WAEC Further Mathematics Past Questions & Answers - Page 74

367.

If \(h(x) = x^{3} - \frac{1}{x^{3}}\), evaluate \(h(a) - h(\frac{1}{a})\)

A.

-1

B.

0

C.

\(2a^{3} - \frac{2}{a^{3}}\)

D.

\(\frac{2}{a^{3}} - 2a^{3}\)

Correct answer is C

\(h(x) = x^{3} - \frac{1}{x^{3}}\)

\(h(a) = a^{3} - \frac{1}{a^{3}}\)

\(h(\frac{1}{a}) = (\frac{1}{a})^{3} - \frac{1}{(\frac{1}{a})^{3}} = \frac{1}{a^{3}} - a^{3}\)

\(h(a) - h(\frac{1}{a}) = (a^{3} - \frac{1}{a^{3}}) - (\frac{1}{a^{3}} - a^{3}) = 2a^{3} - \frac{2}{a^{3}}\)

368.

Given that \(\frac{\mathrm d y}{\mathrm d x} = 3x^{2} - 4\) and y = 6 when x = 3, find the equation for y.

A.

\(x^{3} - 4x - 9\)

B.

\(x^{3} - 4x + 9\)

C.

\(x^{3} + 4x - 9\)

D.

\(x^{3} + 4x + 9\)

Correct answer is A

\(\frac{\mathrm d y}{\mathrm d x} = 3x^{2} - 4\)

\(y = \int (3x^{2} - 4) \mathrm {d} x = x^{3} - 4x + c\)

y = 6 when x = 3

\(6 = 3^{3} - 4(3) + c \implies 6 = 27 - 12 + c\)

\(c = 6 - 15 = -9\)

\(y = x^{3} - 4x - 9\)

369.

A stone is thrown vertically upwards and its height at any time t seconds is \(h = 45t - 9t^{2}\). Find the maximum height reached

A.

45. 25 m

B.

45.50 m

C.

56.00 m

D.

56.25 m

Correct answer is D

\(h = 45t - 9t^{2}\)

\(\frac{\mathrm d h}{\mathrm d t} = 45 - 18t = 0\)

\(45 = 18t \implies t = 2.5 s\)

\(h(2.5) = 45(2.5) - 9(2.5)^{2} = 112.5 - 56.25\)

= \(56.25 m.\)

370.

Two forces \(F_{1} = (7i + 8j)N\) and \(F_{2} = (3i + 4j)N\) act on a particle. Find the magnitude and direction of \(F_{1} - F_{2}\)

A.

\((4\sqrt{2} N, 000°)\)

B.

\((4\sqrt{2} N, 045°)\)

C.

\((4\sqrt{2} N, 090°)\)

D.

\((4\sqrt{2} N, 180°)\)

Correct answer is B

\(F_{1} = (7i + 8j)N ; F_{2} = (3i + 4j)N\)

\(|F_{1} - F_{2}| = |(7i + 8j) - (3i + 4j)| = |4i + 4j|\)

\(|4i + 4j| = \sqrt{4^{2} + 4^{2}} = \sqrt{32} = 4\sqrt{2}\)

\(\tan \theta = \frac{y}{x} = \frac{4}{4} = 1\)

\(\theta = \tan^{-1} 1 = 045°\)

= \((4\sqrt{2} N, 045°)\)