Mark the correct alternative in the following:
The domain of definition of the function
is
5|x|-x2-6≥0
x2-5|x|+6≤0
(|x|-2)(|x|-3)≤0
So, |x|∈[2, 3]
Therefore,x∈[-3, -2]∪[2, 3]
Option C is correct.
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