If $\cfrac{d}{{dx}}f\left( x \right) = 4{x^2} - \cfrac{3}{{{x^4}}}$ ,such that $f\left( 2 \right) = 0$ then, $f\left( x \right)$ is

Option a is correct

$f\left( x \right) = \int {\left( {4{x^3} - \cfrac{3}{{{x^4}}}} \right)dx} = 4\int {{x^3}dx} - 3\int {{x^{ - 4}}dx}$
$= \cfrac{{4{x^4}}}{4} - \cfrac{{3{x^{ - 3}}}}{{ - 3}} + C = {x^4} + \cfrac{1}{{{x^3}}} + C$

$\therefore$ $f\left( 2 \right) = {\left( 2 \right)^4} + \cfrac{1}{{{{\left( 2 \right)}^3}}} + C = 0$ $\Rightarrow$ $C = - 16 - \cfrac{1}{8} = - \cfrac{{129}}{8}$

$\therefore$ $f\left( x \right) = {x^4} + \cfrac{1}{{{x^3}}} - \cfrac{{129}}{8}$

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