A) \[\frac{-1}{13}\]
B) \[\frac{1}{3}\]
C) \[6\]
D) \[-\,6\]
Correct Answer: B
Solution :
(b): Let \[y=\frac{x+2}{2{{x}^{2}}+3x+6}\], then \[2{{x}^{2}}y+\left( 3y-1 \right)x+6y-2=0\] For x to be real, \[{{\left( 3y-1 \right)}^{2}}-8y\left( 6y-2 \right)\ge 0\] Or, \[\left( 1+13y \right)\left( 1-3y \right)\ge 0\] Or, \[\left( 13y+1 \right)\left( 3y-1 \right)\le 0\] Putting each factor equal to zero, we get, \[y\in \left[ \frac{-1}{13},\frac{1}{3} \right]\]; Thus, y will lie between \[-\frac{1}{13}\] and \[\frac{1}{3}\]. Hence the maximum value of y is \[\frac{1}{3}\] and minimum value is \[-\frac{1}{13}\].
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