2009 - JAMB Mathematics Past Questions and Answers - page 9

81
What value of x will make the function x(4 - x) a maximum?
A
4
B
3
C
2
D
1
correct option: c
x(4 - x)

4x - x2

\(\frac{dy}{dx}\) = 4 - 2x

\(\frac{dy}{dx}\) = 0

2x = 4

x = \(\frac{4}{2}\)

= 2
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82
The distance travelled by a particle from a fixed point is given as s = (t3 - t2 - t + 5)cm. Find the minimum distance that the particle can cover from the fixed point.
A
2.3 cm
B
4.0cm
C
5.2
D
g
correct option: b
S = (t3 - t2 - t + 5)cm

\(\frac{ds}{dt}\) = 3t2 - 2t - 1

= (3t + 1)(t - 1)

t = -\(\frac{1}{3}\) or t = 1

Substitute value of t

s = (13 - 12 - 1 + 5)cm

= 4.0 cm
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83
Evaluate \(\int\) sec2\(\theta\) d \(\theta\)
A
sec\(\theta\) tan\(\theta\) + k
B
tan\(\theta\) + k
C
2 sec\(\theta\) + k
D
sec\(\theta\) + k
correct option: b
\(\int\) sec2\(\theta\) d \(\theta\) = tan\(\theta\) + k
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84
\(\begin{array}{c|c} No. of days & 1 & 2 & 3 & 4 & 5 & 6\ \hline No. of students & 20 & 2x & 60 & 40 & x & 50 \end{array}\)
The distribution above shows the number of days a group of 260 students were absent from school in a particular term. How many students were absent for at least four days in the term
A
180
B
120
C
110
D
40
correct option: a
3x + 170 = 260

3x = 260 - 170

3x = 90 \(\to\) x = 30

therefore, 20 + 2x + 60 + 40

= 180
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85
5, 8, 6 and k occurs with frequencies 3, 2, 4 and 1 respectively and have a mean of 5.7. Find the value of k
A
4
B
3
C
2
D
1
correct option: c
\(\begin{array}{c|c} x & f & fx\ \hline 5 & 3 & 15\ 8 & 2 & 16 \ 6 & 4 & 24\ k & 1 & k\ & \sum f & \sum fx = 55 + k\end{array}\)

mean \(\bar{x}\) = \(\frac{55 + k}{10}\)

= 5.7

k = 2
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86
What is the mean deviation of x, 2x, x + 1 and 3x, if their mean is 2?
A
0.5
B
1.0
C
1.5
D
2.0
correct option: d
deviation x - \(\bar{x}\)

x - 2 + 2x - 2 + (x + 1) - 2 + 3x - 2 = 0

x - 2 + 2x - 2 -2x - 2 + 3x - 2 = 0

= 4x - 8

4x = 8

x = \(\frac{8}{4}\)

= 2
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87
in how many ways can 9 people be seated if 3 chairs 3 chairs are available
A
720
B
504
C
336
D
210
correct option: b
\(\frac{n!}{n - r!}\) = \(\frac{9!}{9 - 3!}\)

= \(\frac{9!}{6!}\)

= 9 x 8 x 7

= 502
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88
In how many ways can a delegation of 3 be chosen from 5 men and 3 women, if at least 1 man and 1 woman must be included?
A
15
B
28
C
30
D
45
correct option: b
\(\frac{\begin{pmatrix} 8 \ 3 \end{pmatrix}}{\begin{pmatrix} 4 \ 1 \end{pmatrix}\begin{pmatrix} 4\ 1\end{pmatrix}}\)(Two boys and a girl) OR \(\frac{\begin{pmatrix} 8 \ 3 \end{pmatrix}}{\begin{pmatrix} 4 \ 1 \end{pmatrix}\begin{pmatrix} 4\ 1\end{pmatrix}}\)(Two girls and a boy in the committee)

\(\frac{336}{24}\) or \(\frac{336}{24}\)

14 + 14

= 28
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89
The probability of a student passing any examination is \(\frac{2}{3}\). If the student takes three examinations, what is the probability that he will not pass any of them?
A
\(\frac{2}{3}\)
B
\(\frac{4}{9}\)
C
\(\frac{8}{27}\)
D
\(\frac{1}{27}\)
correct option: d
p(pass) = \(\frac{2}{3}\)

p(fail) = \(\frac{1}{3}\)

p(pass at 5 marks) = \(\frac{1}{3}\) x \(\frac{1}{3}\) x \(\frac{1}{3}\)

= \(\frac{1}{27}\)
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90
\(\begin{array}{c|c} Marks & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9\ \hline No. of students & 3 & 4 & 1 & 0 & 4 & 5 & 2 & 1\end{array}\)

The table above shows the distribution of marks of students in a test. Find the probability of passing the test if the pass mark is 5.
A
\(\frac{3}{5}\)
B
\(\frac{2}{5}\)
C
\(\frac{7}{20}\)
D
\(\frac{1}{5}\)
correct option: a
p (pass at 5mark) = \(\frac{12}{20}\)

= \(\frac{3}{5}\)
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