The domain of the function f(x) = sqrt{log fr | vedclass

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Question

(B) For the function $f(x) = \sqrt{\log \frac{1}{|\sin x|}}$ to be defined,the expression inside the square root must be non-negative:$\log \frac{1}{|

Vedclass · Accepted Answer

$R - \{ n\pi : n \in I \}$

Vedclass · Answer

(B) For the function $f(x) = \sqrt{\log \frac{1}{|\sin x|}}$ to be defined,the expression inside the square root must be non-negative:$\log \frac{1}{|\sin x|} \ge 0$Since $\log 1 = 0$,this implies $\frac{1}{|\sin x|} \ge 1$,which means $|\sin x| \le 1$.Also,the argument of the logarithm must be strictly positive: $\frac{1}{|\sin x|} > 0$,which is always true for $|\sin x| 
eq 0$.Additionally,the denominator cannot be zero: $|\sin x| 
eq 0$.$|\sin x| 
eq 0 \implies \sin x 
eq 0 \implies x 
eq n\pi$ for any integer $n \in I$.Thus,the domain is $R - \{ n\pi : n \in I \}$.

The domain of the function $f(x) = \sqrt{\log \frac{1}{|\sin x|}}$ is

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