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B
x>12
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C
0 < x < 1
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D
x<13 and x>12
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Solution
The correct option is C 0 < x < 1 From the inequality, we have log1/2x−log1/3x=log1/2x−log1/2x.log1/3x.12>0 ⇒log1/2x[1−log32]>0 [Since log1/31/2=log−12−13=log32] ⇒log1/2x>0, As log32<1, 1−log32>0 But 0<base<1 ∴ 0 < x < 1