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.
Evaluate \(\lim \limits_{x \to 1} \frac{1 - x}{x^{2} - 3x + 2}\)
1
\(\frac{1}{2}\)
0
-1
Correct answer is A
\(\lim \limits_{x \to 1} \frac{1 - x}{x^{2} - 3x + 2}\)
\(\frac{1 - x}{x^{2} - 3x + 2} = \frac{-(x - 1)}{(x - 1)(x - 2)}\)
= \(\frac{-1}{x - 2}\)
\(\lim \limits_{x \to 1} \frac{1 - x}{x^{2} - 3x + 2} = \lim \limits_{x \to 1} \frac{-1}{x - 2}\)
= \(\frac{-1}{1 - 2} = \frac{-1}{-1} = 1\)
\(\frac{3}{2}\)
\(\frac{2}{3}\)
\(-\frac{2}{3}\)
\(-\frac{3}{2}\)
Correct answer is D
\(\log_{0.25} 8 = x\)
\(8 = 0.25^{x}\)
\(2^{3} = (2^{-2})^{x} \implies 3 = -2x\)
\(x = -\frac{3}{2}\)
Find the sum of the exponential series \(96 + 24 + 6 +...\)
144
128
72
64
Correct answer is B
\(S_{\infty} = \frac{a}{1 - r}\) (for an exponential series)
\(r = \frac{24}{96} = \frac{6}{24} = \frac{1}{4}\)
\(S_{\infty} = \frac{96}{1 - \frac{1}{4}} = \frac{96}{\frac{3}{4}}\)
= \(\frac{96 \times 4}{3} = 128\)
\(\pm 16\)
\(\pm 8\)
\(\pm 4\)
\(\pm 2\)
Correct answer is C
\(2x^{2} + kx + 5 = 0\)
\(\alpha + \beta = \frac{-b}{a} = \frac{-k}{2}\)
\(\alpha \beta = \frac{c}{a} = \frac{5}{2}\)
\(\alpha^{2} + \beta^{2} = (\alpha + \beta)^{2} - 2\alpha \beta\)
\(-1 = (\frac{-k}{2})^{2} - 2(\frac{5}{2})\)
\(-1 = \frac{k^{2}}{4} - 5 \implies \frac{k^{2}}{4} = 4\)
\(k^{2} = 16 \therefore k = \pm 4\)
Find the equation of the line passing through (0, -1) and parallel to the y- axis.
y = -1
y = 0
x = 0
x = -1
Correct answer is C
No explanation has been provided for this answer.