Given probability distribution is


$$
\begin{aligned}
& \sum_{\mathrm{i}=1}^{\mathrm{n}} \mathrm{P}(\mathrm{X}=\mathrm{x})=1 \\
& 0+\mathrm{k}+2 \mathrm{k}+5 \mathrm{k}^2+2 \mathrm{k}^2+3 \mathrm{k}=1 \\
& 7 \mathrm{k}^2+6 \mathrm{k}-1=0 \\
& 7 \mathrm{k}^2+7 \mathrm{k}-\mathrm{k}-1=0 \\
& 7 \mathrm{k}(\mathrm{k}+1)-1(\mathrm{k}+1)=0 \\
& (\mathrm{k}+1)(7 \mathrm{k}-1)=0 \\
& \mathrm{k}=-1, \frac{1}{7} \\
& \mathrm{E}(\mathrm{X})=\sum_{\mathrm{i}=1}^{\mathrm{n}} \mathrm{x}_{\mathrm{i}} \mathrm{p}\left(\mathrm{x}_{\mathrm{i}}\right)=0 \times 0+2 \mathrm{k}+8 \mathrm{k}+30 \mathrm{k}^2+16 \mathrm{k}^2+30 \mathrm{k} \\
& =46 \mathrm{k}^2+40 \mathrm{k} \\
& \mathrm{Put} \mathrm{k}=\frac{1}{7} \\
& \mathrm{E}(\mathrm{x})=\frac{46}{49}+\frac{40}{7}=\frac{46+280}{49}=\frac{326}{49}
\end{aligned}
$$
So, option (d) is correct