Relations and Functions Class 11 Notes Maths Chapter 2

 Page 1


 
KEY POINTS
? Cartesian Product of two non-empty sets A and B is given by,A 
× B = { (a, b) : a ? A, b ? B} 
? If (a, b) = (x, y), then a = x and b = y 
? Relation R from a non-empty set A to a non-empty set B is a 
subset of A×B. 
? Domain of R = {a : (a, b) ? R} 
? Range of R = { b : (a, b) ? R} 
? Co-domain of R = Set B 
? Range ? Co-domain 
? If n(A) = p, n(B) = q then n(A×B) = pq and number of relations = 
2
pq
 
? Image : If the element x of A corresponds to y ? B under the 
function f, then we say that y is image of x under ‘f ’ 
? f (x) = y 
? If f (x) = y, then x is preimage of y. 
? A relation f from a set A to a set B is said to be a function if 
every element of set A has one and only one image in set B. 
Page 2


 
KEY POINTS
? Cartesian Product of two non-empty sets A and B is given by,A 
× B = { (a, b) : a ? A, b ? B} 
? If (a, b) = (x, y), then a = x and b = y 
? Relation R from a non-empty set A to a non-empty set B is a 
subset of A×B. 
? Domain of R = {a : (a, b) ? R} 
? Range of R = { b : (a, b) ? R} 
? Co-domain of R = Set B 
? Range ? Co-domain 
? If n(A) = p, n(B) = q then n(A×B) = pq and number of relations = 
2
pq
 
? Image : If the element x of A corresponds to y ? B under the 
function f, then we say that y is image of x under ‘f ’ 
? f (x) = y 
? If f (x) = y, then x is preimage of y. 
? A relation f from a set A to a set B is said to be a function if 
every element of set A has one and only one image in set B. 
 
? Df = {x : f(x) is defined} Rf = {f(x) : x ? Df} 
? Let A and B be two non-empty finite sets such that n(A) = p and 
n(B) = q then number of functions from A to B = q
p
. 
? Identity function, f : R ? R; f(x) = x ? x ? R, where R is the set 
of realnumbers. 
Df = R Rf = R 
? Constant function, f : R ? R; f(x) = c ? x ? R where c is a 
constant 
Df = R Rf = {c} 
? Modulus function, f : R ? R; f(x) = |x| ? x ? R 
Df = R 
Rf = R
+
? { 0} = { x : x ? R: x ? 0} 
O X X´ 
Y 
Y´ 
f 
( 
x 
) 
 
= 
 
x
Page 3


 
KEY POINTS
? Cartesian Product of two non-empty sets A and B is given by,A 
× B = { (a, b) : a ? A, b ? B} 
? If (a, b) = (x, y), then a = x and b = y 
? Relation R from a non-empty set A to a non-empty set B is a 
subset of A×B. 
? Domain of R = {a : (a, b) ? R} 
? Range of R = { b : (a, b) ? R} 
? Co-domain of R = Set B 
? Range ? Co-domain 
? If n(A) = p, n(B) = q then n(A×B) = pq and number of relations = 
2
pq
 
? Image : If the element x of A corresponds to y ? B under the 
function f, then we say that y is image of x under ‘f ’ 
? f (x) = y 
? If f (x) = y, then x is preimage of y. 
? A relation f from a set A to a set B is said to be a function if 
every element of set A has one and only one image in set B. 
 
? Df = {x : f(x) is defined} Rf = {f(x) : x ? Df} 
? Let A and B be two non-empty finite sets such that n(A) = p and 
n(B) = q then number of functions from A to B = q
p
. 
? Identity function, f : R ? R; f(x) = x ? x ? R, where R is the set 
of realnumbers. 
Df = R Rf = R 
? Constant function, f : R ? R; f(x) = c ? x ? R where c is a 
constant 
Df = R Rf = {c} 
? Modulus function, f : R ? R; f(x) = |x| ? x ? R 
Df = R 
Rf = R
+
? { 0} = { x : x ? R: x ? 0} 
O X X´ 
Y 
Y´ 
f 
( 
x 
) 
 
= 
 
x
 
? ?Signum function 
1,if x >0
x
, x 0
f : R R ; f (x) = 0,if x = 0 and f (x) = 
x
0,x = 0 — 1,if x < 0
?
?
? ? ?
?
? ?
? ?
?
?
Then  
Df = R 
and Rf = {–1,0,1} 
? Greatest Integer functionf : R ? R; f(x) = [x], x ? R assumes the 
value of the greatest integer, less than or equal to x 
Df = R Rf = Z 
O 
X X´ 
Y 
Y´ 
1 
y  = 1 
y = – 1 
– 1
O 
X X´ 
Y 
Y´ 
Page 4


 
KEY POINTS
? Cartesian Product of two non-empty sets A and B is given by,A 
× B = { (a, b) : a ? A, b ? B} 
? If (a, b) = (x, y), then a = x and b = y 
? Relation R from a non-empty set A to a non-empty set B is a 
subset of A×B. 
? Domain of R = {a : (a, b) ? R} 
? Range of R = { b : (a, b) ? R} 
? Co-domain of R = Set B 
? Range ? Co-domain 
? If n(A) = p, n(B) = q then n(A×B) = pq and number of relations = 
2
pq
 
? Image : If the element x of A corresponds to y ? B under the 
function f, then we say that y is image of x under ‘f ’ 
? f (x) = y 
? If f (x) = y, then x is preimage of y. 
? A relation f from a set A to a set B is said to be a function if 
every element of set A has one and only one image in set B. 
 
