2007 - JAMB Mathematics Past Questions and Answers - page 9
81
The area of a square is 144 sqcm. Find the length of the diagonal.
A
13 cm
B
12√2 cm
C
12 cm
D
11√3 cm
correct option: b
Area of a square = x * x
144 = x2
√144 = x
12 = x
But d2 = 122 x 122
= 144 + 144
= 288
d = √288
= √144 x 2
= 12√2
Users' Answers & Comments144 = x2
√144 = x
12 = x
But d2 = 122 x 122
= 144 + 144
= 288
d = √288
= √144 x 2
= 12√2
82
if X = {all perfect squares less than 40} and y = {all odd numbers from 1 to 15}, find X ∩ Y
A
{9, 25}
B
{3, 9}
C
{1, 9}
D
{9}
correct option: c
x = (1, 4, 9, 16, 25, 36}
y = {1, 3, 5, 7, 9, 11, 13, 15}
X ∩ y = {1, 9}
Users' Answers & Commentsy = {1, 3, 5, 7, 9, 11, 13, 15}
X ∩ y = {1, 9}
83
If y = (1 + x)2, find \(\frac{dy}{dx}\)
A
x - 1
B
2 + 2x
C
1 + 2x
D
2x - 1
correct option: b
If y = (1 + x)2, find \(\frac{dy}{dx}\)
y = (1 + x)2
\(\frac{dy}{dx}\) = 2(1 + x)
= 2 + 2x
Users' Answers & Commentsy = (1 + x)2
\(\frac{dy}{dx}\) = 2(1 + x)
= 2 + 2x
84
If y = x cosx, find \(\frac{dy}{dx}\).
A
cos x - x sinx
B
sinx - x cosx
C
sin x + x cosx
D
cosx + x sin x
correct option: a
If y = x cosx, find \(\frac{dy}{dx}\).
Using product rule \(\frac{dy}{dx}\) = cosx - x sinx
Users' Answers & CommentsUsing product rule \(\frac{dy}{dx}\) = cosx - x sinx
85
Find the size of each exterior angle of a regular octagon.
A
45o
B
40o
C
51o
D
36o
86
Integrate \(\frac{x^2 - \sqrt{x}}{x}\) with respect to x
A
\(\frac{x^2 - \sqrt{x}}{x^2}\) + k
B
\(\frac{x^2 }{2}\) - \(\sqrt{x}\) + k
C
\(\frac{2(x^2 - x)}{3x}\) + k
D
\(\frac{x^2}{2}\) - 2\(\sqrt{x}\) + k
correct option: d
Integrate \(\frac{x^2 - \sqrt{x}}{x}\) with respect to x
∫\(\frac{x^2 - \sqrt{x}}{x}\)dx = ∫(\(\frac{x^2}{x} - \frac {\sqrt{x}}{x}\))dx
= ∫(x - x\(\frac{1}{2}\))dx
= \(\frac{x^2}{2}\) - \(\frac{x^{\frac{1}{2}}}{\frac{1}{2}}\) + k
= \(\frac{x^2}{2}\) - 2\(\sqrt{x}\) + k
Users' Answers & Comments∫\(\frac{x^2 - \sqrt{x}}{x}\)dx = ∫(\(\frac{x^2}{x} - \frac {\sqrt{x}}{x}\))dx
= ∫(x - x\(\frac{1}{2}\))dx
= \(\frac{x^2}{2}\) - \(\frac{x^{\frac{1}{2}}}{\frac{1}{2}}\) + k
= \(\frac{x^2}{2}\) - 2\(\sqrt{x}\) + k
87
Determine the value of \(∫^{\frac{\pi}{2}}_{0}\) (-2 cosx)dx
A
-\(\frac{3}{2}\)
B
-\(\frac{1}{2}\)
C
-2
D
-3
correct option: c
Determine the value of \(∫^{\frac{\pi}{2}}_{0}\) (-2 cosx)dx
