In review — free for everyone. While a book is in review you are one of its reviewers: read it, use it, and tell us what is wrong. When the reports settle, the Class 11 pass is ₹999 for the year and this book’s PDF is ₹299.
Relation or function: the one test
◆FRAME
Here is a rule. Pair each of 1, 2 and 3 with a number: send 1 to 2, send 2 to 3, send 3 to 2. Here is a second rule: send 1 to 2, send 1 to 5, send 2 to 3.
Both are legal pairings, and both are RELATIONS. A relation pairs elements of one set with elements of another, and nothing more is asked of it.
Only the first is a FUNCTION. Look at what the second does with the input 1. It sends 1 to 2, and it also sends 1 to 5. Ask what the rule gives for 1 and there is no single answer. The first rule has no input like that.
That is the whole added constraint. $f: A \to B$ is a function only when every $a \in A$ has exactly one image $f(a) \in B$. Not at least one — exactly one. A rule that hands one input two outputs is still a relation, but it is not a function.
Every input, exactly one output. The whole chapter is built on that one line.
Count lines LEAVING the left-hand dots, and count nothing else. Both pictures are legal pairings and both are relations; what disqualifies the second is that one dot on its left has two lines out, so asking what the rule gives for that input has no single answer. Two lines arriving at the same right-hand dot, which happens in the first picture, breaks nothing.
Before you can choose pairs, you need the set of all pairs to choose from.
The CARTESIAN PRODUCT of A and B is written $A \times B$. It is the set of ALL ordered pairs $(a, b)$, where a comes from A and b comes from B.
The word "ordered" is doing real work. $(a, b)$ and $(b, a)$ are different pairs, unless a and b happen to be equal. So $A \times B$ and $B \times A$ are generally different sets, even though they always have the same size.
That size is easy to count. Each of A’s m elements pairs with each of B’s n elements. The two choices are independent, so $A \times B$ has exactly $m n$ pairs.
The product holds every possible pairing. Which of them you keep is the next idea.
Both grids hold six dots, so whatever makes the two sets different, it is not their size. The highlighted dot is the same two elements in each — but reading it off the left grid gives the pair 2 then p, and off the right grid p then 2. That is all an ordered pair is: a record of which set each element came from.
Worked example
Write out $A \times B$ for $A = \{1, 2\}$, $B = \{x, y, z\}$
$(1,x), (1,y), (1,z)$ start with 1, and pair it with every element of B in turn
$(2,x), (2,y), (2,z)$ now 2 — sweep the second set once for each element of the first
$A \times B$: 6 ordered pairs collecting both sweeps above — nothing is missed and nothing is repeated
$2 \times 3 = 6$ count them: A has 2 elements, B has 3 — the list and the rule agree
Follow one path from left to right and you have written one pair, in the right order, without deciding anything about order. This is the worked example’s own method drawn out rather than its answer: fix a first element, run through all of B, then move on. The count falls out of the shape — three branches hanging off each of two, which is why the size of a Cartesian product multiplies instead of adding.
A RELATION R from A to B is any subset of $A \times B$. That is the whole definition. Choose some of the pairs, and what you have chosen is a relation.
Nothing constrains the choice. An element of A may appear in many pairs, in one pair, or in none at all.
Two sets come out of that choice. The DOMAIN of R is the set of first elements actually used. The RANGE is the set of second elements actually used. Both can be smaller than A and B, because an element that never appears in a chosen pair never shows up.
The same relation can be written three ways. As a set of ordered pairs. As a rule in words, such as "x is a factor of y". Or as an arrow diagram, with one arrow per pair. The pairs are what the relation IS. The other two are ways of naming them.
A relation is a choice of pairs. Domain and range record which elements that choice touched.
The two hollow circles are the point. A relation is a chosen collection of pairs, so it is free to ignore some elements entirely — 5 divides nothing in B, and nothing in A divides 11. Read the domain and range off the solid circles rather than assuming they are the whole of A and B.
Worked example
Find R, its domain, and its range for the relation "is less than" on $A = \{1,2,3,4\}$
$(1,2), (1,3), (1,4)$ 1 is less than 2, 3, and 4
$(2,3), (2,4)$ 2 is less than 3 and 4
$(3,4)$ 3 is less than 4; 4 is less than nothing in A
domain $= \{1, 2, 3\}$ the first elements actually used, collected from the pairs above — 4 is absent, since no element of A is bigger than 4
range $= \{2, 3, 4\}$ the second elements actually used — 1 is absent, since no element of A is smaller than 1; domain and range are read off the pairs you actually have, never off A and B
The exam mistake this prevents is answering "domain = A" because A is what the question named. Domain and range are collected AFTER the pairs are chosen, so either can come out smaller than A; the codomain is the only one of the three that can be written down before any work is done.
$R_1 = \{(1,2), (2,3), (3,2)\}$ check the first elements: 1, 2, 3 — each appears exactly once, so $R_1$ IS a function
2 repeats as a second element that is allowed — the function test says nothing about outputs repeating
$R_2 = \{(1,2), (1,5), (2,3)\}$ check the first elements: 1, 1, 2 — the input 1 appears twice, paired with 2 and with 5
$R_2$ is NOT a function 1 has no single well-defined image; read the first elements only — a repeat there breaks a function, a repeat among second elements breaks nothing
Look only at the left-hand column of each panel, and only at how many arrows leave. The left panel also has two arrows arriving at the same output, which is allowed and is the thing most often mistaken for the failure — it is drawn inside the panel that passes for exactly that reason.
●KEY-TERM
The same check works on a graph, and there it takes a second.
A curve represents a function of x exactly when every VERTICAL line crosses it at most once.
Here is why that is the same check. A vertical line sits at a single x-value, say $x = a$. Every point where it meets the curve is a y-value the curve assigns to that a. One crossing means one output. Two crossings mean two outputs for one input, which is exactly the violation that stops a relation being a function.
Note the wording: at most once, not exactly once. A vertical line that misses the curve altogether just means that x is outside the domain, and that is allowed.
Slide a vertical line across the graph. Two hits anywhere, and it is not a function.
Both curves are parabolas and both are smooth, so neither of those is what the test is about — the only thing that differs is how many marks the guides leave on the vertical axis. Count there, not in the middle of the picture: one mark means one output for that input, two means the definition of a function has already been broken.
