JEE Main 2024 (Online) 1st February Evening ShiftQuestion 1
Let $f(x)=\begin{cases}x-1, & x\text{ is even},\\ 2x, & x\text{ is odd},\end{cases}$ $x\in\mathbf{N}$. If for some $a\in\mathbf{N}$, $f(f(f(a)))=21$, then $\lim_{x\to a^{-}}\left\{\frac{|x|^{3}}{a}-\left[\frac{x}{a}\right]\right\}$, where $[t]$ denotes the greatest integer less than or equal to $t$, is equal to :
Limits & Continuity
Solution
$f(f(f(a)))=21$ (odd). Only $f(22)=21$ gives odd 21 (since $2x$ is even), so $f(f(a))=22$. $f(y)=22$: $y-1=22$ needs $y=23$ even (no); $2y=22$, $y=11$ odd (yes). So $f(a)=11$. $f(a)=11$: $a-1=11\Rightarrow a=12$ even (yes); $2a=11$ impossible. So $a=12$. As $x\to12^-$: $[x/12]=0$, so limit $=\frac{12^3}{12}-0=144$. Option (B).