Nature of Chapter:
1. It’s a continuation of Relation and Function of 11th class. But don’t
worry even if you have not done that much in 11th.
2. Questions on Relations will demand both understanding and solving
skills.
3. Functions here, will demand little bit of Range concept from 11th class,
other than that, we will be learning new concepts here.
4. Some concepts that you will study in this chapter are further explored
in AOD.
Weightage of Sets & Relations (Last 5 years)
2023
2022
2021
2020
2019
Average
JEE Main
3.3 %
2.1 %
1.4 %
1.5 %
0.8 %
1.82 %
Jee Advanced
0%
3%
0%
0%
0%
0.60 %
Weightage of Functions (Last 5 years)
2023
2022
2021
2020
2019
Average
JEE Main
5.6 %
3.6 %
4.4 %
3.8 %
4.2 %
4.32 %
Jee Advanced
6%
0%
0%
5%
0%
2.20 %
Relations & Functions
●
Relations
●
Types of Relations
●
Functions and its classifications
●
Composition of Functions
●
Inverse of a function
●
Functional Equation
Relations & Functions
Critical Topics in the Chapter
●
Types of relations
●
Composition of functions
●
Functional Equation
Relations
Lets first do the basic recall of relations, we covered in 11th.
Relations
Any subset of A × B is defined as a relation from set A to set B,
where A and B are non-empty sets
Relations
Any subset of A × B is defined as a relation from set A to set B,
where A and B are non-empty sets
Remark
Relation from A to A is also called relation on A.
Relations
Description of relations
There are three ways of writing a relation.
Let us do it by examples
Let A = {1, 2, 3} & B = {1, 3, 5, 7}
(1) R = {(1,1), (2, 3), (3, 5)}
Relations
Description of relations
There are three ways of writing a relation.
Let us do it by examples
Let A = {1, 2, 3} & B = {1, 3, 5, 7}
(1) R = {(1,1), (2, 3), (3, 5)}
(or)
(2) R = {(a, b) | b = 2a – 1, a ∈ A and b ∈ B}
Relations
Description of relations
There are three ways of writing a relation.
Let us do it by examples
Let A = {1, 2, 3} & B = {1, 3, 5, 7}
(1) R = {(1,1), (2, 3), (3, 5)}
(or)
(2) R = {(a, b) | b = 2a – 1, a ∈ A and b ∈ B}
(or)
(3) a R b ⇔ b = 2a – 1; a ∈ A and b ∈ B
Q
Write relation on A defined as: xRy ⇔ y = x2 ,
where A = {1, -1, 2, 3, 4}
Q
Write relation on A defined as: xRy ⇔ y = x2 ,
where A = {1, -1, 2, 3, 4}
Solution:
R = {(1, 1), (-1, 1), (2, 4)}
Relations
Result
If n(A) = p and n(B) = q then number of relations that
can be defined from A to B is 2pq.
Q
If A is the set of even natural numbers less than 8
and B is the set of prime numbers less than 7, then
the number of relations from A to B is
A
29
B
92
C
32
D
29-1
Q
If A is the set of even natural numbers less than 8
and B is the set of prime numbers less than 7, then
the number of relations from A to B is
A
29
B
92
C
32
D
29-1
Q
If A is the set of even natural numbers less than 8
and B is the set of prime numbers less than 7, then
the number of relations from A to B is
Solution:
A = {2, 4, 6}, B = {2, 3, 5}
No. of relations from A to B = 23 x 3 =29
Relations
Inverse of Relation
If R is a relation on set A, then the relation R−1 on A ,defined by
R−1 = {(y, x) : (x, y) ∈ R} is called an inverse relation to A.
Clearly,
Domain (R−1) = Range of (R)
Range of (R−1) = domain of (R).
Q
Find the inverse relation R-1 if R is a relation from
{11, 12, 13} to {8, 10, 12} defined by y = x - 3.
Q
Find the inverse relation R-1 if R is a relation from
{11, 12, 13} to {8, 10, 12} defined by y = x - 3.
Solution:
R is a relation from {11, 12, 13} to {8, 10, 12} defined by y = x - 3
Now, y = x - 3
Putting x = 11, 12, 13, we get y = 8, 9, 10 respectively
⇒ {11, 8} ∈ R, (12, 9) ∉ R and (13, 10) ∈ R
Thus, R = {(11, 8), (13, 10)}
⇒ R-1 = {(8, 11), (10, 13)}
Now we start with the Relations Topic of 12th syllabus, which
primarily covers Types of relation.
