BITSAT2019MathematicsApplication of DerivativesActual
The difference of maximum and minimum values of f ( x ) = x 2 e - x is
Options
- Ae
- B1 / e
- C1 - 1 6
- D1 + 1 6
Correct answer
B. 1 / e
Step-by-step solution
We have, f x = x 2 e - x 2 f ' x = 2 x e - x 2 - 2 x 3 e - x 2 f ' x = 2 x e - x 2 1 - x 2 f ' x = 0 , x = 0 , - 1 , 1 f ' - h < 0 , f ' h > 0 , So minimum at x = 0 f ' 1 - h > 0 , f ' 1 + h < 0 , So maximum at x = 1 f ' - 1 - h > 0 , f ' - 1 + h < 0 , So maximum at x = - 1 Minimum f x = f 0 = 0 Maximum f x = e - 1 = 1 e Since, f x ≥ 0 ∀ x So, difference between maximum and minimum values = 1 e - 0 = 1 e