Question

Let \( f(x) \) be defined as follows: 
\[ f(x) = \begin{cases} 3x, & \text{if } x < 0 \\ \min(1+x+\lfloor x \rfloor, 2+x\lfloor x \rfloor), & \text{if } 0 \leq x \leq 2 \\ 5, & \text{if } x > 2 \end{cases} \]
where \(\lfloor . \rfloor\) denotes the greatest integer function. If \(\alpha\) and \(\beta\) are the number of points, where \(f\) is not continuous and is not differentiable, respectively, then \(\alpha + \beta\) equals:

Hint:

When evaluating discontinuity and differentiability for piecewise functions, always check transitions between piecewise segments and integer boundaries within the domain.

Answer 1

Correct Answer: 5

Step 1: Identify discontinuities. The function changes definition at \(x = 0\) and \(x = 2\). Evaluate limits from left and right at these points: \[ \lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} 3x = 0 \] \[ \lim_{x \to 0^+} f(x) = \min(1 + 0 + 0, 2 + 0 \times 0) = 1 \] \[ \lim_{x \to 2^-} f(x) = \min(1 + 2 + 1, 2 + 2 \times 1) = 4 \] \[ \lim_{x \to 2^+} f(x) = 5 \] Discontinuity at \(x = 0\) and \(x = 2\). 

Step 2: Identify points of non-differentiability. Check for differentiability at integer points within \([0, 2]\) and at \(x = 2\), as \(f(x)\) involves the floor function which is non-differentiable at integers: \[ f'(x) \text{ is not defined at } x = 1, 2 \] 

Step 3: Count \(\alpha\) and \(\beta\). \[ \alpha = 2 \text{ (discontinuity at 0 and 2)} \] \[ \beta = 3 \text{ (non-differentiability at 0, 1, and 2)} \] \[ \(\alpha + \beta = 5 \)

Answer 2

Step 1: Understand the given function.
The function \( f(x) \) is defined as:
\[ f(x) = \begin{cases} 3x, & \text{if } x < 0 \\ \min(1 + x + \lfloor x \rfloor, \; 2 + x \lfloor x \rfloor), & \text{if } 0 \leq x \leq 2 \\ 5, & \text{if } x > 2 \end{cases} \]
Here, \( \lfloor x \rfloor \) denotes the greatest integer less than or equal to \( x \). We need to find the number of points where \( f(x) \) is not continuous and not differentiable, and then compute \( \alpha + \beta \).

Step 2: Analyze the function for different intervals.
For \( x < 0 \):
\[ f(x) = 3x \] This is a linear function, hence continuous and differentiable everywhere in \( x < 0 \).

For \( 0 \leq x \leq 2 \):
We have: \[ f(x) = \min(1 + x + \lfloor x \rfloor, \; 2 + x \lfloor x \rfloor) \] The value of \( \lfloor x \rfloor \) changes at integer points 0, 1, and 2. Therefore, we must check at these boundary points for possible discontinuities or non-differentiabilities.

Case 1: For \( 0 \leq x < 1 \), \( \lfloor x \rfloor = 0 \):
\[ f(x) = \min(1 + x, 2) \] So \( f(x) = 1 + x \) for \( x \lt 1 \), as \( 1 + x \lt 2 \). Continuous and differentiable in this region.

Case 2: For \( 1 \leq x < 2 \), \( \lfloor x \rfloor = 1 \):
\[ f(x) = \min(1 + x + 1, 2 + x) \Rightarrow f(x) = \min(x + 2, x + 2) \Rightarrow f(x) = x + 2 \] Hence continuous and differentiable in this interval as well.

Step 3: Check boundary points for continuity and differentiability.
At \( x = 0 \):
Left-hand limit \( f(0^-) = 3(0) = 0 \).
Right-hand limit \( f(0^+) = 1 + 0 + 0 = 1 \).
Since these are not equal, \( f(x) \) is not continuous at \( x = 0 \).

At \( x = 1 \):
Left-hand value \( f(1^-) = 1 + 1 = 2 \).
Right-hand value \( f(1^+) = 1 + 2 = 3 \).
Not continuous at \( x = 1 \).

At \( x = 2 \):
From \( 0 \leq x \leq 2 \), \( f(2) = \min(1 + 2 + 2, 2 + 2(2)) = \min(5, 6) = 5 \).
For \( x > 2 \), \( f(x) = 5 \). Hence continuous at \( x = 2 \), but not differentiable since slope changes suddenly at that point.

Step 4: Identify all discontinuous and non-differentiable points.
- Discontinuous at \( x = 0, 1 \) → \( \alpha = 2 \).
- Non-differentiable at \( x = 0, 1, 2 \) → \( \beta = 3 \).

Step 5: Compute the final result.
\[ \alpha + \beta = 2 + 3 = 5 \]

Final Answer:
\[ \boxed{5} \]