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KVPY Sample Paper KVPY Stream-SX Model Paper-5

  • question_answer
    If \[f(x)={{(x+2019)}^{n}},\] where x is a real variable and n is a positive integer, then the value of \[f(0)+f'(0)+\frac{f''(0)}{2!}+\frac{f'''(0)}{3!}+...+\frac{{{f}^{n-1}}(0)}{(n-1)!}\]

    A) \[{{(2019)}^{n}}\]                   

    B) \[{{(2020)}^{n}}\]

    C) \[{{(2020)}^{n}}-1\]    

    D) \[n\,{{(2019)}^{n}}\]

    Correct Answer: C

    Solution :

    [c]
    \[f(x)={{(x+2019)}^{n}}\]
    \[f(0)={{(2019)}^{n}}\]
    \[f'(x)=n{{(x+2019)}^{n-1}},\]
    \[f'(0)=n{{(2019)}^{n-1}}\]
    \[f''(x)=n(x-1){{(x+2019)}^{n-2}},\]\[f''(0)=n(n-1){{(2019)}^{x-2}}\]
    \[{{f}^{n-1}}(n)=n\,(n-1)...1\]
    \[{{f}^{n-1}}(0)=n(n-1)\times ...1\]
    \[f(0)+f'(0)+\frac{f''0}{2!}+...+\frac{{{f}^{n-1}}(0)}{(n-1)!}\]
    \[{{(2019)}^{n}}+{}^{n}{{C}_{1}}{{(2019)}^{n-1}}+{}^{n}{{C}_{2}}{{(2019)}^{n-2}}\]\[+...+{}^{n}{{C}_{n-1}}2019{{+}^{n}}{{C}_{n-1}}\]
    \[={{(2019+1)}^{n}}-1\]\[={{(2020)}^{n}}-1\]
               


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