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Class XII
Chapter 1 – Relations and Functions
Maths
Exercise 1
...
Also, R is not symmetric as (1, 3) ∈R, but (3, 1) ∉ R
...
[3(1) − 9 ≠ 0]
Hence, R is neither reflexive, nor symmetric, nor transitive
...
∴R is not reflexive
...
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Class XII
Chapter 1 – Relations and Functions
Maths
(1, 6) ∉ R
...
Now, since there is no pair in R such that (x, y) and (y, z) ∈R, then (x, z) cannot belong
to R
...
Hence, R is neither reflexive, nor symmetric, nor transitive
...
(x, x) ∈R
∴R is reflexive
...
[as 2 is not divisible by 4]
∴R is not symmetric
...
Then, y is divisible by x and z is divisible by y
...
⇒ (x, z) ∈R
∴R is transitive
...
(iv) R = {(x, y): x − y is an integer}
Now, for every x ∈ Z, (x, x) ∈R as x − x = 0 is an integer
...
Now, for every x, y ∈ Z if (x, y) ∈ R, then x − y is an integer
...
⇒ (y − x) is an integer
...
Now,
Let (x, y) and (y, z) ∈R, where x, y, z ∈ Z
...
⇒ x − z = (x − y) + (y − z) is an integer
...
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Class XII
Chapter 1 – Relations and Functions
Maths
∴ (x, z) ∈R
∴R is transitive
...
(v) (a) R = {(x, y): x and y work at the same place}
(x, x) ∈ R
∴ R is reflexive
...
⇒ y and x work at the same place
...
∴R is symmetric
...
⇒ x and z work at the same place
...
Hence, R is reflexive, symmetric, and transitive
...
∴ R is reflexive
...
⇒ y and x live in the same locality
...
Now, let (x, y) ∈ R and (y, z) ∈ R
...
⇒ x and z live in the same locality
...
Hence, R is reflexive, symmetric, and transitive
...
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...
Now, let (x, y) ∈R
...
Then, y is not taller than x
...
∴R is not symmetric
...
⇒ x is exactly 7 cm taller than y and y is exactly 7 cm taller than z
...
∴(x, z) ∉R
∴ R is not transitive
...
(d) R = {(x, y): x is the wife of y}
Now,
(x, x) ∉ R
Since x cannot be the wife of herself
...
Now, let (x, y) ∈ R
⇒ x is the wife of y
...
∴(y, x) ∉ R
Indeed if x is the wife of y, then y is the husband of x
...
Let (x, y), (y, z) ∈ R
⇒ x is the wife of y and y is the wife of z
...
Also, this does not imply that x is the wife of z
...
Hence, R is neither reflexive, nor symmetric, nor transitive
...
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Class XII
Chapter 1 – Relations and Functions
Maths
As x cannot be the father of himself
...
Now, let (x, y) ∈R
...
⇒ y cannot be the father of y
...
∴(y, x) ∉ R
∴ R is not symmetric
...
⇒ x is the father of y and y is the father of z
...
Indeed x is the grandfather of z
...
Hence, R is neither reflexive, nor symmetric, nor transitive
...
Answer
R = {(a, b): a ≤ b2}
It can be observed that
∴R is not reflexive
...
∴(4, 1) ∉ R
∴R is not symmetric
...
5) ∈ R
(as 3 < 22 = 4 and 2 < (1
...
25)
But, 3 > (1
...
25
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5) ∉ R
∴ R is not transitive
...
Question 3:
Check whether the relation R defined in the set {1, 2, 3, 4, 5, 6} as
R = {(a, b): b = a + 1} is reflexive, symmetric or transitive
...
A relation R is defined on set A as:
R = {(a, b): b = a + 1}
∴R = {(1, 2), (2, 3), (3, 4), (4, 5), (5, 6)}
We can find (a, a) ∉ R, where a ∈ A
...
It can be observed that (1, 2) ∈ R, but (2, 1) ∉ R
...
Now, (1, 2), (2, 3) ∈ R
But,
(1, 3) ∉ R
∴R is not transitive
Hence, R is neither reflexive, nor symmetric, nor transitive
...
Answer
R = {(a, b); a ≤ b}
Clearly (a, a) ∈ R as a = a
...
Now,
(2, 4) ∈ R (as 2 < 4)
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∴ R is not symmetric
...
Then,
a ≤ b and b ≤ c
⇒a≤c
⇒ (a, c) ∈ R
∴R is transitive
...
Question 5:
Check whether the relation R in R defined as R = {(a, b): a ≤ b3} is reflexive, symmetric
or transitive
...
