Let `f(x) = {(-2sin x, ",","if" x le -(pi)/2),(A sin x + B, ",", "if"-(pi)/2 lt x lt (pi)/2 ),(cos x, ",", "if" x ge (pi)/2 ):}`
Then

Options

A. f(x) is discontinuous for all A and B
B. f(x) is continuous for A = -1 and B = 1
C. f(x) is continuous for A = 1 and B = -1
D. f(x) is continuous for all real values of A, B

Answer

To solve the problem, we need to ensure that the function \( f(x) \) is continuous at the points where the piecewise function changes, specifically at \( x = -\frac{\pi}{2} \) and \( x = \frac{\pi}{2} \).

### Step 1: Check continuity at \( x = -\frac{\pi}{2} \)

We need to find the left-hand limit and the right-hand limit at \( x = -\frac{\pi}{2} \).

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

2. **Right-hand limit** as \( x \to -\frac{\pi}{2}^+ \):
\[ f(x) = A \sin x + B \]
\[ \lim_{x \to -\frac{\pi}{2}^+} f(x) = A \sin\left(-\frac{\pi}{2}\right) + B = A \cdot (-1) + B = -A + B \]

For continuity at \( x = -\frac{\pi}{2} \):
\[ \lim_{x \to -\frac{\pi}{2}^-} f(x) = \lim_{x \to -\frac{\pi}{2}^+} f(x) \]
\[ 2 = -A + B \quad \text{(Equation 1)} \]

### Step 2: Check continuity at \( x = \frac{\pi}{2} \)

Next, we check the limits at \( x = \frac{\pi}{2} \).

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

2. **Right-hand limit** as \( x \to \frac{\pi}{2}^+ \):
\[ f(x) = \cos x \]
\[ \lim_{x \to \frac{\pi}{2}^+} f(x) = \cos\left(\frac{\pi}{2}\right) = 0 \]

For continuity at \( x = \frac{\pi}{2} \):
\[ \lim_{x \to \frac{\pi}{2}^-} f(x) = \lim_{x \to \frac{\pi}{2}^+} f(x) \]
\[ A + B = 0 \quad \text{(Equation 2)} \]

### Step 3: Solve the system of equations

Now we have two equations:
1. \( -A + B = 2 \) (Equation 1)
2. \( A + B = 0 \) (Equation 2)

From Equation 2, we can express \( B \) in terms of \( A \):
\[ B = -A \]

Substituting \( B \) into Equation 1:
\[ -A + (-A) = 2 \]
\[ -2A = 2 \implies A = -1 \]

Now substituting \( A = -1 \) back into Equation 2:
\[ -1 + B = 0 \implies B = 1 \]

### Conclusion

The values of \( A \) and \( B \) are:
\[ A = -1, \quad B = 1 \]