If `f(x)=sin^6x+cos^6x` , then which one of the following is false

Options

A. `f(x)le1`
B. `f(x)le2`
C. `f(x)gt1/4`
D. `f(x)le1/8`

Answer

To solve the problem, we need to analyze the function \( f(x) = \sin^6 x + \cos^6 x \) and determine which of the provided statements is false.

### Step-by-Step Solution:

1. **Rewrite the function**:
\[ f(x) = \sin^6 x + \cos^6 x \]

2. **Use the identity for sum of cubes**:
We can rewrite \( f(x) \) using the identity \( a^3 + b^3 = (a + b)(a^2 - ab + b^2) \):
\[ f(x) = (\sin^2 x + \cos^2 x)(\sin^4 x - \sin^2 x \cos^2 x + \cos^4 x) \]
Since \( \sin^2 x + \cos^2 x = 1 \):
\[ f(x) = \sin^4 x - \sin^2 x \cos^2 x + \cos^4 x \]

3. **Simplify further**:
We can express \( \sin^4 x + \cos^4 x \) as:
\[ \sin^4 x + \cos^4 x = (\sin^2 x + \cos^2 x)^2 - 2\sin^2 x \cos^2 x = 1 - 2\sin^2 x \cos^2 x \]
Thus, we have:
\[ f(x) = 1 - 3\sin^2 x \cos^2 x \]

4. **Find the maximum and minimum values**:
To find the maximum and minimum values of \( f(x) \), we need to analyze \( \sin^2 x \cos^2 x \):
\[ \sin^2 x \cos^2 x = \frac{1}{4} \sin^2(2x) \]
The maximum value of \( \sin^2(2x) \) is 1, hence:
\[ \sin^2 x \cos^2 x \leq \frac{1}{4} \]
Therefore, the minimum value of \( f(x) \) occurs when \( \sin^2 x \cos^2 x = \frac{1}{4} \):
\[ f(x) = 1 - 3 \cdot \frac{1}{4} = 1 - \frac{3}{4} = \frac{1}{4} \]
The maximum value of \( f(x) \) occurs when \( \sin^2 x \cos^2 x = 0 \):
\[ f(x) = 1 - 0 = 1 \]

5. **Conclusion about the function**:
Thus, the function \( f(x) \) has a minimum value of \( \frac{1}{4} \) and a maximum value of \( 1 \):
\[ \frac{1}{4} \leq f(x) \leq 1 \]

6. **Evaluate the statements**:
- \( f(x) \leq 1 \) (True)
- \( f(x) \leq 2 \) (True)
- \( f(x) \geq \frac{1}{4} \) (True)
- \( f(x) \leq \frac{1}{8} \) (False)

### Final Answer:
The false statement is:
\[ \text{D. } f(x) \leq \frac{1}{8} \]