Question

Let \(f : [0, 2] \to R\) be a twice differentiable function such that \(f''(x) \gt 0\), for all \(x \in (0, 2)\). If \(\phi(x) = f(x) + f(2 - x)\), then \(\phi\) is :

• A. decreasing on \((0, 2)\)
• B. decreasing on \((0, 1)\) and increasing on \((1, 2)\)
• C. increasing on \((0, 2)\)
• D. increasing on \((0, 1)\) and decreasing on \((1, 2)\)

Hint:

Differentiate: \(\phi'(x) = f'(x) - f'(2 - x)\), and \(f'' \gt 0\) means \(f'\) is strictly increasing.

Answer 1

The Correct Option is B

\(\phi'(x) = f'(x) - f'(2 - x)\).
Since \(f''(x) \gt 0\) on \((0, 2)\), \(f'\) is strictly increasing.
For \(0 \lt x \lt 1\): \(x \lt 2 - x\), so \(f'(x) \lt f'(2 - x)\) and \(\phi'(x) \lt 0\): \(\phi\) is decreasing on \((0, 1)\).
For \(1 \lt x \lt 2\): \(x \gt 2 - x\), so \(\phi'(x) \gt 0\): \(\phi\) is increasing on \((1, 2)\).
Answer: (2) decreasing on \((0, 1)\) and increasing on \((1, 2)\)