Which one is the correct statement about the | vedclass

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

Question

(C) Given the function $f(x) = \sin 2x$.To determine the intervals of increase and decrease,we find the derivative $f'(x) = \frac{d}{dx}(\sin 2x) = 2

Vedclass · Accepted Answer

$f(x)$ is increasing in $(0, \pi/4)$ and decreasing in $(\pi/4, \pi/2)$

Vedclass · Answer

(C) Given the function $f(x) = \sin 2x$.To determine the intervals of increase and decrease,we find the derivative $f'(x) = \frac{d}{dx}(\sin 2x) = 2 \cos 2x$.For $f(x)$ to be increasing,$f'(x) > 0 \Rightarrow 2 \cos 2x > 0 \Rightarrow \cos 2x > 0$.In the interval $(0, \pi/2)$,$2x$ lies in $(0, \pi)$. $\cos 2x > 0$ when $2x \in (0, \pi/2)$,which means $x \in (0, \pi/4)$.For $f(x)$ to be decreasing,$f'(x) In the interval $(0, \pi/2)$,$\cos 2x Thus,$f(x)$ is increasing in $(0, \pi/4)$ and decreasing in $(\pi/4, \pi/2)$.Therefore,option $(c)$ is correct.

Which one is the correct statement about the function $f(x) = \sin 2x$?

Similar Questions

What type of function is $f(x) = \frac{x - 2}{x + 1}$,where $x \neq -1$?

$y = x^2 e^{-x}$ is increasing on . . . . . . .

Show that the function given by $f(x) = \sin x$ is neither increasing nor decreasing in $(0, \pi)$.

If $f(x) = x e^{x(1 - x)}$,then $f(x)$ is

For what values of $x$ is the function $f(x) = \tan^{-1}(\sin x + \cos x)$ strictly increasing?

Vedclass Test Series

Exam Paper Generator

Online Exam Module