The difference between the greatest between the greatest and least value of the function `f(x)=sin2x-x" on"[-pi//2,pi//6]`, is

Options

A. `(sqrt3+sqrt2)/2`
B. `(sqrt3+sqrt2)/2+pi/6`
C. `(sqrt3)/2+pi/2`
D. `(sqrt3+sqrt2)/2-pi/3`

Explanation

To find the difference between the greatest and least value of the function \( f(x) = \sin(2x) - x \) on the interval \([- \frac{\pi}{2}, \frac{\pi}{6}]\), we will follow these steps:

### Step 1: Find the derivative of the function
First, we need to find the derivative of the function \( f(x) \):
\[ f'(x) = \frac{d}{dx}(\sin(2x) - x) = 2\cos(2x) - 1 \]

**Hint:** The derivative helps us find the critical points where the function may have maximum or minimum values.

### Step 2: Set the derivative to zero
Next, we set the derivative equal to zero to find critical points:
\[ 2\cos(2x) - 1 = 0 \]
This simplifies to:
\[ \cos(2x) = \frac{1}{2} \]

**Hint:** Setting the derivative to zero allows us to find points where the function changes direction (maxima or minima).

### Step 3: Solve for \( x \)
Now, we solve for \( 2x \):
\[ 2x = \frac{\pi}{3} + 2k\pi \quad \text{or} \quad 2x = -\frac{\pi}{3} + 2k\pi \quad (k \in \mathbb{Z}) \]
This gives us:
\[ x = \frac{\pi}{6} + k\pi \quad \text{or} \quad x = -\frac{\pi}{6} + k\pi \]

**Hint:** Finding \( x \) values helps us identify where the function might reach its maximum or minimum.

### Step 4: Determine valid critical points within the interval
We need to check which of these critical points lie within the interval \([- \frac{\pi}{2}, \frac{\pi}{6}]\):
- For \( x = \frac{\pi}{6} \), it is within the interval.
- For \( x = -\frac{\pi}{6} \), it is also within the interval.

**Hint:** Always check if critical points are within the specified domain.

### Step 5: Evaluate the function at the endpoints and critical points
Now we evaluate \( f(x) \) at the endpoints and the critical points:
1. \( f\left(-\frac{\pi}{2}\right) = \sin(-\pi) + \frac{\pi}{2} = 0 + \frac{\pi}{2} = \frac{\pi}{2} \)
2. \( f\left(\frac{\pi}{6}\right) = \sin\left(\frac{\pi}{3}\right) - \frac{\pi}{6} = \frac{\sqrt{3}}{2} - \frac{\pi}{6} \)
3. \( f\left(-\frac{\pi}{6}\right) = \sin\left(-\frac{\pi}{3}\right) + \frac{\pi}{6} = -\frac{\sqrt{3}}{2} + \frac{\pi}{6} \)

**Hint:** Evaluating the function at critical points and boundaries helps us find the maximum and minimum values.

### Step 6: Compare the values to find the greatest and least
Now we compare the values:
- \( f\left(-\frac{\pi}{2}\right) = \frac{\pi}{2} \)
- \( f\left(\frac{\pi}{6}\right) = \frac{\sqrt{3}}{2} - \frac{\pi}{6} \)
- \( f\left(-\frac{\pi}{6}\right) = -\frac{\sqrt{3}}{2} + \frac{\pi}{6} \)

### Step 7: Calculate the difference
Now we find the greatest and least values:
- Greatest value: \( f\left(-\frac{\pi}{2}\right) = \frac{\pi}{2} \)
- Least value: \( f\left(-\frac{\pi}{6}\right) \)

Now, we calculate the difference:
\[ \text{Difference} = \left(\frac{\pi}{2}\right) - \left(-\frac{\sqrt{3}}{2} + \frac{\pi}{6}\right) \]
This simplifies to:
\[ \text{Difference} = \frac{\pi}{2} + \frac{\sqrt{3}}{2} - \frac{\pi}{6} \]

To combine these, we can convert \( \frac{\pi}{2} \) and \( \frac{\pi}{6} \) to a common denominator:
\[ \frac{\pi}{2} = \frac{3\pi}{6} \]
Thus:
\[ \text{Difference} = \left(\frac{3\pi}{6} - \frac{\pi}{6} + \frac{\sqrt{3}}{2}\right) = \frac{2\pi}{6} + \frac{\sqrt{3}}{2} = \frac{\pi}{3} + \frac{\sqrt{3}}{2} \]

### Final Answer
The difference between the greatest and least value of the function \( f(x) \) on the interval \([- \frac{\pi}{2}, \frac{\pi}{6}]\) is:
\[ \frac{\pi}{3} + \frac{\sqrt{3}}{2} \]