Question

Let f(x) = $\frac{1}{x^{2}}$, y(x) = 1/x on [a, b] 0 < a < b Let $\frac{f(b) - f(a)}{g(b) - g(a)}$ = $\frac{f'(c)}{g'(c)}$ for some a < c< b then c is

• A. A.M. of a & b
• B. G.M. of a & b
• C. H.M. of a & b
• D. None

Answer 1

Answer: (C)

H.M. of a & b

Answer 2

$\frac{f(b) - f(a)}{g(b) - g(a)} = \frac{f'(c)}{g'(c)}$

$\frac{1/b^{2} - 1/a^{2}}{1/b - 1/a} = \frac{- 2/c^{3}}{- 1/c^{2}}$̃ $\frac{a^{2} - b^{2}}{a^{2}b^{2}} \times \frac{ab}{a - b} = \frac{2}{c}$

̃ $\frac{a + b}{ab} = \frac{2}{c}$̃ $c = \frac{2ab}{a + b}$