Types of Relations
Types of Relations
(1) Empty Relation: Let A be a set and ɸ ⊂ A ✕ A, it follows that ɸ is a
relation on A which is called the empty relation.
(2) Universal Relation: Let A be a set and A ✕ A ⊆ A ✕ A, it follows that
A ✕ A is a relation on A, which is called the universal relation.
(3) Identity Relation: If every element of A is related to itself only, then
the relation is called identity relation.
Symbolically, the relation IA = {(a, a) : a ∈ A} is called an Identity
relation on A.
Types of Relations
1. Reflexive Relation
Types of Relations
1. Reflexive Relation
A Relation R on a set A is said to be reflexive, if every element of A is related to
itself. Thus if R is reflexive, then (a, a) ∈ R, ∀ a ∈ A.
For example, let A = {1, 2, 3} then R1 = {(x, y) | y ≥ x} is a reflexive relation on A.
Types of Relations
2. Symmetric Relation
Types of Relations
2. Symmetric Relation
A relation R on a set A is said to be a symmetric relation iff
(a, b) ∈ R ⇒ (b, a) ∈ R, ∀ a, b ∈ A
For example, let A = {2, 4, 6} then
R1 = {(2, 4), (2, 6), (4, 4), (4, 2), (6, 2)} is a symmetric relation on A.
Types of Relations
3. Transitive Relation
Types of Relations
3. Transitive Relation
Relation R on set A is transitive if xRy and yRz ⇒ xRz
Types of Relations
3. Transitive Relation
Relation R on set A is transitive if xRy and yRz ⇒ xRz
NOTE
Relation is NOT transitive only when x R y and y R z but
In all other situations R is Transitive.
Types of Relations
3. Transitive Relation
Relation R on set A is transitive if xRy and yRz ⇒ xRz
Let A = {1, 2, 3, 4, 5}
(1) R1 = {(1, 2), (2, 3), (1, 3)}
(2) R2 = {(1, 2), (2, 3), (1, 3), (4, 5), (5, 1)}
(3) R3 = {(1, 3), (2, 4)}
Types of Relations
3. Transitive Relation
Relation R on set A is transitive if xRy and yRz ⇒ xRz
Let A = {1, 2, 3, 4, 5}
(1) R1 = {(1, 2), (2, 3), (1, 3)}
: It is transitive relation.
(2) R2 = {(1, 2), (2, 3), (1, 3), (4, 5), (5, 1)}
: It is not a transitive relation.
(3) R3 = {(1, 3), (2, 4)}
: It is transitive relation.
Q
If a relation R on the set {1, 2, 3} be defined by
(i) R = {(1, 2)}, then R is:
(ii)R = {(1, 1), (2, 2), (3, 3), (1, 2), (2, 3), (1, 3)}. Then R is:
Solution:
Q
Solution:
Let R be the relation in the set {1, 2, 3, 4} given by
R = {(1, 2), (2, 2), (1, 1), (4, 4), (1, 3), (3, 3), (3, 2)}, then R is ___.
IIT 2005
Q
Let R = {(3, 3), (6, 6), (9, 9), (12, 12), (6, 12), (3, 9), (3, 12), (3, 6)}
be a relation on A = {3, 6, 9, 12}. Relation R is
A
Reflexive and transitive only
B
Reflexive only
C
An equivalence relation
D
Reflexive and symmetric only
IIT 2005
Q
Let R = {(3, 3), (6, 6), (9, 9), (12, 12), (6, 12), (3, 9), (3, 12), (3, 6)}
be a relation on A = {3, 6, 9, 12}. Relation R is
A
Reflexive and transitive only
B
Reflexive only
C
An equivalence relation
D
Reflexive and symmetric only
IIT 2005
Q
Let R = {(3, 3), (6, 6), (9, 9), (12, 12), (6, 12), (3, 9), (3, 12), (3, 6)}
be a relation on A = {3, 6, 9, 12}. Relation R is
Solution:
Given A = {3, 6, 9, 12}
Since (3, 3), (6, 6), (9, 9), (12, 12) ∈ R
Thus its reflexive relation (12, 6) ∉ R
Here, (6, 12) ∈ R, but
Thus R is not symmetric
Here, (3, 6) ∈ R and (6, 12) ∈ R, also (3, 12) ∈ R
So relation is transitive also.
