Question
Let f : R → R be given by f x = x - 1 x - 2 x - 5 . Define F x = ∫ 0 x f t d t ,   x > 0 . Then which of the following options is/are correct?
JEE Advanced
Mathematics
Application of Derivatives
2019
JEE Advanced 2019 (Paper 2)
JEE Advanced Mathematics Question (2019) — Solution
Options
- A. F has a local minimum at x = 1
- B. F has a local maximum at x = 2
- C. F x ≠ 0 for all x ∈ 0 , 5
- D. F has two local maxima and one local minimum in 0 , ∞
Answer
C. F x ≠ 0 for all x ∈ 0 , 5
Step-by-step solution
F x = ∫ 0 x f t . d t F ' x = f x = x - 1 x - 2 x - 5 ⇒ F x has maxima at x = 2 and minima at x = 1 and x = 5 now F 2 = ∫ 0 2 x 3 - 8 x 2 + 17 x - 10 . d x = x 4 4 - 8 x 3 3 + 17 x 2 2 - 10 x 0 2 F 2 = - 10 3 which is a maxima in x ∈ 0 , 5 F 0 = 0 Hence F x 0 ∀ x ∈ 0 , 5
Related: Mathematics — Application of Derivatives · All PYQ Banks