SATHEE: Area Question 39

Area Question 39

Question 39

Find all the possible values of $b > 0$ , so that the area of the bounded region enclosed between the parabolas $y = x - b x^{2}$ and $y = \frac{x^{2}}{b}$ is maximum.

(1997C, 5M)

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Solution:

Eliminating $y$ from $y = \frac{x^{2}}{b}$ and $y = x - b x^{2}$ , we get

$\begin{aligned} x^{2} & = b x - b^{2} x^{2} \Rightarrow x & = 0, \frac{b}{1 + b^{2}} \end{aligned}$

Thus, the area enclosed between the parabolas

$\begin{aligned} A & = \int_{0}^{b / (1 + b)^{2}} x - b x^{2} - \frac{x^{2}}{b} d x & = \frac{x^{2}}{2} - \frac{x^{3}}{3} \cdot {\frac{1 + b^{2}}{b}}_{0}^{b / (1 + b)^{2}} = \frac{1}{6} \cdot \frac{b^{2}}{{(1 + b^{2})}^{2}} \end{aligned}$

On differentiating w.r.t. $b$ , we get

$\begin{aligned} \frac{d A}{d b} & = \frac{1}{6} \cdot \frac{{(1 + b^{2})}^{2} \cdot 2 b - 2 b^{2} \cdot (1 + b^{2}) \cdot 2 b}{{(1 + b^{2})}^{4}} & = \frac{1}{3} \cdot \frac{b (1 - b^{2})}{{(1 + b^{2})}^{3}} \end{aligned}$

For maximum value of $A$ , put $\frac{d A}{d b} = 0$

$\Rightarrow b = - 1, 0, 1, since b > 0$

$∴$ We consider only $b = 1$ .

Sign scheme for $\frac{d A}{d b}$ around $b = 1$ is as shown below :

From sign scheme, it is clear that $A$ is maximum at $b = 1$ .

Area Question 39

Question 39

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Area Question 39

Question 39

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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