Let `f(x) = {{:(-2 sin x, x le -pi//2),(a sin x + b, -pi//2 lt x lt pi//2),(cos x, x ge pi//2):}` If f(x) is continuous everywhere then (a,b) =

Options

A. (0,1)
B. (1,1)
C. (-1,1)
D. (-1,0)

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 \( x = -\frac{\pi}{2} \) and \( x = \frac{\pi}{2} \).

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

We need to check the continuity of \( f(x) \) at \( x = -\frac{\pi}{2} \). This means we need to ensure that:

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

Calculating \( f\left(-\frac{\pi}{2}\right) \):

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

Now, calculating the limit from the right:

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

Setting these equal for continuity:

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

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

Next, we check the continuity of \( f(x) \) at \( x = \frac{\pi}{2} \):

\[ f\left(\frac{\pi}{2}\right) = \cos\left(\frac{\pi}{2}\right) = 0 \]

Calculating the limit from the left:

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

Setting these equal for continuity:

\[ 0 = a + b \quad \text{(2)} \]

### Step 3: Solving the equations

Now we have a system of equations:

1. \( -a + b = 2 \)
2. \( a + b = 0 \)

From equation (2), we can express \( b \) in terms of \( a \):

\[ b = -a \]

Substituting \( b = -a \) into equation (1):

\[ -a + (-a) = 2 \implies -2a = 2 \implies a = -1 \]

Now substituting \( a = -1 \) back into equation (2):

\[ -1 + b = 0 \implies b = 1 \]

### Final Result

Thus, the values of \( (a, b) \) are:

\[ (a, b) = (-1, 1) \]