Question:
The interval for which \(2\tan^{-1}x + \sin^{-1}\frac{2x}{1+x^2}\) is independent of \(x\) is
The interval for which \(2\tan^{-1}x + \sin^{-1}\frac{2x}{1+x^2}\) is independent of \(x\) is
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Use the identity for \(\sin^{-1}(2x/(1+x^2))\) in different intervals.
Updated On: Apr 23, 2026
- \(|x|<1\)
- \(|x|>1\)
- \(|x| = 1\)
- \(\phi\)
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The Correct Option is B
Solution and Explanation
Step 1: Formula / Definition}
\[ \sin^{-1}\left(\frac{2x}{1+x^2}\right) = \begin{cases} 2\tan^{-1}x, & |x| \leq 1 \\ \pi - 2\tan^{-1}x, & x>1 \\ -\pi - 2\tan^{-1}x, & x<-1 \end{cases} \]
Step 2: Calculation / Simplification}
For \(|x| \leq 1\): \(2\tan^{-1}x + 2\tan^{-1}x = 4\tan^{-1}x\) (depends on \(x\))
For \(|x|>1\): \(2\tan^{-1}x + (\pi - 2\tan^{-1}x) = \pi\) (constant)
\(\therefore\) Independent of \(x\) when \(|x|>1\).
Step 3: Final Answer
\[ |x|>1 \]
\[ \sin^{-1}\left(\frac{2x}{1+x^2}\right) = \begin{cases} 2\tan^{-1}x, & |x| \leq 1 \\ \pi - 2\tan^{-1}x, & x>1 \\ -\pi - 2\tan^{-1}x, & x<-1 \end{cases} \]
Step 2: Calculation / Simplification}
For \(|x| \leq 1\): \(2\tan^{-1}x + 2\tan^{-1}x = 4\tan^{-1}x\) (depends on \(x\))
For \(|x|>1\): \(2\tan^{-1}x + (\pi - 2\tan^{-1}x) = \pi\) (constant)
\(\therefore\) Independent of \(x\) when \(|x|>1\).
Step 3: Final Answer
\[ |x|>1 \]
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