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JEE Main & Advanced Sample Paper JEE Main Sample Paper-18

  • question_answer
    If \[f(x)=\left\{ \begin{align}   & \frac{x}{\sin x},\,\,x>0 \\  & 2-x,\,\,x\le 0 \\ \end{align} \right.\text{and}\]\[g(x)=\left\{ \begin{matrix}    x+3, & x<1  \\    {{x}^{2}}-2x-2, & 1\le x<2  \\    x-5, & x\ge 2  \\ \end{matrix} \right.\] then the value of \[\underset{x\to 0}{\mathop{\lim }}\,g\,(f(x))\]is

    A)  \[-2\]                                   

    B)  \[-3\]

    C)  \[1\]                                    

    D)  does not exist

    Correct Answer: B

    Solution :

     \[\underset{x\to {{0}^{+}}}{\mathop{\lim }}\,g(f(x))\] \[=g\,(f({{0}^{+}}))=g\,({{1}^{+}})\]         \[\left( \because \,\underset{x\to 0}{\mathop{\lim }}\,\frac{x}{\sin x}={{1}^{+}} \right)\] \[=1-2\,(1)-2=-\,3\] \[\underset{x\to {{0}^{-}}}{\mathop{\lim }}\,g(f(x)=g\,(f({{0}^{-}}))=g({{2}^{+}})\left( \because \underset{x\to {{0}^{-}}}{\mathop{\lim }}\,(2-x)={{2}^{+}} \right)\]\[=2-5=-\,3\] \[\Rightarrow \] \[\underset{x\to 0}{\mathop{\lim }}\,g(f(x))=-\,3\]


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