Let f(x) = begin{cases} x-1, & x text{ is eve | vedclass

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(B) Given $f(x) = \begin{cases} x-1, & x 	ext{ is even} \ 2x, & x 	ext{ is odd} \end{cases}$.We are given $f(f(f(a))) = 21$.Case $1$: If $a$ is eve

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$144$

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(B) Given $f(x) = \begin{cases} x-1, & x 	ext{ is even} \ 2x, & x 	ext{ is odd} \end{cases}$.We are given $f(f(f(a))) = 21$.Case $1$: If $a$ is even,then $f(a) = a-1$ (which is odd). Then $f(f(a)) = 2(a-1) = 2a-2$ (which is even). Then $f(f(f(a))) = (2a-2)-1 = 2a-3$. Setting $2a-3 = 21$,we get $2a = 24$,so $a = 12$.Case $2$: If $a$ is odd,then $f(a) = 2a$ (which is even). Then $f(f(a)) = 2a-1$ (which is odd). Then $f(f(f(a))) = 2(2a-1) = 4a-2$. Setting $4a-2 = 21$,we get $4a = 23$,which has no integer solution for $a$.Thus,$a = 12$.Now,we evaluate $\lim_{x ightarrow 12^{-}} \left( \frac{|x|^3}{12} - \left[ \frac{x}{12} ight] ight)$.As $x ightarrow 12^{-}$,$x$ is slightly less than $12$,so $\frac{x}{12}$ is slightly less than $1$,meaning $\left[ \frac{x}{12} ight] = 0$.Therefore,the limit is $\lim_{x ightarrow 12^{-}} \frac{x^3}{12} - 0 = \frac{12^3}{12} = 12^2 = 144$.

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