Let x*y=x^2+y^3 and (x*1)*1=x*(1*1).

Question

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

Answer: $\dfrac{\pi}{3}$ *(the exact option lettering was not legible in the source, but the numeric value below is fully determined by the given stem)*

Accepted Answer

$x*1=x^2+1$, so $(x*1)*1=(x^2+1)^2+1$. Also $1*1=1^2+1^3=2$, so $x*(1*1)=x*2=x^2+2^3=x^2+8$.

Equating: $(x^2+1)^2+1=x^2+8\ \Rightarrow\ x^4+2x^2+2=x^2+8\ \Rightarrow\ x^4+x^2-6=0$.

Let $u=x^2$: $u^2+u-6=0\Rightarrow(u+3)(u-2)=0\Rightarrow u=2$ (rejecting $u=-3$). So $x^2=2,\ x^4=4$.

$$\frac{x^4+x^2-2}{x^4+x^2+2}=\frac{4+2-2}{4+2+2}=\frac{4}{8}=\frac12.$$

$$2\sin^{-1}\left(\frac12\right)=2\cdot\frac{\pi}{6}=\frac{\pi}{3}.$$

Solutions · 1