The domain of sin^{-1} left[ log_3 left( frac | vedclass

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(A) The given function is $f(x) = \sin^{-1} \left[ \log_3 \left( \frac{x}{3} \right) \right]$.For the function $\sin^{-1}(u)$ to be defined,the argume

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$[1, 9]$

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(A) The given function is $f(x) = \sin^{-1} \left[ \log_3 \left( \frac{x}{3} \right) \right]$.For the function $\sin^{-1}(u)$ to be defined,the argument $u$ must satisfy $-1 \le u \le 1$.Therefore,we must have $-1 \le \log_3 \left( \frac{x}{3} \right) \le 1$.Applying the definition of the logarithm,we exponentiate with base $3$:$3^{-1} \le \frac{x}{3} \le 3^1$.This simplifies to $\frac{1}{3} \le \frac{x}{3} \le 3$.Multiplying the entire inequality by $3$,we get $1 \le x \le 9$.Thus,the domain is $x \in [1, 9]$.

The domain of $\sin^{-1} \left[ \log_3 \left( \frac{x}{3} \right) \right]$ is

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