If x * y = x^{2} + y^{3} and (x * 1) * 1 = x | vedclass

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(B) Given the operation $x * y = x^{2} + y^{3}$.First,evaluate $(x * 1) * 1 = x * (1 * 1)$:$(x * 1) = x^{2} + 1^{3} = x^{2} + 1$.So,$(x * 1) * 1 = (x^

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$\frac{\pi}{3}$

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(B) Given the operation $x * y = x^{2} + y^{3}$.First,evaluate $(x * 1) * 1 = x * (1 * 1)$:$(x * 1) = x^{2} + 1^{3} = x^{2} + 1$.So,$(x * 1) * 1 = (x^{2} + 1) * 1 = (x^{2} + 1)^{2} + 1^{3} = (x^{2} + 1)^{2} + 1$.Now,evaluate $x * (1 * 1)$:$(1 * 1) = 1^{2} + 1^{3} = 2$.So,$x * (1 * 1) = x * 2 = x^{2} + 2^{3} = x^{2} + 8$.Equating both sides:$(x^{2} + 1)^{2} + 1 = x^{2} + 8$$x^{4} + 2x^{2} + 1 + 1 = x^{2} + 8$$x^{4} + x^{2} - 6 = 0$.Let $t = x^{2}$,then $t^{2} + t - 6 = 0$,which factors as $(t + 3)(t - 2) = 0$.Since $x^{2} = t$ must be non-negative,$x^{2} = 2$.Now substitute $x^{2} = 2$ into the expression $2 \sin^{-1}\left(\frac{x^{4} + x^{2} - 2}{x^{4} + x^{2} + 2}ight)$:$x^{4} = (x^{2})^{2} = 2^{2} = 4$.Expression $= 2 \sin^{-1}\left(\frac{4 + 2 - 2}{4 + 2 + 2}ight) = 2 \sin^{-1}\left(\frac{4}{8}ight) = 2 \sin^{-1}\left(\frac{1}{2}ight)$.Since $\sin^{-1}(\frac{1}{2}) = \frac{\pi}{6}$,the value is $2 	imes \frac{\pi}{6} = \frac{\pi}{3}$.

If $x * y = x^{2} + y^{3}$ and $(x * 1) * 1 = x * (1 * 1)$,then the value of $2 \sin^{-1}\left(\frac{x^{4} + x^{2} - 2}{x^{4} + x^{2} + 2}\right)$ is

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