A function f(x) is defined as follows for real x `f(x)={{:(1-x^(2)","," for "xlt1),(0",","for "x=1),(1+x^(2)","," for "x gt1):},` then

Options

A. f(x) is not continuous at x = 1
B. f(x) is continuous at x = 1
C. f(x) is continuous at all real numbers
D. none of these

Answer

To determine the continuity of the function \( f(x) \) defined as follows:

\[ f(x) = \begin{cases} 1 - x^2 & \text{for } x < 1 \\ 0 & \text{for } x = 1 \\ 1 + x^2 & \text{for } x > 1 \end{cases} \]

we need to check the continuity at the point \( x = 1 \).

### Step 1: Determine the Left-Hand Limit at \( x = 1 \)

The left-hand limit is calculated as follows:

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

Substituting \( x = 1 \):

\[ = 1 - (1)^2 = 1 - 1 = 0 \]

### Step 2: Determine the Right-Hand Limit at \( x = 1 \)

The right-hand limit is calculated as follows:

\[ \lim_{x \to 1^+} f(x) = \lim_{x \to 1^+} (1 + x^2) \]

Substituting \( x = 1 \):

\[ = 1 + (1)^2 = 1 + 1 = 2 \]

### Step 3: Evaluate the Function at \( x = 1 \)

The value of the function at \( x = 1 \) is given directly from the definition:

\[ f(1) = 0 \]

### Step 4: Check for Continuity

For the function to be continuous at \( x = 1 \), the following must hold:

\[ \lim_{x \to 1^-} f(x) = \lim_{x \to 1^+} f(x) = f(1) \]

From our calculations:

- Left-hand limit: \( 0 \)
- Right-hand limit: \( 2 \)
- Function value: \( f(1) = 0 \)

Since the left-hand limit \( (0) \) is not equal to the right-hand limit \( (2) \), the function is not continuous at \( x = 1 \).

### Conclusion

Thus, the function \( f(x) \) is **not continuous at \( x = 1 \)**.

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