Q
Consider the following two binary relations on the set
A = {a, b, c}
R1 = {(c, a), (b, b), (a, c), (c, c), (b, c), (a, a), (c, b)} and
R2 = {(a, b), (b, a), (c, c), (c, a), (a, a), (b, b), (a, c)}. Then:
A
both R1 and R2 are not symmetric.
B
R1 is not symmetric but it is transitive
C
R2 is symmetric but it is not transitive
D
both R1 and R2 are transitive.
Q
Consider the following two binary relations on the set
A = {a, b, c}
R1 = {(c, a), (b, b), (a, c), (c, c), (b, c), (a, a), (c, b)} and
R2 = {(a, b), (b, a), (c, c), (c, a), (a, a), (b, b), (a, c)}. Then:
A
both R1 and R2 are not symmetric.
B
R1 is not symmetric but it is transitive
C
R2 is symmetric but it is not transitive
D
both R1 and R2 are transitive.
Q
Consider the following two binary relations on the set
A = {a, b, c}
R1 = {(c, a), (b, b), (a, c), (c, c), (b, c), (a, a), (c, b)} and
R2 = {(a, b), (b, a), (c, c), (c, a), (a, a), (b, b), (a, c)}. Then:
Solution:
Both R1 and R2 are symmetric as for any
(a1, a2) ∈ R1 ⇔ (a2, a1) ∈ R1
And same thing can be verified for R2 as well (a1 ≠ a2).
For checking transitivity, we observe for R2 that
(b, a) ∈ R2, (a, c) ∈ R2 but (b, c) ∉ R2.
Similarly, for R1, (b, c) ∈ R1, (c, a) ∈ R1 but (b, a) ∉ R1.
So neither R1 nor R2 is transitive.
JEE Main 13th Apr, 2023-S2
Q
Let A = {-4, -3, -2, 0, 1, 3, 4} and R = {(a, b) ∈ A x A : b = |a| or b2 = a + 1}
be a relation on A. Then the minimum number of elements, that
must be added to the relation R so that it becomes reflexive and
symmetric, is _______
JEE Main 13th Apr, 2023-S2
Q
Let A = {-4, -3, -2, 0, 1, 3, 4} and R = {(a, b) ∈ A x A : b = |a| or b2 = a + 1}
be a relation on A. Then the minimum number of elements, that
must be added to the relation R so that it becomes reflexive and
symmetric, is _______
Ans: 7
JEE Main 13th Apr, 2023-S2
Q
Let A = {-4, -3, -2, 0, 1, 3, 4} and R = {(a, b) ∈ A x A : b = |a| or b2 = a + 1}
be a relation on A. Then the minimum number of elements, that
must be added to the relation R so that it becomes reflexive and
symmetric, is _______
Solution:
R = [(-4, 4), (-3, 3), (3, -2), (0, 1), (0, 0), (1, 1), (4, 4), (3, 3)}
For reflexive, add ⇒ (-2, -2), (-4, -4), (-3, -3)
For symmetric, add ⇒ (4, -4), (3, -3), (-2, 3), (1, 0)
Q
Consider a relation of whole numbers defined as
x R y ⇔ x ≤ y2. Prove that its reflexive but neither
symmetric nor transitive.
Q
Consider a relation of whole numbers defined as
x R y ⇔ x ≤ y2. Prove that its reflexive but neither
symmetric nor transitive.
Solution:
Clearly, it is reflexive as (a, a) ∈ R as a ≤ a2
But, if (1, 2) ∈ R
Then (2, 1) ∉ R
So it is not symmetric
Now, let us consider
(5, 3) ∈ R, (3, 2) ∈ R
But (5, 2) ∉ R
Thus R is not transitive.
Types of Relations
4. Equivalence Relation
Relation R defined on any set A is equivalence relation if
R is reflexive
R is symmetric
R is transitive
Q
The minimum number of elements that must be added
to the relation R = {(1, 2), (2, 3)} on the set of natural
numbers so that it is an equivalence relation, is_____.
Q
The minimum number of elements that must be added
to the relation R = {(1, 2), (2, 3)} on the set of natural
numbers so that it is an equivalence relation, is_____.
Ans: 7
Solution:
Q
Show that relation R defined on the set A = {0 , 1 , 2 , …. , 12}
given by R = {(a, b) | a − b is multiple of 4} is an equivalence
relation. Also find elements related to 3.
