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 analyze both \(\alpha + \beta\) and \(\alpha - \beta\).

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

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

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

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

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

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

### Step 2: Determine the sign of \(\alpha + \beta\)

In the interval \(0 \leq x \leq \frac{\pi}{2}\):

- \(\sin x \geq 0\) and \(\cos x \geq 0\)
- Therefore, \(\sin x \cos x \geq 0\)

Thus, \(\alpha + \beta\) is positive in the interval \(0 < x < \frac{\pi}{2}\).

### Step 3: Calculate \(\alpha - \beta\)

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

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

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

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

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

### Step 4: Determine the sign of \(\alpha - \beta\)

In the interval \(0 \leq x \leq \frac{\pi}{4}\):

- \(\sin x \geq 0\) and \(\cos x \geq 0\)
- Therefore, \(\sin x \cos x \geq 0\)
- For \(0 \leq x < \frac{\pi}{4}\), \(2x\) ranges from \(0\) to \(\frac{\pi}{2}\), where \(\cos 2x\) is positive.

Thus, \(\alpha - \beta\) is positive in the interval \(0 < x < \frac{\pi}{4}\).

### Conclusion

- \(\alpha + \beta\) is positive in \(0 < x < \frac{\pi}{2}\).
- \(\alpha - \beta\) is positive in \(0 < x < \frac{\pi}{4}\).

### Final Answer

The correct options are A and C.

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