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JEE Main & Advanced Mathematics Functions Question Bank Continuity

  • question_answer
    If \[f(x)=\left\{ \begin{align}   & \,\,\,\,\,\,\,\frac{\sin [x]}{[x]+1},\,\,\text{for}\,x>0 \\  & \frac{\cos \frac{\pi }{2}[x]}{[x]},\,\,\text{for}\,x<0 \\  & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,k,\,\,\text{at}\,x=0 \\ \end{align} \right.\]; where [x] denotes the greatest integer less than or equal to x, then in order that f be continuous at \[x=0\], the value of k is [Kurukshetra CEE 1998]

    A)            Equal to 0

    B)            Equal to 1

    C)            Equal to ?1

    D)            Indeterminate

    Correct Answer: A

    Solution :

               If f is continuous at \[\underset{x\to {{0}^{+}}}{\mathop{\lim }}\,{f}'(x)=\underset{h\to 0}{\mathop{\lim }}\,{f}'(0+h)=\underset{h\to 0}{\mathop{\lim }}\,2(0+h)=0\], then                    \[\underset{x\to {{0}^{-}}}{\mathop{\lim }}\,f(x)=\underset{x\to {{0}^{+}}}{\mathop{\lim }}\,f(x)=f(0)\]\[\Rightarrow \,f(0)=\,\underset{x\to {{0}^{-}}}{\mathop{\lim }}\,f(x)\]                    \[k=\underset{h\to 0}{\mathop{\lim }}\,f(0-h)=\underset{h\to 0}{\mathop{\lim }}\,\,\frac{\cos \frac{\pi }{2}\,[0-h]}{[0-h]}\]                    \[k=\underset{h\to 0}{\mathop{\lim }}\,\,\frac{\cos \frac{\pi }{2}\,[-h]}{[-h]}=\underset{h\to 0}{\mathop{\lim }}\,\,\frac{\cos \frac{\pi }{2}\,[-h-1]}{[-h-1]}\]            \[k=\underset{h\to 0}{\mathop{\lim }}\,\,\frac{\cos \,\left( -\frac{\pi }{2} \right)}{-1}\]; \[k=0\].


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