First slide

Functions (XII)

Question

Let a function $f (x), x \neq 0$ be such that $f (x) + f (\frac{1}{x}) = f (x) . f (\frac{1}{x})$ then f(x) can be

Moderate

A

$1 - x^{2013}$
B

$\sqrt{|x|} + 1$
C

$\frac{π}{2 \tan^{- 1} [x]}$
D

$\frac{2}{1 + k \ln |x|}$

Solution

$\begin{array}{l} (f (\frac{1}{x}) - 1) = \frac{1}{(f (x) - 1)} i . e ., (f (\frac{1}{x}) - 1) \\ is reciprocal of (f (x) - 1) . \\ Now, for f (x) = \frac{\frac{π}{2}}{\tan^{- 1} [x]} \\ f (x) = - 1 = \frac{\cos^{- 1} [x]}{\tan^{- 1} [x]}, f (\frac{1}{x}) - 1 = \frac{\tan^{- 1} [x]}{\cot^{- 1} [x]} \\ Also for f (x) = \frac{2}{1 + k \ln |x|} \\ f (x) - 1 = \frac{1 - k \ln |x|}{1 + k \ln |x|}, f (\frac{1}{x}) - 1 = \frac{1 + k \ln |x|}{1 - k \ln |x|} \end{array}$

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