f(x)=a●b=2cos(x/2)*√2*sin(x/2+π/4)+tan(x/2+π/4)*tan(x/2-π/4)

　　=2√2sin(x/2+π/4)*cos(x/2)+tan(x/2+π/4)*tan(x/2-π/4)

　　=√2(sin(x+π/4)+sin(π/4))+[cos(π/2)-cos(x+π/2)]/[cos(x+π/2)+cos(π/2)]

　　=√2*(sin(x)*cos(π/4)+cos(x)*sin(π/4)+1/2*√2)+[-cos(x+π/2)]/[cos(x+π/2)]

　　=√2*(sin(x)*1/2*√2+cos(x)*1/2*√2+1/2*√2)-1

　　=sin(x)+cos(x)

　　f’(x)=cos(x)-sin(x)

　　则f(x)+f’(x)=2*cos(x)=0,x=π/2

　　即存在实数x∈[0,π]使f(x)+f’(x)=0,且x=π/2