Let f(x) = \ array ll x - 1, xis even, 2x, xis odd, array x ∈ N ....

Given: f\left(x\right) = \left\{$\begin{array}{ll}x - 1; x\text{is even} \\ 2x: x\text{is odd}\end{array}$\right)</p><p>Andf\left(f\left(f\left(a\right)\right)\right) = 21$</p><p>Case-I: If$a$is even</p><p>$⇒ f\left(a\right) = a - 1 = \text{Odd}$</p><p>$⇒ f\left(f\left(a\right)\right) = 2\left(a - 1\right) = \text{Even}$</p><p>$⇒ f\left(f\left(f\left(a\right)\right)\right) = 2a - 3 = 21$</p><p>$⇒ a = 12$</p><p>Case-II: If$a$is odd</p><p>$⇒ f\left(a\right) = 2a = \text{Even}$</p><p>$⇒ f\left(f\left(a\right)\right) = 2a - 1 = \text{Odd}$</p><p>$⇒ f\left(f\left(f\left(a\right)\right)\right) = 4a - 2 = 21$(Not possible)</p><p>Hence,$a = 12$</p><p>Now,$\lim_{x \to {12}^{-}}\left(\dfrac{{\left|x\right|}^{3}}{12} - \left[\dfrac{x}{12}\right]\right) = \lim_{x \to {12}^{-}}\dfrac{{\left|x\right|}^{3}}{12} - \lim_{x \to {12}^{-}}\left[\dfrac{x}{12}\right]$</p><p>$⇒ \lim_{x \to {12}^{-}}\left(\dfrac{{\left|x\right|}^{3}}{12} - \left[\dfrac{x}{12}\right]\right) = 144 - 0$</p><p>$⇒ \lim_{x \to {12}^{-}}\left(\dfrac{{\left|x\right|}^{3}}{12} - \left[\dfrac{x}{12}\right]\right) = 144$