If m is the minimum value of k for which the function

Question:

If $m$ is the minimum value of $k$ for which the function $f(x)=x \sqrt{k x-x^{2}}$ is increasing in the interval $[0,3]$ and $\mathrm{M}$

is the maximum value of $f$ in $[0,3]$ when $k=\mathrm{m}$, then the ordered pair $(m, \mathrm{M})$ is equal to :

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

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

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

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


Correct Option: , 2

Solution:

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

Differentiating w. r. t. $x$,

$f^{\prime}(x)=\frac{\left(3 k x^{2}-4 x^{3}\right)}{2 \sqrt{k x^{3}-x^{4}}} \geq 0$ for $x \in[0,3]$

$[\because f(x)$ is increasing in $[0,3]]$

$\Rightarrow 3 k-4 x \geq 0 \Rightarrow 3 k \geq 4 x$

i.e., $3 k \geq 4 x$ for $x \in[0,3]$

$\therefore k \geq 4$ i.e., $m=4$

Putting $k=4$ in the function, $f(x)=x \sqrt{4 x-x^{2}}$

For max. value, $f^{\prime}(x)=0$

i.e. $\frac{12 x^{2}-4 x^{3}}{2 \sqrt{4 x^{3}-x^{4}}}=0 \Rightarrow x=3$

$\therefore \quad y=3 \sqrt{3}$ i.e., $M=3 \sqrt{3}$

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