Question

The number of points on \([0,2]\) where $f(x)= \begin{cases}x\\{x\\}+1 & 0 \leq x<1 \\ 2-\\{x\\} & 1 \leq x \leq 2\end{cases}$ fails to be continuous or derivable is (A) 0 (B) 1 (C) 2 (D) 3

Short answer

Answer: (C) 2

Step-by-step solution

Check for continuity at x=1

First, we need to check if the limit of the function exists at x=1. The limit from the left side is equal to the limit of the first piece at x=1: \(\lim\limits_{x \to 1^-} (x + 1) = 1+1 = 2\) And the limit from the right side is equal to the limit of the second piece at x=1: \(\lim\limits_{x \to 1^+} (2 - \{x\}) = 2 - \{1\} = 1\) Since the left and right limits are different, the function is discontinuous at x=1. Thus, the function fails to be continuous at 1 point.

Check for derivability at x=1

To check for derivability, we need to find the first derivative of the function for each piece and check if they are equal at x=1. First, find the derivative of the first piece: \(f'(x) = \frac{d}{dx} (x + 1) = 1\) For the second piece, we have: \(f'(x) = \frac{d}{dx} (2 - \{x\}) = 0\) Clearly, the first derivative for both pieces is not equal at x=1. Thus, the function fails to be derivable at 1 point.

Count the number of points

Now that we have found out that the function fails to be continuous and derivable at x=1, we can conclude that the number of points where the function is either discontinuous or non-differentiable is: 1 (discontinuity) + 1 (non-derivability) = 2 Therefore, the correct answer is (C) 2.