Find the value of `a` in order that `f(x)=sqrt(3)sinx-cosx-2a x+b` decreases for all real values of `xdot`

Explanation

To find the value of \( a \) such that the function \( f(x) = \sqrt{3} \sin x - \cos x - 2ax + b \) is monotonically decreasing for all real values of \( x \), we need to ensure that the derivative \( f'(x) \) is less than zero for all \( x \).

### Step 1: Calculate the derivative \( f'(x) \)

The function is given by:
\[ f(x) = \sqrt{3} \sin x - \cos x - 2ax + b \]

To find \( f'(x) \), we differentiate \( f(x) \):
\[ f'(x) = \frac{d}{dx}(\sqrt{3} \sin x) - \frac{d}{dx}(\cos x) - \frac{d}{dx}(2ax) + \frac{d}{dx}(b) \]
Using the derivatives of sine and cosine:
\[ f'(x) = \sqrt{3} \cos x + \sin x - 2a \]

### Step 2: Set the condition for monotonicity

For \( f(x) \) to be monotonically decreasing, we require:
\[ f'(x) < 0 \]
This implies:
\[ \sqrt{3} \cos x + \sin x - 2a < 0 \]
Rearranging gives:
\[ \sqrt{3} \cos x + \sin x < 2a \]

### Step 3: Analyze the left-hand side

The expression \( \sqrt{3} \cos x + \sin x \) can be rewritten using the sine addition formula. We can express it as:
\[ \sqrt{3} \cos x + \sin x = 2 \left( \frac{\sqrt{3}}{2} \cos x + \frac{1}{2} \sin x \right) \]
This can be recognized as:
\[ 2 \sin\left(x + \frac{\pi}{3}\right) \]
Thus, we have:
\[ \sqrt{3} \cos x + \sin x = 2 \sin\left(x + \frac{\pi}{3}\right) \]

### Step 4: Determine the maximum value

The maximum value of \( \sin \) function is 1, hence:
\[ \sqrt{3} \cos x + \sin x \leq 2 \]
This leads us to:
\[ 2 < 2a \]
Dividing both sides by 2 gives:
\[ 1 < a \]
or equivalently:
\[ a > 1 \]

### Conclusion

Thus, the value of \( a \) must be greater than 1 for the function \( f(x) \) to be monotonically decreasing for all real values of \( x \).