ncert solutions for class 12 subject-wise...7/4/2018
Post on 05-Apr-2020
9 Views
Preview:
TRANSCRIPT
NCERT Solutions for Class 12 Subject-wise ● Class 12 Mathematics
● Class 12 Physics
● Class 12 Chemistry
● Class 12 Biology
● Class 12 Accountancy
● Class 12 Business Studies
● Class 12 English
7/4/2018 https://community.toppr.com/content/questions/print/?show_answer=1&show_topic=1&show_solution=1&page=1&qid=421842%2C+4278…
https://community.toppr.com/content/questions/print/?show_answer=1&show_topic=1&show_solution=1&page=1&qid=421842%2C+427819%2C+42… 1/22
#421228
Topic: Types of Relations
Determine whether each of the following relations are reflexive, symmetric and transitive:
(i) Relation in the set defined as
(ii) Relative in the set of natural numbers defined as
(iii) Relation in the set as
(iv) Relative in the set of all integers defined as
(v) Relation R in the set A of human beings in a town at a particular time given by
Solution
(i)
is not reflexive since
Also, is not symmetric as , but
Also, is not transitive as , but .
Hence, is neither reflexive, nor symmetric, nor transitive.
(ii)
It is seen that is not reflexive.
Also .
But, . is not symmetric.
Now, since there is no pair in such that and , then cannot belong to .
is not transitive.
Hence, is neither reflexive, nor symmetric, nor transitive.
(iii)
We know that any number is divisible by itself.
,
is reflexive.
Now, , [ as is divisible by ]
But,
, [ as is not divisible by ]
is not symmetric.
Let . Then, is divisible by and is divisible by .
is divisible by .
R A = {1, 2, 3, . . . , 13, 14}
R = {(x, y) : 3x− y = 0}
R N
R = {(x, y) : y = x+ 5 andx < 4}
R A = {1, 2, 3, 4, 5, 6}
R = {(x, y) : yis divisible byx}
R Z
R = {(x, y) : x− y is an integer}
(a)R = {(x, y) : xandywork at the same place }
(b)R = {(x, y) : xandylive in the same locality}
(c)R = {(x, y) : xis exactly7cm taller thany}
(d)R = {(x, y) : xis wife ofy}
(e)R = {(x, y) : x is father ofy}
A = {1, 2, 3, . . . . 13, 14}
R = {(x, y) : 3x− y = 0}
∴ R = {(1, 3), (2, 6), (3, 9), (4, 12)}
R (1, 1), (2, 2). . . . . . (14, 14) ∉ R
R (1, 3) ∈ R (3, 1) ∉ R.
R (1, 3), (3, 9) ∈ R (1, 9) ∉ R
R
R = {(x, y) : y = x+ 5andx < 4} = {(1, 6), (2, 7), (3, 8)}
(1, 1) ∉ R ⇒ R
(1, 6) ∈ R
(6, 1) ∉ R ∴ R
R (x, y) (y, z) ∈ R (x, z) R
∴ R
R
A = {1, 2, 3, 4, 5, 6}
R = {(x, y) : y is divisible by x}
(x)
⇒ (x, x) ∈ R
∴ R
(2, 4) ∈ R 4 2
(4, 2) ∉ R. 2 4
∴ R
(x, y), (y, z) ∈ R y x z y
∴ z x
7/4/2018 https://community.toppr.com/content/questions/print/?show_answer=1&show_topic=1&show_solution=1&page=1&qid=421842%2C+4278…
https://community.toppr.com/content/questions/print/?show_answer=1&show_topic=1&show_solution=1&page=1&qid=421842%2C+427819%2C+42… 2/22
is transitive.
Hence, is reflexive and transitive but not symmetric.
(iv)
Now, for every as is an integer.
is reflexive.
Now, for every if , then is an integer.
is also an integer.
is an integer.
is symmetric.
Now,
Let and , where .
and are integers.
is an integer.
is transitive.
Hence, is reflexive, symmetric, and transitive.
(v) (a)
is reflexive.
If , then and work at the same place.
and work at the same place.
is symmetric.
Now, let
and work at the same place and and work at the same place.
and work at the same place.
is transitive.
Hence, is reflexive, symmetric and transitive.
(v) (b)
Clearly as and is the same human being.
is reflexive.
If , then and live in the same locality.
and live in the same locality.
is symmetric.
Now, let and .
and live in the same locality and and live in the same locality.
and live in the same locality.
