Question

二次函数f(x)=ax2+bx+c图像过(1,0),是否存在常数a,b,c,使- x≤f(x)≤1/2(1+x2)恒成立

Answer 1

二次函数f(x)=ax2+bx+c图像过(1,0) =>a+b+c=0 =>b=-a-c

- x≤f(x) =>ax^2+(b+1)x+c>=0 =>(b+1)^2-4ac0

代入b的表达式=>1+(a-c)^2-2(a+c)(a-1/2)x^2+bx+(c-1/2)≤0 =>b^2-4(a-1/2)(c-1/2)≤0 且a-1/2(a-c)^2+2(a+c)-12(a-c)^2a=c

分别代回到(1)和(2) =>1-4a