Question

If $f\left(x\right)=x^2-3x-\frac{3}{x}+\frac{1}{x^2}$, then $f\left(\frac{1}{x}\right)$ is

• A. x
• B. $f\left(x\right)$
• C. $f(-x)$
• D. $-f\left(x\right)$

Answer 1

The correct option is B $f\left(x\right)$

Here $f\left(x\right)$ = $x^2-3x-\frac{3}{x}+\frac{1}{x^2}$

so, $f\left(\frac{1}{x}\right)=?$

$putx\to \frac{1}{x}$

$f\left(\frac{1}{x}\right)={\left(\frac{1}{x}\right)}^2-3\left(\frac{1}{x}\right)-\frac{3}{\left(\frac{1}{x}\right)}+\frac{1}{{\left(\frac{1}{x}\right)}^2}$

$=\frac{1}{x^2}-\frac{3}{x}-3x+x^2$

$=x^2-3x-\frac{3}{x}+\frac{1}{x^2}$

$=f\left(x\right)$

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