Let
`Delta(x)=|(sinx, cosx, sin2x+cos2x),(0,1,1),(1,0,-1)|` then `Delta'(x)` vanishes at least once in

Options

A. `(0,pi//2)`
B. `(pi//2,pi)`
C. `(0,pi//4)`
D. `(-pi//2,0)`

Answer

To solve the problem, we need to find the value of the determinant \( \Delta(x) \) and then compute its derivative \( \Delta'(x) \) to determine where it vanishes at least once.

### Step 1: Calculate \( \Delta(x) \)

Given:
\[ \Delta(x) = \begin{vmatrix} \sin x & \cos x & \sin 2x + \cos 2x \\ 0 & 1 & 1 \\ 1 & 0 & -1 \end{vmatrix} \]

Using the determinant formula for a \( 3 \times 3 \) matrix:
\[ \Delta(x) = a(ei - fh) - b(di - fg) + c(dh - eg) \]
where \( a, b, c \) are the elements of the first row, and \( d, e, f, g, h, i \) are the elements of the second and third rows.

Substituting the values:
- \( a = \sin x \), \( b = \cos x \), \( c = \sin 2x + \cos 2x \)
- \( d = 0 \), \( e = 1 \), \( f = 1 \)
- \( g = 1 \), \( h = 0 \), \( i = -1 \)

Calculating each term:
\[ \Delta(x) = \sin x \cdot (1 \cdot (-1) - 1 \cdot 0) - \cos x \cdot (0 \cdot (-1) - 1 \cdot 1) + (\sin 2x + \cos 2x) \cdot (0 \cdot 0 - 1 \cdot 1) \]
\[ = \sin x \cdot (-1) - \cos x \cdot (-1) + (\sin 2x + \cos 2x) \cdot (-1) \]
\[ = -\sin x + \cos x - \sin 2x - \cos 2x \]
Thus, we have:
\[ \Delta(x) = -\sin x + \cos x - \sin 2x - \cos 2x \]

### Step 2: Calculate \( \Delta'(x) \)

Now we differentiate \( \Delta(x) \):
\[ \Delta'(x) = -\cos x - \sin x - 2\cos 2x + \sin 2x \]

### Step 3: Set \( \Delta'(x) = 0 \)

To find where \( \Delta'(x) \) vanishes:
\[ -\cos x - \sin x - 2\cos 2x + \sin 2x = 0 \]

### Step 4: Analyze the intervals

We need to find the intervals where this equation has solutions. We can analyze the behavior of \( \Delta'(x) \) over the intervals \( [0, \pi] \) and \( [\pi, 2\pi] \).

1. **Interval \( [0, \pi] \)**:
- At \( x = 0 \):
\[ \Delta'(0) = -\cos(0) - \sin(0) - 2\cos(0) + \sin(0) = -1 - 0 - 2(1) + 0 = -3 \]
- At \( x = \frac{\pi}{2} \):
\[ \Delta'(\frac{\pi}{2}) = -\cos(\frac{\pi}{2}) - \sin(\frac{\pi}{2}) - 2\cos(\pi) + \sin(\pi) = 0 - 1 + 2(1) + 0 = 1 \]
- At \( x = \pi \):
\[ \Delta'(\pi) = -\cos(\pi) - \sin(\pi) - 2\cos(2\pi) + \sin(2\pi) = 1 - 0 - 2(1) + 0 = -1 \]

Since \( \Delta'(0) < 0 \) and \( \Delta'(\frac{\pi}{2}) > 0 \), by the Intermediate Value Theorem, there is at least one root in \( (0, \frac{\pi}{2}) \).

2. **Interval \( [\pi, 2\pi] \)**:
- At \( x = \frac{3\pi}{2} \):
\[ \Delta'(\frac{3\pi}{2}) = -\cos(\frac{3\pi}{2}) - \sin(\frac{3\pi}{2}) - 2\cos(3\pi) + \sin(3\pi) = 0 + 1 - 2(-1) + 0 = 3 \]
- At \( x = 2\pi \):
\[ \Delta'(2\pi) = -\cos(2\pi) - \sin(2\pi) - 2\cos(4\pi) + \sin(4\pi) = -1 - 0 - 2(1) + 0 = -3 \]

Since \( \Delta'(\frac{3\pi}{2}) > 0 \) and \( \Delta'(2\pi) < 0 \), there is at least one root in \( (\frac{3\pi}{2}, 2\pi) \).

### Conclusion

Thus, \( \Delta'(x) \) vanishes at least once in the intervals \( (0, \frac{\pi}{2}) \) and \( (\frac{3\pi}{2}, 2\pi) \).