If m is the minimum value of k for which the function f(x) = x^k - x^2 is increasing in the interval [0, 3] and M is the maximum value of f in [0, 3] when k = m, then the ordered pair (m, M) is equal to?

Correct option is B (4, 33)
f(x) = x^k - x^2
f'(x) = (3kx - 4x^2)/(2x^k - x^2)
For increasing f'(x) ≥ 0
3kx - 4x^2 ≥ 0
x^k - x^2 < 0
4x^2 - 3kx ≤ 0
x(x - k) < 0
4x(x - 3k/4) ≤ 0
So x ∈ [0, 3k/4] and x ∈ (0, k)
Minimum value of k is m = 4
f(x) = x^k - x^2 = 4^3 - 3^2 = 64 - 9 = 55
f(x) = x^4 - x^2
f'(x) = 4x^3 - 2x = 2x(2x^2 - 1) = 0
x = 0, x = ±1/√2
If k = 4, f(x) = x^4 - x^2
f(3) = 3^4 - 3^2 = 81 - 9 = 72
f(0) = 0
f(1/√2) = (1/√2)^4 - (1/√2)^2 = 1/4 - 1/2 = -1/4
f(3) = 3^4 - 3^2 = 81 - 9 = 72
Let's check the derivative:
f'(x) = kx^(k-1) - 2x
f'(x) ≥ 0 for x ∈ [0, 3]
kx^(k-1) ≥ 2x
k ≥ 2x^(2-k)
If k = 4, 4 ≥ 2x^(-2)
4 ≥ 2/x^2
x^2 ≥ 1/2
x ≥ 1/√2
This is true for x ∈ [0, 3]
When k = 4, f(x) = x^4 - x^2
f'(x) = 4x^3 - 2x = 0
x = 0, x = ±1/√2
f(3) = 3^4 - 3^2 = 81 - 9 = 72
The maximum value occurs at x = 3, which is M = 72
Let's re-examine the solution:
f(x) = x^k - x^2
f'(x) = kx^(k-1) - 2x
For f(x) to be increasing, f'(x) ≥ 0
kx^(k-1) - 2x ≥ 0
kx^(k-2) ≥ 2
k ≥ 2x^(2-k)
If k = 4, 4 ≥ 2x^(-2), x^2 ≥ 1/2, x ≥ 1/√2
This is true for x ∈ [0, 3]
f(3) = 3^4 - 3^2 = 81 - 9 = 72
There must be a mistake in the provided solution. Let's reconsider the interval [0,3]. If k=4, f(x) = x^4 - x^2. f'(x) = 4x^3 - 2x = 2x(2x^2 - 1). f'(x) = 0 when x = 0, x = 1/√2. f(3) = 72. f(0) = 0. f(1/√2) = -1/4. The maximum is at x = 3. Therefore M = 72. However, this is not one of the options. The given solution contains errors.