Domains, reversibility and functional equations: follow the inputs before manipulating the symbols
JEE Main · JEE Advanced85 min study guide11 indexed sections
BY THE END, YOU CAN
Distinguish domain, codomain and range and test injectivity and surjectivity relative to the declared sets.
Determine the domains of compositions and construct inverses on appropriate restrictions.
Use symmetry and periodicity without confusing graph properties with one-to-one behavior.
Solve functional equations by justified substitutions, including the regularity needed for additive functions.
BEFORE YOU START
Real-number inequalities, interval notation and elementary algebra.
Basic graphs of powers, roots, exponentials and trigonometric functions.
Original educational lesson. Work through the examples before revealing the practice answers. Check the current official syllabus separately; historical question frequency does not establish eligibility or predict future questions.
CONNECT THE CONCEPT TO THE EXAM
Selected exam questions & worked solutions
These selected examples are separate from the original practice below. Each identifies its exam, year and wording. They do not represent every paper or every selected year.
No source-verified exam questions have been published for this lesson yet. The worked examples and practice below are original teaching material, not past-exam questions.
01 / UNDERSTAND & APPLY
1. A function includes its input and output sets
A function f:A→B assigns exactly one element of B to every element of A. The set A is its domain and B is its codomain. The range f(A) is the set of outputs actually attained. The codomain is part of the specification; it cannot generally be recovered from a displayed algebraic formula alone. The same rule can be onto one codomain and fail to be onto a larger one.
For example, the rule x↦x² on real inputs has range [0,∞). As a function ℝ→ℝ it misses every negative number. As a function ℝ→[0,∞) it attains every member of its codomain. Neither version is one-to-one because opposite nonzero inputs share their output. Restricting inputs to [0,∞) changes that conclusion.
A relation is not a function of x if one allowed x has two different assigned y values. The circle equation x²+y²=1 does not specify a single real function on [−1,1] until an upper or lower branch is chosen. In contrast, one output occurring at several different inputs is perfectly permissible for a function.
When a question gives only a real formula and asks for its natural domain, use the real inputs for which every operation is defined. Denominators must be nonzero, even roots require nonnegative radicands and logarithms require positive arguments. If the domain is explicitly supplied, respect that choice rather than automatically enlarging it.
ORIGINAL WORKED EXAMPLE
Domain and range require different arguments
Find the natural real domain and range of f(x)=1/(x²−4).
The denominator vanishes at x=±2, so the domain is ℝ excluding −2 and 2.
Write t=x²≥0. Then t−4 belongs to [−4,0) or (0,∞).
Taking reciprocals gives negative outputs (−∞,−1/4] from the first interval and positive outputs (0,∞) from the second.
The endpoint −1/4 is attained at x=0; zero is never attained. Every listed output corresponds to an allowed t and hence at least one real x.
Answer Domain ℝ∖{−2,2}; range (−∞,−1/4]∪(0,∞).
02 / UNDERSTAND & APPLY
2. Count preimages, with the codomain fixed
Injective means f(a)=f(b) forces a=b for all a,b in the domain. Equivalently, an attained output has exactly one preimage. To disprove injectivity, one pair of distinct allowed inputs with the same output suffices. To prove it, a collection of successful numerical tests is not enough; use an argument covering every pair.
Surjective onto B means every b in the declared codomain B has at least one preimage in A. A reliable method is to solve f(x)=b and prove that an allowed solution exists for each b∈B. Bijective means both injective and surjective: every codomain element then has exactly one preimage.
Strict increase or strict decrease on the whole domain proves injectivity. Merely being monotone on separate pieces is insufficient: two pieces can produce overlapping outputs. A function on a finite set of n elements into another set of n elements is injective exactly when it is surjective; this finite equal-cardinality shortcut does not extend to arbitrary infinite sets.
Graphically, a horizontal line meets the graph at most once for injectivity. For surjectivity, every horizontal line corresponding to a codomain value must meet it. These are different questions. The vertical-line test instead checks whether the graph represents a function in the first place.
ORIGINAL WORKED EXAMPLE
Same rule, three different classifications
Classify x² as a map ℝ→ℝ, ℝ→[0,∞), and [0,∞)→[0,∞).
For ℝ→ℝ, the inputs 1 and −1 disprove injectivity, and the output −1 disproves surjectivity.
