Let `f(x)={(x-1)sin1/(x-1)if\ x!=1 0,\ if\ x=1` . Then which one of the following is true? (a) `f` is differentiable at `x=0\ ` and at`\ x-1` (b) `f` is differentiable at `x=0\ ` but not at`\ x=1` (c) `f` is differentiable at `x=0` nor at `x=1` (d) `f` is differentiable at `x=1\ ` but not at`\ x=0`

Options

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

Answer

To determine the differentiability of the function \( f(x) \) defined as:

\[ f(x) = \begin{cases} (x-1) \sin\left(\frac{1}{x-1}\right) & \text{if } x \neq 1 \\ 0 & \text{if } x = 1 \end{cases} \]

we will analyze the differentiability at the points \( x = 0 \) and \( x = 1 \).

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

To check if \( f \) is differentiable at \( x = 1 \), we need to compute the left-hand and right-hand derivatives at that point.

**Left-hand derivative at \( x = 1 \)**:

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

This simplifies to:

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

Since \( \sin\left(\frac{1}{h}\right) \) oscillates between -1 and 1 as \( h \) approaches 0, the limit does not exist.

**Right-hand derivative at \( x = 1 \)**:

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

This simplifies to:

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

Again, since \( \sin\left(\frac{1}{h}\right) \) oscillates between -1 and 1, this limit also does not exist.

Since both the left-hand and right-hand derivatives at \( x = 1 \) do not exist, \( f \) is not differentiable at \( x = 1 \).

### Step 2: Check differentiability at \( x = 0 \)

To check if \( f \) is differentiable at \( x = 0 \), we compute the derivative:

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

This simplifies to:

\[ = \lim_{h \to 0} \frac{(h - 1) \sin\left(\frac{1}{h - 1}\right)}{h} \]

This is a \( 0/0 \) form, so we can apply L'Hôpital's Rule:

Differentiating the numerator and denominator:

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

As \( h \to 0 \), \( \sin\left(\frac{1}{h - 1}\right) \) oscillates, but the term \( (h - 1) \) approaches -1, and thus the limit evaluates to a finite value, indicating that \( f \) is differentiable at \( x = 0 \).

### Conclusion

From the analysis, we conclude that:

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

Thus, the correct option is:

**(b) \( f \) is differentiable at \( x = 0 \) but not at \( x = 1 \)**.