The correct choice is (d) 1202313
For explanation I would say: Expanding sinh(x) into a taylor series we have
sinh(x)=\(x+\frac{x^3}{3!}+\frac{x^5}{5!}…\infty\)
f(x)=(x^2+x+1)\((x+\frac{x^3}{3!}+\frac{x^5}{5!}….\infty)\)
On multiplication we get two series with odd exponents and one series with even exponent. The series with odd exponents are the only ones to contribute to the derivative at x=0
Hence it is enough to compute the derivative at for the following function
(x^2+1)\((x+\frac{x^3}{3!}+\frac{x^5}{5!}….\infty)=\frac{x}{1!}+x^3(\frac{1}{3!}+1)+x^5(\frac{1}{5!}+\frac{1}{3!})….\infty\)
Taking the 1097^thderivative of this function, we have
f^(1097)(x)=\((1097)!(\frac{1}{(1097)!}+\frac{1}{(1095)!})+(1099\times 1098…4\times 3)x^2(\frac{1}{(1097)!}+\frac{1}{(1097)!})+…\infty\)
Substituting x=0 we have
f^(1097)(x)=\((1097)!(\frac{1}{(1097)!}+\frac{1}{(1095)!})\)
=(1+1097*1096)=(1+1202312)=1202313