x=0 f(x)=4? 4 & gt0+a→a & lt; four
x∈(0,? π) f(x)=2 f(x)>x+a→2 & gt; x+a→a & lt; ? π-2? (x+a is an increasing function, the same below)
x=? π f(x)=3? 3 & gt0+a→a & lt; three
x∈(? π,π] f(x)=4/3? f(x)>x+a→4/3 & gt; x+a→a & lt; 4/3-π
x∈(π,3π/2] f(x)=5/6 f(x)>x+a→5/6 & gt; x+a→a & lt; 5/6- 1.5π
x∈(3π/2,2π) f(x)=3/2 f(x)>x+a→3/2 & gt; x+a→a & lt; 3/2-2π
x=2π f(x)=4? 4 & gt2π+a→a & lt; 4-2π
∫min(4,π-2,3,4/3-π,5/6- 1.5π,3/2-2π,4-2π)=3/2-2π
∴ Inequality holds → A