Now,
(1, 2) ∈ R (as 1 < 23 = 8)
But,
(2, 1) ∉ R (as 23 > 1)
∴ R is not symmetric
...
Hence, R is neither reflexive, nor symmetric, nor transitive
...
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Class XII
Chapter 1 – Relations and Functions
Maths
Question 6:
Show that the relation R in the set {1, 2, 3} given by R = {(1, 2), (2, 1)} is symmetric
but neither reflexive nor transitive
...
A relation R on A is defined as R = {(1, 2), (2, 1)}
...
∴ R is not reflexive
...
Now, (1, 2) and (2, 1) ∈ R
However,
(1, 1) ∉ R
∴ R is not transitive
...
Question 7:
Show that the relation R in the set A of all the books in a library of a college, given by R
= {(x, y): x and y have same number of pages} is an equivalence relation
...
R = {x, y): x and y have the same number of pages}
Now, R is reflexive since (x, x) ∈ R as x and x has the same number of pages
...
⇒ y and x have the same number of pages
...
Now, let (x, y) ∈R and (y, z) ∈ R
...
⇒ x and z have the same number of pages
...
Hence, R is an equivalence relation
...
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Class XII
Chapter 1 – Relations and Functions
Maths
Question 8:
Show that the relation R in the set A = {1, 2, 3, 4, 5} given by
, is an equivalence relation
...
But no element of {1, 3, 5} is related to any element of 2, 4}
...
∴R is reflexive
...
∴R is symmetric
...
⇒ (a, c) ∈ R
∴R is transitive
...
Now, all elements of the set {1, 2, 3} are related to each other as all the elements of
this subset are odd
...
Similarly, all elements of the set {2, 4} are related to each other as all the elements of
this subset are even
...
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Class XII
Chapter 1 – Relations and Functions
Maths
Also, no element of the subset {1, 3, 5} can be related to any element of {2, 4} as all
elements of {1, 3, 5} are odd and all elements of {2, 4} are even
...
Question 9:
Show that each of the relation R in the set
, given by
(i)
(ii)
is an equivalence relation
...
Answer
(i)
For any element a ∈A, we have (a, a) ∈ R as
is a multiple of 4
...
Now, let (a, b) ∈ R ⇒
is a multiple of 4
...
Now, let (a, b), (b, c) ∈ R
...
Hence, R is an equivalence relation
...
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Class XII
Chapter 1 – Relations and Functions
Maths
(ii) R = {(a, b): a = b}
For any element a ∈A, we have (a, a) ∈ R, since a = a
...
Now, let (a, b) ∈ R
...
Now, let (a, b) ∈ R and (b, c) ∈ R
...
Hence, R is an equivalence relation
...
Hence, the set of elements related to 1 is {1}
...
Which is
(i) Symmetric but neither reflexive nor transitive
...
(iii) Reflexive and symmetric but not transitive
...
(v) Symmetric and transitive but not reflexive
...
Define a relation R on A as R = {(5, 6), (6, 5)}
...
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(5, 6), (6, 5) ∈ R, but (5, 5) ∉ R
∴R is not transitive
...
(ii)Consider a relation R in R defined as:
R = {(a, b): a < b}
For any a ∈ R, we have (a, a) ∉ R since a cannot be strictly less than a itself
...
∴ R is not reflexive
...
∴ (2, 1) ∉ R
∴ R is not symmetric
...
⇒ a < b and b < c
⇒a
∴R is transitive
...
(iii)Let A = {4, 6, 8}
...
e
...
Relation R is symmetric since (a, b) ∈ R ⇒ (b, a) ∈ R for all a, b ∈ R
...
Hence, relation R is reflexive and symmetric but not transitive
...
∴R is reflexive
...
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Class XII
Chapter 1 – Relations and Functions
Maths
But,
(1, 2) ∉ R (as 13 < 23)
∴ R is not symmetric
...
⇒ a3 ≥ b3 and b3 ≥ c3
⇒ a3 ≥ c 3
⇒ (a, c) ∈ R
∴R is transitive
...
(v) Let A = {−5, −6}
...
Relation R is symmetric as (−5, −6) ∈ R and (−6, −5}∈R
...
Also, (−5, −5) ∈ R
...
Hence, relation R is symmetric and transitive but not reflexive
...
Further, show that the set of all point related to a point P ≠ (0,
0) is the circle passing through P with origin as centre
...
∴R is reflexive
...
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⇒ The distance of point Q from the origin is the same as the distance of point P from the
origin
...
Now,
Let (P, Q), (Q, S) ∈ R
...
