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

  • question_answer
    Let \[f(x)=\int\limits_{0}^{x}{\frac{\cos t}{t}dt,\,\,x>0}\] then \[f(x)\] has [Kurukshetra CEE 2002]

    A) Maxima when \[n=-2,\,-4,\,-6,\,.....\]

    B) Maxima when \[n=-1,\,-3,\,-5,\,....\]

    C) Minima when \[n=0,\,2,\,4,....\]

    D) Minima when \[n=1,3,5....\]

    Correct Answer: B

    Solution :

    • \[f(x)=\int\limits_{0}^{x}{\frac{\cos t}{t}}\,dt\], \[x>0\] Þ \[{f}'(x)=\frac{\cos x}{x}\], \[x>0\]                  
    • Þ\[{f}'(x)=0\Rightarrow \frac{\cos x}{x}=0\]Þ \[x=(2n+1)\text{ }\frac{\pi }{\text{2}}\], for \[n\in z\].           
    • Now \[{f}''(x)=\frac{-x\sin x-\cos x}{{{x}^{2}}}\]           
    • \  \[{f}''[(2n+1)\pi /2]=\frac{-2}{(2n+1)\pi }\,{{(-1)}^{n}}\] = \[\frac{2{{(-1)}^{n+1}}}{(2n+1)\pi }\].           
    • Thus \[{f}''(x)>0\]       \[n=-2,\,-4,\,-6,........\]                     
    • \[{f}''(x)<0\]   \[n=0,\,2,\,4,\,........\]                   
    • \[{f}''(x)>0\]    \[n=1,\,3,\,5........\]                     
    • \[{f}''(x)<0\]  \[n=-1,\,-3,\,-5........\]           
    • Thus \[f(x)\] attain maximum for \[n=-1,\,-3,\,-5\],?. and minimum for \[n=1,\,3,\,5\],?..


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