JEE Main & Advanced JEE Main Solved Paper-2014

  • question_answer
    If f and g are differentiable functions in [0, 1] satisfying f(0) = 2 = g(1), g(0) = 0 and f(1) = 6,then for some \[c\in \}0,1[:\]   JEE Main  Solved  Paper-2014

    A) \[2f'(c)=g'(c)\]  

    B) \[2f'(c)=3g'(c)\]

    C) \[f'(c)=g'(c)\]    

    D) \[f'(c)=2g'(c)\]

    Correct Answer: D

    Solution :

    Given, \[f(0)=2,g(1)=2,g(0)=0,\text{ }f(1)=6\] Let, \[F(x)=f(x)-2g(x)\] \[F(0)=f(0)-2g(0)\] \[F(0)=2-2\times 0\] \[F(0)=2\]                                                            ? (1) \[F(1)=F(1)-2g(1)\] \[F(1)=6-2\times 2\] \[F(1)=2\]                                                            ? (2) F(x) is continuous and differentiable in [0, 1]. \[F(0)=F(1)\] So, according to Rolle?s theorem, there is at least are root between 0 and 1. At which \[F\prime (x)=0.\] \[f\prime (x)-2g\prime (x)=0\] \[f\prime (c)-2g\prime (c)=0\] \[f\prime (c)=2g\prime (c)\]

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