⇒ (x, z) ∈ R
∴ R
R
R = {(x, y) : x− yis an integer}
x ∈ Z, (x, x) ∈ R x− x = 0
∴ R
x, y ∈ Z (x, y) ∈ R x− y
⇒ −(x− y)
⇒ (y − x)
∴ (y, x) ∈ R
⇒ R
(x, y) (y, z) ∈ R x, y, z ∈ Z
⇒ (x− y) (y − z)
⇒ x− z = (x− y) + (y − z)
∴ (x, z) ∈ R
∴ R
R
R = {(x, y) : x and y work at the same place}
⇒ (x, x) ∈ R
∴ R
(x, y) ∈ R x y
⇒ y x
⇒ (y, x) ∈ R
∴ R
(x, y), (y, z) ∈ R
⇒ x y y z
⇒ x z
⇒ (x, z) ∈ R
∴ R
R
R = {(x, y) : x and y live in the same locality}
(x, y) ∈ R x x
∴ R
(x, y) ∈ R x y
⇒ y x
⇒ (y, x) ∈ R
∴ R
(x, y) ∈ R (y, z) ∈ R
⇒ x y y z
⇒ x z
⇒ (x, z) ∈ R
7/4/2018 https://community.toppr.com/content/questions/print/?show_answer=1&show_topic=1&show_solution=1&page=1&qid=421842%2C+4278…
https://community.toppr.com/content/questions/print/?show_answer=1&show_topic=1&show_solution=1&page=1&qid=421842%2C+427819%2C+42… 3/22
is transitive.
Hence, is reflexive, symmetric and transitive.
(v)(c)
Now,
Since human being cannot be taller than himself.
is not reflexive.
Now, let .
is exactly cm taller than
Then, is not taller than .
Indeed if is exactly cm taller than , then is exactly cm shorter than .
is not symmetric.
Now, let
is exactly cm taller than and is exactly cm taller than .
is exactly taller than .
is not transitive.
Hence, is neither reflexive, nor symmetric, nor transitive.
(v)(d)
Now,
Since cannot be the wife of herself.
is not reflexive.
Now, let
is the wife of .
Clearly is not wife of .
Indeed if is the wife of , then is the husband of .
is not transitive.
Let
is the wife of and is the wife of .
This case is not possible. Also, this does not imply that is the wife of .
is not transitive.
Hence, is neither reflexive, nor symmetric, nor transitive.
(v)(e)
As cannot be the father of himself.
is not reflexive.
Now, let
is the father of .
cannot be the father of .
Indeed, is the son or the daughter of .
∴ R
R
R = {(x, y) : x is exactly 7 cm taller than y}
(x, x) ∉ R
x
∴ R
(x, y) ∈ R
⇒ x 7 y
y x
∴ (y, x) ∉ R
x 7 y y 7 x
∴ R
(x, y), (y, z) ∈ R
⇒ x 7 y y 7 z
⇒ x 14 z
∴ (x, z) ∉ R
∴ R
R
R = {(x, y) : x is the wife of y}
(x, x) ∉ R
x
∴ R
(x, y) ∈ R
⇒ x y
y x
∴ (y, x) ∉ R
x y y x
∴ R
(x, y), (y, z) ∈ R
⇒ x y y z
x z
∴ (x, z) ∉ R
∴ R
R
R = {(x, y) : x is the father of y}
(x, x) ∉ R
x
∴ R
(x, y) ∈ R
⇒ x y
⇒ y y
y y
7/4/2018 https://community.toppr.com/content/questions/print/?show_answer=1&show_topic=1&show_solution=1&page=1&qid=421842%2C+4278…
https://community.toppr.com/content/questions/print/?show_answer=1&show_topic=1&show_solution=1&page=1&qid=421842%2C+427819%2C+42… 4/22
is not symmetric.
Now, let and .
is the father of and is the father of .
is not the father of .
Indeed is the grandfather of .
is not transitive.
Hence, is neither reflexive, nor symmetric, nor transitive.
#421245
Topic: Types of Relations
Show that the relation in the set of real numbers, defined as is neither reflexive nor symmetric nor transitive.
Solution
It can be observed that , since .
is not reflexive.
Now, as
But, is not less than .
is not symmetric.
Now,
But,
is not transitive.
Hence, is neither reflexive, nor symmetric, nor transitive.
#421842
Topic: Types of Relations
Check whether the relation defined in the set as is reflexive, symmetric or transitive.
Solution
Let .
A relative is defined on set as:
We can find , where .
For instance,
is not reflexive.
