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To solve this, let's find the first three non-zero terms of the Maclaurin series for \( f(x) = x^2 \cos(5x) \).

### Step-by-Step Solution

1. **Maclaurin Series for \( \cos(x) \):**
   The Maclaurin series for \( \cos(x) \) is:
   \[
   \cos(x) = 1 - \frac{x^2}{2} + \frac{x^4}{24} - \cdots
   \]

2. **Adjust for \( \cos(5x) \):**
   Substitute \( 5x \) in place of \( x \) in the Maclaurin series:
   \[
   \cos(5x) = 1 - \frac{(5x)^2}{2} + \frac{(5x)^4}{24} - \cdots
   \]
   Simplify each term:
   \[
   \cos(5x) = 1 - \frac{25x^2}{2} + \frac{625x^4}{24} - \cdots
   \]

3. **Multiply by \( x^2 \):**
   Now, we multiply \( x^2 \) by each term in \( \cos(5x) \):
   \[
   f(x) = x^2 \cos(5x) = x^2 \left(1 - \frac{25x^2}{2} + \frac{625x^4}{24} - \cdots \right)
   \]
   Expanding this:
   \[
   f(x) = x^2 - \frac{25x^4}{2} + \frac{625x^6}{24} - \cdots
   \]

4. **Identify the First Three Non-Zero Terms:**
   The first three non-zero terms are:
   \[
   x^2 - \frac{25x^4}{2} + \frac{625x^6}{24}
   \]

### Conclusion
The correct answer matches the second option:
\[
x^2 - \frac{25x^4}{2} + \frac{625x^6}{24}
\]

Would you like further details, or do you have any questions about the steps?

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Here are 5 related questions to expand your understanding:

1. How is the Maclaurin series for a function different from the Taylor series?
2. How would the Maclaurin series change if we had \( f(x) = x^3 \cos(5x) \)?
3. What happens to the Maclaurin series if we change \( \cos(5x) \) to \( \sin(5x) \)?
4. Why are we only considering the first three non-zero terms in this problem?
5. How can we use the Maclaurin series for other trigonometric functions, like \( \tan(x) \) or \( \sec(x) \)?

**Tip:** Remember that Maclaurin series are just Taylor series centered at \( x = 0 \), which simplifies calculations for many common functions.

Using the Maclaurin series for cos(x), find the first three non-zero terms of the Maclaurin series for f(x) = x^2 cos(5x).

Explore how to find the first three non-zero terms of the Maclaurin series for f(x) = x^2 cos(5x). This calculus problem covers series expansion techniques and the use of the Maclaurin series for cosine.

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