Let `f(x)={((x-1)^2 sin(1/(x-1))-|x|,; x != 1), (-1,; x=1):}` then which one of the following is true?

Options

A. f is differential at x=0 but not at x=1
B. f is differentiable at x=1 but not at x=0
C. f is neither differentiable at x=0 nor at x=1
D. f is differentiable at x=0 and at x=1

Explanation

To determine the differentiability of the function

\[ f(x) = \begin{cases} (x-1)^2 \sin\left(\frac{1}{x-1}\right) - |x| & \text{if } x \neq 1 \\ -1 & \text{if } x = 1 \end{cases} \]

we need to check the differentiability at \( x = 0 \) and \( x = 1 \).

### Step 1: Check differentiability at \( x = 0 \)

We will use the definition of the derivative:

\[ f'(0) = \lim_{h \to 0} \frac{f(0 + h) - f(0)}{h} \]

First, we need to find \( f(0) \):

\[ f(0) = (0-1)^2 \sin\left(\frac{1}{0-1}\right) - |0| = 1 \cdot \sin(-1) - 0 = \sin(-1) \]

Now, substituting \( h \) into \( f \):

\[ f(h) = (h-1)^2 \sin\left(\frac{1}{h-1}\right) - |h| \]

Thus, we have:

\[ f'(0) = \lim_{h \to 0} \frac{(h-1)^2 \sin\left(\frac{1}{h-1}\right) - |h| - \sin(-1)}{h} \]

As \( h \to 0 \), \( |h| = h \) when \( h \) is positive and \( |h| = -h \) when \( h \) is negative. Therefore, we can analyze the limit separately for \( h \to 0^+ \) and \( h \to 0^- \).

For \( h \to 0^+ \):

\[ f'(0) = \lim_{h \to 0^+} \frac{(h-1)^2 \sin\left(\frac{1}{h-1}\right) - h - \sin(-1)}{h} \]

For \( h \to 0^- \):

\[ f'(0) = \lim_{h \to 0^-} \frac{(h-1)^2 \sin\left(\frac{1}{h-1}\right) + h - \sin(-1)}{h} \]

Both limits involve the term \((h-1)^2 \sin\left(\frac{1}{h-1}\right)\) which oscillates and does not converge as \( h \to 0 \). Therefore, we conclude that:

\[ f'(0) \text{ does not exist.} \]

### Step 2: Check differentiability at \( x = 1 \)

Now we check at \( x = 1 \):

\[ f'(1) = \lim_{h \to 0} \frac{f(1 + h) - f(1)}{h} \]

We know \( f(1) = -1 \). Thus:

\[ f'(1) = \lim_{h \to 0} \frac{(1+h-1)^2 \sin\left(\frac{1}{1+h-1}\right) - (-1)}{h} \]

This simplifies to:

\[ f'(1) = \lim_{h \to 0} \frac{h^2 \sin\left(\frac{1}{h}\right) + 1}{h} = \lim_{h \to 0} \left(h \sin\left(\frac{1}{h}\right) + \frac{1}{h}\right) \]

The term \( h \sin\left(\frac{1}{h}\right) \) approaches 0 as \( h \to 0 \) because \( \sin\left(\frac{1}{h}\right) \) is bounded between -1 and 1. The second term \( \frac{1}{h} \) diverges, indicating that the limit does not exist at \( x = 1 \).

### Conclusion

From our analysis, we find that:

- \( f \) is not differentiable at \( x = 0 \).
- \( f \) is not differentiable at \( x = 1 \).

Thus, the correct answer is:

**f is neither differentiable at \( x = 0 \) nor at \( x = 1 \).**