1992 - JAMB Mathematics Past Questions and Answers - page 2
11
Solve the equation: y - 11y + 24 = 0
A
8, 3
B
64, 9
C
6, 4
D
9, -8
correct option: b
Users' Answers & Comments12
Make t the subject of formula S = ut + \(\frac{1}{2} at^2\)
A
\(\frac{1}{a}\) (-u + \(\sqrt{U^2 - 2as}\))
B
\(\frac{1}{a}\) {u \(\pm\) (U2 - 2as)}
C
\(\frac{1}{a}\) {u \(\pm\) \(\sqrt{2as}\)}
D
\(\frac{1}{a}\) {-u + \(\sqrt{( 2as)}\)}
correct option: a
Given S = ut + \(\frac{1}{2} at^2\)
S = ut + \(\frac{1}{2} at^2\)
∴ 2S = 2ut + at2
= at2 + 2ut - 2s = 0
t = \(\frac{-2u \pm 4u^2 + 2as}{2a}\)
= -2u \(\pi\) \(\frac{\sqrt{u^2 4u^2 + 2as}}{2a}\)
= \(\frac{1}{a}\) (-u + \(\sqrt{U^2 - 2as}\))
Users' Answers & CommentsS = ut + \(\frac{1}{2} at^2\)
∴ 2S = 2ut + at2
= at2 + 2ut - 2s = 0
t = \(\frac{-2u \pm 4u^2 + 2as}{2a}\)
= -2u \(\pi\) \(\frac{\sqrt{u^2 4u^2 + 2as}}{2a}\)
= \(\frac{1}{a}\) (-u + \(\sqrt{U^2 - 2as}\))
13
A man invested a sum of n280.00 partly at 5% and partly at 4%. if the total interest is N12.90 per annum, find the amount invested at 5%
A
14.00
B
120.00
C
140.00
D
160.00
correct option: d
Users' Answers & Comments14
If x + 1 is a factor of x3 + 3x2 + kx + 4, find the value of k
A
6
B
-6
C
8
D
-8
correct option: a
x + 1 is a factor of x3 + 3x2 + kx + 4
Let f(x) = x3 + 3x2 + kx + 4
∴ f(-1) = (-1)3 + 3(-1)2 + k(-1) + 4 = 0
-1 + 3 - k + 4 = 0
∴ k = 6
Users' Answers & CommentsLet f(x) = x3 + 3x2 + kx + 4
∴ f(-1) = (-1)3 + 3(-1)2 + k(-1) + 4 = 0
-1 + 3 - k + 4 = 0
∴ k = 6
15
Resolve \(\frac{3}{x^2 + x - 2}\) into partial fractions
A
\(\frac{1}{x - 1} + \frac{1}{x + 2}\)
B
\(\frac{1}{x + 1} + \frac{1}{x - 2}\)
C
\(\frac{1}{x + 1} - \frac{1}{x - 2}\)
D
\(\frac{1}{x - 2} + \frac{1}{x + 2}\)
correct option: a
\(\frac{3}{x^2 + x - 2}\) = \(\frac{3}{(x - 1)(x + 1)}\)
\(\frac{A}{x - 1}\) + \(\frac{B}{x + 2}\)
A(x + 2) + B(x - 1) = 3
when x = 1, 3A = 3 \(\to\) a = 1
when x = -2, -3B = 3 \(\to\) B = -1
3 = 1 - 1
= \(\frac{1}{x - 1} + \frac{1}{x + 2}\)
Users' Answers & Comments\(\frac{A}{x - 1}\) + \(\frac{B}{x + 2}\)
A(x + 2) + B(x - 1) = 3
when x = 1, 3A = 3 \(\to\) a = 1
when x = -2, -3B = 3 \(\to\) B = -1
3 = 1 - 1
= \(\frac{1}{x - 1} + \frac{1}{x + 2}\)
16
Find all values of x satisfying the inequality -11 \(\leq\) 4 - 3x \(\leq\) 28
A
-5 \(\leq\) x v 8
B
5 \(\leq\) x \(\leq\) 8
C
-8 \(\leq\) x \(\leq\) 5
D
-5 < x \(\leq\) 8
correct option: c
To solve -11 \(\leq\) 4 - 3x \(\leq\) 28
-11 \(\leq\) 4 - 3x also 4 -3x \(\leq\) 28
15 \(\leq\) -3x \(\leq\) 24 = 15 \(\geq\) 3x - 3x \(\geq\) -24
-5 \(\geq\) x, x \(\geq\) -8
i.e. x \(\leq\) 5
∴ -8 \(\leq\) x \(\leq\) 5
Users' Answers & Comments-11 \(\leq\) 4 - 3x also 4 -3x \(\leq\) 28
15 \(\leq\) -3x \(\leq\) 24 = 15 \(\geq\) 3x - 3x \(\geq\) -24
-5 \(\geq\) x, x \(\geq\) -8
i.e. x \(\leq\) 5
∴ -8 \(\leq\) x \(\leq\) 5
17
Find the sum to infinity to the following series 3 + 2 + \(\frac{4}{3}\) + \(\frac{8}{9}\) + \(\frac{16}{17}\) + .....
