Question

If $I=\underset{1}{\overset{2}{\int }}\frac{1}{\sqrt{2x^3-9x^2+12x+4}}dx$ , 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

The correct option is C $\frac{1}{9}<I^2<\frac{1}{8}$

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

$f^{\prime }\left(x\right)=\frac{-(6x^2-18x+12)}{2(2x^3-9x^2+12x+4{)}^{\frac{3}{2}}}$

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

$\Rightarrow f_{min}=f\left(1\right)$ and $f_{max}=f\left(2\right)$

$f\left(1\right)=\frac{1}{\sqrt{2-9+12+4}}=\frac{1}{\sqrt{9}}=\frac{1}{3}$
$f\left(2\right)=\frac{1}{\sqrt{16-36+24+4}}=\frac{1}{\sqrt{8}}$
$\frac{1}{3}<I<\frac{1}{\sqrt{8}}$
$\Rightarrow \frac{1}{9}<I^2<\frac{1}{8}$