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JEE Main & Advanced Mathematics Indefinite Integrals Question Bank Self Evaluation Test - Integrals

  • 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 =

    A) 0

    B) 1, 3

    C) 2

    D) None of these

    Correct Answer: B

    Solution :

    [b] \[\frac{dy}{dx}=f'(x)\] \[\Rightarrow x({{e}^{x}}-1)(x-1){{(x-2)}^{3}}{{(x-3)}^{5}}=0\] Critical points are 0, 1, 2, 3. Consider change of sign of \[\frac{dy}{dx}\] at \[x=3\]. \[x<3,\frac{dy}{dx}\]= negative and \[x>3,\frac{dy}{dx}=\] positive Change if from negative to positive, hence minimum at \[x=3\]. Again minimum and maximum occur alternately. \[\therefore \] 2nd minimum is at \[x=1\].


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