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.

681.

If \((2x^{2} - x - 3)\) is a factor of \(f(x) = 2x^{3} - 5x^{2} - x + 6\), find the other factor

A.

(x - 2)

B.

(x - 1)

C.

(x + 1)

D.

(x + \(\frac{3}{2}\))

Correct answer is A

Divide \((2x^{3} - 5x^{2} - x + 6)\) by \((2x^{2} - x - 3)\) to get the other factor. Be careful of sign conventions.

682.

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

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

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

\(P \cap Q = { }\)

683.

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

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

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

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

684.

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

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

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

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

685.

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 = \(90° - (-67.38°) = 157.38°\)