WAEC Further Mathematics Past Questions & Answers

336.

The velocity $v ms^{-1}$ of a particle moving in a straight line is given by $v = 3t^{2} - 2t + 1$ at time t secs. Find the acceleration of the particle after 3 seconds.

$26 ms^{-2}$

$18 ms^{-2}$

$17 ms^{-2}$

$16 ms^{-2}$

View answer & explanation

Correct answer is D

$v(t) = 3t^{2} - 2t + 1$

$\frac{\mathrm d v}{\mathrm d t} = a(t) = 6t - 2$

$a(3) = 6(3) - 2 = 18 - 2 = 16 ms^{-2}$

337.

A stone is projected vertically with a speed of 10 m/s from a point 8 metres above the ground. Find the maximum height reached. $[g = 10 ms^{-2}]$ .

13 metres

15 metres

18 metres

23 metres

View answer & explanation

Correct answer is A

$v^{2} = u^{2} + 2as$

$v^{2} = u^{2} - 2gs$

$0 = 10^{2} - 2(10)s \implies -100 = -20s$

$s = 5 m + 8 m = 13m$ (8m is the height from where it was thrown)

338.

Which of the following is the semi- interquartile range of a distribution?

$Mode - Median$

$\text{Highest score - Lowest score}$

$\frac{1}{2}(\text{Upper quartile - Median})$

$\frac{1}{2}(\text{Upper quartile - Lower quartile})$

View answer & explanation

Correct answer is D

No explanation has been provided for this answer.

339.

If $P = \begin{vmatrix} 1 & 1 \\ 2 & 1 \end{vmatrix}$ , find $(P^{2} + P)$ .

$\begin{vmatrix} 4 & 3 \\ 6 & 1 \end{vmatrix}$

$\begin{vmatrix} 4 & 3 \\ 6 & 4 \end{vmatrix}$

$\begin{vmatrix} 2 & 2 \\ 6 & 2 \end{vmatrix}$

$\begin{vmatrix} 3 & 2 \\ 6 & 4 \end{vmatrix}$

View answer & explanation

Correct answer is B

$P^{2} = \begin{vmatrix} 1 & 1 \\ 2 & 1 \end{vmatrix} \begin{vmatrix} 1 & 1 \\ 2 & 1 \end{vmatrix}$

$\begin{vmatrix} 1 \times 1 + 1 \times 2 & 1 \times 1 + 1 \times 1 \\ 2 \times 1 + 1 \times 2 & 2 \times 1 + 1 \times 1 \end{vmatrix}$

= $\begin{vmatrix} 3 & 2 \\ 4 & 3 \end{vmatrix} + \begin{vmatrix} 1 & 1 \\ 2 & 1 \end{vmatrix}$

= $\begin{vmatrix} 4 & 3 \\ 6 & 4 \end{vmatrix}$

340.

Evaluate $\int_{1}^{2} [\frac{x^{3} - 1}{x^{2}}] \mathrm {d} x$

0.5

1.0

1.5

2.0

View answer & explanation

Correct answer is B

$\frac{x^{3} - 1}{x^{2}} \equiv x - \frac{1}{x^{2}} = x - x^{-2}$

$\int_{1}^{2} (x - x^{-2}) \mathrm {d} x = (\frac{x^{2}}{2} + \frac{1}{x})|_{1}^{2}$

= $(\frac{2^{2}}{2} + \frac{1}{2}) - (\frac{1^{2}}{2} + \frac{1}{1})$

= $(2 + \frac{1}{2}) - (\frac{1}{2} + 1)$

= $2\frac{1}{2} - 1\frac{1}{2}$

= $1.0$

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