Consider the function
`f(x)={{:(-2sinx,"for",xle-(pi)/(2)),(Asinx+B,"for",-(pi)/(2)lexle(pi)/(2)),(cosx,"for",xge(pi)/(2)):}`
which is continuous at everywhere
The value of A is

Options

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

Answer

To find the value of \( A \) in the piecewise function

\[ f(x) = \begin{cases} -2 \sin x & \text{for } x < -\frac{\pi}{2} \\ A \sin x + B & \text{for } -\frac{\pi}{2} \leq x \leq \frac{\pi}{2} \\ \cos x & \text{for } x > \frac{\pi}{2} \end{cases} \]

which is continuous everywhere, we need to ensure continuity at the points where the function changes its definition, specifically at \( x = -\frac{\pi}{2} \) and \( x = \frac{\pi}{2} \).

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

For continuity at \( x = -\frac{\pi}{2} \), we need:

\[ \lim_{x \to -\frac{\pi}{2}^-} f(x) = \lim_{x \to -\frac{\pi}{2}^+} f(x) = f\left(-\frac{\pi}{2}\right) \]

Calculating the left-hand limit:

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

Calculating the right-hand limit:

\[ \lim_{x \to -\frac{\pi}{2}^+} f(x) = A \sin\left(-\frac{\pi}{2}\right) + B = -A + B \]

Setting these equal for continuity:

\[ 2 = -A + B \quad \text{(Equation 1)} \]

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

For continuity at \( x = \frac{\pi}{2} \), we need:

\[ \lim_{x \to \frac{\pi}{2}^-} f(x) = \lim_{x \to \frac{\pi}{2}^+} f(x) = f\left(\frac{\pi}{2}\right) \]

Calculating the left-hand limit:

\[ \lim_{x \to \frac{\pi}{2}^-} f(x) = A \sin\left(\frac{\pi}{2}\right) + B = A + B \]

Calculating the right-hand limit:

\[ \lim_{x \to \frac{\pi}{2}^+} f(x) = \cos\left(\frac{\pi}{2}\right) = 0 \]

Setting these equal for continuity:

\[ A + B = 0 \quad \text{(Equation 2)} \]

### Step 3: Solve the equations

We now 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 this into Equation 1:

\[ -A + (-A) = 2 \\ -2A = 2 \\ A = -1 \]

### Step 4: Find \( B \)

Now substituting \( A = -1 \) back into Equation 2 to find \( B \):

\[ -1 + B = 0 \\ B = 1 \]

### Conclusion

The values of \( A \) and \( B \) are:

\[ A = -1, \quad B = 1 \]

Thus, the value of \( A \) is \( \boxed{-1} \).