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Functions sit under almost every chapter of JEE maths. Limits, continuity, inverse trigonometric functions and calculus all assume you can say where a function is defined and what values it can take. Yet "find the domain" questions catch strong students, because the method is easy and the conditions are easy to forget. This post gives you a checklist for domain, a method for range, and the rule that makes composite functions safe, with worked examples and a practice routine.
What does the JEE syllabus include?
The JEE (Main) 2026 syllabus lists "relations, functions; one-one, into and onto functions, the composition of functions" in Unit 1, and "real-valued functions, algebra of functions... inverse functions" and "graphs of simple functions" in Unit 7. The NCERT Class 11 chapter on relations and functions defines domain, codomain and range, and the Class 12 chapter defines the composition gof(x) = g(f(x)). Check the syllabus for your exam year.
Why do students find functions harder than they look?
Research suggests the trouble is often the picture in your head, not the definition. Vinner and Dreyfus (1989) compared the definitions of "function" given by 271 college students and 36 junior high school teachers with how they actually handled functions in problems. Mismatches between image and definition were frequent, even among those who stated the formal definition correctly. In other words, knowing the definition does not mean you will use it when a problem looks unfamiliar.
Range is the harder half. In a study of how students read domain and range from graphs, Cho and Moore-Russo (2014) found that, in general, students found the range more difficult to determine than the domain. That matches what happens in JEE: domain questions are checklists, range questions need an argument.
How do you find the domain?
Start from all real numbers and remove whatever breaks the expression:
Denominator: cannot be zero.
Even root: the expression inside must be greater than or equal to zero.
Logarithm: the argument must be strictly positive (and the base positive and not 1).
Inverse sine or cosine: the argument must lie in [-1, 1].
Write each condition, solve each, then take the intersection.
Worked example 1. f(x) = √((x - 1)/(x + 2)).
Denominator: x ≠ -2.
Root: (x - 1)/(x + 2) ≥ 0. The critical points are x = 1 and x = -2. A sign chart gives positive for x < -2, negative between -2 and 1, and positive for x > 1. The fraction is zero at x = 1.
Combine: domain is (-∞, -2) ∪ [1, ∞). The point -2 is excluded and 1 is included.
Worked example 2. f(x) = log(x² - 5x + 6). Need x² - 5x + 6 > 0, that is (x - 2)(x - 3) > 0, so x < 2 or x > 3. Domain: (-∞, 2) ∪ (3, ∞). Note the strict inequality: at x = 2 the argument is zero, and log 0 is not defined.
How do you find the range?
The most reliable method is to set y = f(x), solve for x, and ask which values of y give a real x that lies in the domain.
Worked example 3. f(x) = x/(x² + 1), domain all real numbers.
Set y = x/(x² + 1), so yx² - x + y = 0.
If y = 0, then x = 0 works, so 0 is in the range.
If y ≠ 0, this is a quadratic in x. For real x, the discriminant must be non-negative: 1 - 4y² ≥ 0, so -1/2 ≤ y ≤ 1/2.
Range: [-1/2, 1/2]. The maximum value 1/2 is reached at x = 1, and you can check: 1/(1 + 1) = 1/2.
For functions built from standard pieces, use known ranges instead. For example, sin x lies in [-1, 1], so 3 + 2 sin x lies in [1, 5].
How do composite functions change the domain?
For f(g(x)), x must be in the domain of g, and g(x) must be in the domain of f. Simplifying the formula first can hide the first condition.
Worked example 4. Let f(x) = √x and g(x) = 1 - x².
f(g(x)) = √(1 - x²). Need 1 - x² ≥ 0, so the domain is [-1, 1]. The range is [0, 1].
g(f(x)) = 1 - (√x)² = 1 - x. The formula looks defined for all x, but f(x) = √x needs x ≥ 0. So the domain is [0, ∞), not all real numbers, and the range is (-∞, 1].
This is the trap examiners set: the simplified formula is right, but the domain must come from the original chain.
Common slips and how to avoid them
Using ≥ for a logarithm instead of >.
Including the zero of the denominator in a sign chart answer.
Squaring both sides of an inequality without checking signs.
Simplifying a composite first, then reading the domain from the simplified formula.
Confusing codomain with range. The codomain is what the function is declared to map into; the range is what it actually reaches.
A practice routine
Domain drill: ten functions mixing roots, logs and fractions. Write every condition before solving any.
Range drill: five rational functions by the discriminant method, five using known bounds.
Composite pairs: for five pairs f and g, find the domains of both f(g(x)) and g(f(x)), and explain why they differ.
Graph check: sketch each function, or plot it, and confirm your answer. Our post on using graphing tools to learn maths explains how to do this without skipping the reasoning.
When you are ready for the next step, the same domain thinking drives limits and continuity and inverse trigonometric functions. If a definition such as one-one or onto is still fuzzy, the Fermi Concept Library covers every concept in mathematics, each with its own lesson.
Today, find the domain of g(f(x)) for three composite pairs of your choice, and check each against the original chain rather than the simplified formula.
