Prove

Solution:

It is given that,

$\frac{d}{d x} f(x)=4 x^{3}-\frac{3}{x^{4}}$

$\therefore$ Anti derivative of $4 x^{3}-\frac{3}{x^{4}}=f(x)$

$\therefore f(x)=\int 4 x^{3}-\frac{3}{x^{4}} d x$

$f(x)=4 \int x^{3} d x-3 \int\left(x^{-4}\right) d x$

$\therefore f(x)=4\left(\frac{x^{4}}{4}\right)-3\left(\frac{x^{-3}}{-3}\right)+\mathrm{C}$

$f(x)=x^{4}+\frac{1}{x^{3}}+\mathrm{C}$

Also,

$f(2)=0$

$\therefore f(2)=(2)^{4}+\frac{1}{(2)^{3}}+C=0$

$\Rightarrow 16+\frac{1}{8}+\mathrm{C}=0$

$\Rightarrow \mathrm{C}=-\left(16+\frac{1}{8}\right)$

$\Rightarrow \mathrm{C}=\frac{-129}{8}$

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

Hence, the correct answer is A.