If the function `f(x) = ( a sin x + 2 cos x)/( sin x + cos x )` is increasing for all values of x, then

Options

A. `a gt 1`
B. ` a gt 1 `
C. ` a lt 2`
D. ` a gt 2`

Answer

To determine the conditions under which the function \( f(x) = \frac{a \sin x + 2 \cos x}{\sin x + \cos x} \) is increasing for all values of \( x \), we need to analyze its derivative. A function is increasing if its derivative is non-negative for all \( x \).

### Step 1: Differentiate the function
We will use the quotient rule for differentiation, which states that if \( f(x) = \frac{u}{v} \), then:

\[ f'(x) = \frac{u'v - uv'}{v^2} \]

Here, \( u = a \sin x + 2 \cos x \) and \( v = \sin x + \cos x \).

### Step 2: Find \( u' \) and \( v' \)
We calculate the derivatives:

- \( u' = a \cos x - 2 \sin x \)
- \( v' = \cos x - \sin x \)

### Step 3: Apply the quotient rule
Now we can apply the quotient rule:

\[ f'(x) = \frac{(a \cos x - 2 \sin x)(\sin x + \cos x) - (a \sin x + 2 \cos x)(\cos x - \sin x)}{(\sin x + \cos x)^2} \]

### Step 4: Simplify the numerator
We will simplify the numerator step by step:

1. Expand the first term:
\[ (a \cos x - 2 \sin x)(\sin x + \cos x) = a \cos x \sin x + a \cos^2 x - 2 \sin^2 x - 2 \sin x \cos x \]

2. Expand the second term:
\[ (a \sin x + 2 \cos x)(\cos x - \sin x) = a \sin x \cos x - a \sin^2 x + 2 \cos^2 x - 2 \sin x \cos x \]

3. Combine the two expansions:
\[ \text{Numerator} = a \cos x \sin x + a \cos^2 x - 2 \sin^2 x - 2 \sin x \cos x - (a \sin x \cos x - a \sin^2 x + 2 \cos^2 x - 2 \sin x \cos x) \]

4. Combine like terms:
\[ = (a \cos^2 x - 2 \sin^2 x + a \sin^2 x + 2 \cos^2 x) + (a \cos x \sin x - a \sin x \cos x) - 2 \sin x \cos x \]
\[ = (a + 2 - 2) \cos^2 x + (a - 2) \sin^2 x \]

### Step 5: Set the numerator greater than or equal to zero
For \( f'(x) \geq 0 \), we require:

\[ (a - 2) \sin^2 x + (a + 2) \cos^2 x \geq 0 \]

### Step 6: Analyze the conditions
Since \( \sin^2 x + \cos^2 x = 1 \), we can analyze the coefficients:

1. If \( a - 2 \geq 0 \) (i.e., \( a \geq 2 \)), both terms are non-negative.
2. If \( a + 2 \geq 0 \) (which is always true since \( a \) can be any real number), we focus on the first condition.

### Conclusion
Thus, for the function \( f(x) \) to be increasing for all \( x \), we conclude:

\[ a \geq 2 \]