Question

已知函数f(x)=sin^2x+sinxcosx

(1)求其最小正周期
(2)当0≤x≤π/2时，求其最值及相应的x值
(3)试求不等式f(x)≥1的解集

Answer 1

解：
f(x)=sin²x+sinxcosx

=[1-cos(2x)]/2 +sin(2x)/2
=sin(2x) /2 -cos(2x) /2 +1/2
=(√2/2)sin(2x-π/4)+1/2
最小正周期T=2π/2=π
0≤x≤π/2 -π/4≤2x-π/4≤3π/4 -√2/2≤sin(2x-π/4)≤1
sin(2x-π/4)=1时，f(x)有最大值[f(x)]max=(√2+1)/2
sin(2x-π/4)=-√2/2时，f(x)有最小值[f(x)]min=0
f(x)≥1
(√2/2)sin(2x-π/4)+1/2≥1
sin(2x-π/4)≥√2/2
2kπ+π/4≤2x-π/4≤2kπ+3π/4 (k∈Z)
kπ+π/4≤x≤kπ+π/2 (k∈Z)
x的解集为[kπ+π/4，kπ+π/2] (k∈Z)

Answer 2

f(x)=sinx^2+sinxcosx
=(-1/2)*(1-2sinx^2-1)+(1/2)*2sinxcosx
=1/2+1/2cos2x+1/2sin2x
=(√2/2)*(√2/2cos2x+√2/2sin2x)+1/2
=(√2/2)sin(2x+π/4)+1/2
(1)T=2π/ω=π
(2)当0<=x<=π/2时,0<=2x<=π,π/4<=2x+π/4<=5π/4
所以sin(2x+π/4)的最大值为1，最小值为sin(5π/4)=-√2/2
所以f(x)的最大值为(√2+1)/2,此时2x+π/4=π/2 -> x=π/8
最小值为0,此时2x+π/4=5π/4 -> x=π/2
(3)f(x)>=1 -> (√2/2)sin(2x+π/4)>=1/2 -> sin(2x+π/4)>=√2/2
-> 解得,kπ<=x<=π/4+kπ(k为整数)

Answer 3