Writing $f: A \to B$ fixes two sets, and they play different parts.
The CODOMAIN is B — the set f is DECLARED to map into. It is part of how the function was written down, chosen by whoever defined it.
The RANGE is $\{f(a) : a \in A\}$ — the set of values f actually produces as a runs over the domain. You find it by working out what comes out, not by reading the declaration.
The range is always a subset of the codomain, since every output does land somewhere in B. But it can be a proper subset, when some elements of B are never produced at all.
If the range does equal the codomain, f has a name: it is ONTO. That is a property to be checked, not something the notation hands you.
The codomain is declared. The range is earned.
Both panels declare the same codomain, so the declaration is not what tells them apart — read the vertical axis instead, and read how much of it is solid. The dashed stretch on the left is the whole idea: those outputs were declared and no input reaches them, which is why the range can be smaller than the set f was declared into. On the right nothing is dashed, and that case has a name: ONTO.
Two functions are worth naming first, because they sit at opposite extremes.
The IDENTITY function is $f(x) = x$. It sends every input to itself and changes nothing. Domain is $\mathbb{R}$ and range is $\mathbb{R}$, and the graph is the straight line through the origin at 45 degrees.
A CONSTANT function is $f(x) = c$ for some fixed number c. It sends every input to the same value and ignores x completely. Domain is $\mathbb{R}$, but the range is the single value $\{c\}$, and the graph is a horizontal line.
Compare the two ranges. The identity keeps every distinction between its inputs. The constant function destroys all of them.
One of these preserves everything about its input. The other keeps nothing.
Both functions take in every real number, so the domain tells them apart not at all. Follow the guides to the vertical axis instead: the identity leaves two marks and the constant leaves one, no matter how many inputs are tried. A range can be a whole line or a single point.
▲CONCEPT
A POLYNOMIAL function is built by adding constant multiples of powers of x. One example is $f(x) = a_2 x^2 + a_1 x + a_0$, for fixed numbers $a_2$, $a_1$ and $a_0$.
Its domain is always $\mathbb{R}$. Every real number can be squared, multiplied and added, so no input is ever illegal.
A RATIONAL function is one polynomial divided by another, $f(x) = p(x)/q(x)$.
Here the domain needs care. Division by zero is undefined, so every x with $q(x) = 0$ has to be thrown out. The domain is $\mathbb{R}$ with those points removed.
Finding them means solving $q(x) = 0$. So the domain of a rational function is a small piece of work. It is not something you can write down at a glance.
A polynomial accepts every real number. A rational function accepts every real number except where its bottom vanishes.
Cover the labels and look only at the two lines. They differ in exactly two places, and those two places are the roots of the same quadratic — which is why you can find a rational function’s domain by solving the bottom and nothing else. Ask yourself what the top line would look like if the numerator had been x minus 1 instead of x plus 5; the answer is that it would not change, and neither would the bottom one.
●KEY-TERM
The MODULUS function, written $f(x) = |x|$, gives the distance of x from 0. A distance is never negative, so the output never is either.
It is defined in two pieces. When $x \geq 0$, $f(x) = x$, and nothing changes. When $x < 0$, $f(x) = -x$, which flips the sign.
That second piece surprises people, because of the minus sign sitting in front. Read it as "the negative of a negative number", which is positive. Nothing about the output is negative.
Domain is $\mathbb{R}$, since every real number has a distance from 0. Range is $\{y \in \mathbb{R} : y \geq 0\}$. The graph is a V with its vertex at the origin.
The minus sign in the second piece is what makes the answer positive.
The two arms are the definition’s own two cases, not a shape to memorise. Two inputs on opposite sides of zero land on one output because the modulus keeps the distance and throws away the direction — and once direction is gone there is nothing left to make an output negative.
Worked example
Evaluate $f(x) = |x|$ at $x = 5, -5, 0$
$f(5) = 5$ 5 is not negative, so the first piece applies
$f(-5) = -(-5) = 5$ $-5$ is negative, so the second piece applies and flips the sign
$f(0) = 0$ 0 is not negative, so the first piece applies
$f(5) = f(-5) = 5$ two different inputs share one output, so the modulus function is not one-to-one — perfectly legal, since a function only forbids one input with two outputs, not the reverse
Both brackets read the same, which is where the collapse comes from: once the distance from zero has been taken, nothing is left to tell minus 5 apart from 5. Count the departures and the arrivals separately — three leave the top line, two land on the bottom one. That gap is exactly what "not one-to-one" means, and it is why the output line needs no negative half at all.
●KEY-TERM
The SIGNUM function throws away the size of a number and reports only its sign.
It is defined in three pieces.
$f(x) = 1$ when $x > 0$.
$f(x) = 0$ when $x = 0$.
$f(x) = -1$ when $x < 0$.
Domain is $\mathbb{R}$. Range is the three-element set $\{-1, 0, 1\}$, and nothing else is ever produced.
That small range is the point of the function. An input of 3 and an input of 3000000 both give 1. The output records which side of zero the number lies on, and says nothing at all about how far away it is.
Signum keeps the direction and discards the distance. Modulus does the opposite.
Nothing about the rays changes as the input grows, which is the whole content of discarding the size. The two hollow circles and the solid dot between them are the part that gets lost in a quick sketch: zero is neither positive nor negative, so it gets an output of its own rather than a gap.
●KEY-TERM
The GREATEST INTEGER function is written $f(x) = [x]$, and is also called the floor function. It gives the largest integer that is less than or equal to x.
It rounds DOWN, always. Not toward zero, and not to the nearest integer. Down.
For positive numbers, rounding down and rounding toward zero happen to agree, which is why the difference is easy to miss. For negative numbers they part company, and that is where the mistakes happen.
Domain is $\mathbb{R}$. Range is $ZZ$, the integers — every integer is produced by some input.
One more fact makes the graph clear. For an integer n, $[x] = n$ for every x in the half-open interval $[n, n+1)$. The value holds steady across that whole stretch and then jumps, so the graph is a staircase of flat steps.
Round down, not toward zero. For negative numbers those are two different instructions.
Filled at the left, hollow at the right — draw both ends solid and the picture says the output at 2 is both 1 and 2. The negative side is drawn because that is where rounding down stops agreeing with dropping the decimals: minus 1.4 goes down to minus 2, not up to minus 1.
