Functions and Graphs for CAT 2026: Modulus & Domain
A standalone cheatsheet on functions and graphs for CAT 2026 covering the 5 core function properties, the 3-step domain and range algorithm, modulus graph behavior with critical-point case-splits, even-odd-periodic identification rules, composite and inverse function procedures, and 12 solved CAT-level questions across MCQ and TITA formats. Built to close the algebra-cluster gap with the linear equations, percentages, and ratio guides.
Functions and Graphs for CAT 2026: Modulus & Domain
Functions CAT 2026 questions account for 5 to 7 Quant marks every cycle, yet almost no aspirant carries a clean framework into the exam. Most candidates remember domain rules vaguely, freeze on modulus case-splits, and treat composite questions as one-off puzzles. The result is predictable: 3 to 4 minutes per function question at 50 percent accuracy, when the same problems should take 90 seconds at 80 plus percent. The bottleneck is not intelligence; it is the absence of a function cheatsheet that fits in working memory. Functions show up in 3 to 5 marks of Quant every CAT cycle, with modulus and even-odd questions overrepresented in the TITA bank.
This guide is that cheatsheet for functions CAT 2026: the 5 properties, a 3-step domain and range algorithm, modulus graph and case logic, even-odd-periodic identification, composite and inverse procedures, and 12 solved CAT-level MCQ + TITA questions. It pairs with the algebra cluster on Optima Learn alongside linear equations word problems, percentages, and ratio and proportion.
Functions contribute 5 to 7 marks per CAT cycle. The 5 core properties: domain, range, even-odd, periodic, one-to-one. Domain rules: denominator non-zero, even root ≥ 0, log argument > 0. Modulus |x| has a V-graph; case-split every |expr|. Period of f + g = LCM of periods. Composite (f o g)(x) = f(g(x)); inverse by swap-and-solve. 12 solved Qs below.
5 Function Properties Every CAT 2026 Aspirant Must Know
The functions CAT 2026 syllabus reduces to five recurring properties. The goal is reflex: read the function definition and name the target property in 5 seconds.
| Function Type | Definition | Symmetry / Behavior | CAT Frequency | Example |
|---|---|---|---|---|
| Even | f(−x) = f(x) | Symmetric about y-axis | High | x², |x|, cos x |
| Odd | f(−x) = −f(x) | Symmetric about origin | High | x³, sin x, x|x| |
| Periodic | f(x + T) = f(x), smallest T > 0 | Repeats every period T | Med | sin x (2π), tan x (π), {x} (1) |
| Modulus | |x| = x if x ≥ 0; −x if x < 0 | V-shape, vertex at origin | High | |x − 3|, |x| + |x − 2| |
| One-to-one | f(a) = f(b) ⇒ a = b | Horizontal-line test passes | Med | Linear, ex, x³ |
These five properties span 80 percent of CAT function questions. Knowing all five delivers the full 5 marks every cycle; knowing only two or three delivers 1 to 2.
Build a one-page handwritten chart of these 5 properties before October 2026 and revise twice a week alongside arithmetic formula sheets. By the November mock window, the property-to-question mapping becomes reflexive.
Domain and Range: Finding Them Without a Graph
Domain is the set of inputs where the function is defined; range is the set of outputs. Domain is mechanical (three exclusion rules); range needs a graph or substitution. The 3-step algorithm below finishes in under 60 seconds.
Rule 2: Every even-root expression must be ≥ 0.
Rule 3: Every log argument must be > 0.
Apply each rule separately. Intersect all conditions. Output as union of intervals.
Range has no universal algorithm. Use graph-reading for familiar shapes (modulus, quadratic, trig) and inverse-substitution (write y = f(x), solve for x) when the shape is unclear.
Quadratic ax² + bx + c (a > 0): range is [c − b²/4a, ∞).
Quadratic ax² + bx + c (a < 0): range is (−∞, c − b²/4a].
Modulus |x|: range is [0, ∞).
sin x, cos x: range is [−1, 1].
ex: range is (0, ∞).
