Practicing Success

Target Exam

CUET

Subject

-- Mathematics - Section B1

Chapter

Applications of Derivatives

Question:

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

Options:

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

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

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

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

Correct Answer:

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

Explanation:

We have,

$g(u)=2 \tan ^{-1}\left(e^u\right)-\frac{\pi}{2}$

$\Rightarrow g'(u)=\frac{2 e^u}{1+e^{2 u}}>0$ for all  $u \rightarrow(-\infty, \infty)$

⇒ g is strictly increasing function in $(-\infty, \infty)$.

Now,

$g(u)=2 \tan ^{-1}\left(e^u\right)-\frac{\pi}{2}$

$\Rightarrow g(u)=\tan ^{-1}\left(e^u\right)-\left(\frac{\pi}{2}-\tan ^{-1}\left(e^u\right)\right)$

$\Rightarrow g(u)=\tan ^{-1}\left(e^u\right)-\cot ^{-1}\left(e^u\right)$

$\Rightarrow g(-u)=\tan ^{-1}\left(e^{-u}\right)-\cot ^{-1}\left(e^{-u}\right)$

$\Rightarrow g(-u)=\tan ^{-1}\left(e^{1 / u}\right)-\cot ^{-1}\left(e^{1 / u}\right)$

$\Rightarrow g(-u)=\cot ^{-1}\left(e^u\right)-\tan ^{-1}\left(e^u\right)=-g(u)$

Hence, g(u) is odd and is strictly increasing in $(-\infty, \infty)$