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CET Karnataka Engineering CET - Karnataka Engineering Solved Paper-2004

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

    A)  0                                            

    B)  \[\frac{1}{2}\]

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

    D)  \[-\frac{1}{2}\]

    Correct Answer: A

    Solution :

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


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