Worked example
Evaluate the greatest integer function at $x = 2.7, 5, -2.3$
$[2.7] = 2$ the integers at or below 2.7 are 2, 1, 0, and onward downward — the largest is 2
$[5] = 5$ 5 is itself an integer, and it is at or below 5 — an integer input maps to itself
$[-2.3] = -3$ the candidates at or below -2.3 are -3, -4, and onward — -2 is NOT a candidate, since it is greater than -2.3, not less than or equal to it
rounding toward zero would give $-2$ rounding down gives $-3$ instead, and rounding down is the rule — on the number line, move left until you first land on an integer
Not one connector leans right, and that single fact settles all three answers. Minus 2.3 is the long one and it travels away from zero — which is exactly where rounding down stops agreeing with dropping the decimals, and why the answer is minus 3 rather than minus 2. The integer input is not an exception: 5 makes the same leftward move with nothing left to move.
Given two real functions f and g, four new ones can be built. Each is defined POINTWISE: you evaluate f and g at the same x, then combine the two numbers.
$(f+g)(x) = f(x) + g(x)$.
$(f-g)(x) = f(x) - g(x)$.
$(f g)(x) = f(x) g(x)$.
$(f/g)(x) = f(x)/g(x)$, provided $g(x) \neq 0$.
The formulas are the easy half. The domain is the half that gets dropped.
For a new function to have a value at x, BOTH f and g must have one there. So the domain of the sum, the difference and the product is the OVERLAP of f’s domain and g’s domain.
The quotient narrows it once more. On top of needing both functions defined, it needs $g(x) \neq 0$, so every x where g vanishes comes out as well.
Combine the values pointwise, then intersect the domains. The quotient also loses g’s zeros.
Read down the right-hand column, not across the rows. Three of the four domains are the same line, so the sum, the difference and the product are one rule wearing three signs. The fourth line has a point missing, and finding which point is the extra work: it is not read off the two domains, it comes from solving g(x) = 0.
Worked example
Build $f+g$, $f g$, and $f/g$ for $f(x) = x^2$, $g(x) = x + 1$
$(f+g)(x) = x^2 + x + 1$ add the expressions — both f and g are polynomials with domain $\mathbb{R}$, so the sum is defined everywhere
$(f g)(x) = x^2 (x+1) = x^3 + x^2$ multiply them — again both are defined everywhere, so the domain is $\mathbb{R}$
$(f/g)(x) = x^2/(x+1)$ divide — now the extra condition bites
$x + 1 = 0 => x = -1$ solve $g(x) = 0$ to find what the quotient must exclude — the domain is every real number except $-1$; the formula for the quotient took one line, finding where it is defined took solving an equation, and that is the step that gets marked
The stalk at x = 1 is the whole definition of pointwise: the sum curve’s height there is 1 + 2, read off the two curves below it, and the same reading holds at every other x. The dashed line is the half of the answer that gets dropped — the sum has a perfectly good value there (solid dot), and so do the difference and the product. Only f/g loses x = −1, and it loses it because g is zero, not because −1 is special.
Every idea in this chapter builds on one test: does each input have exactly one output?
A relation may fail that test. A function never does. That single difference is what the chapter is about.
DOMAIN, CODOMAIN and RANGE name the sets the test runs over. What may go in, what is declared to come out, and what actually does.
Six standard functions are named shapes that always pass the test: identity, constant, polynomial and rational, modulus, signum, greatest integer. Knowing them by name means recognising one on sight instead of working it out again.
And the ALGEBRA OF FUNCTIONS builds new functions that still pass, on a domain narrowed to where every piece is defined.
One test, applied everywhere. The rest of the chapter is vocabulary for talking about it.
Read across the row before reading any one panel. The five shapes agree about nothing except the count the test asks for, which is the reason the chapter could name them all as functions in the first place. If you want the harder version of the test, try to find a vertical line anywhere in the signum panel that meets it twice — the two hollow ends at zero exist precisely so that you cannot.
THE TRAP. The codomain and the range are the same set. Whatever B is declared to be is what f produces.
THE REALITY. The codomain is only the declared target. The range is what actually comes out, and it can be a proper subset.
Take $f: \mathbb{R} \to \mathbb{R}$ with $f(x) = x^2$. The codomain is $\mathbb{R}$, because that is how the function was written down. But squaring a real number never gives a negative result, so the range is only $\{y : y \geq 0\}$.
Every negative number sits in the codomain and is never produced. The declaration promised nothing about that. It only fixed where the outputs are allowed to live.
When the two do coincide, the function gets a name: ONTO. That name exists precisely because the two are not the same thing in general.
The codomain is a promise about where outputs may live. The range is the record of where they landed.
A value belongs to the range exactly when its horizontal line meets the curve. The dashed line is the whole argument: minus 1 was declared allowable when the function was written down, and no input ever produces it. Declaring where outputs may land is not the same as saying where they do.
"What may come out" against "what does come out"
Weaker. The question says $f: \mathbb{R} \to \mathbb{R}$, $f(x) = x^2$, so the answer written down is "range $= \mathbb{R}$". The reasoning behind it is never stated out loud, because it feels like reading rather than working: the target set was printed in the question, and printed sets are what answers are made of. Nothing in the line $f: \mathbb{R} \to \mathbb{R}$ is false, and nothing in it has been checked either. Try to produce $-1$ and there is no input to do it with — but that attempt was never made, because writing $\mathbb{R}$ did not feel like a claim.
Stronger. The declared codomain is a promise about where outputs are allowed to land, made before any input is tried. The range is the set of outputs that actually occur, and it can only be found by asking, of each candidate value, whether some $x$ produces it. For $f(x) = x^2$: every square is $\geq 0$, and every $y \geq 0$ is hit by $x = \sqrt{y}$, so the range is $\{y : y \geq 0\}$ — a proper subset of the codomain $\mathbb{R}$. Same function, same line of the question; one answer was read, the other was earned. The two agree exactly when $f$ is onto, which is a separate thing to check.
✕MISCONCEPTION
THE TRAP. $x^2 + y^2 = 1$ defines y as a function of x, since the equation can be solved for y.
THE REALITY. It can be solved — that part is true. Being solvable is not what makes something a function.