For ℝ→[0,∞), every output y≥0 has a real preimage √y. It is surjective but still not injective.
For [0,∞)→[0,∞), if a²=b² with a,b≥0 then a=b. Every y≥0 still has the preimage √y.
The last map is bijective and has inverse √x. Changing the sets is mathematically meaningful, not a cosmetic relabelling.
The composition f∘g sends x first to g(x), then to f(g(x)). It is defined exactly for inputs in the domain of g whose images lie in the domain of f. The order matters, and the formula f(g(x)) must be interpreted with these original restrictions even if algebra later simplifies it.
For functions g:A→B and f:B→C, every input in A is admissible. More generally, if the codomain of g is not contained in the domain of f, use the subset on which g actually lands in that domain. Composition of total functions between specified matching sets is associative: f∘(g∘h)=(f∘g)∘h. This does not make it commutative.
A cancellation can hide a forbidden intermediate value. If g(x)=1/x and f(u)=1/u, the composite formula simplifies to x, but its domain remains x≠0. It is the identity on ℝ∖{0}, not the identity on all real numbers. Domain tracking must therefore precede simplification.
For injective compositions, both factors injective is sufficient. If f∘g is injective, then g must be injective, but f need only be injective on the attained set g(A). Similarly, onto conclusions require attention to the intermediate set rather than a slogan about the composite.
Dom(f∘g)={x∈Dom(g):g(x)∈Dom(f)}
The inner function must exist and its output must be an allowed input to the outer function.
ORIGINAL WORKED EXAMPLE
Reverse the order and rebuild the domain
Let f(x)=√x and g(x)=1/(x−1), on their natural real domains. Find f∘g and g∘f with domains.
For f∘g, x≠1 and 1/(x−1)≥0. The numerator is positive, so this requires x>1.
Thus (f∘g)(x)=√(1/(x−1)) for x>1.
For g∘f, first require x≥0. Then √x must not equal 1, so x≠1.
Thus (g∘f)(x)=1/(√x−1) for x∈[0,∞)∖{1}. At x=4 the two values are 1/√3 and 1, demonstrating that the compositions differ.
Answer f∘g: √(1/(x−1)), x>1; g∘f: 1/(√x−1), x≥0 and x≠1.
04 / UNDERSTAND & APPLY
4. An inverse reverses a bijection
A bijection f:A→B has an inverse f⁻¹:B→A defined by the unique preimage of each output. The identities f⁻¹(f(x))=x for x∈A and f(f⁻¹(y))=y for y∈B must both hold. The inverse-function notation f⁻¹ is not the reciprocal 1/f; those are different operations with different domains.
To find an inverse formula, write y=f(x), solve uniquely for x using the original domain restrictions, and interchange the variable names. The inverse domain is the original range, and its range is the original domain. An algebraic solution with two branches signals a need to restrict the original domain before calling either branch an inverse.
A function that is injective but not onto its stated codomain has an inverse from its range to its domain. It does not have a two-sided inverse defined on the entire larger codomain. For example, eˣ:ℝ→ℝ is injective but not onto; its inverse logarithm has domain (0,∞), the actual range.
The graphs of a bijection and its inverse reflect across y=x because each point (a,b) becomes (b,a). If f and g are composable bijections, the inverse of f∘g is g⁻¹∘f⁻¹: undo the last operation first. Verify this order by substituting rather than trying to memorize a visual pattern.
Full-size scientific diagram. Swipe or scroll horizontally; keyboard users can focus the diagram and use arrow keys.
Equal-scale Cartesian plot. The inverse uses the nonnegative square branch, with both full domains and ranges [0,∞).
ORIGINAL WORKED EXAMPLE
A fractional linear inverse with excluded values
For f(x)=(2x+1)/(x−3), determine its range and inverse.
The domain excludes x=3. Solving y(x−3)=2x+1 gives x(y−2)=3y+1.
When y=2 the original equation would require −6=1, impossible. For each y≠2, the unique solution is x=(3y+1)/(y−2).
This solution never equals 3, since equality would imply 3y+1=3y−6. Thus every y≠2 is genuinely attained.
On ℝ∖{3}→ℝ∖{2}, f is bijective and f⁻¹(x)=(3x+1)/(x−2). Substitution into either composition yields the input on its respective allowed set.
