If `alpha=sin x cos^(3) x and beta=cos x sin^(3) x`, then :

Options

A. `alpha-beta gt 0, " for all " x " in " (0, (pi)/(4))`
B. `alpha-beta lt 0," for all " x " in "(0, (pi)/(4))`
C. `alpha+beta gt 0, " for all " x " in "(0, (pi)/(2))`
D. `alpha+beta lt 0, " for all "x " in " (0, (pi)/(2))`

Answer

To solve the problem where \( \alpha = \sin x \cos^3 x \) and \( \beta = \cos x \sin^3 x \), we need to find \( \alpha + \beta \) and \( \alpha - \beta \).

### Step-by-Step Solution

**Step 1: Calculate \( \alpha + \beta \)**

\[ \alpha + \beta = \sin x \cos^3 x + \cos x \sin^3 x \]

**Step 2: Factor out common terms**

We can factor out \( \sin x \cos x \):

\[ \alpha + \beta = \sin x \cos x (\cos^2 x + \sin^2 x) \]

**Step 3: Use the Pythagorean identity**

Using the identity \( \cos^2 x + \sin^2 x = 1 \):

\[ \alpha + \beta = \sin x \cos x \cdot 1 = \sin x \cos x \]

**Step 4: Determine the positivity of \( \alpha + \beta \)**

In the interval \( [0, \frac{\pi}{2}] \), both \( \sin x \) and \( \cos x \) are positive. Therefore, \( \sin x \cos x > 0 \) for \( x \in (0, \frac{\pi}{2}) \).

Thus, \( \alpha + \beta > 0 \) in the interval \( [0, \frac{\pi}{2}] \).

---

**Step 5: Calculate \( \alpha - \beta \)**

\[ \alpha - \beta = \sin x \cos^3 x - \cos x \sin^3 x \]

**Step 6: Factor out common terms**

Again, we can factor out \( \sin x \cos x \):

\[ \alpha - \beta = \sin x \cos x (\cos^2 x - \sin^2 x) \]

**Step 7: Use the double angle identity**

Using the identity \( \cos^2 x - \sin^2 x = \cos 2x \):

\[ \alpha - \beta = \sin x \cos x \cdot \cos 2x \]

**Step 8: Use the double angle formula for sine**

We know that \( \sin 2x = 2 \sin x \cos x \). Therefore, we can rewrite:

\[ \alpha - \beta = \frac{1}{2} \sin 2x \cdot \cos 2x \]

**Step 9: Determine the positivity of \( \alpha - \beta \)**

In the interval \( [0, \frac{\pi}{4}] \), \( \sin 2x \) and \( \cos 2x \) are both positive. Therefore, \( \sin 2x \cos 2x > 0 \) for \( x \in (0, \frac{\pi}{4}) \).

Thus, \( \alpha - \beta > 0 \) in the interval \( [0, \frac{\pi}{4}] \).

### Final Results

- \( \alpha + \beta = \sin x \cos x \) is positive in \( [0, \frac{\pi}{2}] \).
- \( \alpha - \beta = \frac{1}{2} \sin 2x \cos 2x \) is positive in \( [0, \frac{\pi}{4}] \).