MARKING SCHEME TRIAL EXAM MATHEMATICS T SEMESTER 1 2018

1 (a) 5 5 B1
𝐷𝑔 = (−∞ , ) ∪ ( , ∞ )
2 2

𝑅𝑔 = (−∞ , 0 ) ∪ ( 0 , ∞ ) B1

(b) Since there is only one intersection point between the horizontal line and the graph, B1
the function is one-to-one function.

1
=𝑥
2𝑦 − 5

𝑥(2𝑦 − 5) = 1
M1
1 + 5𝑥
𝑦=
2𝑥

1 + 5𝑥
𝑔−1 (𝑥) =
2𝑥 A1

(c) y
D 1- shape

y=x D 1-
asymptote

D 1 - all
2.5 𝑔−1 (𝑥) correct

g(x)
0 2.5 x

(d) 𝑔 [𝑔−1 (𝑥)]

1 + 5𝑥
= 𝑔[ ]
2𝑥

1
=
1 + 5𝑥 M1
2 [ 2𝑥 ] − 5

1
= A1
1 + 5𝑥 − 5𝑥
𝑥

=𝑥
TOTAL 10 marks
2 (a) 2𝑟 − 1 2𝑟 + 1
−
𝑟(𝑟 − 1) 𝑟(𝑟 + 1)

(𝑟 + 1)(2𝑟 − 1) − (2𝑟 + 1)(𝑟 − 1) M1

=
𝑟(𝑟 − 1)(𝑟 + 1)

2𝑟 2 − 𝑟 + 2𝑟 − 1 − 2𝑟 2 + 2𝑟 − 𝑟 + 1
=
𝑟(𝑟 − 1)(𝑟 + 1)

2𝑟
= A1
𝑟(𝑟 − 1)(𝑟 + 1)

2
=
(𝑟 − 1)(𝑟 + 1)
[ verified ]

(b) 𝑛
2
∑
(𝑟 − 1)(𝑟 + 1)
𝑟=2

𝑛
2𝑟 − 1 2𝑟 + 1
= ∑[ − ] M1
𝑟(𝑟 − 1) 𝑟(𝑟 + 1)
𝑟=2

= ∑[𝑓(𝑟) − 𝑓(𝑟 + 1)] M1

𝑟=2

= 𝑓(2) − 𝑓(𝑛 + 1)

2(2) − 1 2(𝑛 + 1) − 1 M 1 ( sub

= − value)
2(1) (𝑛 + 1)(𝑛 + 1 − 1)

3 2𝑛 + 1
= −
2 𝑛(𝑛 + 1)
A1
∞
(c) 1 2
∑
2 (𝑟 − 1)(𝑟 + 1)
𝑟=2

1 3 2𝑛 + 1
= lim [ − ] M1
2 𝑛→∞ 2 𝑛(𝑛 + 1)

2 1
1 3 𝑛 + 𝑛2
= lim [ − ]
2 𝑛→∞ 2 1 + 1
𝑛
1 3 3 A1
= 2 (2 − 0) = 4

TOTAL 8 MARKS
3 (a) 1 0 0 1 0 0
𝑀2 = ( 1 −2 0) ( 1 −2 0)
−1 3 1 −1 3 1

1 0 0
B1
= (−1 4 0)
1 −3 1
𝑀2 + 𝑎𝑀 + 𝑏𝐼 = 0
1 0 0 1 0 0 1 0 0 0 0 0
(−1 4 0) + 𝑎 ( 1 −2 0) + 𝑏 (0 1 0 ) = (0 0 0) M1
1 −3 1 −1 3 1 0 0 1 0 0 0
1 − 𝑎 = 0,
𝑎=1 A1
1 + 𝑎 + 𝑏 = 0, A1
b = -2

(b) 𝑀2 + 𝑀 − 2𝐼 = 0

𝑀(𝑀2 + 𝑀 − 2𝐼 = 0)

𝑀3 + 𝑀2 − 2𝑀 = 0 M1

𝑀3 + (2𝐼 − 𝑀) − 2𝑀 = 0

𝑀3 = 3𝑀 − 2𝐼 A1

𝑀2 + 𝑀 − 2𝐼 = 0

𝑀−1 (𝑀2 + 𝑀 − 2𝐼 = 0)