Answer Range ℝ∖{2}; f⁻¹(x)=(3x+1)/(x−2), x≠2.
05 / UNDERSTAND & APPLY
5. Symmetry is a statement about all allowed inputs
An even function satisfies f(−x)=f(x), and an odd function satisfies f(−x)=−f(x), on a domain symmetric about zero. Without that domain symmetry the usual even/odd classification is not applicable. An odd function defined at zero must have f(0)=0, but the converse is false: vanishing at zero does not establish oddness.
The zero function is both even and odd on a symmetric domain. Other functions may be neither. When both factors have symmetric domains, even times even and odd times odd are even; even times odd is odd. Sums need matching parity to inherit it. For example, x²+x is neither, despite each summand having a clear parity.
A positive period T means the domain is preserved by the required shifts and f(x+T)=f(x) wherever the statement applies. A nonconstant periodic function on all of ℝ cannot be injective because x and x+T are distinct inputs with equal outputs. A constant function has every positive number as a period and therefore no least positive period.
A claimed fundamental period needs minimality, not only one successful shift. For sin(2x), π is the least positive period. For a transformed function f(ax+b), a known period T of f produces period T/|a| when a≠0; claiming this is fundamental additionally uses minimality for the original function and the full transformed domain.
06 / UNDERSTAND & APPLY
6. A functional equation constrains an entire unknown rule
A functional equation is an identity involving an unknown function at one or more inputs. Its quantifiers and domain are central. An equation for every real x permits substitutions such as 0, −x and x+1. An equation only for x≠0 does not permit x=0. A two-variable identity may allow swapping variables or setting them equal, provided every resulting argument remains allowed.
A good strategy is to search for substitutions that connect a small family of unknown values. An involution such as x↦1/x returns to the original input after two substitutions, making a two-equation linear system possible. Other identities reveal f(0), f(1), symmetry or an additive law. Record these consequences before guessing a formula.
A candidate formula is not a solution until it has been checked in the original identity for every permitted input. Necessary equations can lose information; dividing by a function value can discard zero solutions. Likewise, testing only integers establishes at most integer behavior unless an additional argument reaches all real inputs.
The aim is sometimes a whole family rather than a single rule. A shift equation can leave an arbitrary periodic term. A composition equation can admit many non-linear functions. State any assumed class, such as affine functions or continuous functions, instead of silently restricting the unknown.
ORIGINAL WORKED EXAMPLE
Close a two-input substitution cycle
Find all f:ℝ∖{0}→ℝ satisfying f(x)+2f(1/x)=3x for every x≠0.
Replace x by 1/x, which is permitted and still nonzero: f(1/x)+2f(x)=3/x.
Let A=f(x), B=f(1/x). The system A+2B=3x and 2A+B=3/x has nonzero determinant −3.
Twice the second equation minus the first gives 3A=6/x−3x. Hence f(x)=2/x−x.
Check directly: (2/x−x)+2(2x−1/x)=3x for every x≠0. The linear system proves uniqueness, including the fixed inputs x=1 and x=−1.
Answer The unique solution is f(x)=2/x−x on ℝ∖{0}.
07 / UNDERSTAND & APPLY
7. Additive equations: rational control and the real-line gap
Suppose f:ℝ→ℝ satisfies f(x+y)=f(x)+f(y) for all real x,y. Setting both variables to zero gives f(0)=0. Setting y=−x gives f(−x)=−f(x). Repeated addition then gives f(nx)=nf(x) for integers n, and applying this to x/n gives f(x/n)=f(x)/n for positive integers n.
Consequently f(q)=qf(1) for every rational q. This conclusion is purely algebraic. Extending it from rational inputs to all real inputs requires a further hypothesis. Continuity at zero is one sufficient hypothesis; monotonicity on ℝ is another sufficient hypothesis, although it uses a different squeezing argument.
Under continuity at zero, choose rationals qₙ→x. Additivity gives f(x)−f(qₙ)=f(x−qₙ)→0. Since f(qₙ)=qₙf(1), taking the limit yields f(x)=xf(1). Continuity at every point was not needed as an initial assumption; additivity transports continuity at zero to every other point.
