If `f(x)||2sinx-1|-2cotx|`, then the value of `f'((pi)/(3))` is equal to

Options

A. 0
B. `-(5)/(3)`
C. `(5)/(3)`
D. `(8)/(3)`

Answer

To find the value of \( f'\left(\frac{\pi}{3}\right) \) where \( f(x) = |2\sin x - 1 - 2\cot x| \), we will follow these steps:

### Step 1: Evaluate \( f\left(\frac{\pi}{3}\right) \)

First, we need to evaluate the expression inside the absolute value at \( x = \frac{\pi}{3} \).

\[ f\left(\frac{\pi}{3}\right) = |2\sin\left(\frac{\pi}{3}\right) - 1 - 2\cot\left(\frac{\pi}{3}\right)| \]

Calculating \( \sin\left(\frac{\pi}{3}\right) \) and \( \cot\left(\frac{\pi}{3}\right) \):

\[ \sin\left(\frac{\pi}{3}\right) = \frac{\sqrt{3}}{2}, \quad \cot\left(\frac{\pi}{3}\right) = \frac{1}{\sqrt{3}} \]

Substituting these values into the function:

\[ f\left(\frac{\pi}{3}\right) = |2 \cdot \frac{\sqrt{3}}{2} - 1 - 2 \cdot \frac{1}{\sqrt{3}}| \]
\[ = |\sqrt{3} - 1 - \frac{2}{\sqrt{3}}| \]

### Step 2: Simplify the expression

To simplify \( \sqrt{3} - 1 - \frac{2}{\sqrt{3}} \):

Convert \( \sqrt{3} \) into a common denominator:

\[ = |\frac{3}{\sqrt{3}} - 1 - \frac{2}{\sqrt{3}}| = | \frac{3 - 2 - \sqrt{3}}{\sqrt{3}} | = | \frac{1 - \sqrt{3}}{\sqrt{3}} | \]

Since \( 1 < \sqrt{3} \), we have \( 1 - \sqrt{3} < 0 \), thus:

\[ f\left(\frac{\pi}{3}\right) = -\frac{1 - \sqrt{3}}{\sqrt{3}} = \frac{\sqrt{3} - 1}{\sqrt{3}} \]

### Step 3: Determine the sign of the expression

Since \( f(x) \) is defined as \( |2\sin x - 1 - 2\cot x| \), we need to check the sign of \( 2\sin x - 1 - 2\cot x \) around \( x = \frac{\pi}{3} \).

We found that \( 2\sin\left(\frac{\pi}{3}\right) - 1 - 2\cot\left(\frac{\pi}{3}\right) < 0 \). Therefore, in the neighborhood of \( x = \frac{\pi}{3} \):

\[ f(x) = 2\cot x - 2\sin x + 1 \]

### Step 4: Differentiate \( f(x) \)

Now we differentiate \( f(x) \):

\[ f'(x) = \frac{d}{dx}(2\cot x - 2\sin x + 1) \]

Using the derivatives:

\[ f'(x) = 2(-\csc^2 x) - 2\cos x \]
\[ = -2\csc^2 x - 2\cos x \]

### Step 5: Evaluate \( f'\left(\frac{\pi}{3}\right) \)

Now we substitute \( x = \frac{\pi}{3} \):

\[ f'\left(\frac{\pi}{3}\right) = -2\csc^2\left(\frac{\pi}{3}\right) - 2\cos\left(\frac{\pi}{3}\right) \]

Calculating \( \csc\left(\frac{\pi}{3}\right) \) and \( \cos\left(\frac{\pi}{3}\right) \):

\[ \csc\left(\frac{\pi}{3}\right) = \frac{1}{\sin\left(\frac{\pi}{3}\right)} = \frac{2}{\sqrt{3}}, \quad \cos\left(\frac{\pi}{3}\right) = \frac{1}{2} \]

Substituting these values:

\[ f'\left(\frac{\pi}{3}\right) = -2\left(\frac{2}{\sqrt{3}}\right)^2 - 2\left(\frac{1}{2}\right) \]
\[ = -2\left(\frac{4}{3}\right) - 1 = -\frac{8}{3} - 1 = -\frac{8}{3} - \frac{3}{3} = -\frac{11}{3} \]

Thus, the final answer is:

\[ \boxed{-\frac{11}{3}} \]