Let K be the set of all values of x, where the function ` f(x) = sin |x| - |x| + 2(x-pi) cos |x| ` is not differentiable.
Then, the set K is equal to

Options

A. `{pi}`
B. `{0}`
C. `phi`(an empty set)
D. `{0,pi}`

Answer

To determine the set \( K \) of all values of \( x \) where the function

\[ f(x) = \sin |x| - |x| + 2(x - \pi) \cos |x| \]

is not differentiable, we will analyze the function in two cases based on the definition of the absolute value.

### Step 1: Analyze the function for \( x \geq 0 \)

For \( x \geq 0 \), we have:

\[ |x| = x \]

Thus, the function simplifies to:

\[ f(x) = \sin x - x + 2(x - \pi) \cos x \]

### Step 2: Analyze the function for \( x < 0 \)

For \( x < 0 \), we have:

\[ |x| = -x \]

Thus, the function simplifies to:

\[ f(x) = \sin(-x) - (-x) + 2(x - \pi) \cos(-x) \]

Using the properties of sine and cosine, we know:

\[ \sin(-x) = -\sin x \quad \text{and} \quad \cos(-x) = \cos x \]

So, the function becomes:

\[ f(x) = -\sin x + x + 2(x - \pi) \cos x \]

### Step 3: Check continuity of \( f(x) \)

To check if \( f(x) \) is continuous at \( x = 0 \), we need to evaluate the left-hand limit and right-hand limit at \( x = 0 \):

- As \( x \to 0^+ \):

\[ f(0) = \sin(0) - 0 + 2(0 - \pi) \cos(0) = 0 - 0 - 2\pi = -2\pi \]

- As \( x \to 0^- \):

\[ f(0) = -\sin(0) + 0 + 2(0 - \pi) \cos(0) = 0 + 0 - 2\pi = -2\pi \]

Since both limits equal \( -2\pi \), the function is continuous at \( x = 0 \).

### Step 4: Check differentiability of \( f(x) \)

Now, we need to check the differentiability at \( x = 0 \) by finding the left-hand derivative and right-hand derivative.

- **Right-hand derivative** at \( x = 0 \):

\[ f'(x) = \cos x - 1 + 2\left(\cos x - (x - \pi)\sin x\right) \]

Evaluating at \( x = 0 \):

\[ f'(0^+) = \cos(0) - 1 + 2\left(\cos(0) - (0 - \pi)\sin(0)\right) = 1 - 1 + 2(1 - 0) = 2 \]

- **Left-hand derivative** at \( x = 0 \):

\[ f'(x) = -\cos x + 1 + 2\left(\cos x - (x - \pi)\sin x\right) \]

Evaluating at \( x = 0 \):

\[ f'(0^-) = -\cos(0) + 1 + 2\left(\cos(0) - (0 - \pi)\sin(0)\right) = -1 + 1 + 2(1 - 0) = 2 \]

### Step 5: Conclusion

Since both the left-hand derivative and right-hand derivative at \( x = 0 \) are equal, the function is differentiable at \( x = 0 \).

### Final Result

The function \( f(x) \) is differentiable for all \( x \in \mathbb{R} \). Therefore, the set \( K \) of all values of \( x \) where \( f(x) \) is not differentiable is:

\[ K = \emptyset \]