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A four-variable Boolean function is realized using $4\times 1$ multiplexers as shown in the figure.

The minimized expression for $\text{F(U,V,W,X)}$ is

1. $\left ( UV+\overline{U}\:\overline{V}\right )\overline{W}$
2. $\left ( UV+\overline{U}\:\overline{V}\right )\left (\overline{W}\: \overline{X}+\overline{W}\:X\right )$
3. $\left ( U\:\overline{V}+\overline{U}\:V\right )\overline{W}$
4. $\left ( U\:\overline{V}+\overline{U}\:V\right )\left (\overline{W}\: \overline{X}+\overline{W}\:X\right )$

Letâ€™s assume the output of $1^{st}$ multiplexer is Z.

now according to the output equation of $4\times1$ multiplexer:

$Z=\bar U \bar VI_0+\bar UVI_1+U\bar VI_2+UVI_3$

$\implies Z=0+\bar UV+U\bar V+0)$

$I_0$ ,$I_3$ is connected to ground & $I_2$,$I_4$ is connected to $V_{cc}$

$\left[{\because V_{cc}=1,Ground =0}\right]$

$\implies Z= \bar UV+U\bar V$

$\implies Z=(U\oplus V)$

Now $F=\bar W\bar XI_0+\bar WXI_1+W\bar XI_2+WXI_3$

$\implies F=\bar W\bar X(U\oplus V)+\bar WX(U\oplus V)$

$\implies F=(U\oplus V)(\bar W\bar X+\bar WX)$

$\implies F=(U\oplus V)(\bar W(X+\bar X))$

$\implies F=(U\oplus V)\bar W$

Option C is correct.
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