MHT CET · Maths · Differential Equations
If \(\mathrm{f}^{\prime}(x)=x-\frac{5}{x^5}\) and \(\mathrm{f}(1)=4\), then \(\mathrm{f}(x)\) is
- A \(\frac{x^2}{2}+\frac{9}{4} \frac{1}{x^4}+\frac{5}{4}\)
- B \(\frac{x^2}{2}-\frac{5}{4} \frac{1}{x^4}+\frac{9}{4}\)
- C \(\frac{x^2}{2}+\frac{5}{4} \frac{1}{x^4}+\frac{9}{4}\)
- D \(\frac{x^2}{2}-\frac{9}{4} \frac{1}{x^4}+\frac{5}{4}\)
Answer & Solution
Correct Answer
(C) \(\frac{x^2}{2}+\frac{5}{4} \frac{1}{x^4}+\frac{9}{4}\)
Step-by-step Solution
Detailed explanation
Given \(\mathrm{f}^{\prime}(x)=x-\frac{5}{x^5}\)
\(\therefore \quad\) Integrating both sides, we get
\(\begin{array}{ll}
& \mathrm{f}(x)=\int\left(x-\frac{5}{x^5}\right) \mathrm{d} x \\
& \mathrm{f}(x)=\frac{x^2}{2}+\frac{5}{4} \times \frac{1}{x^4}+\mathrm{c} \\
\therefore \quad & \text { But } \mathrm{f}(1)=4 \\
\therefore \quad & \frac{1}{2}+\frac{5}{4}+\mathrm{c}=4 \\
\therefore \quad & \mathrm{c}=\frac{9}{4} \\
\therefore \quad & \mathrm{f}(x)=\frac{x^2}{2}+\frac{5}{4} \frac{1}{x^4}+\frac{9}{4}
\end{array}\)
\(\therefore \quad\) Integrating both sides, we get
\(\begin{array}{ll}
& \mathrm{f}(x)=\int\left(x-\frac{5}{x^5}\right) \mathrm{d} x \\
& \mathrm{f}(x)=\frac{x^2}{2}+\frac{5}{4} \times \frac{1}{x^4}+\mathrm{c} \\
\therefore \quad & \text { But } \mathrm{f}(1)=4 \\
\therefore \quad & \frac{1}{2}+\frac{5}{4}+\mathrm{c}=4 \\
\therefore \quad & \mathrm{c}=\frac{9}{4} \\
\therefore \quad & \mathrm{f}(x)=\frac{x^2}{2}+\frac{5}{4} \frac{1}{x^4}+\frac{9}{4}
\end{array}\)
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