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 Question:
A function is strictly increasing if its derivative \( f'(x) \) is always positive for all \( x \) in its domain.
Step 2: Key Formula or Approach:
Apply the quotient rule to find \( f'(x) \) and set \( f'(x)>0 \).
Step 3: Detailed Explanation:
Let \( u = k\sin x + 2\cos x \) and \( v = \sin x + \cos x \).
\( f'(x) = \frac{v u' - u v'}{v^2} \).
\( u' = k\cos x - 2\sin x \), \( v' = \cos x - \sin x \).
Numerator of \( f'(x) \):
\( = (\sin x + \cos x)(k\cos x - 2\sin x) - (k\sin x + 2\cos x)(\cos x - \sin x) \)
\( = k\sin x\cos x - 2\sin^2 x + k\cos^2 x - 2\sin x\cos x - [k\sin x\cos x - k\sin^2 x + 2\cos^2 x - 2\sin x\cos x] \)
\( = -2\sin^2 x + k\cos^2 x + k\sin^2 x - 2\cos^2 x \)
\( = (k - 2)\sin^2 x + (k - 2)\cos^2 x = (k - 2)(\sin^2 x + \cos^2 x) = k - 2 \).
For strictly increasing, \( f'(x)>0 \):
\( \frac{k - 2}{(\sin x + \cos x)^2}>0 \).
Since the denominator is always non-negative, we need \( k - 2>0 \Rightarrow k>2 \).
Step 4: Final Answer:
The condition is \( k>2 \).