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

The Correct Option is B

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,

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

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

= \(2. \frac{π}{6}\)

= \(\frac{π}{3}\)