WAEC Further Mathematics Past Questions & Answers - Page 137

681.

If P = \({n^{2} + 1: n = 0,2,3}\) and Q = \({n + 1: n = 2,3,5}\), find P\(\cap\) Q.

A.

{5, 10}

B.

{4, 6}

C.

{1, 3}

D.

{ }

Correct answer is D

\(P = {n^{2} + 1: n = 0,2,3} \therefore P = {1, 5,10}\)

\(Q = {n + 1: n = 2,3,5} \therefore Q = {3, 4, 6}\)

\(P \cap Q = { }\)

682.

Given that \(f(x) = 2x^{2} - 3\) and \(g(x) = x + 1\) where \(x \in R\). Find g o f(x).

A.

\(2(x^{2} - 1)\)

B.

\(2x^{2} + 4x - 1\)

C.

\(2x^{2} + 6x - 1\)

D.

\(3(x^{2} - 1)\)

Correct answer is A

\(f(x) = 2x^{2} - 3; g(x) = x + 1\)

\(g o f(x) = g (2x^{2} - 3)\)

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

683.

The velocity, V, of a particle after t seconds, is \(V = 3t^{2} + 2t - 1\). Find the acceleration of the particle after 2 seconds.

A.

10\(ms^{-2}\)

B.

12\(ms^{-2}\)

C.

14\(ms^{-2}\)

D.

17\(ms^{-2}\)

Correct answer is C

\(accl = \frac{\mathrm d V}{\mathrm d t}\)

\(V = 3t^{2} + 2t - 1 \therefore a = \frac{\mathrm d V}{\mathrm d t} = 6t + 2\)

\(a \text{(after 2 seconds)} = (6\times 2) + 2 = 12+2 = 14ms^{-2}\)

684.

Find the magnitude and direction of the vector \(p = (5i - 12j)\)

A.

(13, 113.38°)

B.

(13, 067.38°)

C.

(13, 025.38°)

D.

(13, 157.38°)

Correct answer is D

\(p = (5i - 12j); |p| = \sqrt{5^{2} + (-12)^{2}}\)

= \(\sqrt{169} = 13\)

\(\tan\theta = \frac{-12}{5} = -2.4 \implies \theta = -67.38°\)

Direction = \(90° - (-67.38°) = 157.38°\)

685.

Solve \(3^{2x} - 3^{x+2} = 3^{x+1} - 27\)

A.

1 or 0

B.

1 or 2

C.

1 or -2

D.

-1 or 2

Correct answer is B

\(3^{2x} - 3^{x+2} = 3^{x+1} - 27\)

= \((3^{x})^{2} - (3^{x}).(3^{2}) = (3^{x}).(3^{1}) - 27\)

Let \(3^{x}\) be B; we have

= \(B^{2} - 9B - 3B + 27 = B^{2} - 12B + 27 = 0\).

Solving the equation, we have B = 3 or 9.

\(3^{x} = 3\) or \(3^{x} = 9\)

\(3^{x} = 3^{1}\) or \(3^{x} = 3^{2}\)

Equating, we have x = 1 or 2.