JEE Main & Advanced Sample Paper JEE Main - Mock Test - 7

  • question_answer
    If \[\Delta (x)=\left| \begin{matrix}    {{e}^{x}} & \sin x  \\    \cos x & In\,(1+{{x}^{2}})  \\ \end{matrix} \right|,\]then the value of \[\underset{x\to 0}{\mathop{\lim }}\,\frac{\Delta (x)}{x}\] is

    A) \[0\]                             

    B) \[2\]

    C) \[-1\]                           

    D) \[-2\]

    Correct Answer: C

    Solution :

    \[\Delta (x)=\left| \begin{matrix}    {{e}^{x}}  \\    \cos x  \\ \end{matrix}\begin{matrix}    \sin x  \\    \,\,\,\,In\,(1+{{x}^{2}})  \\ \end{matrix} \right|\]
    \[{{e}^{x}}\,\,In\,\,(1+{{x}^{2}})-\sin x\,\cos x\]
    So, \[\underset{x\to 0}{\mathop{\lim }}\,\frac{\Delta (x)}{x}=\underset{x\to 0}{\mathop{\lim }}\,\frac{{{e}^{x}}In\,(1+{{x}^{2}})-\sin x\cos x}{x}\]
     \[\underset{x\to 0}{\mathop{\lim }}\,\,\,x{{e}^{x}}\left\{ \frac{In\,(1+{{x}^{2}})}{{{x}^{2}}} \right\}-\underset{x\to 0}{\mathop{\lim }}\,\,\left( \frac{\sin x}{x} \right)\cos x\]
    \[=0\times 1\times 1-1\times 1=-1\]

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