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CET Karnataka Engineering CET - Karnataka Engineering Solved Paper-2011

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

    A)  0                                            

    B)  -1

    C)  1                                            

    D)  e

    Correct Answer: C

    Solution :

    Given,   \[f(x)=\left\{ \begin{matrix}    \frac{\log x}{x-1}, & if\,\,x\ne 1  \\    k\,\,\,\,\,\,\,\,\,\,, & if\,\,x=1  \\ \end{matrix} \right.\,\,at\,\,x=1\] Since, the function is continuous at \[x=1.\]                 Then,   \[f\,\,\,(1)=\underset{x\to 1}{\mathop{lim}}\,\,f(x)\]                                 \[k=\underset{x\to 1}{\mathop{\lim }}\,\,\,\frac{\log x}{x-1}\]   \[\left( \frac{0}{0}\,\,from \right)\]                 Apply L?hospital rule,                                 \[k=\underset{x\to 1}{\mathop{\lim }}\,\frac{1/x}{1}=\frac{1/1}{1}\]                 \[\Rightarrow \]               \[k=1\]


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