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JEE 2023 — Mathematics PYQ

JEE | Mathematics | 2023

\text{The area of the region } \{(x, y) : x^2 \le y \le |x^2 - 4|, y \ge 1\} \text{ is:}

Choose the correct answer:

Explanation

Solution:

1. Curves ki pehchan:

Hame teen conditions di gayi hain:

$y \ge x^2$ (Parabola ke upar ka bhag)

$y \le |x^2 - 4|$ (Isme do case hain: jab $|x| \ge 2$ , tab $y \le x^2 - 4$ aur jab $|x| < 2$ , tab $y \le 4 - x^2$ )

$y \ge 1$ (Horizontal line ke upar ka bhag)

Kyunki $x^2 \le y$ aur $y \le |x^2 - 4|$ , to $x^2 \le |x^2 - 4|$ hona chahiye.

Yadi $x^2 \le x^2 - 4$ , to $0 \le -4$ jo ki galat hai. Iska matlab region sirf $|x| < 2$ ke beech hai, jahan $|x^2 - 4| = 4 - x^2$ hota hai.

2. Intersection Points nikalna:

Symmetry ki wajah se hum $x > 0$ ke liye area nikal kar use $2$ se multiply kar denge.

$x^2 = 4 - x^2 \implies 2x^2 = 4 \implies x^2 = 2 \implies x = \sqrt{2}$ (Curve intersection)

$x^2 = 1 \implies x = 1$ (Parabola aur $y=1$ ka intersection)

$4 - x^2 = 1 \implies x^2 = 3 \implies x = \sqrt{3}$ (Lekin hamara curve $\sqrt{2}$ tak hi bounded hai)

3. Integration setup:

Region $x = 0$ se $x = \sqrt{2}$ tak faila hai, jise do hisson mein banta ja sakta hai:

$x \in [0, 1]$ : Yahan lower boundary $y=1$ hai aur upper $y = 4 - x^2$ .

$x \in [1, \sqrt{2}]$ : Yahan lower boundary $y=x^2$ hai aur upper $y = 4 - x^2$ .

$\text{Total Area} = 2 \left[ \int_{0}^{1} ( (4 - x^2) - 1 ) \, dx + \int_{1}^{\sqrt{2}} ( (4 - x^2) - x^2 ) \, dx \right]$

4. Calculation:

$= 2 \left[ \int_{0}^{1} (3 - x^2) \, dx + \int_{1}^{\sqrt{2}} (4 - 2x^2) \, dx \right]$

Pehla part:

$\left[ 3x - \frac{x^3}{3} \right]_0^1 = 3 - \frac{1}{3} = \frac{8}{3}$

Dusra part:

$\left[ 4x - \frac{2x^3}{3} \right]_1^{\sqrt{2}} = \left( 4\sqrt{2} - \frac{2(\sqrt{2})^3}{3} \right) - \left( 4 - \frac{2}{3} \right)$

$= \left( 4\sqrt{2} - \frac{4\sqrt{2}}{3} \right) - \frac{10}{3} = \frac{8\sqrt{2}}{3} - \frac{10}{3}$

Donon ko jod kar:

$\text{Area} = 2 \left[ \frac{8}{3} + \frac{8\sqrt{2}}{3} - \frac{10}{3} \right]$

$= 2 \left[ \frac{8\sqrt{2} - 2}{3} \right] = \frac{16\sqrt{2} - 4}{3}$

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

Correct Option:

Sahi jawab (2) hai.

$\text{Area} = \frac{4}{3}(4\sqrt{2} - 1)$

JEE
The area of the region enclosed by the curve

Choose the correct answer:

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