JEE Mains · Maths · STD 12 - 2. inverse trigonometric function
\(x * y=x^{2}+y^{3}\) 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 & Solution
Correct Answer
(B) \(\frac{\pi}{3}\)
Step-by-step Solution
Detailed explanation
\(\because( x * 1) * 1= x *(1 * 1)\) \(\left(x^{2}+1\right) * 1=x *(2)\) \(\left(x^{2}+1\right)^{2}+1=x^{2}+8\) \(x^{4}+x^{2}-6=0 \Rightarrow\left(x^{2}+3\right)\left(x^{2}-2\right)=0\) \(x^{2}=2\)…
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