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JEE Mains · Maths · STD 12 - 6. Application of derivatives

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

  1. A \(\left( {5,3\sqrt 6 } \right)\)
  2. B \(\left( {4,3\sqrt 2 } \right)\)
  3. C \(\left( {3,3\sqrt 3 } \right)\)
  4. D \(\left( {4,3\sqrt 3 } \right)\)
Verified Solution

Answer & Solution

Correct Answer

(D) \(\left( {4,3\sqrt 3 } \right)\)

Step-by-step Solution

Detailed explanation

\(f(x)=x \sqrt{k x-x^{2}}\) \(f^{\prime}(x)=\frac{3 k x-4 x^{2}}{2 \sqrt{k x-x^{2}}}\) For increasing \(f(x) \geq 0\) \(k x-x^{2} \geq 0\) \(x^{2}-k x \leq 0\) \(x(x-k) \leq 0\) so \(x \in[0,3)\) +ve \(\boxed{x \geqslant 3}\) minimum value of \(k\) is \(m=4\)…