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JEE Main & Advanced Mathematics Limits, Continuity and Differentiability Question Bank Self Evaluation Test - Continuity and Differentiability

  • question_answer
    Let \[f:[2,7]\to [0,\infty )\] be a continuous and differentiable function. Then, \[(f(7)-f(2))\frac{{{(f(7))}^{2}}+{{(f(2))}^{2}}+f(2)f(7)}{3}\] is, where \[c\in [2,7]\] [2, 7].

    A) \[5{{f}^{2}}(c)f'(c)\]

    B) \[5f'(c)\]

    C) \[f(c)f'(c)\]

    D) None of these

    Correct Answer: A

    Solution :

    [a] Let \[g(x)={{f}^{3}}(x)\] \[\Rightarrow g'(x)=3{{f}^{2}}(x)\cdot f'(x)\] \[\because f:[2,7]\to [0,\infty )\Rightarrow g:[2,7]\to [0,\infty )\] Using Lagrange?s mean value theorem on g(x), We get \[g'(c)=\frac{g(7)-g(2)}{5},c\in [2,\,7]\] \[\Rightarrow 2{{f}^{2}}(c)f'(c)=(f(7)-f(2))\]                         \[\frac{{{(f(7))}^{2}}+{{(f(2))}^{2}}+f(2)f(7)}{3}\]


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