Question

Let $f(x+\frac{1}{x})=x^2+\frac{1}{x^2}(x\ne 0)$, then $f\left(x\right)$ equals:

• A. $x^2-2$
• B. $x^2-1$
• C. $x^2$
• D. None of these

Answer 1

The correct option is A $x^2-2$

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

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

$f(x+\frac{1}{x})=(x+\frac{1}{x}{)}^2-2$

Let $x+\frac{1}{x}=y$

$f\left(y\right)=y^2-2$

$\therefore f\left(x\right)=x^2-2$