PYQ Vault

JEE Mains Maths · Relations and Functions

Functional Equations

Equations for an unknown function: additive and multiplicative rules such as f(x + y) = f(x) + f(y), equations in f(x) and f(1/x) solved by a second substitution, and sums that pair up because f(x) + f(a − x) is constant.

All Relations and Functions notesJEE Mains Maths strategy

Why this matters

Twenty PYQs. Each gives a rule the function obeys rather than a formula, and asks for a value or a sum. Three moves settle almost all of them: put in convenient numbers, substitute to get a second equation, or pair terms from the two ends of a sum. Three ideas cover the page.

Concept 1 of 3: Additive and multiplicative rules

On natural numbers, f(x+y)=f(x)+f(y)f(x+y)=f(x)+f(y) gives f(n)=nf(1)f(n)=nf(1), and f(x+y)=f(x)f(y)f(x+y)=f(x)f(y) gives f(n)=f(1)nf(n)=f(1)^n. When extra terms appear, as in f(x+y)=f(x)+f(y)+2xyf(x+y)=f(x)+f(y)+2xy, put x=y=0x=y=0 first to get f(0)f(0), then either step y=1y=1 repeatedly or assume the polynomial form the question states and match coefficients. f(x+y)+f(x−y)=2f(x)f(y)f(x+y)+f(x-y)=2f(x)f(y) has cosine-type solutions.

Definition

  • f(x+y)=f(x)+f(y)⇒f(n)=nf(1)f(x+y)=f(x)+f(y)\Rightarrow f(n)=nf(1), f(0)=0f(0)=0.
  • f(x+y)=f(x)f(y)f(x+y)=f(x)f(y), f≠0f\ne0 ⇒f(n)=f(1)n\Rightarrow f(n)=f(1)^n, f(0)=1f(0)=1.
  • Extra terms: x=y=0x=y=0 first, then match coefficients.
  • f(x+y)+f(x−y)=2f(x)f(y)f(x+y)+f(x-y)=2f(x)f(y): f(x)=cos⁡(cx)f(x)=\cos(cx) type.

Additive rule

f(x+y)=f(x)+f(y) ⇒ f(n)=n f(1)f(x+y)=f(x)+f(y)\ \Rightarrow\ f(n)=n\,f(1)

Worked example

f(x+y)=f(x)+f(y)f(x+y)=f(x)+f(y) on N\mathbb N and f(1)=4f(1)=4. Find ∑k=15f(k)\sum_{k=1}^{5}f(k).
Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2021 · Paper 23 · Q65Moderate

Example 1 · Relations and Functions · Functional Equations

Let f:N→Nf:N \rightarrow N be a function such that f(m+n)=f(m)+f(n)f(m+n) =f(m) +f(n) for every m,n∈Nm,n\in N. If f(6)=18f(6) = 18, then f(2)⋅f(3)f(2) \cdot f(3) is equal to:

A constant term changes f(0)

In f(x+y)=f(x)+f(y)+cf(x+y)=f(x)+f(y)+c, putting x=y=0x=y=0 gives f(0)=−cf(0)=-c, not 0. Find it before using any pattern.
Drill 6 more on additive and multiplicative rules

Concept 2 of 3: Substitute again, then solve the pair

When the equation links f(x)f(x) with f(g(x))f(g(x)) and g(g(x))=xg(g(x))=x, as for 1x\frac1x, kx\frac kx or 1−x1-x, replace xx by g(x)g(x). That gives a second equation in the same two unknowns; solve the pair as simultaneous equations. A map that returns only after three steps, like 11−x\frac1{1-x}, needs three equations. A polynomial with f(x)f(1x)=f(x)+f(1x)f(x)f\left(\frac1x\right)=f(x)+f\left(\frac1x\right) is 1±xn1\pm x^n.

Definition

  • Partner maps (g∘g=idg\circ g=\mathrm{id}): 1x\frac1x, kx\frac kx, 1−x1-x, −x-x.
  • Replace xx by g(x)g(x); eliminate f(g(x))f(g(x)) between the two equations.
  • f(x)f(1x)=f(x)+f(1x)f(x)f\left(\frac1x\right)=f(x)+f\left(\frac1x\right), ff a polynomial ⇒f(x)=1±xn\Rightarrow f(x)=1\pm x^n.

Two equations, two unknowns

af(x)+bf(g(x))=h(x),af(g(x))+bf(x)=h(g(x))af(x)+bf\big(g(x)\big)=h(x),\quad af\big(g(x)\big)+bf(x)=h\big(g(x)\big)

Worked example

2f(x)+f(1x)=3x2f(x)+f\left(\frac1x\right)=3x. Find f(2)f(2).
Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2025 · 3 Apr 2025 · Q135Moderate

Example 2 · Relations and Functions · Functional Equations

Let ff be a function such that f(x)+3f(24x)f(x) + 3f\left( \frac{24}{x} \right) =4x,x≠0= 4x,x\neq 0. Then f(3)+f(8)f(3) +f(8) is equal to.

A sum may need no solving

If the question asks for f(a)+f(b)f(a)+f(b) where bb is aa's partner, putting x=ax=a and x=bx=b and adding the two equations can give it directly, without finding ff.
Drill 7 more on substitute again, then solve the pair

Concept 3 of 3: Pairing terms when f(x) + f(a − x) is constant

Sums like f(1n)+f(2n)+⋯+f(n−1n)f\left(\frac1n\right)+f\left(\frac2n\right)+\dots+f\left(\frac{n-1}n\right) are set up to pair: check whether f(x)+f(a−x)f(x)+f(a-x) is a constant, with aa the sum of the first and last inputs. Then add the terms in pairs from the two ends. With an odd number of terms the middle one is left over, and it equals half the constant.

Definition

  • f(x)=bxbx+b⇒f(x)+f(1−x)=1f(x)=\frac{b^x}{b^x+\sqrt b}\Rightarrow f(x)+f(1-x)=1.
  • mm terms paired: m2×\frac m2\times constant.
  • Middle term f(a2)=12×f\left(\frac a2\right)=\frac12\times constant.

The standard pair

f(x)=bxbx+b ⇒ f(x)+f(1−x)=1f(x)=\frac{b^x}{b^x+\sqrt b}\ \Rightarrow\ f(x)+f(1-x)=1

Worked example

f(x)=4x4x+2f(x)=\frac{4^x}{4^x+2}. Find ∑k=19f(k10)\sum_{k=1}^{9}f\left(\frac k{10}\right).
Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2025 · 28 Jan 2025 · Q54Moderate

Example 3 · Relations and Functions · Functional Equations

If f(x)=2x2x+2,x∈Rf(x) =\frac{2^{x}}{2^{x}+\sqrt{2}},x \in R, then ∑k=181f(k82)\sum_{k = 1}^{81} f\left( \frac{k}{82} \right) is equal to :

Count the terms

An odd number of terms leaves a middle term unpaired. Add it separately; forgetting it is the usual gap between two options.
Drill 4 more on pairing terms when f(x) + f(a − x) is constant

Summary — formulas & gotchas at a glance

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Formulas (3)

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Test yourself on Relations and Functions

20 past JEE Mains questions from this chapter, timed at 48 minutes and marked the way the exam marks it. You see your score and every answer the moment you finish. Free to start.

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