JEE Main 2022 — Mathematics Calculus

Given $\displaystyle f(\theta )=\mathrm{sin}\theta +{\int }_{-\frac{\pi }{2}}^{\frac{\pi }{2}}(\mathrm{sin}\theta +t\mathrm{cos}\theta )f(t)dt$f(θ)=sinθ+∫−2π​2π​​(sinθ+tcosθ)f(t)dt

$$f(\theta )=\mathrm{sin}\theta +\mathrm{sin}\theta {\int }_{-\frac{\pi }{2}}^{\frac{\pi }{2}}f(t)dt+\mathrm{cos}\theta {\int }_{-\frac{\pi }{2}}^{\frac{\pi }{2}}tf(t)dt$$f(θ)=sinθ+sinθ∫−2π​2π​​f(t)dt+cosθ∫−2π​2π​​tf(t)dt

Let $\displaystyle A={\int }_{-\frac{\pi }{2}}^{\frac{\pi }{2}}f(t)dt,B={\int }_{-\frac{\pi }{2}}^{\frac{\pi }{2}}tf(t)dt$A=∫−2π​2π​​f(t)dt,B=∫−2π​2π​​tf(t)dt

So $f(\theta )=\mathrm{sin}\theta +A\mathrm{sin}\theta +B\mathrm{cos}\theta$f(θ)=sinθ+Asinθ+Bcosθ

i.e. $f(\theta )=(A+1)\mathrm{sin}\theta +B\mathrm{cos}\theta$f(θ)=(A+1)sinθ+Bcosθ

$$A={\int }_{-\frac{\pi }{2}}^{\frac{\pi }{2}}[(A+1)\mathrm{sin}t+B\mathrm{cos}t]dt$$A=∫−2π​2π​​[(A+1)sint+Bcost]dt

$$\Rightarrow A=(A+1){\int }_{-\frac{\pi }{2}}^{\frac{\pi }{2}}\mathrm{sin}tdt+B{\int }_{-\frac{\pi }{2}}^{\frac{\pi }{2}}\mathrm{cos}tdt$$⇒A=(A+1)∫−2π​2π​​sintdt+B∫−2π​2π​​costdt

$$\Rightarrow A=2B\ldots (1)$$⇒A=2B…(1)

$$B={\int }_{-\frac{\pi }{2}}^{\frac{\pi }{2}}t((A+1)\mathrm{sin}t+B\mathrm{cos}t)dt$$B=∫−2π​2π​​t((A+1)sint+Bcost)dt

$$\Rightarrow B={\int }_{-\frac{\pi }{2}}^{\frac{\pi }{2}}t(A+1)\mathrm{sin}tdt$$⇒B=∫−2π​2π​​t(A+1)sintdt

$$\Rightarrow B=(A+1)2{\int }_{0}^{\frac{\pi }{2}}t\mathrm{sin}tdt$$⇒B=(A+1)2∫02π​​tsintdt

$$\Rightarrow B=(A+1)2$$⇒B=(A+1)2

$$\Rightarrow 2A+2-B=0\cdots (2)$$⇒2A+2−B=0⋯(2)

After solving

$$B=-\frac{2}{3},A=-\frac{4}{3}$$B=−32​,A=−34​

$$|{\int }_{0}^{\frac{\pi }{2}}f(\theta )d\theta |=|{\int }_{0}^{\frac{\pi }{2}}(-\frac{1}{3}\mathrm{sin}\theta -\frac{2}{3}\mathrm{cos}\theta )d\theta |$$∣∫02π​​f(θ)dθ∣=∣∫02π​​(−31​sinθ−32​cosθ)dθ∣

$$=|-\frac{1}{3}{\int }_{0}^{\frac{\pi }{2}}\mathrm{sin}\theta d\theta -\frac{2}{3}{\int }_{0}^{\frac{\pi }{2}}\mathrm{cos}\theta d\theta |=1$$=∣−31​∫02π​​sinθdθ−32​∫02π​​cosθdθ∣=1