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

Concept: A function is increasing for all \( x \in \mathbb{R} \) if its derivative is always non-negative: \[ f'(x) \ge 0 \quad \forall x \] Trigonometric expressions like \( A\cos x + B\sin x \) are oscillatory and change sign unless restricted.

Step 1: Differentiate the function. \[ f(x) = k\sin x + 2\cos x \] \[ f'(x) = k\cos x - 2\sin x \]

Step 2: Analyze the sign of \( f'(x) \). The expression: \[ f'(x) = k\cos x - 2\sin x \] is a trigonometric function of the form: \[ R\cos(x + \theta) \] which always oscillates between positive and negative values unless identically zero. Thus, \( f'(x) \) cannot remain non-negative for all \( x \in \mathbb{R} \).

Step 3: Conclusion. Since \( f'(x) \) changes sign for all values of \( k \), the function cannot be increasing for all real \( x \). \[ \Rightarrow \text{No such value of } k \text{ exists} \]