Question

If $f\left(x\right)=x^2-x^{-2}$ then $f\left(\frac{1}{x}\right)$ is equal to

• A. $f\left(x\right)$
• B. $-f\left(x\right)$
• C. $\frac{1}{\text{f(x)}}$
• D. ${\left[\text{f(x)}\right]}^2$

Answer 1

Answer: (B)

$-f\left(x\right)$

We have,

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

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

Put $x=\frac{1}{x}$

So,

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

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

$f\left(\frac{1}{x}\right)=-f\left(x\right)$

Hence, this is the answer.