Question

The number of solutions of the equation ${\sin }^{-1}\left(\left[x^2+\frac{1}{3}\right]\right)+{\cos }^{-1}\left(\left[x^2-\frac{2}{3}\right]\right)=x^2$ for $x\in \left[-1,1\right]$, and $\left[x\right]$ denotes the greatest integer less than or equal to $x$, is:

• A. $0$
• B. $2$
• C. $4$
• D. Infinite

Answer 1

The correct option is A

$0$

Explanation for correct option

There are three case possible for $x\in \left[-1,1\right]$

Case (i): $x\in \left[-1,-\sqrt{\frac{2}{3}}\right]$

${\sin }^{-1}\left(1\right)+{\cos }^{-1}\left(0\right)=x^2$

$\Rightarrow \frac{\pi }{2}+\frac{\pi }{2}=x^2$

$\Rightarrow x=\pm \sqrt{\mathrm{\pi }}$$\to$(Rejected)

Case (ii): $x\in \left[-\sqrt{\frac{2}{3}},\sqrt{\frac{2}{3}}\right]$

${\sin }^{-1}\left(0\right)+{\cos }^{-1}\left(1\right)=x^2$

$\Rightarrow 0+\mathrm{\pi }=x^2$

$\Rightarrow x=\pm \sqrt{\mathrm{\pi }}$$\to$(Rejected)

Case (iii): $x\in \left[\sqrt{\frac{2}{3}},1\right]$

${\sin }^{-1}\left(0\right)+{\cos }^{-1}\left(0\right)=x^2$

$\Rightarrow 0+\mathrm{\pi }=x^2$

$\Rightarrow x=\pm \sqrt{\mathrm{\pi }}$$\to$(Rejected)

Therefore, no solution exists.

Hence, the correct option is $\left(\mathrm{A}\right)$.