Question

Let the function $f$ be defined by $f(x)=\frac{2x+1}{1-3x}$ Then $f^{-1}(x)$ is:

• A. $\frac{x-1}{3x+2}$
• B. $\frac{3x+1}{x-1}$
• C. $\frac{x+1}{3x-2}$
• D. $\frac{2x+1}{1-3x}$
• E. $\frac{1-3x}{2x+1}$

Answer 1

Answer: (A)

Let $y=\frac{2x+1}{1-3x}$ $\Rightarrow$ $y-3xy=2x+1$ $\Rightarrow$ $y-1=(3y+2)x$ $\Rightarrow$ $x=\left(\frac{y-1}{3y+2}\right)$ $\therefore$ $f^{-1}(y)=\frac{y-1}{3y+2}$ $\Rightarrow$ $f^{-1}(x)=\frac{x-1}{3x+2}$