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The problem asks to determine the domain \( D \) of the vector function:

\[
\mathbf{R}(t) = 3t^2 \mathbf{i} - e^{2t} \mathbf{j} + \ln(t+1) \mathbf{k}
\]

### To find the domain:
We analyze the individual components:

1. \( 3t^2 \mathbf{i} \): This is a polynomial, so it's defined for all real numbers \( t \in \mathbb{R} \).
   
2. \( e^{2t} \mathbf{j} \): The exponential function is also defined for all real numbers \( t \in \mathbb{R} \).

3. \( \ln(t+1) \mathbf{k} \): The natural logarithm \( \ln(t+1) \) is only defined when \( t + 1 > 0 \), i.e., \( t > -1 \).

### Conclusion:
The domain of the vector function \( \mathbf{R}(t) \) is \( t > -1 \), or \( D = (-1, \infty) \).

Thus, the correct answer is:
\[
D = (-1, \infty)
\]

Would you like more details on this explanation? Here are some follow-up questions you might consider:

1. How do we find the domain of a logarithmic function in general?
2. What is the domain of the exponential function, and why?
3. How do you compute the domain for vector-valued functions?
4. Can you explain how polynomial functions impact the domain?
5. What happens to the domain when multiple functions are combined?

**Tip:** When finding the domain of a vector function, ensure that all components are defined for the same range of values of \( t \).

Cho hàm véc tơ \(R(t) = 3t^2 \mathbf{i} - e^{2t} \mathbf{j} + \ln(t+1) \mathbf{k}\). Tập xác định của hàm véc tơ \(R(t)\) là gì?

Discover how to find the domain of the vector function \(R(t) = 3t^2 \mathbf{i} - e^{2t} \mathbf{j} + \ln(t+1) \mathbf{k}\). This involves key mathematical concepts such as logarithmic and exponential functions. Perfect for students in Grades 10-12.

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