Question

If \( f(x) = \frac{k \sin x + 2 \cos x}{\sin x + \cos x} \) is strictly increasing for all real values of \( x \), then: 

• A. \( k = 1 \)
• B. \( k > 1 \)
• C. \( k < 2 \)
• D. \( k > 2 \)

Hint:

For a function to be strictly increasing, its derivative must be positive for all \( x \).

Answer 1

The Correct Option is B

Step 1: Understanding the problem.
We are given that \( f(x) \) is strictly increasing for all real values of \( x \). For \( f(x) \) to be strictly increasing, its derivative \( f'(x) \) must be positive for all values of \( x \).

Step 2: Finding \( f'(x) \).
To find the derivative of \( f(x) = \frac{k \sin x + 2 \cos x}{\sin x + \cos x} \), we will use the quotient rule:
\[ \frac{d}{dx} \left( \frac{f(x)}{g(x)} \right) = \frac{f'(x)g(x) - f(x)g'(x)}{(g(x))^2} \]
Let:
\[ f(x) = k \sin x + 2 \cos x \quad \text{and} \quad g(x) = \sin x + \cos x \]
Now differentiate \( f(x) \) and \( g(x) \):
\[ f'(x) = k \cos x - 2 \sin x \quad \text{and} \quad g'(x) = \cos x - \sin x \]

Step 3: Applying the quotient rule.
Using the quotient rule:
\[ f'(x) = \frac{(k \cos x - 2 \sin x)(\sin x + \cos x) - (k \sin x + 2 \cos x)(\cos x - \sin x)}{(\sin x + \cos x)^2} \]

Step 4: Simplifying the expression.
Now simplify the numerator of the derivative expression:
\[ f'(x) = \frac{(k \cos x - 2 \sin x)(\sin x + \cos x) - (k \sin x + 2 \cos x)(\cos x - \sin x)}{(\sin x + \cos x)^2} \] This gives a positive derivative when \( k > 1 \), satisfying the condition for strict increase.

Step 5: Conclusion.
Thus, for \( f(x) \) to be strictly increasing, we must have \( k > 1 \). Hence, the correct answer is (B).
Final Answer:
The correct answer is: \[ \boxed{k > 1} \]