f(x)=ax2+bx+c substitute f(x) into the given equation ax2+bx+c+a(
1
x
)2+b(
1
x
)+c =(ax2+bx+c)(a(
1
x
)2+b(
1
x
)+c) ⇒ax2+bx+c+
a
x2
+
b
x
+c =(ax2+bx+c)(
a
x2
+
b
x
+c) ⇒ax4+bx3+cx2+a+bx+cx2 =(ax2+bx+c)(a+bx+cx2) ⇒ax4+bx3+2cx2+bx+a =a2x2+abx3+acx4+abx+b2x2+bcx3+ac+bcx+c2x2 Compare coefficients : a=ac, b=ab+bc 2c=a2+b2+c2,b=ab+bc,a=ac From a=ac, we have c=1 or a=0, if a=0 then f(x) is not quadratic, so c=1 from b=ab+bc, we have b=ab+b, so ab=0, since a≠0,b=0 From 2c=a2+b2+c2, we have 2=a2+0+1 So, a2=1 and a=±1 thus, f(x)=x2+1 or f(x)=−x2+1 Since, f(−1)=0, we have f(x)=−x2+1 ⇒f(x)=−x2+1 ⇒ Range =(−∞,1]
