# 1998 JAMB Mathematics Past Questions & Answers - page 1

### JAMB CBT App

Study offline with EduPadi JAMB CBT app that has so many features, including thousands of past questions, JAMB syllabus, novels, etc.

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

No further explanations yet...

2
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

Solution
$$\frac{11 + 2}{6x^2 - x - 1}$$ = $$\frac{11 + 2}{3x + 1}$$

= $$\frac{A}{3x + 1}$$ + $$\frac{B}{2x - 1}$$

11x = 2 = A(2x - 1) + B(3x + 1)

put x = $$\frac{1}{2}$$

= -$$\frac{-5}{3}$$

= -$$\frac{-5}{3}$$A $$\to$$ A = 1

∴ $$\frac{11x +2}{6x^2 - x - 1}$$ = $$\frac{1}{3x + 1}$$ + $$\frac{3}{2x - 1}$$

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

Solution
$$\frac{1}{3x}$$ + $$\frac{1}{2}$$ > $$\frac{1}{4x}$$

= $$\frac{2 + 3x}{6x}$$ > $$\frac{1}{4x}$$

= 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 < $$\frac{1}{6}$$

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

Solution
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 -$$\frac{4}{11}$$

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

Solution
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($$\frac{3}{2}$$)$$\frac{1}{3}$$ = $$\sqrt{\frac{3}{2}}$$

6
The kinetic element wit respect to the multiplication shown in the diagram below is $$\begin{array}{c|c} \oplus & p & p & r & s \ \hline p & r & p & r & p \ q & p & q & r & s\ r & r & r & r & r\ s & q & s & r & q\end{array}$$

A
p
B
q
C
r
D
s

Correct Option: D

No further explanations yet...

7
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

Solution
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

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

Solution
$$\begin{vmatrix} x & 1 & 0 \ 1-x & 2 & 3 \ 1 & 1+x & 4\end{vmatrix}$$ = 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

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

Solution
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)

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

Solution
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

### JAMB CBT APP

EduPadi JAMB CBT app that has thousands of past questions and answers, JAMB syllabus, novels, etc. And works offline!