Question

If $\frac{d}{dx}f\left(x\right)=4x^3-\frac{3}{x^4}$ such that $f\left(2\right)=0$. Then $f\left(x\right)$ 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. $x^3+\frac{1}{x^4}-\frac{129}{8}$

Answer 1

The correct option is A $x^4+\frac{1}{x^3}-\frac{129}{8}$

Given,$\frac{d}{dx}f\left(x\right)=4x^3-\frac{3}{x^4}$
The anti derivative of $(4x^3-\frac{3}{x^4})=f\left(x\right)$
Hence,
$f\left(x\right)=\int (4x^3-\frac{3}{x^4})dx$
By taking terms separately, we get
$f\left(x\right)=4\int x^{3}dx-3\int \left(x^{-4}\right)dx=4\left(\frac{x^4}{4}\right)-3\left(\frac{x^{-3}}{-3}\right)+C=x^4+\frac{1}{x^3}+C$
Also, $f\left(2\right)=0$
By substituting $x=2,$ we get
$f\left(2\right)=(2{)}^4+\frac{1}{(2{)}^3}+C=0$
$\Rightarrow 16+\frac{1}{8}+C=0$
$C=-(16+\frac{1}{8})=\left(\frac{-129}{8}\right)$
Hence, $f\left(x\right)=x^4+\frac{1}{x^3}-\frac{129}{8}$