial
No Working/Answer Remarks
marks
1(a) 𝑅𝑓 = {𝑦: 𝑦 ≥ 0} A1 OE
1(b) 𝑓𝑔(𝑥) = 𝑓[𝑔(𝑥)]
10
= 𝑓(𝑥)
10
= √2 ( 𝑥 ) − 5
20 A1
= √𝑥 −5 No ISW
20
√ −5=5
𝑥
20 M1 Correctly squaring their
− 5 = 25 fg(x) and correctly
𝑥
20 isolating their x term
= 30
𝑥 No ISW
2 A1
𝑥=
3
1(c) Let 𝑓 −1 = 𝑦
𝑓(𝑦) = 𝑥
√2𝑦 − 5 = 𝑥
2𝑦 − 5 = 𝑥 2
𝑥 2 +5
𝑦= 2
−1 𝑥 2 +5 A1
∴𝑓 = ,𝑥 ≥ 0
2

D1 D1 – both f and f-1

D1 – All correct
D1 (All
including intersection
correct) points

Partial
No Working/Answer Total marks
marks
2(a) 1 𝑛
𝑈𝑛 = 48 (4)
1 𝑛−1
𝑈𝑛−1 = 48 (4)
1 𝑛
= 4 × 48 (4)
𝑈𝑛
𝑟=𝑈 M1
𝑛−1
1 𝑛
48(4)
= 1 𝑛
4 × 48(4)
1
=4 A1
2(b) 1 1
𝑈1 = 48 (4)
= 12
𝑎
𝑆∞ =
1−𝑟
12
=
1
(1 − 4)
= 16 A1
OE
∑∞
𝑛=4 𝑈𝑛 = 𝑆∞ − ∑31 𝑈𝑛 RHS
12(1−0.253 ) 𝑆∞ − (𝑈1 + 𝑈2 + 𝑈3 )
= 16 − 1−0.25
= 16 − 15.75
A1
= 0.25

2(b) 𝑓(𝑟 + 1) − 𝑓(𝑟) = (𝑟 + 1)(𝑟 + 2)(𝑟 + 3) − 𝑟(𝑟 + 1)(𝑟 + 2)

= (𝑟 + 1)(𝑟 + 2)[(𝑟 + 3) − 𝑟]
= 3(𝑟 + 1)(𝑟 + 2) A1

∑𝑛1 3(𝑟2 + 3𝑟 + 2) = ∑𝑛1 3(𝑟 + 1)(𝑟 + 2)

= ∑𝑛1 𝑓(𝑟 + 1) − 𝑓(𝑟)
B1
= 𝑓(𝑛 + 1) − 𝑓(1)
= (𝑛 + 1)(𝑛 + 2)(𝑛 + 3) − 1(1 + 1)(1 + 2)
M1
= (𝑛 + 1)(𝑛2 + 5𝑛 + 6) − 6
= 𝑛3 + 6𝑛2 + 11𝑛
= 𝑛(𝑛2 + 6𝑛 + 11) A1
𝑛

∑ 3(𝑟2 + 3𝑟 + 2) = 𝑛(𝑛2 + 6𝑛 + 11)

𝑟=1
𝑛 𝑛 𝑛 Must use method of
differences
3 ∑ 𝑟2 = 𝑛3 + 6𝑛2 + 11𝑛 − 9 ∑ 𝑟 − ∑ 6 M1
𝑟=1 𝑟=1 𝑟=1
𝑛
𝑛
3 ∑ 𝑟2 = 𝑛3 + 6𝑛2 + 11𝑛 − 9 ( ) (𝑛 + 1) − 6𝑛
2
𝑟=1
3 1
= 𝑛3 + 2 𝑛2 + 2
𝑛 M1
1
∑ 𝑟2 = ( ) (2𝑛3 + 3𝑛2 + 𝑛)
6
𝑟=1
𝑛
𝑛
∑ 𝑟2 = ( ) (2𝑛2 + 3𝑛 + 1)
6
𝑟=1
𝑛
𝑛 A1
∑ 𝑟2 = ( ) (𝑛 + 1)(2𝑛 + 1) (Deduced)
6
𝑟=1
 Partial Total
No Working/Answer
marks marks
3(a) 3 2 1 1 0 0 𝑅3 ↔ 𝑅1 1 0 3 0 0 1
(1 −2 −1|0 1 0) → (1 −2 −1|0 1 0)
1 0 3 0 0 1 3 2 1 1 0 0

