The number of functions
satisfying , is
satisfying , is
- A.2
- B.3
- C.4
- D.1
Show answer and solution
Correct option: A
Solution
From the recurrence f(n) + (1/n)f(n+1) = 1, we get f(n+1) = n(1 - f(n)).
Starting with f(1) = a (free parameter), we compute:
- f(2) = 1·(1 - a) = 1 - a
- f(3) = 2·(1 - (1-a)) = 2a
- f(4) = 3·(1 - 2a) = 3 - 6a
All four values must be integers with |value| ≤ 8.
The binding constraint is |3 - 6a| ≤ 8, giving -8 ≤ 3 - 6a ≤ 8, so -5/6 ≤ a ≤ 11/6, meaning a ∈ {0, 1} (since a must be an integer satisfying |a| ≤ 8).
Checking: a = 0 gives f = (0, 1, 0, 3) ✓; a = 1 gives f = (1, 0, 2, -3) ✓. Both satisfy all constraints, giving exactly 2 valid functions.
Starting with f(1) = a (free parameter), we compute:
- f(2) = 1·(1 - a) = 1 - a
- f(3) = 2·(1 - (1-a)) = 2a
- f(4) = 3·(1 - 2a) = 3 - 6a
All four values must be integers with |value| ≤ 8.
The binding constraint is |3 - 6a| ≤ 8, giving -8 ≤ 3 - 6a ≤ 8, so -5/6 ≤ a ≤ 11/6, meaning a ∈ {0, 1} (since a must be an integer satisfying |a| ≤ 8).
Checking: a = 0 gives f = (0, 1, 0, 3) ✓; a = 1 gives f = (1, 0, 2, -3) ✓. Both satisfy all constraints, giving exactly 2 valid functions.
[JEE Main 2023 · 25 Jan · Shift 2]