JEE Main 2021 — Mathematics Calculus

Given the function $\displaystyle f(x)=x+{\int }_{0}^{\frac{\pi }{2}}\mathrm{sin}x\cdot \mathrm{cos}yf(y)dy$f(x)=x+∫02π​​sinx⋅cosyf(y)dy

$$f(x)=x+k\mathrm{sin}x...(i)$$f(x)=x+ksinx...(i)

$$f(y)=y+k\mathrm{sin}y...(ii)$$f(y)=y+ksiny...(ii)

$$k={\int }_{0}^{\frac{\pi }{2}}\mathrm{cos}yf(y)dy$$k=∫02π​​cosyf(y)dy

From equations (i)&(ii)

$$f(x)=x+{\int }_{0}^{\frac{\pi }{2}}\mathrm{sin}x\mathrm{cos}y(y+k\mathrm{sin}y)dy$$f(x)=x+∫02π​​sinxcosy(y+ksiny)dy

$$f(x)=x+\mathrm{sin}x{\int }_{0}^{\frac{\pi }{2}}y\mathrm{cos}ydy+\frac{k}{2}\mathrm{sin}x{\int }_{0}^{\frac{\pi }{2}}\mathrm{sin}2ydy$$f(x)=x+sinx∫02π​​ycosydy+2k​sinx∫02π​​sin2ydy

$$f(x)=x+\mathrm{sin}x[y{\int }_{0}^{\frac{\pi }{2}}\mathrm{cos}ydy-{\int }_{0}^{\frac{\pi }{2}}{\frac{d}{dy}y{\int }_{0}^{\frac{\pi }{2}}\mathrm{cos}ydy}dy]+\frac{k}{2}\mathrm{sin}x{\int }_{0}^{\frac{\pi }{2}}\mathrm{sin}2ydy$$f(x)=x+sinx[y∫02π​​cosydy−∫02π​​dyd​y∫02π​​cosydydy]+2k​sinx∫02π​​sin2ydy

$$f(x)=x+\mathrm{sin}x[\frac{\pi }{2}-1]+\frac{k}{2}\mathrm{sin}x\times (1)$$f(x)=x+sinx[2π​−1]+2k​sinx×(1)

$$f(x)=x+\frac{(\pi -2)\mathrm{sin}x}{2}+\frac{k\mathrm{sin}x}{2}...(iii)$$f(x)=x+2(π−2)sinx​+2ksinx​...(iii)

from $(i)$(i) and $(iii)$(iii)

$$k=\frac{\pi }{2}-1+\frac{k}{2}$$k=2π​−1+2k​

$$k=\pi -2$$k=π−2

Hence, $f(x)=x+(\pi -2)\mathrm{sin}x$f(x)=x+(π−2)sinx