A) \[\frac{1}{9}<{{I}^{2}}<\frac{1}{8}\]
B) \[\frac{1}{8}<{{I}^{2}}<\frac{1}{4}\]
C) \[\frac{1}{6}<{{I}^{2}}<\frac{1}{2}\]
D) \[\frac{1}{16}<{{I}^{2}}<\frac{1}{9}\]
Correct Answer: A
Solution :
Let \[f\left( x \right)=2{{x}^{3}}-9{{x}^{2}}+12x+4\] \[\Rightarrow \,\,\,\,\,f'\left( x \right)=6\left( {{x}^{2}}-3x+2 \right)\] \[~\therefore \,\,\,\,\,f\left( x \right)\text{ }decreases\text{ }in\text{ }\left( 1,\text{ }2 \right),\text{ }f\left( 1 \right)=9\] f (2) = 8 \[\Rightarrow \,\,\,\,\,\frac{1}{3}<I<\frac{1}{\sqrt{8}}\] \[\Rightarrow \,\,\,\,\,\frac{1}{9}<{{I}^{2}}<\frac{1}{8}\]
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