1998 - JAMB Mathematics Past Questions and Answers - page 1
1
2x2 + 11x2 + 17x + 6 by 2x + 1
A
x2 + 5x + 6
B
2x2 + 5x - 6
C
2x2 + 5x + 6
D
x2 - 5x + 6
correct option: a
Users' Answers & Comments2
Express in partial fractions \(\frac{11 + 2}{6x^2 - x - 1}\)
A
\(\frac{1}{3x - 1}\) + \(\frac{3}{2x + 1}\)
B
\(\frac{3}{3x + 1}\) - \(\frac{1}{2x - 1}\)
C
\(\frac{3}{3x + 1}\) - \(\frac{1}{2x - 1}\)
D
\(\frac{1}{3x + 1}\) + \(\frac{3}{2x - 1}\)
correct option: d
\(\frac{11 + 2}{6x^2 - x - 1}\) = \(\frac{11 + 2}{3x + 1}\)
= \(\frac{A}{3x + 1}\) + \(\frac{B}{2x - 1}\)
11x = 2 = A(2x - 1) + B(3x + 1)
put x = \(\frac{1}{2}\)
= -\(\frac{-5}{3}\)
= -\(\frac{-5}{3}\)A \(\to\) A = 1
∴ \(\frac{11x +2}{6x^2 - x - 1}\) = \(\frac{1}{3x + 1}\) + \(\frac{3}{2x - 1}\)
Users' Answers & Comments= \(\frac{A}{3x + 1}\) + \(\frac{B}{2x - 1}\)
11x = 2 = A(2x - 1) + B(3x + 1)
put x = \(\frac{1}{2}\)
= -\(\frac{-5}{3}\)
= -\(\frac{-5}{3}\)A \(\to\) A = 1
∴ \(\frac{11x +2}{6x^2 - x - 1}\) = \(\frac{1}{3x + 1}\) + \(\frac{3}{2x - 1}\)
3
If x is a positive real number, find the range of values for which \(\frac{1}{3x}\) + \(\frac{1}{2}\) > \(\frac{1}{4x}\)
A
0 > -\(\frac{1}{6}\)
B
x > 0
C
0 < x < 4
D
0 < x < \(\frac{1}{6}\)
correct option: d
\(\frac{1}{3x}\) + \(\frac{1}{2}\) > \(\frac{1}{4x}\)
= \(\frac{2 + 3x}{6x}\) > \(\frac{1}{4x}\)
= 4(2 + 3x) > 6x = 12x2 - 2x = 0
= 2x(6x - 1) > 0 = x(6x - 1) > 0
Case 1 (-, -) = x < 0, 6x -1 < 0
= x < 0, x < \(\frac{1}{6}\) = x < \(\frac{1}{6}\) (solution)
Case 2 (+, +) = x > 0, 6x -1 > 0 = x > 0, x > \(\frac{1}{6}\)
Combining solutions in cases(1) and (2)
= x > 0, x < \(\frac{1}{6}\) = 0 < x < \(\frac{1}{6}\)
Users' Answers & Comments= \(\frac{2 + 3x}{6x}\) > \(\frac{1}{4x}\)
= 4(2 + 3x) > 6x = 12x2 - 2x = 0
= 2x(6x - 1) > 0 = x(6x - 1) > 0
Case 1 (-, -) = x < 0, 6x -1 < 0
= x < 0, x < \(\frac{1}{6}\) = x < \(\frac{1}{6}\) (solution)
Case 2 (+, +) = x > 0, 6x -1 > 0 = x > 0, x > \(\frac{1}{6}\)
Combining solutions in cases(1) and (2)
= x > 0, x < \(\frac{1}{6}\) = 0 < x < \(\frac{1}{6}\)
4
If p + 1, 2P - 10, 1 - 4p2are three consecutive terms of an arithmetic progression, find the possible values of p
A
-4, 2
B
-3, \(\frac{4}{11}\)
C
-\(\frac{4}{11}\), 2
D
5, -3
correct option: c
2p - 10 = \(\frac{p + 1 + 1 - 4P^2}{2}\) (Arithmetic mean)
= 2(2p - 100 = p + 2 - 4P2)
= 4p - 20 = p + 2 - 4p2
= 4p2 + 3p - 22 = 0
= (p - 2)(4p + 11) = 0
∴ p = 2 or -\(\frac{4}{11}\)
Users' Answers & Comments= 2(2p - 100 = p + 2 - 4P2)
= 4p - 20 = p + 2 - 4p2
= 4p2 + 3p - 22 = 0
= (p - 2)(4p + 11) = 0
∴ p = 2 or -\(\frac{4}{11}\)
5
The sum of the first three terms of a geometric progression is half its sum to infinity. Find the positive common ratio of the progression.
