Question

Consider the function $f\left(x\right)=\frac{x-1}{x+1}$. What is $f\left(2x\right)$ equal to?

• A. $\frac{f\left(x\right)+1}{f\left(x\right)+3}$
• B. $\frac{f\left(x\right)+1}{3f\left(x\right)+1}$
• C. $\frac{3f\left(x\right)+1}{f\left(x\right)+3}$
• D. $\frac{f\left(x\right)+3}{3f\left(x\right)+1}$

Answer 1

The correct option is C $\frac{3f\left(x\right)+1}{f\left(x\right)+3}$

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

By cross multiplying , we get $x-1=xf\left(x\right)+f\left(x\right)$

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

Substituting 2x in place of x in $f\left(x\right)$

$f\left(2x\right)\Rightarrow \frac{2x-1}{2x+1}$

Substituting Equation 1 in the above equation,

$\Rightarrow \frac{2\left(\frac{1+f\left(x\right)}{1-f\left(x\right)}\right)-1}{2(\frac{1+f\left(x\right)}{1-f\left(x\right)})+1}$

After Simplification . we get

$\Rightarrow \frac{2+2f\left(x\right)-1+f\left(x\right)}{2+2f\left(x\right)+1-f\left(x\right)}$

$\Rightarrow \frac{3f\left(x\right)+1}{f\left(x\right)+3}$.