(a) Horizontal asymptote:

Solution:

For \( f(x) = \frac{2(x+3)}{3(x+2)} \), as \( x \to \infty \) or \( x \to -\infty \):

\[ \lim_{x\to\pm\infty} \frac{2(x+3)}{3(x+2)} = \frac{2}{3} \]

Thus, the horizontal asymptote is:

\[ y = \frac{2}{3} \]

[1 mark]

(b)(i) Number of solutions for \( m > 0 \):

Solution:

Set \( f(x) = g(x) \):

\[ \frac{2(x+3)}{3(x+2)} = mx + 1 \]

Cross-multiply:

\[ 2x + 6 = 3mx^2 + 6mx + 3x + 6 \]

Simplify:

\[ 0 = 3mx^2 + (6m + 1)x \]

Factor:

\[ x(3mx + 6m + 1) = 0 \]

Solutions:

1. \( x = 0 \)

2. \( x = -2 – \frac{1}{3m} \) (which is \( < -2 \) for \( m > 0 \))

Thus, there are 2 solutions for \( m > 0 \).

[2 marks]

(b)(ii) Value of \( m \) for one solution:

Solution:

For exactly one solution, the quadratic must have discriminant zero:

\[ (6m + 1)^2 = 0 \implies m = -\frac{1}{6} \]

Thus, the required value is:

\[ m = -\frac{1}{6} \]

[3 marks]

(b)(iii) Range of \( m \) for two solutions with \( x \geq 0 \):

Solution:

We need both solutions \( \geq 0 \):

1. \( x = 0 \) is always a solution

2. For \( x = -2 – \frac{1}{3m} \geq 0 \):

\[ -2 – \frac{1}{3m} \geq 0 \implies -\frac{1}{3m} \geq 2 \implies \frac{1}{m} \leq -6 \]

Since \( m < 0 \), this becomes:

\[ -\frac{1}{6} < m < 0 \]

Thus, the range is:

\[ -\frac{1}{6} < m < 0 \]

[4 marks]