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CET Karnataka Engineering CET - Karnataka Engineering Solved Paper-2009

  • question_answer
    If \[f(x)=1+nx+\frac{n(n-1)}{2}{{x}^{2}}\]\[+\frac{n(n-1)(n-2)}{6}{{x}^{3}}+.....+{{x}^{n}},\]then  \[f''\] (1) is equal to

    A)  \[n(n-1){{2}^{n-1}}\]

    B)  \[(n-1){{2}^{n-1}}\]

    C)  \[n(n-1){{2}^{n-2}}\]   

    D)  \[n(n-1){{2}^{n}}\]

    Correct Answer: C

    Solution :

    Given,   \[f(x)=1+nx+\frac{n(n-1)}{2!}{{x}^{2}}\] \[+\frac{n(n-1)(n-2)}{3!}{{x}^{3}}+......+{{x}^{n}}\]                 \[\Rightarrow \]               \[f(x)={{(1+x)}^{n}}\] \[\Rightarrow \]               \[f'(x)=n{{(1+x)}^{n-1}}\] \[\Rightarrow \]               \[f''(x)=n(n-1){{(1+x)}^{n-2}}\] \[\Rightarrow \]               \[f''(1)=n(n-1){{2}^{n-2}}\]


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