Let y = f(x) be defined parametrically as `y = t^(2) + t|t|, x = 2t - |t|, t in R`. Then, at x = 0,find f(x) and discuss continuity.

Answer

To solve the problem, we need to analyze the parametric equations given for \( y \) and \( x \) in terms of \( t \):

1. **Given Parametric Equations**:
\[ y = t^2 + t|t|, \quad x = 2t - |t|, \quad t \in \mathbb{R} \]

2. **Case Analysis for \( |t| \)**:
We will consider two cases based on the value of \( t \): when \( t \geq 0 \) and when \( t < 0 \).

### Case 1: \( t \geq 0 \)
- Here, \( |t| = t \).
- Substitute into the equations:
\[ x = 2t - t = t \]
\[ y = t^2 + t \cdot t = t^2 + t^2 = 2t^2 \]
- Therefore, for \( t \geq 0 \):
\[ y = 2x^2 \quad \text{(since \( x = t \))} \]

### Case 2: \( t < 0 \)
- Here, \( |t| = -t \).
- Substitute into the equations:
\[ x = 2t - (-t) = 2t + t = 3t \]
\[ y = t^2 + t(-t) = t^2 - t^2 = 0 \]
- Therefore, for \( t < 0 \):
\[ y = 0 \quad \text{(since \( x = 3t \))} \]

### Finding \( f(x) \) at \( x = 0 \)
- From the analysis:
- For \( x \geq 0 \): \( f(x) = 2x^2 \)
- For \( x < 0 \): \( f(x) = 0 \)

- Now we need to evaluate \( f(0) \):
\[ f(0) = 2 \cdot 0^2 = 0 \]

### Continuity Discussion
To check for continuity at \( x = 0 \), we need to verify:
\[ \lim_{x \to 0} f(x) = f(0) \]

1. **Left-hand Limit (LHL)**:
\[ \lim_{x \to 0^-} f(x) = 0 \quad \text{(since \( f(x) = 0 \) for \( x < 0 \))} \]

2. **Right-hand Limit (RHL)**:
\[ \lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} 2x^2 = 2 \cdot 0^2 = 0 \]

3. **Value at \( x = 0 \)**:
\[ f(0) = 0 \]

Since both limits exist and are equal to \( f(0) \):
\[ \lim_{x \to 0} f(x) = f(0) = 0 \]
Thus, \( f(x) \) is continuous at \( x = 0 \).

### Final Answer:
\[ f(x) = \begin{cases} 2x^2 & \text{if } x \geq 0 \\ 0 & \text{if } x < 0 \end{cases} \]
And \( f(x) \) is continuous at \( x = 0 \).