Mathematics 2014 | Study Mode

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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)

Correct option: B
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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

Correct option: D
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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

Correct option: A
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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

\(P = \frac{12Q}{R}\)

Correct option: B
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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

Correct option: A
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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

Correct option: A
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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

Correct option: C
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Question 18
A
\(\sqrt{2}\)
B
\(\sqrt{5}\)
C
3
D
5
Explanation
Common ratio r of the G.P is

\(r = \frac{T_n + 1}{T_n} = \frac{T_2}{T_1}\)

\(r = \frac{\sqrt{10} + 2\sqrt{5}}{\sqrt{10} + \sqrt{5}}\)

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

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

\(\frac{10 - \sqrt{50} + 2\sqrt{50} - 10}{10 - 5}\)

\(\frac{\sqrt{50}}{5}\)

\(\frac{\sqrt{25 \times 2}}{5}\)

\(\frac{5\sqrt{2}}{5}\)

\(\sqrt{2}\)

Correct option: A
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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

Correct option: B
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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

Correct option: C

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