Question

If $f\left(x\right)=\left[{\sin }^{-1}\left(\sin 2x\right)\right]$ ( $[.]$ denotes the greatest integer function), then

• A. ${\int }_0^{\frac{\pi }{2}}f\left(x\right)dx=\frac{\pi }{2}-{\sin }^{-1}\left(\sin 1\right)$
• B. $f\left(x\right)$ is periodic with period $\pi$
• C. $\underset{x\to {\frac{\pi }{2}}^{+}}{lim}f\left(x\right)=-1$
• D. none of these

Answer 1

Answer: (A), (B), (C)

The correct options are
A ${\int }_0^{\frac{\pi }{2}}f\left(x\right)dx=\frac{\pi }{2}-{\sin }^{-1}\left(\sin 1\right)$
B $f\left(x\right)$ is periodic with period $\pi$
C $\underset{x\to {\frac{\pi }{2}}^{+}}{lim}f\left(x\right)=-1$

${\sin }^{-1}\left(\sin 2x\right)=2x,0\le x\le \frac{\pi }{4}$

$=\pi -2x,\frac{\pi }{4}\le x\le \frac{\pi }{2}$

Graph of the function in $[0,\frac{\pi }{2}]$ is as shown in the figure.

Hence, from the graph $f\left(x\right)=\left[{\sin }^{-1}\right(\sin 2x\left)\right]$

will be equal to $0$, from $O$ to $A$.

will be equal to $1$, from $A$ to $B$.

will be equal to $0$, from $B$ to $E$.

From the graph ${\int }_0^{\frac{\pi }{2}}f\left(x\right)dx$ is equal to area of the rectangle $ABCD$.

coordinates of point $A$ are $\left(\frac{1}{2}{\sin }^{-1}\right(\sin 1),0)$

From symmetry of figure, $OA=BE$

Hence $AB=\frac{\pi }{2}-2OA=\frac{\pi }{2}-{\sin }^{-1}\left(\sin 1\right)$

Area of rectangle $ABCD=AB\times AD=[\frac{\pi }{2}-{\sin }^{-1}(\sin 1\left)\right]\times 1=\frac{\pi }{2}-{\sin }^{-1}\left(\sin 1\right)$

Here, the graph is plotted for interval $[0,\frac{\pi }{2}]$ . If we extend it the shape of the graph will repeat itself in every interval of $\pi$. Hence, the function is periodic with period $\pi$

Also, when $x>\frac{\pi }{2}$, the graph of ${\sin }^{-1}\left(\sin 2x\right)$ takes negative values. Hence $\underset{x\to {\frac{\pi }{2}}^{+}}{lim}f\left(x\right)=-1$

Options A, B and C are correct.