Limits Question 2
Question 2 - 2024 (01 Feb Shift 2)
Let $f(x)=\begin{array}{c}x-1, x \text { is even, } \\ 2 x, x \text { is odd, }\end{array} x$. If for some $a \in N, f(f(f(a)))=21$, then $\lim _{x \rightarrow a^{-}}{\frac{|x|^{3}}{a}-\left[\frac{x}{a}\right] }$, where $[t]$ denotes the greatest integer less than or equal to $t$, is equal to :
(1) 121
(2) 144
(3) 169
(4) 225
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Answer (2)
Solution
$f(x)=\begin{array}{cc}x-1 ; & x=\text { even } \\ 2 x ; & x=\text { odd }\end{array}$
$f(f(a)))=21$
C-1: If $a=$ even
$$ f(a)=a-1=\text { odd } $$
$f(f(a))=2(a-1)=$ even
$f(f(f)))=2 a-3=21 \Rightarrow a=12$
C-2: If $a=$ odd
$$ \begin{aligned} & f(a)=2 a=\text { even } \\ & f(f(a))=2 a-1=\text { odd } \\ & f(f(f(a)))=4 a-2=21 \text { (Not possible) } \end{aligned} $$
Hence $a=12$
Now
$$ \begin{aligned} & \lim _{x \rightarrow 12^{-}}\left(\frac{|x|^{3}}{2}-\left[\frac{x}{12}\right]\right) \\ & =\lim _{x \rightarrow 12^{-}} \frac{|x|^{3}}{12}-\lim _{x \rightarrow 12^{-}}\left[\frac{x}{12}\right] \\ & =144-0=144 \end{aligned} $$