Rearrange it: $y^2 = 1 - x^2$, so $y = \pm \sqrt{1 - x^2}$. Read the sign in front. For most values of x there are TWO values of y, not one.
Test it at $x = 0.5$. Then $1 - x^2 = 0.75$, and both $y = \sqrt{0.75}$ and $y = -\sqrt{0.75}$ satisfy the original equation. One input, two outputs.
The graph says the same thing faster. A vertical line at $x = 0.5$ crosses the circle twice. So the circle fails the vertical line test, and is not a function of x.
Being solvable is not being a function. What matters is how many answers the solving gives.
The vertical line crossing twice is only half the argument. Follow the two crossings across to the vertical axis, where they land as two separate outputs for the single input 0.5 — that is the thing the definition of a function forbids, and every vertical line between minus 1 and 1 does the same.
practice If $A = \{1, 2\}$ and $B = \{3, 4, 5\}$, find $A \times B$ and state $n(A \times B)$.
practice Find x and y if $(x + 1, y - 2) = (3, 1)$.
practice If $n(A) = 3$ and $n(B) = 4$, what is $n(A \times B)$?
$7$
$12$
$34$
$43$
practice If $A \times B = \{(1,3), (1,4), (2,3), (2,4)\}$, find A and B.
Answers
$A \times B = \{(1,3),(1,4),(1,5),(2,3),(2,4),(2,5)\}$; $n(A \times B) = 6$
$x = 2$, $y = 3$
$12$
$A = \{1,2\}$; $B = \{3,4\}$
Read the right column first when you are stuck. The three sets look interchangeable from inside a problem — all pairs, sets and functions — so the useful thing to know is which skill is being asked for. The rows are in exercise order for a reason: 2.1 builds the pairs, 2.2 chooses some of them, 2.3 tests what was chosen.
Exercise 2.2 — Relations
practice Let $A = \{1,2,3\}$. Write R = 'less than' as a set of ordered pairs on A.
practice Find the domain and range of the relation $R = \{(1,2),(2,3),(3,4)\}$.
practice If $n(A) = 2$ and $n(B) = 3$, how many relations can be defined from A to B?
$6$
$8$
$32$
$64$
Answers
$\{(1,2),(1,3),(2,3)\}$
Domain $\{1,2,3\}$; range $\{2,3,4\}$
$64$
The three questions get asked at once in practice, and out of order — which is why an answer that is right about the rule can still be wrong about the sets. Each gate asks about a different object: gate 1 about the sets, gate 2 about the pairs, gate 3 about the rule. The right-hand boxes matter as much as the questions: stopping at gate 2 with "a relation, not a function" is a complete answer, not a failure to finish.
Exercise 2.3 — Functions
practice Is $\{(1,2),(2,2),(3,2),(4,3)\}$ a function? State its domain and range.
practice Find the domain of $f(x) = 1/(x - 3)$.
practice Which standard function is $f(x) = [x]$?
Modulus
Signum
Greatest integer
Identity
practice If $f(x) = x^2$ and $g(x) = x - 1$, find $(f - g)(x)$ and $(f g)(x)$.
Answers
Yes — domain $\{1,2,3,4\}$, range $\{2,3\}$
$\mathbb{R} - \{3\}$
Greatest integer
$(f-g)(x) = x^2 - x + 1$; $(f g)(x) = x^3 - x^2$
Miscellaneous — practice set
practice If domain of f is $[0, 5]$ and domain of g is $[2, 8]$, find the domain of $(f+g)$.
practice For $f: \mathbb{R} \to \mathbb{R}$, $f(x) = x^2$, which best describes range vs codomain?
Range = codomain = $\mathbb{R}$
Range is a proper subset of the codomain
Codomain is a proper subset of range
Neither is a subset of the other
practice Is the relation $\{(1,1),(1,2),(2,3)\}$ a function? Explain in one line.
Answers
$[0,5] \cap [2,8] = [2,5]$
Range is a proper subset of the codomain
No — 1 maps to two outputs
Read the middle column before the right one. All three questions are asked of the same two sets, and the only thing that decides the arithmetic is what is being chosen and how many independent choices there are — pairs choose twice, relations choose once per pair, functions choose once per input. The two bars are the reason the chapter exists: nearly all of the relations you can build are not functions.
A domain question about a fraction is a question about its denominator, and the work is a factorisation, not a rule. Both excluded values come out of solving one equation — the numerator never enters it. The two hollow circles on the line are what the answer looks like: not an interval, but the whole line with two single points taken out of it.
$(f g)(x)=x^2-4$; domain of $f/g$ is $\mathbb{R}-{2}$
$x=7$ or $x=-3$
$3 \leq x < 4$
$x^2+y^2=25$
$\mathbb{R}-{2,-2}$
$(f+g)(3)=16$, $(f g)(1)=3$
$n(B)=3$; e.g. $A \times B={(1,p),(1,q),(1,r),(2,p),(2,q),(2,r)}$
domain $=$ range $={-5,-4,-3,0,3,4,5}$; not a function — x=3 gives two y-values
Yes, well-defined; $f(-2)=-4$, $f(3)=9$
range $=2 \cdot ZZ$ (even integers); codomain $ZZ$ is bigger, so range $\neq$ codomain
$-2$
$[3.7+2]=5=[3.7]+2$
$(f/g)(x)=x+1$, domain $\mathbb{R}-{1}$
$27$
Item 19. The trap is answering with a number rather than a stretch. Ask which band the point 4 belongs to instead — every real number belongs to exactly one, which is what makes this a function at all.
Item 26. Set this beside the chapter’s earlier codomain picture, where the unreached part was one solid half of an axis. Here the misses are spread the whole way along and leave no gap you could measure.
Adding a whole number leaves the input sitting exactly as far above its floor as before — both orange bars are 0.7 — so the floor takes the same step and lands on an integer. Check the two arrows are the same length before writing the identity down. At a fractional step like +0.5 the bars come out different, and the identity fails.
The equation names a circle; the condition that x and y be whole numbers keeps twelve of its points and throws the rest away, so the domain is a list of seven values rather than a stretch of the line. Count the dots against the ticks before answering anything else — twelve pairs standing over seven inputs already settles the function question, without checking a single one of them individually.
