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JEE Main & Advanced Mathematics Functions Question Bank Continuity

  • question_answer
    The value of k so that the function \[f(x)=\left\{ \begin{align}   & k(2x-{{x}^{2}}),\ \ \ \text{when}\,x<0 \\  & \,\,\,\,\,\,\,\,\,\cos x,\,\,\,\,\,\,\text{when}\,x\ge \text{0} \\ \end{align} \right.\]is continuous at\[x=0\], is                                      

    A)            1

    B)            2

    C)            4

    D)            None of these

    Correct Answer: D

    Solution :

               \[f(0-)=\underset{x\to 0-}{\mathop{\lim }}\,\,k(2x-{{x}^{2}})=0\]; \[f(0+)=\underset{x\to 0+}{\mathop{\lim }}\,\,\cos x=1\]                    \[\therefore \,\,\,f(0)=\cos x=1\]            Hence no value of k can make \[f(0-)=1.\]


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