JAMB Mathematics Past Questions & Answers - Page 294

1,466.

Given that loga2 = 0.693 and loga3 = 1.097, find loga 13.5

A.

1.404

B.

1.790

C.

2.598

D.

2.790

Correct answer is C

loga 13.5 = loga \(\frac{27}{2}\)

= 3loga 3 - log2a

= 3 x 1.097 - 0.693

= 2.598

1,467.

Find the value of (0.006)3 + (0.004)3 in standard form

A.

2.8 x 10-9

B.

2.8 x 10-7

C.

2.8 x 10-8

D.

2.8 x 10-6

Correct answer is B

(0.006)3 + (0.004)3 = 2.16 x 10-7

(2.16 + 0.64) x 10-7

= 2.8 x 10-7

1,468.

Evaluate 64.764\(^2\) - 35.236\(^2\) correct to 3 significant figures

A.

2960

B.

2950

C.

2860

D.

2850

Correct answer is B

(64.764 - 35.236)(64.764 + 35.236)

= 29.528 x 100

= 2950 to 3 sig. fig.

1,469.

If \(1P03{_4} = 115_{10}\) find P

A.

0

B.

1

C.

2

D.

3

Correct answer is D

\(1 x 4^3 + P x 4^2 + 0 x 4 + 3 = 115_{10}\)

16p + 67 = 115 p = \(\frac{48}{16}\)

= 3

1,470.

For what value of x does 6 sin (2x - 25)o attain its maximum value in the range 0o \(\leq\) x \(\leq\) 180o

A.

12\(\frac{1}{2}\)

B.

32\(\frac{1}{2}\)

C.

57\(\frac{1}{2}\)

D.

147\(\frac{1}{2}\)

Correct answer is C

y = 6 sin (2x - 25)o

\(\frac{dy}{dx}\) = -12 cos(2x - 25o)

\(\frac{dy}{dx}\) = 0

-12 cos x (2x - 25o) = 0

= cos(2x - 25o) = 0

= 2x - 25o = cos-1(0)

= 2x - 25o

= 90o

x = \(\frac{115}{2}\)