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

Option 1)
$\left ( 4,3\sqrt{3} \right )$

Option 2)
$\left ( 3,3\sqrt{3} \right )$

Option 3)
$\left ( 5,3\sqrt{6} \right )$

Option 4)
$\left ( 4,3\sqrt{2} \right )$

Answers (1)

best_answer

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

$f'(x)=3kx-4x^{2}\cdot \frac{1}{2\sqrt{k-x^{2}}}$

For $f'(x)\geqslant 0$

$kx-x^{2}\geqslant 0$

$x^{2}-kx\leq 0$

$x\left ( x-k \right )\leq 0$ so $x\equiv \left [ 0,3 \right ]$

$+ve\: \: \: \: \: x\geqslant 3$

& $3kx-4x^{2}\geqslant 0$

$4x^{2}-3kx\leq 0$

$4x\left ( x-\frac{3k}{4} \right )\leq 0$

$x-\frac{3k}{4}\leq 0$ minimum value of $x=3$

$3-\frac{3k}{4}\leq 0$

$k\geq 4$

minimum value of k is $m=4$

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

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

$=3\sqrt{3},\: \: \: \: \: \: M=3\sqrt{3}$

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

Option 1)

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

Option 2)

$\left ( 3,3\sqrt{3} \right )$

Option 3)

$\left ( 5,3\sqrt{6} \right )$

Option 4)

$\left ( 4,3\sqrt{2} \right )$

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