A
\(\frac{1}{4}\)
B
\(\sqrt{\frac{3}{2}}\)
C
\(\frac{1}{\sqrt{3}}\)
D
\(\frac{1}{\sqrt{2}}\)
correct option: b
Let the G.p be a, ar, ar2, S3 = \(\frac{1}{2}\)S
a + ar + ar2 = \(\frac{1}{2}\)(\(\frac{a}{1 - r}\))
2(1 + r + r)(r - 1) = 1
= 2r3 = 3
= r3 = \(\frac{3}{2}\)
r(\(\frac{3}{2}\))\(\frac{1}{3}\) = \(\sqrt{\frac{3}{2}}\)
Users' Answers & Commentsa + ar + ar2 = \(\frac{1}{2}\)(\(\frac{a}{1 - r}\))
2(1 + r + r)(r - 1) = 1
= 2r3 = 3
= r3 = \(\frac{3}{2}\)
r(\(\frac{3}{2}\))\(\frac{1}{3}\) = \(\sqrt{\frac{3}{2}}\)
6
The kinetic element wit respect to the multiplication shown in the diagram below is \(\begin{array}{c|c} \oplus & p & p & r & s \ \hline p & r & p & r & p
\ q & p & q & r & s\ r & r & r & r & r\ s & q & s & r & q\end{array}\)
\ q & p & q & r & s\ r & r & r & r & r\ s & q & s & r & q\end{array}\)
A
p
B
q
C
r
D
s
correct option: d
Users' Answers & Comments7
The binary operation \(\oplus\) is defined by x \(\ast\) y = xy - y - x for all real values x and y. If x \(\ast\) 3 = 2\(\ast\), find x
A
-1
B
4
C
1
D
5
correct option: c
x \(\ast\) y = xy - y - x, x \(\ast\) 3 = 3x - 3 - x = 2x - 3
2 \(\ast\) x = 2x - x - 2 = x - 2
∴ 2x - 3 = x - 2
x = -2 + 3
= 1
Users' Answers & Comments2 \(\ast\) x = 2x - x - 2 = x - 2
∴ 2x - 3 = x - 2
x = -2 + 3
= 1
8
The determinant of matrix \(\begin{pmatrix} x & 1 & 0 \ 1-x & 2 & 3 \ 1 & 1+x & 4\end{pmatrix}\) in terms of x is
A
-3x2 - 17
B
-3x2 + 9x - 1
C
3x2 + 17
D
3x2 - 9x + 5
correct option: b
\(\begin{vmatrix} x & 1 & 0 \ 1-x & 2 & 3 \ 1 & 1+x & 4\end{vmatrix}\) = x\(\begin{vmatrix}2 & 3 \ 1+x & 4\end{vmatrix}\) - \(\begin{vmatrix}1-x & 3 \ 1 & 4\end{vmatrix}\) = 0
= x[8 - 3(1 + x)] - [4(1 - x)-3] - 0 = x[5 - 3x] - [1 - 4x]
= 5x - 3x2 -1 + 4x
= -3x2 + 9X - 1
Users' Answers & Comments= x[8 - 3(1 + x)] - [4(1 - x)-3] - 0 = x[5 - 3x] - [1 - 4x]
= 5x - 3x2 -1 + 4x
= -3x2 + 9X - 1
9
Let = \(\begin{pmatrix} 1 & 0 \ 0 & 1 \end{pmatrix}\) p = \(\begin{pmatrix} 2 & 3 \ 4 & 5 \end{pmatrix}\) Q = \(\begin{pmatrix} u & 4+u \ -2v & v \end{pmatrix}\) be 2 x 2 matrices such that PQ = 1. Find (u, v)
A
(-\(\frac{5}{2}\) - 1)
B
(-\(\frac{5}{2}\) - \(\frac{3}{2}\))
C
(-\(\frac{5}{6}\) - 1)
D
(\(\frac{5}{2}\) - \(\frac{3}{2}\))
correct option: a
PQ = \(\begin{pmatrix} 2 & 3 \ 4 & 5 \end{pmatrix}\)\(\begin{pmatrix} u & 4+u \ -2v & v \end{pmatrix}\)
= \(\begin{pmatrix} (2u-6v & 2(4+u) +3v)\ 4u-10v & 4(4+u)+5v \end{pmatrix}\)
= \(\begin{pmatrix} 1 & 0 \ 0 & 1 \end{pmatrix}\)
2u - 6v = 1.....(i)
4u - 10v = 0.......(ii)
2(4 + u) + 3v = 0......(iii)
4(4 + u) + 5v = 1......(iv)
2u - 6v = 1 .....(i) x 2
4u - 10v = 0......(ii) x 1
\(\frac{\text{4u - 12v = 0}}{\text{-4u - 10v = 0}}\)
-2v = 2 = v = -1
2u - 6(-1) = 1 = 2u = 5
u = -\(\frac{5}{2}\)
∴ (U, V) = (-\(\frac{5}{2}\) - 1)
Users' Answers & Comments= \(\begin{pmatrix} (2u-6v & 2(4+u) +3v)\ 4u-10v & 4(4+u)+5v \end{pmatrix}\)
= \(\begin{pmatrix} 1 & 0 \ 0 & 1 \end{pmatrix}\)
2u - 6v = 1.....(i)
4u - 10v = 0.......(ii)
2(4 + u) + 3v = 0......(iii)
4(4 + u) + 5v = 1......(iv)
2u - 6v = 1 .....(i) x 2
4u - 10v = 0......(ii) x 1
\(\frac{\text{4u - 12v = 0}}{\text{-4u - 10v = 0}}\)
-2v = 2 = v = -1
2u - 6(-1) = 1 = 2u = 5
u = -\(\frac{5}{2}\)
∴ (U, V) = (-\(\frac{5}{2}\) - 1)
10
a cylindrical drum of diameter 56 cm contains 123.2 litres of oil when full. Find the height of the drum in centimeters
A
12.5
B
25.0
C
45.0
D
50.00
correct option: d
V = \(\pi r^2 h\)
= 123.2 x 1000
= \(\frac{22}{7}\) x 28 x 28 x h
= 123200 = 88 x 28h
= 2464h
∴ h = \(\frac{123200}{2464}\)
= 50cm
N.B: 1 litre = 1 dm3
= 1000cm3
Users' Answers & Comments= 123.2 x 1000
= \(\frac{22}{7}\) x 28 x 28 x h
= 123200 = 88 x 28h
= 2464h
∴ h = \(\frac{123200}{2464}\)
= 50cm
N.B: 1 litre = 1 dm3
= 1000cm3