JEE Main & Advanced Mathematics Limits, Continuity and Differentiability Question Bank Self Evaluation Test - Continuity and Differentiability

  • question_answer
    What is the value of k for which the following function f(x) is continuous for all x? \[f(x)=\left\{ \begin{align}   & \frac{{{x}^{3}}-3x+2}{{{(x-1)}^{2}}},for\,\,x\ne 1 \\  & k\,\,\,\,\,\,\,\,,for\,\,x=1 \\ \end{align} \right.\]

    A) 3

    B) 2

    C) 1

    D) -1

    Correct Answer: A

    Solution :

    [a] Let \[f(x)=\left\{ \begin{matrix}    \frac{{{x}^{3}}-3x+2}{{{(x-1)}^{2}}}, & \forall x\ne 1  \\    k, & \forall x=1  \\ \end{matrix} \right.\] and \[f(x)\] is continuous. \[\therefore \underset{x\to 1}{\mathop{\lim }}\,f(x)=k\] \[\Rightarrow \underset{x\to 1}{\mathop{\lim }}\,\frac{{{x}^{3}}-3x+2}{{{(x-1)}^{2}}}=k\] \[\Rightarrow k=\underset{x\to 1}{\mathop{\lim }}\,\frac{3{{x}^{2}}-3}{2(x-1)}[By\,\,L'\,\,Hospitals\,\,rule]\] \[\Rightarrow k=\underset{x\to 1}{\mathop{\lim }}\,\frac{6x}{2}[By\,\,L'\,\,Hospitals\,\,rule]\] \[\Rightarrow k=3\]


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