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 the integral \(\int^{\frac{\pi}{4}}_{\frac{\pi}{12}} 2 \cos 2x \mathrm {d} x\)
-\(\frac{1}{2}\)
-1
\(\frac{1}{2}\)
1
Correct answer is C
\(\int_{\frac{\pi}{12}} ^{\frac{\pi}{4}} 2 \cos 2x \mathrm {d} x\)
= \([\frac{2 \sin 2x}{2}]|_{\frac{\pi}{12}} ^{\frac{\pi}{4}}\)
= \(\sin 2x |_{\frac{\pi}{12}} ^{\frac{\pi}{4}}\)
= \(\sin 2(\frac{\pi}{4}) - \sin 2(\frac{\pi}{12})\)
= \(\sin \frac{\pi}{2} - \sin \frac{\pi}{6}\)
= \(1 - \frac{1}{2} = \frac{1}{2}\)
37.00\(\pi\)
37.33\(\pi\)
40.00\(\pi\)
42.67\(\pi\)
Correct answer is B
\(\frac{dv}{dt}\) = \(\pi\)(20 - t2)cm2S-1
\(\int\)dv = \(\pi\)(20 - t2)dt
V = \(\pi\) \(\int\)(20 - t2)dt
V = \(\pi\)(20 \(\frac{t}{3}\) - t3) + c
when c = 0, V = (20t - \(\frac{t^3}{3}\))
after t = 2 seconds
V = \(\pi\)(40 - \(\frac{8}{3}\)
= \(\pi\)\(\frac{120 - 8}{3}\)
= \(\frac{112}{3}\)
= 37.33\(\pi\)
Obtain a maximum value of the function f(x) x3 - 12x + 11
-5
-2
2
27
Correct answer is D
f(x) = x3 - 12x + 11
\(\frac{df(x)}{dx)}\) = 3x2 - 12 = 0
∴ 3x2 - 12 = 0 \(\to\) x2m = 4
x = \(\pm\)2, f(+2) = 8 - 24 + 11 = -15
= f(-2) = (-8) + 24 + 11
= 35 - 8 = 27
∴ maximum value = 27
19.5
17.0
14.5
12.5
Correct answer is C
\(\frac{dm}{dt}\) = 4 - 0.6t
\(\int\)dm = \(\int\)(4 - 0.6t)dt
m = \(4t - 0.3t^2 + c\), when t = 0, m = 2g
∴ c = 2
m = \(4t - 0.3t^2 + 2\), when t = 5 minutes
m = \(4(5) - 0.3(5)^2 + 2 = 20 - 7.5 + 2\)
= 14.5
If y = x sin x, Find \(\frac{d^2 y}{d^2 x}\)
2 cosx - x sinx
sinx + x cosx
sinx - x cosx
x sinx - 2 cosx
Correct answer is A
\(y = x \sin x\)
\(\frac{\mathrm d y}{\mathrm d x} = x \cos x + \sin x\)
\(\frac{\mathrm d^{2} y}{\mathrm d x^{2}} = x (- \sin x) + \cos x + \cos x\)
= \(2 \cos x - x \sin x\)