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 \(\int^{1}_{-1}(2x + 1)^2 \mathrm d x\)
3\(\frac{2}{3}\)
4
4\(\frac{1}{3}\)
4\(\frac{2}{3}\)
Correct answer is D
\(\int^{1}_{-1}(2x + 1)^2 \mathrm d x\)
= \(\int^{1}_{-1}(4x^2 + 4x + 1) \mathrm d x\)
= \(\int^{1}_{-1}\)[\(\frac{4x^3}{3} + 2x^2 + x]\)
= [\(\frac{4}{3}\) + 2 + 1] - [\(\frac{-4}{3}\)+2 -1]
= \(\frac{13}{3}\) + \(\frac{1}{3}\)
= \(\frac{14}{3}\)
= \(4 \frac{2}{3}\)
Integrate \(\frac{1 - x}{x^3}\) with respect to x
\(\frac{x - x^2}{x^4}\) + k
\(\frac{4}{x^4} - \frac{3 + k}{x^3}\)
\(\frac{1}{x} - \frac{1}{2x^2}\) + k
\(\frac{1}{3x^2} - \frac{1}{2x}\) + k
Correct answer is C
\(\int \frac{1 - x}{x^3}\)
= \(\int^{1}_{x^3} - \int^{x}_{x^3}\)
= x-3 dx - x-2dx
= \(\frac{1}{2x^2} + \frac{1}{x}\)
Find the point (x, y) on the Euclidean plane where the curve y = 2x2 - 2x + 3 has 2 as gradient
(1, 3)
(2, 7)
(0, 3)
(3, 15)
Correct answer is A
Equation of curve;
y = 2x2 - 2x + 3
gradient of curve;
\(\frac{dy}{dx}\) = differential coefficient
\(\frac{dy}{dx}\) = 4x - 2, for gradient to be 2
∴ \(\frac{dy}{dx}\) = 2
4x - 2 = 2
4x = 4
∴ x = 1
When x = 1, y = 2(1)2 - 2(1) + 3
= 2 - 2 + 3
= 5 - 2
= 3
coordinate of the point where the curve; y = 2x2 - 2x + 3 has gradient equal to 2 is (1, 3)
If y = 3t3 + 2t2 - 7t + 3, find \(\frac{dy}{dt}\) at t = -1
-1
1
-2
2
Correct answer is C
y = 3t3 + 2t2 - 7t + 3
\(\frac{dy}{dt}\) = 9t2 + 4t - 7
When t = -1
\(\frac{dy}{dt}\) = 9(-1)2 + 4(-1) - 7
= 9 - 4 -7
= 9 - 11
= -2
What is the value of sin(-690)?
\(\frac{\sqrt{3}}{2}\)
-\(\frac{\sqrt{3}}{2}\)
\(\frac{-1}{2}\)
\(\frac{1}{2}\)
Correct answer is D
Sin(-690o) = Sin(-360 -3300
sin - 360 = sin 0
∴ sin(-690o) = sin(330o)
Negative angles are measured in clockwise direction
The acute angle equivalent of sin(-330o) = sin(30o)
sin(-330o) = sin(30o)
= \(\frac{1}{2}\)
= 0.5