The given function is $f(x) = \left\{ {\begin{matrix} {{x^2}\left| {\cos {\pi \over x}} \right|,} & {x \ne 0} \\ {0,} & {x = 0} \\ \end{matrix} } \right.$ We need to check the differentiability of this function at x = 0 and x = 2.

Step 1: Differentiability at x = 0

The derivative of f(x) at x = 0 is given by the limit: $f'(0) = \lim_{h \to 0} \frac{f(0+h) - f(0)}{h}$ Substituting the function definition: $f'(0) = \lim_{h \to 0} \frac{h^2 \left| \cos \frac{\pi}{h} \right| - 0}{h}$ $f'(0) = \lim_{h \to 0} h \left| \cos \frac{\pi}{h} \right|$ We know that the value of $|\cos(\theta)|$ is always between 0 and 1, regardless of $\theta$. So, $0 \le \left| \cos \frac{\pi}{h} \right| \le 1$. Multiplying by |h|, we get: $0 \le |h| \left| \cos \frac{\pi}{h} \right| \le |h|$ This can be written as: $0 \le \left| h \cos \frac{\pi}{h} \right| \le |h|$ Now, we take the limit as $h \to 0$: $\lim_{h \to 0} 0 \le \lim_{h \to 0} \left| h \cos \frac{\pi}{h} \right| \le \lim_{h \to 0} |h|$ $0 \le \lim_{h \to 0} \left| h \cos \frac{\pi}{h} \right| \le 0$ By the Squeeze Theorem, we have: $\lim_{h \to 0} \left| h \cos \frac{\pi}{h} \right| = 0$ This implies that $\lim_{h \to 0} h \left| \cos \frac{\pi}{h} \right| = 0$. So, f'(0) = 0. The limit exists and is finite. Therefore, f(x) is differentiable at x = 0.

Step 2: Differentiability at x = 2

The derivative of f(x) at x = 2 is given by the limit: $f'(2) = \lim_{h \to 0} \frac{f(2+h) - f(2)}{h}$ First, we find f(2): $f(2) = 2^2 \left| \cos \frac{\pi}{2} \right| = 4 \times 0 = 0$. Now, we substitute this into the limit definition: $f'(2) = \lim_{h \to 0} \frac{(2+h)^2 \left| \cos\left(\frac{\pi}{2+h}\right) \right| - 0}{h}$ $f'(2) = \lim_{h \to 0} \frac{(2+h)^2}{h} \left| \cos\left(\frac{\pi}{2+h}\right) \right|$ To evaluate this limit, we need to check the left-hand derivative (LHD) and the right-hand derivative (RHD).

Right-Hand Derivative (RHD) at x = 2: We take the limit as $h \to 0^+$. For h > 0, 2+h > 2, which implies $\frac{1}{2+h} < \frac{1}{2}$ and $\frac{\pi}{2+h} < \frac{\pi}{2}$. The angle is in the first quadrant, so $\cos\left(\frac{\pi}{2+h}\right) > 0$. Thus, $\left| \cos\left(\frac{\pi}{2+h}\right) \right| = \cos\left(\frac{\pi}{2+h}\right)$. $RHD = \lim_{h \to 0^+} \frac{(2+h)^2}{h} \cos\left(\frac{\pi}{2+h}\right)$ Using the identity $\cos(\theta) = \sin(\frac{\pi}{2} - \theta)$: $RHD = \lim_{h \to 0^+} \frac{(2+h)^2}{h} \sin\left(\frac{\pi}{2} - \frac{\pi}{2+h}\right) = \lim_{h \to 0^+} \frac{(2+h)^2}{h} \sin\left(\frac{\pi h}{2(2+h)}\right)$ Using the standard limit $\lim_{u \to 0} \frac{\sin u}{u} = 1$: $RHD = \lim_{h \to 0^+} \frac{(2+h)^2}{h} \frac{\sin\left(\frac{\pi h}{2(2+h)}\right)}{\frac{\pi h}{2(2+h)}} \cdot \frac{\pi h}{2(2+h)}$ $RHD = \lim_{h \to 0^+} (2+h)^2 \cdot 1 \cdot \frac{\pi}{2(2+h)} = \lim_{h \to 0^+} \frac{\pi(2+h)}{2} = \frac{2\pi}{2} = \pi$

Left-Hand Derivative (LHD) at x = 2: We take the limit as $h \to 0^-$. For h < 0, 2+h < 2, which implies $\frac{1}{2+h} > \frac{1}{2}$ and $\frac{\pi}{2+h} > \frac{\pi}{2}$. For h approaching 0 from the left, the angle is in the second quadrant, so $\cos\left(\frac{\pi}{2+h}\right) < 0$. Thus, $\left| \cos\left(\frac{\pi}{2+h}\right) \right| = -\cos\left(\frac{\pi}{2+h}\right)$. $LHD = \lim_{h \to 0^-} \frac{(2+h)^2}{h} \left(-\cos\left(\frac{\pi}{2+h}\right)\right)$ $LHD = \lim_{h \to 0^-} -\frac{(2+h)^2}{h} \sin\left(\frac{\pi h}{2(2+h)}\right)$ Following the same procedure as for RHD: $LHD = \lim_{h \to 0^-} -(2+h)^2 \cdot 1 \cdot \frac{\pi}{2(2+h)} = \lim_{h \to 0^-} -\frac{\pi(2+h)}{2} = -\frac{2\pi}{2} = -\pi$

Since $RHD = \pi$ and $LHD = -\pi$, we have $LHD \ne RHD$. Therefore, f(x) is not differentiable at x = 2.

Step 3: Conclusion

From our analysis:

• f(x) is differentiable at x = 0.
• f(x) is not differentiable at x = 2.

This corresponds to option B.