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Home  >>  CBSE XII  >>  Math  >>  Integrals
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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

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

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1 Answer

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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.

 

answered Jan 28, 2013 by sreemathi.v
 
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