JEE Main 2022 (Online) 24th June Morning ShiftQuestion 1
Let $f(\theta)=\sin\theta+\int_{-\pi/2}^{\pi/2}(\sin\theta+t\cos\theta)f(t)\,dt$. Then the value of $\left|\int_{0}^{\pi/2}f(\theta)\,d\theta\right|$ is $\_\_\_\_\_\_$ .
Definite Integration
Answer$1$
Solution
Concept. $f(\theta)=(1+A)\sin\theta+B\cos\theta$ with constants $A=\int f$, $B=\int tf$. Odd/even parts give two equations. Formulas used. (1) $A=\int_{-\pi/2}^{\pi/2}f(t)\,dt$, $B=\int_{-\pi/2}^{\pi/2}tf(t)\,dt$(2) On $[-\frac\pi2,\frac\pi2]$: $\int\sin t=0$, $\int\cos t=2$, $\int t\sin t=2$, $\int t\cos t=0$Step 1 - Equations. Put $f(t)=(1+A)\sin t+B\cos t$ in (1). By (2), $$A=2B,\qquad B=2(1+A).$$ Step 2 - Solve. $B=2+4B$, so $B=-\frac23$, $A=-\frac43$. Then $$f(\theta)=-\frac13\sin\theta-\frac23\cos\theta.$$ Step 3 - Integrate. $\int_0^{\pi/2}f=-\frac13-\frac23=-1$, absolute value $1$. Answer: 1