if I =


If $I=\int_{1}^{2} \frac{d x}{\sqrt{2 x^{3}-9 x^{2}+12 x+4}}$, then:

  1. (1) $\frac{1}{8}

  2. (2) $\frac{1}{9}

  3. (3) $\frac{1}{16}

  4. (4) $\frac{1}{6}

Correct Option: , 2


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

$f^{\prime}(x)=\frac{-1}{2}\left(\frac{\left(6 x^{2}-18 x+12\right)}{\left(2 x^{3}-9 x^{2}-12 x+4\right)^{3 / 2}}\right)$

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

$f(1)=\frac{1}{3}$ and $f(2)=\frac{1}{\sqrt{8}}$

It is increasing function



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