Courses
Courses for Kids
Free study material
Free LIVE classes
More LIVE
Join Vedantu’s FREE Mastercalss

# The function $f:R\to \left( \dfrac{-\pi }{2},\dfrac{\pi }{2} \right)$is given by $f\left( x \right)=2{{\tan }^{-1}}\left( {{e}^{x}} \right)-\dfrac{\pi }{2}$.Then, (a) $f\text{ is even and }{{f}^{'}}\left( x \right)\text{0 for }x>0$(b) $f\text{ is odd and }{{f}^{'}}\left( x \right)\text{0 for all }x\in R$(c) $f\text{ is odd and }{{f}^{'}}\left( x \right)\text{0 for all }x\in R$(d) $f\text{ is neither even nor odd,but }{{f}^{'}}\left( x \right)\text{0 }\forall \text{ }x\in R$ Verified
365.4k+ views
Hint: Find $f\left( -x \right)$and check if $f\left( x \right)$ is odd or even by comparing it to$f\left( -x \right)$. Then find ${{f}^{'}}\left( x \right)$.

We are given $f\left( x \right)=2{{\tan }^{-1}}\left( {{e}^{x}} \right)-\dfrac{\pi }{2}$, $f:R\to \left( \dfrac{-\pi }{2},\dfrac{\pi }{2} \right)$
We have to find if $f\left( x \right)$ is an odd function or even function. We also have to find the nature of ${{f}^{'}}\left( x \right)$.
If $f\left( x \right)=2{{\tan }^{-1}}\left( {{e}^{x}} \right)-\dfrac{\pi }{2}....\left( i \right)$
Therefore, $f\left( -x \right)=2{{\tan }^{-1}}\left( {{e}^{-x}} \right)-\dfrac{\pi }{2}....\left( ii \right)$
We get, $f\left( x \right)+f\left( -x \right)=2{{\tan }^{-1}}\left( {{e}^{-x}} \right)+2{{\tan }^{-1}}\left( {{e}^{-x}} \right)-\dfrac{\pi }{2}-\dfrac{\pi }{2}$
$f\left( x \right)+f\left( -x \right)=2\left( {{\tan }^{-1}}\left( {{e}^{x}} \right)+{{\tan }^{-1}}\left( {{e}^{-x}} \right) \right)-\pi$
Since, we know that ${{\tan }^{-1}}x+{{\tan }^{-1}}y=\dfrac{\pi }{2}$where x > 0, y > 0 and xy = 1
Here we know that ${{e}^{x}}>0,\dfrac{1}{{{e}^{x}}}>0$and ${{e}^{x}}.{{e}^{-x}}={{e}^{x}}.\dfrac{1}{{{e}^{x}}}=1$
Therefore, we get $f\left( x \right)+f\left( -x \right)=2\left[ \dfrac{\pi }{2} \right]-\pi$
$=\pi -\pi =0$
Since, we got $f\left( x \right)+f\left( -x \right)=0$
Therefore, $f\left( x \right)$is an odd function.
Now, taking $f\left( x \right)=2{{\tan }^{-1}}\left( {{e}^{x}} \right)-\dfrac{\pi }{2}$
By differentiating both sides with respect to x.
We get ${{f}^{'}}\left( x \right)=\dfrac{d}{dx}\left[ 2{{\tan }^{-1}}\left( {{e}^{x}} \right)-\dfrac{\pi }{2} \right]$
Since we know that $\dfrac{d}{dx}\left( {{\tan }^{-1}}x \right)=\dfrac{1}{1+{{x}^{2}}}$
$\dfrac{d}{dx}\left( {{e}^{x}} \right)={{e}^{x}}$
$\dfrac{d}{dx}\left( \text{constant} \right)=0$
So, we get ${{f}^{'}}\left( x \right)=\dfrac{2.{{e}^{x}}}{1+{{\left( {{e}^{x}} \right)}^{2}}}$
Therefore, ${{f}^{'}}\left( x \right)=\dfrac{2{{e}^{x}}}{1+{{e}^{2x}}}$
Since, we know that ${{e}^{x}}>0\text{ for }x\in R$
Therefore, ${{e}^{2x}}>0\text{ for }x\in R$
Hence, $\dfrac{2{{e}^{x}}}{1+{{e}^{2x}}}>0\text{ for }x\in R$
Therefore, we get ${{f}^{'}}\left( x \right)>0\text{ for }x\in R$
Therefore, $f$is odd and ${{f}^{'}}\left( x \right)>0\text{ for all }x\in R$

Therefore option (b) is correct.

Note: Students can also observe the odd or even nature of $f\left( x \right)$ by its graph. If $f\left( x \right)$is symmetrical about the x axis then it is even and if $f\left( x \right)$is symmetric about origin, it is odd.
Last updated date: 30th Sep 2023
Total views: 365.4k
Views today: 3.65k