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已知函数y=ax∧2+bx+c的图像经过点(-1,0),且不等式x≤y≤0.5×(1+x∧2)对任意x都成立,求函数y.
人气:493 ℃ 时间:2019-09-03 09:12:23
解答
y=ax∧2+bx+c的图像经过点(-1,0)
0=a-b+c,b=a+c
y<=0.5(1+x^2)
ax^2+bx+c-1/2-1/2x^2<=0
(a-1/2)x^2+bx+c-1/2<=0 对任意x都成立
当a-1/2=0,a=1/2时,bx+c-1/2<=0,则b=0,c=-1/2
y=1/2x^2-1/2
当a-1/2≠0时.a<0,△=b^2-4*(a-1/2)(c-1/2)=(a-c)^2+2(a+c)-1<=0
x≤y对任意x都成立
ax^2+(b-1)x+c>=0 ,a>0,△=(b-1)^2-4*ac=(a-c)^2-2(a+c)+1<=0
两个不等式相加,2(a-c)^2<=0,a=c
ax^2+bx+c-1/2-1/2x^2<=0 代入x=1 ,a+b+c-1<=0
ax^2+(b-1)x+c>=0,代入x=1, a+b+c-1>=0
所以a+b+c-1=0,且b=a+c,a=c
得,a=c=1/4,b=1/2
y=1/4x^2+1/2x+1/4
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