\[ f(x) = \begin{cases} 3x, & \text{if } x < 0 \\ \min(1 + x + \lfloor x \rfloor, 2 + x \lfloor x \rfloor), & \text{if } 0 \leq x \leq 2 \\ 5, & \text{if } x > 2 \end{cases} \]
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 \(x = 2\).
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:
\[ f'(x) \text{ is not defined at } x = 1, 2 \]
\[ \alpha = 2 \quad (\text{discontinuity at 0 and 2}) \]
\[ \beta = 3 \quad (\text{non-differentiability at 0, 1, and 2}) \]
\[ \alpha + \beta = 5 \]
Let \( f : \mathbb{R} \to \mathbb{R} \) be a twice differentiable function such that \( f(x + y) = f(x) f(y) \) for all \( x, y \in \mathbb{R} \). If \( f'(0) = 4a \) and \( f \) satisfies \( f''(x) - 3a f'(x) - f(x) = 0 \), where \( a > 0 \), then the area of the region R = {(x, y) | 0 \(\leq\) y \(\leq\) f(ax), 0 \(\leq\) x \(\leq\) 2 is :
The term independent of $ x $ in the expansion of $$ \left( \frac{x + 1}{x^{3/2} + 1 - \sqrt{x}} \cdot \frac{x + 1}{x - \sqrt{x}} \right)^{10} $$ for $ x>1 $ is: