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

  • question_answer
    If \[f(x)=\left| \begin{matrix}    {{2}^{-x}} & {{e}^{x{{\log }_{e}}2}} & {{x}^{2}}  \\    {{2}^{-3x}} & {{e}^{3x{{\log }_{e}}2}} & {{x}^{4}}  \\    {{2}^{-5x}} & {{e}^{5x{{\log }_{e}}2}} & 1  \\ \end{matrix} \right|,\] then

    A) \[f(x)+f(-x)=0\]

    B) \[f(x)-f(-x)=0\]

    C) \[f(x)+f(-x)=2\]

    D) None of these

    Correct Answer: A

    Solution :

    [a] \[f(x)=\left| \begin{matrix}    {{2}^{-x}} & {{2}^{x}} & {{x}^{2}}  \\    {{2}^{-3x}} & {{2}^{3x}} & {{x}^{4}}  \\    {{2}^{-5x}} & {{2}^{5x}} & 1  \\ \end{matrix} \right|\] \[\therefore f(-x)\left| \begin{matrix}    {{2}^{x}} & {{2}^{-x}} & {{x}^{2}}  \\    {{2}^{3x}} & {{2}^{-3x}} & {{x}^{4}}  \\    {{2}^{5x}} & {{2}^{-5x}} & 1  \\ \end{matrix} \right|=-f(x)\]


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