Question

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

• A. $\frac{\pi}{4}$
• B. $\frac{\pi}{3}$
• C. $\frac{\pi}{2}$
• D. $\frac{\pi}{6}$

Answer 1

Answer: (B)

$\because(x * 1) * 1=x *(1 * 1)$
$\left(x^{2}+1\right) * 1=x *(2)$
$\left(x^{2}+1\right)^{2}+1=x^{2}+8$
$x^{4}+x^{2}-6=0 \Rightarrow\left(x^{2}+3\right)\left(x^{2}-2\right)=0$
$x^{2}=2$
$\Rightarrow 2 \sin ^{-1}\left(\frac{x^{4}+x^{2}-2}{x^{4}+x^{2}+2}\right)=2 \sin ^{-1}\left(\frac{1}{2}\right)$
$=\frac{\pi}{3}$