Exams › JEE Advanced › Maths
Correct answer: 2
f(x) = 3^(alpha*x) + 3^(beta*x). f'(x) = alpha*ln(3)*3^(alpha*x) + beta*ln(3)*3^(beta*x). f''(x) = alpha²*(ln3)²*3^(alpha*x) + beta²*(ln3)²*3^(beta*x). Substituting into 3*f'*ln3 = 2*f + f''*(ln3)²: [3*alpha*(ln3)² - 2 - alpha²*(ln3)²]*3^(alpha*x) + [3*beta*(ln3)² - 2 - beta²*(ln3)²]*3^(beta*x) = 0. Since 3^(alpha*x) and 3^(beta*x) are linearly independent (alpha != beta), each bracket = 0. Let t = ln3, then alpha²*t² - 3*alpha*t² + 2 = 0... wait: coefficient is (3*alpha - alpha²)*t² - 2 = 0 -> alpha² - 3*alpha + 2/t² = 0. Hmm, let u = alpha*ln3: u² - 3u + 2 = 0 -> (u-1)(u-2) = 0 -> u = 1 or u = 2. So alpha*ln3 = 1 or 2, and similarly beta*ln3 = 1 or 2. Since alpha != beta, one gives ln3-value of 1 and the other 2. So alpha = 1/ln3, beta = 2/ln3. alpha + beta = 3/ln3 = 3/ln3... but that's not an integer. Something is off. Let me substitute differently: the equation for each term: 3*(alpha*ln3)*(ln3) - 2 - alpha²*(ln3)² = 0. Let p = alpha*ln3: 3*p*ln3 - 2 - p² = 0 -> no, let p = alpha: 3*p*(ln3)² - 2 - p²*(ln3)² = 0 -> (ln3)²*(3p - p²) = 2 -> p*(3-p) = 2/(ln3)². This doesn't factor nicely. Try: equation is p²*(ln3)² - 3p*(ln3)² + 2 = 0. Let q = p*(ln3): q² - 3q*(ln3) + 2 = 0. Still messy. Alternatively let me try specific values. If alpha = 1, beta = 2 (treating exponents as powers of 3 directly... i.e. 3^x and 3^(2x)): f'(x) = ln3*3^x + 2*ln3*3^(2x). 3f'*ln3 = 3*(ln3)²*3^x + 6*(ln3)²*3^(2x). 2f = 2*3^x + 2*3^(2x). f''*(ln3)² = (ln3)²*3^x + 4*(ln3)²*3^(2x). 2f + f''*(ln3)² = (2 + (ln3)²)*3^x + (2 + 4*(ln3)²)*3^(2x). For this to equal 3f'*ln3 = 3*(ln3)²*3^x + 6*(ln3)²*3^(2x): coefficients of 3^x: 2 + (ln3)² = 3*(ln3)² -> 2 = 2*(ln3)² -> (ln3)² = 1 -> ln3 = 1, impossible. Try alpha and beta as roots of quadratic in the variable itself (not alpha*ln3): from each bracket = 0: alpha²*(ln3)² - 3*alpha*(ln3)² + 2 = 0. This means alpha = [3*(ln3)² +/- sqrt(9*(ln3)⁴ - 8*(ln3)²)] / (2*(ln3)²). alpha + beta = sum of roots = 3*(ln3)² / (ln3)² = 3. So alpha + beta = 3.