If f≤ft(x+displaystyle1/x)=x^2+1/(x^2) then f(x)=

The correct answer is Option (4) → \( x^2-2 \)

Since \( \left(x+\frac{1}{x}\right)^2=x^2+2+\frac{1}{x^2} \), the right side can be written in terms of \( x+\frac{1}{x} \):

    \( \displaystyle x^2+\frac{1}{x^2}=\left(x+\frac{1}{x}\right)^2-2 \)

    \( \displaystyle f\left(x+\frac{1}{x}\right)=\left(x+\frac{1}{x}\right)^2-2 \)

So f squares its input and subtracts 2:

    \( \displaystyle f(x)=x^2-2 \)

Shortcut: Put x = 1: then \( x+\frac{1}{x}=2 \) and \( x^2+\frac{1}{x^2}=2 \), so f(2) = 2. Only \( x^2-2 \) gives 2 at x = 2 (the others give 2.5, 4 and 3).