Question

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

• A. $f\left(x\right)$ is increasing in $\left(0,\frac{\mathrm{\pi }}{2}\right)$ and decreasing in $\left(\frac{\mathrm{\pi }}{2},\mathrm{\pi }\right)$
• B. $f\left(x\right)$ is decreasing in $\left(0,\frac{\mathrm{\pi }}{2}\right)$ and increasing in $\left(\frac{\mathrm{\pi }}{2},\mathrm{\pi }\right)$
• C. $f\left(x\right)$ is increasing in $\left(0,\frac{\mathrm{\pi }}{4}\right)$ and decreasing in $\left(\frac{\mathrm{\pi }}{4},\frac{\mathrm{\pi }}{2}\right)$
• D. The statements (a), (b) and (c) are all correct

Answer 1

The correct option is C

$f\left(x\right)$ is increasing in $\left(0,\frac{\mathrm{\pi }}{4}\right)$ and decreasing in $\left(\frac{\mathrm{\pi }}{4},\frac{\mathrm{\pi }}{2}\right)$

Explanation for the correct option:

Step 1: Calculating differentiation of given function

Given: $f\left(x\right)=\sin 2x$

Calculating the derivative of the function,

$\Rightarrow$$f\text{'}\left(x\right)=2\cos 2x$ $\left[\because \frac{d}{dx}\left(\cos 2x\right)=2\sin 2x\right]$

Step 2: Finding the increasing interval

A function $f\left(x\right)$ is increasing in the interval $\left[a,b\right]$ if $f\text{'}\left(x\right)>0$ in the interval $\left[a,b\right]$.

When $0<\cos \left(2x\right)<1$ , then $f\text{'}\left(x\right)>0$.

We know that,

$0\le \cos \left(2x\right)\le 1$ when $0\le 2x\le \frac{\pi }{2}$

$\Rightarrow$ $0\le \cos \left(2x\right)\le 1$ when $\frac{0}{2}\le x\le \frac{\pi }{2\times 2}$

$\Rightarrow$ $0\le \cos \left(2x\right)\le 1$ when $0\le x\le \frac{\pi }{4}$

So, the function is increasing when $x\in \left(0,\frac{\pi }{4}\right)$

Step 3: Finding the decreasing interval

A function $f\left(x\right)$ is decreasing in the interval $\left[a,b\right]$ if $f\text{'}\left(x\right)<0$ in the interval $\left[a,b\right]$.

When $-1<\cos \left(2x\right)<0$ , then $f\text{'}\left(x\right)<0$.

We know that,

$-1\le \cos \left(2x\right)\le 0$ when $\frac{\pi }{2}\le 2x\le \pi$

$\Rightarrow$ $-1\le \cos \left(2x\right)\le 0$ when $\frac{\pi }{2\times 2}\le x\le \frac{\pi }{2}$

$\Rightarrow$ $-1\le \cos \left(2x\right)\le 0$ when $\frac{\pi }{4}\le x\le \frac{\pi }{2}$

So, the function is decreasing when $x\in \left(\frac{\pi }{4},\frac{\pi }{2}\right)$

Hence, option (C) is correct.