Key Concepts and Formulas

• Continuity of a Function: A function $f(x)$ is continuous at a point $x = c$ if the following three conditions are met:
  1. $f(c)$ is defined.
  2. $\lim_{x \to c} f(x)$ exists.
  3. $\lim_{x \to c} f(x) = f(c)$.
• Existence of Limit: For the limit $\lim_{x \to c} f(x)$ to exist, the left-hand limit and the right-hand limit must be equal: $\lim_{x \to c^-} f(x) = \lim_{x \to c^+} f(x)$.
• Greatest Integer Function: The greatest integer function, denoted by $[x]$, gives the greatest integer less than or equal to $x$. For $0 < x < 1$, $[x] = 0$. For $-1 \le x < 0$, $[x] = -1$.

Step-by-Step Solution

Step 1: Analyze the function definition and continuity requirements. The function $f(x)$ is defined piecewise. For $f(x)$ to be continuous on $\mathbb{R}$, it must be continuous at the points where the definition changes, which are $x = 0$ and $x = 1$. We will use the definition of continuity to set up equations involving $a$ and $b$.

Step 2: Apply the continuity condition at $x = 0$. For continuity at $x = 0$, we need $\lim_{x \to 0^-} f(x) = \lim_{x \to 0^+} f(x) = f(0)$.

• Left-hand limit at $x = 0$: As $x$ approaches $0$ from the left ($x \le 0$), we use the first part of the function definition: $f(x) = \sin x - e^x$. $\lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} (\sin x - e^x) = \sin(0) - e^0 = 0 - 1 = -1$
• Right-hand limit at $x = 0$: As $x$ approaches $0$ from the right ($0 < x < 1$), we use the second part of the function definition: $f(x) = a + [-x]$. $\lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} (a + [-x])$ For $x$ slightly greater than $0$ (e.g., $0.001$), $-x$ is slightly less than $0$ (e.g., $-0.001$). The greatest integer less than or equal to a number slightly less than $0$ is $-1$. So, $[-x] = -1$ for $0 < x < 1$. $\lim_{x \to 0^+} (a + [-x]) = a + (-1) = a - 1$
• Function value at $x = 0$: For $x \le 0$, $f(x) = \sin x - e^x$. $f(0) = \sin(0) - e^0 = 0 - 1 = -1$

Equating the left-hand limit, right-hand limit, and function value at $x=0$: $-1 = a - 1 = -1$ From this, we get $a - 1 = -1$, which implies $a = 0$.

Step 3: Apply the continuity condition at $x = 1$. For continuity at $x = 1$, we need $\lim_{x \to 1^-} f(x) = \lim_{x \to 1^+} f(x) = f(1)$.

• Left-hand limit at $x = 1$: As $x$ approaches $1$ from the left ($0 < x < 1$), we use the second part of the function definition: $f(x) = a + [-x]$. $\lim_{x \to 1^-} f(x) = \lim_{x \to 1^-} (a + [-x])$ For $x$ slightly less than $1$ (e.g., $0.999$), $[-x]$ will be the greatest integer less than or equal to a number slightly less than $-1$ (e.g., $-0.999$). This integer is $-1$. $\lim_{x \to 1^-} (a + [-x]) = a + (-1) = a - 1$ Since we found $a = 0$ from Step 2, this limit is $0 - 1 = -1$.

• Right-hand limit at $x = 1$: As $x$ approaches $1$ from the right ($x \ge 1$), we use the third part of the function definition: $f(x) = 2x - b$. $\lim_{x \to 1^+} f(x) = \lim_{x \to 1^+} (2x - b) = 2(1) - b = 2 - b$

• Function value at $x = 1$: For $x \ge 1$, $f(x) = 2x - b$. $f(1) = 2(1) - b = 2 - b$

Equating the left-hand limit, right-hand limit, and function value at $x=1$: $a - 1 = 2 - b = 2 - b$ We already know $a = 0$. Substituting this into the equation: $0 - 1 = 2 - b$ $-1 = 2 - b$ Solving for $b$: $b = 2 + 1 = 3$

Step 4: Calculate $(a + b)$. We have found $a = 0$ and $b = 3$. $a + b = 0 + 3 = 3$

Common Mistakes & Tips

• Incorrectly evaluating the greatest integer function: Pay close attention to the interval for $[-x]$. For $0 < x < 1$, $-1 < -x < 0$, so $[-x] = -1$. For $x$ approaching $1$ from the left ($0 < x < 1$), $-x$ is between $-1$ and $0$, so $[-x] = -1$.
• Confusing left-hand and right-hand limits: Ensure you are using the correct part of the piecewise function for each limit. For $x \to c^-$, use the definition for $x \le c$ or intervals to the left of $c$. For $x \to c^+$, use the definition for $x \ge c$ or intervals to the right of $c$.
• Algebraic errors: Double-check your algebraic manipulations when solving for $a$ and $b$.

Summary

To ensure the function $f(x)$ is continuous on $\mathbb{R}$, we must enforce continuity at the points where the function definition changes, namely $x=0$ and $x=1$. By applying the definition of continuity ($\lim_{x \to c^-} f(x) = \lim_{x \to c^+} f(x) = f(c)$) at $x=0$, we found the value of $a$. Subsequently, by applying the same continuity condition at $x=1$ and using the value of $a$, we determined the value of $b$. Finally, we summed $a$ and $b$ to obtain the required result.

The final answer is $\boxed{3}$.