Mathematics 2010 | Study Mode

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

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

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$$\frac{2}{3}$$

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

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

Question 56
A
3.1
B
3
C
3.3
D
3.2
Explanation
$$\begin{array} & Marks(x) & freq.(f) & fx \\1 & 2 & 2 \\ 2 & 2 & 4 \\ 3 & 8 & 24 \\ 4 & 4 & 16\\ 5 & 4 & 20 \\ \hline & \sum f = 20 & \sum fx = 66\end{array}$$
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Mean mark ,$$\bar{x}$$ = $$\frac{\sum fx}{\sum f}$$

= $$\frac{66}{20}$$

$$\bar{x}$$ = 3.3

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

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

= $$\frac{1}{8}$$

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

= $$\frac{16}{4}$$

= 4

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

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

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

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

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