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