1993 - JAMB Mathematics Past Questions and Answers - page 2
11
Make x the subject of the relation \(\frac{1 + ax}{1 - ax}\) = \(\frac{p}{q}\)
A
\(\frac{p + q}{a(p - q)}\)
B
\(\frac{p - q}{a(p + q)}\)
C
\(\frac{p - q}{apq}\)
D
\(\frac{pq}{a(p - q)}\)
correct option: b
\(\frac{1 + ax}{1 - ax}\) = \(\frac{p}{q}\) by cross multiplication,
q(1 + ax) = p(1 - ax)
q + qax = p - pax
qax + pax = p - q
∴ x = \(\frac{p - q}{a(p + q)}\)
Users' Answers & Commentsq(1 + ax) = p(1 - ax)
q + qax = p - pax
qax + pax = p - q
∴ x = \(\frac{p - q}{a(p + q)}\)
12
Which of the following is a factor of 15 + 7x - 2x2
A
x + 3
B
x - 3
C
x - 5
D
x + 5
correct option: c
Factorize 15 + 7x - 2x2
(5 - x)(3 + 2x); suppose 15 + 7x = 2x2 = 0
∴ (5 - x)(3 + 2x) = 0
x = 5 or x = -\(\frac{3}{2}\)
Since 5 is a root, then (x - 5) is a factor
Users' Answers & Comments(5 - x)(3 + 2x); suppose 15 + 7x = 2x2 = 0
∴ (5 - x)(3 + 2x) = 0
x = 5 or x = -\(\frac{3}{2}\)
Since 5 is a root, then (x - 5) is a factor
13
Evaluate (x + \(\frac{1}{x}\) + 1)2 - (x + \(\frac{1}{x}\) + 1)2
A
4x2
B
(\(\frac{2}{x}\) + 2)2
C
4
D
4(1 + x)
correct option: d
(x + \(\frac{1}{x}\) + 1)2 - (x + \(\frac{1}{x}\) + 1)2
= (x + \(\frac{1}{x}\) + 1 + x + \(\frac{-1}{x}\) - 1)(x - \(\frac{1}{x}\) + 1 - x + \(\frac{1}{x}\) + 1)
= (2x) (2 + \(\frac{2}{x}\)) = 2x x 2(1 + \(\frac{1}{x}\))
4x (1 + \(\frac{1}{x}\)) = 4x + 4
= 4(1 + x)
Users' Answers & Comments= (x + \(\frac{1}{x}\) + 1 + x + \(\frac{-1}{x}\) - 1)(x - \(\frac{1}{x}\) + 1 - x + \(\frac{1}{x}\) + 1)
= (2x) (2 + \(\frac{2}{x}\)) = 2x x 2(1 + \(\frac{1}{x}\))
4x (1 + \(\frac{1}{x}\)) = 4x + 4
= 4(1 + x)
14
Solve the following simultaneous equation for x. x2 + y - 5 = 0, y - 7x + 3 = 0
A
-2, 4
B
2, 4
C
-1,8
D
1, -8
correct option: d
x2 + y - 5 = 0.....(i)
y - 7x + 3 = 0.........(ii)
y = 7x - 3, substituting the value of y in equation (i)
x2 + (7x - 3) - 5 = 0
x2 + 7x + 3 = 0
(x + 8)(x - 1) = 0
x = -8 or 1
Users' Answers & Commentsy - 7x + 3 = 0.........(ii)
y = 7x - 3, substituting the value of y in equation (i)
x2 + (7x - 3) - 5 = 0
x2 + 7x + 3 = 0
(x + 8)(x - 1) = 0
x = -8 or 1
15
Solve the following equation (3x - 2)(5x - 4) = (3x - 2)2
A
\(\frac{-2}{3}\), 1
B
1
C
\(\frac{2}{3}\), 1
D
\(\frac{2}{3}\), \(\frac{4}{5}\)
correct option: c
(3x - 2)(5x - 4) = (3x - 2)2 = 5x2 - 22x + 6
= 9x2 = 12x + 4
6x2 - 10x + 4 = 0
6x2 - 6x - 4x + 4 = 0
6x(x - 1) -4(x - 1) = (6x - 4)(x -1) = 0
x = 1 or \(\frac{2}{3}\)
Users' Answers & Comments= 9x2 = 12x + 4
6x2 - 10x + 4 = 0
6x2 - 6x - 4x + 4 = 0
6x(x - 1) -4(x - 1) = (6x - 4)(x -1) = 0
x = 1 or \(\frac{2}{3}\)
16
If the function f is defined by f(x + 2) = 2x2 + 7x = 5, find f(-1)
A
-10
B
-8
C
4
D
10
correct option: b
Users' Answers & Comments17
