The domain of f(x) = sin^{-1}left[log_{2}left | vedclass

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(C) The given function is $f(x) = \sin^{-1}\left[\log_{2}\left(\frac{x}{2}\right)\right]$.For the function $\sin^{-1}(u)$ to be defined,the argument $

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$1 \leq x \leq 4$

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(C) The given function is $f(x) = \sin^{-1}\left[\log_{2}\left(\frac{x}{2}\right)\right]$.For the function $\sin^{-1}(u)$ to be defined,the argument $u$ must satisfy $-1 \leq u \leq 1$.Therefore,we must have $-1 \leq \log_{2}\left(\frac{x}{2}\right) \leq 1$.Applying the definition of the logarithm,this inequality is equivalent to $2^{-1} \leq \frac{x}{2} \leq 2^{1}$.This simplifies to $\frac{1}{2} \leq \frac{x}{2} \leq 2$.Multiplying the entire inequality by $2$,we get $1 \leq x \leq 4$.Thus,the domain of the function is $x \in [1, 4]$.

The domain of $f(x) = \sin^{-1}\left[\log_{2}\left(\frac{x}{2}\right)\right]$ is

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