SATHEE: Application of Derivatives 4 Question 3

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

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

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Answer:

Correct Answer: 3. (b)

Solution:

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

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

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

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

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

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

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

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

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

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

Application of Derivatives 4 Question 3

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Application of Derivatives 4 Question 3

Answer:

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

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