The domain of {sin ^{ - 1}}({log _3}x) is | vedclass

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Question

(D) The domain of the function $f(x) = \sin^{-1}(\log_3 x)$ is determined by the condition that the argument of the inverse sine function must lie in

Vedclass · Accepted Answer

$[\frac{1}{3}, 3]$

Vedclass · Answer

(D) The domain of the function $f(x) = \sin^{-1}(\log_3 x)$ is determined by the condition that the argument of the inverse sine function must lie in the interval $[-1, 1]$.Thus,we have the inequality: $-1 \le \log_3 x \le 1$.Since the base of the logarithm is $3$ (which is greater than $1$),we can exponentiate each part of the inequality with base $3$ without changing the direction of the inequality signs:$3^{-1} \le x \le 3^1$.Simplifying the expressions,we get:$\frac{1}{3} \le x \le 3$.Therefore,the domain of the function is $[\frac{1}{3}, 3]$.

The domain of ${\sin ^{ - 1}}({\log _3}x)$ is

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