1994 - JAMB Mathematics Past Questions and Answers - page 2
11
Factorize a2x - b2y - b2x + a2y
A
(a - b)(x + y)
B
(y - x)(a - b)(a + b)
C
(x - y)(a - b)(a + b)
D
(x + y)(a - b)(a + b)
correct option: d
a2x - b2y - b2x + a2y = a2x - b2x - b2y + a2y Rearrange
= x(a2 - b2) + y(a2 - b2)
= (x + y)(a2 - b2)
= (x + y)(a + b)(a + b)
Users' Answers & Comments= x(a2 - b2) + y(a2 - b2)
= (x + y)(a2 - b2)
= (x + y)(a + b)(a + b)
12
Find the values of p and q such that (x - 1)and (x - 3) are factors of px3 + qx2 + 11x - 6
A
-1, -6
B
1, -6
C
1, 6
D
6, -1
correct option: b
Since (x - 1), is a factor, when the polynomial is divided by (x - 1), the remainder = zero
∴ (x - 1) = 0
x = 1
Substitute in the polynomial the value x = 1
= p(1)3 + q(1)2 + 11(1) - 6 = 0
p + q + 5 = 0 .....(i)
Also since x - 3 is a factor, ∴ x - 3 = 0
x = 3
Substitute p(3)3 + q(3)2 + 11(3) - 6 = 0
27p + 9q = -27 ......(2)
Combine eqns. (i) and (ii)
Multiply equation (i) by 9 to eliminate q
9p + 9q = -45
Subt. \(\frac{27p + 9q = -27}{-18p = -18}\)
∴ p = 1
Users' Answers & Comments∴ (x - 1) = 0
x = 1
Substitute in the polynomial the value x = 1
= p(1)3 + q(1)2 + 11(1) - 6 = 0
p + q + 5 = 0 .....(i)
Also since x - 3 is a factor, ∴ x - 3 = 0
x = 3
Substitute p(3)3 + q(3)2 + 11(3) - 6 = 0
27p + 9q = -27 ......(2)
Combine eqns. (i) and (ii)
Multiply equation (i) by 9 to eliminate q
9p + 9q = -45
Subt. \(\frac{27p + 9q = -27}{-18p = -18}\)
∴ p = 1
13
If a = 1, b = 3, solve for x in the equation \(\frac{a}{a - x}\) = \(\frac{b}{x - b}\)
A
\(\frac{4}{3}\)
B
\(\frac{2}{3}\)
C
\(\frac{3}{2}\)
D
\(\frac{3}{4}\)
correct option: c
\(\frac{a}{a - x}\) = \(\frac{b}{x - b}\)
\(\frac{1}{1 - x}\) = \(\frac{3}{x - 3}\)
∴ 3(1 - x) = x - 3
3 - 3x = x - 3
Rearrange 6 = 4x; x = \(\frac {6}{4}\)
= \(\frac{3}{2}\)
Users' Answers & Comments\(\frac{1}{1 - x}\) = \(\frac{3}{x - 3}\)
∴ 3(1 - x) = x - 3
3 - 3x = x - 3
Rearrange 6 = 4x; x = \(\frac {6}{4}\)
= \(\frac{3}{2}\)
14
Solve for r in the following equation \(\frac{1}{r - 1}\) + \(\frac{2}{r + 1}\) = \(\frac{3}{r}\)
A
3
B
4
C
5
D
6
correct option: a
\(\frac{1}{r - 1}\) + \(\frac{2}{r + 1}\) = \(\frac{3}{r}\)
Multiply through by r(r -1) which is the LCM
= (r)(r + 1) + 2(r)(r - 1)
= 3(r - 1)(r + 1)
= r2 + r + 2r2 - 2r
3r2 - 3 = 3r2
r = 3r2 - 3
-r = -3
∴ r = 3
Users' Answers & CommentsMultiply through by r(r -1) which is the LCM
= (r)(r + 1) + 2(r)(r - 1)
= 3(r - 1)(r + 1)
= r2 + r + 2r2 - 2r
3r2 - 3 = 3r2
r = 3r2 - 3
-r = -3
∴ r = 3
15
Find P if \(\frac{x - 3}{(1 - x)(x + 2)}\) = \(\frac{p}{1 - x}\) + \(\frac{Q}{x + 2}\)
A
\(\frac{-2}{3}\)
B
\(\frac{-5}{3}\)
C
\(\frac{5}{3}\)
D
\(\frac{2}{3}\)
correct option: a
\(\frac{x - 3}{(1 - x)(x + 2)}\) = \(\frac{p}{1 - x}\) + \(\frac{Q}{x + 2}\)
Multiply both sides by LCM i.e. (1 - x(x + 2))
∴ x - 3 = p(x + 2) + Q(1 - x)
When x = +1
