If `k=sin^(6)c+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 range of \( k = \sin^6 x + \cos^6 x \), we can follow these steps:

### Step 1: Rewrite the expression
We can express \( k \) in terms of \( \sin^2 x \) and \( \cos^2 x \). Let \( a = \sin^2 x \) and \( b = \cos^2 x \). Since \( a + b = 1 \), we can rewrite \( k \) as:
\[ k = a^3 + b^3 \]

### Step 2: Use the identity for sum of cubes
We can use the identity for the sum of cubes:
\[ a^3 + b^3 = (a + b)(a^2 - ab + b^2) \]
Substituting \( a + b = 1 \):
\[ k = 1 \cdot (a^2 - ab + b^2) = a^2 - ab + b^2 \]

### Step 3: Express \( a^2 + b^2 \)
Using the identity \( a^2 + b^2 = (a + b)^2 - 2ab \):
\[ a^2 + b^2 = 1 - 2ab \]
Thus, we can rewrite \( k \) as:
\[ k = (1 - 2ab) - ab = 1 - 3ab \]

### Step 4: Substitute \( ab \)
Since \( ab = \sin^2 x \cos^2 x = \frac{1}{4} \sin^2 2x \), we can substitute this into our expression for \( k \):
\[ k = 1 - 3 \left(\frac{1}{4} \sin^2 2x\right) = 1 - \frac{3}{4} \sin^2 2x \]

### Step 5: Determine 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 \( \sin^2 2x = 1 \):
\[ k = 1 - \frac{3}{4} \cdot 1 = 1 - \frac{3}{4} = \frac{1}{4} \]

### Conclusion
Thus, the range of \( k \) is:
\[ \frac{1}{4} \leq k \leq 1 \]
So, \( k \) belongs to the interval \( \left[\frac{1}{4}, 1\right] \).