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.
If \((2x^{2} - x - 3)\) is a factor of \(f(x) = 2x^{3} - 5x^{2} - x + 6\), find the other factor
(x - 2)
(x - 1)
(x + 1)
(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.
If P = \({n^{2} + 1: n = 0,2,3}\) and Q = \({n + 1: n = 2,3,5}\), find P\(\cap\) Q.
{5, 10}
{4, 6}
{1, 3}
{ }
Correct answer is D
\(P = {n^{2} + 1: n = 0,2,3} \therefore P = {1, 5,10}\)
\(Q = {n + 1: n = 2,3,5} \therefore Q = {3, 4, 6}\)
\(P \cap Q = { }\)
Given that \(f(x) = 2x^{2} - 3\) and \(g(x) = x + 1\) where \(x \in R\). Find g o f(x).
\(2(x^{2} - 1)\)
\(2x^{2} + 4x - 1\)
\(2x^{2} + 6x - 1\)
\(3(x^{2} - 1)\)
Correct answer is A
\(f(x) = 2x^{2} - 3; g(x) = x + 1\)
\(g o f(x) = g (2x^{2} - 3)\)
= \( 2x^{2} - 3 + 1 = 2x^{2} - 2 = 2(x^{2} - 1)\)
10\(ms^{-2}\)
12\(ms^{-2}\)
14\(ms^{-2}\)
17\(ms^{-2}\)
Correct answer is C
\(accl = \frac{\mathrm d V}{\mathrm d t}\)
\(V = 3t^{2} + 2t - 1 \therefore a = \frac{\mathrm d V}{\mathrm d t} = 6t + 2\)
\(a \text{(after 2 seconds)} = (6\times 2) + 2 = 12+2 = 14ms^{-2}\)
Find the magnitude and direction of the vector \(p = (5i - 12j)\)
(13, 113.38°)
(13, 067.38°)
(13, 025.38°)
(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 = \(90° - (-67.38°) = 157.38°\)