Q.21
Let Z be the set of all integers and f and g are one-one mappings from Z into itself.
If
−−=++=
nodd for n f(n) = g(f(n)) – 1
neven for n g(n) = f(g(n)) + 1
and f(-1) = 3 then
🔍 Functional Equations with One-One Mappings
For students preparing for competitive exams, functional equations with one‑one mappings from integers often look intimidating. In this problem, a functional equation with one‑one mappings from integers is combined with parity (even–odd) conditions on the input, and candidates must evaluate specific statements about the functions f and g. Understanding how to move step‑by‑step through these relations is crucial for solving such functional equation questions efficiently during exams.
1Problem Statement
Let ℤ be the set of all integers and f and g are one‑one mappings from ℤ to itself.
Determine the truth of the options:
(B) f(3) = 2
2Step 1: Use f(1) = 3
1 is odd, so use the second rule g(f(n)) = f(n−1) − 1 for odd n.
Put n = 1:
From here we know g(3) and f(0) are linked by a difference of 1, but we still do not know their individual values.
3Step 2: Express Conditions
- For even n: f(g(n)) = g(n+1) + 1
(n even ⇒ n+1 odd) - For odd n: g(f(n)) = f(n−1) − 1
(n odd ⇒ n−1 even)
Up to this stage, there is still no direct equation fixing particular numeric values like g(2) or f(3); only relations between neighboring points are available.
4Step 3: Check Option (B) f(3) = 2
Assume temporarily that option (B) is true: f(3) = 2.
3 is odd, so apply the odd‑n rule with n = 3:
Key Insight: This relates g(2) and f(2) but does not specify their individual values. No rule links f(2) to f(1) = 3 directly. Since f and g are only constrained by functional equations and injectivity, f(3) is not uniquely determined.
Therefore, option (B) is NOT necessarily true.
5Step 4: Check Option (A) g(2) = 0
Assume temporarily: g(2) = 0.
2 is even, so use even‑n rule with n = 2:
From equation (1): g(3) = f(0) − 1
Equations (1) and (3) are consistent, but nothing forces g(2) = 0. Different injective mappings can satisfy conditions with other g(2) values.
Therefore, option (A) is also NOT necessarily true.
6Logical Conclusion
The functional equations link values only relatively (neighboring integers). With one-one mappings (not necessarily onto), there’s enough freedom for various consistent assignments.
- No deduction uniquely fixes f(3) → (B) cannot be guaranteed
- No deduction uniquely fixes g(2) → (A) cannot be guaranteed
🎯 Exam Strategy Takeaway
Critical Thinking: When functional equations provide relations rather than absolute values, specific numeric claims are usually NOT necessarily true unless uniquely determined by the system.
