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For a, b in R - {0}, let f(x) = a*x² + b*x + a satisfy f(x + 7/4) = f(7/4 - x) for all x in R. Also, the equation a*x² + b*x + a = 7*x + a has only one real solution. Find a + b.

1. 4
2. 5
3. 6
4. 7

Correct answer: 7

Solution

Step 1: f(x + 7/4) = f(7/4 - x) means the parabola is symmetric about x = 7/4. Axis of symmetry of ax²+bx+a is x = -b/(2a) = 7/4, so b = -7a/2. Step 2: The equation ax² + bx + a = 7x + a simplifies to ax² + (b-7)x = 0, i.e., x(ax + b - 7) = 0. This has roots x = 0 and x = (7-b)/a. For only one real solution, these two roots must coincide: (7-b)/a = 0 => b = 7. Step 3: From b = -7a/2 and b = 7: 7 = -7a/2 => a = -2. Step 4: a + b = -2 + 7 = 5.

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