JEE Main 2018 Mathematics - Class 12 JEEM18Maacc7 Solution: For each t , let [t] be the greatest integer less than or equal …

Given,

\mathop {lim}\limits_{x \to {0^ + }} \,\,x\,\left( {\left[ {{1 \over x}} \right] + \left[ {{2 \over x}} \right]} \right. +

\left. {\,.\,.\,.\,.\,.\, + \left[ {{{15} \over x}} \right]} \right)

as we know that

{1 \over x} = \left[ {{1 \over x}} \right] + \left\{ {{1 \over x}} \right\}

\Rightarrow \,\,\,\,\left[ {{1 \over x}} \right] = {1 \over x} - \left\{ {{1 \over x}} \right\}

= \,\,\,\,\mathop {\lim }\limits_{x \to {0^ + }} \,\,x\left[ {{1 \over x} - \left\{ {{1 \over x}} \right\} + {2 \over 2} - \left\{ {{2 \over x}} \right\} + ........{{15} \over x} - \left\{ {{{15} \over x}} \right\}} \right]

= \,\,\,\,\mathop {\lim }\limits_{x \to {0^ + }} \,\left[ {x.{1 \over x} + x.{2 \over x} + .....x.{{15} \over x}} \right]

\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, - \,\left[ {x.\left\{ {{1 \over 2}} \right\} + .. + x.\left\{ {{{15} \over x}} \right\}} \right]

We know

\left\{ {{1 \over x}} \right\}

is fractional part of

{1 \over x}.

So, the range of

\,\left\{ {{1 \over x}} \right\}

0 \le \left\{ {{1 \over x}} \right\} &lt; 1

So,

\mathop {lim}\limits_{x \to {0^ + }} \,\,x\,\left\{ {{1 \over x}} \right\} = 0.

(finite no)

=0

Similarly

\mathop {\lim }\limits_{x \to {0^ + }} x.\left\{ {{2 \over x}} \right\} = 0

= \,\,\,\,\left( {1 + 2 + ... + 15} \right) - \left( {0 + 0...} \right)

= \,\,\,\,{{15 \times 16} \over 2}

= \,\,\,\,120

QJEEM18Maacc7: JEE Main 2018 Mathematics - Class 12 JEEM18Maacc7 Solution: For each t , let [t] be the greatest integer less than or equal …