WAEC Further Mathematics Past Questions & Answers - Page 62

306.

If \(P = \begin{pmatrix} 1 & -2 \\ 3 & 4 \end{pmatrix}\) and \(Q = \begin{pmatrix} -2 & 3 \\ 1 & 0 \end{pmatrix}\), find PQ.

A.

\(\begin{pmatrix} 4 & 1 \\ -2 & 9 \end{pmatrix}\)

B.

\(\begin{pmatrix} -4 & 1 \\ 2 & 9 \end{pmatrix}\)

C.

\(\begin{pmatrix} -4 & 3 \\ -2 & 13 \end{pmatrix}\)

D.

\(\begin{pmatrix} -4 & 3 \\ -2 & 9 \end{pmatrix}\)

Correct answer is D

\(\begin{pmatrix} 1 & -2 \\ 3 & 4 \end{pmatrix} \(\begin{pmatrix} -2 & 3 \\ 1 & 0 \end{pmatrix}\)

\(\begin{pmatrix} (1 \times -2) + (-2 \times 1) & (1 \times 3) + (-2 \times 0) \\ (3 \times -2) + (4 \times 1) & (3 \times 3) + (4 \times 0) \end{pmatrix}\)

= \(\begin{pmatrix} -4 & 3 \\ -2 & 9 \end{pmatrix}\)

307.

Express \(\frac{7\pi}{6}\) radians in degrees.

A.

315°

B.

210°

C.

105°

D.

75°

Correct answer is B

\(\pi = 180°\)

\(\frac{7\pi}{6} = \frac{7 \times 180}{6} \)

= \(210°\)

308.

A rectangle has a perimeter of 24m. If its area is to be maximum, find its dimension.

A.

12, 12

B.

6, 6

C.

4, 8

D.

9, 3

Correct answer is B

\(Perimeter = 2(l + b) = 24\)

\(l + b = 12 \implies l = 12 - b\)

\(Area = (12 - b) \times b = 12b - b^{2}\)

\(\frac{\mathrm d A}{\mathrm d b} = 12 - 2b = 0\) (at maximum)

\(2b = 12 \implies b = 6\)

\(l = 12 - 6 = 6m\)

309.

Evaluate \(\int_{1}^{2} (2 + 2x - 3x^{2}) \mathrm {d} x\).

A.

-2

B.

2

C.

8

D.

10

Correct answer is A

\(\int_{1}^{2} (2 + 2x - 3x^{2}) \mathrm {d} x\)

= \((2x + x^{2} - x^{3})|_{1}^{2}\)

= \((2(2) + 2^{2} - 2^{3}) - (2(1) + 1^{2} - 1^{3})\)

= \(0 - 2 = -2\)

310.

What is the coordinate of the centre of the circle \(5x^{2} + 5y^{2} - 15x + 25y - 3 = 0\)?

A.

\((\frac{15}{2}, -\frac{25}{2})\)

B.

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

C.

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

D.

\((-\frac{15}{2}, \frac{25}{2})\)

Correct answer is B

Equation for a circle: \((x - a)^{2} + (y - b)^{2} = r^{2}\)

Expanding, we have:

\(x^{2} - 2ax + a^{2} + y^{2} - 2by + b^{2} = r^{2}\)

Given: \(5x^{2} + 5y^{2} - 15x + 25y - 3 = 0\)

Divide through by 5,

\(= x^{2} + y^{2} - 3x + 5y - \frac{3}{5} = 0\)

Comparing, we have

\(- 2a = -3; a = \frac{3}{2}\)

\(-2b = 5; b = -\frac{5}{2}\)