Question

\(f(x)\) is a quadratic polynomial satisfying the condition \( f(x) + f\left(\frac{1}{x}\right) = f(x)f\left(\frac{1}{x}\right) \). If \(f(-1)=0\), then the range of \(f\) is

• A. \( [1, \infty) \)
• B. \( [-1,1] \)
• C. \( (-\infty, 1] \)
• D. \( \mathbb{R} \)

Hint:

Rewrite the given equation as \( (f(x)-1)(f(1/x)-1)=1 \). Let \(g(x)=f(x)-1\). Then \(g(x)g(1/x)=1\). For polynomial \(g(x)\), this implies \(g(x)=\pm x^k\). Since \(f(x)\) is quadratic, \(f(x)-1 = \pm x^2\). So \(f(x) = 1 \pm x^2\). Use \(f(-1)=0\) to determine \(f(x) = 1-x^2\). The range of \(y=1-x^2\) is \(y \le 1\).

Answer 1

The Correct Option is C

Given \( f(x) + f\left(\frac{1}{x}\right) = f(x)f\left(\frac{1}{x}\right) \).
This can be rewritten as \( f(x)f\left(\frac{1}{x}\right) - f(x) - f\left(\frac{1}{x}\right) = 0 \).
Adding 1 to both sides: \( f(x)f\left(\frac{1}{x}\right) - f(x) - f\left(\frac{1}{x}\right) + 1 = 1 \).
Factoring, we get \( (f(x)-1)\left(f\left(\frac{1}{x}\right)-1\right) = 1 \).
Let \(g(x) = f(x)-1\).
Then \(g(x)g\left(\frac{1}{x}\right) = 1\).
Since \(f(x)\) is a quadratic polynomial, let \(f(x) = ax^2+bx+c\).
Then \(g(x) = ax^2+bx+(c-1)\).
For \(g(x)g(1/x)=1\) to hold for a polynomial \(g(x)\), \(g(x)\) must be of the form \( \pm x^k \) for some integer \(k\).
Since \(f(x)\) is quadratic, \(g(x) = f(x)-1\) is also a polynomial of degree at most 2.
Thus, the possible forms for \(g(x)\) are \( \pm 1, \pm x, \pm x^2, \pm x^{-1}, \pm x^{-2} \).
For \(f(x)\) to be a quadratic polynomial, \(g(x)=f(x)-1\) must lead to a quadratic \(f(x)\).
This implies \(g(x) = \pm x^2\) or \(g(x)=\pm x^{-2}\) (to potentially become quadratic if multiplied by \(x^2\)), or lower powers that result in \(f(x)\) being quadratic.
The valid polynomial choices for \(g(x)\) that maintain \(f(x)\) as quadratic are \(g(x)=\pm x^2\), \(g(x)=\pm x\), or \(g(x)=\pm 1\).
If \(g(x)=\pm x^2\), then \(f(x)-1 = \pm x^2 \implies f(x) = 1 \pm x^2\).
(Quadratic) If \(g(x)=\pm x\), then \(f(x)-1 = \pm x \implies f(x) = 1 \pm x\).
(Linear, not quadratic as stated for \(f(x)\)) If \(g(x)=\pm 1\), then \(f(x)-1 = \pm 1 \implies f(x) = 2\) or \(f(x)=0\).
(Constant, not quadratic) So, we must have \(f(x) = 1 \pm x^2\).
Given \(f(-1)=0\).
If \(f(x) = 1+x^2\), then \(f(-1) = 1+(-1)^2 = 1+1=2 \neq 0\).
If \(f(x) = 1-x^2\), then \(f(-1) = 1-(-1)^2 = 1-1=0\).
This is the correct function.
The function is \(f(x) = 1-x^2\).
This is a parabola opening downwards, with vertex at \((0,1)\).
The maximum value is \(f(0)=1\).
As \(x \to \pm\infty\), \(f(x) \to -\infty\).
The range of \(f(x)\) is \( (-\infty, 1] \).
\[ \boxed{(-\infty, 1]} \]