2009 - JAMB Mathematics Past Questions and Answers - page 8
71
A chord is drawn 5 cm away from the centre of a circle of radius 13 cm. Calculate the length of the chord.
A
7 cm
B
9 cm
C
12 cm
D
24 cm
correct option: c
132 = 52 + b2
169 = 25 + b2
b2 = 144
b = \(\sqrt{144}\)
= 12 cm
Users' Answers & Comments169 = 25 + b2
b2 = 144
b = \(\sqrt{144}\)
= 12 cm
72
Find the radius of a sphere whose surface area is 154cm2.
A
7.00cm
B
3.50cm
C
3.00cm
D
1.75cm
correct option: b
Surface area of a sphere = 4\(\pi\)r2
4\(\pi\)r2 = 154 cm2
\(\pi\)r2 = \(\frac{154}{4}\)
= 38.5
\(\pi\)r2 = 38.5
r2 = 38.5 + \(\frac{22}{7}\) \(\to\) r2 = 38.5 x \(\frac{7}{22}\)
r2 12.25 \(\to\) r = \(\sqrt{12.25}\)
= 3.5 cm
Users' Answers & Comments4\(\pi\)r2 = 154 cm2
\(\pi\)r2 = \(\frac{154}{4}\)
= 38.5
\(\pi\)r2 = 38.5
r2 = 38.5 + \(\frac{22}{7}\) \(\to\) r2 = 38.5 x \(\frac{7}{22}\)
r2 12.25 \(\to\) r = \(\sqrt{12.25}\)
= 3.5 cm
73
Find the locus of a particle which moves in the first quadrant so that it is equidistant from the lines x=0 and y=0(where k is constant)
A
x + y = 0
B
x - y = 0
C
x + y + k = 0
D
x - y - k = 0
correct option: d
Users' Answers & Comments74
What is the locus of the mid-point of all chords of length 6 cm with a circle of radius 5cm and with centre O?
A
A circle of radius 4 cm and with centre O
B
The perpendicular bisector of the chords
C
a straight line passing through centre O
D
A circle of radius 6 cm and with centre O
correct option: c
Users' Answers & Comments75
What is the value of p if the gradient of the line joining (-1,p) and (p,4) is \(\frac{2}{3}\)?
A
-2
B
-1
C
1
D
2
correct option: d
\(\frac{y_1 - y_o}{x_1 - x_o}\) = m
\(\frac{4 - p}{p + 1}\) = \(\frac{2}{3}\)
2p + 2 = 12 - 3p
5p = 10
p = \(\frac{10}{5}\)
= 2
Users' Answers & Comments\(\frac{4 - p}{p + 1}\) = \(\frac{2}{3}\)
2p + 2 = 12 - 3p
5p = 10
p = \(\frac{10}{5}\)
= 2
76
What is the value of r if the distance between the points (4,2) and (1, r) is 3 units?
