If for non-zero x, if $a f (x) + b f (\frac{1}{x}) = \frac{1}{x} - 25$ where a ≠ b then $\int_{1}^{2} f (x) d x$ is

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Answer 1

Given x as non- zero,

$a f (x) + b f (\frac{1}{x}) = \frac{1}{x} - 25$ ---(1)

Consider x as 1/x

$a f (\frac{1}{x}) + b f (\frac{1}{\frac{1}{x}}) = x - 25$

$a f (\frac{1}{x}) + b f (x) = x - 25$ ---(2)

Multiply equation 1 by a and 2 by b and subtract both

$a^{2} f (x) + a b f (\frac{1}{x}) - a b f (\frac{1}{x}) - b^{2} f (x) = \frac{a}{x} - 25 a - b x + 25 b$

$a^{2} f (x) - b^{2} f (x) = \frac{a}{x} - 25 a - b x + 25 b$

$f (x) = \frac{a}{x (a^{2} - b^{2})} - \frac{b x}{a^{2} - b^{2}} - \frac{25 (a - b)}{a^{2} - b^{2}}$

$\int_{1}^{2} f (x) d x = \frac{1}{a^{2} - b^{2}} [a (\ln 2 - 25) + \frac{47 b}{2}]$