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

  • question_answer
    The function \[f(x)=\int\limits_{-1}^{x}{t({{e}^{t}}-1)(t-1){{(t-2)}^{3}}{{(t-3)}^{5}}}dt\] has a local minimum at x = [IIT 1999]

    A) 0

    B) 1

    C) 2

    D) 3

    Correct Answer: B

    Solution :

    • \[f(x)=\int_{-1}^{x}{t({{e}^{t}}-1)(t-1){{(t-2)}^{3}}{{(t-3)}^{5}}}dt\]                   
    • \[\therefore f'(x)=x({{e}^{x}}-1)(x-1){{(x-2)}^{3}}{{(x-3)}^{5}}\]                   
    • For local minima, slope i.e. \[f'(x)\]should change sign from \[-ve\]to \[+ve\].                   
    • \[f'(x)=0\Rightarrow x=0,\,1,\,2,\,3\]                                
    • If \[x=0-h,\] where h is a very small number, then                   
    • \[f'(x)=(-)(-)(-1)(-1)(-1)=-ve\]                   
    • If \[x=0+h\];  \[f'(x)=(+)(+)(-)(-1)(-1)=-ve\]                   
    • Hence at \[x=0\]neither maxima nor minima.                   
    • If \[x=1-h\] \[f'(x)=(+)(+)(-)(-1)(-1)=-ve\]                   
    • If \[x=1+h\]; \[f'(x)=(+)(+)(+)(-1)(-1)=+ve\]                   
    • Hence at \[x=1\]there is a local minima.                   
    • If \[x=2-h\]; \[f'(x)=(+)(+1)(+)(-)(-)=+ve\]                                
    • If \[x=2+h\]; \[f'(x)=(+)(+)(+)(+)(-1)=-ve\]                   
    • Hence at \[x=2\]there is a local maxima.                   
    • If \[x=3-h\]; \[f'(x)=(+)(+)(+)(+)(-)=-ve\]                   
    • If \[x=3+h\]; \[f'(x)=(+)(+)(+)(+)(+)=+ve\]                   
    • Hence at \[x=3\]there is a local minima.


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