1994 - JAMB Mathematics Past Questions and Answers - page 4
31
What is the value of sin(-690)?
A
\(\frac{\sqrt{3}}{2}\)
B
-\(\frac{\sqrt{3}}{2}\)
C
\(\frac{-1}{2}\)
D
\(\frac{1}{2}\)
correct option: 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
Users' Answers & Commentssin - 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
32
If y = 3t3 + 2t2 - 7t + 3, find \(\frac{dy}{dt}\) at t = -1
A
-1
B
1
C
-2
D
2
correct option: 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
Users' Answers & Comments\(\frac{dy}{dt}\) = 9t2 + 4t - 7
When t = -1
\(\frac{dy}{dt}\) = 9(-1)2 + 4(-1) - 7
= 9 - 4 -7
= 9 - 11
= -2
33
Find the point (x, y) on the Euclidean plane where the curve y = 2x2 - 2x + 3 has 2 as gradient
A
(1, 3)
B
(2, 7)
C
(0, 3)
D
(3, 15)
correct option: 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)
Users' Answers & Commentsy = 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)
34
Evaluate \(\int^{1}_{-1}\)(2x + 1)2dx
A
3\(\frac{2}{3}\)
B
4
C
4\(\frac{1}{3}\)
D
4\(\frac{2}{3}\)
correct option: d
\(\int^{1}_{-1}\)(2x + 1)2dx
= \(\int^{1}_{-1}\)(4x2 + 4x + 1)dx
= \(\int^{1}_{-1}\)[\(\frac{4x^3}{3}\) + 2x2 + c]
= [\(\frac{4}{3}\) + 3 + c] - [4 + \(\frac{1}{3}\) + c]
= \(\frac{8}{3}\) + 3 + -1 - C
= \(\frac{8}{3}\) + 2
= \(\frac{14}{3}\)
= \(\frac{42}{3}\)
Users' Answers & Comments= \(\int^{1}_{-1}\)(4x2 + 4x + 1)dx
= \(\int^{1}_{-1}\)[\(\frac{4x^3}{3}\) + 2x2 + c]
= [\(\frac{4}{3}\) + 3 + c] - [4 + \(\frac{1}{3}\) + c]
= \(\frac{8}{3}\) + 3 + -1 - C
= \(\frac{8}{3}\) + 2
= \(\frac{14}{3}\)
= \(\frac{42}{3}\)
35
\(\begin{array}{c|c} x & 1 & 2 & 3 & 4 & 5 \ \hline f & y + 2 & y - 2 & 2y - 3 & y + 4 & 3y - 4\end{array}\)
This table shows the frequency distribution of a data if the mean is \(\frac{43}{14}\) find y
This table shows the frequency distribution of a data if the mean is \(\frac{43}{14}\) find y
A
1
B
2
C
3
D
4
correct option: b
Users' Answers & Comments36
Find the mean deviation of the set of numbers 4, 5, 9
A
zero
B
2
C
5
D
6
correct option: b
x = \(\frac{\sum x}{N}\)
= \(\frac{18}{3}\)
= 6
\(\begin{array}{c|c} x & x - x & x - x \ \hline 4 & -2 & 2\ 5 & 1 & 1\ 9 & 3 & 3\ \hline & & 6\end{array}\)
M.D = \(\frac{|x - x|}{N}\)
= \(\frac{6}{3}\)
= 2
Users' Answers & Comments= \(\frac{18}{3}\)
= 6
\(\begin{array}{c|c} x & x - x & x - x \ \hline 4 & -2 & 2\ 5 & 1 & 1\ 9 & 3 & 3\ \hline & & 6\end{array}\)
M.D = \(\frac{|x - x|}{N}\)
= \(\frac{6}{3}\)
= 2
37
\(\begin{array}{c|c} \text{Class Interval} & 1 - 5 & 6 - 10 & 11 - 15 & 16 - 20 & 21 - 25 \ \hline Frequency & 6 & 15 & 20 & 7 & 2\end{array}\)
Estimate the median of the frequency distribution above
Estimate the median of the frequency distribution above
A
10\(\frac{1}{2}\)
B
11\(\frac{1}{2}\)
C
12
D
13
correct option: c
Median = L + [\(\frac{\frac{N}{2} - f}{fm}\)]h
N = Sum of frequencies
L = lower class boundary of median class
f = sum of all frequencies below L
fm = frequency of modal class and
h = class width of median class
Median = 11 + [\(\frac{\frac{50}{2} - 21}{20}\)]5
= 11 + (\(\frac{25 - 21}{20}\))5
= 11 + (\(\frac{(4)}{20}\))
11 + 1 = 12
Users' Answers & CommentsN = Sum of frequencies
L = lower class boundary of median class
f = sum of all frequencies below L
fm = frequency of modal class and
h = class width of median class
Median = 11 + [\(\frac{\frac{50}{2} - 21}{20}\)]5
= 11 + (\(\frac{25 - 21}{20}\))5
= 11 + (\(\frac{(4)}{20}\))
11 + 1 = 12
38
\(\begin{array}{c|c} x & 1 & 2 & 3 & 4 & 5 \ \hline f & 2 & 1 & 2 & 1 & 2\end{array}\)
Find the variance of the frequency distribution above
Find the variance of the frequency distribution above
A
\(\frac{3}{2}\)
B
\(\frac{9}{4}\)
C
\(\frac{5}{2}\)
D
3
correct option: b
\(\begin{array}{c|c} x & f & fx & \bar{x} - x & (\bar{x} - x)^2 & f(\bar{x} - x)^2 \ \hline 1 & 2 & 2 & -2 & 4 & 8\ 2 & 1 & 2 & -1 & 1 & 1\ 3 & 2 & 6 & 0 & 0 & 0\ 4 & 1 & 4 & 1 & 1 & 1\ 2 & 2 & 10 & 2 & 4 & 8\ \hline & 8 & 24 & & & 18 \end{array}\)
x = \(\frac{\sum fx}{\sum f}\)
= \(\frac{24}{8}\)
= 3
Variance (62) = \(\frac{\sum f(\bar{x} - x)^2}{\sum f}\)
= \(\frac{18}{8}\)
= \(\frac{9}{4}\)
Users' Answers & Commentsx = \(\frac{\sum fx}{\sum f}\)
= \(\frac{24}{8}\)
= 3
Variance (62) = \(\frac{\sum f(\bar{x} - x)^2}{\sum f}\)
= \(\frac{18}{8}\)
= \(\frac{9}{4}\)
39
In a survey, it was observed that 20 students read newspapers and 35 read novels. If 40 of the students read either newspapers or novels, what is the probability of the students who read both newspapers and novels?
A
\(\frac{1}{2}\)
B
\(\frac{2}{3}\)
C
\(\frac{3}{8}\)
D
\(\frac{3}{11}\)
correct option: c
40 = 20 - x + x + 35 - x
40 = 55 - x
x = 55 - 40
= 15
∴ P(both) \(\frac{15}{40}\)
= \(\frac{3}{8}\)
Users' Answers & Comments40 = 55 - x
x = 55 - 40
= 15
∴ P(both) \(\frac{15}{40}\)
= \(\frac{3}{8}\)
40
A
y = (x - 3)3
B
y = (x + 3)3
C
y = x3 - 27
D
y = -x3 + 27
correct option: c
y = x3 - 27, y = -27 \(\to\) (0, -27)
when y = 0, x = 3 (3, 0)
Users' Answers & Commentswhen y = 0, x = 3 (3, 0)