If $\frac{d}{d x} f(x)=4 x^{3}-\frac{3}{x^{4}}$ such that $f(2)=0$, then $f(x)$ is
(A) $x^{4}+\frac{1}{x^{3}}-\frac{129}{8}$
(B) $x^{3}+\frac{1}{x^{4}}+\frac{129}{8}$
(C) $x^{4}+\frac{1}{x^{3}}+\frac{129}{8}$
(D) $x^{3}+\frac{1}{x^{4}}-\frac{129}{8}$
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.