# The number of solutions of the equation

Question:

The number of solutions of the equation $\sin ^{-1}\left[x^{2}+\frac{1}{3}\right]+\cos ^{-1}\left[x^{2}-\frac{2}{3}\right]=x^{2}$

for $\mathrm{x} \in[-1,1]$, and $[\mathrm{x}]$ denotes the greatest integer less than or equal to $\mathrm{x}$, is :

1. (1) 2

2. (2) 0

3. (3) 4

4. (4) Infinite

Correct Option: , 2

Solution:

Given equation

$\sin ^{-1}\left[x^{2}+\frac{1}{3}\right]+\cos ^{-1}\left[x^{2}-\frac{2}{3}\right]=x^{2}$

Now, $\sin ^{-1}\left[\mathrm{x}^{2}+\frac{1}{3}\right]$ is defined if

$-1 \leq x^{2}+\frac{1}{3}<2 \Rightarrow \frac{-4}{3} \leq x^{2}<\frac{5}{3}$

$\Rightarrow 0 \leq \mathrm{x}^{2}<\frac{5}{3}$

and $\cos ^{-1}\left[x^{2}-\frac{2}{3}\right]$ is defined if

$-1 \leq x^{2}-\frac{2}{3}<2 \Rightarrow \frac{-1}{3} \leq x^{2}<\frac{8}{3}$

$\Rightarrow 0 \leq \mathrm{x}^{2}<\frac{8}{3}$

So, form (1) and (2) we can conclude

$0 \leq x^{2}<\frac{5}{3}$

Case $-1$ if $0 \leq x^{2}<\frac{2}{3}$

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

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

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

but $\pi \notin\left[0, \frac{2}{3}\right)$

$\Rightarrow$ No value of ' $x$ '

Case $-$ II if $\frac{2}{3} \leq x^{2}<\frac{5}{3}$

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

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

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

but $\pi \notin\left[\frac{2}{3}, \frac{5}{3}\right)$

$\Rightarrow$ No value of ' $x$ '

So, number of solutions of the equation is zero.