Verified Solution • 2025–26 Session moderate

QJEEM20Ma2da4: JEE Main 2020 Mathematics - Class 12 JEEM20Ma2da4 Solution: The value of for which , is:

The value of α\alpha for which
4α∫−12e−α∣x∣dx=54\alpha \int\limits_{ - 1}^2 {{e^{ - \alpha \left| x \right|}}dx} = 5 , is:

Multiple Choice Options

  • A log⁡e2{\log _e}2
  • B log⁡e2{\log _e}\sqrt 2
  • C log⁡e(43){\log _e}\left( {{4 \over 3}} \right)
  • D log⁡e(32){\log _e}\left( {{3 \over 2}} \right)

Step-by-Step Verified Solution

4α∫−12e−α∣x∣dx=54\alpha \int\limits_{ - 1}^2 {{e^{ - \alpha \left| x \right|}}dx} = 5

⇒\Rightarrow 4α∫−10eαxdx+4α∫02e−αxdx4\alpha \int\limits_{ - 1}^0 {{e^{\alpha x}}dx} + 4\alpha \int\limits_0^2 {{e^{ - \alpha x}}dx} = 5

⇒\Rightarrow 4α[eαxα]−10+4α[e−αx−α]024\alpha \left[ {{{{e^{\alpha x}}} \over \alpha }} \right]_{ - 1}^0 + 4\alpha \left[ {{{{e^{ - \alpha x}}} \over { - \alpha }}} \right]_0^2 = 5

⇒\Rightarrow 4e−2α+4e−α−34{e^{ - 2\alpha }} + 4{e^{ - \alpha }} - 3 = 0

Let e- α\alpha = t

∴\therefore 4t2 + 4t - 3 = 0

⇒\Rightarrow t = 12{1 \over 2}

∴\therefore e- α\alpha = 12{1 \over 2}

⇒\Rightarrow α\alpha = log⁡e2{\log _e}2

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