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JEE Main & Advanced Mathematics Determinants & Matrices Question Bank Self Evaluation Test - Determinats

  • question_answer
    If \[f(x)=\left| \begin{matrix}    {{x}^{n-1}} & \cos x & \frac{1}{x+3}  \\    0 & \cos \frac{n\pi }{2} & \frac{{{(-1)}^{n}}n!}{{{3}^{n+1}}}  \\    \alpha  & {{\alpha }^{3}} & {{\alpha }^{5}}  \\ \end{matrix} \right|\] then \[\frac{{{d}^{n}}}{d{{x}^{n}}}\,\,{{[f(x)]}_{x=0}}=\]

    A) \[1\]

    B) \[-1\]

    C) \[0\]

    D) None of these

    Correct Answer: C

    Solution :

    [c] We have, \[\frac{{{d}^{n}}}{d{{x}^{n}}}[f(x)]=\left| \begin{matrix}    0 & \cos \left( x+\frac{n\pi }{2} \right) & \frac{{{(-1)}^{n}}n!}{{{(x+3)}^{n+1}}}  \\    0 & \cos \frac{n\pi }{2} & \frac{{{(-1)}^{n}}n!}{{{3}^{n+1}}}  \\    \alpha  & {{\alpha }^{3}} & {{\alpha }^{5}}  \\ \end{matrix} \right|\] \[\therefore \,\,\frac{{{d}^{n}}}{d{{x}^{n}}}{{[f(x)]}_{x=0}}=\left| \begin{matrix}    0 & \cos \frac{n\pi }{2} & \frac{{{(-1)}^{n}}n!}{{{3}^{n+1}}}  \\    0 & \cos \frac{n\pi }{2} & \frac{{{(-1)}^{n}}n!}{{{3}^{n+1}}}  \\    \alpha  & {{\alpha }^{3}} & {{\alpha }^{5}}  \\ \end{matrix} \right|=0\] \[(\because {{R}_{1}}and\,\,{{R}_{2}}are\,\,identical)\]


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