Question

The number of solutions of the equation ${\sin }^{-1}[x^2+\frac{1}{3}]+{\cos }^{-1}[x^2-\frac{2}{3}]=x^2$, for $x\in [-1,1]$ 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 $0$

Given : ${\sin }^{-1}[x^2+\frac{1}{3}]+{\cos }^{-1}[x^2-\frac{2}{3}]=x^2$
As $x\in [-1,1]\Rightarrow x^2\in [0,1]$

When $x^2\in [0,\frac{2}{3})$, then
${\sin }^{-1}\left(0\right)+{\cos }^{-1}(-1)=x^2\Rightarrow x^2=\pi$
Which is not possible.

When $x^2\in [\frac{2}{3},1]$, then
${\sin }^{-1}\left(1\right)+{\cos }^{-1}\left(0\right)=x^2\Rightarrow x^2=\pi$
Which is not possible.
Hence, the given equation has no solution.