JEE 2023 — Mathematics PYQ

Pehle hum dekhte hain ki kab $\sin x$ bada hai aur kab $\cos x$ bada hai. Interval $[-\pi ,\pi ]$ mein dono curves $x=-3\pi /4$ aur $x=\pi /4$ par milte hain.

Step 2: Area calculate karein

Hamein total area nikalna hai, jo ki ${\int }_{-\pi }^{\pi }|f(x)|dx$ hota hai. Curve ki values $x$-axis ke upar aur neeche check karke integral ko break karte hain:

1. Interval $[-\pi ,-3\pi /4]$: Yahan $f(x)=\sin x$ hai aur iski value negative hai.

   $${\text{Area}}_1={\int }_{-\pi }^{-3\pi /4}(-\sin x)dx=[\cos x{]}_{-\pi }^{-3\pi /4}=\cos (-3\pi /4)-\cos (-\pi )=-\frac{1}{\sqrt{2}}-(-1)=1-\frac{1}{\sqrt{2}}$$

2. Interval $[-3\pi /4,-\pi /2]$: Yahan $f(x)=\cos x$ hai aur iski value negative hai.

   $${\text{Area}}_2={\int }_{-3\pi /4}^{-\pi /2}(-\cos x)dx=[-\sin x{]}_{-3\pi /4}^{-\pi /2}=-\sin (-\pi /2)-(-\sin (-3\pi /4))=1-\frac{1}{\sqrt{2}}$$

3. Interval $[-\pi /2,\pi /4]$: Yahan $f(x)=\cos x$ hai aur iski value positive hai.

   $${\text{Area}}_3={\int }_{-\pi /2}^{\pi /4}\cos xdx=[\sin x{]}_{-\pi /2}^{\pi /4}=\sin (\pi /4)-\sin (-\pi /2)=\frac{1}{\sqrt{2}}-(-1)=1+\frac{1}{\sqrt{2}}$$

4. Interval $[\pi /4,\pi ]$: Yahan $f(x)=\sin x$ hai aur iski value positive hai.

   $${\text{Area}}_4={\int }_{\pi /4}^{\pi }\sin xdx=[-\cos x{]}_{\pi /4}^{\pi }=-\cos (\pi )-(-\cos (\pi /4))=1+\frac{1}{\sqrt{2}}$$

Step 3: Total Area

Sabhi parts ko jodne par: