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Two numbers a and b are independently chosen at random with replacement from the set {1, 2, 3, 4, 5, 6}. The probability that lim x->0 ((a^x + b^x)/2)^(2/x) = 6 equals p/q where HCF(p, q) = 1. Find (q - p).
- 5
- 8
- 11
- 14
Correct answer: 8
Solution
The limit lim x->0 ((a^x + b^x)/2)^(2/x) is of the form 1^infinity. Taking logarithm: (2/x)*ln((a^x+b^x)/2). Using Taylor expansion a^x = 1 + x*ln(a) + O(x²): (a^x+b^x)/2 = 1 + x*(ln a + ln b)/2 + O(x²). So ln(...) = x*(ln a + ln b)/2 + O(x²). Multiplying by 2/x: ln a + ln b = ln(ab). Hence L = ab. Need ab = 6. Favorable pairs (a,b) from {1..6}x{1..6} with ab=6: (1,6),(6,1),(2,3),(3,2) = 4 pairs. Probability = 4/36 = 1/9. So p=1, q=9, HCF=1. Answer: q-p = 9-1 = 8.
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