Solution:
A = {0, 1, 2, …., 12}
Transitivity: Let a, b, c ∈ A
R = {(a, b) | a − b is multiple of 4}
Such that (a, b) ∈ R and (b, c) ∈ R
Reflexivity: For any a ∈ A
∴ a − b is multiple of 4
(a, a) ∈ R
and (b − c) is multiple of 4
As, a − a = 0 is multiple of 4.
∴ (a − b) = 4λ, and (b − c) = 4𝜇
So, R is reflexive.
for some λ, 𝜇 ∈ N
Symmetric: Let a, b ∈ A Such that (a, b) ∈ R
∴ a − b is multiple of 4
⇒ a − b = 4λ for some λ ∈ z
∴ (b − a) = 4(−λ)
So, (b − a) is multiple of 4 ⇒ (b − a) ∈ R
It is symmetric
Add these equations
a − b + b − c = 4λ + 4𝜇
a − c = 4(λ + 𝜇)
So, (a − c) is multiple of 4
Thus, it is transitive
So, R is an equivalence relation.
Solution:
Elements related to 3
Let x be the element of A such that (x, 3) ∈ R
Then (x − 3) is multiple of 4
∴ (x − 3) = 0, 4, 8, 12, ….
⇒ x = 3, 7, 11, 15, …
set A = {0, 1, 2, …. , 12}
x = 3, 7, 11
JEE Main 29th Jan, 2023-S2
Q
Let R be a relation defined on N as a R b
if 2a + 3b is a multiple of 5, a, b ∊ N. Then R is
A
an equivalence relation
B
transitive but not symmetric
C
not reflexive
D
symmetric but not transitive
JEE Main 29th Jan, 2023-S2
Q
Let R be a relation defined on N as a R b
if 2a + 3b is a multiple of 5, a, b ∊ N. Then R is
A
an equivalence relation
B
transitive but not symmetric
C
not reflexive
D
symmetric but not transitive
Solution:
JEE Main July 27, 2021 Shift-2
Q
Let Z be the set of Integers and a relation R on Z be
defined by R = {(x, y) ∈ Z × Z : x3 − 3x2y − xy2 + 3y3 = 0}.
Then the relation R is:
A
Symmetric but neither reflexive nor
transitive
B
Reflexive but neither symmetric nor
transitive
C
Reflexive and symmetric, but not
transitive
D
An equivalence relation
JEE Main July 27, 2021 Shift-2
Q
Let Z be the set of Integers and a relation R on Z be
defined by R = {(x, y) ∈ Z × Z : x3 − 3x2y − xy2 + 3y3 = 0}.
Then the relation R is:
A
Symmetric but neither reflexive nor
transitive
B
Reflexive but neither symmetric nor
transitive
C
Reflexive and symmetric, but not
transitive
D
An equivalence relation
JEE Main July 27, 2021 Shift-2
Q
Let Z be the set of Integers and a relation R on Z be
defined by R = {(x, y) ∈ Z × Z : x3 − 3x2y − xy2 + 3y3 = 0}.
Then the relation R is:
Solution:
Types of Relations
Antisymmetric Relation
The relation R is said to be antisymmetric on a set A,
if xRy and yRx hold only when x = y.
FUNCTIONS
Functions
Let A and B be two non-empty sets. Then a function f : A → B is a rule
which associates each element of A to unique element of B.
Notations:
f:A→B
Classification of Functions
Classification of Functions
(a) One-one and many-one
(b) Onto and into
Classification of Functions
(a) One-one and many-one
(b) Onto and into
NOTE
One-one → Injective
Onto → Surjective
One-one and onto → Bijective
Classification of Functions
>
>
f
f
A
A
B
A
B
>
>
f
f
B
A
B
Q
Check whether the following functions are one-one
or many-one.
(a) y = x2 ; x ∈ R
(b) y = x2 ; x ∈ R+
Q
Check whether the following functions are one-one
or many-one.
(a) y = x2 ; x ∈ R
Solution:
Ist Method:
II Method:
Y
Thus
one.
is not one-
X
Q
Check whether the following functions are one-one
or many-one.
(b) y = x2 ; x ∈ R+
Q
Check whether the following functions are one-one
or many-one.
(b) y = x2 ; x ∈ R+
Solution:
Since
Thus for positive values
is positive, thus
Thus, It is one-one.
is increasing in its domain.
