Question

If a function $f(x)$ satisfies the equation \[ f\left(x + \frac{1}{x}\right) = x^2 + \frac{1}{x^2}, \quad x \neq 0 \] then $f(x)$ equals:

• A. $x^2 - 2$ for $x \neq 0$
• B. $x^2 - 2$ for all satisfying $|x| \ge 2$
• C. $x^2 - 2$ for all satisfying $|x|<2$
• D. None of these

Hint:

When $x + \frac{1}{x}$ appears, use the identity $x^2 + \frac{1}{x^2} = (x + \frac{1}{x})^2 - 2$ and check the range via AM–GM or Cauchy–Schwarz.

Answer 1

The Correct Option is B

Let $t = x + \frac{1}{x}$, so $f(t) = x^2 + \frac{1}{x^2}$. We know: \[ x^2 + \frac{1}{x^2} = \left(x + \frac{1}{x}\right)^2 - 2 = t^2 - 2 \] Thus: \[ f(t) = t^2 - 2 \] Now, for real $x$, $t = x + \frac{1}{x}$ satisfies $t \le -2$ or $t \ge 2$ (by AM–GM inequality). Hence the domain of $f(t)$ is $|t| \ge 2$. Therefore: \[ f(x) = x^2 - 2 \quad \text{for all $|x| \ge 2$}. \] \[ \boxed{\text{$x^2 - 2$ for all $|x| \ge 2$}} \]