1993 - JAMB Mathematics Past Questions and Answers - page 1

\(\int \frac{1 - x}{x^3}\)

= \(\int^{1}_{x^3} - \int^{x}_{x^3}\)

= x^-3 dx - x^-2dx

= \(\frac{1}{2x^2} + \frac{1}{x}\)

\(\begin{array}{c|c} 8 & 71 \ 8 & 8 \text{rem} 7\ 8 & 1 \text{rem} 0\end{array}\)

= 107₈

\(\frac{3524}{0.05}\) = 70480

\(\approx\) 70500(3 s.g)

9^{(x - \(\frac{1}{2}\))} 3x² = 3^{2(x - \(\frac{1}{2}\))} = 3x²
∴ 2(x - \(\frac{1}{2}\)) = x²

2x - 1 = x²

hence x² - 2x + 1 = 0

(x - 1)(x - 1) = 0

x = 1

10^y x 5^{(2y - 2)} x 4^{(y - 1)} = 1

but 10^y - (5 x 2)^y = 5^y x 2^y

= (Law of indices)

5^y x 2^y x 5^{(2y - 2)} x 4(y - 1)

\(\frac{1}{√3 - 2}\) - \(\frac{1}{√3 + 2}\)

L.C.M = (3- 2) (3 + 2)

∴ \(\frac{1}{\sqrt{3 - 2}}\) - \(\frac{1}{\sqrt{3 - 2}}\) = \(\frac{\sqrt{3 + 2} - \sqrt{3 - 2}}{\sqrt{3 - 2} + \sqrt{3 - 2}}\)


\(\frac{√3 + 2 - √3 + 2}{3 - 2√3 + 2√3 - 4}\) = \(\frac{4}{3 - 2}\)

= \(\frac{4}{-1}\)

= -4

2log₃y + log₃x² = 4

log₃y² + log₃x² = 4

∴ log₃ (x²y²) = log₃81(correct all to base 4)

x²y² = 81

∴ xy = \(\pm\)9

∴ y = \(\pm\)\(\frac{9}{x}\)

log₅(62.5) - log₅(\(\frac{1}{2}\))

= log₅\(\frac{(62.5)}{\frac{1}{2}}\) - log₅(2 x 62.5)

= log₅(125)

= log₅5³ - 3log₅5

= 3

Principal = N255.00, Interest = 27.00

year = x Rate: 4%

∴ 1 = \(\frac{PRT}{100}\)

27 = \(\frac{225 \times 4 \times T}{100}\)

2700 = 900T

T = 3 years

\(\sqrt{x^2 + 9}\) = x + 1

x² + 9 = (x + 1)² + 1

0 = x² + 2x + 1 - x² - 9

= 2x - 8 = 0

2(x - 4) = 0

x = 4