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Ans is option **(C)**

For the value inside the square root to be real, it should be greater than or equal to zero.

$\therefore$ log$_{e}\frac{4x-x^{2}}{3}\geqslant0$

$\Rightarrow$ $\frac{4x-x^{2}}{3}\geqslant1$

$\Rightarrow$ $\frac{4x-x^{2}}{3}-1\geqslant0$

$\Rightarrow$ $\frac{4x-x^{2}-3}{3}\geqslant0$

Multiplying by 3 on both sides inequality, we get $4x-x^{2}-3\geqslant0$

Multiplying by -1 on both sides inequality, we get $x^{2}-4x+3\leqslant0$

$\Rightarrow$ $(x-1)(x-3)\leqslant0$ $\Rightarrow$ $x\in[1,3]$