Q
Check whether the following function is one-one or
many-one : y = x2 – 5x + 6
Q
Check whether the following function is one-one or
many-one : y = x2 – 5x + 6
Solution:
Given f(x) = x2 − 5x + 6
f(x) = x2 − 5x + 6
f(x1) = f(x2)
f’(x) = 2x − 5
x12 − 5x1 + 6 = x22 − 5x2 + 6
(x12 − x22) − (5x1 − 5x2) = 0
(x1 − x2) (x1 + x2 − 5) = 0
So again
Either x1 − x2 = 0 or x1 − x2 − 5 = 0
This it is not one-one
−
−∞
+
5/2
Thus f(x) is many-one
∞
Q
Check whether the following functions are onto or into.
(a) f : R → [−1, 1] defined as
(b) f : R → [0, 5] defined as f(x) = |3 sin x + 4 cos x|
Q
Check whether the following functions are onto or into.
(a) f : R → [−1, 1] defined as
Solution:
Now x is defined if
Since
Thus f(x) is into.
Q
Check whether the following functions are onto or into.
(b) f : R → [0, 5] defined as f(x) = |3 sin x + 4 cos x|
Q
Check whether the following functions are onto or into.
(b) f : R → [0, 5] defined as f(x) = |3 sin x + 4 cos x|
Solution:
Q
Check whether the following functions is onto or into.
f : N → N defined as
Solution:
For n ∈ even, i.e. 2, 4, 6, 8, 10, ….
,
We will get
For n ∈ odd, i.e. 1, 3, 5, 7, 9, ….., we will get
Thus, we will get all natural numbers.
Thus, function is onto.
Q
If f : (-∞, ∞) → S defined as f(x) = x2 – 4x + 3 is an
onto function then S must be
A
B
[−1, ∞)
C
(−2, ∞)
D
None of these
Q
If f : (-∞, ∞) → S defined as f(x) = x2 – 4x + 3 is an
onto function then S must be
A
B
[−1, ∞)
C
(−2, ∞)
D
None of these
Q
If f : (-∞, ∞) → S defined as f(x) = x2 – 4x + 3 is an
onto function then S must be
Solution:
Here, we know that, for f(x)
Range is
∴ Range of f(x) ∈ [−1, ∞)
Q
Let f : R → [2, ∞] be a function defined as f(x) = x2 - 12ax + 15 - 2a + 36a2.
If f(x) is surjective on R, then the value of a is equal to
A
B
C
D
Q
Let f : R → [2, ∞] be a function defined as f(x) = x2 - 12ax + 15 - 2a + 36a2.
If f(x) is surjective on R, then the value of a is equal to
A
B
C
D
Q
Let f : R → [2, ∞] be a function defined as f(x) = x2 - 12ax + 15 - 2a + 36a2.
If f(x) is surjective on R, then the value of a is equal to
Solution:
Rewriting the given function, we get,
f(x) = (x - 6a)2 + 15 - 2a
∵ f(x) is surjective on R
⇒ 15 - 2a = 2
⇒ 2a = 13
⇒
Composition of functions
Composition of functions
Composition of f(x) and g(x) is denoted as fog(x) and gof(x) and
it is defined as :
(a) fog(x) = f(g(x))
(b) gof(x) = g(f(x))
If f(x) = sin x and g(x) = x2 . Then,
(a) fog(x) =
(b) gof(x) =
Composition of functions
Composition of f(x) and g(x) is denoted as fog(x) and gof(x) and
it is defined as :
(a) fog(x) = f(g(x))
(b) gof(x) = g(f(x))
If f(x) = sin x and g(x) = x2 . Then,
(a) fog(x) = f(g(x))
(b) gof(x) = g(f(x))
= sin(g(x))
= g(f(x))
= sin(x2)
= (f(x))2 = (sin x)2
Q
Let f(x) = x2 and g(x) = 2x , then the solution set of
fog(x) = gof(x) is
A
R
B
{0}
C
{0, 2}
D
None of these
Q
Let f(x) = x2 and g(x) = 2x , then the solution set of
fog(x) = gof(x) is
A
R
B
{0}
C
{0, 2}
D
None of these
Q
Let f(x) = x2 and g(x) = 2x , then the solution set of
fog(x) = gof(x) is
Solution:
Q
A
x
B
1
C
f(x)
D
g(x)
Q
A
x
B
1
C
f(x)
D
g(x)
Q
Solution:
Given, g(x) = 1 + x − [x]
g(x) = 1 + {x}
So, g(x) is always positive as 0 ≤ {x} < 1
Thus, f(g(x)) = 1
Composition of functions
Let us consider f : A → B and g : B → C defined as
f(x) = 3x – 1 and g(x) = 3x where
A = {1, 2, 3},
B = {2, 3, 5, 8} ,
C = {2, 6, 9, 15, 24} then:
gof ( __ ) =
gof ( __ ) =
gof ( __ ) =
Observation
gof : A → C
Composition of functions
Definition
If f : A → B and g : B → C are two functions, then composition of f and g
denoted by gof is defined as a function gof : A → C given by
gof(x) = g(f(x)), ∀ x ∈ A
* Clearly, range of f(x) must be subset of domain of g(x)
Q
Let
find the value of f {f (−2.3)}.
