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KVPY Sample Paper KVPY Stream-SX Model Paper-23

  • question_answer
    If \[\frac{{{a}_{0}}}{n+1}+\frac{{{a}_{1}}}{n}+\frac{{{a}_{2}}}{n-1}+...+\frac{{{a}_{n}}-1}{2}{{a}_{n}}=0,\]then the equation \[{{a}_{0}}{{x}^{n}}+{{a}_{1}}{{x}^{n-1}}+{{a}_{2}}{{x}^{n-2}}+...+{{a}_{n-1}}x+{{a}_{n}}=0\] has

    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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