Solveeit Logo
Login

Question

Question: The function \[f:R\to \left( \dfrac{-\pi }{2},\dfrac{\pi }{2} \right)\]is given by \[f\left( x \righ...

The function f:R→(−π2,π2)f:R\to \left( \dfrac{-\pi }{2},\dfrac{\pi }{2} \right)is given by f(x)=2tan⁡−1(ex)−π2f\left( x \right)=2{{\tan }^{-1}}\left( {{e}^{x}} \right)-\dfrac{\pi }{2}.
Then,
(a) f is even and f′(x)0 for x>0f\text{ is even and }{{f}^{'}}\left( x \right)\text{0 for }x>0
(b) f is odd and f′(x)0 for all x∈Rf\text{ is odd and }{{f}^{'}}\left( x \right)\text{0 for all }x\in R
(c) f is odd and f′(x)0 for all x∈Rf\text{ is odd and }{{f}^{'}}\left( x \right)\text{0 for all }x\in R
(d) f is neither even nor odd,but f′(x)0 ∀ x∈Rf\text{ is neither even nor odd,but }{{f}^{'}}\left( x \right)\text{0 }\forall \text{ }x\in R

Explanation

Solution

Hint: Find f(−x)f\left( -x \right)and check if f(x)f\left( x \right) is odd or even by comparing it tof(−x)f\left( -x \right). Then find f′(x){{f}^{'}}\left( x \right).

We are given f(x)=2tan⁡−1(ex)−π2f\left( x \right)=2{{\tan }^{-1}}\left( {{e}^{x}} \right)-\dfrac{\pi }{2}, f:R→(−π2,π2)f:R\to \left( \dfrac{-\pi }{2},\dfrac{\pi }{2} \right)
We have to find if f(x)f\left( x \right) is an odd function or even function. We also have to find the nature of f′(x){{f}^{'}}\left( x \right).
If f(x)=2tan⁡−1(ex)−π2....(i)f\left( x \right)=2{{\tan }^{-1}}\left( {{e}^{x}} \right)-\dfrac{\pi }{2}....\left( i \right)
Therefore, f(−x)=2tan⁡−1(e−x)−π2....(ii)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(x)+f(−x)=2tan⁡−1(e−x)+2tan⁡−1(e−x)−π2−π2f\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(x)+f(−x)=2(tan⁡−1(ex)+tan⁡−1(e−x))−πf\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⁡−1x+tan⁡−1y=π2{{\tan }^{-1}}x+{{\tan }^{-1}}y=\dfrac{\pi }{2}where x > 0, y > 0 and xy = 1
Here we know that ex>0,1ex>0{{e}^{x}}>0,\dfrac{1}{{{e}^{x}}}>0and ex.e−x=ex.1ex=1{{e}^{x}}.{{e}^{-x}}={{e}^{x}}.\dfrac{1}{{{e}^{x}}}=1
Therefore, we get f(x)+f(−x)=2[π2]−πf\left( x \right)+f\left( -x \right)=2\left[ \dfrac{\pi }{2} \right]-\pi
=π−π=0=\pi -\pi =0
Since, we got f(x)+f(−x)=0f\left( x \right)+f\left( -x \right)=0
Therefore, f(x)f\left( x \right)is an odd function.
Now, taking f(x)=2tan⁡−1(ex)−π2f\left( x \right)=2{{\tan }^{-1}}\left( {{e}^{x}} \right)-\dfrac{\pi }{2}
By differentiating both sides with respect to x.
We get f′(x)=ddx[2tan⁡−1(ex)−π2]{{f}^{'}}\left( x \right)=\dfrac{d}{dx}\left[ 2{{\tan }^{-1}}\left( {{e}^{x}} \right)-\dfrac{\pi }{2} \right]
Since we know that ddx(tan⁡−1x)=11+x2\dfrac{d}{dx}\left( {{\tan }^{-1}}x \right)=\dfrac{1}{1+{{x}^{2}}}
ddx(ex)=ex\dfrac{d}{dx}\left( {{e}^{x}} \right)={{e}^{x}}
ddx(constant)=0\dfrac{d}{dx}\left( \text{constant} \right)=0
So, we get f′(x)=2.ex1+(ex)2{{f}^{'}}\left( x \right)=\dfrac{2.{{e}^{x}}}{1+{{\left( {{e}^{x}} \right)}^{2}}}
Therefore, f′(x)=2ex1+e2x{{f}^{'}}\left( x \right)=\dfrac{2{{e}^{x}}}{1+{{e}^{2x}}}
Since, we know that ex>0 for x∈R{{e}^{x}}>0\text{ for }x\in R
Therefore, e2x>0 for x∈R{{e}^{2x}}>0\text{ for }x\in R
Hence, 2ex1+e2x>0 for x∈R\dfrac{2{{e}^{x}}}{1+{{e}^{2x}}}>0\text{ for }x\in R
Therefore, we get f′(x)>0 for x∈R{{f}^{'}}\left( x \right)>0\text{ for }x\in R
Therefore, ffis odd and f′(x)>0 for all x∈R{{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(x)f\left( x \right) by its graph. If f(x)f\left( x \right)is symmetrical about the x axis then it is even and if f(x)f\left( x \right)is symmetric about origin, it is odd.