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

### Step 1: Differentiate the function
We start by finding the first derivative of the function \( f(x) \).

\[ f'(x) = \frac{d}{dx}(3\sin^4 x - \cos^6 x) \]

Using the chain rule, we differentiate each term:

\[ f'(x) = 12\sin^3 x \cdot \cos x - 6\cos^5 x \cdot (-\sin x) \]

This simplifies to:

\[ f'(x) = 12\sin^3 x \cdot \cos x + 6\cos^5 x \cdot \sin x \]

### Step 2: Set the derivative to zero
To find critical points, we set \( f'(x) = 0 \):

\[ 12\sin^3 x \cdot \cos x + 6\cos^5 x \cdot \sin x = 0 \]

Factoring out common terms:

\[ 6\sin x \cos x (2\sin^2 x + \cos^4 x) = 0 \]

### Step 3: Solve for critical points
From the factored equation, we have two cases:

1. \( \sin x = 0 \)
2. \( 2\sin^2 x + \cos^4 x = 0 \)

For \( \sin x = 0 \):
- This gives \( x = 0, \pi, 2\pi, \ldots \)

For \( 2\sin^2 x + \cos^4 x = 0 \):
- Substitute \( \sin^2 x = 1 - \cos^2 x \):
\[ 2(1 - \cos^2 x) + \cos^4 x = 0 \]
This simplifies to:
\[ \cos^4 x - 2\cos^2 x + 2 = 0 \]
Let \( t = \cos^2 x \):
\[ t^2 - 2t + 2 = 0 \]
The discriminant \( (-2)^2 - 4 \cdot 1 \cdot 2 = 4 - 8 = -4 \) is negative, indicating no real solutions.

Thus, the only critical points are \( x = 0 \) and \( x = \frac{\pi}{2} \).

### Step 4: Determine maximum and minimum values
Now we evaluate \( f(x) \) at the critical points:

1. At \( x = 0 \):
\[ f(0) = 3\sin^4(0) - \cos^6(0) = 3(0) - (1) = -1 \]

2. At \( x = \frac{\pi}{2} \):
\[ f\left(\frac{\pi}{2}\right) = 3\sin^4\left(\frac{\pi}{2}\right) - \cos^6\left(\frac{\pi}{2}\right) = 3(1) - (0) = 3 \]

### Step 5: Calculate the difference
The maximum value is \( 3 \) and the minimum value is \( -1 \). The difference between the maximum and minimum values is:

\[ \text{Difference} = 3 - (-1) = 3 + 1 = 4 \]

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

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