A) exactly one root in (0,1)
B) no root in (0, 1)
C) at least one root in (0, 1)
D) at the most one root in (0, 1)
Correct Answer: C
Solution :
Consider the function |
\[f(x)=\frac{{{a}_{0}}{{x}^{n+1}}}{n+1}+\frac{{{a}_{1}}{{x}^{n}}}{n}+\frac{{{a}_{2}}{{x}^{n-1}}}{n-1}+...+\frac{{{a}_{n-1}}{{x}^{2}}}{2}+{{a}_{n}}x\]\[\Rightarrow \]\[f(0)=f(1)=0\]\[\Rightarrow \]\[f'(x)={{a}_{0}}{{x}^{n}}+{{a}_{1}}{{x}^{n-1}}+{{a}_{2}}{{x}^{n-2}}+...+{{a}_{n-1}}x+{{a}_{n}}\] |
By Rolle's theorem\[f'(x)=0x\in (0,1)\]Hence, at least one solution in (0, 1). |
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