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)

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