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

JEE Main · 2021 · 17 Mar 2021 · Shift 2 · Inverse Trigonometric Functions

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

[x]

denotes the greatest less than or equal to

x

, is:

Answer: (B)

For

x \in [- 1, 1]

the quantity

x^{2}

lies in

[0, 1]

.
If

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

, then

x^{2} + \frac{1}{3} \in [\frac{1}{3}, 1)

and

x^{2} - \frac{2}{3} \in [- \frac{2}{3}, 0)

, so the two greatest-integer values are

0

and

- 1

, and the left side is

sin^{- 1} 0 + cos^{- 1} (- 1) = π

.
If

\frac{2}{3} \leq x^{2} \leq 1

, then

x^{2} + \frac{1}{3} \in [1, \frac{4}{3}]

and

x^{2} - \frac{2}{3} \in [0, \frac{1}{3}]

, so the two values are

1

and

0

, and the left side is

sin^{- 1} 1 + cos^{- 1} 0 = \frac{π}{2} + \frac{π}{2} = π

.
The left side equals

π

for every admissible

x

, while the right side

x^{2}

never exceeds

1

. The number of solutions is

0

JEE Main · 2021 · 17 Mar 2021 · Shift 2 · Single correct option