It can be observed that , but .
is not symmetric.
Now,
But,
is not transitive.
Hence, is neither reflexive, nor symmetric, nor transitive.
#421846
Topic: Types of Relations
∴ (y, x) ∉ R
∴ R
(x, y) ∈ R (y, z) ∈ R
⇒ x y y z
⇒ x z
x z
∴ (x, z) ∉ R
∴ R
R
R R R = {(a, b) : a ≤ }b2
R = {(a, b) : a ≤ }b2
( , ) ∉ R1
2
1
2> =
1
2( )
1
2
2 1
4
∴ R
(1, 4) ∈ R 1 < 42
4 12
∴ (4, 1) ∉ R
∴ R
(3, 2), (2, 1.5) ∈ R
(as3 < = 4and2 < (1.5 = 2.25)22 )2
3 > (1.5 = 2.25)2
∴ (3, 1.5) ∉ R
∴ R
R
R {1, 2, 3, 4, 5, 6} R = {(a, b) : b = a+ 1}
A = {1, 2, 3, 4, 5, 6}
R A
R = {(a, b) : b = a+ 1}
∴ R = {(1, 2), (2, 3), (3, 4), (4, 5), (5, 6)}
(a, a) ∉ R a ∈ A
(1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6) ∉ R
∴ R
(1, 2) ∈ R (2, 1) ∉ R
∴ R
(1, 2), (2, 3) ∈ R
(1, 3) ∉ R
∴ R
R
7/4/2018 https://community.toppr.com/content/questions/print/?show_answer=1&show_topic=1&show_solution=1&page=1&qid=421842%2C+4278…
https://community.toppr.com/content/questions/print/?show_answer=1&show_topic=1&show_solution=1&page=1&qid=421842%2C+427819%2C+42… 5/22
Show that the relative in defined as , is reflexive and transitive but not symmetric.
Solution
Clearly as .
is reflexive.
Now,
But, as is greater than .
is not symmetric.
Now, let .
Then, and
is transitive.
Hence, is reflexive and transitive but not symmetric.
#421847
Topic: Types of Relations
Check whether the relative in defined by is reflexive, symmetric or transitive.
Solution
It is observed that as .
is not reflexive.
Now,
But,
is not symmetric.
We have as and .
But as .
is not transitive.
Hence, is neither reflexive, nor symmetric, nor transitive.
#421848
Topic: Types of Relations
Show that the relative in the set given by is symmetric but neither reflexive nor transitive.
Solution
Let
A relation on is defined as .
It is seen that .
is not reflexive.
Now, as and , then is symmetric.
Now, and
However,
is not transitive.
Hence, is symmetric but neither reflexive nor transitive.
#421849
Topic: Types of Relations
R R R = {(a, b) : a ≤ b}
R = {(a, b); a ≤ b}
(a, a) ∈ R a = a
∴ R
(2, 4) ∈ R(as2 < 4)
(4, 2) ∉ R 4 2
∴ R
(a, b), (b, c) ∈ R
a ≤ b b ≤ c
⇒ a ≤ c
⇒ (a, c) ∈ R
∴ R
R
R R R = {(a, b) : a ≤ }b3
R = {(a, b) : a ≤ }b3
( , ) ∉ R1
2
1
2> =
1
2( )
1
2
3 1
8
∴ R
(1, 2) ∈ R (as 1 < = 8)23
(2, 1) ∉ R(as 2 > )13
∴ R
(3, ) ,( , ) ∈ R3
2
3
2
6
53 < ( )
3
2
3
<3
2( )
6
5
3
(3, ) ∉ R6
53 > ( )6
5
3
∴ R
R
R {1, 2, 3} R = {(1, 2), (2, 1)}
A = {1, 2, 3}
R A R = {(1, 2), (2, 1)}
(1, 1), (2, 2), (3, 3) ∉ R
∴ R
(1, 2) ∈ R (2, 1) ∈ R R
(1, 2) (2, 1) ∈ R
(1, 1) ∉ R
∴ R
R
7/4/2018 https://community.toppr.com/content/questions/print/?show_answer=1&show_topic=1&show_solution=1&page=1&qid=421842%2C+4278…
https://community.toppr.com/content/questions/print/?show_answer=1&show_topic=1&show_solution=1&page=1&qid=421842%2C+427819%2C+42… 6/22
Show that the relation in the set of all the books in a library of a college, given by is an equivalence relation.
Solution
Set is the set of all books in the library of a college.
Now, is reflexive since as and has the same number of pages.
