wht to do?

2\left(\frac{1}{1!.9!}+\frac{1}{3!.7!} \right)+\frac{1}{5!5!} \\\\=\frac{1}{10!}\left( \begin{pmatrix} 10\\1 \end{pmatrix}+\begin{pmatrix} 10\\3 \end{pmatrix}+\begin{pmatrix} 10\\5 \end{pmatrix}+\begin{pmatrix} 10\\7 \end{pmatrix}+\begin{pmatrix} 10\\9 \end{pmatrix}\right) \\\\=\frac{512}{10!}

Now,
x^3-5x^2+33x-19 \\=(x-2)^3+(x-2)^2+25(x-2)+35 \\=-125i-25+125i+35 \\=10

B is the answer.

gr8 jake bhaiya

thnx :D