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 :

Solution

Maths sir · 4 steps

Unwind f three times from 21 to find a, then remember the floor is 0 just left of x = a.

1. Read the rule. An even input gives x - 1, which is odd. An odd input gives 2x, which is even. So an odd output must come from an even input, and an even output must come from an odd input.
2. Work backwards. 21 is odd, so it came from an even number: y - 1 = 21 gives y = 22. Next, 22 is even, so it came from an odd number: 2z = 22 gives z = 11. Then 11 is odd, so it came from an even number: a - 1 = 11 gives a = 12.

   12 →(f) 11 →(f) 22 →(f) 21

3. Left-hand limit. As x → 12⁻, the first term |x|³/12 tends to 1728/12 = 144. Also x/12 stays just below 1, so [x/12] = 0. The curly braces here only group the terms. A fractional-part reading would give a limit of 1, which is not an option.
4. So. The limit is 144 - 0 = 144, which is option (d).

The idea to keep. To solve f(f(f(a))) = k, run the function backwards one layer at a time and use parity to pick the right branch. For a floor at a one-sided limit, check which side of the integer you are on.

Answer from the NTA final answer key. Solution by PrepWiser.