If I= _ 1 ^ 2 d x 2 x^ 3 -9 x^ 2 +12 x+4 , then - Vidyadip

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
$\frac{1}{3} < I < \frac{1}{2}$

✓

Answer

Correct option: A.

$\frac{1}{9} < I^{2} < \frac{1}{8}$

a
$f(x)=\frac{1}{\sqrt{2 x^{3}-9 x^{2}+12 x+4}}$

$f^{\prime}(x)=\frac{-6(x-1)(x-2)}{2\left(2 x^{3}-9 x^{2}+12 x+4\right)^{3 / 2}}$

$\therefore f(\mathrm{x})$ 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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