⇒ The distance of points P and S from the origin is the same
...
Therefore, R is an equivalence relation
...
In other words, if O (0, 0) is the origin and OP = k, then the set of all points related to P
is at a distance of k from the origin
...
Question 12:
Show that the relation R defined in the set A of all triangles as R = {(T1, T2): T1 is similar
to T2}, is equivalence relation
...
Which triangles among T1, T2 and T3
are related?
Answer
R = {(T1, T2): T1 is similar to T2}
R is reflexive since every triangle is similar to itself
...
⇒ T2 is similar to T1
...
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⇒ T1 is similar to T2 and T2 is similar to T3
...
⇒ (T1, T3) ∈ R
∴ R is transitive
...
Now, we can observe that:
∴The corresponding sides of triangles T1 and T3 are in the same ratio
...
Hence, T1 is related to T3
...
What is the set of all elements
in A related to the right angle triangle T with sides 3, 4 and 5?
Answer
R = {(P1, P2): P1 and P2 have same the number of sides}
R is reflexive since (P1, P1) ∈ R as the same polygon has the same number of sides with
itself
...
⇒ P1 and P2 have the same number of sides
...
⇒ (P2, P1) ∈ R
∴R is symmetric
...
⇒ P1 and P2 have the same number of sides
...
⇒ P1 and P3 have the same number of sides
...
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Class XII
Chapter 1 – Relations and Functions
Maths
∴R is transitive
...
The elements in A related to the right-angled triangle (T) with sides 3, 4, and 5 are
those polygons which have 3 sides (since T is a polygon with 3 sides)
...
Question 14:
Let L be the set of all lines in XY plane and R be the relation in L defined as R = {(L1,
L2): L1 is parallel to L2}
...
Find the set of all lines
related to the line y = 2x + 4
...
e
...
Now,
Let (L1, L2) ∈ R
...
⇒ L2 is parallel to L1
...
Now,
Let (L1, L2), (L2, L3) ∈R
...
Also, L2 is parallel to L3
...
∴R is transitive
...
The set of all lines related to the line y = 2x + 4 is the set of all lines that are parallel to
the line y = 2x + 4
...
The line parallel to the given line is of the form y = 2x + c, where c ∈R
...
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...
Choose the correct answer
...
(B) R is reflexive and transitive but not symmetric
...
(D) R is an equivalence relation
...
∴ R is reflexive
...
∴R is not symmetric
...
∴ R is transitive
...
The correct answer is B
...
Choose the
correct answer
...
We have 8 > 6 and also, 6 = 8 − 2
...
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2
Question 1:
Show that the function f: R* → R* defined by
is one-one and onto, where R* is
the set of all non-zero real numbers
...
Onto:
It is clear that for y∈ R*, there exists
such that
∴f is onto
...
Now, consider function g: N → R*defined by
We have,
∴g is one-one
...
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Class XII
Chapter 1 – Relations and Functions
Maths
Further, it is clear that g is not onto as for 1
...
Hence, function g is one-one but not onto
...
∴f is injective
...
But, there does not exist any x in N such that f(x) = x2 = 2
...
Hence, function f is injective but not surjective
...
∴ f is not injective
...
But, there does not exist any element x ∈Z such that f(x) = x2 = −2
...
Hence, function f is neither injective nor surjective
...
∴ f is not injective
...
But, there does not exist any element x ∈ R such that f(x) = x2 = −2
...
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Class XII
Chapter 1 – Relations and Functions
Maths
∴ f is not surjective
...
(iv) f: N → N given by,
f(x) = x3
It is seen that for x, y ∈N, f(x) = f(y) ⇒ x3 = y3 ⇒ x = y
...
Now, 2 ∈ N
...
∴ f is not surjective
Hence, function f is injective but not surjective
...
∴ f is injective
...
But, there does not exist any element x in domain Z such that f(x) = x3 = 2
...
Hence, function f is injective but not surjective
...
Answer
f: R → R is given by,
f(x) = [x]
It is seen that f(1
...
2] = 1, f(1
...
9] = 1
...
2) = f(1
...
2 ≠ 1
...
∴ f is not one-one
...
7 ∈ R
...
Thus, there does not exist any element x
∈ R such that f(x) = 0
...
∴ f is not onto
...
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Answer
f: R → R is given by,
It is seen that
...
∴ f is not one-one
...
It is known that f(x) =
is always non-negative
...
∴ f is not onto
...
Question 5:
Show that the Signum Function f: R → R, given by
is neither one-one nor onto
...
∴f is not one-one
...
