MHT CET · Maths · Indefinite Integration
If \(\frac{\mathrm{d}}{\mathrm{d} x} \mathrm{f}(x)=4 x^3-\frac{3}{x^4}\) such that \(\mathrm{f}(2)=0\), then \(\mathrm{f}(x)\) is equal to
- A \(x^4+\frac{1}{x^3}+\frac{129}{8}\)
- B \(x^4+\frac{1}{x^3}-\frac{129}{8}\)
- C \(x^3+\frac{1}{x^4}+\frac{129}{8}\)
- D \(x^3+\frac{1}{x^4}-\frac{129}{8}\)
Answer & Solution
Correct Answer
(B) \(x^4+\frac{1}{x^3}-\frac{129}{8}\)
Step-by-step Solution
Detailed explanation
Given that \(\mathrm{f}^{\prime}(x)=4 x^3-3 x^{-4}\)
\(\begin{array}{ll}
\therefore & \mathrm{f}(x)=\int\left(4 x^3-3 x^{-4}\right) \mathrm{d} x \\
\therefore & \mathrm{f}(x)=x^4+\frac{1}{x^3}+\mathrm{c}
\end{array}\)
Given that \(\mathrm{f}(2)=0\)
\(\begin{array}{ll}
\therefore & 16+\frac{1}{8}+c=0 \\
\therefore & c=\frac{-129}{8} \\
\therefore & \mathrm{f}(x)=x^4+\frac{1}{x^3}-\frac{129}{8}
\end{array}\)
\(\begin{array}{ll}
\therefore & \mathrm{f}(x)=\int\left(4 x^3-3 x^{-4}\right) \mathrm{d} x \\
\therefore & \mathrm{f}(x)=x^4+\frac{1}{x^3}+\mathrm{c}
\end{array}\)
Given that \(\mathrm{f}(2)=0\)
\(\begin{array}{ll}
\therefore & 16+\frac{1}{8}+c=0 \\
\therefore & c=\frac{-129}{8} \\
\therefore & \mathrm{f}(x)=x^4+\frac{1}{x^3}-\frac{129}{8}
\end{array}\)
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