# Mathematics 2014 | Study Mode

Question 21
A
12
B
10
C
-1
D
-2
##### Explanation
$$0 \begin{vmatrix} 7 & 8 \\ 5 & 4\end{vmatrix}-3 \begin{vmatrix} 1 & 8 \\ 0 & 4\end{vmatrix}+2 \begin{vmatrix} 1 & 7 \\ 0 & 5\end{vmatrix}$$

= 0(28 - 40) - 3(4 - 0) + 2(5 - 0)

= 0(-12) - 3(4) + 2(5)

= 0 - 12 + 10

= -2

Question 22
A
12
B
10
C
9
D
8
##### Explanation
If each interior angle of the polygon is 135o, then each exterior angle is 180o - 135o = 45o. Hence, number of sides =

$$\frac{360^o}{\text{one exterior angle}}$$

$$\frac{360^o}{45^o}$$

= 8

Question 23
A
$$\sqrt{2}$$
B
$$\sqrt{5}$$
C
3
D
5
##### Explanation
Common ratio r of the G.P is

$$r = \frac{T_n + 1}{T_n} = \frac{T_2}{T_1}$$

$$r = \frac{\sqrt{10} + 2\sqrt{5}}{\sqrt{10} + \sqrt{5}}$$

$$r = \frac{\sqrt{10} + 2\sqrt{5}}{\sqrt{10} + \sqrt{5}} \times \frac{\sqrt{10} - \sqrt{5}}{\sqrt{10} - \sqrt{5}}$$

$$= \frac{(\sqrt{10})(\sqrt{10}) + (\sqrt{10})(-\sqrt{5}) + (2\sqrt{5})(\sqrt{10}) + (2\sqrt{5})(-\sqrt{5})}{(\sqrt{10})^2 - (\sqrt{5})^2}$$

$$\frac{10 - \sqrt{50} + 2\sqrt{50} - 10}{10 - 5}$$

$$\frac{\sqrt{50}}{5}$$

$$\frac{\sqrt{25 \times 2}}{5}$$

$$\frac{5\sqrt{2}}{5}$$

$$\sqrt{2}$$

Question 24
A
8.0m
B
7.5m
C
5.0m
D
2.5m
##### Explanation
Using $$V = \pi r^2 h$$

6160 = 22/7 x 28 x 28 x h

$$h = \frac{6160}{22 \times 4 \times 28}$$

$$h = 2.5m$$

Question 25
A
circle with diameter 4m
B
C
semi-circle with diameter 4m
D

Question 26
A
-4, 2
B
4, -2
C
-4, 1
D
4, -1
##### Explanation
Mid point of S(-5, 4) and T(-3, -2) is

$$[\frac{1}{2}(-5 + -3), \frac{1}{2}(4 + 2)]$$

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

$$[\frac{1}{2}(-8), \frac{1}{2}(2)]$$

$$[-4, 1]$$

Question 27
A
5
B
3
C
-3
D
-5
##### Explanation
$$\text{Gradient m} = \frac{y_2 - y_1}{x_2 - x_1}$$

$$\frac{1}{2} = \frac{2 - 4}{1 - x}$$

1 - x = 2(2 - 4)

1 - x = 4 - 8

1 - x = -4

-x = -4 - 1

x = 5

Question 28
A
3x - 4y + 18 = 0
B
3x + 2y - 18 = 0
C
4x + 5y + 3 = 0
D
5x - 2y - 11 = 0
##### Explanation
4x + 3y - 5 = 0 (given)

The equation of the line perpendicular to the given line takes the form 3x - 4y = k

Thus, substitution x = -2 and y = 3 in 3x - 4y = k gives;

3(-2) - 4(3) = k

-6 - 12 = k

k = -18

Hence the required equation is 3x - 4y = -18

3x - 4y + 18 = 0

Question 29
A
$$\frac{25}{13}$$
B
$$\frac{18}{13}$$
C
$$\frac{8}{13}$$
D
$$\frac{5}{13}$$

Question 30
A
8x2 - 2x + 1
B
8x2 - 4x + 1
C
12x2 - 2x + 1
D
12x2 - 4x + 1
##### Explanation
If y = 4x3 - 2x2 + x, then;

$$\frac{\delta y}{\delta x}$$ = 3(4x2) - 2(2x) + 1

= 12x2 - 4x + 1

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