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Class XII
Chapter 1 – Relations and Functions
Maths
Now, as f(x) takes only 3 values (1, 0, or −1) for the element −2 in co-domain R, there
does not exist any x in domain R such that f(x) = −2
...
Hence, the signum function is neither one-one nor onto
...
Show that f is one-one
...
f: A → B is defined as f = {(1, 4), (2, 5), (3, 6)}
...
Hence, function f is one-one
...
Justify your answer
...
...
For any real number (y) in R, there exists
in R such that
∴f is onto
...
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Class XII
Chapter 1 – Relations and Functions
Maths
Hence, f is bijective
...
...
Consider an element −2 in co-domain R
...
Thus, there does not exist any x in domain R such that f(x) = −2
...
Hence, f is neither one-one nor onto
...
Show that f: A × B → B × A such that (a, b) = (b, a) is bijective
function
...
...
Now, let (b, a) ∈ B × A be any element
...
[By definition of f]
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Hence, f is bijective
...
Justify your answer
...
Consider a natural number (n) in co-domain N
...
Then, there exists 4r + 1∈N such that
...
Then,there exists 4r ∈N such that
...
Hence, f is not a bijective function
...
Consider the function f: A → B defined by
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...
Is f one-one and onto? Justify your answer
...
...
Let y ∈B = R − {1}
...
The function f is onto if there exists x ∈A such that f(x) = y
...
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Class XII
Chapter 1 – Relations and Functions
Maths
Hence, function f is one-one and onto
...
Choose the correct answer
...
∴
does not imply that
...
Consider an element 2 in co-domain R
...
∴ f is not onto
...
The correct answer is D
...
Choose the correct answer
...
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Class XII
Chapter 1 – Relations and Functions
Maths
f: R → R is defined as f(x) = 3x
...
⇒ 3x = 3y
⇒x=y
∴f is one-one
...
∴f is onto
...
The correct answer is A
...
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Class XII
Chapter 1 – Relations and Functions
Maths
Exercise 1
...
Write down gof
...
Question 2:
Let f, g and h be functions from R to R
...
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Chapter 1 – Relations and Functions
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Question 3:
Find gof and fog, if
(i)
(ii)
Answer
(i)
(ii)
Question 4:
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What is the inverse of f?
Answer
It is given that
...
Question 5:
State with reason whether following functions have inverse
(i) f: {1, 2, 3, 4} → {10} with
f = {(1, 10), (2, 10), (3, 10), (4, 10)}
(ii) g: {5, 6, 7, 8} → {1, 2, 3, 4} with
g = {(5, 4), (6, 3), (7, 4), (8, 2)}
(iii) h: {2, 3, 4, 5} → {7, 9, 11, 13} with
h = {(2, 7), (3, 9), (4, 11), (5, 13)}
Answer
(i) f: {1, 2, 3, 4} → {10}defined as:
f = {(1, 10), (2, 10), (3, 10), (4, 10)}
From the given definition of f, we can see that f is a many one function as: f(1) = f(2) =
f(3) = f(4) = 10
∴f is not one-one
...
(ii) g: {5, 6, 7, 8} → {1, 2, 3, 4} defined as:
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∴g is not one-one,
Hence, function g does not have an inverse
...
∴Function h is one-one
...
Thus, h is a one-one and onto function
...
Question 6:
Show that f: [−1, 1] → R, given by
is one-one
...
(Hint: For y ∈Range f, y =
, for some x in [−1, 1], i
...
,
)
Answer
f: [−1, 1] → R is given as
Let f(x) = f(y)
...
It is clear that f: [−1, 1] → Range f is onto
...
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Class XII
Chapter 1 – Relations and Functions
Maths
f: [−1, 1] → Range f exists
...
Let y be an arbitrary element of range f
...
Show that f is invertible
...
Answer
f: R → R is given by,
f(x) = 4x + 3
One-one:
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∴ f is a one-one function
...
Therefore, for any y ∈ R, there exists
such that
∴ f is onto
...
Let us define g: R→ R by
...
Show that f is invertible with the inverse
f−1 of given f by
, where R+ is the set of all non-negative real numbers
...
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∴ f is a one-one function
...
Therefore, for any y ∈ R, there exists
such that
...
Thus, f is one-one and onto and therefore, f−1 exists
...
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Class XII
Let
Chapter 1 – Relations and Functions
Maths
...
The range of the principal value branch of
The correct answer is B
...
Show that f is invertible with
...
Let y be an arbitrary element of [−5, ∞)
...
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Let us define g: [−5, ∞) → R+ as
We now have:
∴
and
Hence, f is invertible and the inverse of f is given by
Question 10:
Let f: X → Y be an invertible function
...
