MCQ
If $I=\int\limits_{1}^{2} \frac{d x}{\sqrt{2 x^{3}-9 x^{2}+12 x+4}},$ then
- ✓$\frac{1}{9} < I^{2} < \frac{1}{8}$
- B$\frac{1}{3} < I^{2} < \frac{1}{2}$
- C$\frac{1}{9} < I < \frac{1}{8}$
- D$\frac{1}{3} < I < \frac{1}{2}$
$f^{\prime}(x)=\frac{-6(x-1)(x-2)}{2\left(2 x^{3}-9 x^{2}+12 x+4\right)^{3 / 2}}$
$\therefore f(\mathrm{x})$ is decreasing in $(1,2)$
$f(1)=\frac{1}{3} ; f(2)=\frac{1}{\sqrt{8}}$
$\frac{1}{3}<\mathrm{I}<\frac{1}{\sqrt{8}}$$\Rightarrow \mathrm{I}^{2} \in\left(\frac{1}{9}, \frac{1}{8}\right)$
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