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

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Question

(B) For the function $f(x) = \frac{\sin^{-1}(3 - x)}{\ln(|x| - 2)}$ to be defined,the following conditions must be met:$1$. The argument of $\sin^{-1}

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$(2, 3) \cup (3, 4]$

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(B) For the function $f(x) = \frac{\sin^{-1}(3 - x)}{\ln(|x| - 2)}$ to be defined,the following conditions must be met:$1$. The argument of $\sin^{-1}$ must be in $[-1, 1]$:$-1 \le 3 - x \le 1$$-4 \le -x \le -2$$2 \le x \le 4$So,the domain of the numerator is $[2, 4]$.$2$. The argument of $\ln$ must be positive,and the denominator cannot be zero:$|x| - 2 > 0 \implies |x| > 2 \implies x \in (-\infty, -2) \cup (2, \infty)$.Also,$\ln(|x| - 2) 
eq 0 \implies |x| - 2 
eq 1 \implies |x| 
eq 3 \implies x 
eq 3, x 
eq -3$.Combining these,the domain of the denominator is $(-\infty, -3) \cup (-3, -2) \cup (2, 3) \cup (3, \infty)$.$3$. The domain of $f(x)$ is the intersection of these sets:$[2, 4] \cap ((-\infty, -3) \cup (-3, -2) \cup (2, 3) \cup (3, \infty)) = (2, 3) \cup (3, 4]$.Thus,the correct option is $B$.

Column $I$	Column $II$
$(A)$ If $-1 < x < 1$,then $f(x)$ satisfies	$(p)$ $0 < f(x) < 1$
$(B)$ If $1 < x < 2$,then $f(x)$ satisfies	$(q)$ $f(x) < 0$
$(C)$ If $3 < x < 5$,then $f(x)$ satisfies	$(r)$ $f(x) > 0$
$(D)$ If $x > 5$,then $f(x)$ satisfies	$(s)$ $f(x) < 1$

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

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