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B
(−2+√3,∞)
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C
(−2−√3,−2+√3)
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D
(−∞,−2−√3)∪(−2+√3,∞)
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Solution
The correct option is D(−∞,−2−√3)∪(−2+√3,∞) For f(x)=log5(x2+4x+1) to be defined, we get x2+4x+1>0⇒x2+4x+4−3>0⇒(x+2)2>3⇒x+2>√3,x+2<−√3⇒x∈(−∞,−2−√3)∪(−2+√3,∞)