Question

Let $f\left(x\right)=x\sin x-\frac{1}{2}{\sin }^{2}x$ $\forall x\in (0,\frac{\pi }{2}),$ then which of the following option(s) is/are CORRECT?

• A. $x\sin x-\frac{1}{2}{\sin }^{2}x>0$
• B. $x\sin x-\frac{1}{2}{\sin }^{2}x<0$
• C. $x\sin x-\frac{1}{2}{\sin }^{2}x>\frac{(\pi -1)}{2}$
• D. $x\sin x-\frac{1}{2}{\sin }^{2}x<\frac{(\pi -1)}{2}$

Answer 1

Answer: (A), (D)

$x\sin x-\frac{1}{2}{\sin }^{2}x<\frac{(\pi -1)}{2}$

$\text{Let,}f\left(x\right)=x\sin x-\frac{1}{2}{\sin }^{2}x$
$\Rightarrow f^{\prime }\left(x\right)=x\cos x+\sin x-\sin x\cdot \cos x$
$=\sin x(1-\cos x)+x\cos x$
For $x\in (0,\frac{\pi }{2}),\sin x>0,1-\cos x>0,\cos x>0$
$\Rightarrow f^{\prime }\left(x\right)>0\forall x\in (0,\frac{\pi }{2})$
$\Rightarrow f\left(x\right)$ is strictly increasing in $(0,\frac{\pi }{2})$
The range of $f\left(x\right)$ is $\left(\underset{x\to 0^{+}}{lim}f\right(x),\underset{x\to {\frac{\pi }{2}}^{-}}{lim}f(x\left)\right)=(0,\frac{\pi -1}{2})$
$\Rightarrow 0<x\sin x-\frac{1}{2}{\sin }^{2}x<\frac{\pi -1}{2}$