If `k=sin^(6)x+cos^(6)x,` then k belongs to the interval

Options

A. `[7//8,//54]`
B. `[1//2,//5//8]`
C. `[1//4,1]`
D. none of these

Answer

To find the interval of \( k = \sin^6 x + \cos^6 x \), we can follow these steps:

### Step 1: Rewrite \( k \)
We can rewrite \( k \) using the identity for the sum of cubes. Recall that:
\[ a^3 + b^3 = (a + b)(a^2 - ab + b^2) \]
Let \( a = \sin^2 x \) and \( b = \cos^2 x \). Then:
\[ k = (\sin^2 x)^3 + (\cos^2 x)^3 = (\sin^2 x + \cos^2 x)((\sin^2 x)^2 - \sin^2 x \cos^2 x + (\cos^2 x)^2) \]
Since \( \sin^2 x + \cos^2 x = 1 \), we have:
\[ k = 1 \cdot \left( \sin^4 x + \cos^4 x - \sin^2 x \cos^2 x \right) \]

### Step 2: Simplify \( \sin^4 x + \cos^4 x \)
Using the identity \( a^2 + b^2 = (a + b)^2 - 2ab \):
\[ \sin^4 x + \cos^4 x = (\sin^2 x + \cos^2 x)^2 - 2\sin^2 x \cos^2 x = 1 - 2\sin^2 x \cos^2 x \]
Thus, we can substitute this back into our expression for \( k \):
\[ k = 1 - 2\sin^2 x \cos^2 x - \sin^2 x \cos^2 x = 1 - 3\sin^2 x \cos^2 x \]

### Step 3: Express \( \sin^2 x \cos^2 x \)
We know that:
\[ \sin^2 x \cos^2 x = \left( \frac{1}{2} \sin 2x \right)^2 = \frac{1}{4} \sin^2 2x \]
Substituting this into our expression for \( k \):
\[ k = 1 - 3 \cdot \frac{1}{4} \sin^2 2x = 1 - \frac{3}{4} \sin^2 2x \]

### Step 4: Find the range of \( k \)
The maximum value of \( \sin^2 2x \) is 1, and the minimum value is 0. Therefore:
- When \( \sin^2 2x = 0 \):
\[ k = 1 - \frac{3}{4} \cdot 0 = 1 \]
- When \( \sin^2 2x = 1 \):
\[ k = 1 - \frac{3}{4} \cdot 1 = 1 - \frac{3}{4} = \frac{1}{4} \]

### Step 5: Conclusion
Thus, the values of \( k \) range from \( \frac{1}{4} \) to \( 1 \). Therefore, we conclude:
\[ k \in \left[ \frac{1}{4}, 1 \right] \]