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(C) Given $f(2x+1) = f(1-2x)$. Let $2x+1 = t$,then $x = (t-1)/2$. Substituting this,$f(t) = f(1 - (t-1)) = f(2-t)$. This implies the function $f$ is s

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$4$

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(C) Given $f(2x+1) = f(1-2x)$. Let $2x+1 = t$,then $x = (t-1)/2$. Substituting this,$f(t) = f(1 - (t-1)) = f(2-t)$. This implies the function $f$ is symmetric about the line $x = 1$. Since $f$ is differentiable and $f(2) = f(5) = f(10)$,we use Rolle's Theorem. By symmetry,$f(2) = f(0)$ (since $f(2-2) = f(0)$). Also,$f(2) = f(5)$ implies $f(2-5) = f(-3) = f(5)$. Given $f(2) = f(5) = f(10)$,by symmetry $f(2) = f(0)$,$f(5) = f(-3)$,and $f(10) = f(-8)$. Thus,$f(0) = f(2) = f(5) = f(10) = f(-3) = f(-8)$. Applying Rolle's Theorem between these points: $1$. Between $x = -8$ and $x = -3$,there exists at least one root $c_1 \in (-8, -3)$ such that $f'(c_1) = 0$. $2$. Between $x = -3$ and $x = 0$,there exists at least one root $c_2 \in (-3, 0)$ such that $f'(c_2) = 0$. $3$. Between $x = 0$ and $x = 2$,there exists at least one root $c_3 \in (0, 2)$ such that $f'(c_3) = 0$. $4$. Between $x = 2$ and $x = 5$,there exists at least one root $c_4 \in (2, 5)$ such that $f'(c_4) = 0$. $5$. Between $x = 5$ and $x = 10$,there exists at least one root $c_5 \in (5, 10)$ such that $f'(c_5) = 0$. Since we need roots in $(-5, 10)$,the roots $c_2, c_3, c_4, c_5$ are within this interval. Thus,the minimum number of roots is $4$.

If $f$ is a differentiable function such that $f(2x+1) = f(1-2x)$ for all $x \in R$,then what is the minimum number of roots of the equation $f'(x) = 0$ in the interval $x \in (-5, 10)$,given that $f(2) = f(5) = f(10)$?

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