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.
2480
1240
620
124
Correct answer is B
Given the first and last term of an A.P, the sum of the terms is given by
\(S_{n} = \frac{n}{2} [a + l]\)
where a = first term; l = last term and n = number of terms.
\(\therefore S_{20} = \frac{20}{2} [7 + 117]\)
= \(10 (124)\)
= 1240
At what value of x is the function y = x2 - 2x - 3 minimum?
1
-1
-4
4
Correct answer is A
For y = ax2 - bx + c for minimum y
\(\frac{dy}{dx}\) = 2x - 2
= \(\frac{dy}{dx}\) = 0
∴ 2x - 2 = 0
x = 1
Find the gradient of the line passing through the points (-2, 0) and (0, -4)
2
-4
-2
4
Correct answer is C
Given (-2, 0) and (0, -4)
Gradient = \(\frac{-4 - 0}{0 - (-2)}\)
= \(\frac{-4}{2}\)
= -2
Evaluate x2(x2 - 1)-\(\frac{1}{2}\) - (x2 - 1)\(\frac{1}{2}\)
(x2 - 1)-\(\frac{1}{2}\)
(x2 - 1)1
(x2 - 1)
(x2 - 1)-1
Correct answer is A
x2(x2 - 1)-\(\frac{1}{2}\) - (x2 - 1)\(\frac{1}{2}\) = \(\frac{x^2}{(x^2 - 1)^\frac{1}{2}}\) - \(\frac{(x^2 - 1)^\frac{1}{2}}{1}\)
= \(\frac{x^2 - (x^2 - 1)}{(x^2 - 1) ^\frac{1}{2}}\)
= \(\frac{x^2 - x^2 + 1}{(x^2 - 1)^\frac{1}{2}}\)
= (x2 - 1)-\(\frac{1}{2}\)
Simplify \(\frac{\sqrt{1 + x} + \sqrt{x}}{\sqrt{1 + x} - \sqrt{x}}\)
-2x - 2\(\sqrt{x (1 + x)}\)
1 + 2x + 2\(\sqrt{x (1 + x)}\)
\(\sqrt{x (1 + x)}\)
1 + 2x - 2\(\sqrt{x (1 + x)}\)
Correct answer is B
\(\frac{\sqrt{1 + x} + \sqrt{x}}{\sqrt{1 + x} - \sqrt{x}}\)
= \((\frac{\sqrt{1 + x} + \sqrt{x}}{\sqrt{1 + x} - \sqrt{x}}) (\frac{\sqrt{1 + x} + \sqrt{x}}{\sqrt{1 + x} + \sqrt{x}})\)
= \(\frac{(1 + x) + \sqrt{x(1 + x)} + \sqrt{x(1 + x)} + x}{(1 + x) - x}\)
= \(\frac{1 + 2x + 2\sqrt{x(1 + x)}}{1}\)
= \(1 + 2x + 2\sqrt{x(1 + x)}\)