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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) $x^3+\frac{1}{x^4}-\frac{129}{8}$
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) $x^3+\frac{1}{x^4}-\frac{129}{8}$
Solution:
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Verified Answer
(a) $f(x)=\int\left(4 x^3-\frac{3}{x^4}\right) d x=x^4+\frac{1}{x^3}+\mathrm{C}$
$\therefore f(2)=(2)^4+\frac{1}{(2)^3}+C=0=-\frac{129}{8}$
$\therefore f(2)=(2)^4+\frac{1}{(2)^3}+C=0=-\frac{129}{8}$
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