1995 - JAMB Mathematics Past Questions and Answers - page 4
31
For what value of x is the tangent to the curve y = x2 - 4x + 3 parallel to the x-axis?
A
3
B
2
C
1
D
6
correct option: b
Users' Answers & Comments32
Two variables x and y are such that \(\frac{dy}{dx}\) = 4x - 3 and y = 5 when x = 2. Find y in terms of x
A
2x2 - 3x + 5
B
2x2 - 3x + 3
C
2x2 - 3x
D
4
correct option: b
∫dy = ∫(4x - 3)dx, y = 2x2 - 3x + C
when y = 5, x = 2, C = 3
y = 2x2 - 3x + 3
Users' Answers & Commentswhen y = 5, x = 2, C = 3
y = 2x2 - 3x + 3
33
Find the area bounded by the curve y = 3x2 - 2x + 1, the ordinates x = 1 and x = 3 and the a-axix
A
24
B
22
C
21
D
20
correct option: b
\(\pi\) = \(\int^{3}_{1}(3x^2 - 2x + 1)\)
dx = [x3 - 3x + x + C]
= 22
Users' Answers & Commentsdx = [x3 - 3x + x + C]
= 22
34
\(\begin{array}{c|c} \text{Age in years} & 13 & 14 & 15 & 16 & 17 \ \hline \text{No. of students} & 3 & 10 & 30 & 42 & 15\end{array}\)
The frequency distribution above shows the ages of students in a secondary school. In a pie chart constructed to represent the data, the angles corresponding to the 15 years old is
The frequency distribution above shows the ages of students in a secondary school. In a pie chart constructed to represent the data, the angles corresponding to the 15 years old is
A
27o
B
30o
C
54o
D
108o
correct option: d
Users' Answers & Comments35
The pie chart shows the distribution of students in a secondary school class. If 30 students offered French, how many offered C.R.K?
A
25
B
15
C
10
D
8
36
\(\begin{array}{c|c} class& 1 - 3 & 4 - 6 & 7 - 9\ \hline Frequency & 5 & 8 & 5\end{array}\)
Find the standard deviation of the data using the table above
Find the standard deviation of the data using the table above
A
5
B
\(\sqrt{6}\0
C
\(\frac{5}{3}\)
D
\(\sqrt{5}\)
correct option: a
\(\begin{array}{c|c} \text{class intervals} & Fre(F) & \text{class-marks(x)} & Fx & (x - x)& (x - x)^2 & F(x - x)^2 \ \hline 1 - 3 & 5 & 10 & 10 & -3 & 9 & 90\ 4 - 6 & 8 & 40 & 40 & 0 & 0 & 0 \ 7 - 9 & 5 & 40 & 40 & 3 & 9 & 360 \ \hline & 18 & & 90 & & & 450 \end{array}\)
x = \(\frac{\sum fx}{\sum f}\)
= \(\frac{90}{18}\)
= 5
S.D = \(\frac{\sum f(x - x)^2}{\sum f}\)
= \(\frac{450}{18}\)
= \(\sqrt{25}\)
= 5
Users' Answers & Commentsx = \(\frac{\sum fx}{\sum f}\)
= \(\frac{90}{18}\)
= 5
S.D = \(\frac{\sum f(x - x)^2}{\sum f}\)
= \(\frac{450}{18}\)
= \(\sqrt{25}\)
= 5
37
The variance of the scores 1, 2, 3, 4, 5 is
A
1, 2
B
1, 4
C
2.0
D
3.0
correct option: c
Users' Answers & Comments38
\(\begin{array}{c|c} \text{Class Interval} & Frequency & \text{Class boundaries} & Class Mid-point \ \hline 1.5 - 1.9 & 2 & 1.45 - 1.95 & 1.7\ 2.0 - 2.4 & 21 & 1.95 - 2.45 & 2.2\ 2.5 - 2.9 & 4 & 2.45 - 2.95 & 2.7 \ 3.0 - 2.9 & 15 & 2.95 - 3.45 & 3.2\ 3.5 - 3.9 & 10 & 3.45 - 3.95 & 3.7\ 4.0 - 4.4 & 5 & 3.95 - 4.45 & 4.2\ 4.5 - 4.9 & 3 & 4.45 - 4.95 & 4.7\end{array}\)
Find the mode of the distribution above to find the mode of the distribution.
Find the mode of the distribution above to find the mode of the distribution.
A
3.2
B
3.3
C
3.7
D
4.2
correct option: b
Mode = a + (b - a)(fm - Fb)
2Fm - Fa - Fb
= 3.0 + \(\frac{(3.4 - 3)(15 - 4)}{2(15) - 4 - 10}\)
= 3 + \(\frac{(6.4)(11)}{30 - 14}\)
= 3 + \(\frac{4.4}{16}\)
= 3 + 0.275
= 3.275
= 3.3cm
Users' Answers & Comments2Fm - Fa - Fb
= 3.0 + \(\frac{(3.4 - 3)(15 - 4)}{2(15) - 4 - 10}\)
= 3 + \(\frac{(6.4)(11)}{30 - 14}\)
= 3 + \(\frac{4.4}{16}\)
= 3 + 0.275
= 3.275
= 3.3cm
39
\(\begin{array}{c|c} \text{Class Interval} & Frequency & \text{Class boundaries} & Class Mid-point \ \hline 1.5 - 1.9 & 2 & 1.45 - 1.95 & 1.7\ 2.0 - 2.4 & 21 & 1.95 - 2.45 & 2.2\ 2.5 - 2.9 & 4 & 2.45 - 2.95 & 2.7 \ 3.0 - 2.9 & 15 & 2.95 - 3.45 & 3.2\ 3.5 - 3.9 & 10 & 3.45 - 3.95 & 3.7\ 4.0 - 4.4 & 5 & 3.95 - 4.45 & 4.2\ 4.5 - 4.9 & 3 & 4.45 - 4.95 & 4.7\end{array}\)
The median of the distribution above is
The median of the distribution above is
A
4.0
B
3.4
C
3.2
D
3.0
correct option: b
Median = \(\frac{a + b}{fm}\) (\(\frac{1}{2} \sum f - CF_b\))
= 2.95 + \(\frac{0.5}{15}\)(2.-7)
= 2.95 + \(\frac{0.5}{15}\) x 13
= 2.95 + 0. 43
= 3.38
= 3.4
Users' Answers & Comments= 2.95 + \(\frac{0.5}{15}\)(2.-7)
= 2.95 + \(\frac{0.5}{15}\) x 13
= 2.95 + 0. 43
= 3.38
= 3.4
40
Let p be a probability function on set S, where S = (a1, a2, a3, a4). Find P(a1) if P(a2) = \(\frac{1}{3}\), p(a3) = \(\frac{1}{6}\) and p(a4) = \(\frac{1}{5}\)
A
\(\frac{7}{3}\)
B
\(\frac{2}{3}\)
C
\(\frac{1}{3}\)
D
\(\frac{3}{10}\)
correct option: d
Users' Answers & Comments