The interval in which the function f(x) = ope | vedclass

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

Question

(C) To find the interval where $f(x) = \operatorname{Tan}^{-1}(\sin x + \cos x)$ is increasing,we find its derivative $f'(x)$.$f'(x) = \frac{1}{1 + (\

Vedclass · Accepted Answer

$\left(-\frac{3\pi}{4}, \frac{\pi}{4}\right)$

Vedclass · Answer

(C) To find the interval where $f(x) = \operatorname{Tan}^{-1}(\sin x + \cos x)$ is increasing,we find its derivative $f'(x)$.$f'(x) = \frac{1}{1 + (\sin x + \cos x)^2} \cdot \frac{d}{dx}(\sin x + \cos x)$.$f'(x) = \frac{\cos x - \sin x}{1 + (\sin x + \cos x)^2}$.For $f(x)$ to be an increasing function,we must have $f'(x) > 0$.Since the denominator $1 + (\sin x + \cos x)^2$ is always positive,we only need the numerator to be positive:$\cos x - \sin x > 0$.$\cos x > \sin x$.Dividing by $\cos x$ (assuming $\cos x > 0$),we get $1 > 	an x$,which implies $	an x This inequality holds when $x \in \left(-\frac{\pi}{2}, \frac{\pi}{4}ight)$.Comparing this with the given options,the interval $\left(-\frac{3\pi}{4}, \frac{\pi}{4}ight)$ contains the region where the function is increasing,and specifically,the function increases within the range provided in option $C$.

The interval in which the function $f(x) = \operatorname{Tan}^{-1}(\sin x + \cos x)$ is an increasing function,is

Similar Questions

The set of real values of $x$ for which $f(x) = \frac{x}{\log x}$ is increasing,is

The function $f(x) = x^x$ is a decreasing function in the interval of . . . . . . . $(x \in R^{+})$

Let $f(x) = e^x - e^{-x} + \cos x$,then $f(x)$ is

The function $f(x) = \cos |x| - 2ax + b$ increases on the entire real line. What is the range of $a$?

The interval in which the function $f(x) = 2x^2 - \log x$,for $x > 0$ decreases is

Vedclass Test Series

Exam Paper Generator

Online Exam Module