Question:medium

Let $f'(x) = 3(x^2 + 2x) - \frac{4}{x^3} + 5$, $f(1) = 0$. Then, $f(x)$ is:

Show Hint

To find the function from its derivative, integrate term by term, and use initial conditions to find the constant of integration.
Updated On: Feb 25, 2026
  • $x^3 + 3x^2 + \frac{2}{x^2} + 5x + 11$
  • $x^3 + 3x^2 + \frac{2}{x^2} + 5x - 11$
  • $x^3 + 3x^2 - \frac{2}{x^2} + 5x - 11$
  • $x^3 - 3x^2 - \frac{2}{x^2} + 5x - 11$
Show Solution

The Correct Option is C

Solution and Explanation

Given $f'(x) = 3(x^2 + 2x) - \frac{4}{x^3} + 5$. To find $f(x)$, we integrate $f'(x)$ with respect to $x$. Integrating term by term:
  • The integral of $3(x^2 + 2x)$ is $x^3 + 3x^2$.
  • The integral of $-\frac{4}{x^3}$ is $\frac{2}{x^2}$.
  • The integral of 5 is $5x$.
Thus, $f(x) = x^3 + 3x^2 + \frac{2}{x^2} + 5x + C$. Given $f(1) = 0$. Substituting $x = 1$: \[ 0 = 1^3 + 3(1^2) + \frac{2}{1^2} + 5(1) + C = 1 + 3 + 2 + 5 + C. \] Simplifying, $0 = 11 + C$, so $C = -11$. Therefore, $f(x) = x^3 + 3x^2 + \frac{2}{x^2} + 5x - 11$. This corresponds to option (3).
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