Consider the function
`f(x)={{:(ax-2" for "-2ltxlt-1),(-1" for "-1lexle1),(a+2(x-1)^(2)" for "1ltxlt2):}`
What is the value of a for which f(x) is continuous at `x=-1 and x=1`?

Options

A. `-1`
B. 1
C. 0
D. 2

Answer

To determine the value of \( a \) for which the function \( f(x) \) is continuous at \( x = -1 \) and \( x = 1 \), we need to ensure that the left-hand limit and right-hand limit at these points are equal to the function value at those points.

### Step 1: Analyze the function at \( x = -1 \)

For \( x \) approaching \(-1\) from the left (denoted as \( x \to -1^- \)):
- The function is given by \( f(x) = ax - 2 \).

Thus, the left-hand limit at \( x = -1 \) is:
\[ \lim_{h \to 0} f(-1 - h) = \lim_{h \to 0} (a(-1 - h) - 2) = \lim_{h \to 0} (-a - ah - 2) = -a - 2 \]

For \( x = -1 \) (the function value):
- The function value is \( f(-1) = -1 \).

Setting the left-hand limit equal to the function value:
\[ -a - 2 = -1 \]

### Step 2: Solve for \( a \)

Rearranging the equation:
\[ -a = -1 + 2 \\ -a = 1 \\ a = -1 \]

### Step 3: Analyze the function at \( x = 1 \)

For \( x \) approaching \( 1 \) from the left (denoted as \( x \to 1^- \)):
- The function value is \( f(x) = -1 \).

Thus, the left-hand limit at \( x = 1 \) is:
\[ \lim_{h \to 0} f(1 - h) = -1 \]

For \( x = 1 \) (the function value):
- The function value is \( f(1) = -1 \).

Setting the left-hand limit equal to the function value:
\[ -1 = -1 \]

This condition holds true, confirming continuity at \( x = 1 \).

### Conclusion

The value of \( a \) for which \( f(x) \) is continuous at both \( x = -1 \) and \( x = 1 \) is:
\[ \boxed{-1} \]