Let the function $g:(-\infty, \infty) \rightarrow\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$ be given by $g(u)=2 \tan ^{-1}\left(e^{u}\right)-\frac{\pi}{2}$. Then, $g$ is

(C) odd and is strictly increasing in $(-\infty, \infty)$

(A) even and is strictly increasing in $(0, \infty)$

(B) odd and is strictly decreasing in $(-\infty, \infty)$

(D) neither even nor odd, but is strictly increasing in $(-\infty, \infty)$

JEE Advanced · 2008 · 13 Apr 2008 · Paper 2 · Inverse Trigonometric Functions

Let the function

g : (- \infty, \infty) \to (- \frac{π}{2}, \frac{π}{2})

be given by

g (u) = 2 tan^{- 1} (e^{u}) - \frac{π}{2}

. Then,

g

Answer: (C)

We have

g (u) = 2 tan^{- 1} (e^{u}) - \frac{π}{2}

. Using the identity

2 tan^{- 1} (x) = tan^{- 1} (x) + cot^{- 1} (x) + tan^{- 1} (x) - cot^{- 1} (x)

, we can write

g (u) = tan^{- 1} (e^{u}) - cot^{- 1} (e^{u})

. Check:

g (- u) = tan^{- 1} (e^{- u}) - cot^{- 1} (e^{- u}) = cot^{- 1} (e^{u}) - tan^{- 1} (e^{u}) = - g (u)

, so

g

is odd. The derivative is

g^{'} (u) = \frac{e ^{u}}{1 + e ^{2 u}} + \frac{e ^{u}}{1 + e ^{2 u}} = \frac{2 e ^{u}}{1 + e ^{2 u}} > 0

for all

u

, so

g

is strictly increasing on

(- \infty, \infty)

JEE Advanced · 2008 · 13 Apr 2008 · Paper 2 · Single correct option