Mark every denominator (non-zero), every even root (≥ 0), every log (argument > 0). One condition per line.
Each yields an interval or union; use sign charts for quadratics and the case-split for modulus inequalities.
Intersect (AND) all conditions; write the domain as a union of intervals: e.g., (−∞, −2) ∪ (1, 3) ∪ (3, ∞).
Square-root condition is ≥ 0 (includes zero); log condition is > 0 (excludes zero). Mixing these shifts the domain boundary by one endpoint — enough to fail a TITA integer-count question.
One PYQ-style log-and-sqrt trap CAT keeps reusing: f(x) = √(log(x − 2)). Aspirants apply only the root rule (x − 2 ≥ 0) and write x ≥ 2. The correct chain is log(x − 2) ≥ 0 AND x − 2 > 0, which forces x − 2 ≥ 1, so the domain is [3, ∞). Skipping the inner log positivity step costs the mark on roughly one in three CAT 2026 domain questions.
The Modulus Function: Graph Behavior, Cases and 3 CAT Traps
The modulus f(x) = |x| is among the highest-frequency CAT function topics: a V-graph with vertex at the origin, slopes ±1, and the visual anchor for every modulus question.
|x − a| = x − a if x ≥ a; |x − a| = a − x if x < a
|x| = k ⇒ x = k or x = −k (only if k ≥ 0)
|x| < k ⇒ −k < x < k (only if k > 0)
|x| > k ⇒ x > k or x < −k
For sums of moduli like |x − 1| + |x + 2|, use the critical-point split: mark sign-change points (x = 1, x = −2), split the line into intervals, resolve each modulus per interval. The function becomes piecewise linear with extrema at the corners.
Interval x < −2: f = (1 − x) + (−2 − x) = −1 − 2x
Interval −2 ≤ x ≤ 1: f = (1 − x) + (x + 2) = 3
Interval x > 1: f = (x − 1) + (x + 2) = 2x + 1
Minimum value = 3, achieved on the entire middle interval.
Three modulus traps recur: forgetting |x| = k needs k ≥ 0, dropping a case (three moduli need four intervals), and misreading the V vertex (|x − 3| sits at x = 3).
For |x − 5| = −3, aspirants write x = 8 or x = 2. Both are wrong: |anything| is non-negative, so the equation has no solution. CAT plants this inside multi-step chains where missing it invalidates the whole answer.
Even, Odd and Periodic Functions: Identification and CAT Use
Even: f(−x) = f(x) (y-axis symmetric). Odd: f(−x) = −f(x) (origin symmetric). Periodic: f(x + T) = f(x) for the smallest T > 0. These collapse complex expressions and enable fast MCQ elimination.
Even × Even = Even · Odd × Odd = Even
Even × Odd = Odd
Composite: (Even o Even) = Even; (Odd o Odd) = Odd; (Even o Odd) = Even.
Every function f can be written as (even part) + (odd part) where even part = [f(x) + f(−x)]/2 and odd part = [f(x) − f(−x)]/2.
Periodic sums follow LCM logic: f + g has period LCM(T1, T2). Example: sin x (2π) + cos 2x (π) has period 2π.
tan x, cot x: period π
sin 2x, cos 2x: period π
sin (x/2): period 4π
{x} (fractional part): period 1
|sin x|: period π (modulus halves the period)
Want a topic-priority map showing which CAT Quant topics deliver the highest marks-per-hour at your level?
Build My Quant Topic PriorityComposite and Inverse Functions: The Substitution Engine
Composite and inverse questions test substitution discipline. Compute (f o g)(x) by substituting g(x) for every x in f; order matters since (f o g) generally differs from (g o f). The CAT test: substitute cleanly in 60 seconds.
(g o f)(x) = g(f(x)) · Substitute f(x) wherever x appears in g.
Domain of (f o g) = {x : x ∈ dom(g) and g(x) ∈ dom(f)}
Example: f(x) = 2x + 1, g(x) = x². (f o g)(x) = 2x² + 1. (g o f)(x) = (2x + 1)².
