JEE Main 2023 (Online) 29th January Morning ShiftQuestion 1
Let $f(x)=x+\frac{a}{\pi^{2}-4}\sin x+\frac{b}{\pi^{2}-4}\cos x$, $x\in R$ be a function which satisfies $f(x)=x+\int_{0}^{\pi/2}\sin(x+y)f(y)\,dy$. Then $(a+b)$ is equal to
Definite Integration
Solution
Concept. $\sin(x+y)=\sin x\cos y+\cos x\sin y$, so $f(x)=x+p\sin x+q\cos x$ where $p,q$ are two definite integrals. Putting this form back gives two linear equations in $p,q$. Formulas used. (1) $p=\int_0^{\pi/2}\cos y\,f(y)\,dy$, $q=\int_0^{\pi/2}\sin y\,f(y)\,dy$(2) $\int_0^{\pi/2}y\cos y\,dy=\frac\pi2-1$, $\int_0^{\pi/2}y\sin y\,dy=1$, $\int_0^{\pi/2}\sin y\cos y\,dy=\frac12$, $\int_0^{\pi/2}\cos^2y\,dy=\int_0^{\pi/2}\sin^2y\,dy=\frac\pi4$Step 1 - Equation for $p$. Put $f(y)=y+p\sin y+q\cos y$ in (1). By (2), $$p=\frac\pi2-1+\frac p2+\frac{\pi q}4\ \Longrightarrow\ 2p-\pi q=2\pi-4.$$ Step 2 - Equation for $q$. $$q=1+\frac{\pi p}4+\frac q2\ \Longrightarrow\ 2q-\pi p=4.$$ Step 3 - Solve. From Step 1, $p=\pi-2+\frac{\pi q}2$. Put in Step 2: $4q-\pi(2\pi-4)-\pi^2q=8$, so $$q=-\frac{2\pi^2-4\pi+8}{\pi^2-4},\qquad p=\pi-2+\frac{\pi q}2=\frac{8-8\pi}{\pi^2-4}.$$ Step 4 - Compare. $p=\frac{a}{\pi^2-4}$, $q=\frac{b}{\pi^2-4}$, so $a=8-8\pi$, $b=-2\pi^2+4\pi-8$: $$a+b=-2\pi^2-4\pi=-2\pi(\pi+2).$$ Answer: (B)