Question

当x∈[0,π/2],则实数f(x)=[sin(x+π/2)+1] [cos(π/2-x)+1]的最小值为（ ）.

Answer 1

f(x)=[sin(x+π/2)+1] [cos(π/2-x)+1]

=(cosx+1)(sinx+1)

=cosxsinx+cosx+sinx+1

令cosx+sinx=t,sinx+cosx=√2sin(x+π/4)∈[1,√2]

t²=1+2cosxsinx,cosxsinx=（t²-1）/2

即

f=(t²-1)/2+t+1

=1/2t²+t+1/2

=1/2(t+1)²

所以

最小值=1/2 (1+1)²=2.