Question

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

• A. $x^2$
• B. $x^2-1$
• C. $x^2-2$
• D. N.O.T

Answer 1

Answer: (C)

$x^2-2$

We are given

$f(x+\frac{1}{2})=x^2+\frac{1}{x^2}$ (where $x\ne 0$)

$=x^2+\frac{1}{x^2}+2-2$.

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

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

So, simply put $x+\frac{1}{x}\to x$

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