SATHEE: Area Question 2

Area Question 2

Question 2

If the area (in sq units) of the region $Missing or unrecognized delimiter for \left$ is $a \sqrt{2} + b$ , then $a - b$ is equal to

(2019 Main, 12 April I) (a) $\frac{10}{3}$ (b) 6 (c) $\frac{8}{3}$ (d) $- \frac{2}{3}$

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Solution:

Given region is $Missing or unrecognized delimiter for \left$

Now, for point $P$ , put value of $y = 1 - x$ to $y^{2} = 4 x$ , we get $(1 - x)^{2} = 4 x \Rightarrow x^{2} + 1 - 2 x = 4 x$

$\begin{aligned} \Rightarrow & x^{2} - 6 x + 1 = 0 \Rightarrow x & = \frac{6 \pm \sqrt{36 - 4}}{2} & = 3 \pm 2 \sqrt{2} . \end{aligned}$

Since, $x$ -coordinate of $P$ less than $x$ -coordinate of point $A (1, 0)$ .

$∴ x = 3 - 2 \sqrt{2}$

Now, required area

$= \int_{0}^{3 - 2 \sqrt{2}} 2 \sqrt{x} d x + \int_{3 - 2 \sqrt{2}}^{1} (1 - x) d x$

$=2\left|\frac{x^{3 / 2}}{3 / 2}\right|{0}^{3-2 \sqrt{2}}+x-{\frac{x^{2}}{2}}{3-2 \sqrt{2}}$

$= \frac{4}{3} (3 - 2 \sqrt{2})^{32} + 1 - \frac{1}{2} - (3 - 2 \sqrt{2}) + \frac{(3 - 2 \sqrt{2})^{2}}{2}$

$= \frac{4}{3} {[(\sqrt{2} - 1)^{2}]}^{3 / 2} + \frac{1}{2} - 3 + 2 \sqrt{2} + \frac{1}{2} (9 + 8 - 12 \sqrt{2})$

$= \frac{4}{3} (\sqrt{2} - 1)^{3} - \frac{5}{2} + 2 \sqrt{2} + \frac{17}{2} - 6 \sqrt{2}$

$= \frac{4}{3} (2 \sqrt{2} - 3 (2) + 3 (\sqrt{2}) - 1) - 4 \sqrt{2} + 6$

$= \frac{4}{3} (5 \sqrt{2} - 7) - 4 \sqrt{2} + 6 = \frac{8 \sqrt{2}}{3} - \frac{10}{3}$

$= a \sqrt{2} + b$

So, on comparing $a = \frac{8}{3}$ and $b = - \frac{10}{3}$

(given)

$∴ a - b = \frac{8}{3} + \frac{10}{3} = 6$

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