Let phi(x) = f(x) + f(2a - x),x in [0, 2a] an | vedclass

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(B) Given $\phi(x) = f(x) + f(2a - x)$.Taking the derivative with respect to $x$,we get $\phi^{\prime}(x) = f^{\prime}(x) - f^{\prime}(2a - x)$.We are

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decreasing on $[0, a]$

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(B) Given $\phi(x) = f(x) + f(2a - x)$.Taking the derivative with respect to $x$,we get $\phi^{\prime}(x) = f^{\prime}(x) - f^{\prime}(2a - x)$.We are given $f^{\prime \prime}(x) > 0$ for all $x \in [0, a]$,which implies that $f^{\prime}(x)$ is a strictly increasing function.For $x \in [0, a]$,we have $x Since $f^{\prime}(x)$ is strictly increasing,$x Therefore,$\phi^{\prime}(x) = f^{\prime}(x) - f^{\prime}(2a - x) Since $\phi^{\prime}(x) < 0$ on $[0, a]$,$\phi(x)$ is a decreasing function on $[0, a]$.

Let $\phi(x) = f(x) + f(2a - x)$,$x \in [0, 2a]$ and $f^{\prime \prime}(x) > 0$ for all $x \in [0, a]$. Then $\phi(x)$ is

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