Question:

If m is the minimum value of k for which the function \(f(x) = x\sqrt{kx - x^2}\) 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 :

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Compute \(f'(x)=\frac{x(3k-4x)}{2\sqrt{kx-x^2}}\) and require it to be non-negative on (0,3), so \(3k\ge 4x\) for all x up to 3.
Updated On: Nov 5, 2026
  • \(\left(5, 3\sqrt{6}\right)\)
  • \(\left(4, 3\sqrt{3}\right)\)
  • \(\left(4, 3\sqrt{2}\right)\)
  • \(\left(3, 3\sqrt{3}\right)\)
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The Correct Option is B

Solution and Explanation

\(f'(x)=\sqrt{kx-x^2}+\frac{x(k-2x)}{2\sqrt{kx-x^2}}=\frac{2(kx-x^2)+kx-2x^2}{2\sqrt{kx-x^2}}=\frac{x(3k-4x)}{2\sqrt{kx-x^2}}\).
For \(x\in(0,3)\), \(f'\ge0\) needs \(3k\ge4x\) for all such x, i.e. \(3k\ge12\Rightarrow k\ge4\). (The domain condition \(kx-x^2\ge0\) on [0,3] only needs \(k\ge3\), which is weaker.) So \(m=4\).
With \(k=4\), f is increasing on [0,3], so its maximum is at \(x=3\): \(M=3\sqrt{12-9}=3\sqrt3\).
Ordered pair \((4,3\sqrt3)\).
Answer: (2) \((4, 3\sqrt{3})\)
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