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JEE Main & Advanced Mathematics Functions Question Bank Continuity

  • question_answer
    If \[f(x)=\left\{ \begin{align}   & \,\,\,\,\,\,\,\,\,\frac{1-\cos 4x}{{{x}^{2}}},\ \ \text{when}\,x<0 \\  & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,a,\,\,\,\text{when}\,\,x=0 \\  & \frac{\sqrt{x}}{\sqrt{(16+\sqrt{x})}-4},\,\,\text{when}\,x>0 \\ \end{align} \right.\],            is continuous at \[x=0\], then the value of 'a' will be                                                                                 [IIT 1990; AMU 2000]

    A)            8

    B)            ?8

    C)            4

    D)            None of these

    Correct Answer: A

    Solution :

               \[\underset{x\to 0-}{\mathop{\lim }}\,f(x)=\underset{x\to 0-}{\mathop{\lim }}\,\left( \frac{2\,{{\sin }^{2}}2x}{{{(2x)}^{2}}} \right)\,4=8\]                    \[\underset{x\to 0+}{\mathop{\lim }}\,f(x)=\underset{x\to 0+}{\mathop{\lim }}\,\sqrt{16+\sqrt{x}+4}=8\]. Hence \[a=8\].


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