$T h e d o m a i n o f t h e f u n c t i o n f (x) = \sqrt{\sin^{- 1} (\log_{2} x)} + \sin^{- 1} (\frac{1 + x^{2}}{2 x}) + \sqrt{\cos (\sin x)} i s$

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${1}$

$N o t d e f i n e d f o r a n y r e a l x .$

${x : 1 \leq x \leq 2}$

${- 1, 1}$

answer is A.

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Detailed Solution

$\begin{array}{l} f (x) = \sqrt{\sin^{- 1} (\log_{e}^{x})} + \sin^{- 1} (\frac{1 + x^{2}}{2 x}) + \sqrt{\cos \sin x} i s d e f i n e d \\ i f f \sin^{- 1} (\log_{2}^{x}) \geq 0 \Rightarrow \log_{2}^{x} \geq 0 \Rightarrow x \geq 1 \\ a n d - 1 \leq \frac{x^{2} + 1}{2 x} \leq 1 \\ \Rightarrow | \frac{x^{2} + 1}{2 x} | \leq 1 \Rightarrow \frac{{| x |}^{2} + 1}{2 | x |} \leq 1 \\ \Rightarrow {| x |}^{2} + 1 \leq 2 | x | \\ \Rightarrow {| x |}^{2} - 2 | x | + 1 \leq 0 \\ \Rightarrow {(| x | - 1)}^{2} \leq 0 \\ ∴ | x | - 1 = 0 \Rightarrow | x | = 1 \Rightarrow x = \pm 1 \\ a n d \cos (\sin x) \geq 0 \forall x \in R \\ ∴ D o m a i n = {1} \end{array}$