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BCECE Engineering BCECE Engineering Solved Paper-2015

  • question_answer
    Let \[f(x)={{({{x}^{3}}+2)}^{30}}\] .If \[{{f}^{n}}(x)\] is a polynomial of degree 20, where \[{{f}^{n}}(x)\] denotes the nth order derivative of \[f(x)\] with respect to x, then the value of n is

    A) 60                          

    B) 40     

    C) 70                          

    D) 50

    Correct Answer: C

    Solution :

    We have, \[f(x)={{({{x}^{3}}+2)}^{30}}\] Clearly, it is a polynomial of degree 90. It is given that \[{{f}^{n}}(x)\]is a polynomial of degree 20. \[\therefore \]  \[90-n=20\] \[\Rightarrow \]               \[n=70\]


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