−𝑅1 + 𝑅2 → 𝑅2 1 0 3 0 0 1
→ (0 −2 −4|0 1 −1) B1 OE
−3𝑅1 + 𝑅3 → 𝑅3
0 2 −8 1 0 −3

1
− 𝑅2 → 𝑅2 1 0 3 0 0 1
2 1 1
→ (0 1 2 |0 − 2 2 )
0 2 −8 1 0 −3

−2𝑅2 +𝑅3 → 𝑅3 1 0 3 0 0 1
1 1
→ (0 1 2 |0 − 2 2 )
0 0 −12 1 1 −4

0 0 1 M1
1
− 𝑅3 → 𝑅3 1 0 3 1 1
1 2| 0
→
12
(0 − 2 2)
0 0 1−1 −1 1
12 12 3

1 1
0
−2𝑅3 +𝑅2 → 𝑅2 1 0 0 41 4
1 1
→ −3𝑅
3 + 𝑅1 → 𝑅1
0 1 0|| 6 −3 −6
0 0 1 1 1 1 A1
− 12 − 12 3
( )
1 1
0 7
4 4
1 1 1
∴ 𝑀−1 = − − A1
6 3 6
1 1 1
−
( 12 −
12 3 )

3𝑥 + 2𝑦 + 𝑧 = 7
𝑥 − 2𝑦 − 𝑧 = 1
𝑥 + 3𝑧 = 11
3 2 1 𝑥 7
(1 −2 −1) (𝑦) = ( 1 )
𝑧 B1
1 0 3 11
1 1
0
𝑥 4 4 7
1 1 1
(𝑦) = 6
− 3
− 6
( 1)
𝑧 1 1 1 11
− 12 − 12 3
( )
2 M1
= (−1)
3

∴ 𝑥 = 2, 𝑦 = −1, 𝑧=3 A1
 Partial
No Working/Answer Total marks
marks
4 (a) 𝑖(𝑧 + 7) + 3(𝑧 ∗ − 𝑖) = 𝑖(𝑥 + 𝑦𝑖 + 7) + 3(𝑥 − 𝑦𝑖 − 𝑖) B1 For z* = x – yi
= 𝑖𝑥 + 𝑦𝑖 2 + 7𝑖 + 3𝑥 − 3𝑦𝑖 − 3𝑖
= 3𝑥 − 𝑦 + 𝑖(𝑥 − 3𝑦 + 4) A1

𝑖(𝑧 + 7) + 3(𝑧 ∗ − 𝑖) = 0
3𝑥 − 𝑦 + 𝑖(𝑥 − 3𝑦 + 4) = 0
3𝑥 − 𝑦 = 0 and 𝑥 − 3𝑦 + 4 = 0 M1 4
Sub y = 3𝑥 into 𝑥 − 3𝑦 + 4 = 0
𝑥 − 9𝑥 + 4 = 0
1
𝑥=2
1 3
When 𝑥 = 2 , 𝑥 = 2 1 3
A1 Must seen 𝑧 = 2 + 2 𝑖
1 3
∴𝑧= + 𝑖
2 2
(b) √5 + 𝑎𝑖 √5 + 𝑎𝑖 1 − √5𝑖
= × B1
1 + √5𝑖 1 + √5𝑖 1 − √5𝑖
√5 − 5𝑖 + 𝑎𝑖 + 𝑎√5
=
1+5
√5(1 + 𝑎) + 𝑖(𝑎 − 5)
= M1
6 4
For real number, 𝑎 − 5 = 0
M1
𝑎=5
√5(1+5)
The real number = 6
A1
= √5

Partial Total marks /

No Working/Answer
marks remarks
5 𝑥 2 + 𝑦 2 + 2𝑥 − 6𝑦 − 40 = 0
(𝑥 + 1)2 − 1 + (𝑦 − 3)2 − 9 = 40
(𝑥 + 1)2 + (𝑦 − 3)2 = 50 A1
Cente C (−1, 3) Radius ,r = 5√2 A1A1
At point P (4,k)
𝑥 2 + 𝑦 2 + 2𝑥 − 6𝑦 − 40 = 0
16 + 𝑘 2 + 8 − 6𝑘 − 40 = 0 8
𝑘 2 − 6𝑘 − 16 = 0 M1
(𝑘 + 2)(𝑘 − 8) = 0
𝑘 = −2 or 𝑘 = 8 A1

