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A question says f(x + y) = f(x) + f(y) + 2xy for all real x and y, f is differentiable and f′(0) = 3, then asks for f(x). There is no formula to look up. There is a short set of moves that cracks almost every equation of this kind, and this post walks through them with four standard types and their traps.
Where is this in the syllabus?
Functional equations are not named in either syllabus. They are a way of testing topics that are. The JEE (Main) 2026 syllabus covers functions in Unit 1 and limits, continuity and differentiability in Unit 7. The JEE Advanced 2026 brochure lists "even and odd functions", "composition of functions" and "derivative of a function". Check the syllabus for your exam year.
What are the standard moves?
Put x = y = 0 to find f(0).
Put y = −x to learn whether f is odd or even.
Put y = x, then repeat to reach f(2x), f(3x) and so on.
If f is differentiable, write f′(x) as the limit of [f(x + h) − f(x)]/h and use the equation to replace f(x + h).
Guess the family, then verify your answer in the original equation.
Type 1: f(x + y) = f(x) + f(y)
x = y = 0 gives f(0) = 2f(0), so f(0) = 0.
y = −x gives f(x) + f(−x) = f(0) = 0, so f is odd.
Repeating y = x gives f(nx) = n f(x) for whole numbers n.
If f(1) = 3, then f(n) = 3n for every integer n, and f(1/2) = 3/2 because 2f(1/2) = f(1). The pattern says f(x) = 3x, but you can only conclude that for all real x if f is continuous. Georg Hamel's 1905 paper in Mathematische Annalen, on a basis of all numbers and the discontinuous solutions of this equation, showed that discontinuous solutions exist too. That is why JEE questions always add a condition such as continuity or differentiability.
Type 2: f(x + y) = f(x) f(y)
x = y = 0 gives f(0) = f(0)², so f(0) is 0 or 1.
If f(0) = 0, then f(x) = f(x + 0) = f(x) f(0) = 0 for every x. That is the zero function.
Otherwise f(0) = 1, and f(x) f(−x) = 1, so f is never zero. Also f(x) = f(x/2)², so f(x) > 0.
If f is differentiable, f′(x) = f(x) f′(0), whose solutions are exponentials. With f(1) = 2, f(x) = 2ˣ, and the sum f(1) + f(2) + f(3) + f(4) + f(5) = 2 + 4 + 8 + 16 + 32 = 62.
Type 3: f(xy) = f(x) + f(y) for x > 0
x = y = 1 gives f(1) = 0.
y = 1/x gives f(1/x) = −f(x).
This is the logarithm's rule, and with differentiability the solutions are multiples of ln x. Watch the domain: the equation is only given for positive x, so you cannot put x = 0.
Worked example: the first-principles trick
f(x + y) = f(x) + f(y) + 2xy for all real x, y; f is differentiable and f′(0) = 3. Find f(x).
Put x = y = 0: f(0) = 2f(0), so f(0) = 0.
Write the derivative: f′(x) = lim as h → 0 of [f(x + h) − f(x)]/h.
Use the equation with y = h: f(x + h) − f(x) = f(h) + 2xh.
So f′(x) = lim [f(h)/h] + 2x. Since f(0) = 0, f(h)/h = [f(h) − f(0)]/h, which tends to f′(0) = 3.
Hence f′(x) = 3 + 2x, and integrating gives f(x) = 3x + x² + C. Since f(0) = 0, C = 0.
Answer: f(x) = x² + 3x.
Verify: (x + y)² + 3(x + y) = (x² + 3x) + (y² + 3y) + 2xy. Both sides agree.
Which mistakes cost marks?
Concluding f(x) = kx without a continuity condition. Integers and rationals follow from the equation; all real numbers need the extra condition.
Missing the zero solution in Type 2, or dividing by f(0) before checking it is not zero.
Using values outside the domain, such as x = 0 in Type 3.
Not verifying. Many wrong options in multiple-choice questions fail the original equation.
Forgetting the constant when integrating f′(x), and not using f(0) to fix it.
For the building blocks, see our posts on domain, range and composite functions and periodic, even and odd functions.
If you get stuck choosing the first substitution, the Fermi Problem Library has worked problems with a step-by-step solver that explains the reasoning rather than printing an answer.
A practice routine
Write the five moves on a card and keep it beside you for the first week.
Solve two equations of each type, starting each with x = y = 0.
Do one first-principles question per session.
Verify every final answer with two random values.
Log each mistake as substitution, domain, missing case or calculus slip.
Try this tonight: if f(x + y) = f(x) + f(y) + xy, f is differentiable and f′(0) = 1, show that f(x) = x²/2 + x and verify it.
