Question

Let \[ f(x) = \cos(|\pi - x|) + (x - \pi) \sin |x| \quad \text{and} \quad g(x) = x^2 \text{ for } x \in \mathbb{R}. \] If \( h(x) = f(g(x)) \), then}

• A. \( h \) is not differentiable at \( x = 0 \)
• B. \( h'( \sqrt{2} ) = 0 \)
• C. \( h'(x) = 0 \) has a solution in \( (-\pi, \pi) \)
• D. There exists \( x_0 \in (-\pi, \pi) \) such that \( h(x_0) = x_0 \)

Hint:

When working with absolute value functions in compositions, be cautious of points where the function may not be differentiable, especially at the points where the argument inside the absolute value changes sign.

Answer 1

The Correct Option is B, C, D

Step 1: Analyze \( f(x) \).
We are given \( f(x) = \cos(|\pi - x|) + (x - \pi) \sin |x| \). The function \( f(x) \) is piecewise defined, and the absolute value functions \( |\pi - x| \) and \( |x| \) cause discontinuities at \( x = 0 \) and \( x = \pi \).
Step 2: Compute \( h(x) = f(g(x)) \).
The composition \( h(x) = f(g(x)) = f(x^2) \) is evaluated for each piece of \( f(x) \), noting that \( g(x) = x^2 \) affects the function differently depending on whether \( x \) is positive or negative.
Step 3: Check differentiability at \( x = 0 \).
At \( x = 0 \), \( g(x) = x^2 \) takes the value 0, so \( h(x) = f(x^2) \) becomes \( f(0) \), which involves terms that are not differentiable due to the absolute value functions involved in \( f(x) \). Therefore, \( h(x) \) is not differentiable at \( x = 0 \).
Step 4: Conclusion.
Thus, the correct answer is \( \boxed{(A)} \).