Shortest distance, d = √50 − 1

=7 M1A1
 Partial Total marks /
No Working/Answer
marks remarks
6(a) 1 3 𝑎
𝐮 = [2] , 𝐯 = [−4] 𝐰=[ 7 ]
2 2 −2
𝑖 𝑗 𝑘
𝐮 × 𝐯 = |1 2 2 | 2
3 −4 2
2 2 1 2 1 2
= 𝒊| |− 𝒋| | + 𝒌| |
−4 2 3 2 3 −4
= 𝒊(4 + 8) − 𝒋(2 − 6) + 𝒌(−4 − 6) Correct vector product
= 12𝒊 + 𝟒𝒋 − 10𝒌 M1
12 𝑎
(𝐮 × 𝐯) ∙ 𝐰 = ( 4 ) ∙ ( 7 )
−10 −2
= 12a + 28 + 20
= 12a + 48 A1 COA
(b)(i) 12a + 48 = 0
Sets their expression
𝑎 = −4 A1
equal to 0
(ii) 1 3 −4
[2] = 𝑐 [−4] + 𝑑 [ 7 ] B1
2 2 −2
1 = 3𝑐 − 4𝑑 − − − − − − − (1)
A1
2 = 2𝑐 − 2𝑑 − − − − − − − (2) Both c and d correct
𝑐 = 3, 𝑑 = 2 A1
A1 A1 – correct linear
∴ 𝐮 = 3𝐯 + 2𝐰
combination stated
Section B
Partial Total marks /
No Working/Answer
marks remarks
7(a) Let the amount to be paid each year be RMx.
After one year, the debt = RM(400,000 + 4% interest – x)
If 𝑃𝑛 is the debt after n years, B1
𝑃1 = 400,000 + (0.04)(400,000) − 𝑥 M1
= 400,000(1.04) – x
𝑃2 = 𝑃1 + 0.04𝑃1 − 𝑥 M1
= 𝑃1 + 0.04𝑃1 − 𝑥
= 1.04𝑃1 − 𝑥
= 1.04[400,000(1.04) – 𝑥] − 𝑥
= 400,000(1.04)2 – 1.04x−𝑥
𝑃3 = 𝑃2 + 0.04𝑃2 − 𝑥 M1
= 400,000(1.04)3 – 1.042 x−1.04𝑥 − 𝑥
⋮
𝑃30 = 400,000(1.04)30 – 1.0429 x−1.0428 𝑥 − ⋯ − 𝑥
= 400,000(1.04)30 – x(1 + 1.04 + 1.042 + ⋯ + 1.0429 ) 9
1.0430 −1
= 400,000(1.04)30 – 𝑥 ( 1.04−1 )
= 400,000(1.04)30 – 25𝑥(1.0430 − 1) M1
The loan is settled if 𝑃30 = 0 B1
400,000(1.04)30 – 25𝑥(1.0430 − 1) = 0
25𝑥(1.0430 − 1) = 400,000(1.04)30
400,000(1.04)30 M1
𝑥=
25((1.0430 − 1)
= 23132 A1
23132
Monthly installment = 12
= 1927.7
The amount to be paid each month is RM1927.7 A1

7(b) 𝑏+𝑎 1 1 B1
√ = (𝑏 + 𝑎)2 (𝑐 + 𝑎)−2
𝑐+𝑎
1 1
1 1 −
𝑎 2 − 𝑎 2
= 𝑏 (1 + 𝑏) 𝑐
2 2 (1 + 𝑐 ) M1
1
𝑏 2 𝑎 𝑎
= ( 𝑐 ) (1 + 2𝑏) (1 − 2𝑐)
1
𝑏 2 𝑎 𝑎
= ( 𝑐 ) (1 + 2𝑏 − 2𝑐) M1
1 1 1 𝑏
= [1 + 2 𝑎 (𝑏 − 𝑐 )] √𝑐 A1
4
By taking a = 1, b = 100, c = 400
𝑏+𝑎 1 1 1 𝑏
√ = [1 + 2 𝑎 (𝑏 − 𝑐 )] √ 𝑐
𝑐+𝑎
100+1 1 1 1 100
√ = [1 + 2 (100 − 400)] √400
400+1
M1

101 1 13 1
√ = [1 + 2 (400)] (2)
401
803
= 1600 A1
 Partial
No Working/Answer Total marks / remarks
marks
8(a) 𝐩 = 5i + 10j 𝐪 = 3i + 4j 𝐫 = –i + 6j
(i) ⃗⃗⃗⃗⃗
If PQ and QR are perpendicular, then 𝑃𝑄 ∙ 𝑄𝑅 ⃗⃗⃗⃗⃗ = 0
⃗⃗⃗⃗⃗
𝑃𝑄 = ⃗⃗⃗⃗⃗ ⃗⃗⃗⃗⃗⃗
𝑃𝑂 + 𝑂𝑄
= (−5𝐢 − 10𝐣) + (3𝐢 + 𝟒𝐣)
= −2𝐢 − 6𝐣 M1

