1997 - JAMB Mathematics Past Questions and Answers - page 2

11
Find the minimum value of X2 - 3x + 2 for all real values of x
A
-\(\frac{1}{4}\)
B
-\(\frac{1}{2}\)
C
\(\frac{1}{4}\)
D
\(\frac{1}{2}\)
correct option: a
y = X2 - 3x + 2, \(\frac{dy}{dx}\) = 2x - 3

at turning pt, \(\frac{dy}{dx}\) = 0

∴ 2x - 3 = 0

∴ x = \(\frac{3}{2}\)

\(\frac{d^2y}{dx^2}\) = \(\frac{d}{dx}\)(\(\frac{d}{dx}\))

= 270

∴ ymin = 2\(\frac{3}{2}\) - 3\(\frac{3}{2}\) + 2

= \(\frac{9}{4}\) - \(\frac{9}{2}\) + 2

= -\(\frac{1}{4}\)
Users' Answers & Comments
12
Make F the subject of the formula t = \(\sqrt{\frac{v}{\frac{1}{f} + \frac{1}{g}}}\)
A
\(\frac{gv-t^2}{gt^2}\)
B
\(\frac{gt^2}{gv-t^2}\)
C
\(\frac{v}{\frac{1}{t^2} - \frac{1}{g}}\)
D
\(\frac{gv}{t^2 - g}\)
correct option: b
t = \(\sqrt{\frac{v}{\frac{1}{f} + \frac{1}{g}}}\)

t2 = \(\frac{v}{\frac{1}{f} + \frac{1}{g}}\)

= \(\frac{vfg}{ftg}\)

\(\frac{1}{f} + \frac{1}{g}\) = \(\frac{v}{t^2}\)

= (g + f)t2 = vfg

gt2 = vfg - ft2

gt2 = f(vg - t2)

f = \(\frac{gt^2}{gv-t^2}\)
Users' Answers & Comments
13
What value of g will make the expression 4x2 - 18xy + g a perfect square?
A
9
B
\(\frac{9y^2}{4}\)
C
81y^2
D
\(\frac{18y^2}{4}\)
correct option: d
4x2 - 18xy + g = g \(\to\) (\(\frac{18y}{4}\))2

= \(\frac{18y^2}{4}\)
Users' Answers & Comments
14
Find the value of k if \(\frac{5 + 2r}{(r + 1)(r - 2)}\) expressed in partial fraction is \(\frac{k}{r - 2}\) + \(\frac{L}{r + 1}\) where K and L are constants
A
3
B
2
C
1
D
-1
correct option: a
5 + 2r = k(r + 1) + L(r - 2)

but r - 2 = 0 and r = 2

9 = 3k

k = 3
Users' Answers & Comments
15
Let f(x) = 2x + 4 and g(x) = 6x + 7 here g(x) > 0. Solve the inequality \(\frac{f(x)}{g(x)}\) < 1
A
x < - \(\frac{3}{4}\)
B
x > - \(\frac{4}{3}\)
C
x > - \(\frac{3}{4}\)
D
x > - 12
correct option: c
\(\frac{f(x)}{g(x)}\) < 1

∴ \(\frac{2x +4}{6x + 7}\) < 1

= 2x + 4 < 6x + 7

= 6x + 7 > 2x + 4

= 6x - 2x > 4 - 7

= 4x > -3

∴ x > -\(\frac{3}{4}\)
Users' Answers & Comments
16
Find the range of values of x which satisfies the inequality 12x2 < x + 1
A
-\(\frac{1}{4}\) < x < \(\frac{1}{3}\)
B
\(\frac{1}{4}\) < x < -\(\frac{1}{3}\)
C
-\(\frac{1}{3}\) < x < \(\frac{1}{4}\)
D
-\(\frac{1}{4}\) < x < - \(\frac{1}{3}\)
correct option: a
12x2 < x + 1

12 - x - 1 < 0

12x2 - 4x + 3x - 1 < 0

4x(3x - 1) + (3x - 1) < 0

Case 1 (+, -)

4x + 1 > 0, 3x - 1 < 0

x > -\(\frac{1}{4}\)x < \(\frac{1}{3}\)

Case 2 (-, +) 4x + 1 < 0, 3x - 1 > 0

-\(\frac{1}{4}\) < x < - \(\frac{1}{3}\)
Users' Answers & Comments
17
Sn is the sum of the first n terms of a series given by Sn = n2n - 1. Find the nth term
A
4n + 1
B
4n - 1
C
2n + 1
D
2n - 1
correct option: d
Users' Answers & Comments
18
The nth term of a sequence is given 31 - n , find the sum of the first terms of the sequence.
A
\(\frac{13}{9}\)
B
1
C
\(\frac{1}{3}\)
D
\(\frac{1}{9}\)
correct option: a
Tn = 31 - n

S3 = 31 - 1 + 31 - 2 + 31 - 3

= 1 + \(\frac{1}{3}\) + \(\frac{1}{9}\)

= \(\frac{13}{9}\)
Users' Answers & Comments
19
Two binary operations \(\ast\) and \(\oplus\) are defines as m \(\ast\) n = mn - n - 1 and m \(\oplus\) n = mn + n - 2 for all real numbers m, n.

Find the value of 3 \(\oplus\) (4 \(\ast\) 50)
A
60
B
57
C
54
D
42
correct option: c
m \(\ast\) n = mn - n - 1, m \(\oplus\) n = mn + n - 2

3 \(\oplus\) (4 \(\ast\) 5) = 3 \(\oplus\) (4 x 5 - 5 - 1) = 3 \(\oplus\) 14

3 \(\oplus\) 14 = 3 x 14 + 14 - 2

= 54
Users' Answers & Comments
20
If X \(\ast\) Y = X + Y - XY, find x when (x \(\ast\) 2) + (x \(\ast\) 3) = 68
A
24
B
22
C
-12
D
-21
correct option: d
x \(\ast\) y = x + y - xy

(x \(\ast\) 2) + (x \(\ast\) 3) = 68

= x + 2 - 2x + x + 3 - 3x

= 86

3x = 63

x = -21
Users' Answers & Comments
Please share this, thanks: