Question

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

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

Answer 1

Answer: (B)

$\frac{π}{3}$

Answer 2

Given $x * y$ = $x^2 + y^3$ and $(x * 1) * 1$ =$x * (1 * 1)$

So, $(x^2 + 1) * 1 = x * 2$

$⇒ (x^2 + 1)^2 + 1 = x^2 + 8$

$⇒ x^4 + 2x^2 + 2 = x^2 + 8$

$⇒ (x^2)^2 + x^2 – 6 = 0$

$∴$ $(x^2 + 3)(x^2 – 2) = 0$

$∴ x^2 = 2$

Now,

$2sin^{-1}\bigg(\frac{x^4+x^2-2}{x^4+x2+2}\bigg)$

$= 2sin^{-1}\bigg(\frac{4}{8}\bigg)$

= $2. \frac{π}{6}$

= $\frac{π}{3}$