Application Of Derivatives Ques 83
83. 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: 83.(b)
Solution:
Formula:
Maxima and Minima of functions of one variable :
- 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] ……(ii) \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 $(m, M)=(4,3 \sqrt{3})$
BETA
