Question

The function \( f(x) \) is given by:

\[ f(x) = \begin{cases} x \lfloor x \rfloor & \text{if } 0 \leq x < 2 \\ (x - 1)x & \text{if } 2 \leq x < 3 \end{cases} \]

The function is:

• A. differentiable at \( x = 2 \)
• B. not differentiable at \( x = 2 \)
• C. continuous at \( x = 2 \)
• D. None of these

Hint:

For differentiability at a point, the function must be continuous at that point. If it is not continuous, it cannot be differentiable.

Answer 1

The Correct Option is B

Step 1: First, we need to check if the function is continuous at \( x = 2 \). - For \( 0 \leq x < 2 \), the function is \( f(x) = x[x] \), where \( [x] \) represents the greatest integer function. - For \( 2 \leq x < 3 \), the function is \( f(x) = (x - 1)x \). 
Step 2: Check if the function is continuous at \( x = 2 \): 
- Left-hand limit at \( x = 2 \) is: \[ \lim_{x \to 2^-} f(x) = 2[2] = 2 \times 2 = 4. \] - Right-hand limit at \( x = 2 \) is: \[ \lim_{x \to 2^+} f(x) = (2 - 1) \times 2 = 1 \times 2 = 2. \] Since the left-hand and right-hand limits are not equal, the function is not continuous at \( x = 2 \). 
Step 3: Since the function is not continuous at \( x = 2 \), it cannot be differentiable there. Therefore, the correct answer is: \[ \boxed{{B. not differentiable at } x = 2}. \]