1992 - JAMB Mathematics Past Questions and Answers - page 4
31
Evaluate \(\frac{(x - 2)(x^2 + 3x - 2)}{x^2 - 4}\)
A
7
B
2
C
3
D
4
correct option: b
Users' Answers & Comments32
If y = x sin x, Find \(\frac{d^2 y}{d^2 x}\)
A
2 cosx - x - x sinx
B
sinx + x cosx
C
sinx - x cosx
D
x sinx - 2 cosx
correct option: a
Users' Answers & Comments33
Ice forms on a refrigerator ice-box at the rate of (4 - 06t)g per minute after t minutes. If initially there are 2g of ice in the box, find the mass of ice formed in 5 minutes
A
19.5
B
17.0
C
14.5
D
12.5
correct option: c
\(\frac{dm}{dt}\) = 4 - 0.6t
\(\int\)dm = \(\int\)(4 - 0.6t)dt
m = 4t - 0.3t2 + c, when t = 0, m = 2g
∴ c = 2
m = 4t - 0.3t2 + 2, when t = 5 minutes
m = 4(5) - 0.3(5)2 + 2 = 20 - 7.5
= 14.5
Users' Answers & Comments\(\int\)dm = \(\int\)(4 - 0.6t)dt
m = 4t - 0.3t2 + c, when t = 0, m = 2g
∴ c = 2
m = 4t - 0.3t2 + 2, when t = 5 minutes
m = 4(5) - 0.3(5)2 + 2 = 20 - 7.5
= 14.5
34
Obtain a maximum value of the function f(x) x3 - 12x + 11
A
-5
B
-2
C
2
D
27
correct option: 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
Users' Answers & Comments\(\frac{df(x)}{dx)}\) = 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
35
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 option: 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\(\pi\)
Users' Answers & Comments\(\int\)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\(\pi\)
36
Evaluate the integral \(\int^{\frac{\pi}{4}}_{\frac{\pi}{12}}\) 2 cos 2x dx
A
-\(\frac{1}{2}\)
B
-1
C
\(\frac{1}{2}\)
D
1
correct option: c
let I = \(\int^{\frac{\pi}{4}}_{\frac{\pi}{12}}\) 2 cos 2x dx
= 2(sin 2x)\(\frac{\pi}{4}\) (sin 2x)\(\frac{\pi}{4}\)\(\frac{\pi}{12}\)
(2)\(\frac{\pi}{12}\)
= -1 - \(\frac{1}{2}\)
= \(\frac{1}{2}\)
Users' Answers & Comments= 2(sin 2x)\(\frac{\pi}{4}\) (sin 2x)\(\frac{\pi}{4}\)\(\frac{\pi}{12}\)
(2)\(\frac{\pi}{12}\)
= -1 - \(\frac{1}{2}\)
= \(\frac{1}{2}\)
37
A store keeper checked his stock of five commodities and arrived at the following statics
\(\begin{array}{c|c} Comoditiy & Quantity\ \hline F & 125\ G & 113\ H & 108\ K & 216 \ M & 68\end{array}\)
What angle will commodity H represent on a pie chart?
\(\begin{array}{c|c} Comoditiy & Quantity\ \hline F & 125\ G & 113\ H & 108\ K & 216 \ M & 68\end{array}\)
What angle will commodity H represent on a pie chart?
A
216o
B
108o
C
68o
D
54o
correct option: d
H will represent \(\frac{108}{720}\) x \(\frac{360^o}{1}\) = 54o
Users' Answers & Comments38
\(\begin{array}{c|c} x & 2 & 4 & 6 & 8\ \hline f & 4 & y & 6 & 5 \end{array}\)
If the mean of the above frequency distribution is 5.2, find y
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 option: 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
Users' Answers & Comments= \(\frac{5.2}{1}\)
= \(\frac{8 + 4y + 36 + 40}{4 + y + 6 + 5}\)
= \(\frac{5.2}{1}\)
= \(\frac{84 + 4y}{15 + y}\)
= 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
39
\(\begin{array}{c|c} \text{No. of children} & 0 & 1 & 2 & 3 & 4 & 5 & 6 \ \hline \text{No. of families} & 7 & 11 & 6 & 7 & 7 & 5 & 3 \end{array}\)
Find the mode and median respectively of the distribution above
Find the mode and median respectively of the distribution above
A
2, 1
B
1, 2
C
1, 5
D
5, 2
correct option: b
Users' Answers & Comments40
If the scores of 3 students in a test are 5, 6 and 7, find the standard deviation of their scores
A
\(\frac{2}{3}\)
B
\(\frac{2}{3}\sqrt{3}\)
C
\(\sqrt{\frac{2}{3}}\)
D
\(\sqrt{\frac{3}{2}}\)
correct option: c
(x) = \(\frac{5 + 6 + 7}{3}\)
= \(\frac{18}{3}\)
= 6
\(\begin{array}{c|c} scores(X) & \text{d = (x - x) deviation} & (deviation)^2\\hline 5 & 5 - 6 & 1\ 6 & 6 - 6 & 0 \ 7 & 7 - 6 & 1\ \hline & & 2\end{array}\)
S.D \(\sqrt{\frac{\sum d^2}{n}}\) where d = deviation = (x - x)
= \(\sqrt{\frac{2}{3}}\)
Users' Answers & Comments= \(\frac{18}{3}\)
= 6
\(\begin{array}{c|c} scores(X) & \text{d = (x - x) deviation} & (deviation)^2\\hline 5 & 5 - 6 & 1\ 6 & 6 - 6 & 0 \ 7 & 7 - 6 & 1\ \hline & & 2\end{array}\)
S.D \(\sqrt{\frac{\sum d^2}{n}}\) where d = deviation = (x - x)
= \(\sqrt{\frac{2}{3}}\)