A) [2, 4]
B) (2, 3) È (3, 4]
C) [2,\[\infty \])
D) \[(-\infty ,\ -3)\cup [2,\ \infty )\]
Correct Answer: B
Solution :
\[f(x)=\frac{{{\sin }^{-1}}(3-x)}{\log \left[ |x|-2 \right]}\] Let \[g(x)={{\sin }^{-1}}(3-x)\] Þ \[-1\le 3-x\le 1\] Domain of \[g(x)\] is [2, 4] and let \[h(x)=\log \left[ |x|-2 \right]\] Þ \[|x|-2>0\] Þ \[|x|\,>2\] Þ \[x<-2\] or \[x>2\] Þ \[(-\infty ,\,-2)\cup (2,\,\infty )\] we know that \[(f/g)(x)=\]\[\frac{f(x)}{g(x)}\forall x\in {{D}_{1}}\cap {{D}_{2}}-\left\{ x\in R:g(x)=0 \right\}\] \ Domain of \[f(x)=(2,\,4]-\{3\}\]\[=(2,\,3)\cup (3,\,4]\].
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