The domain of the function f(x) = frac{sin^{- | vedclass

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(B) For the function $f(x)$ to be defined,the expression inside the square root must be strictly positive and the argument of $\sin^{-1}$ must be in t

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$[2, 3)$

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(B) For the function $f(x)$ to be defined,the expression inside the square root must be strictly positive and the argument of $\sin^{-1}$ must be in the interval $[-1, 1]$.$1$. For the denominator: $9 - x^2 > 0$ $\Rightarrow x^2 $2$. For the numerator: $-1 \leq x - 3 \leq 1 \Rightarrow 2 \leq x \leq 4 \dots (ii)$.$3$. Taking the intersection of $(i)$ and $(ii)$,we get $2 \leq x Thus,the domain is $[2, 3)$.

The domain of the function $f(x) = \frac{\sin^{-1}(x-3)}{\sqrt{9-x^2}}$ is

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