Without a regularity condition, additivity alone does not force f(x)=cx for all real x. Non-linear additive functions exist, though constructing them goes beyond elementary algebra. In a problem that omits regularity, report the rational consequence or an additive family rather than silently adding continuity.
ORIGINAL WORKED EXAMPLE
Use the missing hypothesis explicitly
Let f:ℝ→ℝ be additive and continuous at zero, with f(1)=3. Determine f(√2).
Additivity gives f(q)=3q for every rational q.
Choose a rational sequence qₙ converging to √2. Then √2−qₙ→0, and continuity at zero gives f(√2−qₙ)→f(0)=0.
By additivity, f(√2)−3qₙ=f(√2−qₙ). Therefore f(√2)=lim 3qₙ=3√2.
The same argument gives f(x)=3x for every real x, and that formula satisfies both additivity and continuity.
Answer f(√2)=3√2; in fact f(x)=3x for all real x.
08 / UNDERSTAND & APPLY
8. Recognize free families and stated function classes
The equation f(x+1)=f(x)+2 on ℝ determines how values repeat across integer shifts, but it does not determine the values on one unit interval. Subtract the predictable linear part: p(x)=f(x)−2x then satisfies p(x+1)=p(x). Conversely every 1-periodic p gives a solution f(x)=2x+p(x). Even continuity leaves infinitely many such solutions.
An involution satisfies f(f(x))=x on a set mapped into itself. It is automatically injective: equal outputs remain equal after applying f and therefore have equal inputs. It is also onto: each x is the image of f(x). Thus every involution is a bijection and is its own inverse.
Classifying affine involutions is easier than classifying all involutions. For f(x)=ax+b, expansion gives f(f(x))=a²x+b(a+1). Coefficient comparison is justified because the identity holds for every real x. These coefficient conditions describe affine solutions only; other involutions may exist on suitable domains.
ORIGINAL WORKED EXAMPLE
Two equations, two kinds of solution set
Describe all real functions satisfying f(x+1)=f(x)+2, and all affine real functions satisfying f(f(x))=x.
For the shift equation, define p(x)=f(x)−2x. Direct subtraction gives p(x+1)=p(x), and the converse substitution verifies every such periodic p.
Examples include f(x)=2x and f(x)=2x+sin(2πx). Both are continuous, so continuity does not make this equation unique.
For the affine composition equation, a²=1 and b(a+1)=0. If a=1 then b=0. If a=−1, b is unrestricted.
Checking shows the identity f(x)=x and every reflection f(x)=−x+b are affine involutions. No claim about all non-affine functions follows from assuming the affine form.
Answer Shift family: 2x+p(x), with p 1-periodic. Affine involutions: x or −x+b for any real b.
09 / UNDERSTAND & APPLY
9. Original practice: sets, composition and inverses
Give the domain with each formula. For classification questions, supply a proof or an explicit counterexample rather than only a label.
ORIGINAL PRACTICE · 1
Find the natural domain of √(3−x)/(x+1).
Reveal answer & reasoning
(−∞,3]∖{−1}.
The square root requires x≤3, while the denominator forbids −1. Both conditions apply simultaneously.
ORIGINAL PRACTICE · 2
Is f:ℝ→(0,∞), f(x)=eˣ, bijective?
Reveal answer & reasoning
Yes.
Strict increase proves injectivity, and for every y>0 the real input ln y maps to y. The positive codomain is essential.
ORIGINAL PRACTICE · 3
Find the inverse of f(x)=3x−5 on ℝ.
Reveal answer & reasoning
f⁻¹(x)=(x+5)/3.
Solve y=3x−5 for x. Both compositions simplify to the identity on ℝ.
ORIGINAL PRACTICE · 4
For f(x)=x² and g(x)=x+1 on ℝ, find f∘g and g∘f.
Reveal answer & reasoning
(x+1)² and x²+1, respectively.
Apply the inner function first. The first expands to x²+2x+1, generally different from the second.
ORIGINAL PRACTICE · 5
If f(x)=1/x on ℝ∖{0}, what are f∘f and its domain?
Reveal answer & reasoning
The identity x, with domain ℝ∖{0}.
The first application is undefined at zero, so simplification cannot restore that excluded input.
ORIGINAL PRACTICE · 6
Classify x³+x as even, odd or neither on ℝ.
