Let f is defined by the following function
`f(x) = {(sinx, x lt 0), (x^2, 0lex lt 1),(2-x,1 le x le 2),(x-3,xge 2):}`
For what value of x, f is not continuous ?

Options

A. 1 only
B. 2 only
C. 0 and 2 only
D. 0,1, and 2

Answer

To determine the points at which the function \( f(x) \) is not continuous, we need to analyze the function piece by piece, particularly at the points where the definition of the function changes. The function is defined as follows:

\[ f(x) = \begin{cases} \sin x & \text{if } x < 0 \\ x^2 & \text{if } 0 \leq x < 1 \\ 2 - x & \text{if } 1 \leq x \leq 2 \\ x - 3 & \text{if } x > 2 \end{cases} \]

### Step 1: Identify the points of interest
The function changes its definition at \( x = 0 \), \( x = 1 \), and \( x = 2 \). We will check the continuity at these points.

### Step 2: Check continuity at \( x = 0 \)

1. **Calculate \( f(0) \)**:
\[ f(0) = 0^2 = 0 \]

2. **Calculate \( \lim_{x \to 0^-} f(x) \)**:
\[ \lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} \sin x = \sin(0) = 0 \]

3. **Calculate \( \lim_{x \to 0^+} f(x) \)**:
\[ \lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} x^2 = 0^2 = 0 \]

4. **Check if all limits are equal**:
\[ \lim_{x \to 0^-} f(x) = f(0) = \lim_{x \to 0^+} f(x) = 0 \]
Since all values are equal, \( f \) is continuous at \( x = 0 \).

### Step 3: Check continuity at \( x = 1 \)

1. **Calculate \( f(1) \)**:
\[ f(1) = 2 - 1 = 1 \]

2. **Calculate \( \lim_{x \to 1^-} f(x) \)**:
\[ \lim_{x \to 1^-} f(x) = \lim_{x \to 1^-} (2 - x) = 2 - 1 = 1 \]

3. **Calculate \( \lim_{x \to 1^+} f(x) \)**:
\[ \lim_{x \to 1^+} f(x) = \lim_{x \to 1^+} x^2 = 1^2 = 1 \]

4. **Check if all limits are equal**:
\[ \lim_{x \to 1^-} f(x) = f(1) = \lim_{x \to 1^+} f(x) = 1 \]
Since all values are equal, \( f \) is continuous at \( x = 1 \).

### Step 4: Check continuity at \( x = 2 \)

1. **Calculate \( f(2) \)**:
\[ f(2) = 2 - 2 = 0 \]

2. **Calculate \( \lim_{x \to 2^-} f(x) \)**:
\[ \lim_{x \to 2^-} f(x) = \lim_{x \to 2^-} (2 - x) = 2 - 2 = 0 \]

3. **Calculate \( \lim_{x \to 2^+} f(x) \)**:
\[ \lim_{x \to 2^+} f(x) = \lim_{x \to 2^+} (x - 3) = 2 - 3 = -1 \]

4. **Check if all limits are equal**:
\[ \lim_{x \to 2^-} f(x) = 0, \quad f(2) = 0, \quad \lim_{x \to 2^+} f(x) = -1 \]
Since \( \lim_{x \to 2^+} f(x) \neq f(2) \), \( f \) is not continuous at \( x = 2 \).

### Conclusion
The function \( f(x) \) is not continuous at \( x = 2 \).

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