Relations and Functions

1. Check this before you revise anything

Two gaps between what coaching material assumes and what this chapter actually contains — in opposite directions.

The one-one/onto classification has moved to Class 12, where it belongs with inverse functions. Drilling it here is revising the wrong year's syllabus. Exponential and logarithmic functions run the other way — CBSE examines them under this chapter, but the 2026-27 NCERT text never defines either one. Section 6 below covers the minimum you need, taught here because the textbook does not.

2. What this chapter covers

3. The Cartesian product — pairing, in order

$A\times B=\{(a,b):a\in A,b\in B\}$

Every element of $A$ pairs with every element of $B$, in that order. Order inside each pair matters completely — $(a,b)$ and $(b,a)$ are different objects unless $a=b$.

Worked, mirroring the textbook's own Example 3. $A=\{1,2,3\}$, $B=\{3,4\}$, $C=\{4,5,6\}$. Since $B\cap C=\{4\}$: $A\times (B\cap C)=\{(1,4),(2,4),(3,4)\}$. Computing $A\times B$ and $A\times C$ separately and intersecting gives the identical set — a direct check that $A\times (B\cap C)=(A\times B)\cap (A\times C)$ for this case, exactly the kind of identity Exercise 2.1 asks you to verify in general.

Triples work the same way: $A\times A\times A=\{(a,b,c):a,b,c\in A\}$. $\mathbb{R}\times \mathbb{R}$ is every point in the coordinate plane; $\mathbb{R}\times \mathbb{R}\times \mathbb{R}$ is every point in three-dimensional space.

4. Relations: domain, codomain, and range

A relation $R$ from set $A$ to set $B$ is simply a subset of $A\times B$, chosen by describing some relationship between the first and second element of each pair.

Worked, mirroring the textbook's own Example 7. $A=\{1,2,3,4,5,6\}$, $R=\{(x,y):y=x+1\}$. Roster form: $R=\{(1,2),(2,3),(3,4),(4,5),(5,6)\}$. Domain $=\{1,2,3,4,5\}$ — note $6$ is missing, since $6+1=7\notin A$. Range $=\{2,3,4,5,6\}$. Codomain $=A=\{1,2,3,4,5,6\}$, the full set, whether or not every element gets used.

Counting relations. Since a relation is just a subset of $A\times B$, and a set with $n$ elements has $2^n$ subsets, the number of possible relations from $A$ to $B$ is:

$2^{n(A)\times n(B)}$

Worked: proving properties of a relation, mirroring the textbook's own Example 19

Let $R$ be a relation on $\mathbb{Q}$ (the rationals) defined by $R=\{(a,b):a,b\in \mathbb{Q},a-b\in \mathbb{Z}\}$. Show: (i) $(a,a)\in R$ for every $a$, (ii) $(a,b)\in R⟹(b,a)\in R$, and (iii) $(a,b)\in R$ and $(b,c)\in R⟹(a,c)\in R$.

(i) $a-a=0$, and $0\in \mathbb{Z}$, so $(a,a)\in R$ for every rational $a$.

(ii) If $(a,b)\in R$ then $a-b\in \mathbb{Z}$. Since the negative of any integer is still an integer, $b-a=-(a-b)\in \mathbb{Z}$ too, so $(b,a)\in R$.

(iii) If $(a,b)\in R$ and $(b,c)\in R$, then $a-b\in \mathbb{Z}$ and $b-c\in \mathbb{Z}$. Adding them: $a-c=(a-b)+(b-c)$, a sum of two integers, which is itself an integer. So $(a,c)\in R$.

All three properties hold here — a genuine contrast worth noticing. The Miscellaneous Exercise asks the same three questions about a different relation (defined by $a=b^2$ on the naturals), where every single one of them turns out false. Whether a relation satisfies these properties depends entirely on how it is defined, never on some general rule.

5. Functions: relations with a strict promise

A function $f$ from $A$ to $B$ is a relation where every element of $A$ has exactly one image in $B$ — no element left out, and no element pointing to two different places.

Every function is a relation. Not every relation is a function. Checking which of the three examples below are functions is a genuinely common exam question:

If $f(a)=b$, $b$ is the image of $a$, and $a$ is the preimage of $b$. A real function is one whose domain and range are both subsets of $\mathbb{R}$.

