If \(\frac{1}{p}\) = \(\frac{a^2 + 2ab + b^2}{a - b}\) and \(\frac{1}{q}\) = \(\frac{a + b}{a^2 - 2ab + b^2}\) Find \(\frac{p}{q}\)

  • A \(\frac{a + b}{a - b}\)
  • B \(\frac{1}{a^2 - b^2}\)
  • C \(\frac{a - b}{a + b}\)
  • D a2 - b2
jamb 1988mathematics

The correct answer is B. \(\frac{1}{a^2 - b^2}\)

\(\frac{1}{p} = \frac{a^{2} + 2ab + b^{2}}{a - b}\)

\(\frac{1}{q} = \frac{a + b}{a^{2} - 2ab + b^{2}}\)

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

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

\(\therefore p = \frac{a - b}{(a + b)^{2}}\)

\(\frac{p}{q} = p \times \frac{1}{q} = \frac{a - b}{(a + b)^{2}} \times \frac{a + b}{(a - b)^{2}}\)

= \(\frac{1}{(a + b)(a - b)}\)

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

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