The function f(x), that satisfies the condition f(x)=x+π/2∫0sinx.cosyf(y)dy is:
(A) x+23(π–2)sinx
(B) x+(π+2)sinx
(C) x+π2sinx
(D) x+(π–2)sinx
Solution
We have, f(x)=x+π/2∫0sinx.cosyf(y)dy=x+sinxπ/2∫0cosyf(y)dy=x+sinx.k
Where, k=π/2∫0cosyf(y)dy=π/2∫0cosy(y+siny.k)dy
∴k=π/2∫0ycosydy+kπ/2∫0cosysinydy
⇒k=ysiny|π/20–π/2∫0sinydy+k2π/2∫0sin2ydy=π2+cosy|π/20–k2.12cos2y|π/20
⇒k=π2–1–k4(–1–1)=π2–1+k2
⇒k2=π2–1=π–22
∴k=π–2
Now, f(x)=x+sinx.k=x+(π–2)sinx
Answer: (D)