(Hint: suppose g1 and g2 are two inverses of f
...
Use one-one ness of f)
...
Also, suppose f has two inverses (say
)
...
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Hence, f has a unique inverse
...
Find f−1 and
show that (f−1)−1 = f
...
Thus, the inverse of f exists and f−1 = g
...
e
...
If we define h: {1, 2, 3} → {a, b, c} as
h(1) = a, h(2) = b, h(3) = c, then we have:
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∴
Thus, the inverse of g exists and g−1 = h ⇒ (f−1)−1 = h
...
Hence, (f−1)−1 = f
...
Show that the inverse of f−1 is f, i
...
,
(f−1)−1 = f
...
Then, there exists a function g: Y → X such that gof = IXand fog = IY
...
Now, gof = IXand fog = IY
⇒ f−1of = IXand fof−1= IY
Hence, f−1: Y → X is invertible and f is the inverse of f−1
i
...
, (f−1)−1 = f
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Class XII
Chapter 1 – Relations and Functions
f: R → R is given as
Maths
...
Question 14:
Let
be a function defined as
...
Then, there exists x ∈
such that
Let us define g: Range
as
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e
...
Hence, the inverse of f is the map g: Range
, which is given by
The correct answer is B
...
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Class XII
Chapter 1 – Relations and Functions
Maths
Exercise 1
...
In the event that * is not a binary operation, give justification for this
...
It is not a binary operation as the image of (1, 2) under * is 1 * 2 = 1 − 2
= −1 ∉ Z+
...
It is seen that for each a, b ∈ Z+, there is a unique element ab in Z+
...
Therefore, * is a binary operation
...
It is seen that for each a, b ∈ R, there is a unique element ab2 in R
...
Therefore, * is a binary operation
...
It is seen that for each a, b ∈ Z+, there is a unique element |a − b| in Z+
...
Therefore, * is a binary operation
...
* carries each pair (a, b) to a unique element a * b = a in Z+
...
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(i) On Z, define a * b = a − b
(ii) On Q, define a * b = ab + 1
(iii) On Q, define a * b
(iv) On Z+, define a * b = 2ab
(v) On Z+, define a * b = ab
(vi) On R − {−1}, define
Answer
(i) On Z, * is defined by a * b = a − b
...
∴1 * 2 ≠ 2 * 1; where 1, 2 ∈ Z
Hence, the operation * is not commutative
...
(ii) On Q, * is defined by a * b = ab + 1
...
It can be observed that:
(1 * 2) * 3 = (1 × 2 + 1) * 3 = 3 * 3 = 3 × 3 + 1 = 10
1 * (2 * 3) = 1 * (2 × 3 + 1) = 1 * 7 = 1 × 7 + 1 = 8
∴(1 * 2) * 3 ≠ 1 * (2 * 3) ; where 1, 2, 3 ∈ Q
Therefore, the operation * is not associative
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(iii) On Q, * is defined by a * b
It is known that:
ab = ba a, b ∈ Q
⇒
a, b ∈ Q
⇒ a * b = b * a a, b ∈ Q
Therefore, the operation * is commutative
...
(iv) On Z+, * is defined by a * b = 2ab
...
It can be observed that:
∴(1 * 2) * 3 ≠ 1 * (2 * 3) ; where 1, 2, 3 ∈ Z+
Therefore, the operation * is not associative
...
It can be observed that:
and
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It can also be observed that:
∴(2 * 3) * 4 ≠ 2 * (3 * 4) ; where 2, 3, 4 ∈ Z+
Therefore, the operation * is not associative
...
It can also be observed that:
∴ (1 * 2) * 3 ≠ 1 * (2 * 3) ; where 1, 2, 3 ∈ R − {−1}
Therefore, the operation * is not associative
...
Write the operation table of the operation∨
...
Thus, the operation table for the given operation ∨ can be given as:
∨
1
2
3
4
5
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(i) Compute (2 * 3) * 4 and 2 * (3 * 4)
(ii) Is * commutative?
(iii) Compute (2 * 3) * (4 * 5)
...
Therefore, the operation * is
commutative
...
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∴(2 * 3) * (4 * 5) = 1 * 1 = 1
Question 5:
Let*′ be the binary operation on the set {1, 2, 3, 4, 5} defined by a *′ b = H
...
F
...
Is the operation *′ same as the operation * defined in Exercise 4 above? Justify
your answer
...
C
...
