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Explanation:

Given, logx (x2 + 12) = 4
⇒ x2 + 12 = x4 
⇒ a + 12 = a2     [Take x2 = a]
⇒ a2 – a – 12 = 0
⇒ (a – 4)(a + 3) = 0
⇒ a = -3 or 4.        [a = x2 cannot be negative hence -3 is rejected]
⇒ x2 = 4 
⇒ x = ± 2        [x = -2 is rejected as log is not defined for negative numbers]
⇒ x = 2

Also, 3logy x = 1
⇒ 3logy 2 = 1
logy 23 = 1
⇒ 23 = y1 
⇒ y = 8

∴ x + y = 2 + 8 = 10

Hence, option (d).

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