The value of a for which the function `f (x) = sin x-Cos x-ax+b `decreases for all real values of x is given by

Options

A. ` a ge sqrt(2)`
B. ` a ge 1`
C. ` a lt sqrt(2)`
D. ` a lt 1`

Answer

To find the value of \( a \) for which the function \( f(x) = \sin x - \cos x - ax + b \) decreases for all real values of \( x \), we need to analyze the derivative of the function.

### Step 1: Differentiate the function
We start by differentiating \( f(x) \):
\[ f'(x) = \frac{d}{dx}(\sin x - \cos x - ax + b) \]
Using the rules of differentiation, we get:
\[ f'(x) = \cos x + \sin x - a \]

### Step 2: Set the derivative less than or equal to zero
For the function to be decreasing for all \( x \), we need:
\[ f'(x) \leq 0 \]
This implies:
\[ \cos x + \sin x - a \leq 0 \]
Rearranging gives:
\[ \cos x + \sin x \leq a \]

### Step 3: Analyze the expression \( \cos x + \sin x \)
The maximum value of \( \cos x + \sin x \) can be found using the Cauchy-Schwarz inequality or by recognizing that:
\[ \cos x + \sin x = \sqrt{2} \sin\left(x + \frac{\pi}{4}\right) \]
The maximum value of \( \sin \) is 1, hence:
\[ \cos x + \sin x \leq \sqrt{2} \]

### Step 4: Set the maximum value to find \( a \)
For the inequality \( \cos x + \sin x \leq a \) to hold for all \( x \), we must have:
\[ a \geq \sqrt{2} \]

### Conclusion
Thus, the value of \( a \) for which the function \( f(x) \) decreases for all real values of \( x \) is:
\[ \boxed{\sqrt{2}} \]