If `f(x)` is a continuous function such that `f(x)={(2Sin(-(pi)/2x),,x lt -1),(abs(ax^2+x+b),,-1lexle1),(Sinpix , , xgt1):}` . Find a+b

Options

A. -3
B. 3
C. -1
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 \( x = -1 \)

We need to check the left-hand limit and the right-hand limit at \( x = -1 \).

1. **Left-hand limit as \( x \to -1^- \)**:
\[ f(x) = 2 \sin\left(-\frac{\pi}{2} x\right) \]
Substituting \( x = -1 \):
\[ f(-1) = 2 \sin\left(-\frac{\pi}{2} \cdot (-1)\right) = 2 \sin\left(\frac{\pi}{2}\right) = 2 \cdot 1 = 2 \]

2. **Right-hand limit as \( x \to -1^+ \)**:
\[ f(x) = |ax^2 + x + b| \]
Substituting \( x = -1 \):
\[ f(-1) = |a(-1)^2 + (-1) + b| = |a - 1 + b| \]

For continuity at \( x = -1 \):
\[ \lim_{x \to -1^-} f(x) = \lim_{x \to -1^+} f(x) \]
Thus,
\[ 2 = |a - 1 + b| \tag{1} \]

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

Next, we check the continuity at \( x = 1 \).

1. **Left-hand limit as \( x \to 1^- \)**:
\[ f(x) = |ax^2 + x + b| \]
Substituting \( x = 1 \):
\[ f(1) = |a(1)^2 + 1 + b| = |a + 1 + b| \]

2. **Right-hand limit as \( x \to 1^+ \)**:
\[ f(x) = \sin(\pi x) \]
Substituting \( x = 1 \):
\[ f(1) = \sin(\pi \cdot 1) = \sin(\pi) = 0 \]

For continuity at \( x = 1 \):
\[ \lim_{x \to 1^-} f(x) = \lim_{x \to 1^+} f(x) \]
Thus,
\[ |a + 1 + b| = 0 \tag{2} \]

### Step 3: Solve the Equations

From equation (2):
\[ |a + 1 + b| = 0 \implies a + 1 + b = 0 \implies a + b = -1 \tag{3} \]

From equation (1):
\[ 2 = |a - 1 + b| \implies a - 1 + b = 2 \text{ or } a - 1 + b = -2 \]

**Case 1**: \( a - 1 + b = 2 \)
\[ a + b - 1 = 2 \implies a + b = 3 \tag{4} \]

**Case 2**: \( a - 1 + b = -2 \)
\[ a + b - 1 = -2 \implies a + b = -1 \tag{5} \]

### Step 4: Combine Results

From equations (3) and (5), we see that both lead to the same result:
\[ a + b = -1 \]

### Final Result

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