Q24 / 66 · JEE Advanced 2008 (Paper 2) · Mathematics

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

  • A
    even and is strictly increasing in $(0, \infty)$
  • B
    odd and is strictly decreasing in $(-\infty, \infty)$
  • Correct answer
    C
    odd and is strictly increasing in $(-\infty, \infty)$
  • D
    Neither even nor odd, but is strictly increasing in $(-\infty, \infty)$

Solution

$\therefore \quad g(u)=2 \tan ^{-1}\left(e^u\right)-\frac{\pi}{2}$ for $\quad u \in(-\infty, \infty)$ and $$ \begin{aligned} g(-u) & =2 \tan ^{-1}\left(e^{-u}\right)-\frac{\pi}{2} \\ & =2\left(\cot ^{-1}\left(e^u\right)\right)-\frac{\pi}{2} \\ & =2\left(\frac{\pi}{2}-\tan ^{-1}\left(e^u\right)\right)-\frac{\pi}{2} \\ & =\frac{\pi}{2}-2 \tan ^{-1}\left(e^u\right)=-g(u) \end{aligned} $$ $\therefore \quad g(-u)=-g(u)$ $\Rightarrow g(u)$ is an odd function.

Hence, the correct answer is option: 'C'.

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