The number of points at which the function `f(x) =|x-0.5|+|x-1| + tan x` does not have a derivative in the interval (0,2) is:

Options

A. 1
B. 2
C. 3
D. 4

Answer

To determine the number of points at which the function \( f(x) = |x - 0.5| + |x - 1| + \tan x \) does not have a derivative in the interval \( (0, 2) \), we will analyze each component of the function separately.

### Step 1: Identify points of non-differentiability for the absolute value functions

The function \( f(x) \) consists of two absolute value functions: \( |x - 0.5| \) and \( |x - 1| \). These functions are not differentiable at the points where their arguments equal zero.

- For \( |x - 0.5| \):
- It is not differentiable at \( x = 0.5 \).

- For \( |x - 1| \):
- It is not differentiable at \( x = 1 \).

### Step 2: Identify points of non-differentiability for the tangent function

The function \( \tan x \) is not differentiable where it is undefined. The tangent function is undefined at odd multiples of \( \frac{\pi}{2} \).

- In the interval \( (0, 2) \):
- The only point where \( \tan x \) is undefined is at \( x = \frac{\pi}{2} \).

### Step 3: Compile all points of non-differentiability

Now we compile all the points where \( f(x) \) is not differentiable:

1. From \( |x - 0.5| \): \( x = 0.5 \)
2. From \( |x - 1| \): \( x = 1 \)
3. From \( \tan x \): \( x = \frac{\pi}{2} \)

### Step 4: Count the total points

Now we count the total number of points of non-differentiability in the interval \( (0, 2) \):

- \( x = 0.5 \)
- \( x = 1 \)
- \( x = \frac{\pi}{2} \) (approximately 1.57, which lies in the interval)

Thus, there are a total of **3 points** where the function does not have a derivative in the interval \( (0, 2) \).

### Final Answer

The number of points at which the function \( f(x) \) does not have a derivative in the interval \( (0, 2) \) is **3**.

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