Correct : a

To find the correct asymptotic bound for T(n), let us solve the given linear homogeneous recurrence relation step-by-step using its characteristic equation.
1. Set up the Characteristic Equation: The recurrence relation for n ≥ 2 is:
T(n) = 5T(n-1) - 6T(n-2)

We assume a solution of the form T(n) = rⁿ. Substituting this into the relation gives the quadratic characteristic equation:
r² - 5r + 6 = 0
2. Find the Roots of the Equation: Factoring the quadratic polynomial:
(r - 2)(r - 3) = 0

The characteristic roots are r₁ = 2 and r₂ = 3.
3. Write the General Solution: Since the roots are real and distinct, the general solution is a linear combination of both roots raised to the power of n:
T(n) = c₁(2ⁿ) + c₂(3ⁿ)
4. Solve for Constants using Initial Conditions: We use the given base cases to evaluate c₁ and c₂:
• For n = 0:
    T(0) = c₁(2⁰) + c₂(3⁰) → c₁ + c₂ = 1
• For n = 1:
    T(1) = c₁(2¹) + c₂(3¹) → 2c₁ + 3c₂ = 2

Multiply the first equation by 2:
2c₁ + 2c₂ = 2

Subtracting this from the second equation:
(2c₁ + 3c₂) - (2c₁ + 2c₂) = 2 - 2 → c₂ = 0

Substituting c₂ = 0 back into the first equation:
c₁ + 0 = 1 → c₁ = 1
5. Find the Exact Expression and Asymptotic Complexity: Substituting our calculated constants c₁ = 1 and c₂ = 0 back into the general solution formula:
T(n) = 1 · (2ⁿ) + 0 · (3ⁿ) = 2ⁿ

Since the exact form reduces directly to 2ⁿ, its tight asymptotic bound is:
T(n) = Θ(2ⁿ)
Conclusion: The statement T(n) = Θ(2ⁿ) is true, matching option (a).