First slide

Introduction to ITF

Question

$Let f (x) be the derivative of \tan^{- 1} (\frac{x}{\sqrt{x^{2} + 1} + 1}) for all$ $x \in R & g (x) = \frac{1}{f (x)} + \frac{1}{f (x^{2})}; for all x \in R .$ $Let h^{- 1} (x) represent inverse function of h (x) then, which of the following is$ (are) TRUE?

Difficult

A

$g (x) is an even function$
B

$Range of g (x) is [4, \infty)$
C

$(g^{- 1})' (8) = \frac{1}{12}$
D

$(g^{- 1})' (8) = \frac{- 1}{12}$

Solution

$\begin{array}{l} f (x) = \frac{d}{d x} (\tan^{- 1} \frac{x}{\sqrt{x^{2} + 1} + 1}) \\ Let x = \tan θ \Rightarrow θ \in (- \frac{π}{2}, \frac{π}{2}) \\ \tan^{- 1} (\frac{\tan θ}{\sec θ + 1}) = \tan^{- 1} (\tan \frac{θ}{2}) \\ \Rightarrow \frac{θ}{2} = \frac{\tan^{- 1} x}{2} \Rightarrow f (x) = \frac{1}{2 (1 + x^{2})} \\ g (x) = \frac{1}{f (x)} + \frac{1}{f (x^{2})} = 2 (1 + x^{2}) + 2 (1 + x^{4}) \\ = 2 (x^{4} + x^{2} + 2) \\ \underline{Range} : [4, \infty) \\ Since g is many one g^{- 1} does not exists. \end{array}$

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