If m is the minimum value of

Question:

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

  1. $(4,3 \sqrt{2})$

  2. $(4,3 \sqrt{3})$

  3. $(3,3 \sqrt{3})$

  4. $(5,3 \sqrt{6})$


Correct Option: , 2

Solution:

$f(x)=x \sqrt{k x-x^{2}}$

$f^{\prime}(x)=\frac{3 k x-4 x^{2}}{2 \sqrt{k x-x^{2}}}$

$f(x)=x \sqrt{k x-x^{2}}$

$=3 \sqrt{4 \times 3-3^{2}}=3 \sqrt{3}, M=3 \sqrt{3}$

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