JEE · Mathematics
Understand functions as mappings, correctly identify domain, codomain and range, classify mappings and determine when composition or an inverse is valid.
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In short
A function assigns exactly one output to every input in its domain. Domain tells you which inputs are permitted, codomain declares the target set and range contains the outputs actually achieved.
Before composing or inverting functions, check these sets explicitly. A function has an inverse from its codomain back to its domain only when it is bijective.
Official JEE documents define content scope. They do not publish chapter weightage, so none is asserted here.
Sources: JEE Main 2026 syllabus and JEE Advanced 2026 syllabus, linked in the sources section below.
Never begin with algebraic manipulation before determining the domain.
The reasoning order is domain, then mapping behaviour, then representation, then operation, then validation. Determine where an expression is defined before simplifying it algebraically.
Injective, surjective and bijective are separate, checkable conditions.
A function is injective when different inputs have different outputs, surjective when every codomain element is attained, and bijective when it is both. In standard school usage, an into function has range that is a proper subset of the codomain.
An inverse f⁻¹: B→A exists only when f is bijective with a fixed domain and codomain.
A one-to-one function can also be inverted onto its range if the codomain is correspondingly restricted. Writing an inverse expression before checking one-one and onto status is a common source of error.
(g∘f)(x) = g(f(x)) requires x in the domain of f and f(x) in the domain of g.
Composition is evaluated inner function first. In general g∘f is not equal to f∘g, so composition order must be preserved throughout the calculation.
Even and odd tests require a domain symmetric under x ↦ -x.
f(-x) = f(x) defines an even function and f(-x) = -f(x) defines an odd function, both only meaningful on a domain compatible with sign reversal of x. Not every function is even or odd.
Consider f: R→R, f(x) = x². Domain and codomain are both R. Does an inverse function exist?
Method: test injectivity before trying to algebraically write √x. f(1) = f(-1) = 1, so f is not injective. Therefore f: R→R is not invertible.
Restrict instead to f: [0,∞)→[0,∞). Now f is one-one and onto, so f⁻¹(x) = √x for x ≥ 0.
Validity check: f⁻¹(f(x)) = √(x²) = x, because the restricted original domain requires x ≥ 0.
A one-to-one function can also be inverted onto its range if the codomain is correspondingly restricted.
Treating range and codomain as synonyms.
Knowledge gapWhy it happens
Codomain is the declared target set; range is the subset of the codomain actually attained, and the two can differ.
How it is corrected
State the codomain from the problem and compute the range separately before comparing them.
Cancelling a square while forgetting the ± sign.
Execution errorWhy it happens
Taking a square root of both sides of an equation introduces two possible sign cases unless the domain has already been restricted.
How it is corrected
Track the domain restriction that removes one sign case, or keep both cases until the domain is checked.
Writing an inverse before testing one-one and onto.
Decision / selection errorWhy it happens
An inverse function exists only when the original function is bijective on its stated domain and codomain.
How it is corrected
Prove injectivity and surjectivity, or explicitly restrict domain and codomain, before writing any inverse expression.
Reversing composition order.
Execution errorWhy it happens
g∘f and f∘g apply the functions in different orders and are generally not equal.
How it is corrected
Evaluate the inner function first and keep the composition order exactly as written.
Evaluating a function outside its domain.
Execution errorWhy it happens
An expression that is undefined at a given input cannot be substituted into further calculation.
How it is corrected
Check the domain before substituting any specific value or composing with another function.
Calling an into function onto.
Decision / selection errorWhy it happens
An into function has range that is a proper subset of the codomain, so not every codomain element is attained.
How it is corrected
Explicitly show that every codomain element has a preimage before calling a function onto.
Testing even or odd on a non-symmetric domain.
Execution errorWhy it happens
The even and odd tests require a domain that is symmetric under x ↦ -x; otherwise f(-x) may not even be defined.
How it is corrected
Confirm the domain is symmetric under sign change before applying the even or odd test.
FAQ
Straight answers about how Rank Sarthi fits into serious exam preparation.
A function has an inverse from its codomain back to its domain only when it is bijective, that is both one-one and onto, on the stated domain and codomain.
Codomain is the declared target set B in f: A→B. Range is the set {f(x): x in A} of outputs actually attained, and it is always a subset of the codomain.
A function is onto, or surjective, when every element of the codomain is attained by some element of the domain.
f(x) = x² on R is not injective because f(1) = f(-1) = 1, so it is not invertible on R. Restricting the domain and codomain to [0,∞) makes it bijective, giving f⁻¹(x) = √x for x ≥ 0.
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