JEE 2023 — Mathematics PYQ

Solving:

Step 1: Intersection points nikalna

Circle $x^2+y^2=21$ aur Parabola $y^2=4x$ ka intersection:

$x^2+4x-21=0$

$(x+7)(x-3)=0$

$x=3$ (Kyunki $x\ge 1$)

Jab $x=3$, tab $y^2=12⟹y=\pm 2\sqrt{3}$.

Step 2: Area ($\Delta$) ki calculation

Area x-axis ke upar aur niche symmetric hai, isliye:

$\Delta =2\left[{\int }_1^3\sqrt{4x}dx+{\int }_3^{\sqrt{21}}\sqrt{21-x^2}dx\right]$

Pehla part:

$2{\int }_1^{3}2\sqrt{x}dx=4{\left[\frac{x^{3/2}}{3/2}\right]}_1^3=\frac{8}{3}(3\sqrt{3}-1)=8\sqrt{3}-\frac{8}{3}$

Doosra part:

$2{\int }_3^{\sqrt{21}}\sqrt{21-x^2}dx=2{\left[\frac{x}{2}\sqrt{21-x^2}+\frac{21}{2}{\sin }^{-1}\frac{x}{\sqrt{21}}\right]}_3^{\sqrt{21}}$

$={\left[x\sqrt{21-x^2}+21{\sin }^{-1}\frac{x}{\sqrt{21}}\right]}_3^{\sqrt{21}}$

$=(0+21\cdot \frac{\pi }{2})-(3\sqrt{12}+21{\sin }^{-1}\frac{3}{\sqrt{21}})$

$=\frac{21\pi }{2}-6\sqrt{3}-21{\sin }^{-1}\sqrt{\frac{3}{7}}$

$\Delta =(8\sqrt{3}-\frac{8}{3})+(\frac{21\pi }{2}-6\sqrt{3}-21{\sin }^{-1}\sqrt{\frac{3}{7}})$

$\Delta =2\sqrt{3}-\frac{8}{3}+21(\frac{\pi }{2}-{\sin }^{-1}\sqrt{\frac{3}{7}})$

Kyunki $\frac{\pi }{2}-{\sin }^{-1}\theta ={\cos }^{-1}\theta$:

$\Delta =2\sqrt{3}-\frac{8}{3}+21{\cos }^{-1}\sqrt{\frac{3}{7}}$

Kyunki ${\cos }^{-1}\sqrt{\frac{3}{7}}={\sin }^{-1}\frac{2}{\sqrt{7}}$ (Triangle rule se):

$\Delta =2\sqrt{3}-\frac{8}{3}+21{\sin }^{-1}\frac{2}{\sqrt{7}}$

Step 3: Final value nikalna

Hame chahiye: $\frac{1}{2}\left(\Delta -21{\sin }^{-1}\frac{2}{\sqrt{7}}\right)$

$=\frac{1}{2}\left(2\sqrt{3}-\frac{8}{3}+21{\sin }^{-1}\frac{2}{\sqrt{7}}-21{\sin }^{-1}\frac{2}{\sqrt{7}}\right)$

$=\frac{1}{2}\left(2\sqrt{3}-\frac{8}{3}\right)$

$=\sqrt{3}-\frac{4}{3}$

Sahi Option: (2)