# Mathematics 2014 | Study Mode

Question 11
A
(2y - 3x) (y + 6x)
B
(2y - 3x) (y - 6x)
C
(2y + 3x) (y - 6x)
D
(3y + 2x) (y - 6x)
##### Explanation
2y2 - 15xy + 18x2

2y2 - 12xy - 3xy + 18x2

2y(y - 6x) - 3x(y - 6x)

(2y - 3x) (y - 6x)

Question 12
A
-6
B
-4
C
O
D
1
##### Explanation
if y - 1 is a factor of y3 + 4y2 + ky - 6, then

f(1) = (1)3 + 4(1)2 + k(1) - 6 = 0 (factor theorem)

1 + 4 + k - 6 = 0

5 - 6 + k = 0

-1 + k = 0

k = 1

Question 13
A
18
B
12
C
9
D
6
##### Explanation
$$y \propto w^2$$

y = kw2

8 = k(2)2

8 = k(4)

k = 8/4

k = 2

Thus y = 2w2
When w = 3, y = 2(3)2

y = 2 x 9 = 18

Question 14
A
$$P = \frac{Q}{12R}$$
B
$$P = \frac{12Q}{R}$$
C
$$P = 12QR$$
D
$$P = \frac{12}{QR}$$
##### Explanation
$$P \propto \frac{Q}{R}$$

$$P = K \frac{Q}{R}$$

When Q = 36, R = 16, P = 27

Then substitute into the equation

$$27 = K \frac{36}{16}$$

$$K = \frac{27 \times 16}{36}$$

$$K = 12$$

So the equation connecting P, Q and R is

$$P = \frac{12Q}{R}$$

Question 15
A
-3 < x < 1
B
x < -3 or x > 1
C
-3 < x < 5
D
x < -3 or x > 5
##### Explanation
Consider the range -3 < x < -1

= { -2, -1, 0}, for instance

When x = -2,

$$\frac{-2 - 5}{-2 + 3} < -1$$

$$\frac{-7}{1} < -1$$

When x = -1,

$$\frac{-1 - 5}{-1 + 3} < -1$$

$$\frac{-6}{2} < -1$$

= -3 < -1

When x = 0,

$$\frac{0 - 5}{0 + 3} < -1$$

$$\frac{- 5}{3} < -1$$

Hence -3 < x < 1

Question 16
A
$$x \geq 4$$
B
$$x \leq 3$$
C
$$x \geq -3$$
D
$$x \leq -4$$
##### Explanation
$$\frac{x}{2} + \frac{3}{4} \leq \frac{5x}{6} - \frac{7}{12}$$

$$12\frac{x}{2} + 12\frac{3}{4} \leq 12\frac{5x}{6} - 12\frac{7}{12}$$

6x + 9 $$\leq$$ 10x - 7

6x - 10x $$\leq$$ - 7 - 9

-4x $$\leq$$ -16

-4x/-4 $$\geq$$ -16/-4

x $$\geq$$ 4

Question 17
A
89
B
75
C
73
D
69
##### Explanation
a + 3d = 13 .......... (1)
a + 9d = 31 .......... (2)

(2) - (1): 6d = 18

d = 18/6 = 3

From (1), a + 3(3) = 13

a + 9 = 13

a = 13 - 9 = 4

Hence,
T24 = a + 23d
T24 = 4 + 23(3)
T24 = 4 + 69
T24 = 73

Question 18
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 19
A
3,4
B
3,-4
C
-3,4
D
-3,-4
##### Explanation
x * y = xy
x * 2 = 12 - x

Thus by comparison,

x = x, y = 2

But x * y = x * 2

xy = 12 - x

x2 = 12 - x

x2 + x - 12 = 0

x2 + 4x - 3x - 12 = 0

x(x + 4) - 3(x + 4) = 0

(x - 3)(x + 4) = 0

x - 3 = 0 or x + 4 = 0

So x = 3 or x = -4

Question 20
A
8
B
5
C
3
D
2
##### Explanation
$$\begin{pmatrix} 5 & -6 \\ 2 & -7 \end{pmatrix}\begin{pmatrix} 5 \\ 2 \end{pmatrix} = \begin{pmatrix} 7 \\ -11 \end{pmatrix}$$

By matrices multiplication;

5x - 6y = 7 ........(1)
2x - 7y = -11 ......(2)
2 x (1): 10x - 12y = 14 .......(3)
5 x (2): 10x - 35y = -55 ......(4)

(3) - (4): 23y = 69

y = 69/23 = 3

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