6. Standard functions and their graphs

Worked, mirroring the textbook's own Example 15. $f(x)=1/x$, $x\in \mathbb{R}-\{0\}$. As $x$ shrinks toward $0$ from the positive side, $f(x)$ grows without bound; as $x$ shrinks toward $0$ from the negative side, $f(x)$ plunges without bound. Neither the domain nor the range ever includes $0$ — $f(x)=1/x$ can equal any real number except zero, and $x$ can be any real number except zero.

Exponential and logarithmic functions — CBSE requires these, NCERT does not teach them here

An exponential function $f(x)=a^x$ (with $a>0$, $a\ne 1$) has domain $\mathbb{R}$ and range $(0,\infty )$ — output is always positive, never zero, however negative $x$ gets. It always passes through $(0,1)$, since $a^0=1$ for any valid $a$. It increases everywhere if $a>1$, and decreases everywhere if $0<a<1$.

A logarithmic function $f(x)={\log }_{a}x$ (same restriction on $a$) is defined only for $x>0$, so its domain is $(0,\infty )$ and its range is all of $\mathbb{R}$. It always passes through $(1,0)$, since ${\log }_{a}1=0$ for any valid base.

They are mirror images of each other — ${\log }_{a}x$ reverses exactly what $a^x$ does, which is why one's domain is the other's range, and vice versa.

7. The algebra of real functions

For two real functions $f,g$ sharing a common domain $X$:

$(f+g)(x)=f(x)+g(x)(f-g)(x)=f(x)-g(x)$ $(fg)(x)=f(x)g(x)\left(\frac{f}{g}\right)(x)=\frac{f(x)}{g(x)},g(x)\ne 0$

A scalar multiple $(\alpha f)(x)=\alpha f(x)$ works the same way, for any real number $\alpha$.

Worked, mirroring the textbook's own Example 16. $f(x)=x^2$, $g(x)=2x+1$.

$(f+g)(x)=x^2+2x+1(f-g)(x)=x^2-2x-1$ $(fg)(x)=x^2(2x+1)=2x^3+x^2\left(\frac{f}{g}\right)(x)=\frac{x^2}{2x+1},x\ne -\frac{1}{2}$

The quotient's domain needs the extra exclusion — $f$ and $g$ are each defined for all of $\mathbb{R}$, but $f/g$ additionally excludes wherever $g(x)=0$, since division by zero is never defined.

Worked: reconstructing a linear function from a few of its values

Textbook Example 20. $f=\{(1,1),(2,3),(0,-1),(-1,-3)\}$ is known to be a linear function from $\mathbb{Z}$ to $\mathbb{Z}$ — that is, $f(x)=mx+c$ for some constants $m,c$. Find $f(x)$.

A linear function is fully determined by any two of its values, since two points pin down both unknowns. Using $(0,-1)$: $f(0)=c=-1$. Using $(1,1)$: $f(1)=m+c=1$, so $m=1-c=1-(-1)=2$.

Check against the remaining two given points, since only two were actually needed: $f(2)=2(2)-1=3$ — matches $(2,3)$; $f(-1)=2(-1)-1=-3$ — matches $(-1,-3)$. Both extra points confirm the answer without being required to find it — a useful habit whenever a problem hands you more data than the minimum needed.

$f(x)=2x-1$

Summary

• $A\times B=\{(a,b):a\in A,b\in B\}$; order inside each pair matters, so generally $A\times B\ne B\times A$.
• A relation from $A$ to $B$ is any subset of $A\times B$; domain is the first elements used, range is the second elements used, codomain is all of $B$.
• The number of relations from $A$ to $B$ is $2^{n(A)\times n(B)}$, since a relation is just a subset of $A\times B$.
• A function is a relation where every element of the domain has exactly one image — no element skipped, none pointing two places at once.
• Standard functions (identity, constant, polynomial, rational, modulus, signum, greatest integer) are classified by the shape of their rule, not by injective/surjective behaviour — that classification belongs to Class 12.
• Exponential ($a^x$) and logarithmic (${\log }_{a}x$) functions are examinable here under CBSE despite the current NCERT text never defining them; they are mirror images of each other, with domain and range swapped.
• $(f\pm g)(x)$, $(fg)(x)$ and $(f/g)(x)$ combine two functions pointwise; a quotient additionally excludes wherever the denominator function is zero.