A) \[{{x}^{2}}-7x+3=0\]
B) \[{{x}^{2}}-20x+66=0\]
C) \[{{x}^{2}}-17x+66=0\]
D) \[{{x}^{2}}-18x+60=0\]
Correct Answer: C
Solution :
\[f(x)=\left\{ \begin{align} & {{x}^{2}}-3,\,\,2<x<3 \\ & 2x+5,\,3<x<4 \\ \end{align} \right.\] \\[\underset{x\to {{3}^{-}}}{\mathop{\lim }}\,f(x)=\underset{x\to {{3}^{-}}}{\mathop{\lim }}\,({{x}^{2}}-3)=6\] and \[\underset{x\to {{3}^{+}}}{\mathop{\lim }}\,f(x)=\underset{x\to {{3}^{+}}}{\mathop{\lim }}\,(2x+5)=11\] Hence, the required equation will be \[{{x}^{2}}-\](sum of roots) x+ (Product of roots) = 0 i.e., \[{{x}^{2}}-17x+66=0\].
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