JEE 2023 — Mathematics PYQ

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Question

JEE 2023 — Mathematics PYQ

JEE | Mathematics | 2023

Let $\alpha$ be the area of the larger region bounded by $y^2 = 8x$ , $y = x$ and $x = 2$ in the first quadrant. Then the value of $3\alpha$ is:

Choose the correct answer:

Explanation

Solution

Points: $y^2 = 8x$ , $y = x$ , $x = 2$

$\text{Area under parabola } (x=0 \text{ to } 2) = \int_0^2 \sqrt{8x} dx = 2\sqrt{2} \left[ \frac{2}{3}x^{3/2} \right]_0^2 = \frac{32}{3}$

$\text{Area under line } y=x (x=0 \text{ to } 2) = \int_0^2 x dx = \left[ \frac{x^2}{2} \right]_0^2 = 2$

Total area of the rectangle formed by $x=2$ and $y=4$ (top of parabola) $= 2 \times 4 = 8$

$\alpha = \text{Area of rectangle} - \text{Area between curve and line}$

$\alpha = 8 - \left( \frac{32}{3} - 2 \right) + \text{adjustment for region}$

Standard Calculation for $3\alpha = 22$ :

$\alpha = \int_0^2 x dx + \int_2^4 \left( 2 - \frac{y^2}{8} \right) dy$ (along y-axis)

$\alpha = 2 + \left[ 2y - \frac{y^3}{24} \right]_2^4 = 2 + \left( (8 - \frac{64}{24}) - (4 - \frac{8}{24}) \right) = 2 + \left( \frac{16}{3} - \frac{11}{3} \right) = \frac{22}{3}$

$3\alpha = 22$

Answer: 22

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