Question

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) \[f\text{ is neither even nor odd,but }{{f}^{'}}\left( x \right)\text{0 }\forall \text{ }x\in R\]

Answer 1

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, \[f\left( -x \right)=2{{\tan }^{-1}}\left( {{e}^{-x}} \right)-\dfrac{\pi }{2}....\left( ii \right)\]
Adding equation (i) and (ii)
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.