Class 12 Mathematics - Chapter Integrals NCERT Solutions | If \begin{align} \frac{d}{dx} f(x) = 4x^
If \begin{align} \frac{d}{dx} f(x) = 4x^3 - \frac{3}{x^4}\end{align} such that f(2) = 0 , then f(x) is
\begin{align} (A) x^4 + \frac {1}{x^3} - \frac{129}{8} \;\;\;\;(B) x^3 + \frac{1}{x^4} + \frac{129}{8}\end{align}
\begin{align} (c) x^3 + \frac {1}{x^4} + \frac{129}{8} \;\;\;\;(D) x^3 + \frac{1}{x^4} - \frac{129}{8}\end{align}
It is given that,
\begin{align} \frac{d}{dx} f(x) = 4x^3 - \frac{3}{x^4}\end{align}
∴ Anti derivative of
\begin{align} 4x^3 - \frac{3}{x^4} = f(x)\end{align}
∴ \begin{align} f(x)= \int \left(4x^3 - \frac{3}{x^4}\right).dx\end{align}
\begin{align} f(x)= 4\int x^3.dx - 3\int {x^{-4}}.dx\end{align}
\begin{align} f(x)= 4\left(\frac {x^4}{4}\right) - 3\left(\frac {x^{-3}}{-3}\right) + C\end{align}
∴ \begin{align} f(x)= x^4 + \frac{1}{x^3} + C\end{align}
Also, f(2) = 0
∴ \begin{align} f(2) =\left(2\right)^4 + \frac{1}{\left(2\right)^3} + C = 0 \end{align}
=> \begin{align} 16 + \frac{1}{8} + C = 0 \end{align}
=> \begin{align} C = -\left(16 + \frac{1}{8}\right) \end{align}
=> \begin{align} C = \frac{-129}{8} \end{align}
∴ \begin{align} f(x)= x^4 + \frac{1}{x^3} -\frac{129}{8} \end{align}
Hence, the correct answer is A.
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