A rule written on two lines is still one function, and the question is where the two lines could quarrel. Look only at the boundary: everywhere else exactly one line applies, so nothing there can go wrong. Two things are worth checking on any piecewise rule you meet — whether every input is covered by some line, and whether the lines agree wherever both could apply.
f(x) = $x^2$ - 9 and g(x) = 1/(x+3). What is the domain of the product function (fg)(x) = f(x) . g(x)?
R ∖{-3} — where g is undefined.
R — the (x+3) factor cancels away.
R ∖{3, -3} — wrongly excludes x=3 too.
R ∖{3} — excludes the wrong point.
Check your answer
✓ R ∖{-3} — where g is undefined. — (A) g(x)=1/(x+3) is undefined at x=-3, so the product (fg)(x) is undefined there too, however the algebraic expression simplifies afterward. $f(x)=x^2-9$ is defined everywhere, so -3 is the only exclusion.
R — the (x+3) factor cancels away. — Simplifying $(x^2-9)/(x+3)$ to x-3 is only a shortcut for computing values — the original product was never defined at x=-3, since g(x)=1/(x+3) has no value there, and simplification cannot restore that.
R ∖{3, -3} — wrongly excludes x=3 too. — $f(x)=x^2-9$ equals 0 at x=3, but a function equalling 0 somewhere says nothing about its domain — f itself is defined at x=3, and every other real number.
R ∖{3} — excludes the wrong point. — The exclusion comes from g(x)=1/(x+3), which is undefined at x=-3, not from any root of f — x=3 is a perfectly valid input to both f and g.
A = {1, 2} and B = {3, 4, 5}. How many subsets does A × B have, including the empty set and A × B itself?
36
64
12
8
Check your answer
36 — The number of subsets of a set with k elements is $2^k$, not $k^2$ — with 6 pairs in A x B, that is $2^6=64$, not $6^2=36.$
✓ 64 — (B) A x B has |A| times |B| = 2 times 3 = 6 ordered pairs. A set with 6 elements has $2^6$ = 64 subsets, since each pair is independently included or excluded.
12 — Counting subsets is exponential, not a doubling — each of the 6 pairs is independently in or out, giving $2^6$ choices total, not 2 times 6.
8 — A x B has |A| times |B| = 2 times 3 = 6 elements in total, and every one of them contributes its own in/out choice — the exponent has to be all 6, not just |B|=3.
Solve [2x - 1] = 3 for real x, where [.] denotes the greatest integer (floor) function. Give the complete solution set.
x = 2 only
1.5 <= x < 2
2 <= x < 2.5
2 < x <= 2.5
Check your answer
x = 2 only — [y]=3 does not mean y=3 exactly — it means y lies anywhere in the half-open interval [3,4). Solving 2x-1=3 alone finds only one point of that interval, x=2, and misses the rest.
1.5 <= x < 2 — The equation is [2x-1]=3, not [2x]=3 — the -1 inside the bracket has to carry through the whole inequality before dividing by 2, not be dropped.
✓ 2 <= x < 2.5 — (C) [2x-1]=3 means 3 <= 2x-1 < 4. Adding 1 gives 4 <= 2x < 5, and dividing by 2 gives 2 <= x < 2.5 — the left end included, the right end excluded.
2 < x <= 2.5 — The greatest integer function satisfies n <= y < n+1 for [y]=n — the left end is included and the right end is not, the opposite of what this option uses.
Solve |x - 2| = 3x + 1 for real x, keeping only genuinely valid solutions.
x = 1/4 and x = -3/2
x = -3/2
no solution exists
x = 1/4
Check your answer
x = 1/4 and x = -3/2 — Solving each case’s linear equation gives x=-1.5 (from x>=2) and x=0.25 (from x<2), but x=-1.5 does not satisfy x>=2 — it must be rejected, not kept alongside the valid root.
x = -3/2 — x=-1.5 came from the case x>=2, but -1.5 is not >=2, so it fails its own case assumption — and it also makes the right-hand side 3x+1 negative, which a non-negative modulus can never equal. x=0.25 is the one genuinely valid root.
no solution exists — Solving |x-2|=3x+1 case by case does produce a valid answer — x=0.25 — not a contradiction; a sign slip in expanding -(x-2) is what leads to a false ‘no solution’ conclusion.
✓ x = 1/4 — (D) For x>=2, x-2=3x+1 gives x=-1.5, which fails x>=2 and is rejected. For x<2, -(x-2)=3x+1 gives x=0.25, which does satisfy x<2 and gives a non-negative right-hand side, 1.75=1.75. Only x=1/4 survives.
Solve sgn(x^2 - 9) = 1 for real x, where sgn denotes the signum function.
x < -3 or x > 3
-3 < x < 3
x = -3 or x = 3
x > 3 only
Check your answer
✓ x < -3 or x > 3 — (A) sgn(y)=1 exactly when y>0. $x^2-9>0$ factors as (x-3)(x+3)>0, true when x<-3 or x>3 — both branches, since $x^2-9$ is positive outside the interval [-3,3].
-3 < x < 3 — sgn(y)=1 means y is strictly positive, y>0 — not y<0. -3<x<3 is exactly where $x^2-9<0$, the opposite condition.
x = -3 or x = 3 — sgn(y)=1 requires y>0, a strict inequality — x=-3 or x=3 are where $x^2-9=0$ exactly, which gives sgn=0, not sgn=1.
x > 3 only — $x^2-9>0$ holds on two separate rays, x<-3 as well as x>3 — dropping the left branch misses half of the valid solutions.
(f + g)(x) = 3x + 5 and (f - g)(x) = x + 1 for all real x. Find f(x) and g(x).
f(x) = 3x + 5, g(x) = x + 1
f(x) = 2x + 3, g(x) = x + 2
f(x) = 2x + 3, g(x) = 2x + 4
f(x) = x + 2, g(x) = 2x + 3
Check your answer
f(x) = 3x + 5, g(x) = x + 1 — 3x+5 and x+1 are (f+g)(x) and (f-g)(x) — combinations of f and g, not f and g individually. Recovering f and g requires solving the two equations as a system.
