Further Mathematics questions and answers

Further Mathematics Questions and Answers

Test your knowledge of advanced level mathematics with this aptitude test. This test comprises Further Maths questions and answers from past JAMB and WAEC examinations.

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

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

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

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

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

\(\theta = \tan^{-1} 1 = 045°\)

= \((4\sqrt{2} N, 045°)\)