Divide the expression x3 + 7x2 - x - 7 by -1 + x2
A
-x3 + 7x2 - x - 7
B
-x3 = 7x + 7
C
x - 7
D
x + 7
correct option: d
Users' Answers & Comments18
Simplify \(\frac{1}{p}\) - \(\frac{1}{q}\) + \(\frac{p}{q}\) - \(\frac{q}{p}\)
A
\(\frac{1}{p - q}\)
B
\(\frac{-1}{p + q}\)
C
\(\frac{1}{pq}\)
D
\(\frac{1}{pq(p - q)}\)
correct option: b
\(\frac{1}{p}\) - \(\frac{1}{q}\) + \(\frac{p}{q}\) - \(\frac{q}{p}\) = \(\frac{q - p}{pq}\) + \(\frac{p^2 - q^2}{pq}\)
\(\frac{q - p}{pq}\) x \(\frac{pq}{p^2q^2}\) = \(\frac{q - p}{p^2 q^2}\)
\(\frac{-(p - q)}{(p + q)(p - q)}\)
= \(\frac{-1}{p + q}\)
Users' Answers & Comments\(\frac{q - p}{pq}\) x \(\frac{pq}{p^2q^2}\) = \(\frac{q - p}{p^2 q^2}\)
\(\frac{-(p - q)}{(p + q)(p - q)}\)
= \(\frac{-1}{p + q}\)
19
Solve the inequality y2 - 3y > 18
A
-3 < y < 6
B
y < -3 or y > 6
C
y > -3 or y > 6
D
y < 3 or y < 6
correct option: a
y2 - 3y > 18 = 3y - 18 > 0
y2 - 6y + 3y - 18 > 0 = y(y - 6) + 3 (y - 6) > 0
= (y + 3) (y - 6) > 0
Case 1 (+, +) \(\to\) (y + 3) > 0, (y - 6) > 0
= y > -3 y > 6
Case 2 (-, -) \(\to\) (y + 3) < 0, (y - 6) < 0
= y < -3, y < 6
Combining solution in case 1 and Case 2
= x < -3y < 6
= -3 < y < 6
Users' Answers & Commentsy2 - 6y + 3y - 18 > 0 = y(y - 6) + 3 (y - 6) > 0
= (y + 3) (y - 6) > 0
Case 1 (+, +) \(\to\) (y + 3) > 0, (y - 6) > 0
= y > -3 y > 6
Case 2 (-, -) \(\to\) (y + 3) < 0, (y - 6) < 0
= y < -3, y < 6
Combining solution in case 1 and Case 2
= x < -3y < 6
= -3 < y < 6
20
If x is negative, what is the range of values of x within which \(\frac{x + 1}{3}\) > \(\frac{1}{X + 3}\)
A
3 < x < 4
B
-4 < x < -3
C
-2 < x < -1
D
-3 < x < 0
correct option: b
\(\frac{x + 1}{3}\) > \(\frac{1}{X + 3}\) = \(\frac{x + 1}{3}\) > \(\frac{x + 3}{X + 3}\)
= (x + 1)(x + 3)2 > 3(x + 3) = (x + 1)[x2 + 6x + 9] > 3(x + 3)
x3 + 7x2 + 15x + 9 > 3x + 9 = x3 + 7x2 + 12x > 0
= x(x + 3)9x + 4) > 0
Case 1 (+, +, +) = x > 0 , x + 3 > 0, x + 4 > 0
= x > -4 (solution only)
Case 2 (+, -, -) = x > 0, x + 4 < 0
= x > 0, x < -3, x < -4 = x < -3(solution only)
Case 3 (-, +, -) = x < 0, x > -3, x < -4 = x < -0, -4 < x < 3(solutions)
Case 4 (-, -, +) = x < 0, x + 3 < 0, x + 4 > 0
= x < 0, x < -5, x > -4 = x < -0, -4 < x < -3(solution)
combining the solutions -4 < x < -3
Users' Answers & Comments= (x + 1)(x + 3)2 > 3(x + 3) = (x + 1)[x2 + 6x + 9] > 3(x + 3)
x3 + 7x2 + 15x + 9 > 3x + 9 = x3 + 7x2 + 12x > 0
= x(x + 3)9x + 4) > 0
Case 1 (+, +, +) = x > 0 , x + 3 > 0, x + 4 > 0
= x > -4 (solution only)
Case 2 (+, -, -) = x > 0, x + 4 < 0
= x > 0, x < -3, x < -4 = x < -3(solution only)
Case 3 (-, +, -) = x < 0, x > -3, x < -4 = x < -0, -4 < x < 3(solutions)
Case 4 (-, -, +) = x < 0, x + 3 < 0, x + 4 > 0
= x < 0, x < -5, x > -4 = x < -0, -4 < x < -3(solution)
combining the solutions -4 < x < -3