AP EAMCET · Maths · Functions
The range of the real valued function \(f(x)=\sin ^{-1}\left(\frac{1+x^2}{2 x}\right)+\cos ^{-1}\left(\frac{2 x}{1+x^2}\right)\) is
- A \(\left\{\frac{\pi}{2}\right\}\)
- B \(\mathbf{R}\)
- C \(\mathbf{Q}\)
- D \(\left\{-\frac{\pi}{2}, \frac{\pi}{2}\right\}\)
Answer & Solution
Correct Answer
(A) \(\left\{\frac{\pi}{2}\right\}\)
Step-by-step Solution
Detailed explanation
\(f(x)=\sin ^{-1}\left(\frac{1+x^2}{2 x}\right)+\cos ^{-1}\left(\frac{2 x}{1+x^2}\right)\) We know, domain of \(\sin ^{-1}(x)\) is \([-1,1]\) \(\therefore-1 \leq \frac{1+x^2}{2 x} \leq 1 \text { and }-1 \leq \frac{2 x}{1+x^2} \leq 1\) When,…
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