Let `f(x)={x^2|(cos)pi/x|, x!=0 and 0,x=0,x in RR,` then `f` is

Options

A. differentiable both at x = 0 and at x = 2
B. differentiable at x = 0 but not differentiable at x = 2
C. not differentiable at x = 0 but differentiable at x = 2
D. differentiable neither at x = 0 nor at x = 2

Answer

To determine the differentiability of the function
\[ f(x) = \begin{cases} x^2 |\cos(\frac{\pi}{x})| & \text{if } x \neq 0 \\ 0 & \text{if } x = 0 \end{cases} \]
we will analyze the left-hand derivative and right-hand derivative at the point \( x = 0 \) and also check the differentiability at \( x = 2 \).

### Step 1: Finding the left-hand derivative at \( x = 0 \)

The left-hand derivative at \( x = 0 \) is given by:
\[ f'(0^-) = \lim_{h \to 0^-} \frac{f(-h) - f(0)}{-h} \]
Substituting \( f(-h) \) and \( f(0) \):
\[ f'(0^-) = \lim_{h \to 0^-} \frac{(-h)^2 |\cos(\frac{\pi}{-h})| - 0}{-h} = \lim_{h \to 0^-} \frac{h^2 |\cos(\frac{\pi}{h})|}{-h} \]
This simplifies to:
\[ f'(0^-) = \lim_{h \to 0^-} -h |\cos(\frac{\pi}{h})| \]
Since \( |\cos(\frac{\pi}{h})| \) is bounded between 0 and 1, we have:
\[ f'(0^-) = \lim_{h \to 0^-} -h \cdot \text{(bounded value)} = 0 \]

### Step 2: Finding the right-hand derivative at \( x = 0 \)

The right-hand derivative at \( x = 0 \) is given by:
\[ f'(0^+) = \lim_{h \to 0^+} \frac{f(h) - f(0)}{h} \]
Substituting \( f(h) \) and \( f(0) \):
\[ f'(0^+) = \lim_{h \to 0^+} \frac{h^2 |\cos(\frac{\pi}{h})| - 0}{h} = \lim_{h \to 0^+} h |\cos(\frac{\pi}{h})| \]
Again, since \( |\cos(\frac{\pi}{h})| \) is bounded, we have:
\[ f'(0^+) = \lim_{h \to 0^+} h \cdot \text{(bounded value)} = 0 \]

### Step 3: Conclusion at \( x = 0 \)

Since both the left-hand and right-hand derivatives at \( x = 0 \) are equal:
\[ f'(0^-) = f'(0^+) = 0 \]
Thus, \( f \) is differentiable at \( x = 0 \).

### Step 4: Finding the derivatives at \( x = 2 \)

Now, we check the differentiability at \( x = 2 \).

#### Right-hand derivative at \( x = 2 \):
\[ f'(2^+) = \lim_{h \to 0^+} \frac{f(2+h) - f(2)}{h} \]
Substituting \( f(2+h) \) and \( f(2) \):
\[ f'(2^+) = \lim_{h \to 0^+} \frac{(2+h)^2 |\cos(\frac{\pi}{2+h})| - 4 \cdot 0}{h} \]
This simplifies to:
\[ f'(2^+) = \lim_{h \to 0^+} \frac{(2+h)^2 |\cos(\frac{\pi}{2+h})|}{h} \]
As \( h \to 0 \), \( |\cos(\frac{\pi}{2+h})| \) approaches 0, leading to:
\[ f'(2^+) = 0 \]

#### Left-hand derivative at \( x = 2 \):
\[ f'(2^-) = \lim_{h \to 0^-} \frac{f(2-h) - f(2)}{-h} \]
Substituting \( f(2-h) \):
\[ f'(2^-) = \lim_{h \to 0^-} \frac{(2-h)^2 |\cos(\frac{\pi}{2-h})| - 4 \cdot 0}{-h} \]
This simplifies to:
\[ f'(2^-) = \lim_{h \to 0^-} \frac{(2-h)^2 |\cos(\frac{\pi}{2-h})|}{-h} \]
Again, as \( h \to 0 \), \( |\cos(\frac{\pi}{2-h})| \) approaches 0, leading to:
\[ f'(2^-) = 0 \]

### Step 5: Conclusion at \( x = 2 \)

Since both the left-hand and right-hand derivatives at \( x = 2 \) are equal:
\[ f'(2^-) = f'(2^+) = 0 \]
Thus, \( f \) is differentiable at \( x = 2 \).

### Final Conclusion

The function \( f(x) \) is differentiable at both \( x = 0 \) and \( x = 2 \).