Question

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

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

Hint:

When expressions contain \(x-\frac{1}{x}\) and \(x^2+\frac{1}{x^2}\), use the identity \(\left(x-\frac{1}{x}\right)^2=x^2+\frac{1}{x^2}-2\).

Answer 1

The Correct Option is B

Step 1: Substitute t = x - 1/x.
Then \(x^2+1/x^2=t^2+2\), so \(\dfrac{f(x)}{g(x)}=\dfrac{t^2+2}{t}=t+\dfrac{2}{t}\).

Step 2: Minimise h(t) = t + 2/t for t > 0 and t < 0.
For \(t>0\): by AM-GM, \(t+2/t\geq 2\sqrt{2}\), with equality at \(t=\sqrt{2}\). For \(t<0\): \(t+2/t\leq -2\sqrt{2}\). The local minimum (smallest positive value) is \(2\sqrt{2}\). \[ \boxed{2\sqrt{2}} \]