Question
Let f(x)=x^ 2025 -x^ 2000 , x [0,1] and the minimum value of the function f(x) in the interval [0,1] be (80)^ 80 (n)^ -81 . Then n is equal to
JEE Main
Mathematics
Application of Derivatives
2026
JEE Main 2026 (22 January Shift 1)
JEE Main Mathematics Question (2026) — Solution
Options
- A. -40
- B. -41
- C. -80
- D. -81
Answer
D. -81
Step-by-step solution
f(x) = x^ 2025 - x^ 2000 , f'(x) = x^ 1999 (2025x^ 25 - 2000) = 0. Critical point: x^ 25 = 2000 2025 = 80 81 , i.e., x = ( 80 81 )^ 1/25 . f(0) = f(1) = 0. At the critical point: f = x^ 2000 (x^ 25 -1) = ( 80 81 )^ 80 ( 80 81 -1 ) = - 80^ 80 81^ 81 . This equals (80)^ 80 (n)^ -81 , so n^ -81 = -81^ -81 n = -81.
Related: Mathematics — Application of Derivatives · All PYQ Banks