Given expression:
$xP+\bar{x}Q=f$
Or, $x(y \: op_1 \: z) + \bar{x}(y \: op_2 \: z) = f \quad \to (1)$ $${\begin{array}{|c|c|c|c|}\hline
\textbf{X}& \textbf{Y}& \textbf{Z}&\bf{ Output(f)}\\\hline
0&0&0&0 \\\hline 0&0&1&0\\ \hline 0&1&0&0 \\ \hline 0&1&1&1\\ \hline 1&0&0& 0 \\ \hline 1&0&1&1 \\ \hline 1&1&0& 1 \\ \hline 1&1&1&1\\ \hline
\end{array}}\qquad{\begin{array}{|c|c|c|c|c|}\hline
\textbf{x\\yz}& \textbf{00}& \textbf{01}&\bf{ 11}& \textbf{10} \\\hline
0&...&...&1&... \\\hline 1&...&1&1&1\\ \hline
\end{array}}$$ $f \implies xz+xy+yz$
$\quad \implies xz+xy+(x+\bar{x})yz $
$\quad \implies xz+xy+xyz+\bar{x}yz$
$\quad \implies x(z+y+yz)+\bar{x}yz$
$\quad \implies x[z+y(1+z)]+\bar{x}(y \bullet z)$
$\quad \implies x(z+y) + \bar{x}(y \bullet z) \quad \to (2)$
Comparing $(1)$ and $(2)$ we get,
$OP_1 = + (OR)$
$OP_2 = \bullet (AND)$
Correct Answer: $D$