The operation table for the operation *′ can be given as:
*′
1
2
3
4
5
1
1
1
1
1
1
2
1
2
1
2
1
3
1
1
3
1
1
4
1
2
1
4
1
5
1
1
1
1
5
We observe that the operation tables for the operations * and *′ are the same
...
Question 6:
Let * be the binary operation on N given by a * b = L
...
M
...
Find
(i) 5 * 7, 20 * 16 (ii) Is * commutative?
(iii) Is * associative? (iv) Find the identity of * in N
(v) Which elements of N are invertible for the operation *?
Answer
The binary operation * on N is defined as a * b = L
...
M
...
(i) 5 * 7 = L
...
M
...
C
...
C
...
C
...
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(iii) For a, b, c ∈ N, we have:
(a * b) * c = (L
...
M of a and b) * c = LCM of a, b, and c
a * (b * c) = a * (LCM of b and c) = L
...
M of a, b, and c
∴(a * b) * c = a * (b * c)
Thus, the operation * is associative
...
C
...
of a and 1 = a = L
...
M
...
(v) An element a in N is invertible with respect to the operation * if there exists an
element b in N, such that a * b = e = b * a
...
C
...
C
...
Thus, 1 is the only invertible element of N with respect to the operation *
...
C
...
of a and b a binary operation?
Justify your answer
...
C
...
of a and b
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Class XII
Chapter 1 – Relations and Functions
4
4
4
12
4
20
5
5
10
15
20
Maths
5
It can be observed from the obtained table that:
3 * 2 = 2 * 3 = 6 ∉ A, 5 * 2 = 2 * 5 = 10 ∉ A, 3 * 4 = 4 * 3 = 12 ∉ A
3 * 5 = 5 * 3 = 15 ∉ A, 4 * 5 = 5 * 4 = 20 ∉ A
Hence, the given operation * is not a binary operation
...
C
...
of a and b
...
C
...
of a and b
It is known that:
H
...
F
...
C
...
of b and a a, b ∈ N
...
For a, b, c ∈ N, we have:
(a * b)* c = (H
...
F
...
C
...
of a, b, and c
a *(b * c)= a *(H
...
F
...
C
...
of a, b, and c
∴(a * b) * c = a * (b * c)
Thus, the operation * is associative
...
But this relation is not true for any a ∈ N
...
Question 9:
Let * be a binary operation on the set Q of rational numbers as follows:
(i) a * b = a − b (ii) a * b = a2 + b2
(iii) a * b = a + ab (iv) a * b = (a − b)2
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Answer
(i) On Q, the operation * is defined as a * b = a − b
...
It can also be observed that:
Thus, the operation * is not associative
...
For a, b ∈ Q, we have:
∴a * b = b * a
Thus, the operation * is commutative
...
(iii) On Q, the operation * is defined as a * b = a + ab
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Chapter 1 – Relations and Functions
Maths
It can be observed that:
Thus, the operation * is not commutative
...
(iv) On Q, the operation * is defined by a * b = (a − b)2
...
It can be observed that:
Thus, the operation * is not associative
...
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(vi) On Q, the operation * is defined as a * b = ab2
It can be observed that:
Thus, the operation * is not commutative
...
Hence, the operations defined in (ii), (iv), (v) are commutative and the operation defined
in (v) is associative
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Class XII
Chapter 1 – Relations and Functions
Maths
Question 10:
Find which of the operations given above has identity
...
However, there is no such element e ∈ Q with respect to each of the six operations
satisfying the above condition
...
Question 11:
Let A = N × N and * be the binary operation on A defined by
(a, b) * (c, d) = (a + c, b + d)
Show that * is commutative and associative
...
Answer
A=N×N
* is a binary operation on A and is defined by:
(a, b) * (c, d) = (a + c, b + d)
Let (a, b), (c, d) ∈ A
Then, a, b, c, d ∈ N
We have:
(a, b) * (c, d) = (a + c, b + d)
(c, d) * (a, b) = (c + a, d + b) = (a + c, b + d)
[Addition is commutative in the set of natural numbers]
∴(a, b) * (c, d) = (c, d) * (a, b)
Therefore, the operation * is commutative
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Class XII
Chapter 1 – Relations and Functions
Maths
Therefore, the operation * is associative
...
e
...
Therefore, the operation * does not have any identity element
...
Justify
...
(ii) If * is a commutative binary operation on N, then a * (b * c) = (c * b) * a
Answer
(i) Define an operation * on N as:
a*b=a+b
a, b ∈ N
Then, in particular, for b = a = 3, we have:
3*3=3+3=6≠3
Therefore, statement (i) is false
...
H
...
