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.
Given that \(f '(x) = 3x^{2} - 6x + 1\) and f(3) = 5, find f(x).
\(f(x) = x^{3} - 3x^{2} + x + 20\)
\(f(x) = x^{3} - 3x^{2} + x + 31\)
\(f(x) = x^{3} - 3x^{2} + x + 2\)
\(f(x) = x^{3} - 3x^{2} + x - 13\)
Correct answer is C
\(f ' (x) = 3x^{2} - 6x + 1\)
\(f(x) = \int (3x^{2} - 6x + 1) \mathrm {d} x\)
= \(x^{3} - 3x^{2} + x + c\)
\(f(3) = 5 = 3^{3} - 3(3^{2}) + 3 + c\)
\(27 - 27 + 3 + c = 5 \implies 3 + c = 5\)
\(c = 2\)
\(f(x) = x^{3} - 3x^{2} + x + 2\)
2
\(\frac{1}{2}\)
\(-\frac{1}{2}\)
-2
Correct answer is B
\((1 + kx)^{8} = ^{8}C_{0}(1^{8})(kx)^{0} + ^{8}C_{1}(1^{7})(kx)^{1} + ...\)
The 5th term = \(^{8}C_{5 - 1}(1^{4})(kx)^{4}\)
= \(^{8}C_{4} (kx)^{4}\)
\(\implies 70k^{4} = \frac{35}{8}\)
\(k^{4} = \frac{\frac{35}{8}}{70}\)
\(k^{4} = \frac{1}{16}\)
\(k = \frac{1}{2}\)
Find the derivative of \(3x^{2} + \frac{1}{x^{2}}\)
\(6x + 2x^{2}\)
\(6x + \frac{1}{2x}\)
\(6x - \frac{2}{x^{3}}\)
\(6x - \frac{1}{2x}\)
Correct answer is C
\(y = 3x^{2} + \frac{1}{x^{2}} \equiv y = 3x^{2} + x^{-2}\)
\(\frac{\mathrm d y}{\mathrm d x} = 6x - 2x^{-3}\)
= \(6x - \frac{2}{x^{3}}\)
7
2
-2
-7
Correct answer is B
\(a * b = a^{2} + b + ab\)
\(5 * x = 5^{2} + x + 5x = 37\)
\(25 + 6x = 37 \implies 6x = 12\)
\(x = 2\)
Find the locus of points which is equidistant from P(4, 5) and Q(-6, -1).
3x - 5y + 13 = 0
3x - 5y - 7 = 0
5x - 3y + 7 = 0
5x + 3y - 1 = 0
Correct answer is D
Midpoint between P(4, 5) and Q(-6, -1) = \(\frac{4 - 6}{2}, \frac{5 - 1}{2} = (-1, 2)\)
Gradient of PQ = \(\frac{5 - (-1)}{4 - (-6)} = \frac{3}{5}\)
Gradient of line perpendicular to PQ = \(\frac{-1}{\frac{3}{5}} = -\frac{5}{3}\)
\(line = \frac{y - 2}{x + 1} = \frac{-5}{3}\)
\(3y - 6 = -5x - 5\)
\(5x + 3y - 6 + 5 = 0 \implies 5x + 3y - 1 = 0\)