A) 0
B) 1
C) \[\frac{1}{2011}\]
D) can't be determined
Correct Answer: A
Solution :
We have, \[...(x-2008)\,(x-2009)\,x\] which is an identity. Also, for \[x=-1\] \[-1=k\,(-2)\,(-3)\,(-4)\,(-5)\,....\,(-2009)\,(-2010)\,(-1)\]\[\Rightarrow \] \[k=-\frac{1}{(2010)!}\] For \[x=2010,\] \[\Rightarrow \] \[\,p(2010)=0\]You need to login to perform this action.
You will be redirected in
3 sec