Mathematics questions and answers

Mathematics Questions and Answers

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.

2,291.

\(\begin{array}{c|c} x & 2 & 4 & 6 & 8\\ \hline f & 4 & y & 6 & 5 \end{array}\)
If the mean of the above frequency distribution is 5.2, find y

A.

2, 1

B.

1, 2

C.

1, 5

D.

5, 2

Correct answer is C

Mean \(\bar{x}\) = \(\frac{\sum fx}{\sum f}\)

= \(\frac{5.2}{1}\)

= \(\frac{8 + 4y + 36 + 40}{4 + y + 6 + 5}\)

= \(\frac{5.2}{1}\)

= \(\frac{84 + 4y}{15 + y}\)

= 5.2(15 + y)

= 84 + 4y

= 5.2 x 15 + 5.2y

= 84 + 4y

= 78 + 5.2y

= 84 = 4y

= 5.2y - 4y

= 84 - 78

1.2y = 6

y = \(\frac{6}{1.2}\)

= \(\frac{60}{12}\)

= 5

2,293.

Evaluate the integral \(\int^{\frac{\pi}{4}}_{\frac{\pi}{12}} 2 \cos 2x \mathrm {d} x\)

A.

-\(\frac{1}{2}\)

B.

-1

C.

\(\frac{1}{2}\)

D.

1

Correct answer is C

\(\int_{\frac{\pi}{12}} ^{\frac{\pi}{4}} 2 \cos 2x \mathrm {d} x\)

= \([\frac{2 \sin 2x}{2}]|_{\frac{\pi}{12}} ^{\frac{\pi}{4}}\)

= \(\sin 2x |_{\frac{\pi}{12}} ^{\frac{\pi}{4}}\)

= \(\sin 2(\frac{\pi}{4}) - \sin 2(\frac{\pi}{12})\)

= \(\sin \frac{\pi}{2} - \sin \frac{\pi}{6}\)

= \(1 - \frac{1}{2} = \frac{1}{2}\)

2,294.

A student blows a balloon and its volume increases at a rate of \(\pi\)(20 - t2)cm3S-1 after t seconds. If the initial volume is 0 cm3, find the volume of the balloon after 2 seconds

A.

37.00\(\pi\)

B.

37.33\(\pi\)

C.

40.00\(\pi\)

D.

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\(\pi\)

2,295.

Obtain a maximum value of the function f(x) x3 - 12x + 11

A.

-5

B.

-2

C.

2

D.

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