JEE 2023 — Mathematics PYQ

Humein equation di gayi hai:

Humein pata hai ki $\sin (x+y)=\sin x\cos y+\cos x\sin y$. Isse integral mein rakhte hain:

Maan lijiye:

$C_1={\int }_0^{\pi /2}\cos yf(y)dy$ aur $C_2={\int }_0^{\pi /2}\sin yf(y)dy$.

Toh function ban gaya: $f(x)=x+C_1\sin x+C_2\cos x$.

2. $C_1$ aur $C_2$ ki values nikalna

Ab hum $f(x)$ ki is value ko $C_1$ aur $C_2$ ki definitions mein wapas rakhenge.

$C_1$ ke liye:

Integration solve karne par:

• $\int y\cos y=[y\sin y+\cos y{]}_0^{\pi /2}=(\frac{\pi }{2}-1)$

• $\int \sin y\cos y=[\frac{{\sin }^{2}y}{2}{]}_0^{\pi /2}=\frac{1}{2}$

• $\int {\cos }^{2}y=[\frac{y}{2}+\frac{\sin 2y}{4}{]}_0^{\pi /2}=\frac{\pi }{4}$

Toh equation bani: $C_1=(\frac{\pi }{2}-1)+\frac{C_1}{2}+\frac{C_2\pi }{4}⟹\frac{C_1}{2}-\frac{C_2\pi }{4}=\frac{\pi -2}{2}$ --- (Eq. i)

$C_2$ ke liye:

Isi tarah solve karne par humein dusri equation milegi:

$\frac{C_2}{2}-\frac{C_1\pi }{4}=1$ --- (Eq. ii)

3. $(a+b)$ ki value nikalna

Sawaal mein function diya hai: $f(x)=x+\frac{a}{{\pi }^2-4}\sin x+\frac{b}{{\pi }^2-4}\cos x$.

Iska matlab:

• $C_1=\frac{a}{{\pi }^2-4}⟹a=C_1({\pi }^2-4)$

• $C_2=\frac{b}{{\pi }^2-4}⟹b=C_2({\pi }^2-4)$

  Isliye: $a+b=(C_1+C_2)({\pi }^2-4)$

Equations (i) aur (ii) ko add karne par:

$(\frac{C_1}{2}+\frac{C_2}{2})-\frac{\pi }{4}(C_1+C_2)=\frac{\pi -2}{2}+1$

$\frac{C_1+C_2}{2}(1-\frac{\pi }{2})=\frac{\pi }{2}$

$(C_1+C_2)\frac{(2-\pi )}{4}=\frac{\pi }{2}⟹(C_1+C_2)=\frac{2\pi }{2-\pi }$

Ab value put karte hain:

$a+b=\left(\frac{2\pi }{2-\pi }\right)({\pi }^2-4)$

$a+b=\left(\frac{2\pi }{2-\pi }\right)(\pi -2)(\pi +2)$

Kyunki $(\pi -2)=-(2-\pi )$, toh:

$a+b=-2\pi (\pi +2)$

Sahi vikalp (Correct option) hai: (2) $-2\pi (\pi +2)$