Question

求一道数学题，求过程





Answer 1

10.ax^2+bx+c=x,
ax^2+(b-1)x+c=0无实根，
<==>△1=(b-1)^2-4ac<0,
①f[f(x)]-x=f(ax^2+bx+c)-x=a(ax^2+bx+c)^2+b(ax^2+bx+c)+c-x
=a[ax^2+(b-1)x+c]^2+ax^2+2ax[ax^2+(b-1)x+c]+b[ax^2+(b-1)x+c}+(b-1)x+c
=[ax^2+(b-1)x+c]{a[ax^2+(b-1)x+c]+2ax+b+1}=0,
化为ax^2+(b-1)x+c=0，或a^2x^2+(ab+a)x+ac+b+1=0,
△2=(ab+a)^2-4a^2(ac+b+1)=a^2[b^2+2b+1-4ac-4b-4]=a^2[(b-1)^2-4ac-4]<0,
①正确，弃B.
②a>0时ax^2+(b-1)x+c>0,a^2x^2+(ab+a)x+ac+b+1>0,
∴f[f(x)]-x>0,②正确，弃C.
④a+b+c=0,b=-a-c,
△1=(a+c+1)^2-4ac=(a-c)^2+2(a+c)+1<0,
∴a+c<-1/2,ac>0,
∴a<0,
∴ax^2+(b-1)x+c<0,a^2x^2+(ab+a)x+ac+b+1>0,
∴f[f(x)]-x<0,④正确，弃D,选A.