JEE Advanced
Mathematics
Area Under Curves
2020
JEE Advanced 2020 (Paper 1)
JEE Advanced Mathematics Question (2020) — Solution
Question
Let the functions f : R → R and g : R → R be defined by f x = e x − 1 − e − x − 1 and g x = 1 2 e x − 1 + e 1 − x . Then the area of the region in the first quadrant bounded by the curves y = f x ,   y = g x and x = 0 is
Options
- A. 2 − 3 + 1 2 e − e − 1
- B. 2 + 3 + 1 2 e − e − 1
- C. 2 − 3 + 1 2 e + e − 1
- D. 2 + 3 + 1 2 e + e − 1
Answer
A. 2 − 3 + 1 2 e − e − 1
Step-by-step solution
f x = e x − 1 − e − x − 1 = 0 , x ≤ 1 e x − 1 − e 1 - x , x > 1 and g x = 1 2 e x − 1 + e 1 − x At x = 0 , f 0 = 0 & g 0 = e + e - 1 2 At x = 1 , f 1 = 0 & g 1 = 1 2 e 1 - 1 + e 1 - 1 = 1 For point of intersection of the two curves for x > 1 , f x = g x ⇒ e x − 1 − e 1 - x = 1 2 e x − 1 + e 1 − x ⇒ 2 e x − 1 − 2 e 1 - x = e x − 1 + e 1 − x ⇒ e x − 1 = 3 e 1 − x ⇒ e x − 1 2 = 3 ⇒ x = 1 + ln 3 ∴ Required area = Area of the shaded region = ∫ 0 1 g x d x + ∫ 1 1 + ln 3 g x − f x d x = ∫ 0 1 1 2 e x − 1 + e 1 − x d x + ∫ 1 1 + ln 3 1 2 e x − 1 + e 1 − x − e x − 1 − e 1 − x d x = 1 2 ∫ 0 1 e x − 1 + e 1 − x d x + 1 2 ∫ 1 1 + ln 3 3 e 1 − x - e x − 1 d x = 1 2 e x − 1 - e 1 − x 0 1 − 1 2 3 e 1 − x + e x − 1 1 1 + ln 3 = 1 2 e − e − 1 − 1 2 2 3 − 4 = e − e − 1 2 + 2 − 3
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