where [ . ] denotes the GIF. Then
Q
Let
find the value of f {f (−2.3)}.
Solution:
where [ . ] denotes the GIF. Then
Q
Let f be a function defined by
x ≠ 3, 2 ;
f k(x) denote the composition of f with itself taken k times
i.e., f 3(x) = f(f(f(x))), then which one of the following in NOT
correct ?
A
f 2009(2009) =
2009
B
C
D
f 2012(2012) = 2012
Q
Let f be a function defined by
x ≠ 3, 2 ;
f k(x) denote the composition of f with itself taken k times
i.e., f 3(x) = f(f(f(x))), then which one of the following in NOT
correct ?
A
f 2009(2009) =
2009
B
C
D
f 2012(2012) = 2012
Solution:
JEE Main 25th June 2022, S-1
Q
Let f : R → R be a function defined by
If the function
g(x) = f(f(f(x)) + f(f(x)), then the greatest integer
less than or equal to g(1) is___
JEE Main 25th June 2022, S-1
Q
Let f : R → R be a function defined by
If the function
g(x) = f(f(f(x)) + f(f(x)), then the greatest integer
less than or equal to g(1) is___
Ans : 2
Solution:
Q
then find f(g(x)) and find its domain and range.
Solution:
Inverse of function
Inverse of function
Consider a function f : {1, 2, 3} → {3, 6, 9} defined as f(x) = 3x
f
f(1) = 3
f(2) = 6
1
3
f(3) = 9
2
6
3
9
A
B
Inverse of function
Consider g : {3, 6, 9} → {1, 2, 3} defined as
f(1) = 3
then
g(3)= 1
f
f(2) = 6
f(3) = 9
g(6)= 2
1
3
2
6
3
9
A
B
g
g(9)= 3
Inverse of function
Try to observe in previous example
fog(x) = x ∀ x (i.e., fog is identity function from B to B i.e. IB) and
gof(x) = x ∀ x (i.e., gof is identity function from A to A i.e. IA)
Inverse of function
Definition
A function f : X → Y is said to be invertible, if there exists a function
g : Y → X such that gof = Ix and fog = Iy. The function g(x) is called
inverse of f(x) and is denoted as f-1(x)
Inverse of function
Algorithm to find Inverse of function
Step 1: Write y = f(x) and replace x ↔ y
Step 2: Find value of y. This is f-1(x)
Q
Find inverse function of following:
Q
Find inverse function of following:
Solution:
Q
Find inverse function of following:
Q
Find inverse function of following:
Solution:
Squaring both sides
Q
Solution:
Find inverse function of following:
JEE Main 2020
Q
The inverse of function
is:
A
B
C
D
JEE Main 2020
Q
The inverse of function
is:
A
B
C
D
Solution:
Q
If f(x) = (ax2 + b)3, then find the function g such that
f (g (x )) = g (f (x)).
Q
If f(x) = (ax2 + b)3, then find the function g such that
f (g (x )) = g (f (x)).
Solution:
Q
If f : [1, ∞) → [2, ∞) is given by
then f-1(x) is :
A
B
C
D
NOTE
Inverse of a function is unique.
Q
If f : [1, ∞) → [2, ∞) is given by
then f-1(x) is :
A
B
C
D
Solution:
Q
If f : [1, ∞) → [1, ∞) is defined as 2x(x - 1) , then f-1(x) is
A
B
C
D
Not defined
Q
If f : [1, ∞) → [1, ∞) is defined as 2x(x - 1) , then f-1(x) is
A
B
C
D
Not defined
Solution:
Now,
For x = 4, we get
Rejected
Thus
Q
For what values of ⍺ the function y = ⍺x + 3 is inverse of itself.
