The domain of the function f(x) = sqrt{log_{0 | vedclass

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(D) We have $f(x) = \sqrt{\log_{0.5} x!}$.For $f(x)$ to be defined,the expression inside the square root must be non-negative:$\log_{0.5} x! \geq 0$.S

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$\{0, 1\}$

Vedclass · Answer

(D) We have $f(x) = \sqrt{\log_{0.5} x!}$.For $f(x)$ to be defined,the expression inside the square root must be non-negative:$\log_{0.5} x! \geq 0$.Since the base of the logarithm is $0.5$ (which is between $0$ and $1$),the inequality sign reverses when we remove the logarithm:$x! \leq (0.5)^0$.$x! \leq 1$.The factorial $x!$ is defined for non-negative integers $x$. We check the values:For $x = 0$,$0! = 1 \leq 1$ (True).For $x = 1$,$1! = 1 \leq 1$ (True).For $x = 2$,$2! = 2 
ot\leq 1$ (False).Thus,the domain is $\{0, 1\}$.

The domain of the function $f(x) = \sqrt{\log_{0.5} x!}$ is

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