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Step-by-Step Solution
Step 1: Express the function piecewise
The given function is
$f(x) = \sin|x| - |x| + 2(x - \pi)\cos|x|.$
To analyze its differentiability, we consider two cases for $x$:
Case I: $x \ge 0 \implies |x| = x.$
Then
$f(x) = \sin x - x + 2(x - \pi)\cos x.$
Case II: $x < 0 \implies |x| = -x.$
Then
$f(x) = \sin(-x) - (-x) + 2(x - \pi)\cos(-x).$
Using $\sin(-x) = -\sin x$ and $\cos(-x) = \cos x,$ we get
$f(x) = -\sin x + x + 2(x - \pi)\cos x.$
Step 2: Find the derivatives for $x \neq 0$
Away from $x = 0,$ the function is a combination of sine, cosine, polynomials, and absolute value (but treated piecewise). Within each region ($x > 0$ and $x < 0$), there is no further absolute value “corner,” so the derivative can be computed directly:
Case I: $x > 0$
$f(x) = \sin x - x + 2(x - \pi)\cos x.$
This is a standard combination of differentiable functions for $x > 0,$ so $f'(x)$ exists and is well-defined for all $x > 0.$
Case II: $x < 0$
$f(x) = -\sin x + x + 2(x - \pi)\cos x.$
Again, a standard combination of sine, cosine, and polynomials for $x < 0,$ so $f'(x)$ exists and is well-defined for all $x < 0.$
Step 3: Check differentiability at $x = 0$
The main point of concern is $x = 0,$ where the expression involves $|x|.$ To check differentiability, we compute the left-hand derivative (from $x < 0$) and the right-hand derivative (from $x > 0$) and see if they match.
Right-hand limit ($x \to 0^+$):
For $x > 0,$ $f(x) = \sin x - x + 2(x - \pi)\cos x.$ We find
$$
f'(0^+) = \lim_{x \to 0^+} \frac{f(x) - f(0)}{x - 0}.
$$
However, an easier way is to directly differentiate for $x>0$ and then substitute $x=0$:
$$
f'(x) = \cos x - 1 + 2\cos x - 2(x - \pi)\sin x.
$$
Evaluating at $x=0$:
$$
f'(0^+) = \cos(0) - 1 + 2\cos(0) - 2(0 - \pi)\sin(0) = 1 - 1 + 2\cdot 1 - 0 = 2.
$$
Left-hand limit ($x \to 0^-$):
For $x < 0,$ $f(x) = -\sin x + x + 2(x - \pi)\cos x.$ Differentiating for $x 0$). The factor $(x-\pi)$ does not introduce any non-differentiability. Consequently, $f(x)$ remains differentiable at $x = \pi.$
Step 5: Conclude on non-differentiability
The checks show no point in the real line causes $f(x)$ to be non-differentiable. Therefore, the set
$$
K = \{ \text{all real } x \mid f(x) \text{ is not differentiable} \}
$$
is empty, i.e. $K = \varnothing.$
Final Answer
$K = \varnothing.$
