解由知点p1(x1,1994)和p2(x2,1994)关于直线x=(x1+x2)/2对称
又由函数f(x)=ax^2+bx+c的对称轴为x=-b/2a
即(x1+x2)/2=-b/2a
即x1+x2=-b/a
故f(X1+X2)
=f(-b/a)
=a(-b/a)^2+b×(-b/a)+c
=b^2/a-b^2/a+c
=c
即f(X1+X2)=c.
解由知点p1(x1,1994)和p2(x2,1994)关于直线x=(x1+x2)/2对称
又由函数f(x)=ax^2+bx+c的对称轴为x=-b/2a
即(x1+x2)/2=-b/2a
即x1+x2=-b/a
故f(X1+X2)
=f(-b/a)
=a(-b/a)^2+b×(-b/a)+c
=b^2/a-b^2/a+c
=c
即f(X1+X2)=c.