Let `f:RrarrR` be defined as `f(x)={{:(2sin(-(pix)/2)",",if xlt-1),(|ax^2+x+b|",",if -1lexle1),(sin(pix)",",if xgt1):}`
If f(x) is continuous on R , then a+b equals

Options

A. -3
B. -1
C. 3
D. 1

Answer

To solve the problem, we need to ensure that the function \( f(x) \) is continuous at the points where the definition of the function changes, specifically at \( x = -1 \) and \( x = 1 \).

### Step 1: Check Continuity at \( x = -1 \)

The function is defined as:
- \( f(x) = 2\sin\left(-\frac{\pi x}{2}\right) \) for \( x < -1 \)
- \( f(x) = |ax^2 + x + b| \) for \( -1 \leq x \leq 1 \)

To ensure continuity at \( x = -1 \), we need:
\[ \lim_{x \to -1^-} f(x) = \lim_{x \to -1^+} f(x) = f(-1) \]

Calculating \( \lim_{x \to -1^-} f(x) \):
\[ \lim_{x \to -1^-} f(x) = 2\sin\left(-\frac{\pi(-1)}{2}\right) = 2\sin\left(\frac{\pi}{2}\right) = 2 \]

Calculating \( f(-1) \):
\[ f(-1) = |a(-1)^2 + (-1) + b| = |a - 1 + b| \]

Setting the limits equal for continuity:
\[ |a - 1 + b| = 2 \]

This gives us two equations:
1. \( a - 1 + b = 2 \) (Equation 1)
2. \( a - 1 + b = -2 \) (Equation 2)

### Step 2: Check Continuity at \( x = 1 \)

For \( x = 1 \):
- \( f(x) = |ax^2 + x + b| \) for \( -1 \leq x \leq 1 \)
- \( f(x) = \sin(\pi x) \) for \( x > 1 \)

Again, we need:
\[ \lim_{x \to 1^-} f(x) = \lim_{x \to 1^+} f(x) = f(1) \]

Calculating \( \lim_{x \to 1^-} f(x) \):
\[ \lim_{x \to 1^-} f(x) = |a(1)^2 + 1 + b| = |a + 1 + b| \]

Calculating \( f(1) \):
\[ f(1) = \sin(\pi \cdot 1) = 0 \]

Setting the limits equal for continuity:
\[ |a + 1 + b| = 0 \]

This implies:
\[ a + 1 + b = 0 \quad \text{(Equation 3)} \]

### Step 3: Solve the Equations

Now we have three equations:
1. \( a - 1 + b = 2 \)
2. \( a - 1 + b = -2 \)
3. \( a + 1 + b = 0 \)

From Equation 1:
\[ a + b = 3 \quad \text{(from } a - 1 + b = 2\text{)} \]

From Equation 2:
\[ a + b = 1 \quad \text{(from } a - 1 + b = -2\text{)} \]

From Equation 3:
\[ a + b = -1 \quad \text{(from } a + 1 + b = 0\text{)} \]

### Step 4: Determine Valid Values

Since \( a + b \) cannot simultaneously equal 3, 1, and -1, we must check which equations are consistent.

From Equation 3:
\[ a + b = -1 \]

This is the only valid equation that can be satisfied with the continuity conditions.

### Conclusion

Thus, the value of \( a + b \) is:
\[ \boxed{-1} \]