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JEE Main & Advanced AIEEE Solved Paper-2003

  • question_answer
    If \[f(x)={{x}^{n}}\], then the value of \[f(1)-\frac{f'(1)}{1!}+\frac{f''(1)}{2!}-\frac{f'''(1)}{3!}+...+\frac{{{(-1)}^{n}}{{f}^{n}}(1)}{n!}\]     AIEEE  Solved  Paper-2003

    A)                         \[{{2}^{n}}\]      

    B)                       \[{{2}^{n-1}}\]                                  

    C) 0                             

    D) 1

    Correct Answer: C

    Solution :

    Since,        \[f(x)={{x}^{n}}\]             \[\Rightarrow \]                                 \[f'(x)=n{{x}^{n-1}}\Rightarrow \,f'(1)=n\]             \[f''(x)=n(n-1)\,{{x}^{n-2}}\Rightarrow f''(1)=n\,(n-1)\]                                 ????????.                                 ????????.                                 \[f''(x)=n(n-1)\,(n-2)\,....\,\,2.1\]             \[\Rightarrow \]   \[f''(1)=n\,(n-1)\,(n-2)\,....\,2.1\]                                                     We have,             \[f(1)-\frac{f'(1)}{1!}+\frac{f''(1)}{2!}-\frac{f'''(1)}{3!}+....+\frac{{{(-1)}^{n}}{{f}^{n}}(1)}{n!}\]             \[=1-\frac{n}{1!}+\frac{n(n-1)}{2!}-\frac{n(n-1)\,(n-2)}{3!}+...\]                                                 \[+\frac{{{(-1)}^{n}}n(n-1)\,(n-2)\,....\,2.1}{n!}\]             \[={{(1-1)}^{n}}=0\]


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