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Let f(x) be defined as: f(x) = 3*(1 + |tan x|)^(a / |tan x|) for -1/2 < x < 0; f(x) = b for x = 0; f(x) = 3*(1 + |sin x| / 3)^(6 / x) for 0 < x <= 2/3. If f is continuous at x = 0, which of the following is correct?

1. a + ln(b/3) = 4
2. a * ln(b/3) = -4
3. a + b < 0
4. a - ln(b/3) = 4

Correct answer: a + ln(b/3) = 4

Solution

Left-hand limit: lim_(x->0⁻) 3*(1+|tan x|)^(a/|tan x|). Let t = |tan x|, as x->0⁻, t->0⁺. Limit = 3 * [lim_(t->0+)(1+t)^(1/t)]^a = 3*e^a. Right-hand limit: lim_(x->0⁺) 3*(1+|sin x|/3)^(6/x). Let u = |sin x|/3. As x->0⁺, u->0⁺. lim (1+u)^(6/x) = exp[6/x * ln(1+u)] ~ exp[6/x * u] = exp[6/x * sin(x)/3] = exp[6/(3) * sin(x)/x] = exp[2*1] = e². Right limit = 3*e². For continuity: 3*e^a = b = 3*e². From left = right: a=2, b=3*e². Check options: a + ln(b/3) = 2 + ln(e²) = 2 + 2 = 4. Option A: 4. CORRECT. Option B: a*ln(b/3) = 2*2 = 4 ≠ -4. WRONG. Option C: a+b = 2+3e² > 0. WRONG. Option D: a - ln(b/3) = 2 - 2 = 0 ≠ 4. WRONG. Answer: option A.

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