JAMB Mathematics Past Questions & Answers - Page 305

\(\frac{1}{3x}\) + \(\frac{1}{2}\) > \(\frac{1}{4x}\)

= \(\frac{2 + 3x}{6x}\) > \(\frac{1}{4x}\)

= 4(2 + 3x) > 6x = 12x² - 2x = 0

= 2x(6x - 1) > 0 = x(6x - 1) > 0

Case 1 (-, -) = x < 0, 6x -1 < 0

= x < 0, x < \(\frac{1}{6}\) = x < \(\frac{1}{6}\) (solution)

Case 2 (+, +) = x > 0, 6x -1 > 0 = x > 0, x > \(\frac{1}{6}\)

Combining solutions in cases(1) and (2)

= x > 0, x < \(\frac{1}{6}\) = 0 < x < \(\frac{1}{6}\)

\(\frac{11x + 2}{6x^2 - x - 1}\) = \(\frac{11x + 2}{(3x + 1)(2x - 1)}\)

= \(\frac{A}{3x + 1}\) + \(\frac{B}{2x - 1}\)

11x + 2 = A(2x - 1) + B(3x + 1)

put x = \(\frac{1}{2}\)

\(\frac{15}{2} = \frac{5}{2}B\)

B = 3.

put x = \(-\frac{1}{3}\)

\(-\frac{5}{3} = \frac{-5}{3}\)A \(\implies\) A = 1

∴ \(\frac{11x +2}{6x^2 - x - 1}\) = \(\frac{1}{3x + 1}\) + \(\frac{3}{2x - 1}\)

No explanation has been provided for this answer.

\(\frac{x + a}{x - a}\) = m

x + a = mx - ma
a + ma = mx - x
a(m + 1) = x(m - 1)

\(\frac{a}{x}\) = \(\frac{m - 1}{m + 1}\)

\(\sqrt{x}\) - \(\sqrt{(x - 2)}\) - 1 = 0

= \(\sqrt{x}\) - \(\sqrt{(x - 2)}\) = 1

= (\(\sqrt{x}\) - \(\sqrt{(x - 2)}\))² = 1

= x - 2 \(\sqrt{x(x - 2)}\) + x -2 = 1

= (2x - 3)² = [2 \(\sqrt{x(x - 4)}\)]²

= 4x² - 12x + 9

= 4(x² - 2x)

= 4x² - 12x + 9

= 4x² - 8x

4x = 9

x = \(\frac{9}{4}\)