Let $x * y = x^{2} + y^{3}$ and $(x * 1) * 1 = x * (1 * 1)$. Then a value of $2 \sin^{-1}\left(\frac{x^{4}+x^{2}-2}{x^{4}+x^{2}+2}\right)$ is

JEE Main · 2022 · 24 Jun 2022 · Shift 2 · Inverse Trigonometric Functions

Let

x * y = x^{2} + y^{3}

and

(x * 1) * 1 = x * (1 * 1)

.

Then a value of

2 sin^{- 1} (\frac{x ^{4} + x ^{2} - 2}{x ^{4} + x ^{2} + 2})

Answer: (B)

Given the operation

x * y = x^{2} + y^{3}

, we use the condition

(x * 1) * 1 = x * (1 * 1)

.

First, compute

x * 1 = x^{2} + 1

and

1 * 1 = 1^{2} + 1^{3} = 2

.

Then

(x^{2} + 1) * 1 = (x^{2} + 1)^{2} + 1

and

x * 2 = x^{2} + 8

.

Equating:

(x^{2} + 1)^{2} + 1 = x^{2} + 8

Expanding:

x^{4} + 2 x^{2} + 1 + 1 = x^{2} + 8

x^{4} + x^{2} - 6 = 0

Factoring:

(x^{2} + 3) (x^{2} - 2) = 0

Since

x^{2} \geq 0

, we have

x^{2} = 2

.

Substituting into the expression:

2 sin^{- 1} (\frac{4 + 2 - 2}{4 + 2 + 2}) = 2 sin^{- 1} (\frac{4}{8}) = 2 sin^{- 1} (\frac{1}{2}) = 2 \cdot \frac{π}{6} = \frac{π}{3}

JEE Main · 2022 · 24 Jun 2022 · Shift 2 · Single correct option