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Step-by-Step Solution
Step 1: Understand the piecewise definition of the function
The function is defined as:
$$
f(x) =
\begin{cases}
\sin x - e^x & \text{if } x \le 0,\\
a + [-x] & \text{if } 0 < x < 1,\\
2x - b & \text{if } x \ge 1.
\end{cases}
$$
We need to ensure that $f(x)$ is continuous for all real $x$. Continuity at the boundary points $x=0$ and $x=1$ must be carefully checked.
Step 2: Continuity at $x=0$
For $f(x)$ to be continuous at $x=0$, the left-hand limit (as $x \to 0^-$) must equal the right-hand limit (as $x \to 0^+$), which must also equal the value of $f(0)$.
2.1: Left-hand limit at $x=0$
When $x \le 0$,
$$f(x) = \sin x - e^x.$$
So, at $x=0$,
$$f(0^-) = \sin(0) - e^0 = 0 - 1 = -1.$$
Also, $f(0)$ itself (since $x \le 0$ applies at $x=0$) is $-1.$
2.2: Right-hand limit at $x=0$
When $0 < x < 1$,
$$f(x) = a + [-x].$$
As $x \to 0^+$, $-x \to 0^-$, so the greatest integer less than or equal to $-x$ approaches $-1$ (since $-x$ is slightly negative for $x$ just greater than 0). Thus,
$$[-x] \bigg|_{x \to 0^+} = -1.$$
Therefore,
$$ f(0^+) = a + (-1) = a - 1.$$
2.3: Equate these for continuity at $x=0$
For continuity at $x=0$, we set
$$ f(0^-) = f(0^+) \implies -1 = a - 1. $$
Hence,
$$ a = 0. $$
Step 3: Continuity at $x=1$
Next, we ensure continuity at $x = 1$ by matching the left-hand limit (from $0 < x < 1$) to the right-hand limit (from $x \ge 1$), and also to $f(1)$.
3.1: Left-hand limit at $x=1$
For $0 < x < 1$,
$$ f(x) = a + [-x]. $$
When $x = 1^-$, $-x = -1$, so
$$[-1] = -1.$$
Since we found $a=0$,
$$ f(1^-) = 0 + (-1) = -1.$$
3.2: Right-hand limit at $x=1$
For $x \ge 1$,
$$ f(x) = 2x - b. $$
At $x=1$,
$$ f(1^+) = 2(1) - b = 2 - b.$$
But $f(1)$ must be the same from the right side definition at $x=1$, giving $f(1)=2-b$ as well.
3.3: Equate these for continuity at $x=1$
Continuity demands
$$ f(1^-) = f(1^+) \implies -1 = 2 - b. $$
Hence,
$$ b = 3. $$
Step 4: Find $a + b$
From the previous steps, we have $a = 0$ and $b = 3$. Therefore,
$$ a + b = 0 + 3 = 3. $$
Final Answer
The value of $(a + b)$ is $3$.
