Answer

Verified

486.6k+ 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, \[f\left( -x \right)=2{{\tan }^{-1}}\left( {{e}^{-x}} \right)-\dfrac{\pi }{2}....\left( ii \right)\]

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.

Recently Updated Pages

what is the correct chronological order of the following class 10 social science CBSE

Which of the following was not the actual cause for class 10 social science CBSE

Which of the following statements is not correct A class 10 social science CBSE

Which of the following leaders was not present in the class 10 social science CBSE

Garampani Sanctuary is located at A Diphu Assam B Gangtok class 10 social science CBSE

Which one of the following places is not covered by class 10 social science CBSE

Trending doubts

Which places in India experience sunrise first and class 9 social science CBSE

Which are the Top 10 Largest Countries of the World?

Fill the blanks with the suitable prepositions 1 The class 9 english CBSE

How do you graph the function fx 4x class 9 maths CBSE

In Indian rupees 1 trillion is equal to how many c class 8 maths CBSE

The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths

Give 10 examples for herbs , shrubs , climbers , creepers

Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE

Why is there a time difference of about 5 hours between class 10 social science CBSE