QJEEM20Ma0b89: JEE Main 2020 Mathematics - Class 12 JEEM20Ma0b89 Solution: Let [t] denote the greatest integer less than or equal to t.…

Question

Verified Solution • 2025–26 Session moderate

Let [t] denote the greatest integer less than or equal to t.
Then the value of

\int\limits_1^2 {\left| {2x - \left[ {3x} \right]} \right|dx}

is ______.

Accepted Answer

\int\limits_1^2 {\left| {2x - \left[ {3x} \right]} \right|dx}

= \int\limits_1^2 {\left| {2x - \left( {3x - \left\{ {3x} \right\}} \right)} \right|dx}

= \int\limits_1^2 {\left| {\left\{ {3x} \right\} - x} \right|dx}

We know, 0

\le

\left\{ {3x} \right\} &lt; 1

and x > 1

\therefore \left\{ {3x} \right\} - x &lt; 0

So,

\left| {\left\{ {3x} \right\} - 2} \right| = - \left[ {\left\{ {3x} \right\} - x} \right]

= \int\limits_1^2 {\left( {x - \left\{ {3x} \right\}} \right)dx}

= \left[ {{{{x^2}} \over 2}} \right]_1^2 - \int\limits_{3 \times {1 \over 3}}^{6 \times {1 \over 3}} {\left\{ {3x} \right\}dx}

[Period of {x} = 1, so period of {3x} =

{1 \over 3}

]

= \left( {2 - {1 \over 2}} \right) - \left( {6 - 3} \right)\int\limits_0^{{1 \over 3}} {3xdx}

= {3 \over 2} - 3 \times 3\left[ {{{{x^2}} \over 2}} \right]_0^{{1 \over 3}}

= {3 \over 2} - {9 \over 2}\left[ {{1 \over 9} - 0} \right]

= {3 \over 2} - {1 \over 2}

= 1

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