Verified Solution • 2025–26 Session moderate

QJEEM06Ma7edf: JEE Main 2006 Mathematics - Class 12 JEEM06Ma7edf Solution: The value of where denotes the greatest integer not exceeding is

The value of \int\limits_1^a {\left[ x \right]} f'\left( x \right)dx,a > 1 where [x]{\left[ x \right]} denotes the greatest integer not exceeding xx is

Multiple Choice Options

  • A af(a)−{f(1)+f(2)+.............f([a])}af\left( a \right) - \left\{ {f\left( 1 \right) + f\left( 2 \right) + .............f\left( {\left[ a \right]} \right)} \right\}
  • B [a]f(a)−{f(1)+f(2)+...........f([a])}\left[ a \right]f\left( a \right) - \left\{ {f\left( 1 \right) + f\left( 2 \right) + ...........f\left( {\left[ a \right]} \right)} \right\}
  • C [a]f([a])−{f(1)+f(2)+...........f(a)}\left[ a \right]f\left( {\left[ a \right]} \right) - \left\{ {f\left( 1 \right) + f\left( 2 \right) + ...........f\left( a \right)} \right\}
  • D af([a])−{f(1)+f(2)+.............f(a)}af\left( {\left[ a \right]} \right) - \left\{ {f\left( 1 \right) + f\left( 2 \right) + .............f\left( a \right)} \right\}

Step-by-Step Verified Solution

Let a=k+ha = k + h where kk is an integer such that

[a]=k\left[ a \right] = k and 0 \le h < 1

∴\therefore ∫1a[x]f′(x)dx=∫121f′(x) dx\int\limits_1^a {\left[ x \right]f'\left( x \right)dx = \int\limits_1^2 {1f'\left( x \right)} } \,dx

            +∫232f′(x)dx+....\,\,\,\,\,\,\,\,\,\,\,\, + \int\limits_2^3 {2f'\left( x \right)dx + } ....

            ∫k−1k(k−1)dx+∫kk+hkf′(x)dx\,\,\,\,\,\,\,\,\,\,\,\,\int\limits_{k - 1}^k {\left( {k - 1} \right)dx + \int\limits_k^{k + h} {kf'\left( x \right)} } dx

{f(2)−f(1)}+2{f(3)−f(2)}\left\{ {f\left( 2 \right) - f\left( 1 \right)} \right\} + 2\left\{ {f\left( 3 \right) - f\left( 2 \right)} \right\}

             \,\,\,\,\,\,\,\,\,\,\,\,\, +3{f(4)−f(3)}+ ........ + 3\left\{ {f\left( 4 \right) - f\left( 3 \right)} \right\} + \,........\,

             \,\,\,\,\,\,\,\,\,\,\,\,\, +(k−1){f(k)−f(k−1)}+ \left( {k - 1} \right)\left\{ {f\left( k \right) - f\left( {k - 1} \right)} \right\}

             \,\,\,\,\,\,\,\,\,\,\,\,\, +k{f(k+h)−f(k)}+ k\left\{ {f\left( {k + h} \right) - f\left( k \right)} \right\}

=−f(1)−f(2)−f(3).......= - f\left( 1 \right) - f\left( 2 \right) - f\left( 3 \right).......

             \,\,\,\,\,\,\,\,\,\,\,\,\, −f(k)+kf(k+h)- f\left( k \right) + kf\left( {k + h} \right)

=[a]f(a)−{f(1)+f(2)= \left[ a \right]f\left( a \right) - \left\{ {f\left( 1 \right) + f\left( 2 \right)} \right.

             \,\,\,\,\,\,\,\,\,\,\,\,\, +f(3)+........f([a])}\left. { + f\left( 3 \right) + ........f\left( {\left[ a \right]} \right)} \right\}

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