Step 1: Understanding the Question:
For a function to be increasing, its first derivative must be greater than or equal to zero, i.e., \( f'(x) \geq 0 \). We need to find the derivative of the given function and apply this condition.
Step 2: Key Formula or Approach:
We will use the quotient rule for differentiation: If \( f(x) = \frac{u(x)}{v(x)} \), then \( f'(x) = \frac{u'(x)v(x) - u(x)v'(x)}{[v(x)]^2} \).
Here, \( u(x) = k\sin x + 2\cos x \) and \( v(x) = \sin x + \cos x \).
Step 3: Detailed Explanation:
First, find the derivatives of \( u(x) \) and \( v(x) \):
\( u'(x) = k\cos x - 2\sin x \)
\( v'(x) = \cos x - \sin x \)
Now, apply the quotient rule:
\[ f'(x) = \frac{(k\cos x - 2\sin x)(\sin x + \cos x) - (k\sin x + 2\cos x)(\cos x - \sin x)}{(\sin x + \cos x)^2} \]
Expand the numerator:
\[ \text{Numerator} \] \[= (k\cos x\sin x + k\cos^2 x - 2\sin^2 x - 2\sin x\cos x) - (k\sin x\cos x - k\sin^2 x + 2\cos^2 x - 2\cos x\sin x) \]
\[ = k\cos^2 x - 2\sin^2 x - k\sin x\cos x + k\sin^2 x - 2\cos^2 x + 2\sin x\cos x - (k-2)\sin x \cos x \]
This is getting complicated. Let's simplify differently.
\[ \text{Numerator}\] \[ = (k\cos x\sin x + k\cos^2 x - 2\sin^2 x - 2\sin x\cos x) - (k\sin x\cos x - k\sin^2 x + 2\cos^2 x - 2\sin x\cos x) \]
Let's group the terms:
\[ = (k\cos^2 x + k\sin^2 x) - (2\sin^2 x + 2\cos^2 x) \]
\[ = k(\cos^2 x + \sin^2 x) - 2(\sin^2 x + \cos^2 x) \]
Using the identity \( \sin^2 x + \cos^2 x = 1 \):
\[ \text{Numerator} = k(1) - 2(1) = k - 2 \]
So, the derivative is:
\[ f'(x) = \frac{k - 2}{(\sin x + \cos x)^2} \]
For \( f(x) \) to be increasing, \( f'(x) \geq 0 \).
The denominator \( (\sin x + \cos x)^2 \) is always non-negative.
Therefore, the condition simplifies to the numerator being non-negative:
\[ k - 2 \geq 0 \]
\[ k \geq 2 \]
Step 4: Final Answer:
The value of \( k \) for which the function is increasing is \( k \geq 2 \).