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JEE Main & Advanced JEE Main Paper Phase-I (Held on 08-1-2020 Morning)

  • question_answer
    If c is a point at which Rolle's theorem holds for the function, \[f(x)=lo{{g}_{e}}\left( \frac{{{x}^{2}}+\alpha }{7x} \right)\] in the interval [3, 4], where \[\alpha \in R,\] Then f"is equal to  [JEE MAIN Held On 08-01-2020 Morning]

    A) \[-\frac{1}{24}\]

    B) \[\frac{1}{12}\]

    C) \[\frac{\sqrt{3}}{7}\]     

    D) \[-\frac{1}{12}\]

    Correct Answer: B

    Solution :

    [b] For application of Roll?s theorem\[f(3)=f(4)\] \[\frac{9+\alpha }{21}=\frac{16+\alpha }{28}\Rightarrow 36+4\alpha =48+3\alpha \Rightarrow \alpha =12\] \[\Rightarrow \frac{2c}{{{c}^{2}}+12}=\frac{1}{c}\Rightarrow 2{{c}^{2}}={{c}^{2}}+12\Rightarrow {{c}^{2}}=12\] \[f''(x)=\frac{({{x}^{2}}+12)2-2x(2x)}{{{({{x}^{2}}+12)}^{2}}}+\frac{1}{{{x}^{2}}}\] \[f''(c)=\frac{(24)(2)-4(12)}{{{(12+12)}^{2}}}+\frac{1}{12}=\frac{1}{12}\]


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