The correct option is C (−π2,π2)
f(x)−λπ2∫0sinx(cost⋅f(t)) dt=sinx
⇒f(x)−λsinxπ2∫0cost⋅f(t)dt=sinx
⇒f(x)−Asinx=sinx
where, A=λπ2∫0cost⋅f(t)dt
f(x)=(A+1)sinx⋯(i)
⇒A=λπ2∫0cost(A+1)sintdt
⇒A=λ(A+1)2π2∫0sin2t dt
⇒A=λ(A+1)2[−cos2t2]π20
⇒A=λ(A+1)2
⇒A=λ2−λ
Now, from equation (i)
⇒f(x)=(λ2−λ+1)sinx
⇒f(x)=(22−λ)sinx
Since, λ>2 and sinx increases in (−π2,π2)
Hence, f(x) decreases in (−π2,π2)