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.
Find the angle between \((5i + 3j)\) and \((3i - 5j)\)
180°
90°
45°
0°
Correct answer is B
\(a . b = |a||b|\cos \theta\)
\(\cos \theta = \frac{a . b}{|a||b|}\)
= \( \frac{(5i + 3j).(3i - 5j)}{(\sqrt{5^2 + 3^2})(\sqrt{3^{2} + (-5)^{2}})}\)
= \(\frac{0}{34} = 0\)
\(\theta = \cos^{-1} 0 = 90°\)
\(4\sqrt{2}\)
\(6\sqrt{2}\)
\(2\sqrt{10}\)
\(4\sqrt{10}\)
Correct answer is C
\(BC = BA + AC\)
Given, \(AB\), then \(BA = - AB\)
= \(AB = \begin{pmatrix} 4 \\ 3 \end{pmatrix} \implies BA = \begin{pmatrix} -4 \\ -3 \end{pmatrix}\)
\(\therefore BC = \begin{pmatrix} -4 \\ -3 \end{pmatrix} + \begin{pmatrix} 2 \\ -3 \end{pmatrix}\)
= \(\begin{pmatrix} -2 \\ -6 \end{pmatrix}\)
\(|BC| = \sqrt{(-2)^{2} + (-6)^{2}} = \sqrt{40} \)
= \(2\sqrt{10}\)
Integrate \((x - \frac{1}{x})^{2}\) with respect to x.
\(\frac{1}{3}(x - \frac{1}{x})^{3} + c\)
\(\frac{x^{3}}{3} - x\sqrt{\frac{1}{x^{3}}} + c\)
\(\frac{x^{3}}{3} - 2x + \frac{1}{x^{3}} + c\)
\(\frac{x^3}{3} - 2x - \frac{1}{x} + c\)
Correct answer is D
\((x - \frac{1}{x})^{2} = x^2 - 2 + \frac{1}{x^2}\)
\(\int (x^2 + \frac{1}{x^2} - 2) \mathrm {d} x\)
= \(\int (x^2 + x^{-2} - 2) \mathrm {d} x\)
= \(\frac{x^3}{3} - 2x - \frac{1}{x}\)
If \(Px^{2} + (P+1)x + P = 0\) has equal roots, find the values of P.
\(\text{-1 and }\frac{-1}{3}\)
\(\text{1 and }\frac{-1}{3}\)
\(\text{-1 and }\frac{1}{3}\)
\(\text{1 and }\frac{1}{3}\)
Correct answer is B
For equal roots, \(b^{2} - 4ac = 0\)
From the equation, \(a = P, b = (P+1), c = P\)
\((P+1)^{2} - 4(P)(P) = P^{2} + 2P + 1 - 4P^{2} = 0\)
\(-3P^{2} + 2P + 1 = 0 \implies 3P^{2} - 2P - 1 = 0\)
\(3P^{2} - 3P + P - 1 = 0\)
\(3P(P - 1) + 1(P - 1) = 0\)
\(P = \text{1 or }\frac{-1}{3}\)
| Age in years | 10 - 14 | 15 - 19 | 20 - 24 | 25 - 29 | 30 - 34 |
| Frequency | 6 | 8 | 14 | 10 | 12 |
Find the mean of the distribution.
23.4
23.6
24.3
24.6
Correct answer is B
| Age in years |
Classmark (x) |
Frequency (f) |
fx |
| 10 - 14 | 12 | 6 | 72 |
| 15 - 19 | 17 | 8 | 136 |
| 20 - 24 | 22 | 14 | 308 |
| 25 - 29 | 27 | 10 | 270 |
| 30 - 34 | 32 | 12 | 384 |
| Total | 50 | 1170 |
\(Mean = \frac{\sum fx}{\sum f}\)
= \(\frac{1170}{50} = 23.4\)