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JEE Main & Advanced JEE Main Paper (Held on 12-4-2019 Morning)

  • question_answer
    Let \[f:R\to R\]be a continuously differentiable function wuch that \[f\left( 2 \right)=6\] and \[f\left( 2 \right)=\frac{1}{48}.\] If \[\int_{6}^{f(x)}{4{{t}^{3}}}dt=(x-2)g(x),\]then\[\underset{x\to 2}{\mathop{\lim }}\,g(x)\]is equal to:                         [JEE Main Held on 12-4-2019 Morning]

    A) 24                               

    B) 36

    C) 12                               

    D) 18

    Correct Answer: D

    Solution :

    \[\underset{x\to 2}{\mathop{\lim }}\,g(x)=\underset{x\to 2}{\mathop{\lim }}\,\frac{\int\limits_{6}^{f(x)}{4{{t}^{3}}dt}}{x-2}\]\[=\underset{x\to 2}{\mathop{\lim }}\,\frac{4.{{f}^{3}}(x).f'(x)}{1}\]               \[=4{{f}^{3}}(2)f'(2)=18\]


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