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.
5
4
3
2
Correct answer is D
\(S_{n} = \frac{n}{2}(2a + (n - 1)d \)
\(\frac{n}{2}(2a + (n - 1) d = n^{2} + 2n\)
\(n(2a + (n - 1) d = 2n^{2} + 4n\)
\(2an + n^{2}d - nd = 2n^{2} + 4n\)
Comparing the equations, d = 2 (coefficient of \(n^{2}\)).
\(f^{-1} : x \to \frac{2x + 3}{x - 1}, x \neq 1\)
\(f^{-1} : x \to \frac{x + 3}{x + 2}, x \neq -2\)
\(f^{-1} : x \to \frac{x - 1}{2x + 3}, x \neq -\frac{3}{2}\)
\(f^{-1}: x \to \frac{x - 2}{x + 3}, x \neq -3\)
Correct answer is A
\(f(x) = \frac{x + 3}{x - 2}\)
\(f(y) = \frac{y + 3}{y - 2}\)
Let f(y) = x,
\(x = \frac{y + 3}{y - 2}\)
\(x(y - 2) = y + 3\)
\(xy - y = 2x + 3 \implies y(x - 1) = 2x + 3\)
\(y = \frac{2x + 3}{x - 1}\)
Find the range of values of x for which \(2x^{2} + 7x - 15 > 0\).
\(x < -\frac{3}{2}\) or \(x > 5\)
\(x < -5\) or \(x > \frac{3}{2}\)
\(-\frac{3}{2} < x < 5\)
\(-5 < x < \frac{3}{2}\)
Correct answer is B
\(2x^{2} + 7x - 15 > 0 \implies 2x^{2} - 3x + 10x - 15 > 0\)
\(x(2x - 3) + 5(2x - 3) > 0\)
\((x + 5)(2x - 3) > 0\)
For their product to be positive, they are either both +ve or -ve.
\(x + 5 > 0 \implies x > -5\)
\(2x - 3 > 0 \implies 2x > 3\)
\(x > \frac{3}{2}\)
Check:
\(x > -5: x = -3\)
\(2(-3)^{2} + 7(-3) - 15 = 18 - 21 - 15 = -18 < 0\) (Not satisfied)
\(\therefore x < -5\)
\(x > \frac{3}{2}: x = 2\)
\(2(2^{2}) + 7(2) - 15 = 8 + 14 - 15 = 7 > 0\) (Satisfied)
Simplify \(\frac{\sqrt{3} + \sqrt{48}}{\sqrt{6}}\)
\(3\sqrt{2}\)
\(5\sqrt{2}\)
\(\frac{5\sqrt{2}}{2}\)
\(\frac{3\sqrt{2}}{2}\)
Correct answer is C
\(\frac{\sqrt{3} + \sqrt{48}}{\sqrt{6}} = \frac{\sqrt{3} + 4\sqrt{3}}{\sqrt{6}}\)
\(\frac{5\sqrt{3}}{\sqrt{6} = \frac{5\sqrt{3} \times \sqrt{6}}{\sqrt{6} \times \sqrt{6}}\)
\(\frac{5\sqrt{18}{6} = \frac{15\sqrt{2}}{6}\)
= \(\frac{5\sqrt{2}}{2}\)
\(\frac{\sqrt{3} \tan x - 1}{\sqrt{3} + \tan x}\)
\(\frac{\sqrt{3} \tan x}{\sqrt{3} + \tan x}\)
\(\frac{\sqrt{3} \tan x + 1}{\sqrt{3} - \tan x}\)
\(\frac{3 \tan x - 1}{\sqrt{3} - \tan x}\)
Correct answer is A
No explanation has been provided for this answer.