The function f(x) = tan^{-1}(sin x + cos x) i | vedclass

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

Question

(D) Given $f(x) = 	an^{-1}(\sin x + \cos x)$.To find where the function is increasing,we calculate the derivative $f'(x)$.$f'(x) = \frac{1}{1 + (\sin

Vedclass · Accepted Answer

$(-\frac{\pi}{2}, \frac{\pi}{4})$

Vedclass · Answer

(D) Given $f(x) = 	an^{-1}(\sin x + \cos x)$.To find where the function is increasing,we calculate the derivative $f'(x)$.$f'(x) = \frac{1}{1 + (\sin x + \cos x)^2} \cdot (\cos x - \sin x)$.$f'(x) = \frac{\cos x - \sin x}{1 + (\sin^2 x + \cos^2 x + 2\sin x \cos x)} = \frac{\cos x - \sin x}{2 + \sin 2x}$.For the function to be increasing,we require $f'(x) > 0$.Since $2 + \sin 2x > 0$ for all $x$,we must have $\cos x - \sin x > 0$.$\cos x > \sin x$.Dividing by $\cos x$ (assuming $\cos x > 0$),we get $1 > 	an x$,which implies $x Considering the domain of the function,the interval where $f'(x) > 0$ is $(-\frac{\pi}{2}, \frac{\pi}{4})$.

The function $f(x) = \tan^{-1}(\sin x + \cos x)$ is an increasing function in

Similar Questions

Let $f(x)=e^x-x$ and $g(x)=x^2-x, \forall x \in R$. Then the set of all $x \in R$,where the function $h(x)=(fog)(x)$ is increasing is

The function $f(x) = ax + b$ is strictly increasing for all real $x$ if

The equation $x^3+x-1=0$ has

Prove that the function $f$ given by $f(x)=x^{2}-x+1$ is neither strictly increasing nor decreasing on $(-1,1).$

The function $f(x) = 1 - e^{-\frac{x^2}{2}}$ is .......

Vedclass Test Series

Exam Paper Generator

Online Exam Module