Problem 11 - Olympiad

For x < 2, f(x) = (x^2 - 4) / (x - 2) = (x - 2)(x + 2) / (x - 2) = x + 2. So lim_(x→2⁻) f(x) = 2 + 2 = 4. For continuity at x = 2, we need lim_(x→2⁺) f(x) = f(2) = 4. For x ≥ 2, f(x) = ax + b, so lim_(x→2⁺) f(x) = 2a + b. We need 2a + b = 4. Additionally, f(2) must equal the left-hand limit: a(2) + b = 4, which is the same condition. There are infinitely many pairs (a, b) satisfying 2a + b = 4. However, among the given choices, only a = 4, b = -4 satisfies 2(4) + (-4) = 8 - 4 = 4. Choice A: a=4, b=0 gives 2(4)+0=8 ≠ 4. Choice B: a=4, b=-4 gives 8-4=4 ✓. Choice C: a=2, b=0 gives 4+0=4, which also works. Wait, choice C also satisfies. Let me check: a=2, b=0 gives 2(2)+0=4. So both B and C satisfy. The question should have a unique answer. Let me adjust: perhaps the question intends for f to be differentiable as well, or there is an additional condition. Since the question asks only for continuity, both B and C are valid. Let me revise the choices so only one is correct. If I set a = 4, b = -4, that works. If I set a = 2, b = 0, that also works. Let me change the choices: [a=4,b=0], [a=4,b=-4], [a=1,b=2], [a=3,b=-2]. Then only a=4,b=-4 gives 2(4)+(-4)=4, and a=3,b=-2 gives 6-2=4, which also works. Hmm. Let me use a unique pair: a = 3, b = -2 gives 6-2=4. a = 4, b = -4 gives 8-4=4. Both work. The issue is that 2a + b = 4 has infinitely many solutions. To make the question have a unique answer, I need an additional constraint. Let me add that f is differentiable at x = 2. Then the left-hand derivative must equal the right-hand derivative. For x < 2, f(x) = x + 2, so f'(x) = 1. Thus f'(2⁻) = 1. For x ≥ 2, f'(x) = a, so f'(2⁺) = a. For differentiability, a = 1. Then 2(1) + b = 4, so b = 2. That gives a = 1, b = 2. Let me revise the question to ask for continuity only but adjust choices to have a unique answer: use a = 4, b = -4 as the intended answer and make other choices incorrect. Actually, since 2a + b = 4, if I pick choices where only one satisfies, I need to be careful. Let me set choices as: [a=1, b=1], [a=4, b=-4], [a=2, b=3], [a=3, b=0]. Only a=4, b=-4 gives 2(4)+(-4)=4. a=3,b=0 gives 6 ≠ 4. a=2,b=3 gives 4+3=7 ≠ 4. a=1,b=1 gives 2+1=3 ≠ 4. So choice B is unique. Let me update.