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STD 12 - 1. relation and function — JEE STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceJEESTD 12 - 1. relation and function4 Marks
Question
Let $f(x)=\left\{\begin{array}{l}x-1, x \text { is even, } \\ 2 x, x \text { is odd, }\end{array}\right.$. 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 :
✓

Answer

$f(x)=\left\{\begin{array}{cc}x-1 ; & x=\text { even } \\ 2 x ; & x=\text { odd }\end{array}\right.$
$ f(f(\mathrm{a})))=21 $
$ \text { C $-$ 1: If } a=\text { even } $
$ f(\mathrm{a})=\mathrm{a}-1=\text { odd } $
$ f(a))=2(a-1)=\text { even } $
$ f(f(f(a)))=2 a-3=21 $
$\Rightarrow a=12 $
$ \text { C $-$ 2: If } a=\text { odd } $
$ f(\mathrm{a})=2 \mathrm{a}=\text { even } $
$ f(f(\mathrm{a}))=2 \mathrm{a}-1=\text { odd } $
$ f(f(f(a)))=4 a-2=21 \ ($Not possible$)$
Hence $\mathrm{a}=12$
Now
$ \lim _{x \rightarrow 12^{-}}\left(\frac{|x|^3}{12}-\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 .$

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