Question 1193841
**1. Define the Probability Density Function (PDF)**

* **f(x) = { ax, 0 ≤ x ≤ 1 
             { b(x-1)², 1 < x ≤ 2 
             { 0, otherwise**

**2. Conditions for a Valid PDF**

* **Non-negativity:** f(x) ≥ 0 for all x 
    * This condition is satisfied since:
        * For 0 ≤ x ≤ 1: ax ≥ 0 if a ≥ 0 
        * For 1 < x ≤ 2: b(x-1)² ≥ 0 for any value of b

* **Normalization:** The total area under the PDF curve must equal 1. 
    * This means: 
        * ∫₀¹ ax dx + ∫₁² b(x-1)² dx = 1

**3. Calculate the Integrals**

* ∫₀¹ ax dx = [a(x²/2)]₀¹ = a/2
* ∫₁² b(x-1)² dx = b ∫₀¹ y² dy (where y = x-1) = b[y³/3]₀¹ = b/3

**4. Apply the Normalization Condition**

* a/2 + b/3 = 1

**5. Determine the Relationship between a and b**

* a/2 + b/3 = 1 
* 3a + 2b = 6 
* **3a = 6 - 2b** 

**6. Additional Constraint (Optional)**

* To uniquely determine the values of 'a' and 'b', we would typically need an additional constraint. For example, we could specify the mean or variance of the random variable X.

**7. Solve for a and b**

* Without further constraints, we can express 'a' in terms of 'b' using the equation 3a = 6 - 2b. 
* For example:
    * If we assume b = 1, then a = (6 - 2)/3 = 4/3

**Therefore:**

* **a = (6 - 2b) / 3**
* **b can be any non-negative value.**

**Note:** To find specific values for 'a' and 'b', you would need additional information about the distribution, such as its mean, variance, or other relevant properties.