Question

Let \( \phi(x) = f(x) + f(2a - x) \), \( x \in [0, 2a] \), and \( f''(x) > 0 \) for all \( x \in [0, a] \). Then \( \phi(x) \) is:

• A. increasing on \( [0,a] \).
• B. decreasing on \( [0,a] \).
• C. increasing on \( [0,2a] \).
• D. decreasing on \( [0,2a] \).

Hint:

If \( f'' > 0 \): \begin{itemize} \item Function is convex. \item First derivative is increasing. \item Useful in symmetry expressions. \end{itemize}

Answer 1

The Correct Option is B

Concept: Given: \[ \phi(x) = f(x) + f(2a-x) \] Differentiate: \[ \phi'(x) = f'(x) - f'(2a-x) \] Step 1: Use convexity condition. Since \( f''(x) > 0 \), function is convex. Thus: \[ f'(x) \text{ is increasing} \] Step 2: Compare slopes. For \( x \in [0,a] \): \[ x \le a \Rightarrow 2a-x \ge a \] Since \( f'(x) \) is increasing: \[ f'(x) \le f'(2a-x) \] Thus: \[ \phi'(x) \le 0 \] Step 3: Conclusion. \[ \phi'(x) < 0 \Rightarrow \phi(x) \text{ decreasing on } [0,a] \]