The domain of the real-valued function f(x) = | vedclass

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(D) For the function $f(x)$ to be defined, the following conditions must be satisfied:$1$. The expression inside the square root in the numerator must

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$[4, \infty)$

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(D) For the function $f(x)$ to be defined, the following conditions must be satisfied:$1$. The expression inside the square root in the numerator must be non-negative: $\log_{10}\left(\frac{x}{x-2}\right) \geq 0$.This implies $\frac{x}{x-2} \geq 10^0$, so $\frac{x}{x-2} \geq 1$.$\frac{x}{x-2} - 1 \geq 0$ $\Rightarrow \frac{x - (x-2)}{x-2} \geq 0$ $\Rightarrow \frac{2}{x-2} > 0$.This holds when $x-2 > 0$, i.e.,$x > 2$.$2$. The expression inside the square root in the denominator must be strictly positive: $[x]^2 - 5[x] + 6 > 0$.$([x]-2)([x]-3) > 0$.This implies $[x]  3$.If $[x]  3$, then $x \geq 4$.$3$. Combining the conditions $x > 2$ and ($x < 2$ or $x \geq 4$), we get $x \in [4, \infty)$.

The domain of the real-valued function $f(x) = \frac{\sqrt{\log_{10}\left(\frac{x}{x-2}\right)}}{\sqrt{[x]^2-5[x]+6}}$ is (where $[x]$ denotes the greatest integer function):

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