JAMB Mathematics Past Questions & Answers - Page 307

log₄(y - 1) + log₄(\(\frac{1}{2}\)x) = 1

log₄(y - 1)(\(\frac{1}{2}\)x) \(\to\) (y - 1)(\(\frac{1}{2}\)x) = 4 ........(1)

log₂(y + 1) + log₂x = 2

log₂(y + 1)x = 2 \(\to\) (y + 1)x = 2₂ = 4.....(ii)

From equation (ii) x = \(\frac{4}{y + 1}\)........(iii)

put equation (iii) in (i) = y (y - 1)[\(\frac{1}{2}(\frac{4}{y - 1}\))] = 4

= 2y - 2

= 4y + 4

2y = -6

y = -3

x = \(\frac{4}{-3 + 1}\)

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

X = 2

therefore x = -2, y = -3

c^{\(\frac{1}{3}\)} = a^{\(\frac{1}{2}\)}b

= a^{\(\frac{1}{2}\)}b x a^-2

= a^{-\(\frac{3}{2}\)}

= (c^{\(\frac{1}{3}\)})³

= (a^{-\(\frac{3}{2}\)})^{\(\frac{1}{3}\)}

c = a^{-\(\frac{1}{2}\)}

[\(\frac{1}{0.03}\) + \(\frac{1}{0.024}\)]

= [\(\frac{1}{0.03 \times 0.024}\)]^-1

= [\(\frac{0.024}{0.003}\)]^-1

= \(\frac{0.03}{0.024}\)

= \(\frac{30}{24}\) = 1.25

1011₂ + x₇ = 25₁₀ = 1011₂ = 1 x 2³ + 0 x 2² + 1 x 2¹ + 1 x 2^o

= 8 + 0 + 2 + 1

= 11₁₀

x₇ = 25₁₀ - 11₁₀

= 14₁₀

\(\begin{array}{c|c}
7 & 14 \\ 7 & 2 R 0 \\ & 0 R 2
\end{array}\)

X = 20₇

Equation: x\(^3\) - 2x\(^2\) - 5x + 6 = 0.

First, bring out a\(_n\) which is the coefficient of x\(^3\) = 1.

Then, a\(_0\) which is the coefficient void of x = 6.

The factors of a\(_n\) = 1; The factors of a\(_0\) = 1, 2, 3 and 6.

The numbers to test for the roots are \(\pm (\frac{a_0}{a_n})\).

= \(\pm (1, 2, 3, 6)\).

Test for +1: 1\(^3\) - 2(1\(^2\)) - 5(1) + 6 = 1 - 2 - 5 + 6 = 0.

Therefore x = 1 is a root of the equation.

Using long division method, \(\frac{x^3 - 2x^2 - 5x + 6}{x - 1}\) = x\(^2\) - x - 6.

x\(^2\) - x - 6 = (x - 3)(x + 2).

x = -2, 3.

\(\therefore\) The roots of the equation = 1, -2 and 3.