A) \[(-2,0)\]
B) \[(-2,2)\]
C) \[(-5,5)\]
D) \[(0,\infty )\]
Correct Answer: D
Solution :
[d] We have \[{{5}^{x+2}}>{{\left( \frac{1}{25} \right)}^{\frac{1}{x}}}\Rightarrow {{5}^{x+2}}>{{5}^{-\frac{2}{x}}}\Rightarrow x+2>-\frac{2}{x}\] \[[\because \,\,\,base\,\,\,5>1]\] \[\Rightarrow x+2+\frac{2}{x}>0\Rightarrow \frac{{{x}^{2}}+2x+2}{x}>0\Rightarrow \frac{1}{x}>0\] \[[\because {{x}^{2}}+2x+2>0\forall x\in R]\] \[\Rightarrow x>0\,\,or\,\,x\in (0,\,\,\infty )\]
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