here is an analysis of each of the proposed recursive definitions:

**a) f(0) = 0, f(n) = 2f(n – 2) for n ≥ 1**

This definition is valid. To prove this, we can use induction.

**Base case:** n = 1: f(1) = 2f(-1) = 2 * 0 = 0.

**Inductive step:** Assume that f(k) = 0 for some k ≥ 1. Then, f(k + 1) = 2f(k – 1) = 2 * 0 = 0.

Therefore, by induction, f(n) = 0 for all n ≥ 1.

**b) f(0) = 1, f(n) = f(n – 1) – 1 for n ≥ 1**

This definition is valid. To prove this, we can use induction.

**Base case:** n = 1: f(1) = f(0) – 1 = 1 – 1 = 0.

**Inductive step:** Assume that f(k) = 1 – k for some k ≥ 1. Then, f(k + 1) = f(k) – 1 = (1 – k) – 1 = -k.

Therefore, by induction, f(n) = 1 – n for all n ≥ 1.

**c) f(0) = 2, f(1) = 3, f(n) = f(n – 1) – 1 for n ≥ 2**

This definition is valid. To prove this, we can use induction.

**Base case:** n = 2: f(2) = f(1) – 1 = 3 – 1 = 2.

**Inductive step:** Assume that f(k) = k + 1 for some k ≥ 2. Then, f(k + 1) = f(k) – 1 = (k + 1) – 1 = k.

Therefore, by induction, f(n) = k + 1 for all n ≥ 2.

**d) f(0) = 1, f(1) = 2, f(n) = 2f(n – 2) for n ≥ 2**

This definition is valid. To prove this, we can use induction.

**Base case:** n = 2: f(2) = 2f(0) = 2 * 1 = 2.

**Inductive step:** Assume that f(k) = 2^k for some k ≥ 2. Then, f(k + 1) = 2f(k – 2) = 2 * 2^(k – 2) = 2^k.

Therefore, by induction, f(n) = 2^n for all n ≥ 2.

**e) f(0) = 1, f(n) = 3f(n – 1) if n is odd and n ≥ 1 and f(n) = 9f(n – 2) if n is even and n ≥ 2**

This definition is not valid. The definition does not provide a clear way to determine the value of f(n) for all n ≥ 1. For example, the value of f(3) cannot be determined using the given definition.