Question

For $f\left(x\right)={\sin }^{2}x,x\in (0,\pi )$ point of inflection is

• A. $\frac{\pi }{6}$
• B. $\frac{2\pi }{4}$
• C. $\frac{3\pi }{4}$
• D. $\frac{4\pi }{3}$

Answer 1

The correct option is C $\frac{3\pi }{4}$

$f\left(x\right)={\sin }^{2}x$

$f^{\prime }\left(x\right)=\sin 2x$

$f^{\prime \prime }\left(x\right)=2\cos 2x$

$2\cos 2x=0$

$\Rightarrow x=\frac{\pi }{4},\frac{3\pi }{4}$

Hence, there are two points in $(0,\pi )$

$\frac{3\pi }{4}$ is the correct answer.