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.
\(\frac{d}{dx}\) cos(3x\(^2\) - 2x) is equal to
-sin(6x - 2)
-sin(3x2 - 2x)dx
(6x - 2) sin(3x2 - 2x)
-(6x - 2)sin(3x2 - 2x)
Correct answer is D
Let \(3x^{2} - 2x = u\)
\(y = \cos u \implies \frac{\mathrm d y}{\mathrm d u} = - \sin u\)
\(\frac{\mathrm d u}{\mathrm d x} = 6x - 2\)
\(\therefore \frac{\mathrm d y}{\mathrm d x} = (6x - 2) . - \sin u\)
= \(- (6x - 2) \sin (3x^{2} - 2x)\)
Differentiate \(\frac{6x^3 - 5x^2 + 1}{3x^2}\) with respect to x
\(\frac{2 + 2}{3x^3}\)
2 + \(\frac{1}{6x}\)
2 - \(\frac{2}{3x^3}\)
\(\frac{1}{5}\)
Correct answer is C
\(\frac{6x^3 - 5x^2 + 1}{3x^2}\)
let y = 3x2
y = \(\frac{6x^3}{3x^2}\) - \(\frac{6x^2}{3x^2}\) + \(\frac{1}{3x^2}\)
Y = 2x - \(\frac{5}{3}\) + \(\frac{1}{3x^2}\)
\(\frac{dy}{dx}\) = 2 + \(\frac{1}{3}\)(-2)x-3
= 2 - \(\frac{2}{3x^3}\)
In a triangle XYZ, if < ZYZ is 60, XY = 3cm and YZ = 4cm, calculate the length of the sides XZ.
√23cm
√13cm
2√5cm
2√3cm
Correct answer is B
(XZ)2 = 32 + 42 - 2 x 3 x 4 cos60o
= 25 - 24\(\frac{1}{2}\)
XZ = √13cm
\(\frac{20}{√3}\)m
\(\frac{40}{√3}\)m
20√3m
40√3m
Correct answer is D
\(\frac{40}{x}\) = tan 30°
x = \(\frac{40}{tan 30}\)
= \(\frac{40}{1\sqrt{3}}\)
= 40√3m
3√10
3√5
√26
√13
Correct answer is D
2x - y .....(i)
x + y.....(ii)
from (i) y = 2x - 4
from (ii) y = -x + 2
2x - 4 = -x + 2
x = 2
y = -x + 2
= -2 + 2
= 0
\(x_1\) = 2
\(y_1\) = 0
\(x_2\) = 4
\(y_2\) = 3
Hence, dist. = \(\sqrt{(y_2 - y_1)^2 + (x_2 - x_1)^2}\)
= \(\sqrt{(3 - 0)^2 + (4 - 2)^2}\)
= \(\sqrt{3^2 + 2^2}\)
= \(\sqrt{13}\)