Determine a & b so that f is continuous at `x = pi/2`
whre `f(x) = [{:((1-sin^(3)x)/(3 cos^(2)x), if, x lt pi/2),(a, if, x=pi/2),((b(1-sinx))/(pi-2x)^(2), if, x gt pi/2):}`

Answer

To determine the values of \( a \) and \( b \) so that the function \( f \) is continuous at \( x = \frac{\pi}{2} \), we need to ensure that:

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

Given the function:

\[ f(x) = \begin{cases} \frac{1 - \sin^3 x}{3 \cos^2 x} & \text{if } x < \frac{\pi}{2} \\ a & \text{if } x = \frac{\pi}{2} \\ \frac{b(1 - \sin x)}{(\pi - 2x)^2} & \text{if } x > \frac{\pi}{2} \end{cases} \]

### Step 1: Calculate \( \lim_{x \to \frac{\pi}{2}^-} f(x) \)

We need to evaluate:

\[ \lim_{x \to \frac{\pi}{2}^-} \frac{1 - \sin^3 x}{3 \cos^2 x} \]

Substituting \( x = \frac{\pi}{2} \):

\[ 1 - \sin^3\left(\frac{\pi}{2}\right) = 1 - 1 = 0 \]
\[ 3 \cos^2\left(\frac{\pi}{2}\right) = 3 \cdot 0 = 0 \]

This gives us a \( \frac{0}{0} \) form, so we apply L'Hôpital's Rule.

### Step 2: Apply L'Hôpital's Rule

Differentiate the numerator and denominator:

- Numerator: \( \frac{d}{dx}(1 - \sin^3 x) = -3 \sin^2 x \cos x \)
- Denominator: \( \frac{d}{dx}(3 \cos^2 x) = -6 \cos x \sin x \)

Now we have:

\[ \lim_{x \to \frac{\pi}{2}^-} \frac{-3 \sin^2 x \cos x}{-6 \cos x \sin x} = \lim_{x \to \frac{\pi}{2}^-} \frac{3 \sin x}{2} \]

Substituting \( x = \frac{\pi}{2} \):

\[ \lim_{x \to \frac{\pi}{2}^-} \frac{3 \sin\left(\frac{\pi}{2}\right)}{2} = \frac{3 \cdot 1}{2} = \frac{3}{2} \]

Thus, we have:

\[ \lim_{x \to \frac{\pi}{2}^-} f(x) = \frac{3}{2} \]

### Step 3: Set \( f\left(\frac{\pi}{2}\right) \)

From the definition of \( f \):

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

### Step 4: Set the limits equal

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

\[ a = \frac{3}{2} \]

### Step 5: Calculate \( \lim_{x \to \frac{\pi}{2}^+} f(x) \)

Now we evaluate:

\[ \lim_{x \to \frac{\pi}{2}^+} \frac{b(1 - \sin x)}{(\pi - 2x)^2} \]

Substituting \( x = \frac{\pi}{2} \):

\[ 1 - \sin\left(\frac{\pi}{2}\right) = 1 - 1 = 0 \]
\[ (\pi - 2 \cdot \frac{\pi}{2})^2 = 0^2 = 0 \]

This also gives us a \( \frac{0}{0} \) form, so we apply L'Hôpital's Rule again.

### Step 6: Apply L'Hôpital's Rule Again

Differentiate the numerator and denominator:

- Numerator: \( \frac{d}{dx}(b(1 - \sin x)) = -b \cos x \)
- Denominator: \( \frac{d}{dx}((\pi - 2x)^2) = -4(\pi - 2x) \)

Now we have:

\[ \lim_{x \to \frac{\pi}{2}^+} \frac{-b \cos x}{-4(\pi - 2x)} = \lim_{x \to \frac{\pi}{2}^+} \frac{b \cos x}{4(\pi - 2x)} \]

Substituting \( x = \frac{\pi}{2} \):

\[ \cos\left(\frac{\pi}{2}\right) = 0 \quad \text{and} \quad \pi - 2\left(\frac{\pi}{2}\right) = 0 \]

Again, we have a \( \frac{0}{0} \) form. We apply L'Hôpital's Rule again.

### Step 7: Apply L'Hôpital's Rule Again

Differentiate again:

- Numerator: \( \frac{d}{dx}(b \cos x) = -b \sin x \)
- Denominator: \( \frac{d}{dx}(-4(\pi - 2x)) = 8 \)

Now we have:

\[ \lim_{x \to \frac{\pi}{2}^+} \frac{-b \sin x}{8} \]

Substituting \( x = \frac{\pi}{2} \):

\[ \lim_{x \to \frac{\pi}{2}^+} \frac{-b \cdot 1}{8} = -\frac{b}{8} \]

### Step 8: Set the limits equal

For continuity:

\[ -\frac{b}{8} = a \]

Substituting \( a = \frac{3}{2} \):

\[ -\frac{b}{8} = \frac{3}{2} \]

### Step 9: Solve for \( b \)

Multiplying both sides by -8:

\[ b = -12 \]

### Final Values

Thus, the values of \( a \) and \( b \) that make \( f \) continuous at \( x = \frac{\pi}{2} \) are:

\[ a = \frac{3}{2}, \quad b = -12 \]