Mathematics 2014 | Study Mode

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Question 21
A
-6
B
-2
C
3
D
6
Explanation
\(\begin{vmatrix}
-x & 12 \\
-1 & 4
\end{vmatrix} = - 12\)

-4x - (-1)12 = -12

-4x + 12 = -12

-4x = -12 - 12

-4x = - 24

x = 6

Correct option: D
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Question 22
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

Correct option: D
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Question 23
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

Correct option: D
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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

\(h = \frac{6160}{22 \times 4 \times 28} \)

\(h = 2.5m \)

Correct option: D
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Question 25
A
circle with diameter 4m
B
circle with radius 4m
C
semi-circle with diameter 4m
D
semi-circle with radius 4m
Explanation

Explanation not provided, please join discuss, your contributions are welcomed


Correct option: B
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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

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

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

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

\([-4, 1]\)

Correct option: C
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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

Correct option: A
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Question 28
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;

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

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

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

\([\frac{3}{8}, \frac{-3}{2}]\)

Correct option: A
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Question 29
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

Correct option: A
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Question 30
A
\(\frac{25}{13}\)
B
\(\frac{18}{13}\)
C
\(\frac{8}{13}\)
D
\(\frac{5}{13}\)
Explanation

Explanation not provided, please join discuss, your contributions are welcomed


Correct option: B

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