JEE 2023 — Mathematics PYQ

• Pehli curve hai $y=\sqrt{4-(x-1{)}^2}$. Dono side square karne par:

  $y^2=4-(x-1{)}^2⟹(x-1{)}^2+y^2=2^2$.

  Yeh ek circle ki equation hai jiska center $(1,0)$ aur radius $2$ hai. Kyunki $y=\sqrt{\dots }$ positive hai, toh yeh sirf upper semi-circle hai. Yeh $x$-axis ko $x=1-2=-1$ aur $x=1+2=3$ par touch karta hai.

• Doosri curve hai ek seedhi rekha (straight line): $y=2x$.

2. Intersection Point nikalen:

Dono equations ko barabar rakhen:

$2x=\sqrt{4-(x-1{)}^2}$

$4x^2=4-(x^2-2x+1)$

$5x^2-2x-3=0$

Solving this: $(5x+3)(x-1)=0⟹x=1$ (Kyunki $y\ge 0$ hai, isliye $x$ positive hona chahiye line ke liye).

Toh intersection point $(1,2)$ hai.

3. Area of Region A nikalen:

Region A ki condition hai: $y\ge 0$ aur $2x\le y\le \sqrt{4-(x-1{)}^2}$.

• Jab $x\in [-1,0]$ hai, toh $2x$ negative hai aur $y\ge 0$ hai, isliye yahan boundary $0\le y\le \text{circle}$ hogi.

• Jab $x\in [0,1]$ hai, toh $2x\le y\le \text{circle}$ hai.

• $\text{Area}(A)={\int }_{-1}^0\sqrt{4-(x-1{)}^2}dx+{\int }_0^1(\sqrt{4-(x-1{)}^2}-2x)dx$

• $\text{Area}(A)={\int }_{-1}^1\sqrt{4-(x-1{)}^2}dx-{\int }_0^{1}2xdx$

  ${\int }_{-1}^1\sqrt{4-(x-1{)}^2}dx$ ek quadrant (chauthaai hissa) ka area hai: $\frac{1}{4}\pi (2{)}^2=\pi$.

  ${\int }_0^{1}2xdx=[x^2{]}_0^1=1$.

  Isliye, $\text{Area}(A)=\pi -1$.

4. Area of Region B nikalen:

Region B ki condition hai: $0\le y\le \min \{2x,\sqrt{4-(x-1{)}^2}\}$.

• Jab $x\in [0,1]$ hai, toh $\min (2x,\text{circle})=2x$.

• Jab $x\in [1,3]$ hai, toh $\min (2x,\text{circle})=\text{circle}$.

• $\text{Area}(B)={\int }_0^{1}2xdx+{\int }_1^3\sqrt{4-(x-1{)}^2}dx$

  Yahan ${\int }_1^3\sqrt{4-(x-1{)}^2}dx$ bhi ek quadrant ka area hai: $\frac{1}{4}\pi (2{)}^2=\pi$.

  Isliye, $\text{Area}(B)=1+\pi$.

5. Ratio nikalen:

$\text{Ratio}=\frac{\text{Area}(A)}{\text{Area}(B)}=\frac{\pi -1}{\pi +1}$

Sahi vikalp (Correct option) hai: (3) $\frac{\pi -1}{\pi +1}$