A) \[x\in \left( -\frac{3}{2},\frac{1}{4} \right)\]
B) \[x\in \left( -\frac{3}{2},\frac{1}{4} \right]\]
C) \[x\in \left[ -\frac{3}{2},\frac{1}{4} \right)\]
D) \[x<\frac{1}{4}\]
Correct Answer: A
Solution :
\[\sqrt{9{{x}^{2}}+6x+1}<(2-x)\] \[\Rightarrow \]\[\sqrt{{{(3x+1)}^{2}}}<(2-x)\] \[\Rightarrow \]\[\pm \,(3x+1)<2-x\] Taking positive sign \[3x+1<2-x\] \[\Rightarrow \] \[x<\frac{1}{4}\] Taking negative sign \[-3x-1<2-x\] \[\Rightarrow \] \[-2x<3\] \[\Rightarrow \] \[x>-\frac{3}{2}\] \[\therefore \] \[x\in \left( -\frac{3}{2},\frac{1}{4} \right)\]
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