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. `[1, 2]`

Answer

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 the sum of cubes:
\[ f(x) = \sin^6 x + \cos^6 x = (\sin^2 x + \cos^2 x)(\sin^4 x - \sin^2 x \cos^2 x + \cos^4 x) \]
Since \( \sin^2 x + \cos^2 x = 1 \), we have:
\[ f(x) = 1 \cdot (\sin^4 x - \sin^2 x \cos^2 x + \cos^4 x) \]

### Step 2: Simplify further
We can use the identity \( a^2 + b^2 = (a + b)^2 - 2ab \) to simplify \( \sin^4 x + \cos^4 x \):
\[ \sin^4 x + \cos^4 x = (\sin^2 x + \cos^2 x)^2 - 2\sin^2 x \cos^2 x = 1 - 2\sin^2 x \cos^2 x \]
Thus,
\[ f(x) = 1 - 3\sin^2 x \cos^2 x \]

### Step 3: Use the double angle identity
Using the double angle identity, we can express \( \sin^2 x \cos^2 x \) as:
\[ \sin^2 x \cos^2 x = \frac{1}{4} \sin^2(2x) \]
Substituting this into our function gives:
\[ f(x) = 1 - 3 \cdot \frac{1}{4} \sin^2(2x) = 1 - \frac{3}{4} \sin^2(2x) \]

### Step 4: Determine the range of \( \sin^2(2x) \)
The function \( \sin^2(2x) \) ranges from 0 to 1. Therefore:
- When \( \sin^2(2x) = 0 \):
\[ f(x) = 1 - \frac{3}{4} \cdot 0 = 1 \]
- When \( \sin^2(2x) = 1 \):
\[ f(x) = 1 - \frac{3}{4} \cdot 1 = 1 - \frac{3}{4} = \frac{1}{4} \]

### Step 5: Conclusion
Thus, the range of \( f(x) \) is:
\[ \left[\frac{1}{4}, 1\right] \]
The correct option is (a) \([1/4, 1]\).

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