Detailed Solution:

(a) Asymptotes:

(i) Vertical asymptote:
Occurs when denominator equals zero: \[ x + 1 = 0 \Rightarrow x = -1 \]

(ii) Horizontal asymptote:
For large \( |x| \), the function behaves like: \[ \frac{2x}{x} = 2 \] So \( y = 2 \) is the horizontal asymptote.

(b) Graph Sketch:

Key features to include:
1. Vertical asymptote at \( x = -1 \) (dashed vertical line)
2. Horizontal asymptote at \( y = 2 \) (dashed horizontal line)
3. x-intercept when \( 2x – 1 = 0 \Rightarrow x = \frac{1}{2} \)
4. y-intercept at \( f(0) = -1 \)
5. Hyperbolic shape with branches in upper-right and lower-left quadrants

(c) Solving Inequality:

Solve \( 0 < \frac{2x – 1}{x + 1} < 2 \):
First part \( \frac{2x – 1}{x + 1} > 0 \):
Critical points at \( x = -1 \) and \( x = \frac{1}{2} \)
Solution: \( x < -1 \) or \( x > \frac{1}{2} \)
Second part \( \frac{2x – 1}{x + 1} < 2 \): \[ \frac{2x – 1 – 2(x + 1)}{x + 1} < 0 \] \[ \frac{-3}{x + 1} < 0 \Rightarrow x + 1 > 0 \Rightarrow x > -1 \]
Combining both conditions: \( x > \frac{1}{2} \)

(d) Absolute Value Inequality:

Solve \( 0 < \frac{2|x| – 1}{|x| + 1} < 2 \):
First find where numerator equals zero: \[ 2|x| – 1 = 0 \Rightarrow |x| = \frac{1}{2} \Rightarrow x = \pm \frac{1}{2} \]
The inequality holds when \( |x| > \frac{1}{2} \), so: \[ x < -\frac{1}{2} \text{ or } x > \frac{1}{2} \]

Markscheme:

(a)
(i) \( x = -1 \) A1
(ii) \( y = 2 \) A1
[2 marks]

(b)
• Correct hyperbolic shape with two branches A1
• Correct asymptotes shown A1
• Correct intercepts at \( (\frac{1}{2}, 0) \) and \( (0, -1) \) A1
[3 marks]

(c)
• Correct solution \( x > \frac{1}{2} \) A2
Note: Accept equivalent interval notation
[2 marks]

(d)
• Correct solution \( x < -\frac{1}{2} \) or \( x > \frac{1}{2} \) A1
Note: Accept equivalent notation
[1 mark]

Total: [8 marks]