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J & K CET Engineering J and K - CET Engineering Solved Paper-2010

  • question_answer
    If the function \[f(x)=\left\{ \begin{matrix}    \frac{{{x}^{3}}-8}{{{x}^{2}}-4}, & if & x\ne 2  \\    k, & if & x=2  \\ \end{matrix} \right.\] is continuous at \[x=2,\]then the value of k is

    A)  \[2\]

    B)  \[0\]

    C)  \[3\]

    D)  \[4\]

    Correct Answer: C

    Solution :

    \[f(x)=\left\{ \begin{matrix}    \frac{{{x}^{3}}-8}{{{x}^{2}}-4}, & if & x\ne 2  \\    k, & if & x=2  \\ \end{matrix} \right.\] Since, \[f(x)\] is continuous at \[x=2,\] then \[f(2-h)=f(2+h)=f(2),\] ie, LHL = RHL =  value of function at \[x=2,\] Now,   \[LHL=\underset{h\to 0}{\mathop{\lim }}\,\,f(x)=\underset{h\to 0}{\mathop{\lim }}\,\,f(2-h)\] \[=\underset{h\to 0}{\mathop{\lim }}\,\,\frac{{{(2-h)}^{3}}-8}{{{(2-h)}^{2}}-4}\] \[=\underset{h\to 0}{\mathop{\lim }}\,\,\frac{3{{(2-h)}^{2}}}{2(2-h)}\] (By L Hospital rule) \[=\underset{h\to 0}{\mathop{\lim }}\,\,\,\frac{3}{2}\,(2-h)\] \[=\frac{3}{2}\,.2=3\] \[=f(2)=k\] \[\therefore \] \[k=3\]


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