Let a function f(x) be definded by `f(x) =(x-|x-1|)/(x)`
which of the following is not true ?

Options

A. Discontinuous at x=0
B. Discontinuous at x=1
C. Not differentiable at x=0
D. Not differentiable at x=1

Answer

To solve the problem, we will analyze the function \( f(x) = \frac{x - |x - 1|}{x} \) and determine the properties of continuity and differentiability at specific points.

### Step 1: Analyze the function based on the definition of absolute value

The absolute value function \( |x - 1| \) can be defined piecewise:
- For \( x \geq 1 \), \( |x - 1| = x - 1 \)
- For \( x < 1 \), \( |x - 1| = -(x - 1) = 1 - x \)

Thus, we can rewrite \( f(x) \) in two cases:

1. **Case 1: \( x \geq 1 \)**
\[ f(x) = \frac{x - (x - 1)}{x} = \frac{1}{x} \]

2. **Case 2: \( x < 1 \)**
\[ f(x) = \frac{x - (1 - x)}{x} = \frac{2x - 1}{x} = 2 - \frac{1}{x} \]

### Step 2: Determine the function at specific points

- **At \( x = 0 \)**: The function is not defined since \( f(0) = \frac{0 - |0 - 1|}{0} \) leads to division by zero.

- **At \( x = 1 \)**:
\[ f(1) = \frac{1 - |1 - 1|}{1} = \frac{1 - 0}{1} = 1 \]

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

To check continuity at \( x = 1 \), we need to evaluate the left-hand limit and right-hand limit as \( x \) approaches 1.

- **Left-hand limit as \( x \to 1^- \)**:
\[ \lim_{x \to 1^-} f(x) = \lim_{x \to 1^-} \left( 2 - \frac{1}{x} \right = 2 - 1 = 1 \]

- **Right-hand limit as \( x \to 1^+ \)**:
\[ \lim_{x \to 1^+} f(x) = \lim_{x \to 1^+} \frac{1}{x} = 1 \]

Since both limits equal \( f(1) = 1 \), \( f(x) \) is continuous at \( x = 1 \).

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

To check differentiability at \( x = 1 \), we need to find the derivatives from both sides.

- **For \( x < 1 \)**:
\[ f'(x) = \frac{d}{dx} \left( 2 - \frac{1}{x} \right) = \frac{1}{x^2} \]

- **For \( x > 1 \)**:
\[ f'(x) = \frac{d}{dx} \left( \frac{1}{x} \right) = -\frac{1}{x^2} \]

Evaluating the derivatives at \( x = 1 \):
- Left-hand derivative \( f'(1^-) = 1 \)
- Right-hand derivative \( f'(1^+) = -1 \)

Since the left-hand and right-hand derivatives are not equal, \( f(x) \) is not differentiable at \( x = 1 \).

### Conclusion

1. \( f(x) \) is not defined at \( x = 0 \) (not continuous).
2. \( f(x) \) is continuous at \( x = 1 \).
3. \( f(x) \) is not differentiable at \( x = 1 \).

### Answer

The statement that is **not true** is that \( f(x) \) is differentiable at \( x = 1 \).