JEE Main & Advanced Mathematics Determinants & Matrices Question Bank System of linear equations, Some special determinants, differentiation and integration of determinants

  • question_answer
    If \[\Delta (x)=\left| \,\begin{matrix}    {{x}^{n}} & \sin x & \cos x  \\    n! & \sin \frac{n\pi }{2} & \cos \frac{n\pi }{2}  \\    a & {{a}^{2}} & {{a}^{3}}  \\ \end{matrix}\, \right|,\] then the value of \[\frac{{{d}^{n}}}{d{{x}^{n}}}[\Delta (x)]\] at \[x=0\]is

    A) - 1

    B) 0

    C) 1

    D) Dependent of a

    Correct Answer: B

    Solution :

      \[\frac{{{d}^{n}}}{d{{x}^{n}}}[\Delta (x)]=\left| \begin{matrix}    \frac{{{d}^{n}}}{d{{x}^{n}}}{{x}^{n}} & \frac{{{d}^{n}}}{d{{x}^{n}}}\sin x & \frac{{{d}^{n}}}{d{{x}^{n}}}\cos x  \\    n! & \sin \left( \frac{n\pi }{2} \right) & \cos \left( \frac{n\pi }{2} \right)  \\    a & {{a}^{2}} & {{a}^{3}}  \\ \end{matrix} \right|\]     \[=\left| \,\,\begin{matrix}    n! & \sin \left( x+\frac{n\pi }{2} \right) & \cos \left( x+\frac{n\pi }{2} \right)  \\    n! & \sin \left( \frac{n\pi }{2} \right) & \cos \left( \frac{n\pi }{2} \right)  \\    a & {{a}^{2}} & {{a}^{3}}  \\ \end{matrix}\, \right|\] \[\Rightarrow \]  \[{{[{{\Delta }^{n}}(x)]}_{x=0}}=\left| \,\begin{matrix}    n\,! & \sin \,\left( 0+\frac{n\pi }{2} \right) & \cos \,\left( 0+\frac{n\pi }{2} \right)  \\    n\,! & \sin \,\left( \frac{n\pi }{2} \right) & \cos \,\left( \frac{n\pi }{2} \right)  \\    a & {{a}^{2}} & {{a}^{3}}  \\ \end{matrix}\, \right|\,=\,0\]                                                         {Since \[{{R}_{1}}\equiv {{R}_{2}}\]}.


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