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The image you've uploaded contains a mathematical assignment with several questions. I'll help by providing step-by-step solutions and explanations for each of the tasks. Here's an outline of how we can approach this:

### 1. Find the values of \( x \) that satisfy the inequalities:
   - (a) \( \left|\frac{2x + 1}{x - 1}\right| \leq 2 \)
   - (b) \( |x + 1|\sqrt{x} < \sqrt{x^2 + 3} \)
   - (c) \( |x - 2| + |x| \geq 2 \)
   - (d) \( x + 1 \leq \sqrt{x^2 + 3x - 4} \)
   - (e) \( |2x + 1| < 5 - |2x| \)

### 2. If \( \frac{x + a}{x - 2} \leq 2 \) has the solution set \( \{x \mid -4 \leq x \leq 2\} \), find the value of \( a \).

### 3. Determine the domain of the following functions:
   - (a) \( f(u) = \frac{u^3}{8} \)
   - (b) \( g(x) = \frac{x}{x^2 - 1} \)
   - (c) \( h(t) = \sqrt{x^2 - 4} \)

### 4. Determine \( (f \circ g)(x) \) and \( (g \circ f)(x) \) and their domains, where:
   - \( f(x) = \sqrt{x^2 - 1} \)
   - \( g(x) = |1 + x| \)

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I'll begin by solving the inequalities in **Question 1** step-by-step.

#### 1(a) Solve \( \left|\frac{2x + 1}{x - 1}\right| \leq 2 \):

We can break this inequality into two parts:
1. \( \frac{2x + 1}{x - 1} \leq 2 \)
2. \( \frac{2x + 1}{x - 1} \geq -2 \)

Let's solve each case individually:

- **Case 1**: \( \frac{2x + 1}{x - 1} \leq 2 \)
   \[
   \frac{2x + 1}{x - 1} \leq 2
   \]
   Multiply both sides by \( (x - 1) \), making sure to reverse the inequality if \( x - 1 \) is negative.

- **Case 2**: \( \frac{2x + 1}{x - 1} \geq -2 \)
   Similarly, solve for \( x \).

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Should I continue solving this inequality, or would you like to focus on another part of the question? Feel free to ask for more details!

Tentukan nilai x yang memenuhi pertidaksamaan berikut, tentukan nilai a, tentukan domain fungsi, dan tentukan (f ∘ g)(x) dan (g ∘ f)(x) beserta domainnya dari f(x) = √(x^2 - 1) dan g(x) = |1 + x|.

This problem covers solving algebraic inequalities, finding the domain of functions, and determining the composition of functions f(x) = √(x^2 - 1) and g(x) = |1 + x|. Detailed steps provided for Grades 10-12.

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