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(C) Given $f(2x + 1) = f(1 - 2x)$.Let $t = 2x + 1$,then $x = (t - 1)/2$. Substituting this into the $RHS$: $f(t) = f(1 - 2((t - 1)/2)) = f(1 - (t - 1)

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$4$

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(C) Given $f(2x + 1) = f(1 - 2x)$.Let $t = 2x + 1$,then $x = (t - 1)/2$. Substituting this into the $RHS$: $f(t) = f(1 - 2((t - 1)/2)) = f(1 - (t - 1)) = f(2 - t)$.This implies $f$ is symmetric about the line $x = 1$.Given $f(2) = f(5) = f(10)$.Using symmetry $f(x) = f(2 - x)$:$f(2) = f(0)$$f(5) = f(2 - 5) = f(-3)$$f(10) = f(2 - 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: $x = -8, -3, 0, 2, 5$.By Rolle's Theorem,between any two roots of $f(x) = c$,there exists at least one root of $f'(x) = 0$.Between $(-8, -3)$,$(-3, 0)$,$(0, 2)$,and $(2, 5)$,there are at least $4$ roots of $f'(x) = 0$.Since the interval is $(-5, 10)$,we check the roots in this range:Roots exist in $(-3, 0)$,$(0, 2)$,and $(2, 5)$. These $3$ roots are definitely in $(-5, 10)$.Additionally,since $f(5) = f(10)$,there is at least one root in $(5, 10)$.Thus,there are at least $3 + 1 = 4$ roots in $(-5, 10)$.

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

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