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Let varphi(x)=f(x)+f(2a-x) and f"(x)>0 f...

Let `varphi(x)=f(x)+f(2a-x)` and `f"(x)>0` for all `x in [0,\ a]` . Then, `varphi(x)` (a) increases on `[0,\ a]` (b) decreases on `[0,\ a]` (c) increases on `[-a ,\ 0]` (d) decreases on `[a ,\ 2a]`

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To solve the problem, we need to analyze the function \( \varphi(x) = f(x) + f(2a - x) \) given that \( f''(x) > 0 \) for all \( x \in [0, a] \). This condition implies that \( f'(x) \) is increasing on the interval \([0, a]\). ### Step-by-Step Solution: 1. **Differentiate \( \varphi(x) \)**: We start by finding the first derivative of \( \varphi(x) \): \[ \varphi'(x) = f'(x) + f'(2a - x) \cdot (-1) = f'(x) - f'(2a - x) ...
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