= (c * b) * a
= (b * c) * a [* is commutative]
= a * (b * c) [Again, as * is commutative]
= L
...
S
...
Question 13:
Consider a binary operation * on N defined as a * b = a3 + b3
...
(A) Is * both associative and commutative?
(B) Is * commutative but not associative?
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For, a, b, ∈ N, we have:
a * b = a3 + b3 = b3 + a3 = b * a [Addition is commutative in N]
Therefore, the operation * is commutative
...
Hence, the operation * is commutative, but not associative
...
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Find the function g: R → R such that g o f = f
o g = 1R
...
One-one:
Let f(x) = f(y), where x, y ∈R
...
Onto:
For y ∈ R, let y = 10x + 7
...
Therefore, f is one-one and onto
...
Let us define g: R → R as
Now, we have:
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Question 2:
Let f: W → W be defined as f(n) = n − 1, if is odd and f(n) = n + 1, if n is even
...
Find the inverse of f
...
Answer
It is given that:
f: W → W is defined as
One-one:
Let f(n) = f(m)
...
⇒n−m=2
However, this is impossible
...
∴Both n and m must be either odd or even
...
It is clear that any odd number 2r + 1 in co-domain N is the image of 2r in domain N
and any even number 2r in co-domain N is the image of 2r + 1 in domain N
...
Hence, f is an invertible function
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Class XII
Chapter 1 – Relations and Functions
Maths
Now, when n is odd:
And, when n is even:
Similarly, when m is odd:
When m is even:
∴
Thus, f is invertible and the inverse of f is given by f—1 = g, which is the same as f
...
Question 3:
If f: R → R is defined by f(x) = x2 − 3x + 2, find f(f(x))
...
Question 4:
Show that function f: R → {x ∈ R: −1 < x < 1} defined by f(x) =
, x ∈R is one-one
and onto function
...
Suppose f(x) = f(y), where x, y ∈ R
...
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Class XII
Chapter 1 – Relations and Functions
Maths
It can be observed that if x is positive and y is negative, then we have:
Since x is positive and y is negative:
x>y⇒x−y>0
But, 2xy is negative
...
Thus, the case of x being positive and y being negative can be ruled out
...
When x and y are both positive, we have:
When x and y are both negative, we have:
∴ f is one-one
...
If y is negative, then there exists
If y is positive, then there exists
such that
such that
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Hence, f is one-one and onto
...
Answer
f: R → R is given as f(x) = x3
...
⇒ x3 = y3 … (1)
Now, we need to show that x = y
...
x3 ≠ y3
However, this will be a contradiction to (1)
...
Question 6:
Give examples of two functions f: N → Z and g: Z → Z such that g o f is injective but g is
not injective
...
We first show that g is not injective
...
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Class XII
Chapter 1 – Relations and Functions
Maths
g(1) =
∴ g(−1) = g(1), but −1 ≠ 1
...
Now, gof: N → Z is defined as
...
⇒
Since x and y ∈ N, both are positive
...
(Hint: Consider f(x) = x + 1 and
Answer
Define f: N → N by,
f(x) = x + 1
And, g: N → N by,
We first show that g is not onto
...
It is clear that this element is not an image
of any of the elements in domain N
...
Now, gof: N → N is defined by,
Then, it is clear that for y ∈ N, there exists x = y ∈ N such that gof(x) = y
...
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Define the relation R in P(X) as follows:
For subsets A, B in P(X), ARB if and only if A ⊂ B
...
∴R is reflexive
...
This cannot be implied to B ⊂ A
...
∴ R is not symmetric
...
⇒A⊂C
⇒ ARC
∴ R is transitive
...
Question 9:
Given a non-empty set X, consider the binary operation *: P(X) × P(X) → P(X) given by
A * B = A ∩ B A, B in P(X) is the power set of X
...
Answer
It is given that
...
Thus, X is the identity element for the given binary operation *
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Class XII
Chapter 1 – Relations and Functions
Maths
This case is possible only when A = X = B
...
Hence, the given result is proved
...
Answer
Onto functions from the set {1, 2, 3, … ,n} to itself is simply a permutation on n symbols
1, 2, …, n
...
Question 11:
Let S = {a, b, c} and T = {1, 2, 3}
...
(i) F = {(a, 3), (b, 2), (c, 1)} (ii) F = {(a, 2), (b, 1), (c, 1)}
Answer
S = {a, b, c}, T = {1, 2, 3}
(i) F: S → T is defined as:
F = {(a, 3), (b, 2), (c, 1)}
⇒ F (a) = 3, F (b) = 2, F(c) = 1
Therefore, F−1: T → S is given by
F−1 = {(3, a), (2, b), (1, c)}
...
