∫αlog4ex−1dx=6π Let ex−1=t2exdx=2tdt=∫t2+12dt=2tan−1t=2tan−1(ex−1)αloge4=2[tan−13−tan−1eα−1]=6π=3π−tan−1eα−1=12π⇒tan−1eα−1=4πeα=2e−α=21x2−(2+21)x+1=02x2−5x+2=0
Let ∫_α^log _e 4 d x √ e^x-1= π 6. Then e^α and e^-α are the roots of the equation :
JEE Main 2024 — Mathematics Calculus
Let ∫αloge4ex−1dx=6π. Then eα and e−α are the roots of the equation :
Held on 8 Apr 2024 · Verified 6 Jul 2026.
Options
- A
x2+2x−8=0
- B
x2−2x−8=0
- C
2x2−5x+2=0
- D
2x2−5x−2=0
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