WAEC Further Mathematics Past Questions & Answers - Page 101

501.

The angle subtended by an arc of a circle at the centre is \(\frac{\pi}{3} radians\). If the radius of the circle is 12cm, calculate the perimeter of the major arc.

A.

\(4(6 + 5\pi)\)

B.

\(4(6 + 2\pi)\)

C.

\(4(3 + 3\pi)\)

D.

\(4(3 + 5\pi)\)

Correct answer is A

The angle subtended by the minor arc = \(\frac{\pi}{3} radians\)

The angle subtended by the major arc = \(2\pi - \frac{\pi}{3} = \frac{5\pi}{3}\)

Perimeter of the major arc = \(r\theta + 2r\)

= \(12 \times \frac{5\pi}{3} + 2(12) = 20\pi + 24\)

= \(4(5\pi + 6)\)

502.

Find the coordinates of the point which divides the line joining P(-2, 3) and Q(4, 9) internally in the ratio 2 : 3.

A.

\((5\frac{2}{3}, \frac{2}{5})\)

B.

\((\frac{2}{5}, 5\frac{2}{5})\)

C.

\((\frac{2}{5}, 2\frac{2}{5})\)

D.

\((\frac{-2}{5}, 5\frac{2}{5})\)

Correct answer is B

\((x = \frac{nx_{1} + mx_{2}}{n + m}, y = \frac{ny_{1} + my_{2}}{n + m})\)

Given P(-2, 3) and Q(4, 9),

\((\frac{2(4) + 3(-2)}{2 + 3}, \frac{2(9) + 3(3)}{2 + 3})\)

= \((\frac{2}{5}, \frac{27}{5})\)

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

503.

Evaluate \(\int_{0}^{2} (8x - 4x^{2}) \mathrm {d} x\).

A.

\(-16\)

B.

\(\frac{-16}{3}\)

C.

\(\frac{16}{3}\)

D.

\(16\)

Correct answer is C

\(\int_{0}^{2} (8x - 4x^{2}) \mathrm {d} x  = (\frac{8x^{1 + 1}}{2} - \frac{4x^{2+1}}{3})|_{0}^{2}\)

= \((4x^{2} - \frac{4x^{3}}{3}) |_{0}^{2}\)

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

= \(16 - \frac{32}{3} = \frac{16}{3}\)

504.

An object is thrown vertically upwards from the top of a cliff with a velocity of \(25ms^{-1}\). Find the time, in seconds, when it is 20 metres above the cliff. \([g = 10ms^{-2}]\).

A.

0 and 1

B.

0 and 4

C.

0 and 5

D.

1 and 4

Correct answer is D

\(s = ut + \frac{at^{2}}{2}\)

This movement is against gravity, so it is negative.

\(s = ut - \frac{gt^{2}}{2}\)

\(s = 20m, u = 25ms^{-1}\)

\(20 = 25t - \frac{10t^{2}}{2} \implies 20 = 25t - 5t^{2}\)

\(5t^{2} - 25t + 20 = 0 \)

\(5t^{2} - 5t - 20t + 20 = 0 \implies 5t(t - 1) - 20(t - 1) = 0\)

\(5t - 20 = \text{0 or t - 1 = 0}\)

\(t = \text{1 or 4}\)

505.

Given that \(P = \begin{pmatrix} y - 2 & y - 1 \\ y - 4 & y + 2 \end{pmatrix}\) and |P| = -23, find the value of y.

A.

-4

B.

-3

C.

-1

D.

2

Correct answer is B

\(P = begin{pmatrix} y - 2 & y - 1 \\ y - 4 & y + 2 \end{pmatrix}\)

\(|P| = (y - 2)(y + 2) - (y - 1)(y - 4) = (y^{2} - 4) - (y^{2} - 5y + 4) = -23\)

\(5y - 8 = -23 \implies 5y = -23 + 8 = -15\)

\(y = \frac{-15}{5} = -3\)