Basic Functional Equations
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Direct answer
A functional equation is a puzzle in which the unknown is a rule, not a number — an equation satisfied by f for all permitted inputs, from which the function itself must be recovered. The recognisable archetypes: f(x + y) = f(x) + f(y) (Cauchy's equation, whose continuous solutions are f(x) = kx), f(x + y) = f(x)·f(y) (exponential f(x) = a^x), f(xy) = f(x) + f(y) (logarithmic), and f(x)·f(1/x) = f(x) + f(1/x) (yielding f(x) = 1 ± xⁿ). The method is substitution-driven: insert x = 0, y = 0, x = y, y = 1/x; deduce f(0), f(1), oddness or evenness; guess the closed form and verify — the guess-then-prove discipline JEE marking demands.
What you must remember
- Cauchy archetype: f(x + y) = f(x) + f(y) with continuity gives f(x) = kx; setting y = x gives f(2x) = 2f(x), and integer inputs satisfy f(n) = n·f(1) — the standard opening move.
- Exponential archetype: f(x + y) = f(x)f(y) with f not identically zero gives f(x) = a^x with f(1) = a; also f(x) = f(x/2)² ≥ 0.
- Logarithmic archetype: f(xy) = f(x) + f(y) gives f(x) = k·ln x on positive reals, f(1) = 0 always.
- The reciprocal classic: f(x)·f(1/x) = f(x) + f(1/x) rearranges to [f(x) − 1][f(1/x) − 1] = 1, whose natural solutions are f(x) = 1 ± xⁿ.
- Standard substitutions: x = y = 0; x = 1, y = 1; replacing y by 1/x or −x; putting y = x; each targets the equation's structure — addition suggests x + y moves, products suggest xy and 1/x moves.
- Verification is part of the answer: any candidate f must be checked in the original equation over the whole domain — a function fitting sampled values can still fail elsewhere, and JEE options are built on such near-misses.
Solving f(x)·f(1/x) = f(x) + f(1/x)
Rearrange: f(x)·f(1/x) − f(x) − f(1/x) = 0, so adding 1 to both sides factors as [f(x) − 1]·[f(1/x) − 1] = 1. This structure screams reciprocation: if f(x) − 1 = xⁿ then f(1/x) − 1 = x^(−n), and the product is 1 for every n. Hence f(x) = 1 + xⁿ for any real n is a solution family — and equally f(x) = 1 − xⁿ works, since the two minus signs multiply to plus. Verify concretely with f(x) = 1 + x²: f(2) = 5, f(1/2) = 1.25; product 6.25, sum 6.25 ✓. With the extra condition f(2) = 5, the family collapses: 1 + 2ⁿ = 5 gives n = 2, uniquely.
The same problem illustrates the guess-then-prove economy. Sampling: x = 1 gives f(1)² = 2f(1), so f(1) = 0 or 2 — both consistent (n = 0... f(1) = 2 matches 1 + 1ⁿ = 2; f(1) = 0 matches 1 − 1ⁿ = 0). The sampled values alone cannot finish; only the factored form [f(x) − 1][f(1/x) − 1] = 1 pins the family, and only substitution confirms it everywhere. That three-beat rhythm — sample, factor, verify — is the entire technique in miniature.
Guess, then prove
JEE Main keeps functional equations short: identify the archetype from the equation's shape and produce f, usually with one auxiliary condition to fix constants. JEE Advanced is where the topic becomes a signature — multi-step derivations that begin with f(x + y) = f(x) + f(y) + xy or f(x)f(y) = f(x) + f(y) + xy-type hybrids, which are handled by defining g(x) = f(x) − c to absorb the constant term and reduce to an archetype. The dominant trap is over-assuming: solving Cauchy's equation without the continuity (or monotonicity, or boundedness) hypothesis is incomplete — wild non-linear solutions exist, and a rigorous answer states which hypothesis the question supplied. The second trap is verification-free guessing: a polynomial that matches three sampled values but fails the equation for general x, sitting in the options, waiting. The third is domain blindness: f(xy) = f(x) + f(y) presumes positive inputs; allowing x = 0 forces f(0) considerations that collapse the solution. Write the equation down, choose substitutions that mirror its structure, and finish with the check — every mark lives in that last step.
Frequently asked questions
What are the standard archetypes of functional equations?
f(x + y) = f(x) + f(y) → f(x) = kx (with continuity); f(x + y) = f(x)f(y) → f(x) = a^x; f(xy) = f(x) + f(y) → logarithmic — recognising the shape is half the solution.
How do you solve f(x)·f(1/x) = f(x) + f(1/x)?
Rewrite as [f(x) − 1][f(1/x) − 1] = 1, giving the family f(x) = 1 ± xⁿ; an extra value condition such as f(2) = 5 fixes n = 2.
Which substitutions should be tried first?
x = 0, y = 0, x = y, and replacing y by 1/x or −x, chosen to mirror the equation's structure — addition forms invite x + y and −x, product forms invite xy and 1/x.
Why is continuity needed for f(x + y) = f(x) + f(y) to force f(x) = kx?
Because without regularity (continuity, monotonicity or boundedness on an interval), pathological non-linear solutions exist; the hypothesis is what collapses the family to straight lines.
Why must a guessed function be verified?
Because agreement at sampled points does not guarantee agreement everywhere; verification in the original equation is the only proof, and examiners deliberately include near-miss options.