ISI-2014-23

Question

Answer 1

Let f(x)=(x−ω1)(x−ω2)⋯(x−ω9).

Let g(x)=f(x)∑ (j=1 to 9) 1/(x−ωj).

Then g(x)=f ′(x) and the value we want is g(1)=f ′(1).

Let h(x)=(x−1)*f(x)=x^10−1.

Then h′′(x)=2f ′(x)+(x−1)f ′′(x) and also of course h′′(x)=90x^8.

Therefore g(x)=f ′(1)=(1/2)*h′′(1)=45.