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JEE Main & Advanced Mathematics Applications of Derivatives Question Bank Rolle's theorem Lagrange's mean value theorem

  • question_answer
    The function \[f(x)={{(x-3)}^{2}}\] satisfies all the conditions of mean value theorem in [3, 4]. A point on\[y={{(x-3)}^{2}}\], where the tangent is parallel to the chord joining (3, 0) and (4, 1) is

    A)            \[\left( \frac{7}{2},\frac{1}{2} \right)\]

    B)            \[\left( \frac{7}{2},\frac{1}{4} \right)\]

    C)            (1, 4)

    D)            (4, 1)

    Correct Answer: B

    Solution :

               Let the point be \[({{x}_{1}},\,{{y}_{1}}).\] Therefore \[{{y}_{1}}={{({{x}_{1}}-3)}^{2}}\] ?..(i)            Now slope of the tangent at \[({{x}_{1}},\,{{y}_{1}})\] is \[2({{x}_{1}}-3),\] but it is equal to 1. Therefore, \[2({{x}_{1}}-3)=1\Rightarrow {{x}_{1}}=\frac{7}{2}\]            \[\therefore {{y}_{1}}={{\left( \frac{7}{2}-3 \right)}^{2}}=\frac{1}{4}\]. Hence the point is \[\left( \frac{7}{2},\frac{1}{4} \right)\].


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