Which of the following fiunction is not differentiable at x=1?

Options

A. `f(x) =(x^(2)-1)|(x-1)(x-2)|`
B. `f(x) = sin (|x-1|)-|x-1|`
C. `f(x) = tan (|x-1|)+|x-1|`
D. none of these

Answer

To determine which of the given functions is not differentiable at \( x = 1 \), we will analyze each function step by step.

### Step 1: Analyze the First Function
The first function is given as:
\[ f(x) = (x^2 - 1) |x - 1| (x - 2) \]
We can break this down:
- The term \( |x - 1| \) has a potential point of non-differentiability at \( x = 1 \).
- However, \( x^2 - 1 = (x - 1)(x + 1) \) is a polynomial and differentiable everywhere.
- The product of differentiable functions is differentiable unless there is a point where one of the functions is not differentiable.

At \( x = 1 \):
\[ f(1) = (1^2 - 1) |1 - 1| (1 - 2) = 0 \]
The left-hand derivative and right-hand derivative at \( x = 1 \) exist, hence \( f(x) \) is differentiable at \( x = 1 \).

### Step 2: Analyze the Second Function
The second function is:
\[ f(x) = \sin(|x - 1|) - |x - 1| \]
At \( x = 1 \):
- For \( x > 1 \), \( |x - 1| = x - 1 \) and \( f(x) = \sin(x - 1) - (x - 1) \).
- For \( x < 1 \), \( |x - 1| = 1 - x \) and \( f(x) = \sin(1 - x) - (1 - x) \).

Calculating the derivatives:
- Right-hand limit as \( h \to 0 \):
\[ \lim_{h \to 0} \frac{\sin(h) - h}{h} = 0 \]
- Left-hand limit as \( h \to 0 \):
\[ \lim_{h \to 0} \frac{\sin(-h) + h}{-h} = 0 \]
Both limits are equal, hence \( f(x) \) is differentiable at \( x = 1 \).

### Step 3: Analyze the Third Function
The third function is:
\[ f(x) = \tan(|x - 1|) + |x - 1| \]
At \( x = 1 \):
- For \( x > 1 \), \( |x - 1| = x - 1 \) and \( f(x) = \tan(x - 1) + (x - 1) \).
- For \( x < 1 \), \( |x - 1| = 1 - x \) and \( f(x) = \tan(1 - x) + (1 - x) \).

Calculating the derivatives:
- Right-hand limit as \( h \to 0 \):
\[ \lim_{h \to 0} \frac{\tan(h) + h}{h} = 2 \]
- Left-hand limit as \( h \to 0 \):
\[ \lim_{h \to 0} \frac{\tan(-h) + (1 - (1 - h))}{-h} = -2 \]
The limits are not equal, hence \( f(x) \) is not differentiable at \( x = 1 \).

### Conclusion
The function that is not differentiable at \( x = 1 \) is the third option.

### Final Answer
The function that is not differentiable at \( x = 1 \) is:
\[ \text{Option 3: } f(x) = \tan(|x - 1|) + |x - 1| \]