Question

Let \[ f(x)= \begin{cases} (x-1)\sin\!\left(\dfrac{1}{x-1}\right), & x \neq 1 \\ 0, & x = 1 \end{cases} \] Then which one of the following is true?

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

Hint:

Functions of the form \(x\sin(1/x)\) are differentiable at zero.

Answer 1

The Correct Option is C

Step 1: At \(x=1\), limit definition of derivative exists.
Step 2: At \(x=0\), function oscillates, derivative does not exist.