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.
Factorize completely \((x^2 + x)^2 - (2x + 2)^2\)
(x + 1)(x + 2)(x - 2)
(x + 1) 2(x + 2) (x - 2)
(x + 1)2 (x + 2)2
(x + 1)2 (x + 2)(x - 2)
Correct answer is D
\((x^{2} + x)^{2} - (2x + 2)^{2}\)
= \((x^{2} + x + 2x + 2)(x^{2} + x - (2x + 2))\)
= \((x^{2} + 3x + 2)(x^{2} - x - 2)\)
= \(((x + 1)(x + 2))((x + 1)(x - 2))\)
= \((x + 1)^{2} (x + 2)(x - 2)\)
If (g(y)) = \(\frac{y - 3}{11}\) + \(\frac{11}{y^2 - 9}\). what is g(y + 3)?
\(\frac{y}{11} + \frac{11}{y(y + 6)}\)
\(\frac{y}{11} + \frac{11}{y(y + 3)}\)
\(\frac{y + 30}{11} + \frac{11}{y(y + 3)}\)
\(\frac{y + 3}{11} + \frac{11}{y(y - 6)}\)
Correct answer is A
\(g(y) = \frac{y - 3}{11} + \frac{11}{y^{2} - 9}\)
\(\therefore g(y + 3) = \frac{(y + 3) - 3}{11} + \frac{11}{(y + 3)^{2} - 9}\)
\(g(y + 3) = \frac{y}{11} + \frac{11}{y^{2} + 6y + 9 - 9}\)
\(g(y + 3) = \frac{y}{11} + \frac{11}{y^{2} + 6y}\)
= \(\frac{y}{11} + \frac{11}{y(y + 6)}\)
If a = -3, b = 2, c = 4, evaluate \(\frac{a^3 - b^3 - c^{\frac{1}{2}}}{b - a - c}\)
37
\(\frac{-37}{5}\)
\(\frac{37}{5}\)
-37
Correct answer is D
\(\frac{a^{3} - b^{3} - c^{\frac{1}{2}}}{b - a - c} = \frac{(-3)^{3} - (2)^{3} - 4^{\frac{1}{2}}}{2 - (-3) - 4}\)
= \(\frac{-37}{1} = -37\)
If x varies inversely as the cube root of y and x = 1 when y = 8, find y when x = 3
\(\frac{1}{3}\)
\(\frac{2}{3}\)
\(\frac{8}{27}\)
\(\frac{4}{9}\)
Correct answer is C
\(x \propto \frac{1}{\sqrt[3]{y}} \implies x = \frac{k}{\sqrt[3]{y}}\)
When y = 8, x = 1
\(1 = \frac{k}{\sqrt[3]{8}} \implies 1 = \frac{k}{2}\)
\(k = 2\)
\(\therefore x = \frac{2}{\sqrt[3]{y}}\)
When x = 3,
\(3 = \frac{2}{\sqrt[3]{y}} \implies \sqrt[3]{y} = \frac{2}{3}\)
\(y = (\frac{2}{3})^{3} = \frac{8}{27}\)
\(\frac{a + b}{a - b}\)
\(\frac{1}{a^2 - b^2}\)
\(\frac{a - b}{a + b}\)
a2 - b2
Correct answer is B
\(\frac{1}{p} = \frac{a^{2} + 2ab + b^{2}}{a - b}\)
\(\frac{1}{q} = \frac{a + b}{a^{2} - 2ab + b^{2}}\)
\(\frac{1}{p} = \frac{(a + b)^{2}}{a - b}\)
\(\frac{1}{q} = \frac{a + b}{(a - b)^{2}}\)
\(\therefore p = \frac{a - b}{(a + b)^{2}}\)
\(\frac{p}{q} = p \times \frac{1}{q} = \frac{a - b}{(a + b)^{2}} \times \frac{a + b}{(a - b)^{2}}\)
= \(\frac{1}{(a + b)(a - b)}\)
= \(\frac{1}{a^{2} - b^{2}}\)