JEE Main & Advanced JEE Main Online Paper (Held On 08 April 2018)

  • question_answer
    For each \[t\in \mathbf{R},\] let[t] be the greatest integer less than or equal to t. Then \[\underset{x\to 0+}{\mathop{\lim }}\,x\left( \left[ \frac{1}{x} \right]+\left[ \frac{2}{x} \right]+......+\left[ \frac{15}{x} \right] \right)\] [JEE Main Online 08-04-2018]

    A)  is equal to 120.

    B)  does not exist (in R).

    C)  is equal to 0.

    D)  is equal to 15.

    Correct Answer: A

    Solution :

    \[\underset{x\to {{0}^{+}}}{\mathop{\lim }}\,x\left[ \frac{1}{x}-\left\{ \frac{1}{x} \right\}+\frac{2}{x}-\left\{ \frac{2}{x} \right\}+.....+\frac{15}{x}-\left\{ \frac{15}{x} \right\} \right]\] \[=120-\underset{x\to {{0}^{+}}}{\mathop{\lim }}\,x\underbrace{\left[ \left\{ \frac{1}{x} \right\}+\left\{ \frac{2}{x} \right\}+.....+\left\{ \frac{15}{x} \right\} \right]}_{finite}\] \[=120-0=120\].


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