A) \[\frac{{{a}_{0}}}{n}+\frac{{{a}_{1}}}{n-1}+....+{{a}_{n-1}}=0\]
B) \[\frac{{{a}_{0}}}{n-1}+\frac{{{a}_{1}}}{n-2}+....+{{a}_{n-2}}=0\]
C) \[n{{a}_{0}}+\left( n-1 \right){{a}_{1}}+....+{{a}_{n-1}}=0\]
D) \[\frac{{{a}_{0}}}{n+1}+\frac{{{a}_{1}}}{n}+.....+{{a}_{n}}=0\]
Correct Answer: D
Solution :
Consider the function f defined by \[\because \] Since, f(x) is a polynomial, so it is continuous and differentiable for all x. Consequently, f(x) is continuous in the closed interval [0, 1] and differentiable in the open Also, f(0) = 0 and \[\alpha ={{\omega }^{2}}\theta \]i.e. f(0) = f(1) Thus al the three conditions of Rolle?s theorem are satisfied. Hence there is atleast one value of x in the open interval (0, 1), where f?(x) = 0 i.e. \[\Rightarrow \]
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