If frac{d}{d x} f(x)=4 x^{3}-frac{3}{x^{4}} s | vedclass

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(C) Given that,$\frac{d}{d x} f(x)=4 x^{3}-\frac{3}{x^{4}}$.To find $f(x)$,we integrate both sides with respect to $x$:$f(x) = \int \left( 4 x^{3} - 3

Vedclass · Accepted Answer

$x^{4}+\frac{1}{x^{3}}-\frac{129}{8}$

Vedclass · Answer

(C) Given that,$\frac{d}{d x} f(x)=4 x^{3}-\frac{3}{x^{4}}$.To find $f(x)$,we integrate both sides with respect to $x$:$f(x) = \int \left( 4 x^{3} - 3 x^{-4} \right) dx$$f(x) = 4 \int x^{3} dx - 3 \int x^{-4} dx$$f(x) = 4 \left( \frac{x^{4}}{4} \right) - 3 \left( \frac{x^{-3}}{-3} \right) + C$$f(x) = x^{4} + \frac{1}{x^{3}} + C$Given $f(2) = 0$,we substitute $x = 2$:$f(2) = (2)^{4} + \frac{1}{(2)^{3}} + C = 0$$16 + \frac{1}{8} + C = 0$$C = -\left( 16 + \frac{1}{8} \right) = -\frac{128+1}{8} = -\frac{129}{8}$Therefore,$f(x) = x^{4} + \frac{1}{x^{3}} - \frac{129}{8}$.Thus,the correct option is $C$.

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

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