𝑀 + 𝐼 − 2𝑀−1 = 0

1
𝑀−1 = (𝑀 + 𝐼)
2 A1

TOTAL 7 MARKS
4 (a) Choose V= ( 0, 0 )
Let F (a, 0 ) and A ( a, 5 )
A at parabola ,
Substitute A into equation,
𝑦 2 = 4𝑎𝑥
52 = 4𝑎(𝑎) M1
5
𝑎= M1
2
Equation of parabola is
5
𝑦 2 = 4( )𝑥
2
𝑦 2 = 10𝑥 A1
(b) When x = 11 cm from vertex
𝑦 2 = 10(11)
𝑦 = √110 M1
|𝐶𝐷| = 2√110 A1
TOTAL 5 MARKS
5 (a) 𝑧̅1 = 3 + 3𝑖
|𝑧̅1 | = √32 + 32 = 3√2 M1
3 M1
Argument of 𝑧̅1 = tan−1 (3)
= 0.785 𝑟𝑎𝑑
𝑧̅1 𝑖𝑛 𝑝𝑜𝑙𝑎𝑟 𝑓𝑜𝑟𝑚 A1
𝑧̅1 = 3√2 (cos 0.785 + 𝑖 sin 0.785)

𝒊̅3
+ [̅̅̅̅̅̅̅̅
(b) ̅̅̅𝑧
𝑧1 2 (3+3𝑖)(3+2𝑖) −𝑖
13
+ [−𝑧 ] = 13 −(3+2𝑖)
]
2

3+15𝑖 ̅̅̅̅̅̅̅̅̅̅̅̅̅̅
𝑖(3−2𝑖)
= 13
+ [(3+2𝑖)(3−2𝑖)] M1

3+15𝑖 ̅̅̅̅̅
2+3𝑖 M1
= +[ ]
13 13

3+15𝑖+2−3𝑖
=
13
5+12𝑖
= A1
13

TOTAL 6 MARKS
6 (a) 2
⃗⃗⃗⃗⃗
𝐴𝐵= (−2) M1
1
⃗⃗⃗⃗⃗
The line is parallel to 𝐴𝐵
The vector equation of line is
𝒓 = 𝒊 + 2𝒋 + 𝜆(2𝒊 − 2𝒋 + 𝒌)
A1
Or 𝒓 = 3𝒊 + 𝒌 + (2𝒊 − 2𝒋 + 𝒌)

(b) 𝒓∙𝒏=𝒄∙𝒏
2 1 2
𝒓 ∙ (−2) = (1) ∙ (−2) M1
1 4 1

= 2-2+4
= 4
Equation of plane is 2𝑥 − 2𝑦 + 𝑧 = 4 A1
(c) Equation line: 𝒓 = 𝒊 + 2𝒋 + 𝜆(2𝒊 − 2𝒋 + 𝒌) ---------------- ①
Equation plane: 2𝑥 − 2𝑦 + 𝑧 = 4 ---------------- ②
Substitute ① into ②,
2(1+ 2𝜆)-2(2-2 𝜆) + 𝜆 = 4 M1
9𝜆=6
2 A1
𝜆=3
7 2 2 A1
Position vector of N is 𝒊 + 3𝒋 + 3𝒌
3

7 4 M1
1 3 3
⃗⃗⃗⃗⃗ = (1) , 𝑂𝑁
⃗⃗⃗⃗⃗⃗ = 2 ⃗⃗⃗⃗⃗ = 1
𝑂𝐶 then 𝐶𝑁 −3
3
4 2 10
−
(3) ( 3)

4 2 1 2 10 2
⃗⃗⃗⃗⃗ | = √( ) + (− ) + (− )
|𝐶𝑁
3 3 3
A1
 117
= √ 9
= √13 (shown)

TOTAL 9 MARKS
7 (a)