✓ f(x) = 2x + 3, g(x) = x + 2 — (B) Adding the two given equations eliminates g, 2f(x)=4x+6, so f(x)=2x+3. Subtracting eliminates f, 2g(x)=2x+4, so g(x)=x+2. Checking, (f+g)(x)=3x+5 and (f-g)(x)=x+1, both matching the given data.
f(x) = 2x + 3, g(x) = 2x + 4 — Adding the two given equations gives 2f(x)=4x+6, so f(x)=2x+3 — correct. Subtracting gives 2g(x)=2x+4, which must also be divided by 2 to get g(x)=x+2, not left as 2x+4.
f(x) = x + 2, g(x) = 2x + 3 — The system does solve to f(x)=2x+3 and g(x)=x+2 — but this option reports f(x)=x+2 and g(x)=2x+3, reversing which name goes with which solution.
Does the equation $y^2$ = x - 1 define y as a function of x? Check specifically what happens at x = 5.
Yes — one y for every x >= 1.
Yes — rewritten as y = (x-1)/y.
No — x = 5 gives y = 2 and y = -2.
No — no solutions exist for x < 1.
Check your answer
Yes — one y for every x >= 1. — At x=5, $y^2=4$ has two solutions, y=2 and y=-2, not just y=2 — both pairs (5,2) and (5,-2) satisfy the original equation, so x=5 genuinely has two outputs.
Yes — rewritten as y = (x-1)/y. — $y^2=x-1$ cannot be validly rearranged to isolate y this way — dividing by y is not a legitimate algebraic step here, and it does not change the underlying fact that two y-values solve the equation at x=5.
✓ No — x = 5 gives y = 2 and y = -2. — (C) At x=5, $y^2=4$, so y=2 and y=-2 both satisfy the equation — one input, two outputs, which the one-input-one-output test rules out. The equation does not define a function of x.
No — no solutions exist for x < 1. — A restricted domain, on its own, is completely fine for a function — plenty of genuine functions are undefined for x<1. The actual problem is that x=5 pairs with two different y-values, not that some x-values are excluded.
f is R → R, defined by f(x) = $x^2$ + 2. What is the range of f, and is f onto (does its range equal its codomain)?
Range = R; f is onto.
Range = [2, ∞); f is onto.
Range = (-∞, 2]; f is not onto.
Range = [2, ∞); f is not onto.
Check your answer
Range = R; f is onto. — Codomain and range answer different questions — the codomain is the declared target, R here, while the range is what f actually outputs, [2,infinity). They need not match, and here they genuinely do not.
Range = [2, ∞); f is onto. — The range [2,infinity) is correct, but it is a proper subset of the codomain R — for instance 0 is in the codomain but never an output of f. A declared codomain does not automatically equal the actual range.
Range = (-∞, 2]; f is not onto. — $x^2$ is never negative, so $x^2+2$ is never less than 2 — the range has a minimum of 2 and extends upward, [2,infinity), not a maximum of 2 extending downward.
✓ Range = [2, ∞); f is not onto. — (D) $x^2\geq 0$ for every real x, so $f(x)=x^2+2\geq 2$, giving range [2,infinity). The codomain is the whole of R, which is strictly larger — for example 0 is never an output — so f is not onto.
A = {1, 2, 3} and B = {4, 5}. How many different relations R from A to B contain the pair (1, 4)?
32
64
2
16
Check your answer
✓ 32 — (A) A x B has 3 times 2 = 6 pairs. Fixing (1,4) as included leaves the other 5 pairs each freely in or out, giving $2^5=32$ relations that contain (1,4).
64 — 64 counts every possible relation from A to B, including the many that leave out (1,4) entirely — the question asks only for relations that DO contain it, a smaller count.
2 — Fixing (1,4) as included settles only one of the 6 pairs — the other 5 pairs in A x B are each still freely in or out, giving $2^5$ choices, not just the 2 choices for (1,4) alone.
16 — A x B has |A| times |B| = 3 times 2 = 6 pairs total, not 5 — fixing one pair leaves 5 free pairs, so the count is $2^5=32$, not $2^4=16.$
Solve $[x]^2$ - 5[x] + 6 = 0 for real x, where [.] is the greatest integer function. Give the complete solution set.
2 <= x < 3
2 <= x < 4
x = 2 or x = 3
3 <= x < 4
Check your answer
2 <= x < 3 — Letting n=[x], $n^2-5n+6=0$ factors as (n-2)(n-3)=0, giving TWO roots, n=2 and n=3 — both must be kept, not just n=2. n=3 contributes the interval [3,4).
✓ 2 <= x < 4 — (B) Letting n=[x], the equation becomes $n^2-5n+6=0$, which factors as (n-2)(n-3)=0, giving n=2 or n=3. [x]=2 means 2<=x<3, and [x]=3 means 3<=x<4. Together these intervals join into 2<=x<4.
x = 2 or x = 3 — [x]=2 does not mean x=2 alone — it means x lies anywhere in [2,3), a whole interval of solutions, not one point. The same holds for [x]=3, giving [3,4).
3 <= x < 4 — The quadratic (n-2)(n-3)=0 has two roots, n=2 and n=3 — both must be kept, not just n=3. n=2 contributes the interval [2,3).
The equation is solved in the bracket, but the answer is asked for in x, and that is where the marks go. Substituting for the bracket turns a greatest-integer equation into an ordinary quadratic, so every root of that quadratic is one whole stretch of x, not one value — three bars here for two roots, because the last one is what the two together cover. Ask how many roots the quadratic had before writing any interval down.
JEE-application problems — continued
Solve |2x - 3| <= 5 for real x.
x <= -1 or x >= 4
-4 <= x <= 4
-1 <= x <= 4
-2 <= x <= 8
Check your answer
x <= -1 or x >= 4 — |y|<=k gives a single interval, -k<=y<=k — the ‘or’ disjunction pattern belongs to |y|>=k instead, which describes two separate rays, not this bounded case.
-4 <= x <= 4 — |2x-3|<=5 expands directly to -5<=2x-3<=5 — the -3 stays inside the compound inequality and is resolved by adding 3 throughout, not by first moving it outside the modulus bars.
✓ -1 <= x <= 4 — (C) |2x-3|<=5 means -5<=2x-3<=5. Adding 3 throughout gives -2<=2x<=8, and dividing every part by 2 gives -1<=x<=4.
