Q87 / 150 · MHT CET 2024 (10 May Shift 2) · Mathematics
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$x^4+\frac{1}{x^3}+\frac{129}{8}$
- Correct answerB$x^4+\frac{1}{x^3}-\frac{129}{8}$
- C$x^3+\frac{1}{x^4}+\frac{129}{8}$
- D$x^3+\frac{1}{x^4}-\frac{129}{8}$
Solution
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 $\mathrm{f}(2)=0$ $\begin{array}{ll} \therefore & 16+\frac{1}{8}+c=0 \\ \therefore & c=\frac{-129}{8} \\ \therefore & \mathrm{f}(x)=x^4+\frac{1}{x^3}-\frac{129}{8} \end{array}$
Given that $\mathrm{f}(2)=0$ $\begin{array}{ll} \therefore & 16+\frac{1}{8}+c=0 \\ \therefore & c=\frac{-129}{8} \\ \therefore & \mathrm{f}(x)=x^4+\frac{1}{x^3}-\frac{129}{8} \end{array}$
Hence, the correct answer is option: 'B'.