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.
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.