If a function satisfies the condition $f(x+\frac{1}{x}) = x^2 + \frac{1}{x^2} , x \ne 0 $ then domain of f (x) is

Problem: If a function satisfies the condition $f(x+\frac{1}{x}) = x^2 + \frac{1}{x^2} , x \ne 0 $ then domain of f (x) is

(A) [-2 , 2]

(B) (-∞ , 2] ∪ [2 , ∞ )

(C) (0 , ∞)

(D) none of these

Sol: $f(x+ \frac{1}{x}) = (x+\frac{1}{x})^2 – 2 $

⇒ f(y) = y² –2, where y = x + 1/x

for x > 0, y = x + 1/x  ≥ 2 and for x < 0, y = x + 1/x  ≤ –2

Hene (B) is the correct answer.