(+1) - 3 = p(+1 + 2) + Q(1 - 1)
-2 = 3p + 0(Q)
3p = -2
∴ p = \(\frac{-2}{3}\)
Users' Answers & CommentsMultiply both sides by LCM i.e. (1 - x(x + 2))
∴ x - 3 = p(x + 2) + Q(1 - x)
When x = +1
(+1) - 3 = p(+1 + 2) + Q(1 - 1)
-2 = 3p + 0(Q)
3p = -2
∴ p = \(\frac{-2}{3}\)
16
Find the range of values of x for which \(\frac{1}{x}\) > 2 is true
A
x < \(\frac{1}{2}\)
B
x < 0 or x < \(\frac{1}{2}\)
C
0 < x < \(\frac{1}{2}\)
D
1 < x < 2
correct option: c
\(\frac{1}{x}\) > 2 = \(\frac{x}{x^2}\) > 2
x > 2x2
= 2x2 < x
= 2x2 - x < 0
= x(2x - 10 < 0
Case 1(+, -) = x > 0, 2x - 1 < 0
x > 0, x < \(\frac{1}{2}\) (solution)
Case 2(-, 4) = x < 0, 2x - 1 > 0
x < 0, x , \(\frac{1}{2}\) = 0
Users' Answers & Commentsx > 2x2
= 2x2 < x
= 2x2 - x < 0
= x(2x - 10 < 0
Case 1(+, -) = x > 0, 2x - 1 < 0
x > 0, x < \(\frac{1}{2}\) (solution)
Case 2(-, 4) = x < 0, 2x - 1 > 0
x < 0, x , \(\frac{1}{2}\) = 0
17
If the 6th term of an arithmetic progression is 11 and the first term is 1, find the common difference.
A
\(\frac{12}{5}\)
B
\(\frac{5}{3}\)
C
-2
D
2
correct option: d
In an AP, Tn = a + (n - 1)d
T6 = a + 5d = 11
The first term = a = 1
∴ T6 = 1 + 5d = 11
5d = 11 - 1
5d = 10
∴ d = 2
Users' Answers & CommentsT6 = a + 5d = 11
The first term = a = 1
∴ T6 = 1 + 5d = 11
5d = 11 - 1
5d = 10
∴ d = 2
18
Find the value of log10 r + log10 r2 + log10 r4 + log10 r8 + log10 r16 + log10 r32 = 63
A
10-1
B
10o
C
10
D
102
correct option: c
log10 r + log10 r2 + log10 r4 + log10 r8 + log10 r16 + log10 r32 = 63
log10r63 = 63
63 = 1063
∴ r = 10
Users' Answers & Commentslog10r63 = 63
63 = 1063
∴ r = 10
19
Find the nth term of the sequence 3, 6, 10, 15, 21.....
A
\(\frac{n(n - 1)}{2}\)
B
\(\frac{n(n + 1)}{2}\)
C
\(\frac{(n + 1)(n + 2)}{2}\)
D
n(2n + 1)
correct option: c
\(\frac{(n + 1)(n + 2)}{2}\)
If n = 1, the expression becomes 3
n = 2, the expression becomes 6
n = 4, the expression becomes 15
n = 5, the expression becomes 21
Users' Answers & CommentsIf n = 1, the expression becomes 3
n = 2, the expression becomes 6
n = 4, the expression becomes 15
n = 5, the expression becomes 21
20
A binary operation \(\oplus\) is defines on the set of all positive integers by a \(\oplus\) b = ab for all positive integers a, b. Which of the following properties does NOT hold?
A
closure
B
identity
C
positive
D
inverse
correct option: d
a \(\oplus\) b = ab
The set of all national rules Q, is closed under the operations, additions, subtraction, multiplication and division. Since a \(\oplus\) b = ab; b \(\oplus\) a = ba = ab
The number 1 is the identity element under multiplication
Users' Answers & CommentsThe set of all national rules Q, is closed under the operations, additions, subtraction, multiplication and division. Since a \(\oplus\) b = ab; b \(\oplus\) a = ba = ab
The number 1 is the identity element under multiplication