若函数y= -(sinα -m)²+m ² -2m- 1(0小于等于α小于等于π/2)的最大值为负值,求m的取值范围
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若函数y= -(sinα -m)²+m ² -2m- 1(0小于等于α小于等于π/2)的最大值为负值,求m的取值范围
若函数y= -(sinα -m)²+m ² -2m- 1(0小于等于α小于等于π/2)的最大值为负值,求m的取值范围
若函数y= -(sinα -m)²+m ² -2m- 1(0小于等于α小于等于π/2)的最大值为负值,求m的取值范围
0≤α≤π/2,那么0≤sinα≤1,
①当m<0时,sinα=0时,y取到最大值-(0-m)²+m²-2m-1=-2m-1<0得m>-1/2,结合前提m<0,即-1/2<m<0满足题意;
②当0≤m≤1时,sinα=m时,y取到最大值m² -2m- 1=(m-1)^2-2<0得1-√2<m<1+√2,结合前提0≤m≤1,即0≤m≤1;
③当m>1时,sinα=1时,y取到最大值-(1-m)²+m²-2m-1=-2<0恒成立,结合前提m>1得m>1;
综上,得m>-1/2.