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.
-4
-3
-1
2
Correct answer is B
\(P = begin{pmatrix} y - 2 & y - 1 \\ y - 4 & y + 2 \end{pmatrix}\)
\(|P| = (y - 2)(y + 2) - (y - 1)(y - 4) = (y^{2} - 4) - (y^{2} - 5y + 4) = -23\)
\(5y - 8 = -23 \implies 5y = -23 + 8 = -15\)
\(y = \frac{-15}{5} = -3\)
Given that \(\frac{\mathrm d y}{\mathrm d x} = \sqrt{x}\), find y.
\(2x^{\frac{3}{2}} + c\)
\(\frac{2}{3}x^{\frac{3}{2}} + c\)
\(\frac{3}{2}x^{\frac{3}{2}} + c\)
\(\frac{2}{3}x^{2} + c\)
Correct answer is B
\(\frac{\mathrm d y}{\mathrm d x} = \sqrt{x} = x^{\frac{1}{2}}\)
\(y = \int x^{\frac{1}{2}} \mathrm {d} x\)
= \(\frac{x^{\frac{1}{2} + 1}}{\frac{1}{2} + 1} + c = \frac{x^{\frac{3}{2}}}{\frac{3}{2}} + c \)
= \(\frac{2}{3}x^{\frac{3}{2}} + c\)
Find the range of values of x for which \(x^{2} + 4x + 5\) is less than \(3x^{2} - x + 2\)
\(x > \frac{-1}{2}, x > 3\)
\(x < \frac{-1}{2}, x > 3\)
\(\frac{-1}{2} \leq x \leq 3\)
\(\frac{-1}{2} < x < 3\)
Correct answer is B
No explanation has been provided for this answer.
\(\frac{1}{3}\)
\(\frac{1}{2}\)
\(2\)
\(3\)
Correct answer is B
\(T_{n} = ar^{n - 1}\)
\(T_{4} = ar^{4 - 1} = ar^{3} = 192\)
\(T_{9} = ar^{9 - 1} = ar^{8} = 6\)
Dividing \(T_{9}\) by \(T_{4}\),
\(r^{8 - 3} = \frac{6}{192}\)
\(r^{5} = \frac{1}{32} = (\frac{1}{2})^{5}\)
\(r = \frac{1}{2}\)
Differentiate \(x^{2} + xy - 5 = 0\)
\(\frac{-(2x + y)}{x}\)
\(\frac{(2x - y)}{x}\)
\(\frac{-x}{2x + y}\)
\(\frac{(2x + y)}{x}\)
Correct answer is A
\(\frac{\mathrm d}{\mathrm d x}(x^2 + xy - 5) = \frac{\mathrm d (x^{2})}{\mathrm d x} + \frac{\mathrm d (xy)}{\mathrm d x} - \frac{\mathrm d (5)}{\mathrm d x} = 0\)
= \(2x + x\frac{\mathrm d y}{\mathrm d x} + y = 0\)
\(\implies x\frac{\mathrm d y}{\mathrm d x} = -(2x + y)\)
\(\frac{\mathrm d y}{\mathrm d x} = \frac{-(2x + y)}{x}\)