If y = f(x) defined parametrically by `x = 2t - |t - 1| and y = 2t^(2) + t|t|`, then

Options

A. (a)f(x) is continuous for all `x in R`
B. (b)f(x) is continuous for all `x in R - {2}`
C. (c)f(x) is differentiable for all `x in R`
D. (d)f(x) is differentiable for all `x in R - {2}`

Answer

To solve the problem step by step, we will analyze the parametric equations given for \( x \) and \( y \) in terms of \( t \), and then determine the continuity and differentiability of the function \( f(x) \).

### Step 1: Analyze the Parametric Equations

We have:
- \( x = 2t - |t - 1| \)
- \( y = 2t^2 + t|t| \)

### Step 2: Consider Cases for \( |t - 1| \)

The expression \( |t - 1| \) can be simplified based on the value of \( t \):
1. **Case 1**: \( t \geq 1 \)
- Here, \( |t - 1| = t - 1 \)
- Thus, \( x = 2t - (t - 1) = 2t - t + 1 = t + 1 \)
- For \( y \), since \( t \geq 1 \), \( |t| = t \):
\[ y = 2t^2 + t^2 = 3t^2 \]

2. **Case 2**: \( 0 \leq t < 1 \)
- Here, \( |t - 1| = 1 - t \)
- Thus, \( x = 2t - (1 - t) = 2t + t - 1 = 3t - 1 \)
- For \( y \), since \( t \geq 0 \), \( |t| = t \):
\[ y = 2t^2 + t^2 = 3t^2 \]

3. **Case 3**: \( t < 0 \)
- Here, \( |t - 1| = 1 - t \)
- Thus, \( x = 2t - (1 - t) = 2t + t - 1 = 3t - 1 \)
- For \( y \), since \( t < 0 \), \( |t| = -t \):
\[ y = 2t^2 - t^2 = t^2 \]

### Step 3: Express \( y \) in Terms of \( x \)

Now we have three cases for \( (x, y) \):
1. For \( t \geq 1 \):
- \( x = t + 1 \) implies \( t = x - 1 \)
- Thus, \( y = 3(x - 1)^2 = 3(x^2 - 2x + 1) = 3x^2 - 6x + 3 \)

2. For \( 0 \leq t < 1 \):
- \( x = 3t - 1 \) implies \( t = \frac{x + 1}{3} \)
- Thus, \( y = 3\left(\frac{x + 1}{3}\right)^2 = \frac{(x + 1)^2}{3} = \frac{x^2 + 2x + 1}{3} \)

3. For \( t < 0 \):
- \( x = 3t - 1 \) implies \( t = \frac{x + 1}{3} \)
- Thus, \( y = \left(\frac{x + 1}{3}\right)^2 = \frac{(x + 1)^2}{9} = \frac{x^2 + 2x + 1}{9} \)

### Step 4: Check Continuity and Differentiability

To check continuity and differentiability:
- The function \( f(x) \) must be continuous at the points where the cases meet, specifically at \( t = 0 \) and \( t = 1 \).

1. **At \( t = 0 \)**:
- From \( t < 0 \): \( y = \frac{(0 + 1)^2}{9} = \frac{1}{9} \)
- From \( 0 \leq t < 1 \): \( y = \frac{(0 + 1)^2}{3} = \frac{1}{3} \)
- These values do not match, indicating a discontinuity.

2. **At \( t = 1 \)**:
- From \( 0 \leq t < 1 \): \( y = \frac{(1 + 1)^2}{3} = \frac{4}{3} \)
- From \( t \geq 1 \): \( y = 3(1)^2 = 3 \)
- These values also do not match, indicating another discontinuity.

### Conclusion

The function \( f(x) \) is not continuous at \( x = -1 \) and \( x = 2 \), which means it is not differentiable at these points either.

### Final Answer

The function \( f(x) \) is continuous and differentiable for all \( x \in \mathbb{R} \) except at \( x = -1 \) and \( x = 2 \).