The inverse f-inverse(x) reverses f, and only one-to-one functions have one. The CAT algorithm is mechanical: write y = f(x), swap x and y, then solve for y. Verify f(f-inverse(x)) = x.
Step 2: Swap x and y.
Step 3: Solve for y; this is f−1(x).
Example: f(x) = (3x − 5)/2. y = (3x − 5)/2. Swap: x = (3y − 5)/2. Solve: 2x = 3y − 5, so f−1(x) = (2x + 5)/3.
Pair the composite-inverse drill with the improve CAT Quant score weekly mock-analysis loop; the geometry formulas guide covers locus problems where composites reappear, and the CAT 2026 marking scheme shows why this 5-mark topic moves the percentile needle.
12 Solved CAT-Level Function Questions (TITA + MCQ)
The 12 questions cover domain, range, modulus, even-odd, periodicity, composite, and inverse. Target: 90 to 120 seconds per medium question, 150 to 180 for harder modulus and composite ones.
Find the domain of f(x) = √(x − 2) + 1/(x − 5).
x − 2 ≥ 0 and x ≠ 5 ⇒ [2, 5) ∪ (5, ∞).
If f(x) = x² − 4x + 7, find the range of f on the real line.
Complete the square: f(x) = (x − 2)² + 3, minimum 3. Range = [3, ∞).
How many integer values of x satisfy |x − 3| + |x + 1| ≤ 8?
Critical points x = −1, 3 split the line; combining the three case bounds gives −3 ≤ x ≤ 5, integers −3 through 5. Count = 9.
Which of the following is an odd function: (a) x² + cos x, (b) x³ + sin x, (c) x|x| + 1, (d) ex?
Only (b) gives f(−x) = −x³ − sin x = −f(x). Answer: (b).
Find the period of f(x) = sin(3x) + cos(2x).
LCM of 2π/3 (sin 3x) and π (cos 2x) = 2π.
If f(x) = 2x + 1 and g(x) = x² − 3, find (f o g)(2).
g(2) = 1, then f(1) = 3.
If f(x) = (5x + 3)/2, find f−1(x).
Swap and solve: 2x = 5y + 3 ⇒ f−1(x) = (2x − 3)/5.
Find the domain of f(x) = log(x² − 5x + 6).
(x − 2)(x − 3) > 0 ⇒ Domain = (−∞, 2) ∪ (3, ∞).
How many real values of x satisfy |2x − 5| = |x + 4|?
Two cases give x = 9 and x = 1/3, both valid. Solutions = 2.
If f(x) = x + 2 and g(x) = 3x, is (f o g)(x) equal to (g o f)(x)?
(f o g) = 3x + 2 vs (g o f) = 3x + 6. Answer: No.
Find the minimum value of f(x) = |x − 1| + |x − 4| + |x − 7|.
Minimum at the middle point x = 4: f(4) = 3 + 0 + 3 = 6.
Write f(x) = ex as the sum of an even and an odd function.
Even (ex+e−x)/2 + odd (ex−e−x)/2 = cosh x + sinh x.
- Identify even, odd, modulus, or composite structure in 5 seconds
- For domain, scan denominators, even roots, logs; apply the 3 rules
- For range, recognise the standard shape (linear, quadratic, modulus, exponential)
- For multiple moduli, write the critical-point split immediately
- For composites, substitute the entire inner expression; simplify last
- Define the function type in 5 seconds: even, odd, modulus, composite, or piecewise.
- Apply the 3 domain exclusion rules independently, then intersect.
- Write the case-split or critical-point chart before any modulus algebra.
- Period of f + g equals LCM of individual periods.
- Preserve the inner expression intact in composites; simplify last.
- Use swap-and-solve for inverses; verify f(f−1(x)) = x.
Functions reward discipline more than insight: memorise the properties, drill 50 problems, lock in the 5 to 7 marks.
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