-2 <= x <= 8 — -5<=2x-3<=5 gives -2<=2x<=8 after adding 3 throughout — but 2x still needs dividing by 2 on every part of the inequality, giving -1<=x<=4, not left as -2<=2x<=8’s numbers unscaled.
f(x) = (x + 1) / $(x^2 - 5x + 6)$. Find the domain of f.
R ∖{-1}
R ∖{5, 6}
[2, 3]
R ∖{2, 3}
Check your answer
R ∖{-1} — The numerator, x+1, equals 0 at x=-1 — but that only means f(-1)=0, a perfectly valid output. The domain restriction comes from the DENOMINATOR being 0, not the numerator.
R ∖{5, 6} — $x^2-5x+6$ must be factored to find where it equals 0: (x-2)(x-3)=0, giving x=2 and x=3 — the coefficients 5 and 6 are not themselves the roots.
[2, 3] — The denominator vanishing at x=2 and x=3 REMOVES those two points from an otherwise unrestricted domain of all real numbers — it does not restrict the domain down to only the interval between them.
✓ R ∖{2, 3} — (D) The denominator $x^2-5x+6$ factors as (x-2)(x-3), which equals 0 at x=2 and x=3. Division by 0 is undefined there, so the domain is every real number except 2 and 3.
f(x) = x (the identity function) and g(x) = c (a constant function) satisfy (f + g)(3) = 10. Find c, then find (f - g)(-2).
c = 7; (f - g)(-2) = -9
c = 7; (f - g)(-2) = 9
c = 13; (f - g)(-2) = -15
c = 7; (f - g)(-2) = -7
Check your answer
✓ c = 7; (f - g)(-2) = -9 — (A) (f+g)(3)=f(3)+g(3)=3+c=10, so c=7. Then (f-g)(-2)=f(-2)-g(-2)=-2-7=-9, using f(-2)=-2 since f is the identity function.
c = 7; (f - g)(-2) = 9 — f(x)=x, so f(-2)=-2, not +2 — (f-g)(-2)=f(-2)-g(-2)=-2-7=-9, not -2 minus (-7)=5’s sign pattern flipped to 9.
c = 13; (f - g)(-2) = -15 — (f+g)(3)=f(3)+g(3)=3+c, since f is the identity function, f(3)=3 — not 3-c. Setting 3+c=10 gives c=7, not the 13 this sign confusion would produce.
c = 7; (f - g)(-2) = -7 — (f-g)(-2) means f(-2) MINUS g(-2), both terms needed — f(-2)=-2 must be included, not dropped in favour of just -c=-7.
A curve is defined implicitly by $y^3$ = x. Does this curve pass the vertical line test, i.e. does it define y as a function of x?
No — a vertical line crosses this curve twice, just as it does for $y^2$ = x, since both involve y raised to a power greater than 1.
Yes — for every real x, exactly one real y satisfies $y^3$ = x, since cubing is one-to-one over the reals.
No — the curve fails the test because x is written as a function of y $(y^3 = x)$ rather than y as a function of x.
Yes, but only for x >= 0, since negative x-values have no real cube root.
Check your answer
No — a vertical line crosses this curve twice, just as it does for $y^2$ = x, since both involve y raised to a power greater than 1. — Squaring is not one-to-one (both 2 and -2 square to 4), which is why $y^2=x$ fails the test — but cubing IS one-to-one over the reals (only 2 cubes to 8), so $y^3=x$ behaves completely differently and does pass.
✓ Yes — for every real x, exactly one real y satisfies $y^3$ = x, since cubing is one-to-one over the reals. — (B) Cubing is one-to-one over the real numbers — every real number has exactly one real cube root, positive, negative, or zero. So for every x, y = cube root of x is the unique solution, and the curve passes the vertical line test everywhere.
No — the curve fails the test because x is written as a function of y $(y^3 = x)$ rather than y as a function of x. — The vertical line test only checks how many y-values share one x-value on the graph — it does not care how the equation happens to be written down; $y^3=x$ graphs to exactly one y per x regardless of which variable was isolated first.
Yes, but only for x >= 0, since negative x-values have no real cube root. — Real numbers have real cube roots at every value, including negative ones — for example the cube root of -8 is -2, since $(-2)^3=-8.$ The domain is all of R, not just x>=0.
A = {a, b} and B = {c, d, e}. How many non-empty relations are there from A to B?
64
6
63
32
Check your answer
64 — 64 is the total number of relations from A to B, including the empty relation (choosing none of the 6 pairs) — the question asks for NON-empty relations, so the empty one must be subtracted out.
6 — 6 is just the number of individual pairs in A x B — a relation is any SUBSET of those 6 pairs, and there are $2^6=64$ such subsets in total, not 6.
✓ 63 — (C) A x B has 2 times 3 = 6 pairs, so there are $2^6=64$ relations from A to B in total, including the empty one. Removing the single empty relation leaves 64-1=63 non-empty relations.
32 — Excluding the empty relation removes exactly ONE relation from the total, giving 64-1=63 — it does not shrink the exponent itself, which would incorrectly halve the whole count to $2^5=32.$
If A × B = B × A and both A and B are non-empty, what must be true of A and B — and if |A| = 3, what is |A × B|?
A and B must be disjoint; |A × B| = 6.
A is a subset of B; |A × B| = 3.
No such A, B exist unless one of them is empty; |A × B| is undefined.
A = B; |A × B| = 9.
Check your answer
A and B must be disjoint; |A × B| = 6. — Disjointness has nothing to do with A x B = B x A — in fact if A and B were disjoint but different, taking any a in A and b in B would give (a,b) in A x B but (a,b) not in B x A (since a is not in B), so the two products would differ, not match.
A is a subset of B; |A × B| = 3. — A x B = B x A forces the full equality A=B, not just A being a subset of B — picking any a in A and b in B and using the given equality shows a must be in B and b must be in A, so the two sets coincide exactly.
No such A, B exist unless one of them is empty; |A × B| is undefined. — Taking A=B always satisfies A x B = B x A trivially, for any non-empty set — for example A=B={1,2,3} clearly gives A x B = B x A. Such sets certainly exist, and neither has to be empty.
✓ A = B; |A × B| = 9. — (D) For any a in A and b in B, (a,b) is in A x B, so by A x B=B x A it is also in B x A, meaning a is in B and b is in A. This holds for every element of both sets, forcing A=B. With |A|=3, then |B|=3 too, so |A x B|=3 times 3=9.
f is R → R, defined by f(x) = |x| - 3. What is the range of f, and does range equal codomain?
