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JEE Main & Advanced Mathematics Functions Question Bank Differentiability

  • question_answer
    If \[f(x)=\left\{ \begin{align}   & a{{x}^{2}}+b;\,\,x\le 0 \\  & \,\,\,\,\,\,\,\,\,{{x}^{2}};x>0\, \\ \end{align} \right.\] possesses derivative at \[x=0\], then

    A)            \[a=0,b=0\]                               

    B)            \[a>0,=0\]

    C)            \[a\in R,=0\]

    D)            None of these

    Correct Answer: C

    Solution :

     \[f(x)\]possesses derivative at \[x=0\], so it is both continuous and differentiable at \[x=0\]. Now \[f(0+0)=0\], \[f(0-0)=b,f(0)=b\], \[\therefore \,\,b=0\]            Also \[Rf'(0)=0,Lf'(0)=0,\forall a\in R\]            \ \[f'(0)=0\] if \[b=0\].


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