2011 - JAMB Mathematics Past Questions and Answers - page 4
31
In a right angled triangle, if tan \(\theta\) = \(\frac{3}{4}\). What is cos\(\theta\) - sin\(\theta\)?
A
\(\frac{2}{3}\)
B
\(\frac{3}{5}\)
C
\(\frac{1}{5}\)
D
\(\frac{4}{5}\)
correct option: c
tan\(\theta\) = \(\frac{3}{4}\)
from Pythagoras tippet, the hypotenus is T
i.e. 3, 4, 5.
then sin \(\theta\) = \(\frac{3}{5}\) and cos\(\theta\) = \(\frac{4}{3}\)
cos\(\theta\) - sin\(\theta\)
\(\frac{4}{5}\) - \(\frac{3}{5}\) = \(\frac{1}{5}\)
Users' Answers & Commentsfrom Pythagoras tippet, the hypotenus is T
i.e. 3, 4, 5.
then sin \(\theta\) = \(\frac{3}{5}\) and cos\(\theta\) = \(\frac{4}{3}\)
cos\(\theta\) - sin\(\theta\)
\(\frac{4}{5}\) - \(\frac{3}{5}\) = \(\frac{1}{5}\)
32
A man walks 100 m due West from a point X to Y, he then walks 100 m due North to a point Z. Find the bearing of X from Z.
A
195o
B
135o
C
225o
D
045o
correct option: b
tan\(\theta\) = \(\frac{100}{100}\) = 1
\(\theta\) = tan-1(1) = 45o
The bearing of x from z is N45oE or 135o
Users' Answers & Comments\(\theta\) = tan-1(1) = 45o
The bearing of x from z is N45oE or 135o
33
The derivatives of (2x + 1)(3x + 1) is
A
12x + 1
B
6x + 5
C
6x + 1
D
12x + 5
correct option: d
(2x + 1)(3x + 1) IS
2x + 1 \(\frac{d(3x + 1)}{d}\) + (3x + 1) \(\frac{d(2x + 1)}{d}\)
2x + 1 (3) + (3x + 1) (2)
6x + 3 + 6x + 2 = 12x + 5
Users' Answers & Comments2x + 1 \(\frac{d(3x + 1)}{d}\) + (3x + 1) \(\frac{d(2x + 1)}{d}\)
2x + 1 (3) + (3x + 1) (2)
6x + 3 + 6x + 2 = 12x + 5
34
\(\begin{array}{c|c} Class Intervals & 0 - 2 & 3 - 5 & 6 - 8 & 9 - 11 & \ \hline Frequency & 3 & 2 & 5 & 3 &\end{array}\)
Find the mode of the above distribution.
Find the mode of the above distribution.
A
9
B
8
C
10
D
7
correct option: d
Mode = L1 + (\(\frac{D_1}{D_1 + D_2}\))C
D1 = frequency of modal class - frequency of the class before it
D1 = 5 - 2 = 3
D2 = frequency of modal class - frequency of the class that offers it
D2 = 5 - 3 = 2
L1 = lower class boundary of the modal class
L1 = 5 - 5
C is the class width = 8 - 5.5 = 3
Mode = L1 + (\(\frac{D_1}{D_1 + D_2}\))C
= 5.5 + \(\frac{3}{2 + 3}\)C
= 5.5 + \(\frac{3}{5}\) x 3
= 5.5 + \(\frac{9}{5}\)
= 5.5 + 1.8
= 7.3 \(\approx\) = 7
Users' Answers & CommentsD1 = frequency of modal class - frequency of the class before it
D1 = 5 - 2 = 3
D2 = frequency of modal class - frequency of the class that offers it
D2 = 5 - 3 = 2
L1 = lower class boundary of the modal class
L1 = 5 - 5
C is the class width = 8 - 5.5 = 3
Mode = L1 + (\(\frac{D_1}{D_1 + D_2}\))C
= 5.5 + \(\frac{3}{2 + 3}\)C
= 5.5 + \(\frac{3}{5}\) x 3
= 5.5 + \(\frac{9}{5}\)
= 5.5 + 1.8
= 7.3 \(\approx\) = 7
35
Find the value of x at the minimum point of the curve y = x3 + x2 - x + 1
A
\(\frac{1}{3}\)
B
-\(\frac{1}{3}\)
C
1
D
-1
correct option: d
y = x3 + x2 - x + 1