⃗⃗⃗⃗⃗ = 𝑄𝑂
𝑄𝑅 ⃗⃗⃗⃗⃗⃗ + 𝑂𝑅
⃗⃗⃗⃗⃗ ⃗⃗⃗⃗⃗ and 𝑅𝑄
⃗⃗⃗⃗⃗ correct
Both 𝑃𝑄
= (−3𝐢 − 𝟒𝐣) + (−𝐢 + 𝟔𝐣) M1
= −4𝐢 + 2𝐣
4
⃗⃗⃗⃗⃗ = (−2) ∙ (−4)
⃗⃗⃗⃗⃗ ∙ 𝑄𝑅
𝑃𝑄
−6 2
= 8 + (−12)
= −4 M1
⃗⃗⃗⃗⃗
Since 𝑃𝑄 ∙ 𝑄𝑅 ⃗⃗⃗⃗⃗ ≠ 0, PQ and QR are not perpendicular to
A1
each other.

(ii) Position vector of X

1
⃗⃗⃗⃗⃗
𝑂𝑋 = ⃗⃗⃗⃗⃗
𝑂𝑃 + (𝑃𝑄 ⃗⃗⃗⃗⃗ )
2
1
= (5𝐢 + 10𝐣) + (−2𝐢 − 6𝐣)
2
= (5𝐢 + 10𝐣) + (−𝐢 − 3𝐣)
= 4𝐢 + 7𝐣 M1
A1
Position vector of Y 3
1
⃗⃗⃗⃗⃗
𝑂𝑌 = 𝑂𝑄 ⃗⃗⃗⃗⃗⃗ + (𝑄𝑅
⃗⃗⃗⃗⃗ )
2
1
= (3𝐢 + 𝟒𝐣) + (−4𝐢 + 2𝐣)
2
= (3𝐢 + 𝟒𝐣) + (−2𝐢 + 𝐣) A1
= 𝐢 + 5𝐣

(a) 𝑃𝑄𝑅 ∶
(iii) Angle between ⃗⃗⃗⃗⃗ 𝑄𝑃 and 𝑄𝑅 ⃗⃗⃗⃗⃗
⃗⃗⃗⃗⃗ ⃗⃗⃗⃗⃗ = |𝑄𝑃
𝑄𝑃 ∙ 𝑄𝑅 ⃗⃗⃗⃗⃗ |𝑐𝑜𝑠𝑃𝑄𝑅
⃗⃗⃗⃗⃗ ||𝑄𝑅 B1
⃗⃗⃗⃗⃗ ∙ 𝑄𝑅
𝑄𝑃 ⃗⃗⃗⃗⃗
𝑐𝑜𝑠𝑃𝑄𝑅 =
⃗⃗⃗⃗⃗ ||𝑄𝑅
|𝑄𝑃 ⃗⃗⃗⃗⃗ |
2 −4 M1
( )∙( )
= 6 2
√2 + 6 √(−4)2 + 22
2 2 4

−8 + 12
=
√40 √20 M1
= 0.1414
𝑃𝑄𝑅 = 𝑐𝑜𝑠 −1 0.1414
A1
= 81.87°
8(b) ⃗⃗⃗⃗⃗
𝐴𝐵 = ⃗⃗⃗⃗⃗ ⃗⃗⃗⃗⃗
𝐴𝑂 + 𝑂𝐵
−4 8
= (−3) + ( 7 )
2 −8
4
=( 4 ) M1
−6

⃗⃗⃗⃗⃗ = 𝐵𝑂
𝐵𝐶 ⃗⃗⃗⃗⃗ + 𝑂𝐶
⃗⃗⃗⃗⃗
−8 16
= (−7) + ( 15 )
8 −20
8 4
M1
=( 8 )
−12
8
⃗⃗⃗⃗⃗ = ( 8 )
𝐵𝐶
−12
4
⃗⃗⃗⃗⃗
𝐵𝐶 = 2 ( 4 )
−6
⃗⃗⃗⃗⃗
𝐵𝐶 = 2𝐴𝐵 ⃗⃗⃗⃗⃗ M1
⃗⃗⃗⃗⃗
Since 𝐵𝐶 = 2𝐴𝐵 ⃗⃗⃗⃗⃗ , and B is the common point, therefore A, A1
B, and C are collinear.