If `f(x)={{:(x",",0lexle1),(x+a",",xgt1):}` then

Options

A. f is discontinuous at x=1
B. f is differentiable at x=1
C. f is continuous but not differentible at x=1
D. none of these

Answer

To determine the continuity and differentiability of the function \( f(x) \) at \( x = 1 \), we will analyze the given piecewise function:

\[ f(x) = \begin{cases} x & \text{for } 0 \leq x \leq 1 \\ x + a & \text{for } x > 1 \end{cases} \]

### Step 1: Evaluate \( f(1) \)

First, we need to find the value of \( f(1) \) using the first piece of the function since \( 1 \) falls within the interval \( [0, 1] \).

\[ f(1) = 1 \]

### Step 2: Evaluate the limit as \( x \) approaches \( 1 \) from the left

Next, we will find the left-hand limit of \( f(x) \) as \( x \) approaches \( 1 \):

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

### Step 3: Evaluate the limit as \( x \) approaches \( 1 \) from the right

Now, we will find the right-hand limit of \( f(x) \) as \( x \) approaches \( 1 \):

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

### Step 4: Check for continuity at \( x = 1 \)

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

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

From our calculations:

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

Setting these equal gives:

\[ 1 + a = 1 \]

This simplifies to:

\[ a = 0 \]

### Step 5: Conclusion on continuity

If \( a \neq 0 \), then \( \lim_{x \to 1^+} f(x) \neq f(1) \), which means \( f(x) \) is discontinuous at \( x = 1 \). Therefore, if \( a \neq 0 \), \( f(x) \) is discontinuous at \( x = 1 \).

### Step 6: Check for differentiability at \( x = 1 \)

Since \( f(x) \) is discontinuous at \( x = 1 \) when \( a \neq 0 \), it cannot be differentiable at that point.

### Final Answer

Thus, the function \( f(x) \) is discontinuous at \( x = 1 \) if \( a \neq 0 \), and hence not differentiable at that point.

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