If `f(x)=sin^6x+cos^6x ,` then range of `f(x)` is `[1/4,1]` (b) `[1/4,3/4]` (c) `[3/4,1]` (d) none of these

Options

A. `[1/4,1]`
B. `[1/4,3/4]`
C. `[3/4,1]`
D. None of these

Explanation

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

### Step 1: Rewrite the Function
We can express \( f(x) \) in a different form using the identity for cubes:
\[ f(x) = \sin^6 x + \cos^6 x = (\sin^2 x)^3 + (\cos^2 x)^3 \]
Let \( a = \sin^2 x \) and \( b = \cos^2 x \). Then, we have \( a + b = 1 \).

### Step 2: Use the Sum of Cubes Formula
Using the formula for the sum of cubes:
\[ a^3 + b^3 = (a + b)(a^2 - ab + b^2) \]
Substituting \( a + b = 1 \):
\[ f(x) = 1(a^2 - ab + b^2) = a^2 - ab + b^2 \]

### Step 3: Express \( a^2 + b^2 \)
We can express \( a^2 + b^2 \) in terms of \( a + b \):
\[ a^2 + b^2 = (a + b)^2 - 2ab = 1 - 2ab \]
So, we can rewrite \( f(x) \):
\[ f(x) = (1 - 2ab) - ab = 1 - 3ab \]

### Step 4: Find the Range of \( ab \)
Since \( a = \sin^2 x \) and \( b = \cos^2 x \), we know:
\[ ab = \sin^2 x \cos^2 x = \frac{1}{4} \sin^2(2x) \]
The maximum value of \( \sin^2(2x) \) is 1, so:
\[ \text{Maximum of } ab = \frac{1}{4} \]
The minimum value of \( ab \) is 0 (when either \( \sin^2 x = 0 \) or \( \cos^2 x = 0 \)).

### Step 5: Determine the Range of \( f(x) \)
Now substituting the bounds of \( ab \) into \( f(x) \):
- When \( ab = 0 \):
\[ f(x) = 1 - 3(0) = 1 \]
- When \( ab = \frac{1}{4} \):
\[ f(x) = 1 - 3\left(\frac{1}{4}\right) = 1 - \frac{3}{4} = \frac{1}{4} \]

Thus, the range of \( f(x) \) is:
\[ \left[\frac{1}{4}, 1\right] \]

### Conclusion
The correct option is:
\[ \text{(a) } \left[\frac{1}{4}, 1\right] \]