Hence, F is not invertible i
...
, F−1 does not exist
...
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Class XII
Chapter 1 – Relations and Functions
Maths
Consider the binary operations*: R ×R → and o: R × R → R defined as
and
a o b = a, a, b ∈ R
...
Further, show that a, b, c ∈ R, a*(b o c) = (a * b) o (a * c)
...
Does o distribute
over *? Justify your answer
...
For a, b ∈ R, we have:
∴a * b = b * a
∴ The operation * is commutative
...
Now, consider the operation o:
It can be observed that 1 o 2 = 1 and 2 o 1 = 2
...
Let a, b, c ∈ R
...
Now, let a, b, c ∈ R, then we have:
a * (b o c) = a * b =
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Now,
1 o (2 * 3) =
(1 o 2) * (1 o 3) = 1 * 1 =
∴1 o (2 * 3) ≠ (1 o 2) * (1 o 3) (where 1, 2, 3 ∈ R)
The operation o does not distribute over *
...
Show that the empty set Φ is the identity for the operation * and all
the elements A of P(X) are invertible with A−1 = A
...
Answer
It is given that *: P(X) × P(X) → P(X) is defined as
A * B = (A − B) ∪ (B − A) A, B ∈ P(X)
...
Then, we have:
A * Φ = (A − Φ) ∪ (Φ − A) = A ∪ Φ = A
Φ * A = (Φ − A) ∪ (A − Φ) = Φ ∪ A = A
∴A * Φ = A = Φ * A
...
Now, an element A ∈ P(X) will be invertible if there exists B ∈ P(X) such that
A * B = Φ = B * A
...
Hence, all the elements A of P(X) are invertible with A
−1
= A
...
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Class XII
Chapter 1 – Relations and Functions
Maths
Show that zero is the identity for this operation and each element a ≠ 0 of the set is
invertible with 6 − a being the inverse of a
...
The operation * on X is defined as:
An element e ∈ X is the identity element for the operation *, if
Thus, 0 is the identity element for the given operation *
...
i
...
,
a = −b or b = 6 − a
But, X = {0, 1, 2, 3, 4, 5} and a, b ∈ X
...
∴b = 6 − a is the inverse of a a ∈ X
...
e
...
Question 15:
Let A = {−1, 0, 1, 2}, B = {−4, −2, 0, 2} and f, g: A → B be functions defined by f(x) =
x2 − x, x ∈ A and
...
(Hint: One may note that two function f: A → B and g: A → B such
that f(a) = g(a) a ∈A, are called equal functions)
...
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Class XII
Chapter 1 – Relations and Functions
Maths
It is given that A = {−1, 0, 1, 2}, B = {−4, −2, 0, 2}
...
It is observed that:
Hence, the functions f and g are equal
...
Then number of relations containing (1, 2) and (1, 3) which are
reflexive and symmetric but not transitive is
(A) 1 (B) 2 (C) 3 (D) 4
Answer
The given set is A = {1, 2, 3}
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Chapter 1 – Relations and Functions
Maths
R = {(1, 1), (2, 2), (3, 3), (1, 2), (1, 3), (2, 1), (3, 1)}
This is because relation R is reflexive as (1, 1), (2, 2), (3, 3) ∈ R
...
But relation R is not transitive as (3, 1), (1, 2) ∈ R, but (3, 2) ∉ R
...
Hence, the total number of desired relations is one
...
Question 17:
Let A = {1, 2, 3}
...
The smallest equivalence relation containing (1, 2) is given by,
R1 = {(1, 1), (2, 2), (3, 3), (1, 2), (2, 1)}
Now, we are left with only four pairs i
...
, (2, 3), (3, 2), (1, 3), and (3, 1)
...
Also,
for transitivity we are required to add (1, 3) and (3, 1)
...
This shows that the total number of equivalence relations containing (1, 2) is two
...
Question 18:
Let f: R → R be the Signum Function defined as
and g: R → R be the Greatest Integer Function given by g(x) = [x], where [x] is
greatest integer less than or equal to x
...
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Chapter 1 – Relations and Functions
Maths
f: R → R is defined as
Also, g: R → R is defined as g(x) = [x], where [x] is the greatest integer less than or
equal to x
...
Then, we have:
[x] = 1 if x = 1 and [x] = 0 if 0 < x < 1
...
Hence, fog and gof do not coincide in (0, 1]
...
e
...
Hence, the total number of binary operations on the set {a, b} is 24 i
...
, 16
...
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