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 :
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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