# 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

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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.

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.

Last updated date: 30th Sep 2023

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