Step 1: Understanding the Concept:
We need to determine if the function is injective (one-one) and/or surjective (onto).
A function is one-one if $f(x_1) = f(x_2)$ implies $x_1 = x_2$.
A function is onto if for every $y$ in the codomain, there exists an $x$ in the domain such that $f(x) = y$.
Step 2: Key Formula or Approach:
For one-one: Set $\frac{2x_1+3}{3x_1+4} = \frac{2x_2+3}{3x_2+4}$ and solve.
For onto: Express $x$ in terms of $y$ ($y = f(x)$) and find the range of possible $y$ values. Compare it with the typical codomain (real numbers $\mathbb{R}$) or interpret the given options.
Step 3: Detailed Explanation:
Check for One-One (Injectivity):
Let $f(x_1) = f(x_2)$.
\[ \frac{2x_1+3}{3x_1+4} = \frac{2x_2+3}{3x_2+4} \]
Cross-multiply:
\[ (2x_1+3)(3x_2+4) = (2x_2+3)(3x_1+4) \]
\[ 6x_1x_2 + 8x_1 + 9x_2 + 12 = 6x_1x_2 + 8x_2 + 9x_1 + 12 \]
Cancel common terms from both sides:
\[ 8x_1 + 9x_2 = 8x_2 + 9x_1 \]
\[ 9x_2 - 8x_2 = 9x_1 - 8x_1 \]
\[ x_2 = x_1 \]
Since $f(x_1) = f(x_2)$ implies $x_1 = x_2$, the function is one-one.
Check for Onto (Surjectivity):
Let $y = f(x) = \frac{2x+3}{3x+4}$.
We will solve for $x$ in terms of $y$:
\[ y(3x + 4) = 2x + 3 \]
\[ 3xy + 4y = 2x + 3 \]
\[ 3xy - 2x = 3 - 4y \]
\[ x(3y - 2) = 3 - 4y \]
\[ x = \frac{3 - 4y}{3y - 2} \]
For $x$ to be a real number, the denominator cannot be zero.
So, $3y - 2 \neq 0 \Rightarrow y \neq \frac{2}{3}$.
Also, we must ensure $x \neq -\frac{4}{3}$. If $\frac{3 - 4y}{3y - 2} = -\frac{4}{3}$, then $9 - 12y = -12y + 8 \Rightarrow 9 = 8$, which is absurd. So $x$ will never be $-\frac{4}{3}$.
The range of the function is all real numbers except $\frac{2}{3}$.
If the codomain is assumed to be $\mathbb{R}$, the function is not onto. However, option (C) specifically states "onto for $y \neq \frac{2}{3}$", meaning it is a bijection to its range $\mathbb{R} \setminus \{\frac{2}{3}\}$. This matches our findings perfectly.
Step 4: Final Answer:
The function is onto for $y \neq \frac{2}{3}$ and one-one.