If \(f(x) = 2x^3 - 15x^2+36x+1,\ x \in [1,5]\), then the absolute minimum value of \(f(x)\) is:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → 24
On a closed interval, the absolute minimum is the smallest value of f at the critical points and the end points.
\(f'(x) = 6x^2 - 30x + 36\)
\(= 6(x^2 - 5x + 6) = 6(x - 2)(x - 3)\)
So the critical points are x = 2 and x = 3, both in [1, 5]. The values are:
\(f(1) = 2 - 15 + 36 + 1 = 24\)
\(f(2) = 16 - 60 + 72 + 1 = 29\)
\(f(3) = 54 - 135 + 108 + 1 = 28\)
\(f(5) = 250 - 375 + 180 + 1 = 56\)
The smallest value is 24, at the end point x = 1.
Why the other options are incorrect
Option (1) 22 — Incorrect. f does not take the value 22 at any critical point or end point; the smallest of these values is 24.
Option (2) 28 — Incorrect. 28 is the local minimum value, at x = 3, but the end point x = 1 gives the smaller value 24.
Option (4) 29 — Incorrect. 29 is f(2), the local maximum value.
NCERT (2026-27 edition):
Maxima and Minima
