# Mathematics 2014 | Study Mode

Question 31
A
$$\frac{1}{3} \sin 3x$$
B
$$-\frac{1}{3} \sin 3x$$
C
3 sin 3x
D
-3 sin 3x
##### Explanation
y = cos 3x

Let u = 3x so that y = cos u

Now, $$\frac{\delta y}{\delta x} = 3$$,

$$\frac{\delta y}{\delta x} = -sin u$$

By the chain rule,

$$\frac{\delta y}{\delta x} = \frac{\delta y}{\delta u} \times \frac{\delta u}{\delta x}$$

$$\frac{\delta y}{\delta x} = (-\sin u) (3)$$

$$\frac{\delta y}{\delta x} = -3 \sin u$$

$$\frac{\delta y}{\delta x} = -3 \sin 3x$$

Question 32
A
4
B
1
C
-1
D
-4
##### Explanation
y = x2 - 2x - 3,

Then $$\frac{\delta y}{\delta x} = 2x - 2$$

But at minimum point,$$\frac{\delta y}{\delta x} = 0$$,

Which means 2x - 2 = 0

2x = 2

x = 1.

Hence the minimum value of y = x2 - 2x - 3 is;

ymin = (1)2 - 2(1) - 3

ymin = 1 - 2 - 3

ymin = -4

Question 33
A
cos 2x + k
B
$$\frac{1}{2}$$cos 2x + k
C
$$-\frac{1}{2}$$cos 2x + k
D
-cos 2x + k
##### Explanation
$$\int \sin 2x dx = \frac{1}{2} (-\cos 2x) + k$$

$$- \frac{1}{2} \cos 2x + k$$

Question 34
A
$$\frac{1}{12} (2x + 3)^6 + k$$
B
$$\frac{1}{3} (2x + 3)^{\frac{1}{2}} + k$$
C
$$\frac{1}{3} (2x + 3)^{\frac{3}{2}} + k$$
D
$$\frac{1}{12} (2x + 3)^{\frac{3}{4}} + k$$
##### Explanation
$$\int (2x + 3)^{\frac{1}{2}} \delta x$$

let u = 2x + 3, $$\frac{\delta y}{\delta x} = 2$$

$$\delta x = \frac{\delta u}{2}$$

Now $$\int (2x + 3)^{\frac{1}{2}} \delta x = \int u^{\frac{1}{2}}.{\frac{\delta x}{2}}$$

$$= \frac{1}{2} \int u^{\frac{1}{2}} \delta u$$

$$= \frac{1}{2} u^{\frac{3}{2}} \times \frac{2}{3} + k$$

$$= \frac{1}{3} u^{\frac{3}{2}} + k$$

$$= \frac{1}{3} (2x + 3)^{\frac{3}{2}} + k$$

Question 35
A
$$\begin{pmatrix} 3 & -3 \\ 8 & 2 \end{pmatrix}$$
B
$$\begin{pmatrix} 3 & 3 \\ 8 & 2 \end{pmatrix}$$
C
$$\begin{pmatrix} -2 & 2 \\ 2 & 2 \end{pmatrix}$$
D
$$\begin{pmatrix} -2 & 3 \\ 3 & 2 \end{pmatrix}$$
##### Explanation
y - 4x + 3 = 0

When y = 0, 0 - 4x + 3 = 0

Then -4x = -3

x = 3/4

So the line cuts the x-axis at point (3/4, 0).

When x = 0, y - 4(0) + 3 = 0

Then y + 3 = 0

y = -3

So the line cuts the y-axis at the point (0, 3)

Hence the midpoint of the line y - 4x + 3 = 0, which lies between the x-axis and the y-axis is;

$$[\frac{1}{2}(x_1 + x_2), \frac{1}{2}(y_1 + y_2)]$$

$$[\frac{1}{2}(\frac{3}{4} + 0), \frac{1}{2}(0 + -3)]$$

$$[\frac{1}{2}(\frac{3}{4}), \frac{1}{2}(-3)]$$

$$[\frac{3}{8}, \frac{-3}{2}]$$

Question 36
A
t
B
-t
C
2
D
-2
##### Explanation
Mean x = $$\frac{\sum x}{n}$$

= [(2 - t) + (4 + t) + (3 - 2t) + (2 + t) + (t - 1) $$\div$$] 5

= [11 - 1 + 3t - 3t] $$\div$$ 5

= 10 $$\div$$ 5

= 2

Question 37
A
1
B
2
C
3
D
4

Question 38
A
6
B
5
C
4
D
3
##### Explanation
First arrange the numbers in order of magnitude;
1,2,3,3,4,5,5,5,5,6,7,8,9,9,10

Hence the median = 5

Question 39
A
$$\sqrt{2}$$
B
$$\sqrt{3}$$
C
$$\sqrt{6}$$
D
$$\sqrt{10}$$
##### Explanation
Mean x = $$\frac{\sum x}{n}$$

$$= \frac{5 + 4 + 3 + 2 + 1}{5}$$

$$= \frac{15}{5}$$

= 3

$$\begin{array}{c|c} x & d = x - 3 & d^2 \\ \hline 5 & 2 & 4 \\ 4 & 1 & 1 \\ 3 & 0 & 0 \\ 2 & -1 & 1 \\ 1 & -2 & 4 \\ \hline & & \sum d^2 + 10 \end{array}$$

Hence, standard deviation;

$$= \sqrt{\frac{\sum d^2}{n}} = \sqrt{\frac{10}{5}}$$

$$= \sqrt{2}$$

Question 40
A
$$\frac{7!}{3!}$$
B
$$\frac{7!}{4!}$$
C
$$\frac{7!}{3!4!}$$
D
$$\frac{7!}{2!5!}$$
##### Explanation
A team of 2 girls can be selected from 7 girls in $$^7C_3$$

$$= \frac{7!}{(7 - 3)! 3!}$$

$$= \frac{7!}{4! 3!} ways$$

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