Question

Let $x\ast y=x^2+y^3$ and $(x\ast 1)\ast 1=x\ast (1\ast 1)$. 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)

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