JEE Main & Advanced JEE Main Paper (Held On 09-Jan-2019 Evening)

  • question_answer
    For each\[x\text{ }\in \text{ }R\], let [x] be the greatest integer less than or equal to x. Then\[\underset{x\to {{0}^{-}}}{\mathop{\lim }}\,\,\frac{x([x]+\left| x \right|)\,sin\,[x]}{\left| x \right|}\] [JEE Main Online Paper (Held On 09-Jan-2019 Evening]

    A) 1                                             

    B)               0

    C)               \[sin\text{ }1\]                           

    D)               \[-sin\text{ }1\]

    Correct Answer: D

    Solution :

    \[\underset{x\to {{0}^{-}}}{\mathop{\lim }}\,\,\frac{x([x]+\left| x \right|)\,\,sin\,[x]}{\left| x \right|}\] \[\underset{h\to 0}{\mathop{\lim }}\,\,\frac{(0-h)([0-h]+\left| 0-h \right|)sin\,[0-h]}{\left| 0-h \right|}\] \[=\,\,\,\underset{h\to 0}{\mathop{\lim }}\,\,\frac{-h(-1+h)sin(-1)}{h}\] \[=\,\,\,\underset{h\to 0}{\mathop{\lim }}\,\,(-1+h)\,sin(1)\,\,=\,\,-sin1\]


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