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 the range of this function. We will then evaluate the given statements to find which one is false.

### Step 1: Rewrite the function using algebraic identities
We can use the identity for the sum of cubes:
\[ a^3 + b^3 = (a + b)(a^2 - ab + b^2) \]
Let \( a = \sin^2 x \) and \( b = \cos^2 x \). Then:
\[ f(x) = a^3 + b^3 = (\sin^2 x)^3 + (\cos^2 x)^3 \]
Using the identity, we have:
\[ f(x) = (\sin^2 x + \cos^2 x)((\sin^2 x)^2 - \sin^2 x \cos^2 x + (\cos^2 x)^2) \]
Since \( \sin^2 x + \cos^2 x = 1 \), we can simplify:
\[ f(x) = 1 \cdot ((\sin^2 x)^2 - \sin^2 x \cos^2 x + (\cos^2 x)^2) \]

### Step 2: Simplify the expression
Now, we need to simplify \( (\sin^2 x)^2 + (\cos^2 x)^2 - \sin^2 x \cos^2 x \):
\[ (\sin^2 x)^2 + (\cos^2 x)^2 = \sin^4 x + \cos^4 x \]
Using the identity \( \sin^4 x + \cos^4 x = (\sin^2 x + \cos^2 x)^2 - 2\sin^2 x \cos^2 x \):
\[ = 1 - 2\sin^2 x \cos^2 x \]
Thus, we can rewrite \( f(x) \):
\[ f(x) = 1 - 3\sin^2 x \cos^2 x \]

### Step 3: Substitute \( \sin^2 x \cos^2 x \)
Using the identity \( \sin^2 x \cos^2 x = \frac{1}{4} \sin^2(2x) \):
\[ f(x) = 1 - \frac{3}{4} \sin^2(2x) \]

### Step 4: Determine the range of \( f(x) \)
The maximum value of \( \sin^2(2x) \) is 1, and the minimum value is 0. Thus:
- Maximum of \( f(x) \):
\[ f_{\text{max}} = 1 - \frac{3}{4} \cdot 0 = 1 \]
- Minimum of \( f(x) \):
\[ f_{\text{min}} = 1 - \frac{3}{4} \cdot 1 = 1 - \frac{3}{4} = \frac{1}{4} \]

### Step 5: Conclusion about the range
The range of \( f(x) \) is:
\[ \frac{1}{4} \leq f(x) \leq 1 \]

### Step 6: Evaluate the statements
Now we can evaluate the provided statements:
1. \( f(x) < 1 \) - True (since \( f(x) \) can be at most 1)
2. \( f(x) \leq 2 \) - True (since \( f(x) \) is always less than or equal to 1)
3. \( f(x) > \frac{1}{4} \) - True (since the minimum value is \( \frac{1}{4} \))
4. \( f(x) < \frac{1}{8} \) - False (since the minimum value is \( \frac{1}{4} \))

### Final Answer
The false statement is:
4. \( f(x) < \frac{1}{8} \)