The difference between the greatest and least value of the function `f(x)=cosx+1/2cos2x-1/3cos3x` is

Options

A. `(4)/(3)`
B. 1
C. `(9)/(4)`
D. `(1)/(6)`

Answer

To find the difference between the greatest and least values of the function \( f(x) = \cos x + \frac{1}{2} \cos 2x - \frac{1}{3} \cos 3x \), we will follow these steps:

### Step 1: Differentiate the function
We start by differentiating the function \( f(x) \) with respect to \( x \):
\[ f'(x) = -\sin x - \frac{1}{2} \cdot 2\sin 2x + \frac{1}{3} \cdot 3\sin 3x \]
This simplifies to:
\[ f'(x) = -\sin x - \sin 2x + \sin 3x \]

### Step 2: Set the derivative to zero
To find the critical points, we set the derivative equal to zero:
\[ -\sin x - \sin 2x + \sin 3x = 0 \]
This can be rearranged to:
\[ \sin 3x = \sin x + \sin 2x \]

### Step 3: Use trigonometric identities
Using the identity \( \sin a + \sin b = 2 \sin \left( \frac{a+b}{2} \right) \cos \left( \frac{a-b}{2} \right) \), we can express \( \sin x + \sin 2x \):
\[ \sin 2x + \sin x = 2 \sin \left( \frac{3x}{2} \right) \cos \left( \frac{x}{2} \right) \]
Thus, we have:
\[ \sin 3x = 2 \sin \left( \frac{3x}{2} \right) \cos \left( \frac{x}{2} \right) \]

### Step 4: Find critical points
The critical points occur when:
1. \( \sin x = 0 \) → \( x = n\pi \) for \( n \in \mathbb{Z} \)
2. \( \sin 2x = 0 \) → \( x = \frac{n\pi}{2} \) for \( n \in \mathbb{Z} \)
3. \( \sin 3x = 0 \) → \( x = \frac{n\pi}{3} \) for \( n \in \mathbb{Z} \)

### Step 5: Evaluate the function at critical points
Now we evaluate \( f(x) \) at the critical points \( x = 0, \frac{\pi}{2}, \pi, \frac{3\pi}{2}, 2\pi \).

1. For \( x = 0 \):
\[ f(0) = \cos(0) + \frac{1}{2} \cos(0) - \frac{1}{3} \cos(0) = 1 + \frac{1}{2} - \frac{1}{3} = \frac{6}{6} + \frac{3}{6} - \frac{2}{6} = \frac{7}{6} \]

2. For \( x = \pi \):
\[ f(\pi) = \cos(\pi) + \frac{1}{2} \cos(2\pi) - \frac{1}{3} \cos(3\pi) = -1 + \frac{1}{2} + \frac{1}{3} = -1 + \frac{3}{6} + \frac{2}{6} = -\frac{1}{6} \]

3. For \( x = 2\pi \):
\[ f(2\pi) = \cos(2\pi) + \frac{1}{2} \cos(4\pi) - \frac{1}{3} \cos(6\pi) = 1 + \frac{1}{2} - \frac{1}{3} = \frac{7}{6} \]

4. For \( x = \frac{\pi}{2} \):
\[ f\left(\frac{\pi}{2}\right) = \cos\left(\frac{\pi}{2}\right) + \frac{1}{2} \cos(\pi) - \frac{1}{3} \cos\left(\frac{3\pi}{2}\right) = 0 - \frac{1}{2} + 0 = -\frac{1}{2} \]

5. For \( x = \frac{3\pi}{2} \):
\[ f\left(\frac{3\pi}{2}\right) = \cos\left(\frac{3\pi}{2}\right) + \frac{1}{2} \cos(3\pi) - \frac{1}{3} \cos\left(\frac{9\pi}{2}\right) = 0 + \frac{1}{2} + 0 = \frac{1}{2} \]

### Step 6: Determine maximum and minimum values
From the evaluations:
- Maximum value is \( \frac{7}{6} \)
- Minimum value is \( -\frac{1}{2} \)

### Step 7: Calculate the difference
The difference between the greatest and least values is:
\[ \text{Difference} = \frac{7}{6} - \left(-\frac{1}{2}\right) = \frac{7}{6} + \frac{3}{6} = \frac{10}{6} = \frac{5}{3} \]

### Final Answer
The difference between the greatest and least value of the function \( f(x) \) is \( \frac{5}{3} \).

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