Verified Solution • 2025–26 Session moderate

QJEEM21Mac325: JEE Main 2021 Mathematics - Class 12 JEEM21Mac325 Solution: If the integral , where , , are integers and [x] denotes the…

If the integral

∫010[sin⁡2πx]ex−[x]dx=αe−1+βe−12+γ\int_0^{10} {{{[\sin 2\pi x]} \over {{e^{x - [x]}}}}} dx = \alpha {e^{ - 1}} + \beta {e^{ - {1 \over 2}}} + \gamma , where α\alpha , β\beta , γ\gamma are integers and [x] denotes the greatest integer less than or equal to x, then the value of α\alpha + β\beta + γ\gamma is equal to :

Multiple Choice Options

  • A 0
  • B 10
  • C 20
  • D 25

Step-by-Step Verified Solution

Given integral

∫010[sin⁡2πx]ex−[x]dx=10∫01[sin⁡2πx]e{x}dx\int\limits_0^{10} {{{[\sin 2\pi x]} \over {{e^{x - [x]}}}}dx = 10\int\limits_0^1 {{{[\sin 2\pi x]} \over {{e^{\{ x\} }}}}dx} } (using property of definite in.)

=10[∫01/20.dx+∫1/21−1exdx]= 10\left[ {\int\limits_0^{1/2} {0.dx} + \int\limits_{1/2}^1 {{{ - 1} \over {{e^x}}}dx} } \right]

= −10[e−x−1]1/21=10[e−1−e−1/2]- 10\left[ {{{{e^{ - x}}} \over { - 1}}} \right]_{1/2}^1 = 10\left[ {{e^{ - 1}} - {e^{ - 1/2}}} \right]

=10e−1−10e−1/2= 10{e^{ - 1}} - 10{e^{ - 1/2}}

comparing with the given relation,

α\alpha = 10, β\beta = −- 10, γ\gamma = 0

α\alpha + β\beta + γ\gamma = 0

Therefore, the correct answer is (A).

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