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.
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.
\(26 ms^{-2}\)
\(18 ms^{-2}\)
\(17 ms^{-2}\)
\(16 ms^{-2}\)
Correct answer is D
\(v(t) = 3t^{2} - 2t + 1\)
\(\frac{\mathrm d v}{\mathrm d t} = a(t) = 6t - 2\)
\(a(3) = 6(3) - 2 = 18 - 2 = 16 ms^{-2}\)
13 metres
15 metres
18 metres
23 metres
Correct answer is A
\(v^{2} = u^{2} + 2as\)
\(v^{2} = u^{2} - 2gs\)
\(0 = 10^{2} - 2(10)s \implies -100 = -20s\)
\(s = 5 m + 8 m = 13m\) (8m is the height from where it was thrown)
Which of the following is the semi- interquartile range of a distribution?
\(Mode - Median\)
\(\text{Highest score - Lowest score}\)
\(\frac{1}{2}(\text{Upper quartile - Median})\)
\(\frac{1}{2}(\text{Upper quartile - Lower quartile})\)
Correct answer is D
No explanation has been provided for this answer.
If \(P = \begin{vmatrix} 1 & 1 \\ 2 & 1 \end{vmatrix}\), find \((P^{2} + P)\).
\(\begin{vmatrix} 4 & 3 \\ 6 & 1 \end{vmatrix}\)
\(\begin{vmatrix} 4 & 3 \\ 6 & 4 \end{vmatrix}\)
\(\begin{vmatrix} 2 & 2 \\ 6 & 2 \end{vmatrix}\)
\(\begin{vmatrix} 3 & 2 \\ 6 & 4 \end{vmatrix}\)
Correct answer is B
\( P^{2} = \begin{vmatrix} 1 & 1 \\ 2 & 1 \end{vmatrix} \begin{vmatrix} 1 & 1 \\ 2 & 1 \end{vmatrix}\)
\(\begin{vmatrix} 1 \times 1 + 1 \times 2 & 1 \times 1 + 1 \times 1 \\ 2 \times 1 + 1 \times 2 & 2 \times 1 + 1 \times 1 \end{vmatrix}\)
= \(\begin{vmatrix} 3 & 2 \\ 4 & 3 \end{vmatrix} + \begin{vmatrix} 1 & 1 \\ 2 & 1 \end{vmatrix}\)
= \(\begin{vmatrix} 4 & 3 \\ 6 & 4 \end{vmatrix}\)
Evaluate \(\int_{1}^{2} [\frac{x^{3} - 1}{x^{2}}] \mathrm {d} x\)
0.5
1.0
1.5
2.0
Correct answer is B
\(\frac{x^{3} - 1}{x^{2}} \equiv x - \frac{1}{x^{2}} = x - x^{-2}\)
\(\int_{1}^{2} (x - x^{-2}) \mathrm {d} x = (\frac{x^{2}}{2} + \frac{1}{x})|_{1}^{2}\)
= \((\frac{2^{2}}{2} + \frac{1}{2}) - (\frac{1^{2}}{2} + \frac{1}{1})\)
= \((2 + \frac{1}{2}) - (\frac{1}{2} + 1)\)
= \(2\frac{1}{2} - 1\frac{1}{2}\)
= \(1.0\)