Let f(x)= … x∈ N. If for some a∈ N,f(f(f(a)))=21, then undersetx→ a^- lim |x|^3 a-[ x a], where [t] denotes the greatest integer less than or equal…

Given: f(x)={\begin{matrix}x-1;x\text{is even} \\ 2x:x\text{is odd}\end{matrix}

And $f(f(f(a)))=21$f(f(f(a)))=21

Case-I: If $a$a is even

$$\Rightarrow f(a)=a-1=\text{Odd}$$⇒f(a)=a−1=Odd

$$\Rightarrow f(f(a))=2(a-1)=\text{Even}$$⇒f(f(a))=2(a−1)=Even

$$\Rightarrow f(f(f(a)))=2a-3=21$$⇒f(f(f(a)))=2a−3=21

$$\Rightarrow a=12$$⇒a=12

Case-II: If $a$a is odd

$$\Rightarrow f(a)=2a=\text{Even}$$⇒f(a)=2a=Even

$$\Rightarrow f(f(a))=2a-1=\text{Odd}$$⇒f(f(a))=2a−1=Odd

$\Rightarrow f(f(f(a)))=4a-2=21$⇒f(f(f(a)))=4a−2=21 (Not possible)

Hence, $a=12$a=12

Now,$\underset{x\rightarrow {12}^{-}}{\mathrm{lim}}(\frac{{|x|}^{3}}{12}-[\frac{x}{12}])=\underset{x\rightarrow {12}^{-}}{\mathrm{lim}}\frac{{|x|}^{3}}{12}-\underset{x\rightarrow {12}^{-}}{\mathrm{lim}}[\frac{x}{12}]$x→12−lim​(12∣x∣3​−[12x​])=x→12−lim​12∣x∣3​−x→12−lim​[12x​]

$$\Rightarrow \underset{x\rightarrow {12}^{-}}{\mathrm{lim}}(\frac{{|x|}^{3}}{12}-[\frac{x}{12}])=144-0$$⇒x→12−lim​(12∣x∣3​−[12x​])=144−0

$$\Rightarrow \underset{x\rightarrow {12}^{-}}{\mathrm{lim}}(\frac{{|x|}^{3}}{12}-[\frac{x}{12}])=144$$⇒x→12−lim​(12∣x∣3​−[12x​])=144