Question

Let \( f(x) = x^2 + \frac{1}{x^2} \) and \( g(x) = x - \frac{1}{x} \) for \( x \in \mathbb{R} \setminus \{-1, 0, 1\} \). Then, the local minimum of \( \frac{f(x)}{g(x)} \) is:

• A. \( -3 \)
• B. \( 2\sqrt{2} \)
• C. \( -2\sqrt{2} \)
• D. \( 3 \)

Hint:

Functional Extremum}
Substitute with symmetry: Let \( t = x - \frac{1}{x} \)
Transform expression to a single variable to simplify calculus
Use derivatives to locate local minima

Answer 1

The Correct Option is B

We have: \[ f(x) = x^2 + \frac{1}{x^2}, \quad g(x) = x - \frac{1}{x} \Rightarrow \frac{f(x)}{g(x)} = \frac{x^2 + \frac{1}{x^2}}{x - \frac{1}{x}}. \] Let \( t = x - \frac{1}{x} \), then: \[ x^2 + \frac{1}{x^2} = t^2 + 2. \] Hence, \[ \frac{f(x)}{g(x)} = \frac{t^2 + 2}{t}. \] Minimize \( \frac{t^2 + 2}{t} \). Using calculus: \[ y = \frac{t^2 + 2}{t}, \quad \frac{dy}{dt} = \frac{t(t^2 + 2)' - (t^2 + 2)}{t^2} = \frac{2t^2 - (t^2 + 2)}{t^2} = \frac{t^2 - 2}{t^2}. \] Setting derivative zero: \[ t^2 - 2 = 0 \Rightarrow t = \pm\sqrt{2}. \] Compute: \[ \frac{t^2 + 2}{t} = \frac{4}{\sqrt{2}} = 2\sqrt{2}. \]