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 ?
? ?
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