How good are you with figures and formulas? Find out with these Mathematics past questions and answers. This Test is useful for both job aptitude test candidates and students preparing for JAMB, WAEC, NECO or Post UTME.
Evaluate \(\lim \limits_{x \to 2} \frac{(x - 2)(x^2 + 3x - 2)}{x^2 - 4}\)
7
2
3
4
Correct answer is B
\(\lim \limits_{x \to 2} \frac{(x - 2)(x^2 + 3x - 2)}{x^2 - 4}\)
\(\frac{(x - 2)(x^{2} + 3x - 2)}{x^{2} - 4} = \frac{(x - 2)(x^{2} + 3x - 2)}{(x - 2)(x + 2)}\)
= \(\frac{(x^{2} + 3x - 2)}{x + 2}\)
\(\therefore \lim \limits_{x \to 2} \frac{(x - 2)(x^2 + 3x - 2)}{x^2 - 4} = \lim \limits_{x \to 2} \frac{x^{2} + 3x - 2}{x + 2}\)
= \(\frac{2^{2} + 3(2) - 2}{2 + 2}\)
= \(\frac{4 + 6 - 2}{4} = 2\)
Given that \(\theta\) is an acute angle and sin \(\theta\) = \(\frac{m}{n}\), find cos \(\theta\)
\(\frac{\sqrt{n^2 - m^2}}{m}\)
\(\frac{\sqrt{(n + m)(n - m)}}{n}\)
\(\frac{m}{\sqrt{n^2 - m^2}}\)
\(\sqrt{\frac{n}{n^2 - m^2}}\)
Correct answer is B
sin \(\theta\) = \(\frac{m}{n}\)
Opp = m; Hyp = n
Adj = \(\sqrt{n^{2} - m^{2}}\)
\(\cos \theta = \frac{\sqrt{n^{2} - m^{2}}}{n}\)
= \(\frac{\sqrt{(n + m)(n - m)}}{n}\)
Find then equation line through (5, 7) parallel to the line 7x + 5y = 12
5x + 7y = 120
7x + 5y = 70
x + y = 7
15x + 17y = 90
Correct answer is B
Equation (5, 7) parallel to the line 7x + 5y = 12
5Y = -7x + 12
y = \(\frac{-7x}{5}\) + \(\frac{12}{5}\)
Gradient = \(\frac{-7}{5}\)
∴ Required equation = \(\frac{y - 7}{x - 5}\) = \(\frac{-7}{5}\) i.e. 5y - 35 = -7x + 35
5y + 7x = 70
What is the perpendicular distance of a point (2, 3) from the line 2x - 4y + 3 = 0?
\(\frac{\sqrt{5}}{2}\)
\(\frac{\sqrt{5}}{20}\)
\(\frac{5}{\sqrt{13}}\)
6
Correct answer is A
2x - 4y + 3 = 0
Required distance = \(\frac{(2 \times 2) + 3(-4) + 3}{\sqrt{2^2} + (-4)^2}\)
= \(\frac{4 - 12 + 3}{\sqrt{20}}\)
= \(\frac{-5}{-2\sqrt{5}}\)
= \(\frac{\sqrt{5}}{2}\)
The locus of a point which is equidistant from two given fixed points is the
perpendicular bisector of the straight line joining them
parallel line to the straight line joining them
transverse to the straight line joining them
angle bisector of 90o which the straight line joining them makes with the horizontal
Correct answer is A
No explanation has been provided for this answer.