The function defined by $\mathrm{f}(x)=\frac{2 x+3}{3 x+4}, x \neq-\frac{4}{3}$ is

To test the injectivity of $f(x) = \frac{2x + 3}{3x + 4}$, where $x \neq -\frac{4}{3}$, suppose $f(x_1) = f(x_2)$. This implies $\frac{2x_1 + 3}{3x_1 + 4} = \frac{2x_2 + 3}{3x_2 + 4}$.

Cross-multiplying gives $(2x_1 + 3)(3x_2 + 4) = (2x_2 + 3)(3x_1 + 4)$.

Expanding both sides yields $6x_1x_2 + 8x_1 + 9x_2 + 12 = 6x_1x_2 + 8x_2 + 9x_1 + 12$.

Subtracting $6x_1x_2 + 12$ from both sides gives $8x_1 + 9x_2 = 8x_2 + 9x_1$, which simplifies to $x_1 = x_2$, confirming the function is one-one.

To analyze surjectivity, let $y = f(x)$ and solve for $x$: $y(3x + 4) = 2x + 3$, so $3xy + 4y = 2x + 3$.

Rearranged, this becomes $x(3y - 2) = 3 - 4y$, giving $x = \frac{3 - 4y}{3y - 2}$ when $3y - 2 \neq 0$.

Hence, $y \neq \frac{2}{3}$, indicating the function is not onto if the codomain is $\mathbb{R}$, but is onto if the codomain is $\mathbb{R} \setminus \{\frac{2}{3}\}$.

The function is one-one, and onto if and only if the codomain excludes $\frac{2}{3}$, matching option C.