Reveal answer & reasoning
Odd.
Replacing x by −x yields −x³−x=−(x³+x), and the domain is symmetric.
ORIGINAL PRACTICE · 7
Can a function on ℝ be both injective and have positive period 4?
Reveal answer & reasoning
No.
For any x, periodicity gives f(x+4)=f(x) at distinct real inputs, contradicting injectivity.
10 / UNDERSTAND & APPLY
10. Original practice: substitutions and verification
Every identity in this section holds for all inputs in the stated domain. Include a final substitution check and identify any regularity hypothesis you use.
ORIGINAL PRACTICE · 1
Find all f:ℝ→ℝ satisfying f(x)+f(−x)=2x for every real x.
Reveal answer & reasoning
There is no solution.
Replacing x by −x leaves the left side unchanged but makes the right side −2x. At x=1 this would require the same sum to equal both 2 and −2.
ORIGINAL PRACTICE · 2
Find f:ℝ→ℝ if f(x)+2f(−x)=3x for every real x.
Reveal answer & reasoning
f(x)=−3x.
The reflected equation is f(−x)+2f(x)=−3x. Twice it minus the original yields 3f(x)=−9x. Checking gives −3x+6x=3x.
ORIGINAL PRACTICE · 3
An additive function has f(1)=5. Determine f(−3/2) without a continuity assumption.
Reveal answer & reasoning
−15/2.
Additivity forces f(q)=qf(1) for every rational q, including −3/2. No extension to irrational inputs is needed.
ORIGINAL PRACTICE · 4
An additive function is continuous at zero and f(2)=8. Find f(π).
Reveal answer & reasoning
4π.
Additivity gives f(2)=2f(1), so f(1)=4. Continuity at zero extends rational linearity to all real inputs.
ORIGINAL PRACTICE · 5
Give two different continuous solutions of f(x+1)=f(x)+1.
Reveal answer & reasoning
For example, f(x)=x and f(x)=x+sin(2πx).
The added sine is 1-periodic, so both functions increase by exactly 1 under a unit shift. This establishes nonuniqueness even under continuity.
ORIGINAL PRACTICE · 6
Find all affine functions f(x)=ax+b satisfying f(f(x))=4x+3.
Reveal answer & reasoning
f(x)=2x+1 or f(x)=−2x−3.
Coefficient comparison gives a²=4 and b(a+1)=3. For a=2, b=1; for a=−2, b=−3. Substituting either function twice gives 4x+3.
ORIGINAL PRACTICE · 7
If f maps a set A into itself and f(f(x))=x, prove f is onto A.
Reveal answer & reasoning
Every y∈A has preimage f(y).
Because f(y)∈A and f(f(y))=y, the required input exists for each codomain element. This uses the self-map condition.
11 / UNDERSTAND & APPLY
11. A compact audit of function reasoning
Begin by writing the domain and codomain. For a range problem, prove both that no other outputs occur and that every proposed output is attained. For a composite, test each intermediate input before simplifying. For an inverse, establish uniqueness and remember that input and output sets exchange roles.
Keep three distinct ideas separate: a function gives one output per input; injectivity gives at most one input per output; surjectivity gives at least one input per codomain element. A formula can satisfy one, two or all of these relevant requirements depending on its declared sets.
In a functional equation, write the quantifiers beside the domain, choose substitutions that remain allowed, and check whether a step divides by something that could vanish. Once a candidate is derived, verify the original identity rather than only the simplified necessary conditions.
The most consequential hidden assumption is regularity. Rational values of an additive function do not determine all real values without additional information. Similarly, a recurrence under a unit shift leaves freedom within a period. Recognizing the correct size of the solution family is part of solving the problem.
KEEP THE BIG IDEAS
Quick revision
A function is a rule together with its domain and codomain.
Composition requires both the inner input and the outer input to be admissible.
A two-sided inverse reverses a bijection and exchanges domain with range.
Functional equations are identities over stated domains, so substitutions and verification must respect every quantifier.
Additivity yields rational linearity; real linearity needs a suitable regularity hypothesis.
The teaching notes, worked explanations and original practice are written for this lesson. Separately labeled selected exam questions identify their exam, year, source and whether wording is paraphrased or an exact excerpt. Educational references do not certify current exam coverage.