Let g: (-infty, infty) to (-frac{pi}{2}, frac | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) For the function $g(x) = 2 	an^{-1}(e^x) - \frac{\pi}{2}$,we find its derivative: $g'(x) = 2 \cdot \frac{1}{1 + (e^x)^2} \cdot e^x = \frac{2e^x}{

Vedclass · Accepted Answer

An odd function and strictly increasing on $(-\infty, \infty)$.

Vedclass · Answer

(C) For the function $g(x) = 2 	an^{-1}(e^x) - \frac{\pi}{2}$,we find its derivative: $g'(x) = 2 \cdot \frac{1}{1 + (e^x)^2} \cdot e^x = \frac{2e^x}{1 + e^{2x}}$. Since $e^x > 0$ for all $x \in (-\infty, \infty)$,it follows that $g'(x) > 0$. Thus,$g(x)$ is a strictly increasing function on $(-\infty, \infty)$. Now,check for parity: $g(-x) = 2 	an^{-1}(e^{-x}) - \frac{\pi}{2} = 2 	an^{-1}(\frac{1}{e^x}) - \frac{\pi}{2}$. Using the identity $	an^{-1}(\frac{1}{u}) = \frac{\pi}{2} - 	an^{-1}(u)$ for $u > 0$: $g(-x) = 2(\frac{\pi}{2} - 	an^{-1}(e^x)) - \frac{\pi}{2} = \pi - 2 	an^{-1}(e^x) - \frac{\pi}{2} = \frac{\pi}{2} - 2 	an^{-1}(e^x) = -g(x)$. Since $g(-x) = -g(x)$,the function $g(x)$ is an odd function.

Let $g: (-\infty, \infty) \to (-\frac{\pi}{2}, \frac{\pi}{2})$ be defined by $g(x) = 2 \tan^{-1}(e^x) - \frac{\pi}{2}$. Then $g(x)$ is...

Similar Questions

The function $f(x) = \sin x - \cos x$ is ........

Draw the graph of the function $f: R \rightarrow R$ defined by $f(x) = x^{3}, x \in R$.

How many positive real numbers $x$ satisfy the equation $x^3-3|x|+2=0$?

The equation $6^{x}+8^{x}=10^{x}$ has

If $[x]^2-5[x]+6=0$,where $[x]$ denotes the greatest integer function,then

Vedclass Test Series

Exam Paper Generator

Online Exam Module