If `a=sinxcos^3x` and `b=cosxsin^3x` 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, we need to calculate \( \alpha - \beta \) and \( \alpha + \beta \) where \( \alpha = \sin x \cos^3 x \) and \( \beta = \cos x \sin^3 x \).

### Step 1: Calculate \( \alpha - \beta \)

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

### Step 2: Factor the expression

We can factor the expression as follows:

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

### Step 3: Use the double angle identities

Recall the double angle identities:
- \( \sin 2x = 2 \sin x \cos x \)
- \( \cos 2x = \cos^2 x - \sin^2 x \)

Thus, we can rewrite the expression:

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

### Step 4: Rewrite using sine of a double angle

Using the identity \( \sin 2x \cos 2x = \frac{1}{2} \sin 4x \):

\[ \alpha - \beta = \frac{1}{4} \sin 4x \]

### Step 5: Analyze the function \( \alpha - \beta \)

The function \( \frac{1}{4} \sin 4x \) oscillates between -0.25 and 0.25. We need to check the values of this function in the interval \( [0, \frac{\pi}{4}] \).

- At \( x = 0 \):
\[ \alpha - \beta = \frac{1}{4} \sin 0 = 0 \]

- At \( x = \frac{\pi}{4} \):
\[ \alpha - \beta = \frac{1}{4} \sin \pi = 0 \]

Since \( \sin 4x \) is positive in the interval \( [0, \frac{\pi}{4}] \), \( \alpha - \beta \) is positive in this range.

### Step 6: Calculate \( \alpha + \beta \)

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

### Step 7: Factor the expression

We can factor the expression similarly:

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

### Step 8: Simplify using Pythagorean identity

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

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

### Step 9: Use the double angle identity

Again, using the double angle identity:

\[ \alpha + \beta = \frac{1}{2} \sin 2x \]

### Step 10: Analyze the function \( \alpha + \beta \)

The function \( \frac{1}{2} \sin 2x \) oscillates between -0.5 and 0.5. We need to check the values of this function in the interval \( [0, \frac{\pi}{2}] \).

- At \( x = 0 \):
\[ \alpha + \beta = \frac{1}{2} \sin 0 = 0 \]

- At \( x = \frac{\pi}{2} \):
\[ \alpha + \beta = \frac{1}{2} \sin \pi = 0 \]

Since \( \sin 2x \) is positive in the interval \( [0, \frac{\pi}{2}] \), \( \alpha + \beta \) is positive in this range.

### Conclusion

From the analysis:
- \( \alpha - \beta \) is positive in the interval \( [0, \frac{\pi}{4}] \).
- \( \alpha + \beta \) is positive in the interval \( [0, \frac{\pi}{2}] \).