\(∫^{\frac{\pi}{2}}_{0}\) (-2 cosx)dx = -2 sin x\(∫^{\frac{\pi}{2}} _{0}\)
= 2(sin 0 - sin\(\frac{\pi}{2}\))
= 2(0 - 1)
= -2
Users' Answers & Comments\(∫^{\frac{\pi}{2}}_{0}\) (-2 cosx)dx = -2 sin x\(∫^{\frac{\pi}{2}} _{0}\)
= 2(sin 0 - sin\(\frac{\pi}{2}\))
= 2(0 - 1)
= -2
88
Find the value of x for which the function f(x) = 2x3 - x2 - 4x + 4 has a maximum value
A
\(\frac{2}{3}\)
B
1
C
-1
D
-\(\frac{2}{3}\)
correct option: b
f(x) = 2x3 - x2 - 4x + 4
f(x) = 6x2 - 2x - 4 at turning point, f1(x) = 0
6x2 - 2x - 4 = 0, 3x2 - x - 2 = 0, 3x2 - 3x + 2x - 2 = 0
(3x + 2)(x - 1) = 0, x = -\(\frac{2}{3}\) or 1
f11(x) = 12x - 2,
when x = \(\frac{2}{3}\), f11(x) = 12(-\(\frac{2}{3}\)) - 2 = -10 < 0
\(\to\) f(x) is maximum @ x = -\(\frac{2}{3}\)
when x = 1, f11(x) = 12(1)- 2 = 10 > 0
\(\to\) f(x) is maximum @ x = 1
Users' Answers & Commentsf(x) = 6x2 - 2x - 4 at turning point, f1(x) = 0
6x2 - 2x - 4 = 0, 3x2 - x - 2 = 0, 3x2 - 3x + 2x - 2 = 0
(3x + 2)(x - 1) = 0, x = -\(\frac{2}{3}\) or 1
f11(x) = 12x - 2,
when x = \(\frac{2}{3}\), f11(x) = 12(-\(\frac{2}{3}\)) - 2 = -10 < 0
\(\to\) f(x) is maximum @ x = -\(\frac{2}{3}\)
when x = 1, f11(x) = 12(1)- 2 = 10 > 0
\(\to\) f(x) is maximum @ x = 1
89
The volume of a hemispherical bowl is 718\(\frac{2}{3}\) cm3. Find its radius.
A
4.0cm
B
7.0cm
C
3.8cm
D
5.6cm[\(\pi\) = \(\frac{22}{7}\)]
correct option: b
Volume of hemisphere = \(\frac{2}{3}\pi r^3\)
\(\frac{2}{3}\pi r^3\) = 718\(\frac{2}{3}\)
\(\frac{2}{3}\) x \(\frac{22}{7} r^3\)
= \(\frac{2156}{3}\)
r3 = \(\frac{2156}{3}\) x \(\frac{3}{2}\) x \(\frac{7}{22}\)
r3 = 49 x 7
r = 3\(\sqrt{343}\)
= 7cm3
Users' Answers & Comments\(\frac{2}{3}\pi r^3\) = 718\(\frac{2}{3}\)
\(\frac{2}{3}\) x \(\frac{22}{7} r^3\)
= \(\frac{2156}{3}\)
r3 = \(\frac{2156}{3}\) x \(\frac{3}{2}\) x \(\frac{7}{22}\)
r3 = 49 x 7
r = 3\(\sqrt{343}\)
= 7cm3
90
A man 40m from the foot of a tower observes the angle of elevation of the tower to be 30o, Determine the height of the tower
A
40m
B
40\(\sqrt{3}\)m
C
20m
D
\(\frac{40\sqrt{3}}{3}\)
correct option: d
tan 30o = \(\frac{h}{40m}\)
h = 40 tan 30o
= 40 x \(\frac{1}{\sqrt{30}}\)m
by rationalizing
h = \(\frac{40\sqrt{3}}{3}\)m
Users' Answers & Commentsh = 40 tan 30o
= 40 x \(\frac{1}{\sqrt{30}}\)m
by rationalizing
h = \(\frac{40\sqrt{3}}{3}\)m