Mathematics 2010 | Study Mode

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Question 51
A
6(2x + 1)
B
3(2x + 1)
C
6(2x + 1)2
D
2(2x + 1)2
Explanation
If y = (2x + 1)3, then

Let u = 2x + 1 so that, y = u3

\(\frac{dy}{du}\) = 3u2 and \(\frac{dy}{dx}\) = 2

Hence by the chain rule,

\(\frac{dy}{dx}\) = \(\frac{dy}{du}\) x \(\frac{du}{dx}\)

= 3u2 x 2

= 6u2

= 6(2x + 1)2

Correct option: C
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Question 52
A
sin x - x cosx
B
sinx + x cosx
C
sinx - cosx
D
sinx + cosx
Explanation
If y = x sinx, then

Let u = x and v = sinx

\(\frac{du}{dx}\) = 1 and \(\frac{dv}{dx}\) = cosx

Hence by the product rule,

\(\frac{dy}{dx}\) = v \(\frac{du}{dx}\) + u\(\frac{dv}{dx}\)

= (sin x) x 1 + x cosx

= sinx + x cosx

Correct option: B
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Question 53
A
4\(\frac{5}{6}\)
B
6\(\frac{2}{3}\)
C
1\(\frac{5}{6}\)
D
2\(\frac{5}{6}\)
Explanation
\(\int^{2}_{0}(x^3 + x^2)\)dx = \(\int^{2}_{0}\)(\(\frac{x^4}{4} + {\frac {x^3}{3}}\))

= (\(\frac{2^4}{4} + {\frac {2^3}{3}}\)) - (\(\frac{0^4}{4} + {\frac {0^3}{3}}\))

= (\(\frac{16}{4} + {\frac {8}{3}}\)) - 0

= \(\frac{80}{12}

= {\frac {20}{3}}\) or 6\(\frac{2}{3}\)

Correct option: B
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Question 54
A
6
B
2
C
9
D
7
Explanation
From the table, the number of students who failed the test is given as:

3 + 1 + 5 = 9

Correct option: C
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Question 55
A
16
B
20
C
13
D
15
Explanation
from the table, the number of students who took the test is 2 + 2 + 8 + 4 + 4 = 20

Correct option: B
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Question 56
A
3.1
B
3
C
3.3
D
3.2
Explanation
\(\begin{array} & Marks(x) & freq.(f) & fx \\1 & 2 & 2 \\ 2 & 2 & 4 \\ 3 & 8 & 24 \\ 4 & 4 & 16\\ 5 & 4 & 20 \\ \hline & \sum f = 20 & \sum fx = 66\end{array}\)
___________________________________

Mean mark ,\(\bar{x}\) = \(\frac{\sum fx}{\sum f}\)

= \(\frac{66}{20}\)

\(\bar{x}\) = 3.3

Correct option: C
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Question 57
A
100
B
200
C
30
D
50
Explanation
A committee of 2 women and 3 men can be chosen from 6 men and 5 women, in \(^{5}C_{2}\) x \(^{6}C_{3}\) ways

= \(\frac{5!}{(5 - 2)!2!} \times {\frac{6!}{(6 - 3)!3!}}\)

= \(\frac{5!}{3!2!} \times {\frac{6!}{3 \times 3!}}\)

= \(\frac{5 \times 4 \times 3!}{3! \times 2!} \times {\frac{6 \times 5 \times 4 \times 3!}{3! \times 3!}}\)

= \(\frac{5 \times 4}{1 \times 2} \times {\frac{6 \times 5 \times 4}{1 \times 2 \times 3}}\)

= 10 x \(\frac{6 \times 20}{6}\)

= 200

Correct option: B
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Question 58
A
\(\frac{1}{2}\)
B
\(\frac{1}{3}\)
C
\(\frac{1}{9}\)
D
\(\frac{1}{8}\)
Explanation
P(H) = \(\frac{1}{2}\) and P(T) = \(\frac{1}{2}\)

Using the binomial prob. distribution,

(H + T)3 = H3 + 3H2T1 + 3HT2 + T3

Hence the probability that three heads show in a toss of the three coins is H3

= (\(\frac{1}{2}\))3

= \(\frac{1}{8}\)

Correct option: D
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Question 59
A
6
B
10
C
\(\frac{2}{5}\)
D
\(\frac{5}{2}\)
Explanation
\(\begin{array}& x & x - \bar{x} & (x - \bar{x})^2 \\2 & -2 & 4 \\ 3 & -1 & 1 \\ 5 & 1 & 1 \\ 6 & 2 & 4\\ \hline \sum x = 16 & & \sum (x - \bar{x}^2) = 0 \end{array}\)
___________________________________

\(\bar{x}\) = \(\frac{\sum x }{N}\)

= \(\frac{16}{4}\)

= 4

S = \(\sqrt{\frac {(x - \bar{x})^2}{N}}\)

= \(\sqrt{\frac {(10)}{4}}\)

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

Correct option: D
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Question 60
A
N37550
B
N40400
C
N41400
D
42400
Explanation
Total cost = N34,000 + N2,000

= N36,000

15% = N115

\(\frac{115}{100}\) x N36,000

= N41,400

Correct option: C

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