Application of Derivatives 4 Question 3

####3. 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 $M$ is the maximum value of $f$ in the interval $[0,3]$ when $k=m$, then the ordered pair $(m, M)$ is equal to

(2019 Main, 12 April I)

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

(b) $(4,3 \sqrt{3})$

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

(d) $(5,3 \sqrt{6})$

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Answer:

Correct Answer: 3. (b)

Solution:

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

the function $f(x)$ is defined if $k x-x^{2} \geq 0$

$ \begin{array}{rrr} \Rightarrow & x^{2}-k x & \leq 0 \\ \Rightarrow & x & \in[0, k] \end{array} $

because it is given that $f(x)$ is increasing in interval $x \in[0,3]$, so $k$ should be positive.

Now, on differentiating the function $f(x)$ w.r.t. $x$, we get

$ \begin{aligned} f^{\prime}(x) & =\sqrt{k x-x^{2}}+\frac{x}{2 \sqrt{k x-x^{2}}} \times(k-2 x) \\ & =\frac{2\left(k x-x^{2}\right)+k x-2 x^{2}}{2 \sqrt{k x-x^{2}}}=\frac{3 k x-4 x^{2}}{2 \sqrt{k x-x^{2}}} \end{aligned} $

as $f(x)$ is increasing in interval $x \in[0,3]$, so

$ \begin{aligned} & f^{\prime}(x) \geq 0 \forall x \in(0,3) \\ & \Rightarrow \quad 3 k x-4 x^{2} \geq 0 \\ & \Rightarrow \quad 4 x^{2}-3 k x \leq 0 \\ & \Rightarrow 4 x x-\frac{3 k}{4} \leq 0 \Rightarrow x \in 0, \frac{3 k}{4} \text { (as } k \text { is positive) } \end{aligned} $

So, $\quad 3 \leq \frac{3 k}{4} \Rightarrow k \geq 4$

$\Rightarrow$ Minimum value of $k=m=4$

and the maximum value of $f$ in $[0,3]$ is $f(3)$.

$\because f$ is increasing function in interval $x \in[0,3]$

$\because M=f(3)=3 \sqrt{4 \times 3-3^{2}}=3 \sqrt{3}$

Therefore, ordered pair $(m, M)=(4,3 \sqrt{3})$



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