A) \[\left( 1,\,\,\frac{1}{2} \right)\]
B) \[\left( \frac{1}{2},\,2 \right)\]
C) \[\left( 2,\,\,\frac{-1}{2} \right)\]
D) \[\left( \frac{-2}{3},\,\,\frac{-1}{6} \right)\]
Correct Answer: D
Solution :
\[\frac{dy}{dx}=\frac{a}{x}+2bx+1\]Þ\[{{\left( \frac{dy}{dx} \right)}_{x=1}}=a+2b+1=0\]Þ \[a=-2b-1\] and \[{{\left( \frac{dy}{dx} \right)}_{x=2}}=\frac{a}{2}+4b+1=0\] Þ \[\frac{-2b-1}{2}+4b+1=0\]Þ \[-b+4b+\frac{1}{2}=0\] Þ \[3b=\frac{-1}{2}\] Þ \[b=\frac{-1}{6}\] and \[a=\frac{1}{3}-1=\frac{-2}{3}\].
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