A) 2
B) 3
C) 4
D) 5
Correct Answer: D
Solution :
We have,\[{{x}^{2}}+9<{{(x+3)}^{2}}<8x+25\] \[\Rightarrow \]\[{{x}^{2}}+9<{{x}^{2}}+6x+9<8x+25\] \[\Rightarrow \]\[{{x}^{2}}+9<{{x}^{2}}+6x+9\] and\[{{x}^{2}}+6x+9<8x+25\] \[\Rightarrow \]\[6x>0\]and\[{{x}^{2}}-2x-16<0\] \[\Rightarrow \]\[x>0\]and\[1-\sqrt{17}<x<1+\sqrt{17}\] \[\Rightarrow \]\[0<x<1+\sqrt{17}\] \[\therefore \]x = 1, 2, 3, 4, 5
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