Let and have the same number of pages.
and have the same number of pages.
is symmetric.
Now, let and
and and have the same number of pages and and have the same number of pages.
and have the same numbers of pages.
is transitive.
Hence, is an equivalence relation.
#421850
Topic: Types of Relations
Show that the relative in the set given by , is an equivalence relation. Show that all the elements of are related to
each other and all the elements of are related to each other. But no element of is related to any element of .
Solution
It is clear that for any element , we have (which is even).
is reflexive.
Let
is even.
is also even.
is symmetric.
Now, let and .
is even and is even.
is even and is even.
is even. [Sum of two even integers is even]
is even.
is transitive.
Hence, is an equivalence relation.
Now, all elements of the set are related to each other as all the elements of this subset are odd. Thus, the modulus of the difference between any two elements will be
even.
Similarly, all elements of the set are related to each other as all the elements of this subset are even.
Also, no element of the subset can be related to any element of as all elements of are odd and all elements of are even. Thus, the modulus of
the difference between the two elements (from each of these two subsets) will not be even.
Hence, is an equivalence relation.
#421854
Topic: Types of Relations
R A R = {(x, y) : x and yhave same number of pages}
A
R = {(x, y) : x and y have the same number of pages}
R (x, x) ∈ R x x
(x, y) ∈ R ⇒ x y
⇒ y x
⇒ (y, x) ∈ R
∴ R
(x, y) ∈ R (y, z) ∈ R
⇒ x y y z
⇒ x z
⇒ (x, z) ∈ R
∴ R
R
R A = {1, 2, 3, 4, 5} R = {(a, b) : |a− b| is even} {1, 3, 5}
{2, 4} {1, 3, 5} {2, 4}
A = {1, 2, 3, 4, 5}
R = {(a, b) : |a− b|is even}
a ∈ A |a− a| = 0
∴ R
(a, b) ∈ R
⇒ |a− b|
⇒ | − (a− b)| = |b− a|
⇒ (b, a) ∈ R
∴ R
(a, b) ∈ R (b, c) ∈ R
⇒ |a− b| |b− c|
⇒ (a− c) (b− c)
⇒ (a− c) = (a− b) + (b− c)
⇒ |a− c|
⇒ (a, c) ∈ R
∴ R
R
{1, 2, 3}
{2, 4}
{1, 3, 5} {2, 4} {1, 3, 5} {2, 4}
R
7/4/2018 https://community.toppr.com/content/questions/print/?show_answer=1&show_topic=1&show_solution=1&page=1&qid=421842%2C+4278…
https://community.toppr.com/content/questions/print/?show_answer=1&show_topic=1&show_solution=1&page=1&qid=421842%2C+427819%2C+42… 7/22
Show that each of the relation in the set , given by
(i)
(ii)
is an equivalence relation. Find the set of all elements related to in each case.
Solution
(i)
For any element , we have as is a multiple of .
is reflexive.
Now, let is a multiple of .
is a multiple of .
is symmetric.
Now, let
is a multiple of and is a multiple of .
is a multiple of and is a multiple of .
is a multiple of .
is a multiple of .
.
is transitive.
Hence, is an equivalence relation.
The set of elements related to is since
is a multiple of ,
is a multiple of , and
is a multiple of
Hence, is an equivalence relation.
(ii)
For any element , we have , since .
is reflexive.
Now, let .
is symmetric.
Now, let and .
and
is transitive.
Hence, is an equivalence relation.
The elements in that are related to will be those elements from set which are equal to .
Hence, the set of elements related to is .
