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

Let $f = \{(1, 1), (2, 3), (0, -1), (-1, -3)\}$ be a linear function from $\mathbb{Z}$ into $\mathbb{Z}$. Find $f(x)$.

Let $R$ be the set of real numbers. Define the real function $f: R \rightarrow R$ by $f(x) = x + 10$ and sketch the graph of this function.

The function $f$ is defined by $f(x) = \begin{cases} 1 - x, & x < 0 \\ 1, & x = 0 \\ x + 1, & x > 0 \end{cases}$ Draw the graph of $f(x)$.

The minimum value of $|x| + |x + \frac{1}{2}| + |x - 3| + |x - \frac{5}{2}|$ is

If $f(x) = \frac{1}{\sqrt{x+2 \sqrt{2x-4}}} + \frac{1}{\sqrt{x-2 \sqrt{2x-4}}}$ for $x > 2$,then $f(11)$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module