The difference between the maximum and minimum value of the function `f(x)=3sin^4x-cos^6x` is :

Options

A. `3/2`
B. `5/2`
C. `3`
D. `4`

Answer

To find the difference between the maximum and minimum values of the function \( f(x) = 3\sin^4 x - \cos^6 x \), we will follow these steps:

### Step 1: Rewrite the function
We start by rewriting the function in a more manageable form:
\[ f(x) = 3\sin^4 x - \cos^6 x \]
Using the identity \( \cos^2 x = 1 - \sin^2 x \), we can express \( \cos^6 x \) as:
\[ \cos^6 x = (1 - \sin^2 x)^3 \]
Thus, we can rewrite \( f(x) \) as:
\[ f(x) = 3\sin^4 x - (1 - \sin^2 x)^3 \]

### Step 2: Expand \( (1 - \sin^2 x)^3 \)
Now we expand \( (1 - \sin^2 x)^3 \):
\[ (1 - \sin^2 x)^3 = 1 - 3\sin^2 x + 3\sin^4 x - \sin^6 x \]
Substituting this back into \( f(x) \):
\[ f(x) = 3\sin^4 x - (1 - 3\sin^2 x + 3\sin^4 x - \sin^6 x) \]
This simplifies to:
\[ f(x) = 3\sin^4 x - 1 + 3\sin^2 x - 3\sin^4 x + \sin^6 x \]
Combining like terms gives:
\[ f(x) = \sin^6 x + 3\sin^2 x - 1 \]

### Step 3: Analyze the function
To find the maximum and minimum values of \( f(x) \), we can analyze the terms:
- The term \( \sin^6 x \) ranges from 0 to 1.
- The term \( 3\sin^2 x \) ranges from 0 to 3.

### Step 4: Find maximum value
The maximum value of \( f(x) \) occurs when both \( \sin^6 x \) and \( 3\sin^2 x \) are at their maximum:
\[ \sin^6 x = 1 \quad \text{and} \quad 3\sin^2 x = 3 \]
Thus, the maximum value is:
\[ f(x)_{\text{max}} = 1 + 3 - 1 = 3 \]

### Step 5: Find minimum value
The minimum value occurs when \( \sin^6 x = 0 \) and \( 3\sin^2 x = 0 \):
\[ f(x)_{\text{min}} = 0 + 0 - 1 = -1 \]

### Step 6: Calculate the difference
Finally, we calculate the difference between the maximum and minimum values:
\[ \text{Difference} = f(x)_{\text{max}} - f(x)_{\text{min}} = 3 - (-1) = 3 + 1 = 4 \]

### Final Answer
The difference between the maximum and minimum value of the function \( f(x) \) is \( \boxed{4} \).