#421860
Topic: Types of Relations
R A = {x ∈ Z : 0 ≤ x ≤ 12}
R = {(a, b) : |a− b|is a multiple of 4}
R = {(a, b) : a = b}
1
A = {x ∈ Z : 0 ≤ x ≤ 12} = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12}
R = {(a, b) : |a− b|is a multiple of 4}
a ∈ A (a, a) ∈ R |a− a| = 0 4
∴ R
(a, b) ∈ R ⇒ |a− b| 4
⇒ | − (a− b)| = |b− a| 4
⇒ (b, a) ∈ R
∴
(a, b), (b, c) ∈ R
⇒ |a− b| 4 |b− c| 4
⇒ (a− b) 4 (b− c) 4
⇒ (a− c) = (a− b) + (b− c) 4
⇒ |a− c| 4
⇒ (a, c) ∈ R
∴ R
R
1 {1, 5, 9}
|1 − 1| = 0 4
|5 − 1| = 4 4
|9 − 1| = 8 4
R
A = {x ∈ Z : 0 ≤ x ≤ 12} = {0, 1, 2, 3, 4, . . . . . . , 11, 12}
R = {(a, b) : a = b} = {(0, ), (1, 1), (2, 2), . . . . . . . . , (11, 11), (12, 12)}
a ∈ A (a, a) ∈ R a = a
∴ R
(a, b) ∈ R
⇒ a = b
⇒ b = a
⇒ (b, a) ∈ R
∴ R
(a, b) ∈ R (b, c) ∈ R
⇒ a = b b = c
⇒ a = c
⇒ (a, c) ∈ R
∴ R
R
R 1 A 1
1 {1}
7/4/2018 https://community.toppr.com/content/questions/print/?show_answer=1&show_topic=1&show_solution=1&page=1&qid=421842%2C+4278…
https://community.toppr.com/content/questions/print/?show_answer=1&show_topic=1&show_solution=1&page=1&qid=421842%2C+427819%2C+42… 8/22
Given an example of a relation. Which is
(i) Symmetric but neither reflexive nor transitive.
(ii) Transitive but neither reflexive nor symmetric.
(iii) Reflexive and symmetric but not transitive.
(iv) Reflexive and transitive but not symmetric.
(v) Symmetric and transitive but not reflexive.
Solution
(i)
Let
Define a relation on as
Relation is not reflexive as .
Now, as and also , is symmetric.
, but
is not transitive.
Hence, relation is symmetric but not reflexive or transitive.
(ii)
Consider a relation in defined as
For any , we have since cannot be strictly less than a itself. In fact, .
is not reflexive.
Now,
But, is not less than .
is not symmetric.
Now, let .
and
is transitive.
Hence, relation is transitive but not reflexive and symmetric.
(iii)
Let
Define a relation on as:
Relation is reflexive since for every
Relation is symmetric since for all .
Relation is not transitive since , but .
Hence, relation is reflexive and symmetric but not transitive.
(iv)
Define a relation in as:
Clearly as .
is reflexive.
Now,
But,
is not symmetric.
A = {5, 6, 7}
R A R = {(5, 6), (6, 5)}
R (5, 5), (6, 6), (7, 7) ∉ R
(5, 6) ∈ R (6, 5) ∈ R R
⇒ (5, 6), (6, 5) ∈ R (5, 5) ∉ R
∴ R
R
R R
R = {(a, b) : a < b}
a ∈ R (a, a) ∉ R a a = a
∴ R
(1, 2) ∈ R(as1 < 2)
2 1
∴ (2, 1) ∉ R
∴ R
(a, b), (b, c) ∈ R
⇒ a < b b < c
⇒ a < c
⇒ (a, c) ∈ R
∴ R
R
A = {4, 6, 8}
R A
A = {(4, 4), (6, 6), (8, 8), (4, 6), (6, 4), (6, 8), (8, 6)}
R {a ∈ A, (a, a) ∈ Ri. e. , (4, 4), (6, 6), (8, 8)} ∈ R
R (a, b) ∈ R ⇒ (b, a) ∈ R a, b ∈ R
R (4, 6), (6, 8) ∈ R (4, 8) ∉ R
R
R R
R = {(a, b) : ≥ }a3 b3
(a, a) ∈ R =a3 a3
∴ R
(2, 1) ∈ R(as ≥ )23 13
(1, 2) ∉ R(as < )13 23
∴ R
7/4/2018 https://community.toppr.com/content/questions/print/?show_answer=1&show_topic=1&show_solution=1&page=1&qid=421842%2C+4278…
https://community.toppr.com/content/questions/print/?show_answer=1&show_topic=1&show_solution=1&page=1&qid=421842%2C+427819%2C+42… 9/22
Let
and
is transitive.
Hence, relation is reflexive and transitive but not symmetric.
(v)
Let .
Define a relation on as:
Relation is not reflexive as
Relation is symmetric as and
It is seen that .
Also, .
the relation is transitive.
Hence, relation is symmetric and transitive but not reflexive.
#423434
Topic: Types of Relations
Show that the relation in the set of points in a plane given by
, is an equivalence relation. Further, show that the
set of all point related to a point is the circle passing through with origin as centre.
Solution
Clearly, since the distance of point from the origin is always the same as the distance of the same point from the origin.
is reflexive.
Now, let .
the distance of point from the origin is the same as the distance of point from the origin.
The distance of point from the origin is the same as the distance of point from the origin.
is symmetric.