A
1
B
2
C
3
D
4
correct option: b
Applying the formula
(x - a)2 + (y - b)2 = r2
(4 - 1)2 + (2 - r)2 = (3)2
(3)2 + (2 - r)2 = (3)2
9 + ( 4 - 4r + r2) = 9
13 - 4r + r2 = 9
r2 - 4r + 13 - 9 = 0
(r2 - 2r) - (2r + 4) = 0
r(r - 2) -2(r - 2) = 0
(r - 2) (r - 2)
r = 2 or -2
Users' Answers & Comments(x - a)2 + (y - b)2 = r2
(4 - 1)2 + (2 - r)2 = (3)2
(3)2 + (2 - r)2 = (3)2
9 + ( 4 - 4r + r2) = 9
13 - 4r + r2 = 9
r2 - 4r + 13 - 9 = 0
(r2 - 2r) - (2r + 4) = 0
r(r - 2) -2(r - 2) = 0
(r - 2) (r - 2)
r = 2 or -2
77
Find the value of sin 45o - cos 30o
A
\(\frac{2 + \sqrt{6}}{4}\)
B
\(\frac{ \sqrt{2} + \sqrt{3}}{4}\)
C
\(\frac{ \sqrt{2} + \sqrt{3}}{6}\)
D
\(\frac{ \sqrt{2} - \sqrt{3}}{2}\)
correct option: d
sin 45o - cos 30o
\(\frac{1}{\sqrt{2}}\) - \(\frac{\sqrt{3}}{2}\)
= \(\frac{2 - \sqrt{2} \times \sqrt{3}}{2\sqrt{2}})\)
= \(\frac{2 - \sqrt{6} \times \sqrt{2}}{2\sqrt{2} \sqrt{2}}\)
= \(\frac{2\sqrt{2} - \sqrt{12}}{4}\)
= \(\frac{2\sqrt{2} - 2\sqrt{3}}{4}\)
= \(\frac{2(\sqrt{2} - \sqrt{3})}{4}\)
= \(\frac{\sqrt{2} - \sqrt{3}}{2}\)
Users' Answers & Comments\(\frac{1}{\sqrt{2}}\) - \(\frac{\sqrt{3}}{2}\)
= \(\frac{2 - \sqrt{2} \times \sqrt{3}}{2\sqrt{2}})\)
= \(\frac{2 - \sqrt{6} \times \sqrt{2}}{2\sqrt{2} \sqrt{2}}\)
= \(\frac{2\sqrt{2} - \sqrt{12}}{4}\)
= \(\frac{2\sqrt{2} - 2\sqrt{3}}{4}\)
= \(\frac{2(\sqrt{2} - \sqrt{3})}{4}\)
= \(\frac{\sqrt{2} - \sqrt{3}}{2}\)
78
A cliff on the bank of a river is 300 meters high. If the angle of depression of a point on the opposite side of the river is 60o, find the width of the river.
A
75√3m
B
100√3m
C
200√3m
D
100m
correct option: b
Tan 60o = \(\frac{300}{x}\)
x Tan 60o = 300
x√3 = 300
x = \(\frac{300}{\sqrt{3}}\) x \(\frac{\sqrt{3}}{\sqrt{3}}\)
x = \(\frac{300\sqrt{3}}{3}\) x 100\(\sqrt{3}\)
Users' Answers & Commentsx Tan 60o = 300
x√3 = 300
x = \(\frac{300}{\sqrt{3}}\) x \(\frac{\sqrt{3}}{\sqrt{3}}\)
x = \(\frac{300\sqrt{3}}{3}\) x 100\(\sqrt{3}\)
79
If y = 3 cos 4x, \(\frac{dy}{dx}\) equals
A
6 sin 8x
B
-24 sin 4x
C
12 sin 4x
D
-12 sin 4x
correct option: d
If y = 3 cos 4x, \(\frac{dy}{dx}\) equals
\(\frac{dy}{dx}\) = 3(4 - sin 4x)
= -12 sin 4x
Users' Answers & Comments\(\frac{dy}{dx}\) = 3(4 - sin 4x)
= -12 sin 4x
80
If s = (2 + 3t)(5t - 4), find \(\frac{dy}{dx}\) when t = \(\frac{4}{5}\) sec
A
0 units per sec.
B
15 units per seconds
C
22 units per seconds
D
26 units per seconds
correct option: c
s = (2 + 3t)(5t - 4), find \(\frac{dy}{dx}\) when t = \(\frac{4}{5}\) sec
= 10t - 8 + 15t2 - 12t
\(\frac{ds}{dt}\) = 15t2 - 2t - 8
= 30t - 2 ; t = \(\frac{4}{5}\)
= 30t - 2
= 30 x \(\frac{4}{5}\) - 2 \(\to\) 24 - 2
= 22 sec
Users' Answers & Comments= 10t - 8 + 15t2 - 12t
\(\frac{ds}{dt}\) = 15t2 - 2t - 8
= 30t - 2 ; t = \(\frac{4}{5}\)
= 30t - 2
= 30 x \(\frac{4}{5}\) - 2 \(\to\) 24 - 2
= 22 sec