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JCECE Engineering JCECE Engineering Solved Paper-2008

  • question_answer
    Let\[f(x)=\left\{ \begin{matrix}    \frac{\sin \pi x}{5x}, & x\ne 0  \\    k, & x=0  \\ \end{matrix} \right.\], if \[f(x)\] is continuous at\[x=0\], then \[k\] is equal to

    A) \[\frac{\pi }{5}\]                                              

    B) \[\frac{5}{\pi }\]

    C) \[1\]                                     

    D) \[0\]

    Correct Answer: A

    Solution :

    Since, \[f(x)\] is continuous at\[x=0\] \[\therefore \]  \[\underset{x\to 0}{\mathop{\lim }}\,f(x)=f(0)\] \[\Rightarrow \]               \[\underset{x\to 0}{\mathop{\lim }}\,\frac{\sin \pi x}{5x}=k\] \[\Rightarrow \]               \[\lim \left( \frac{\sin \pi x}{\pi x} \right)\frac{\pi }{5}=k\] \[\Rightarrow \]               \[(1)\frac{\pi }{5}=k\]     \[\left( \because \,\,\underset{x\to 0}{\mathop{\lim }}\,\frac{\sin x}{x}=1 \right)\] \[\Rightarrow \]               \[k=\frac{\pi }{5}\]


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