Q
For what values of ⍺ the function y = ⍺x + 3 is inverse of itself.
Solution:
By comparing coefficients
According to question
Taking common
Remark
Previous question can also be asked as:
Let f(x) = ⍺x + 3, then for what values of ‘⍺’, f(f(x)) = x , ∀ x.
JEE Main 25th Jan 2023 S-1
Q
For some a, b, c ∈ N, let f (x) = ax -3 and g(x) = xb + c, x ∊ R.
If
is equal to
, then (f o g)(ac) + (g o f)(b)
JEE Main 25th Jan 2023 S-1
Q
For some a, b, c ∈ N, let f (x) = ax -3 and g(x) = xb + c, x ∊ R.
If
is equal to
Ans : 2039
, then (f o g)(ac) + (g o f)(b)
JEE Main 25th Jan 2023 S-1
Q
For some a, b, c ∈ N, let f (x) = ax -3 and g(x) = xb + c, x ∊ R.
If
is equal to
Solution:
, then (f o g)(ac) + (g o f)(b)
Inverse of function
Remark
(fog)-1 = g-1 of-1
Inverse of function
NOTE
A function is invertible iff it is bijective.
Inverse of function
NOTE
A function is invertible iff it is bijective.
In this case f-1 is not a function.
Inverse of function
NOTE
A function is invertible iff it is bijective.
In this case f-1 is not a function.
JEE Main 25th July, 2021 S2
Q
Consider function f : A → B and g : B → C (A, B, C ⊆ R)
such that (gof)-1 exists then:
A
f and g both are one-one
B
f and g both are onto
C
f is one-one and g is onto
D
f is onto and g is one-one
JEE Main 25th July, 2021 S2
Q
Consider function f : A → B and g : B → C (A, B, C ⊆ R)
such that (gof)-1 exists then:
A
f and g both are one-one
B
f and g both are onto
C
f is one-one and g is onto
D
f is onto and g is one-one
JEE Main 25th July, 2021 S2
Q
Consider function f : A → B and g : B → C (A, B, C ⊆ R)
such that (gof)-1 exists then:
Solution:
∴ (gof)-1 exist ⇒ gof is bijective
⇒ ‘f’ must be one-one and ‘g’ must be ONTO
Inverse of function
NOTE
(a) Graph of y = f-1(x) is reflection of graph of y = f(x) about y = x.
For example.
(i) ex and ln x
(ii) y = x2 ; x ≥ 0 and
Inverse of function
NOTE
(a) Graph of y = f-1(x) is reflection of graph of y = f(x) about y = x.
For example.
(i) ex and ln x
(ii) y = x2 ; x ≥ 0 and
(b) Solutions of f(x) = x and f-1(x) = x are same.
Q
If
is defined as f(x) = x2 - 3x + 4, then
find solution of x = f -1 (x).
Q
If
is defined as f(x) = x2 - 3x + 4, then
find solution of x = f -1 (x).
Solution:
Solution of x = f -1 (x) is same as that of solution of x = f(x)
Functional Equation
Functional Equation
Q
Q
If f(x + y) = f(x) + f(y) ∀ x, y ∈ R and f(1) = 5 then find
If f : R ➝ R is a function satisfying f(x + y) = f(xy) for all x, y ∈ R
and
Q
If
Q
If g(x) g(y) = g(x) + g(y) + g(xy) - 2 ∀ x, y and g(2) = 5 then
, then find f(x)
find g(x) if its given to be a polynomial.
Slide 149
1
@kaliaiit@gmail.com
Sir, is this slide needed?
_Reassigned to Arvind Kalia_
anurag tiwari, 22-09-2023
1
yes
Arvind Kalia, 22-09-2023
Functional Equation
Result
Functional Equation
NOTE
In (e), f(x) is a polynomial is very important condition.
Normally it acts as indication of application of (e) result
Q
If f(x + y) = f(x) + f(y) ∀ x, y ∈ R and f(1) = 5 then find
Q
If f(x + y) = f(x) + f(y) ∀ x, y ∈ R and f(1) = 5 then find
Solution:
f(x) = Kx also f(1) = 5
⸫ 5 = K(1) ⇒ K = 5
⸫ f(x) = 5x
Now,
JEE Main 29th Jan, 2023 - S1
Q
Suppose f is a function satisfying f(x + y) = f(x) + f(y) for all
x, y ∈ N and
. If
then m is equal to____.
