A) \[a+b+e=c+d\]
B) \[a+b+c=d+e\]
C) \[b+c+d=a+e\]
D) None of these
Correct Answer: D
Solution :
Let, \[f(x)=a{{x}^{4}}+b{{x}^{3}}+c{{x}^{2}}+dx+e\] Since, \[({{x}^{2}}-1)\]i.e., \[(x-1)(x+1)\]is a factor of \[f(x).\]Therefore, by factor theorem, \[f(1)=0\]and \[f(-1)=0\] \[\Rightarrow \]\[a+b+c+d+e=0\]and \[a-b+c-d+e=0\] \[\Rightarrow \]\[a+b+c=-(d+e)\]and \[a+c+e=b+d\]
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