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
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![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](/uploads/image/z/3757307-59-7.jpg?t=%E5%B7%B2%E7%9F%A5%E5%87%BD%E6%95%B0f%28x%29%3Dsin%5E2x%2Bsinxcosx%281%29%E6%B1%82%E5%85%B6%E6%9C%80%E5%B0%8F%E6%AD%A3%E5%91%A8%E6%9C%9F%282%29%E5%BD%930%E2%89%A4x%E2%89%A4%CF%80%2F2%E6%97%B6%2C%E6%B1%82%E5%85%B6%E6%9C%80%E5%80%BC%E5%8F%8A%E7%9B%B8%E5%BA%94%E7%9A%84x%E5%80%BC%283%29%E8%AF%95%E6%B1%82%E4%B8%8D%E7%AD%89%E5%BC%8Ff%28x%29%E2%89%A51%E7%9A%84%E8%A7%A3%E9%9B%86)
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
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
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
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)
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...
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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)
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