Question

f(x) = x\(^{2025}\) - x\(^{2000}\), x \(\in\) [0, 1], then minimum value of f(x) is :

• A. (80)\(^{400}\). (81)\(^{-395}\)((80)\(^5\) – (81)\(^5\))
• B. (80)\(^{300}\). (81)\(^{-295}\)((80)\(^5\) – (81)\(^5\))
• C. (80)\(^{-395}\). (81)\(^{400}\)((80)\(^5\) – (81)\(^5\))
• D. (80)\(^{-395}\). (81)\(^{400}\)((80)\(^5\) – (81)\(^5\))

Hint:

To find the absolute extrema of a function on a closed interval, always compare the function values at the critical points inside the interval and at the endpoints of the interval.

Answer 1

The Correct Option is A

Step 1: Understanding the Question:
We need to find the absolute minimum value of the function f(x) = x\(^{2025}\) - x\(^{2000}\) on the closed interval [0, 1]. We can use calculus to find critical points and then evaluate the function at these points and the interval endpoints.

Step 2: Finding Critical Points:
First, find the derivative of f(x) and set it to zero. \[ f'(x) = 2025x^{2024} - 2000x^{1999} \] Set \(f'(x) = 0\): \[ 2025x^{2024} - 2000x^{1999} = 0 \] Factor out the lowest power of x, which is \(x^{1999}\): \[ x^{1999}(2025x^{25} - 2000) = 0 \] This gives two potential critical points: \(x=0\) (which is an endpoint) and \(2025x^{25} - 2000 = 0\).
Solving for x from the second part: \[ x^{25} = \frac{2000}{2025} = \frac{25 \times 80}{25 \times 81} = \frac{80}{81} \] \[ x = \left(\frac{80}{81}\right)^{1/25} \] This critical point is between 0 and 1, so it is within our interval.

Step 3: Evaluating the Function:
We evaluate f(x) at the endpoints (0 and 1) and at the critical point \(x_c = (80/81)^{1/25}\).
At \(x=0\): \(f(0) = 0^{2025} - 0^{2000} = 0\).
At \(x=1\): \(f(1) = 1^{2025} - 1^{2000} = 1 - 1 = 0\).
At \(x_c = \left(\frac{80}{81}\right)^{1/25}\): \[ f(x_c) = x_c^{2025} - x_c^{2000} = x_c^{2000}(x_c^{25} - 1) \] Substitute the values we found: \[ f(x_c) = \left(\left(\frac{80}{81}\right)^{1/25}\right)^{2000} \left(\frac{80}{81} - 1\right) \] \[ f(x_c) = \left(\frac{80}{81}\right)^{80} \left(-\frac{1}{81}\right) = -\frac{80^{80}}{81^{80} \cdot 81} = -\frac{80^{80}}{81^{81}} \]
Step 4: Final Answer:
The values of the function are 0, 0, and a negative value \(-\frac{80^{80}}{81^{81}}\). The minimum value is therefore \(-\frac{80^{80}}{81^{81}}\).