? Df = {x : f(x) is defined} Rf = {f(x) : x ? Df} 
? Let A and B be two non-empty finite sets such that n(A) = p and 
n(B) = q then number of functions from A to B = q
p
. 
? Identity function, f : R ? R; f(x) = x ? x ? R, where R is the set 
of realnumbers. 
Df = R Rf = R 
? Constant function, f : R ? R; f(x) = c ? x ? R where c is a 
constant 
Df = R Rf = {c} 
? Modulus function, f : R ? R; f(x) = |x| ? x ? R 
Df = R 
Rf = R
+
? { 0} = { x : x ? R: x ? 0} 
O X X´ 
Y 
Y´ 
f 
( 
x 
) 
 
= 
 
x
 
? ?Signum function 
1,if x >0
x
, x 0
f : R R ; f (x) = 0,if x = 0 and f (x) = 
x
0,x = 0 — 1,if x < 0
?
?
? ? ?
?
? ?
? ?
?
?
Then  
Df = R 
and Rf = {–1,0,1} 
? Greatest Integer functionf : R ? R; f(x) = [x], x ? R assumes the 
value of the greatest integer, less than or equal to x 
Df = R Rf = Z 
O 
X X´ 
Y 
Y´ 
1 
y  = 1 
y = – 1 
– 1
O 
X X´ 
Y 
Y´ 
 
? f : R ? R, f(x) = x
2
 
Df = R Rf = [0, ? ? 
? f : R ? R, f(x) = x
3
 
Df = R Rf = R 
? Exponential function, f : R ? R ; f(x) = a
x
, a > 0, a ? 1 
X´ 
O 
X 
Y 
Y´ 
O 
X X´ 
Y 
Y´ 
O 
X X´ 
Y 
Y´
2 
1 
–2 –1 1 2 3 4 
–1
–2
3 
Page 5


 
KEY POINTS
? Cartesian Product of two non-empty sets A and B is given by,A 
× B = { (a, b) : a ? A, b ? B} 
? If (a, b) = (x, y), then a = x and b = y 
? Relation R from a non-empty set A to a non-empty set B is a 
subset of A×B. 
? Domain of R = {a : (a, b) ? R} 
? Range of R = { b : (a, b) ? R} 
? Co-domain of R = Set B 
? Range ? Co-domain 
? If n(A) = p, n(B) = q then n(A×B) = pq and number of relations = 
2
pq
 
? Image : If the element x of A corresponds to y ? B under the 
function f, then we say that y is image of x under ‘f ’ 
? f (x) = y 
? If f (x) = y, then x is preimage of y. 
? A relation f from a set A to a set B is said to be a function if 
every element of set A has one and only one image in set B. 
 
? Df = {x : f(x) is defined} Rf = {f(x) : x ? Df} 
? Let A and B be two non-empty finite sets such that n(A) = p and 
n(B) = q then number of functions from A to B = q
p
. 
? Identity function, f : R ? R; f(x) = x ? x ? R, where R is the set 
of realnumbers. 
Df = R Rf = R 
? Constant function, f : R ? R; f(x) = c ? x ? R where c is a 
constant 
Df = R Rf = {c} 
? Modulus function, f : R ? R; f(x) = |x| ? x ? R 
Df = R 
Rf = R
+
? { 0} = { x : x ? R: x ? 0} 
O X X´ 
Y 
Y´ 
f 
( 
x 
) 
 
= 
 
x
 
? ?Signum function 
1,if x >0
x
, x 0
f : R R ; f (x) = 0,if x = 0 and f (x) = 
x
0,x = 0 — 1,if x < 0
?
?
? ? ?
?
? ?
? ?
?
?
Then  
Df = R 
and Rf = {–1,0,1} 
? Greatest Integer functionf : R ? R; f(x) = [x], x ? R assumes the 
value of the greatest integer, less than or equal to x 
Df = R Rf = Z 
O 
X X´ 
Y 
Y´ 
1 
y  = 1 
y = – 1 
– 1
O 
X X´ 
Y 
Y´ 
 
? f : R ? R, f(x) = x
2
 
Df = R Rf = [0, ? ? 
? f : R ? R, f(x) = x
3
 
Df = R Rf = R 
? Exponential function, f : R ? R ; f(x) = a
x
, a > 0, a ? 1 
X´ 
O 
X 
Y 
Y´ 
O 
X X´ 
Y 
Y´ 
O 
X X´ 
Y 
Y´
2 
1 
–2 –1 1 2 3 4 
–1
–2
3 
 
Df = R Rf= (0, ?) 
0 < a < 1 a > 1 
? Natural exponential function, f(x) = e
x
 
1 1 1
1 ... , 2 3
1! 2! 3!
e e ? ? ? ? ? ? ? ?
? Logarithmic functions, f : (0, ?) ? R ; f(x)lo g
a
x, a > 0, a ? 1
Df = (0, ?) 
R
f = R 
? Natural logarithrnic function f(x) = logex or log x. 
? Let f : X ? R and g : X ? R be any two real functions where x ? 
R then 
(f ± g) (x) = f(x) ± g(x) ? x ? X 
(fg) (x) = f(x) g(x) ? x ? X 
X 
Y 
Y’ 
X’ 
X 
Y 
Y’ 
X
’ 
(0 ,  1) 
(0 ,  1) 
O O 
?
?
? ? 0
f ? x
f
X provided g ? x
g ? x
? ?
?
g
?
? x ? ? ? x ?
? ?