Differentiability Analysis of Piecewise Absolute Cosine Function

To determine the differentiability of the function

$$f(x) = \begin{cases} \left| x^2 \cos \left(\frac{\pi}{x}\right) \right|, & x \neq 0 \\ 0, & x = 0 \end{cases}$$

at $x = 0$ and $x = 2$, we analyze each point using the fundamental definition of differentiability.

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1. Differentiability at $x = 0$

By definition, the derivative of $f(x)$ at $x = 0$ is given by: $f'(0) = \lim_{h \to 0} \frac{f(0+h) - f(0)}{h}$

Substituting $f(0) = 0$ and $f(h) = \left| h^2 \cos \left(\frac{\pi}{h}\right) \right| = h^2 \left| \cos \left(\frac{\pi}{h}\right) \right|$ for $h \neq 0$: $f'(0) = \lim_{h \to 0} \frac{h^2 \left| \cos \left(\frac{\pi}{h}\right) \right| - 0}{h} = \lim_{h \to 0} h \left| \cos \left(\frac{\pi}{h}\right) \right|$

Since $0 \le \left| \cos \left(\frac{\pi}{h}\right) \right| \le 1$ for all $h \neq 0$, we have: $-\left| h \right| \le h \left| \cos \left(\frac{\pi}{h}\right) \right| \le \left| h \right|$

Applying the Squeeze (Sandwich) Theorem: $\lim_{h \to 0} \left(-\left| h \right|\right) = 0 \quad \text{and} \quad \lim_{h \to 0} \left| h \right| = 0$

Therefore, $f'(0) = \lim_{h \to 0} h \left| \cos \left(\frac{\pi}{h}\right) \right| = 0$

Since $f'(0)$ exists and is finite, $f$ is differentiable at $x = 0$.

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2. Differentiability at $x = 2$

Let $g(x) = x^2 \cos\left(\frac{\pi}{x}\right)$ so that $f(x) = |g(x)|$ near $x = 2$.

At $x = 2$: $g(2) = 2^2 \cos\left(\frac{\pi}{2}\right) = 4 \times 0 = 0$

Now, we evaluate the derivative of $g(x)$ at $x = 2$: $g'(x) = 2x \cos\left(\frac{\pi}{x}\right) + x^2 \left( -\sin\left(\frac{\pi}{x}\right) \right) \cdot \left( -\frac{\pi}{x^2} \right) = 2x \cos\left(\frac{\pi}{x}\right) + \pi \sin\left(\frac{\pi}{x}\right)$

Evaluating at $x = 2$: $g'(2) = 2(2) \cos\left(\frac{\pi}{2}\right) + \pi \sin\left(\frac{\pi}{2}\right) = 4(0) + \pi(1) = \pi \neq 0$

Now, let's find the Left-Hand Derivative (LHD) and Right-Hand Derivative (RHD) of $f(x)$ at $x = 2$:

• Left-Hand Derivative ($f'(2^-)$): For $x < 2$ (close to $2$), $\frac{\pi}{x} > \frac{\pi}{2}$, which implies $\cos\left(\frac{\pi}{x}\right) < 0$, so $g(x) < 0$. Thus, $f(x) = |g(x)| = -g(x)$. $f'(2^-) = \lim_{x \to 2^-} \frac{-g(x) - 0}{x - 2} = -g'(2) = -\pi$

• Right-Hand Derivative ($f'(2^+)$): For $x > 2$ (close to $2$), $\frac{\pi}{x} < \frac{\pi}{2}$, which implies $\cos\left(\frac{\pi}{x}\right) > 0$, so $g(x) > 0$. Thus, $f(x) = |g(x)| = g(x)$. $f'(2^+) = \lim_{x \to 2^+} \frac{g(x) - 0}{x - 2} = g'(2) = \pi$

Since $f'(2^-) \neq f'(2^+)$ ($-\pi \neq \pi$), $f$ is non-differentiable at $x = 2$.

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Conclusion

$f$ is differentiable at $x = 0$ but non-differentiable at $x = 2$.

Hence, the correct choice is B.