Browse Questions

# If $\frac{\large d}{\large dx}f(x)=4x^3-\frac{\large 3}{\large x^4}$ such that $f(2)=0$.Then $f(x)$ is

$\begin{array}{1 1}(A)\;x^4+\frac{1}{x^3}-\frac{129}{8} & (B)\;x^3+\frac{1}{x^4}+\frac{129}{8}\\(C)\;x^4+\frac{1}{x^3}+\frac{129}{8} & (D)\;x^3+\frac{1}{x^4}-\frac{129}{8}\end{array}$

Toolbox:
• $\frac{d}{dx}[F(x)+c]=f(x)$.
Given $\frac{d}{dx}f(x)=4x^3-\frac{3}{x^4}.$

$\int 4x^3-\frac{3}{x^4}=f(x)+c.$

$\;\;\;\;=4\begin{bmatrix}\frac{x^{3+1}}{3+1}\end{bmatrix}-3\begin{bmatrix}\frac{x^{-4+1}}{-4+1}\end{bmatrix}$.

$\;\;\;=4\frac{x^4}{4}-3\bigg(\frac{1}{3x^3}\bigg)$.

$f(x)=x^4+\frac{1}{x^3}+c.$

To find the value of c,let us take f(2)=0.Hence $2^4+\frac{1}{2^3}+c=0$.

$\Rightarrow 16+\frac{1}{8}=-c$

$c=\frac{-129}{8}.$

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

Hence A is the correct answer.