JEE Main20241 Feb 2024Morning ShiftMathematicsArea Under CurvesActual
The area enclosed by the curves x y + 4 y = 16 and x + y = 6 is equal to:
Options
- A28 − 30 log e 2
- B30 − 28 log e 2
- C30 − 32 log e 2
- D32 − 30 log e 2
Correct answer
C. 30 − 32 log e 2
Step-by-step solution
Given: x y + 4 y = 16 , x + y = 6 ⇒ y x + 4 = 16 , x + y = 6 ⇒ 6 - x x + 4 = 16 ⇒ 6 x + 24 - x 2 - 4 x - 16 = 0 ⇒ - x 2 + 2 x + 8 = 0 ⇒ x 2 - 2 x - 8 = 0 ⇒ x - 4 x + 2 = 0 ⇒ x = 4 , - 2 So, the required area is given by, A = ∫ − 2 4 6 − x − 16 x + 4 d x ⇒ A = 6 x - x 2 2 - 16 log x + 4 - 2 4 ⇒ A = 24 - 8 - 16 log 8 - - 12 - 2 - 16 log 2 ⇒ A = 16 - 16 log 8 + 14 + 16 log 2 ⇒ A = 30 − 32 ln 2