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JEE Main & Advanced Sample Paper JEE Main - Mock Test - 19

  • question_answer
    Let \[f(x)\] be a non-constant twice differentiable function on R such that \[f(2+x)=f(2-x)\] and \[f'\left( \frac{1}{2} \right)=f'(1)=0\] Then minimum number of roots of the equation \[f''(x)=0\]in \[(0,4)\]is

    A) \[3\]                       

    B)        \[4\]             

    C) \[5\]                       

    D)        \[6\]

    Correct Answer: B

    Solution :

          [b] \[f(2+x)=f(2-x)\] \[f'(2+x)=-f'(2-x)\] When \[x=0,\]then \[f'(2)=0\]. When \[x=-1,\]then \[f'(1)=-f'(3)=0\]. When \[x=\frac{-3}{2},\]then \[f'\left( \frac{1}{2} \right)=-f'\left( \frac{7}{2} \right)=0.\] \[\therefore \,\,\,\,\,f'\left( \frac{1}{2} \right)=f'(1)=f'(2)=f'(3)=f'\left( \frac{7}{2} \right)=0\]. Using Rolle's Theorem on f(x), minimum number of roots of \[f''(x)=0\]is 4.   


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