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.
\(\begin{pmatrix} 3 \\ -1 \end{pmatrix}\)
\(\begin{pmatrix} -5 \\ -1 \end{pmatrix}\)
\(\begin{pmatrix} -5 \\ 1 \end{pmatrix}\)
\(\begin{pmatrix} -5 \\ -9 \end{pmatrix}\)
Correct answer is B
\(4 \begin{pmatrix} 3 \\ 2 \end{pmatrix} + 3 \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} -3 \\ 5 \end{pmatrix}\)
\(\begin{pmatrix} 12 \\ 8 \end{pmatrix} + \begin{pmatrix} 3x \\ 3y \end{pmatrix} = \begin{pmatrix} -3 \\ 5 \end{pmatrix}\)
\(\begin{pmatrix} 3x \\ 3y \end{pmatrix} = \begin{pmatrix} -3 - 12 \\ 5 - 8 \end{pmatrix}\)
\(\begin{pmatrix} 3x \\ 3y \end{pmatrix} = \begin{pmatrix} -15 \\ -3 \end{pmatrix}\)
\(\begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} -5 \\ -1 \end{pmatrix}\)
12
11
10
8
Correct answer is A
\(Mean : \frac{x + 4 + z + t}{4} = 5 \implies x + 4 + z + t = 20\)
\(\implies x + z + t = 16 ... (1)\)
\(Median : \frac{4 + z}{2} = z \implies 4 + z = 2z\)
\(4 = z\)
From 1,
\(\implies x + 4 + t = 16 \)
\(x + t = 12\)
Find, in surd form, the value of \(\cos 165\).
\(\frac{1}{4}(\sqrt{6} + \sqrt{2})\)
\(\frac{1}{4}(\sqrt{6} - \sqrt{2})\)
\(-\frac{1}{4}(\sqrt{6} - \sqrt{2})\)
\(-\frac{1}{4}(\sqrt{6} + \sqrt{2})\)
Correct answer is D
\(\cos 165 = -\cos (180 - 165) = -\cos 15\)
\(\cos 15 = \cos (45 - 30)\)
\(\cos (x - y) = \cos x \cos y + \sin x \sin y\)
\(\cos (45 - 30) = \cos 45 \cos 30 + \sin 45 \sin 30\)
= \((\frac{\sqrt{2}}{2})(\frac{\sqrt{3}}{2}) + (\frac{\sqrt{2}}{2})(\frac{1}{2})\)
= \(\frac{1}{4}(\sqrt{6} + \sqrt{2})\)
\(\therefore \cos 165 = -\frac{1}{4}(\sqrt{6} + \sqrt{2})\)
\((-1, 5)\)
\((-\frac{1}{2}, 3\frac{1}{2})\)
\((0, 4\frac{1}{2})\)
\((1\frac{1}{2}, 5)\)
Correct answer is C
No explanation has been provided for this answer.
\(\frac{3}{5}\)
\(\frac{1}{3}\)
\(\frac{1}{6}\)
\(\frac{2}{15}\)
Correct answer is C
For independent events A and B, \(P(A \cap B) = P(A) \times P(B)\)
\(\frac{1}{15} = \frac{2}{5} \times P(B)\)
\(P(B) = \frac{1}{15} \div \frac{2}{5}\)
\(P(B) = \frac{1}{6}\)