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J & K CET Engineering J and K - CET Engineering Solved Paper-2005

  • question_answer
    Let \[f(x)=\left\{ \begin{matrix}    (1/2)\,\{g(x)+(x)\}\,\,sin\,(x),\,x\ge 1  \\    \sin \,x/x,\,x<1  \\ \end{matrix} \right.\] where \[g(x)=\left\{ \begin{matrix}    1, & if & x>0  \\    -1, & if & x<0  \\    0, & if & x=0  \\ \end{matrix} \right.\] Then, \[\underset{x\to 1}{\mathop{\lim }}\,\,\,f(x)\] is equal to

    A)  \[0\]     

    B)  \[2\]

    C)  \[sin\text{ }1\]          

    D)  None of these

    Correct Answer: B

    Solution :

    Given,  \[f(x)=\left\{ \begin{align}   & \frac{1}{2}\{g(x)+(x)\}\,\sin \,x,\,x\ge 1 \\  & \frac{\sin \,x}{x},\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,x<1 \\ \end{align} \right.\] and \[g(x)=\left\{ \begin{matrix}    1, & if & x>0  \\    -1, & if & x<0  \\    0, & if & x=0  \\ \end{matrix} \right.\] Now,   \[RHL=\underset{x\to {{1}^{+}}}{\mathop{\lim }}\,\frac{1}{2}\{g(x)+(x)\}\,\sin x\] \[=\underset{x\to {{1}^{+}}}{\mathop{\lim }}\,\frac{1}{2}\{1+x)\,\sin x\] \[=\frac{1}{2}.(1+1)\,sin1\] \[=\,sin1\] and  \[LHL=\underset{x\to {{1}^{-}}}{\mathop{\lim }}\,\frac{\sin \,x}{x}\] \[=\frac{\sin \,(-1)}{(-1)}\] \[=\frac{-\sin \,1}{-1}\] \[=\sin 1\] Since,  \[RHL=LHL=\sin \,1\] \[\therefore \]    \[\underset{x\to 1}{\mathop{\lim }}\,f(x)=\sin 1\]


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