1991 - JAMB Mathematics Past Questions and Answers - page 4
31
Evaluate x2(x2 - 1)-\(\frac{1}{2}\) - (x2 - 1)\(\frac{1}{2}\)
A
(x2 - 1)-\(\frac{1}{2}\)
B
(x2 - 1)1
C
(x2 - 1)
D
(x2 - 1)-1
correct option: a
x2(x2 - 1)-\(\frac{1}{2}\) - (x2 - 1)\(\frac{1}{2}\) = \(\frac{x^2}{(x^2 - 1)^\frac{1}{2}}\) - \(\frac{(x^2 - 1)^\frac{1}{2}}{1}\)
= \(\frac{x^2 - (x^2 - 1)}{(x^2 - 1) ^\frac{1}{2}}\)
= \(\frac{x^2 - x^2 + 1}{(x^2 - 1)^\frac{1}{2}}\)
= (x2 - 1)-\(\frac{1}{2}\)
Users' Answers & Comments= \(\frac{x^2 - (x^2 - 1)}{(x^2 - 1) ^\frac{1}{2}}\)
= \(\frac{x^2 - x^2 + 1}{(x^2 - 1)^\frac{1}{2}}\)
= (x2 - 1)-\(\frac{1}{2}\)
32
Find the gradient of the line passing through the points (-2, 0) and (0, -4)
A
2
B
-4
C
-2
D
4
correct option: c
Given (-2, 0) and (0, -4)
Gradient = \(\frac{-4 - 0}{0 - (-2)}\)
= \(\frac{-4}{2}\)
= -2
Users' Answers & CommentsGradient = \(\frac{-4 - 0}{0 - (-2)}\)
= \(\frac{-4}{2}\)
= -2
33
At what value of x is the function y = x2 - 2x - 3 minimum?
A
1
B
-1
C
-4
D
4
correct option: a
For y = ax2 - bx + c for minimum y
\(\frac{dy}{dx}\) = 2x - 2
= \(\frac{dy}{dx}\) = 0
∴ 2x - 2 = 0
x = 1
Users' Answers & Comments\(\frac{dy}{dx}\) = 2x - 2
= \(\frac{dy}{dx}\) = 0
∴ 2x - 2 = 0
x = 1
34
Find the sum of the 20 term in an arithmetic progression whose first term is 7 and last term is 117
A
2480
B
1240
C
620
D
124
correct option: b
Users' Answers & Comments35
The area of a square is 144 sq cm. Find the length of its diagonal
A
11\(\sqrt{3cm}\)
B
12cm
C
12\(\sqrt{2cm}\)
D
13cm
correct option: c
BD = \(\sqrt{x^2 + x^2}\)
= \(\sqrt{12^2 + 12^2}\)
= \(\sqrt{144 + 144}\)
= 2(144)
= 12\(\sqrt{2cm}\)
Users' Answers & Comments= \(\sqrt{12^2 + 12^2}\)
= \(\sqrt{144 + 144}\)
= 2(144)
= 12\(\sqrt{2cm}\)
36
One angle of a rhombus is 60o. The shorter of the two diagonals is 8cm long. Find the length of the longer one.
A
8\(\sqrt{3}\)
B
\(\frac{16}{\sqrt{3}}\)
C
\(\sqrt{3}\)
D
\(\frac{10}{\sqrt{3}}\)
correct option: a
\(\frac{x}{4}\) = \(\frac{\tan 60}{1}\)
x = 4 tan60
= 4\(\sqrt{3}\)
BD = 2x
= 8\(\sqrt{3}\)
Users' Answers & Commentsx = 4 tan60
= 4\(\sqrt{3}\)
BD = 2x
= 8\(\sqrt{3}\)
37
If the exterior angles of a pentagon are xo, (x + 5)o, (x + 10), (x + 15)o and (x + 20)o, find x
A
118o
B
72o
C
62o
D
36o
correct option: c
Users' Answers & Comments38
A flagstaff stands on the top of a vertical tower. A man standing 60 m away from the tower observes that the angles of elevation of the top and bottom of the flagstaff are 64o and 62o respectively. Find the length of the flagstaff.
A
60 (tan 62o - tan 64o)
B
60 (cot 64o - cot 62o)
C
60 (cot 62o - cot 64o)
D
60 (tan 64o - tan 62o)
correct option: d
\(\frac{BC}{60}\) = \(\frac{tan 62}{1}\)
BC = 60 tan 62
\(\frac{AC}{60}\) = \(\frac{tan 62}{1}\)
AC = 60 tan 64
AB = AC - BC
= 60(tan 64o - tan 62o)
Users' Answers & CommentsBC = 60 tan 62
\(\frac{AC}{60}\) = \(\frac{tan 62}{1}\)
AC = 60 tan 64
AB = AC - BC
= 60(tan 64o - tan 62o)
39
Simplify cos2 x (sec2x + sec2 x tan2x)
A
tan x
B
tanx secx
C
\(\frac{1 + tan^2x}{sec^2x}\)
D
cosec2x
correct option: c
cos2 x (sec2x + sec2 x tan2x)
= \(\frac{cos^2x}{cos^2x}\) x \(\frac{cos^2x sin^2x}{cos^2x cos^2x}\)
= \(\frac{1 + sin^2x}{cos^2x}\)
= \(\frac{1 + tan^2x}{sec^2x}\)
Users' Answers & Comments= \(\frac{cos^2x}{cos^2x}\) x \(\frac{cos^2x sin^2x}{cos^2x cos^2x}\)
= \(\frac{1 + sin^2x}{cos^2x}\)
= \(\frac{1 + tan^2x}{sec^2x}\)
40
If cos x = \(\sqrt{\frac{a}{b}}\) find coses x
A
\(\frac{b}{\sqrt{b - a}}\)
B
\(\sqrt{\frac{b}{a}}\)
C
\(\sqrt{\frac{b}{b - a}}\)
D
\(\sqrt{\frac{b - a}{a}}\)
correct option: c
cosx = \(\sqrt{\frac{a}{b}}\)
y2 + \(\sqrt{(a)^2}\) = \(\sqrt{(b)^2}\) by pythagoras
y2 = b - a
∴ y = b - a
cosec x = \(\frac{1}{sin x}\) = \(\frac{1}{y}\)
\(\frac{b}{y}\) = \(\frac{\sqrt{b}}{\sqrt{b - a}}\)
= \(\sqrt{\frac{b}{b - a}}\)
Users' Answers & Commentsy2 + \(\sqrt{(a)^2}\) = \(\sqrt{(b)^2}\) by pythagoras
y2 = b - a
∴ y = b - a
cosec x = \(\frac{1}{sin x}\) = \(\frac{1}{y}\)
\(\frac{b}{y}\) = \(\frac{\sqrt{b}}{\sqrt{b - a}}\)
= \(\sqrt{\frac{b}{b - a}}\)