Mathematics 2014 | Study Mode

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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,

\(\frac{\delta y}{\delta x} = \frac{\delta y}{\delta u} \times \frac{\delta u}{\delta x}\)

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

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

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

Correct option: D
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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

Correct option: D
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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
\(\int \sin 2x dx = \frac{1}{2} (-\cos 2x) + k\)

\(- \frac{1}{2} \cos 2x + k\)

Correct option: C
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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 \(\int (2x + 3)^{\frac{1}{2}} \delta x = \int u^{\frac{1}{2}}.{\frac{\delta x}{2}}\)

\( = \frac{1}{2} \int u^{\frac{1}{2}} \delta u\)

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

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

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

Correct option: C
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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;

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

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

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

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

Correct option: A
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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

Correct option: C
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Question 37
A
1
B
2
C
3
D
4
Explanation

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


Correct option: D
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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

Correct option: B
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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;

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


\( = \sqrt{2}\)

Correct option: A
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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 \(^7C_3\)

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

\( = \frac{7!}{4! 3!} ways\)

Correct option: C

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