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(A) Given $f(x) = \sin^{-1} ( \frac{1 + x^2}{2 x} ) + \cos^{-1} ( \frac{2 x}{1 + x^2} )$.For $\sin^{-1} ( \frac{1 + x^2}{2 x} )$ to be defined,we must

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$ \{ \frac{\pi}{2} \} $

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(A) Given $f(x) = \sin^{-1} ( \frac{1 + x^2}{2 x} ) + \cos^{-1} ( \frac{2 x}{1 + x^2} )$.For $\sin^{-1} ( \frac{1 + x^2}{2 x} )$ to be defined,we must have $| \frac{1 + x^2}{2 x} | \leq 1$.Since $1 + x^2 \geq 2|x|$ for all $x \in R$,the condition $| \frac{1 + x^2}{2 x} | \leq 1$ holds only when $|x| = 1$,i.e.,$x = 1$ or $x = -1$.If $x = 1$,$f(1) = \sin^{-1}(1) + \cos^{-1}(1) = \frac{\pi}{2} + 0 = \frac{\pi}{2}$.If $x = -1$,$f(-1) = \sin^{-1}(-1) + \cos^{-1}(-1) = -\frac{\pi}{2} + \pi = \frac{\pi}{2}$.Thus,the domain is $\{ -1, 1 \}$ and the range is $\{ \frac{\pi}{2} \}$.

The range of the real valued function $f(x) = \sin^{-1} ( \frac{1 + x^2}{2 x} ) + \cos^{-1} ( \frac{2 x}{1 + x^2} )$ is

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