A) \[\left( \frac{1}{13},\frac{1}{3} \right)\]
B) \[\left[ -\frac{1}{13},\frac{1}{3} \right]\]
C) \[\left( -\frac{1}{3},\frac{1}{13} \right)\]
D) none of these
Correct Answer: B
Solution :
Let \[y=\frac{x+2}{2{{x}^{2}}+3x+6}\] \[\Rightarrow \] \[y(2{{x}^{2}}+3x+6)=x+2\] \[\Rightarrow \] \[2y\,{{x}^{2}}+(3y-1)x+6y-2=0\] Since, \[x\]is real, therefore discriminant \[{{b}^{4}}-4ac\ge 0\] \[\Rightarrow \]\[{{(3y-1)}^{2}}-4(2y)(6y-2)\ge 0\] \[\Rightarrow \]\[(3y-1)[3y-1-16y]\ge 0\] \[\Rightarrow \]\[(13y+1)(3y-1)\le 0\] \[\Rightarrow \]\[-\frac{1}{13}\le y\le \frac{1}{3}\] \[\therefore \]Maximum value of \[y\] is \[\frac{1}{3}\]and minimum value is \[-\frac{1}{13}.\]
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