A) \[-5\le x\le 1\]
B) \[-5\le x\] and\[x>1\]
C) \[-4<x\le 1\]
D) \[\phi \]
E) \[0\le x\le 1\]
Correct Answer: C
Solution :
\[f(x)=\sqrt{5-4x-{{x}^{2}}}+{{x}^{2}}\log (x+4)\] Note that\[\sqrt{5-4x-{{x}^{2}}}\]is not defined when \[1<x\le 5\] and\[log\text{ (}x+4)\]is not defined when \[\text{(}x+4)\le 0\] \[\Rightarrow \] \[x\le -4\] \[\therefore \]Required domain is \[-4<x\le 1\] and \[x>5\]
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