JEE Main 2021 · 25 July · morning · Single correct · +4 / −1
Let g : N → N be defined as
g(3n + 1) = 3n + 2,
g(3n + 2) = 3n + 3,
g(3n + 3) = 3n + 1, for all n ≥ 0.
Then which of the following statements is true?
- (a) There exists an onto function f : N → N such that fog = f
- (b) There exists a one-one function f : N → N such that fog = f
- (c) gogog = g
- (d) There exists a function : f : N → N such that gof = f
Answer and solution
Answer (a) There exists an onto function f : N → N such that fog = f
Solution
Maths sir · 3 steps
See g as a rotation inside each block of three, and every option becomes quick to check.
- What g does. Split ℕ into blocks {3n+1, 3n+2, 3n+3}. Inside each block, g sends 3n+1 → 3n+2 → 3n+3 → 3n+1, so it is a 3-cycle with no fixed point.
- What fog means. f(g(x)) = f(x) for every x means f takes the same value at x, at g(x) and at g(g(x)). So f must be constant on each block, and the simplest choice gives block n the value n+1.
f(3n+1) = f(3n+2) = f(3n+3) = n+1
Option Check Verdict (a) onto f with f ° g = f The f above is constant on each block, and block n gives the value n+1, so every natural number is hit True (b) one-one f with f ° g = f f(1) = f(g(1)) = f(2), yet 1 ≠ 2 False (c) g ° g ° g = g Three turns of a 3-cycle return each number to itself, so g ° g ° g is the identity, and g(1) = 2 ≠ 1 False (d) some f with g ° f = f g(f(x)) = f(x) needs f(x) to be a fixed point of g, and g has none False - So. Only (a) holds: f(x) = ⌈ x/3 ⌉ is onto and satisfies f ° g = f.
The idea to keep. Spot the cycle in a piecewise map first. Then f ° g = f means f is constant on each cycle, and g ° f = f means f only takes values that g fixes.
Answer from the NTA final answer key. Solution by PrepWiser.