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CEE Kerala Engineering CEE Kerala Engineering Solved Paper-2008

  • question_answer
    If\[f(x)=\left\{ \begin{matrix}    \frac{1-\cos x}{x}, & x\ne 0  \\    k, & x=0  \\ \end{matrix} \right.\]is continuous at\[x=0,\]then the value of\[k\]is

    A)  \[0\]                    

    B)         \[1/2\]

    C)  \[\frac{1}{4}\]                  

    D)         \[-\frac{1}{2}\]

    E)  None of these

    Correct Answer: A

    Solution :

    Given, \[f(x)=\left\{ \begin{matrix}    \frac{1-\cos x}{x}, & x\ne 0  \\    k, & x=0  \\ \end{matrix} \right.\] \[\underset{x\to 0}{\mathop{\lim }}\,f(x)=\underset{x\to 0}{\mathop{\lim }}\,\frac{1-\cos x}{x}\] \[=\underset{x\to 0}{\mathop{\lim }}\,\frac{2{{\sin }^{2}}\frac{x}{2}}{4\left( \frac{x}{2} \right)}.x=0\]and\[f(0)=k\] \[\because \]Function is continuous at\[x=0\]. \[\therefore \] \[\underset{x\to 0}{\mathop{\lim }}\,f(x)=f(0)\Rightarrow k=0\]


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