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JEE Main & Advanced Mathematics Applications of Derivatives Question Bank Critical Thinking

  • question_answer
    The function \[f(x)=\frac{\text{ln}(\pi +x)}{\text{ln}(e+x)}\] is   [IIT 1995]

    A) Increasing on \[\left[ 0,\,\infty  \right)\]

    B) Decreasing on \[\left[ 0,\,\infty  \right)\]

    C)  Decreasing on \[\left[ 0,\frac{\pi }{e} \right)\]and increasing on \[\left[ \frac{\pi }{e},\infty  \right)\]

    D)  Increasing on \[\left[ 0,\frac{\pi }{e} \right)\] and decreasing on \[\left[ \frac{\pi }{e},\infty  \right)\]

    Correct Answer: B

    Solution :

    • Let \[f(x)=\frac{\ln (\pi +x)}{\ln (e+x)}\]           
    • \[\therefore f'(x)=\frac{\ln (e+x)\times \frac{1}{\pi +x}-\ln (\pi +x)\frac{1}{e+x}}{{{\ln }^{2}}(e+x)}\]                       
    • \[=\frac{(e+x)\ln (e+x)-(\pi +x)\ln (\pi +x)}{{{\ln }^{2}}(e+x)\times (e+x)(\pi +x)}\]           
    • \[\Rightarrow f'(x)<0\]for all \[x\ge 0\ ,\ \ \{\because \pi >e\}\]           
    • Hence \[f(x)\]is decreasing in \[[0,\infty )\].


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