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To solve problem 1, we need to determine the constants \(a\) and \(b\) so that the function \(f(x)\) is differentiable at \(x = 1\). The function is defined as:

\[
f(x) = 
\begin{cases}
x^2 - ax & \text{if } x \leq 1 \\
ax + b & \text{if } x > 1
\end{cases}
\]

For \(f(x)\) to be differentiable at \(x = 1\), it must be continuous and have the same derivative from both sides at \(x = 1\).

### Step 1: Continuity at \(x = 1\)
The continuity condition requires:

\[
\lim_{x \to 1^-} f(x) = \lim_{x \to 1^+} f(x) = f(1).
\]

Calculating these limits:

1. From the left (\(x \leq 1\)):
   
   \[
   f(1) = 1^2 - a \cdot 1 = 1 - a.
   \]

2. From the right (\(x > 1\)):

   \[
   f(1) = a \cdot 1 + b = a + b.
   \]

Equating both sides for continuity:

\[
1 - a = a + b. \quad \text{(Equation 1)}
\]

### Step 2: Differentiability at \(x = 1\)
The differentiability condition requires:

\[
\lim_{x \to 1^-} f'(x) = \lim_{x \to 1^+} f'(x).
\]

Finding the derivatives:

1. For \(x \leq 1\):

   \[
   f'(x) = \frac{d}{dx}(x^2 - ax) = 2x - a.
   \]

   So, 

   \[
   f'(1) = 2 \cdot 1 - a = 2 - a.
   \]

2. For \(x > 1\):

   \[
   f'(x) = \frac{d}{dx}(ax + b) = a.
   \]

Equating both derivatives for differentiability:

\[
2 - a = a. \quad \text{(Equation 2)}
\]

### Step 3: Solving the Equations
Now, solve Equations 1 and 2 simultaneously:

From Equation 2:

\[
2 - a = a \implies 2 = 2a \implies a = 1.
\]

Substituting \(a = 1\) into Equation 1:

\[
1 - 1 = 1 + b \implies 0 = 1 + b \implies b = -1.
\]

Therefore, \(a + b = 1 + (-1) = 0\).

### Final Answer:
\[
\boxed{0 \text{ (Option C)}}
\]

Would you like to see more details or have any questions? 

### Related Questions:
1. What is the difference between continuity and differentiability?
2. How do you check if a function is differentiable at a point?
3. Can a function be continuous but not differentiable? Provide an example.
4. What happens if \(a\) and \(b\) do not satisfy the differentiability condition?
5. How can you graphically verify if a function is differentiable at a given point?

### Tip:
Always ensure continuity before checking differentiability; differentiability implies continuity, but not the other way around!

The function f(x) = { x^2 - ax if x <= 1, ax + b if x > 1, where a and b are constants. If f is differentiable at x = 1, then a + b = ?

Learn how to solve the differentiability of a piecewise function where f(x) = x^2 - ax for x ≤ 1 and f(x) = ax + b for x > 1. This problem covers key concepts of continuity, differentiability, and limits. Ideal for AP Calculus students.

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