Range = [-3, infinity); range does not equal codomain, so f is not onto.
Range = R; range equals codomain, so f is onto.
Range = (-infinity, -3]; range does not equal codomain, so f is not onto.
Range = [-3, infinity); since the codomain was declared as R, range automatically equals codomain.
Check your answer
✓ Range = [-3, infinity); range does not equal codomain, so f is not onto. — (A) |x|>=0 for every real x, so f(x)=|x|-3>=-3, giving range [-3,infinity). The codomain is all of R, strictly larger — for example -5 is never an output — so range does not equal codomain and f is not onto.
Range = R; range equals codomain, so f is onto. — |x| is never negative, so |x|-3 is never less than -3 — the range is bounded below at -3 and extends upward, [-3,infinity), not the whole of R.
Range = (-infinity, -3]; range does not equal codomain, so f is not onto. — |x|>=0 for every real x, so |x|-3>=-3 — the range extends UPWARD from -3, not downward to it. (-infinity,-3] has the direction reversed.
Range = [-3, infinity); since the codomain was declared as R, range automatically equals codomain. — Declaring the codomain to be R does not force the range to match it — the range is whatever f’s rule actually produces, [-3,infinity), a proper subset of the declared codomain R.
R = {(1,2), (2,4), (3,2), (4,8)} is proposed as a relation from {1,2,3,4} to {2,4,8}. Is R a function? If so, is it onto its codomain {2,4,8}?
No — output 2 repeats for inputs 1, 3.
Yes, and onto — {2,4,8} is fully used.
Yes, but not onto — codomain not fully used.
Yes, but not onto — two inputs share an output.
Check your answer
No — output 2 repeats for inputs 1, 3. — A repeated OUTPUT is completely fine for a function — two different inputs are allowed to share one output. The one-input-one-output test only forbids one INPUT having two different outputs, which never happens here.
✓ Yes, and onto — {2,4,8} is fully used. — (B) Each of 1, 2, 3, 4 appears exactly once as a first coordinate, so R is a function. Its outputs are {2,4,8}, exactly matching the codomain {2,4,8}, with every codomain element hit at least once (2 from x=1 and x=3, 4 from x=2, 8 from x=4) — so R is onto.
Yes, but not onto — codomain not fully used. — Onto asks whether every element of the CODOMAIN, {2,4,8}, is actually produced as an output — it has nothing to do with whether domain elements like 1, 3, 4 themselves reappear among the outputs.
Yes, but not onto — two inputs share an output. — Two inputs sharing an output (1 and 3 both giving 2) is exactly what makes R fail to be one-to-one — it says nothing about onto-ness, which only asks whether every codomain element gets hit at least once, and here every one of 2, 4, 8 does.
Evaluate [3.7] + [-3.7], where [.] denotes the greatest integer (floor) function.
-4
6
-1
1
Check your answer
-4 — The expression is [3.7]+[-3.7], a SUM of two terms. Reporting just [-3.7]=-4 silently drops the first term [3.7]=3 — the sum is 3+(-4)=-1, not -4 alone.
6 — The expression is [3.7]+[-3.7], with a genuine negative sign on the second term — computing [3.7]+[3.7]=3+3=6 silently drops that minus sign.
✓ -1 — (C) [3.7] is the greatest integer not exceeding 3.7, which is 3. [-3.7] is the greatest integer not exceeding -3.7, which is -4 (since -4<=-3.7<-3). Adding, 3+(-4)=-1.
1 — [.] is the FLOOR function, not ordinary rounding — [3.7]=3 (the greatest integer not exceeding 3.7, not rounded up to 4), and [-3.7]=-4 (the greatest integer not exceeding -3.7, not truncated toward zero to -3).
Solve $|x^2$ - 4| = x + 2 for real x. Give the complete solution set.
x = -2, 3 only
x = 1 only
x = -2, 1, 3, -1
x = -2, 1, 3
Check your answer
x = -2, 3 only — The case $x^2-4<0$ (when -2<x<2) also needs solving — it contributes x=1, a genuine third solution missed by only checking where $x^2-4$ is already non-negative.
x = 1 only — The case $x^2-4\geq 0$ (when x<=-2 or x>=2) also needs solving — it contributes x=-2 and x=3, both genuine solutions missed by only checking where $x^2-4$ is negative.
x = -2, 1, 3, -1 — Rearranging $-(x^2-4)=x+2$ correctly gives $-x^2+4=x+2$, so $x^2+x-2=0$, factoring as (x+2)(x-1)=0. A sign slip to $x^2-x-2=0$ instead factors as (x-2)(x+1)=0, producing the spurious extra root x=-1, which does not actually satisfy the original equation.
✓ x = -2, 1, 3 — (D) Case $x^2-4\geq 0$ (x<=-2 or x>=2), $x^2-4=x+2$ gives $x^2-x-6=0=(x-3)(x+2)$, so x=3 or x=-2, both valid in this case. Case $x^2-4<0$ (-2<x<2), $-(x^2-4)=x+2$ gives $x^2+x-2=0=(x+2)(x-1)$, so x=-2 (rejected, outside the open interval) or x=1 (valid). All three, x=-2, 1, 3, check directly in the original equation.
Every pair of A x B is a switch a relation sets to in or out, which is why the number of relations is 2 raised to the number of switches. Being told a pair is included settles one switch and leaves the other five alone — 2⁵ = 32, not 2, and not the whole 2⁶ = 64. Count the columns before reaching for the exponent; six of them is n(A)·n(B), and reading five there is what produces 16.
When a modulus equation is set, count the crossings before trusting the algebra. Dropping the bars solves only the case the expression looks like, and that case cannot see the arch — the crossing at x = 1 lives entirely between the two folds. The plot is also the cheapest check on the finished answer: three roots on paper should be three intersections on the picture, and if they are not, one case was skipped or one candidate should have been rejected.
Onto asks one question, and it is about the right-hand column: does every declared output have something arriving? Both of these are functions, and in both, two inputs land on the same output — that shared landing is a fact about the left and decides nothing here. Move one arrowhead and 8 goes hollow; the range drops below the codomain, and only then does onto fail.