Question

If $\frac{dx}{\sqrt{2x^{3}-9x^{2}+12x+4}}$, then :

• A. $\frac{1}{6} < I^{2} < \frac{1}{2}$
• B. $\frac{1}{8} < I^{2} < \frac{1}{4}$
• C. $\frac{1}{9} < I^{2} < \frac{1}{8}$
• D. $\frac{1}{16} < I^{2} < \frac{1}{9}$

Answer 1

Answer: (C)

$f\left(x\right) = \frac{1}{\sqrt{2x^{3}-9x^{2}+12x+4}}$
$f'\left(x\right) = \frac{-6\left(x-1\right)\left(x-2\right)}{2\left(2x^{3}-9x^{2}+12x+4\right)^{3/2}}$
$\therefore ƒ\left(x\right)$ is decreasing in $\left(1,2\right)$
$f\left(1\right) = \frac{1}{3} ; f\left(2\right) = \frac{1}{\sqrt{8}}$
$\frac{1}{3} < I < \frac{1}{\sqrt{8}} \Rightarrow I^{2} \in \left(\frac{1}{9}, \frac{1}{8}\right)$