\(\frac{dy}{dx}\) = \(\frac{d(x^3)}{dx}\) + \(\frac{d(x^2)}{dx}\) - \(\frac{d(x)}{dx}\) + \(\frac{d(1)}{dx}\)
\(\frac{dy}{dx}\) = 3x2 + 2x - 1 = 0
\(\frac{dy}{dx}\) = 3x2 + 2x - 1
At the maximum point \(\frac{dy}{dx}\) = 0
3x2 + 2x - 1 = 0
(3x2 + 3x) - (x - 1) = 0
3x(x + 1) -1(x + 1) = 0
(3x - 1)(x + 1) = 0
therefore x = \(\frac{1}{3}\) or -1
For the maximum point
\(\frac{d^2y}{dx^2}\) < 0
\(\frac{d^2y}{dx^2}\) 6x + 2
when x = \(\frac{1}{3}\)
\(\frac{dx^2}{dx^2}\) = 6(\(\frac{1}{3}\)) + 2
= 2 + 2 = 4
\(\frac{d^2y}{dx^2}\) > o which is the minimum point
when x = -1
\(\frac{d^2y}{dx^2}\) = 6(-1) + 2
= -6 + 2 = -4
-4 < 0
therefore, \(\frac{d^2y}{dx^2}\) < 0
the maximum point is -1
Users' Answers & Comments\(\frac{dy}{dx}\) = \(\frac{d(x^3)}{dx}\) + \(\frac{d(x^2)}{dx}\) - \(\frac{d(x)}{dx}\) + \(\frac{d(1)}{dx}\)
\(\frac{dy}{dx}\) = 3x2 + 2x - 1 = 0
\(\frac{dy}{dx}\) = 3x2 + 2x - 1
At the maximum point \(\frac{dy}{dx}\) = 0
3x2 + 2x - 1 = 0
(3x2 + 3x) - (x - 1) = 0
3x(x + 1) -1(x + 1) = 0
(3x - 1)(x + 1) = 0
therefore x = \(\frac{1}{3}\) or -1
For the maximum point
\(\frac{d^2y}{dx^2}\) < 0
\(\frac{d^2y}{dx^2}\) 6x + 2
when x = \(\frac{1}{3}\)
\(\frac{dx^2}{dx^2}\) = 6(\(\frac{1}{3}\)) + 2
= 2 + 2 = 4
\(\frac{d^2y}{dx^2}\) > o which is the minimum point
when x = -1
\(\frac{d^2y}{dx^2}\) = 6(-1) + 2
= -6 + 2 = -4
-4 < 0
therefore, \(\frac{d^2y}{dx^2}\) < 0
the maximum point is -1
36
Evaluate \(\int^{1}_{2}\)(3 - 2x)dx
A
33m
B
5
C
2
D
6
correct option: c
\(\int^{1}_{0}\)(3 - 2x)dx
[3x - x2]o
[3(1) - (1)2] - [3(0) - (0)2]
(3 - 1) - (0 - 0) = 2 - 0
= 2
Users' Answers & Comments[3x - x2]o
[3(1) - (1)2] - [3(0) - (0)2]
(3 - 1) - (0 - 0) = 2 - 0
= 2
37
Find \(\int^{1}_{0}\) cos4 x dx
A
\(\frac{3}{4}\) sin 4x + k
B
-\(\frac{1}{4}\) sin 4x + k
C
-\(\frac{3}{4}\) sin 4x + k
D
\(\frac{1}{4}\) sin 4x + k
correct option: d
\(\int^{1}_{0}\) cos4 x dx
let u = 4x
\(\frac{dy}{dx}\) = 4
dx = \(\frac{dy}{4}\)
\(\int^{1}_{0}\)cos u. \(\frac{dy}{4}\) = \(\frac{1}{4}\)\(\int\)cos u du
= \(\frac{1}{4}\) sin u + k
= \(\frac{1}{4}\) sin4x + k
Users' Answers & Commentslet u = 4x
\(\frac{dy}{dx}\) = 4
dx = \(\frac{dy}{4}\)
\(\int^{1}_{0}\)cos u. \(\frac{dy}{4}\) = \(\frac{1}{4}\)\(\int\)cos u du
= \(\frac{1}{4}\) sin u + k
= \(\frac{1}{4}\) sin4x + k
38
The sum of four consecutive integers is 34. Find the least of these numbers
A
7
B
6
C
8
D
5
correct option: a
Let the numbers be a, a + 1, a + 2, a + 3
a + a + 1 + a + 2 + a + 3 = 34
4a = 34 - 6
4a = 28
a = \(\frac{28}{4}\)
= 7
The least of these numbers is a = 7
Users' Answers & Commentsa + a + 1 + a + 2 + a + 3 = 34
4a = 34 - 6
4a = 28
a = \(\frac{28}{4}\)
= 7
The least of these numbers is a = 7
39
\(\begin{array}{c|c} No & 0 & 1 & 2 & 3 & 4 & 5 \ \hline Frequency & 1 & 4 & 3 & 8 & 2 & 5 \end{array}\). From the table above, find the median and range of the data respectively.
A
(8,5)
B
(3, 5)
C
(5 , 8)
D
(5 , 3)
correct option: b
Median = \(\frac{\sum fx}{\sum f}\)
\(\begin{array}{c|c}
No & 0 & 1 & 2 & 3 & 4 & 5 \ F & 1 & 4 & 3 & 8 & 2 & 5 \ fx & 0 & 4 & 6 & 24 & 8 & 25 \end{array}\)
\(\sum fx\) = 0 + 4 + 6 + 24 + 8 + 25 = 67
\(\sum f\) = 23
Median = \(\frac{\sum fx}{\sum f}\) = \(\frac{67}{23}\) = 2.913
= \(\approx\) 3
Range = 5 - 0 = 5
(3, 5)
Users' Answers & Comments\(\begin{array}{c|c}
No & 0 & 1 & 2 & 3 & 4 & 5 \ F & 1 & 4 & 3 & 8 & 2 & 5 \ fx & 0 & 4 & 6 & 24 & 8 & 25 \end{array}\)
\(\sum fx\) = 0 + 4 + 6 + 24 + 8 + 25 = 67
\(\sum f\) = 23
Median = \(\frac{\sum fx}{\sum f}\) = \(\frac{67}{23}\) = 2.913
= \(\approx\) 3
Range = 5 - 0 = 5
(3, 5)
40
\(\begin{array}{c|c}
Class Interval & 3 - 5 & 6 - 8 & 9 - 11 \ \hline Frequency & 2 & 2 & 2 \end{array}\). Find the standard deviation of the above distribution.
Class Interval & 3 - 5 & 6 - 8 & 9 - 11 \ \hline Frequency & 2 & 2 & 2 \end{array}\). Find the standard deviation of the above distribution.
A
√5
B
√6
C
√7
D
√2
correct option: b
\(\begin{array}{c|c}Class Interval & 3 - 3 & 6 - 8 & 9 - 11 \ x & 4 & 7 & 10 \ f & 2 & 2 & 2 \ f - x & 8 & 14 & 20 \ |x - \bar{x}|^2 & 9 & 0 & 9 \ |x - \bar{x}|^2 & 18 0 & 18 \end{array}\)
\(\bar{x}\) = \(\frac {\sum fx}{\sum f}\)
= \(\frac {8 + 14 + 20}{2 + 2 + 2}\)
= \(\frac{42}{6}\)
\(\bar{x}\) = 7
S.D = \(\sqrt\frac{\sum f(x - \bar{x})^2}{\sum f}\)
= \(\sqrt\frac{18 + 0 + 18}{6}\)
= \(\sqrt\frac{36}{6}\)
= \(\sqrt {6}\)
Users' Answers & Comments\(\bar{x}\) = \(\frac {\sum fx}{\sum f}\)
= \(\frac {8 + 14 + 20}{2 + 2 + 2}\)
= \(\frac{42}{6}\)
\(\bar{x}\) = 7
S.D = \(\sqrt\frac{\sum f(x - \bar{x})^2}{\sum f}\)
= \(\sqrt\frac{18 + 0 + 18}{6}\)
= \(\sqrt\frac{36}{6}\)
= \(\sqrt {6}\)