求使函数y=x^2+ax-2/x^2-x+1的值域为(-∝,2)的a的取值范围
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![求使函数y=x^2+ax-2/x^2-x+1的值域为(-∝,2)的a的取值范围](/uploads/image/z/3799242-18-2.jpg?t=%E6%B1%82%E4%BD%BF%E5%87%BD%E6%95%B0y%3Dx%5E2%2Bax-2%2Fx%5E2-x%2B1%E7%9A%84%E5%80%BC%E5%9F%9F%E4%B8%BA%28-%E2%88%9D%2C2%29%E7%9A%84a%E7%9A%84%E5%8F%96%E5%80%BC%E8%8C%83%E5%9B%B4)
求使函数y=x^2+ax-2/x^2-x+1的值域为(-∝,2)的a的取值范围
求使函数y=x^2+ax-2/x^2-x+1的值域为(-∝,2)的a的取值范围
求使函数y=x^2+ax-2/x^2-x+1的值域为(-∝,2)的a的取值范围
因为分母=x^2-x+1=(x-1/2)^2+3/4.故可知,函数定义域是R,且分母恒为正.又由题设值域知,对任意实数x,恒有f(x)-2<0,即恒有{[x^2+ax-2]/[x^2-x+1]}-2<0===>-[x^2-(a+2)x+4]/[x^2-x+1]<0===>[x^2-(a+2)x+4]/[x^2-x+1]>0===>由分母恒大于0,得:对任意实数x,恒有x^2-(a+2)x+4>0===>(a+2)^2-16<0===>-6
y=(x^2+ax-2)/(x^2-x+1)
=[(x^2-x+1)+(a+1)x-3]/(x^2-x+1)
=1+[(a+1)x-3]/(x^2-x+1)
值域为(-∝,2),则
1+[(a+1)x-3]/(x^2-x+1)<2;
[(a+1)x-3]/(x^2-x+1)<1;
∵根据判别式判定,x^2-x+1>0;
则(a+...
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y=(x^2+ax-2)/(x^2-x+1)
=[(x^2-x+1)+(a+1)x-3]/(x^2-x+1)
=1+[(a+1)x-3]/(x^2-x+1)
值域为(-∝,2),则
1+[(a+1)x-3]/(x^2-x+1)<2;
[(a+1)x-3]/(x^2-x+1)<1;
∵根据判别式判定,x^2-x+1>0;
则(a+1)x-3
x∈R,则根据判别式来判定,(a+2)^2-4×4=a^2+4a-12<0.
∴-4
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