1. **Problem statement:** Given the function $f : A \to B$ where $A = \mathbb{R} - \{\frac{1}{2}\}$, $B = \mathbb{R} - \{\frac{1}{2}\}$, and $$f(x) = \frac{5x - 3}{2x + 1},$$ show that $f$ is one-one (injective) and find its inverse $f^{-1}$. 2. **Check if $f$ is one-one:** A function is one-one if $f(x_1) = f(x_2)$ implies $x_1 = x_2$. Assume $f(x_1) = f(x_2)$: $$\frac{5x_1 - 3}{2x_1 + 1} = \frac{5x_2 - 3}{2x_2 + 1}.$$ Cross-multiply: $$(5x_1 - 3)(2x_2 + 1) = (5x_2 - 3)(2x_1 + 1).$$ Expand both sides: $$10x_1 x_2 + 5x_1 - 6x_2 - 3 = 10x_1 x_2 + 5x_2 - 6x_1 - 3.$$ Simplify by subtracting $10x_1 x_2$ and $-3$ from both sides: $$5x_1 - 6x_2 = 5x_2 - 6x_1.$$ Bring all terms to one side: $$5x_1 - 6x_2 - 5x_2 + 6x_1 = 0,$$ $$11x_1 - 11x_2 = 0,$$ $$11(x_1 - x_2) = 0,$$ which implies $$x_1 = x_2.$$ Thus, $f$ is one-one. 3. **Find the inverse $f^{-1}$:** Start with $$y = \frac{5x - 3}{2x + 1}.$$ Solve for $x$ in terms of $y$: $$y(2x + 1) = 5x - 3,$$ $$2xy + y = 5x - 3,$$ Bring all $x$ terms to one side: $$2xy - 5x = -y - 3,$$ Factor $x$: $$x(2y - 5) = -y - 3,$$ $$x = \frac{-y - 3}{2y - 5}.$$ Therefore, $$f^{-1}(y) = \frac{-y - 3}{2y - 5}.$$ 4. **Domain and range considerations:** Since $A = \mathbb{R} - \{\frac{1}{2}\}$, the function is defined for all real $x$ except $x = \frac{1}{2}$ (which makes denominator zero). Similarly, the range excludes values making denominator of $f^{-1}$ zero, i.e., $2y - 5 = 0 \Rightarrow y = \frac{5}{2}$. But since $B = \mathbb{R} - \{\frac{1}{2}\}$, the function's codomain excludes $\frac{1}{2}$, consistent with the function's behavior. **Final answers:** - $f$ is one-one. - The inverse function is $$f^{-1}(x) = \frac{-x - 3}{2x - 5}.$$