SATHEE: Area Question 22

Area Question 22

Question 22

If the straight line $x = b$ divide the area enclosed by $y = (1 - x)^{2}, y = 0$ and $x = 0$ into two parts $R_{1} (0 \leq x \leq b)$ and $R_{2} (b \leq x \leq 1)$ such that $R_{1} - R_{2} = \frac{1}{4}$ . Then, $b$ equals (a) $\frac{3}{4}$ (b) $\frac{1}{2}$ (c) $\frac{1}{3}$ (d) $\frac{1}{4}$

(2011)

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Solution:

Here, area between 0 to $b$ is $R_{1}$ and $b$ to 1 is $R_{2}$ .

$∴ \int_{0}^{b} (1 - x)^{2} d x - \int_{b}^{1} (1 - x)^{2} d x = \frac{1}{4}$

$$ \Rightarrow \quad \frac{(1-x)^{3}}{-3}{ }{0}^{b}-{\frac{(1-x)^{3}}{-3}}{b}^{1}=\frac{1}{4} $$

$\Rightarrow - \frac{1}{3} [(1 - b)^{3} - 1] + \frac{1}{3} [0 - (1 - b)^{3}] = \frac{1}{4}$

$\Rightarrow - \frac{2}{3} (1 - b)^{3} = - \frac{1}{3} + \frac{1}{4} = - \frac{1}{12} \Rightarrow (1 - b)^{3} = \frac{1}{8}$

$\Rightarrow (1 - b) = \frac{1}{2} \Rightarrow b = \frac{1}{2}$

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Area Question 22

Question 22

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