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Let f(x) = 3^(alpha*x) + 3^(beta*x) where alpha != beta. If the relation 3 * f'(x) * log₃(e) = 2*f(x) + f''(x) * (log₃(e))² holds for all x, find the value of alpha + beta.

1. 3
2. 2
3. -3
4. 6

Correct answer: 2

Solution

Let k = log₃(e) = 1/ln(3). Then f'(x) = alpha*ln(3)*3^(alpha*x) + beta*ln(3)*3^(beta*x) = (alpha/k)*3^(alpha*x) + (beta/k)*3^(beta*x). Similarly f''(x) = (alpha²/k²)*3^(alpha*x)... wait, f''(x) = alpha²*(ln3)² * 3^(alpha*x) + beta²*(ln3)²*3^(beta*x) = (alpha²/k²... no). Let me use ln3 directly. f'(x) = alpha*ln3 * 3^(alpha*x) + beta*ln3 * 3^(beta*x). f''(x) = alpha²*(ln3)² * 3^(alpha*x) + beta²*(ln3)² * 3^(beta*x). Given relation: 3 * f'(x) * (1/ln3) = 2*f(x) + f''(x)*(1/ln3)². Substituting: 3*(alpha*ln3*3^(alpha*x)+beta*ln3*3^(beta*x))*(1/ln3) = 2*(3^(alpha*x)+3^(beta*x)) + (alpha²*(ln3)²*3^(alpha*x)+beta²*(ln3)²*3^(beta*x))/(ln3)². Simplifying LHS: 3*alpha*3^(alpha*x) + 3*beta*3^(beta*x). RHS: (2+alpha²)*3^(alpha*x) + (2+beta²)*3^(beta*x). Equating coefficients: 3*alpha = 2+alpha² and 3*beta = 2+beta². So alpha² - 3*alpha + 2 = 0 => (alpha-1)(alpha-2) = 0 => alpha = 1 or 2. Similarly beta = 1 or 2. Since alpha != beta, one is 1 and the other is 2. alpha + beta = 3.

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