Let \(f(\mathrm{x})=\left\{\begin{array}{c}\mathrm{x}-1, \mathrm{x} \text { is even, } \\ 2 \mathrm{x}, \quad \mathrm{x} \text { is odd, }\end{array} \mathrm{x} \in \mathrm{N}\right.\). If for some \(\mathrm{a} \in \mathrm{N}, f(f(f(\mathrm{a})))=21\), then \(\lim _{\mathrm{x} \rightarrow \mathrm{a}^{-}}\left\{\frac{|\mathrm{x}|^{3}}{\mathrm{a}}-\left[\frac{\mathrm{x}}{\mathrm{a}}\right]\right\}\), where \([t]\) denotes the greatest integer less than or equal to \(t\), is equal to :

JEE Main · 2024 · 01 Feb 2024 · Shift 2 · Limits

Let

f (x) = {x - 1, x is even, 2 x, x is odd, 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 :

Answer: (B)

f (x) = x - 1

x

is even,

f (x) = 2 x

x

is odd, for

x \in N

.

Case

a

even:

f (a) = a - 1

(odd).

f (f (a)) = 2 (a - 1)

(even).

f (f (f (a))) = 2 (a - 1) - 1 = 2 a - 3

. Setting

2 a - 3 = 21

gives

a = 12

, which is indeed even — consistent.

Case

a

odd:

f (a) = 2 a

(even).

f (f (a)) = 2 a - 1

(odd).

f (f (f (a))) = 2 (2 a - 1) = 4 a - 2

. Setting

4 a - 2 = 21

gives

a = \frac{23}{4}

, not a natural number — rejected.

So

a = 12

.

Now evaluate

x \to 1 2^{-} lim {\frac{∣ x ∣ ^{3}}{a} - [\frac{x}{a}]}

with

a = 12

.

As

x \to 1 2^{-}

with

x > 0

\frac{∣ x ∣ ^{3}}{12} \to \frac{1 2 ^{3}}{12} = 1 2^{2} = 144

continuously.

For

[\frac{x}{12}]

: as

x \to 1 2^{-}

\frac{x}{12} \to 1^{-}

(a value just below

1

), so the greatest integer part is

0

.

The limit equals

144 - 0 = 144

JEE Main · 2024 · 01 Feb 2024 · Shift 2 · Single correct option