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