D 1- shape
D 1-
reflection
D 1- all
correct

1
|cos 2𝑥| =
2
𝜋 𝜋 𝜋 𝜋 M1
2𝑥 = , 𝜋 − , 𝜋 + , 2𝜋 −
3 3 3 3

𝜋 𝜋 2𝜋 5𝜋
𝑥= , , , M1
6 3 3 6

Solution set of x is
𝜋 𝜋 2𝜋 5𝜋
{𝑥: 0 < 𝑥 < , <𝑥< , <𝑥<𝜋 } A1
6 3 3 6

(b) cos(𝑥 + 45° ) = cos 𝑥 cos 45° − sin 𝑥 sin 45°

(i) √2 √2 B1
= cos 𝑥 − sin 𝑥
2 2
√2 √2
cos(𝑥 + 45° ) − √2 sin 𝑥 = cos 𝑥 − sin 𝑥 − √2 sin 𝑥
2 2
√2 3√2 B1
= cos 𝑥 − sin 𝑥
2 2
√2 3√2
Let cos 𝑥 − sin 𝑥  𝑅 cos(𝑥 + 𝛼)
2 2
 𝑅 cos 𝑥 cos 𝛼 − 𝑅 sin 𝑥 sin 𝛼
√2
𝑅 cos 𝛼 = --------------①
2
3√2 M1
𝑅 sin 𝛼 = ---------------②
2
√2 3√2
①2 + ②2 : 𝑅2 = ( )2+ ( )2
2 2
R = √5 = 2.236
3√2
② 2
: tan 𝛼 = √2
=3
①
2 A 1 (both)
𝛼 = 71.57°
cos(𝑥 + 45° ) − √2 sin 𝑥 = 2.236 cos(𝑥 + 71.57° ) A1

(ii) cos( 𝑥 + 45° ) − √2 sin 𝑥 = 2

°
2.236 cos ( 𝑥 + 71.57 ) = 2
° 500
cos ( 𝑥 + 71.57 ) = M1
559
 Basic angle = 26.56° M1
° ° ° M1
( 𝑥 + 71.57 ) = 333.44 , 386.56
° ° A1
𝑥 = 261.87 , 314.99

TOTAL 15 MARKS
8 (a) 6𝑥 + 2𝑦 + 3𝑧 = 3850 B1
4𝑥 + 𝑦 + 5𝑧 = 3200 B1
𝑥 + 𝑦 + 𝑧 = 1000 B1

(b) 𝑃𝑋 = 𝑄
6 2 3 𝑥 3850 B1
(4 1 5) (𝑦) = (3200)
1 1 1 𝑧 1000
6 2 3 3850
(4 1 5|3200) B1
1 1 1 1000
1 1 1 1000
𝑅1 ↔ 𝑅3 (4 1 5|3200)
M1
6 2 3 3850
1 1 1 1000
4𝑅1 − 𝑅2 → 𝑅2 (0 3 −1| 800 )
6 2 3 3850
1 1 1 1000
6𝑅1 − 𝑅3 → 𝑅3 (0 3 −1| 800 )
0 4 3 2150 M1

1 1 1 1000
4𝑅2 − 3𝑅3 → 𝑅3 (0 3 −1 | 800 )
0 0 −13 −3250

−13𝑧 = −3250
z = 250, M1
3𝑦 − (250) = 800
𝑦 = 350 , A1
𝑥 + 350 + 250 = 1000
𝑥 = 400 A 1(all)
𝑥 = 400, 𝑦 = 350 , z = 250

(c) 6𝑥 + 15𝑦 + 3𝑧 = 3850 B1

4𝑥 + 𝑦 + 5𝑧 = 3200
𝑥 + 𝑦 + 𝑧 = 1000
6 15 3
𝐿𝑒𝑡 𝑃 = (4 1 5) M1
1 1 1 M1
|𝑃| = 6 |1 5| − 15 |4 5| + 3 |4 1|
1 1 1 1 1 1
= 6(−4) − 15(−1) + 3(3)=0 A1
Since |𝑃| = 0 , P is singular , thus the system of linear equation have no unique solution.
It is not possible for company A to award 15 employees for their creativity instead of 2 with A1
the same total amount of money.
TOTAL 15 MARKS