JEE Main 29th Jan, 2023 - S1
Q
Suppose f is a function satisfying f(x + y) = f(x) + f(y) for all
x, y ∈ N and
. If
then m is equal to____.
Ans : 10
Solution:
JEE Main 2020
Q
If f(x + y) = f(x) f(y) and
x, y ∈ N, where N
is the set of all natural numbers, then the value of is
A
B
C
D
Recall: f(x + y) = f(x) × f(y) ⇒ f(x) = ax
JEE Main 2020
Q
If f(x + y) = f(x) f(y) and
x, y ∈ N, where N
is the set of all natural numbers, then the value of is
A
B
C
D
Solution:
Q
If
, then find f(x)
Q
If
Solution:
Replace
Solving both we get
, then find f(x)
Q
If 2f(x) + f(1 - x) = x2 ∀ x then find f(x)
Q
If 2f(x) + f(1 - x) = x2 ∀ x then find f(x)
Solution:
2f(x) + f(1 - x) = x2
Replace x ➝ 1 - x
2f(1 - x) + f(x) = (1 - x)2
Solving both we get
3f(x) = 2x2 - (1 - x)2
3f(x) = x2 + 2x - 1
Q
∀ x ∈ R - {0},where f(x) be
a polynomial function and f(5) = 126, then f(3) =
A
28
B
26
C
27
D
25
Recall:
and f(x) is a polynomial then f(x) = ±xn + 1
Q
∀ x ∈ R - {0},where f(x) be
a polynomial function and f(5) = 126, then f(3) =
A
28
B
26
C
27
D
25
Q
∀ x ∈ R - {0},where f(x) be
a polynomial function and f(5) = 126, then f(3) =
Solution:
Q
If ‘f’ is polynomial such that
where (x ≠ 0, ± 1) and f(3) = 28, then find value of
Q
If ‘f’ is polynomial such that
where (x ≠ 0, ± 1) and f(3) = 28, then find value of
Ans: 5
Solution:
Q
If g(x) g(y) = g(x) + g(y) + g(xy) - 2 ∀ x, y and g(2) = 5
then find g(x), if its given to be a polynomial.
Recall:
and f(x) is a polynomial then f(x) = ±xn + 1
Q
If g(x) g(y) = g(x) + g(y) + g(xy) - 2 ∀ x, y and g(2) = 5
then find g(x), if its given to be a polynomial.
Solution:
Put x = 1, y = 2, we get
g(1) g(1) = g(1) + g(2) + g(2) - 2
5g(1) = g(1) + 5 + 5 - 2
4g(1) = 8
g(1) = 2
But g(2) = 5
Now, put
⸫n=2
g(x) = x2 + 1
Q
If f(x + y) + f(x - y) = 3f(x) f(y) ∀ x, y and f(0) ≠ 0 then
prove that f(x) is even.
Q
If f(x + y) + f(x - y) = 3f(x) f(y) ∀ x, y and f(0) ≠ 0 then
prove that f(x) is even.
Solution:
Put y = 0, we get
f(x) + f(x) = 3f(x) f(0)
2f(x) = 3f(x) f(0)
Now, put x = 0, we get
⸫ f(x) is even
Q
If f(k + x) = f(k - x) and f(2k + x) = -f(2k - x) for k > 0
then prove that f(x) is periodic.
Solution:
f(k + x) = f(k - x)
Now, f(2k + x) = -f(2k - x)
Put x = k + x in (1)
f(-(x - 2k)) = -f(x - 2k)
∴ f(2k + x) = f(-x)
∴ f(2k + x) = f(x - 2k)
Also, put x = k - x in (1)
Replace x ➝ x + 2k
We get, f(2k - x) = f(x) ……..(3)
f(x + 4k) = f(x)
But f(2k + x) = -f(2k - x)
Thus f(x) is periodic with period ‘4k’
Using (1) and (2)
∴ f(-x) = -f(x)
Means f(x) is odd function.
JEE Main 24th Jan, 2023 - S2
Q
is equal to
A
2011
B
2010
C
1010
D
1011
JEE Main 24th Jan, 2023 - S2
Q
is equal to
A
2011
B
2010
C
1010
D
1011
Solution: