2012 - JAMB Mathematics Past Questions and Answers - page 3
21
If \(\begin{vmatrix} 5 & 3 \ x & 2 \end{vmatrix}\) = \(\begin{vmatrix} 3 & 5 \ 4 & 5 \end{vmatrix}\), find the value of x
A
3
B
4
C
5
D
7
correct option: c
\(\begin{vmatrix} 5 & 3 \ x & 2 \end{vmatrix}\) = \(\begin{vmatrix} 3 & 5 \ 4 & 5 \end{vmatrix}\)
10 - 3x = 15 - 20
-3x = 15 - 20 - 10
-3x = -15
x = 5
Users' Answers & Comments10 - 3x = 15 - 20
-3x = 15 - 20 - 10
-3x = -15
x = 5
22
Given that I3 is a unit matrix of order 3, find |I3|
A
-1
B
O
C
1
D
2
correct option: c
Recall that a unit matrice is a diagonal matrix in which the elements in the leading diagonal is unity. Therefore,
I3 = \(\begin{pmatrix} 1 & 0 & 0 \ 0 & 1 & 0 \ 0 & 0 & 1 \end{pmatrix}\)
I3 = \(+1\begin{vmatrix} 1 & 0 \ 0 & 1 \end{vmatrix} - 0\begin{vmatrix} 0 & 0 \ 0 & 1 \end{vmatrix} + 0 \begin{vmatrix} 0 & 1 \ 0 & 0 \end{vmatrix} \)
I3 = +1(1 - 0) - 0(0 - 0) + 0(0 - 0)
= 1(1)
= 1
Users' Answers & CommentsI3 = \(\begin{pmatrix} 1 & 0 & 0 \ 0 & 1 & 0 \ 0 & 0 & 1 \end{pmatrix}\)
I3 = \(+1\begin{vmatrix} 1 & 0 \ 0 & 1 \end{vmatrix} - 0\begin{vmatrix} 0 & 0 \ 0 & 1 \end{vmatrix} + 0 \begin{vmatrix} 0 & 1 \ 0 & 0 \end{vmatrix} \)
I3 = +1(1 - 0) - 0(0 - 0) + 0(0 - 0)
= 1(1)
= 1
23
The angles of a polygon are given by x, 2x, 3x, 4x and 5x respectively. Find the value of x.
A
24o
B
30o
C
33o
D
36o
correct option: d
Users' Answers & Comments24
Calculate the volume of a cuboid of length 0.76cm, breadth 2.6cm and height 0.82cm.
A
3.92cm3
B
2.13cm3
C
1.97cm3
D
1.62cm3
correct option: d
Volume of cuboid = L x b x h
= 0.76cm x 2.6cm x 0.82cm
= 1.62cm3
Users' Answers & Comments= 0.76cm x 2.6cm x 0.82cm
= 1.62cm3
25
The locus of a point equidistant from the intersection of lines 3x - 7y + 7 = 0 and 4x - 6y + 1 = 0 is a
A
line parallel to 7x + 13y + 8 = 0
B
circle
C
semicircle
D
bisector of the line 7x + 13y + 8 = 0
correct option: a
Users' Answers & Comments26
The gradient of the straight line joining the points P(5, -7) and Q(-2, -3) is
A
\(\frac{1}{2}\)
B
\(\frac{2}{5}\)
C
\(-\frac{4}{7}\)
D
\(-\frac{2}{3}\)
correct option: c
PQ = \(\frac{y_1 - y_0}{x_1 - x_0}\) = \(\frac{-3 - (-7)}{-2 - 5}\) = \(\frac{-3 + 7}{-2 - 5}\) = \(\frac{4}{-7}\)
Users' Answers & Comments27
The distance between the point (4, 3) and the intersection of y = 2x + 4 and y = 7 - x is
A
\(\sqrt{13}\)
B
\(3\sqrt{2}\)
C
\(\sqrt{26}\)
D
\(10\sqrt{5}\)
correct option: b
P1 (4, 3), P2 (x, y)
y = 2x + 4 .....(1)
y = 7 - x .....(2)
Substitute (2) in (1)
7 - x = 2x + 4
7 - 4 = 2x + x
3 = 3x
x = 1
Substitute in eqn (2)
y = 7 - x
y = 7 - 1
y = 6
P2 (1, 6)
Distance between 2 points is given as
D = \(\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\)
D = \(\sqrt{(1 - 4)^2 + (6 - 3)^2}\)
D = \(\sqrt{(-3)^2 + (3)^2}\)
D = \(\sqrt{9 + 9}\)
D = \(\sqrt{18}\)
D = \(\sqrt{9 \times 2}\)
D = \(3\sqrt{2}\)
Users' Answers & Commentsy = 2x + 4 .....(1)
y = 7 - x .....(2)
Substitute (2) in (1)
7 - x = 2x + 4
7 - 4 = 2x + x
3 = 3x
x = 1
Substitute in eqn (2)
y = 7 - x
y = 7 - 1
y = 6
P2 (1, 6)
Distance between 2 points is given as
D = \(\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\)
D = \(\sqrt{(1 - 4)^2 + (6 - 3)^2}\)
D = \(\sqrt{(-3)^2 + (3)^2}\)
D = \(\sqrt{9 + 9}\)
D = \(\sqrt{18}\)
D = \(\sqrt{9 \times 2}\)
D = \(3\sqrt{2}\)
28
Find the equation of the line through the points (-2, 1) and (-\(\frac{1}{2}\), 4)
A
y = 2x - 3
B
y = 2x + 5
C
y = 3x - 2
D
y = 2x + 1
correct option: b
\(\frac{y - y_1}{x + x_1}\) = \(\frac{y_2 - y_1}{x - x_1}\)
\(\frac{y - 1}{x + 2}\) = \(\frac{4 - 1}{-\frac{1}{2} + 2}\)
= \(\frac{y - 1}{x + 2}\) = \(\frac{3}{\frac{3}{2}}\)
y = 2x + 5
Users' Answers & Comments\(\frac{y - 1}{x + 2}\) = \(\frac{4 - 1}{-\frac{1}{2} + 2}\)
= \(\frac{y - 1}{x + 2}\) = \(\frac{3}{\frac{3}{2}}\)
y = 2x + 5
29
If angle \(\theta\) is 135o, evaluate cos\(\theta\)
A
\(\frac{1}{2}\)
B
\(\frac{\sqrt{2}}{2}\)
C
\(-\frac{\sqrt{2}}{2}\)
D
\(-\frac{1}{2}\)
correct option: b
\(\theta\) = 135o
Cos 135o = Cos(90 + 45)o
= cos90ocos45o - sin90osin45o
= 0cos45o - (1 x \(\frac{\sqrt{2}}{2}\))
= \(-\frac{\sqrt{2}}{2}\)
Users' Answers & CommentsCos 135o = Cos(90 + 45)o
= cos90ocos45o - sin90osin45o
= 0cos45o - (1 x \(\frac{\sqrt{2}}{2}\))
= \(-\frac{\sqrt{2}}{2}\)
30
If y = x2 - \(\frac{1}{x}\), find \(\frac{\delta y}{\delta x}\)
A
2x - \(\frac{1}{x^2}\)
B
2x + x2
C
2x - x2
D
2x + \(\frac{1}{x^2}\)
correct option: d
y = x2 - \(\frac{1}{x}\)
y = x2 - x-1
\(\frac{\delta y}{\delta x}\) = 2x + x-2
\(\frac{\delta y}{\delta x}\) = 2x + \(\frac{1}{x^2}\)
Users' Answers & Commentsy = x2 - x-1
\(\frac{\delta y}{\delta x}\) = 2x + x-2
\(\frac{\delta y}{\delta x}\) = 2x + \(\frac{1}{x^2}\)