First slide

Functions (XII)

Question

$If a f (x + 1) + b f (\frac{1}{x + 1}) = x, x \neq - 1, a \neq b then f (2) is equl to$

Difficult

A

$\frac{2 a + b}{2 (a^{2} - b^{2})}$
B

$\frac{a}{a^{2} - b^{2}}$
C

$\frac{a + 2 b}{a^{2} - b^{2}}$
D

none of these

Solution

$\begin{array}{l} a f (x + 1) + b f (\frac{1}{x + 1}) = (x + 1) - 1 \\ Replacing x + 1 by \frac{1}{x + 1}, we get \\ a f (\frac{1}{x + 1}) + b f (x + 1) = \frac{1}{x + 1} - 1 \\ Eq . (1) \times a - Eq . (2) \times b \\ \Rightarrow (a^{2} - b^{2}) f (x + 1) = a (x + 1) - a - \frac{b}{x + 1} + b \\ Putting x = 1, (a^{2} - b^{2}) f (2) = 2 a - a - \frac{b}{2} + b = a + \frac{b}{2} = \frac{2 a + b}{2} \end{array}$

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