A
1270
B
190
C
18
D
9
correct option: d
3 + 2 + \(\frac{4}{3}\) + \(\frac{8}{9}\) + \(\frac{16}{17}\) + .....
a = 3
r = \(\frac{2}{3}\)
s \(\alpha\) = \(\frac{a}{1 - r}\) = \(\frac{3}{1 - \frac{2}{3}}\)
= \(\frac{3}{\frac{1}{3}}\)
= 3 x 3
= 9
Users' Answers & Commentsa = 3
r = \(\frac{2}{3}\)
s \(\alpha\) = \(\frac{a}{1 - r}\) = \(\frac{3}{1 - \frac{2}{3}}\)
= \(\frac{3}{\frac{1}{3}}\)
= 3 x 3
= 9
18
What is the n-th term of the sequence 2, 6, 12, 20...?
A
4n - 2
B
2(3n - 1)
C
n2 + n
D
n2 + 3n + 2
correct option: c
Given that 2, 6, 12, 20...? the nth term = n2 + n
check: n = 1, u1 = 2
n = 2, u2 = 4 + 2 = 6
n = 3, u3 = 9 + 3 = 12
∴ n = 4, u4 = 16 + 4 = 20
Users' Answers & Commentscheck: n = 1, u1 = 2
n = 2, u2 = 4 + 2 = 6
n = 3, u3 = 9 + 3 = 12
∴ n = 4, u4 = 16 + 4 = 20
19
For an arithmetical sequence, the first term is 2 and the common difference is 3. Find the sum of the first 11 terms
A
157
B
187
C
197
D
200
correct option: b
a = 2, d = 3 and n = 11
To find Sn/sub> = \(\frac{n}{2}\) [2a + (n - 1) \(\delta\)]
= \(\frac{11}{2}\) [2(2) + (11 - 1) 3]
= \(\frac{11}{2}\)n [4 + 10(3)]
= \(\frac{11}{2}\)(34)
= 11 x 17
= 187
Users' Answers & CommentsTo find Sn/sub> = \(\frac{n}{2}\) [2a + (n - 1) \(\delta\)]
= \(\frac{11}{2}\) [2(2) + (11 - 1) 3]
= \(\frac{11}{2}\)n [4 + 10(3)]
= \(\frac{11}{2}\)(34)
= 11 x 17
= 187
20
If the binary operation \(\ast\) is defined by m \(\ast\) n = mn + m + n for any real number m and n, find the identity of the elements under this operation
A
e = 1
B
e = -1
C
e = -2
D
e = 0
correct option: b
m \(\ast\) n = mn + m + n
m \(\ast\) e = me + m + e, e \(\ast\) m = e
∴ me + m + e, m(e + 1)e - e = 0
e + 1 = 0
∴ e = -1
Users' Answers & Commentsm \(\ast\) e = me + m + e, e \(\ast\) m = e
∴ me + m + e, m(e + 1)e - e = 0
e + 1 = 0
∴ e = -1