If is the minimum value of for which the function is increasing in the interval and is the maximum value of in the interval when , then the ordered pair is equal to
- A.
- B.
- C.
- D.
Show answer and solution
Correct option: B
Solution
For f(x) = x√(kx - x²) to be defined on [0,3], we need kx - x² ≥ 0, i.e., x(k-x) ≥ 0, so k ≥ 3 (domain must cover [0,3]).
Rewrite f(x) = x√(x(k-x)) = x^(3/2)·√(k-x).
f'(x) = (3/2)x^(1/2)·√(k-x) + x^(3/2)·(-1)/(2√(k-x))
= √x/(2√(k-x)) · [3(k-x) - x]
= √x/(2√(k-x)) · (3k - 4x)
For f to be increasing on [0,3], we need f'(x) ≥ 0 for all x ∈ [0,3]. The factor √x/(2√(k-x)) ≥ 0, so we need 3k - 4x ≥ 0 for all x ∈ [0,3]. The worst case is x = 3: 3k - 12 ≥ 0 → k ≥ 4.
So minimum k = m = 4.
With k = 4: f(x) = x√(4x - x²) = x·√(x(4-x)).
f'(x) = 0 when 3(4) - 4x = 0 → x = 3, so f is increasing on [0,3] with maximum at x
Rewrite f(x) = x√(x(k-x)) = x^(3/2)·√(k-x).
f'(x) = (3/2)x^(1/2)·√(k-x) + x^(3/2)·(-1)/(2√(k-x))
= √x/(2√(k-x)) · [3(k-x) - x]
= √x/(2√(k-x)) · (3k - 4x)
For f to be increasing on [0,3], we need f'(x) ≥ 0 for all x ∈ [0,3]. The factor √x/(2√(k-x)) ≥ 0, so we need 3k - 4x ≥ 0 for all x ∈ [0,3]. The worst case is x = 3: 3k - 12 ≥ 0 → k ≥ 4.
So minimum k = m = 4.
With k = 4: f(x) = x√(4x - x²) = x·√(x(4-x)).
f'(x) = 0 when 3(4) - 4x = 0 → x = 3, so f is increasing on [0,3] with maximum at x
[JEE Main 2019 · 12 April · Shift I]