Let f ( x ) = { x - 1 , x is even , 2 x , x is odd , x ? N . If for some a ? N , f ( f ( f ( a ) ) ) = 21 , then lim x ? a - { | x | 3 a - [ x a ] } , where [ t ] denotes the greatest integer less than or equal to t , is equal to:

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Solution

Q.
Let $f (x) = {\begin{matrix} x - 1, & x is even, \\ 2 x, & x is odd, \end{matrix} x \in N$ . If for some $a \in N, f (f (f (a))) = 21$ , then $\lim_{x \to a^{-}} {\frac{| x |^{3}}{a} - [\frac{x}{a}]},$ where $[t]$ denotes the greatest integer less than or equal to $t,$ is equal to: [2024]

1 169

2 121

3 225

4 144

Ans.
(4)

There are two cases arise:

Case I : Let $a$ is even $∴$ $f (a) = a - 1$ (odd)

$\Rightarrow f (f (a)) = f (a - 1) = 2 a - 2$ (even)

$\Rightarrow f (f (f (a))) = f (2 a - 2) = 2 a - 2 - 1 = 2 a - 3$

$\Rightarrow 21 = 2 a - 3 \Rightarrow a = 12$

$So, \lim_{x \to a^{-}} {\frac{| x |^{3}}{a} - [\frac{x}{a}]} = \lim_{x \to 12^{-}} {\frac{| x^{3} |}{12} - [\frac{x}{12}]}$

$= 144$ $(∵ x < 12)$

Case II : Let $a$ is odd

$∴ f (a) = 2 a (even) \Rightarrow f (f (a)) = f (2 a) = 2 a - 1 (odd)$

$\Rightarrow f (f (f (a))) = f (2 a - 1) = 2 (2 a - 1)$

$\Rightarrow 21 = 4 a - 2 \Rightarrow a = \frac{23}{4} \notin N$

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