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(B) Given $f(2x+1) = f(1-2x)$.Let $2x+1 = 1+t$,then $2x = t$,so $x = t/2$.Substituting this,$f(1+t) = f(1-t)$,which implies $f$ is symmetric about the

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$3$

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(B) Given $f(2x+1) = f(1-2x)$.Let $2x+1 = 1+t$,then $2x = t$,so $x = t/2$.Substituting this,$f(1+t) = f(1-t)$,which implies $f$ is symmetric about the line $x = 1$.Given $f(2) = f(5) = f(10)$.Using symmetry $f(1+x) = f(1-x)$:For $x=1$,$f(2) = f(0)$.For $x=4$,$f(5) = f(-3)$.For $x=9$,$f(10) = f(-8)$.Since $f(2) = f(5) = f(10)$,we have $f(-8) = f(-3) = f(0) = f(2) = f(5) = f(10)$.We have $5$ distinct points where the function values are equal: $x_1 = -8, x_2 = -3, x_3 = 0, x_4 = 2, x_5 = 5$.By Rolle's Theorem,between any two roots of $f(x) = k$,there exists at least one root of $f'(x) = 0$.Between $(-8, -3), (-3, 0), (0, 2), (2, 5)$,there are at least $4$ roots of $f'(x) = 0$.Since the interval is $(-5, 10)$,we consider the roots within this range.The roots found are in $(-8, 5)$,so the roots in $(-5, 10)$ are at least $3$ (specifically between $(-3, 0), (0, 2), (2, 5)$).However,checking the symmetry and the values,the minimum number of roots is $3$.

If $f$ is a differentiable function such that $f(2x+1) = f(1-2x)$ for all $x \in R$,then find 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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