JEE Main202124 Feb 2021Evening ShiftMathematicsDifferential EquationsActual
Let f x be a differentiable function defined on 0 , 2 such that f ' x = f ' 2 - x for all x ∈ 0 , 2 , f 0 = 1 and f 2 = e 2 . Then the value of ∫ 0 2 f x d x is
Options
- A2 1 + e 2
- B1 + e 2
- C1 - e 2
- D2 1 - e 2
Correct answer
B. 1 + e 2
Step-by-step solution
f ' x = f ' 2 - x f x = - f 2 - x + c put x = 0 f 0 = - f 2 + c c = f 0 + f 2 = 1 + e 2 so, f x + f 2 - x = 1 + e 2 I = ∫ 0 2 f x d x I = ∫ 0 2 f 2 - x d x 2 I = ∫ 0 2 f x + f 2 - x d x 2 I = 1 + e 2 ∫ 0 2 d x I = 1 + e 2