Question

Find the value of \( k \) if the function \( f(x) = k\sin x + 2\cos x \) is increasing for all \( x \)

• A. Possible for some \( k \)
• B. \( k > 2 \)
• C. Impossible for all \( x \in \mathbb{R} \)
• D. \( k = 2 \)

Hint:

Any expression of the form \( a\cos x + b\sin x \) always varies between \( \pm \sqrt{a^2 + b^2} \), so it cannot stay strictly positive or negative for all \( x \).

Answer 1

The Correct Option is C

Step 1: Understanding the Question:
A function \(f(x)\) is considered monotonically increasing over its entire domain if its first derivative \(f'(x)\) is non-negative for all \(x\).
Step 2: Key Formula or Approach:
The derivative of a sum of trigonometric functions is found using standard differentiation rules:
\[ \frac{d}{dx}(\sin x) = \cos x \text{ and } \frac{d}{dx}(\cos x) = -\sin x \] The range of the function \( a\cos x + b\sin x \) is \([-\sqrt{a^2+b^2}, \sqrt{a^2+b^2}]\).
Step 3: Detailed Explanation:
Differentiating the given function:
\[ f'(x) = k\cos x - 2\sin x \] For the function to be increasing for all \(x\), we need:
\[ k\cos x - 2\sin x \geq 0 \text{ for all } x \in \mathbb{R} \] However, the expression \( k\cos x - 2\sin x \) is a periodic wave that oscillates between \( -\sqrt{k^2 + (-2)^2} \) and \( \sqrt{k^2 + (-2)^2} \).
Because this range always includes negative values (unless \(k\) and the coefficient are both zero, which is not the case here), there will always be intervals where \(f'(x)<0\).
Step 4: Final Answer:
Since the derivative must take negative values periodically, it is impossible for the function to be increasing for all real \(x\).