Now, let .
The distance of points and from the origin is the same and also, the distance of points and from the origin is the same.
The distance of points and from the origin is the same.
.
is transitive.
Therefore, is an equivalence relation.
The set of all points related to will be those points whose distance from the origin is the same as the distance of point from the origin.
In other words, if is the origin and , then the set of all points related to is at a distance of from the origin.
Hence, this set of points forms a circle with the centre as the origin and this circle passes through point .
#423437
Topic: Types of Relations
Show that the relation defined in the set of all triangles as , is equivalence relation. Consider three right angle triangles with sides
with sides and with sides . Which triangles among and are related?
Solution
(a, b), (b, c) ∈ R
⇒ ≥a3 b3 ≥b3 c3
⇒ ≥a3 c3
⇒ (a, c) ∈ R
∴ R
R
A = {−5, −6}
R A
R = {(−5, −6), (−6, −5), (−5, −5)}
R (−6, −6) ∉ R
R (−5, −6) ∈ R (−6, −5) ∈ R
(−5, −6), (−6, −5) ∈ R
(−5, −5) ∈ R
∴ R
R
R A
R = {(P ,Q) : distance of the point P from the origin is same as the distance of the point Q from the origin}
P ≠ (0, 0) P
R = {(P ,Q) : distance of point P from the origin is the same as the distance of point Q from the origin}
(P ,P ) ∈ R P P
∴ R
(P ,Q) ∈ R
⇒ P Q
⇒ Q P
⇒ (Q,P ) ∈ R
∴ R
(P ,Q), (Q, S) ∈ R
⇒ P Q Q S
⇒ P S
⇒ (P , S) ∈ R
∴ R
R
P ≠ (0, 0) P
O(0, 0) OP = k P k
P
R A R = {( , ) : is similar to }T1 T2 T1 T2 T1
3, 4, 5, T2 5, 12, 13 T3 6, 8, 10 ,T1 T2 T3
7/4/2018 https://community.toppr.com/content/questions/print/?show_answer=1&show_topic=1&show_solution=1&page=1&qid=421842%2C+4278…
https://community.toppr.com/content/questions/print/?show_answer=1&show_topic=1&show_solution=1&page=1&qid=421842%2C+427819%2C+4… 10/22
is reflexive since every triangle is similar to itself.
Further, if , then is similar to .
is similar to .
is symmetric.
Now,
Let .
is similar to and is similar to .
is similar to .
is transitive.
Thus, is an equivalence relation.
Now, we can observe that:
The corresponding sides of triangles and are in the same ratio.
Then, triangle is similar to triangle .
Hence, is related to .
#423440
Topic: Types of Relations
Show that the relation defined in the set of all polygons as , is an equivalence relation. What is the set of all
elements in related to the right angle triangle with sides and ?
Solution
is reflexive since as the same polygon has the same number of sides with itself.
Let .
and have the same number of sides.
and have the same number of sides.
is symmetric.
Now,
Let .
and have the same number of sides. Also, and have the same number of sides.
and have the same number of sides.
is transitive.
Hence, is an equivalence relation.
The elements in related to the right-angled triangle with sides and are those polygons which have sides (since is a polygon with sides).
Hence, the set of all elements in related to triangle is the set of all triangles.
#423441
Topic: Types of Relations
Let be the set of all lines in plane and be the relation in defined as . Show that is an equivalence relation. Find the set of
all lines related to the line .
Solution
R = {( , ) : is similar to }T1 T2 T1 T2
R
( , ) ∈ RT1 T2 T1 T2
⇒ T2 T1
⇒ ( , ) ∈ RT2 T1
∴ R
( , ), ( , ) ∈ RT1 T2 T2 T3
⇒ T1 T2 T2 T3
⇒ T1 T3
⇒ ( , ) ∈ RT1 T3
∴ R
R
= = (= )3
6
4
8
5
10
1
2
∴ T1 T3
T1 T3
T1 T3
R A R = {( , ) : and have same number of sides}P1 P2 P1 P2
A T 3, 4 5
R = {( , ) : and have same the number of sides}P1 P2 P1 P2
R ( , ) ∈ RP1 P1
( , ) ∈ RP1 P2
⇒ P1 P2
⇒ P2 P1
⇒ ( , ) ∈ RP2 P1
∴ R
( , ), ( , ) ∈ RP1 P2 P2 P3
⇒ P1 P2 P2 P3
⇒ P1 P3
⇒ ( , ) ∈ RP1 P3
∴ R
R
A (T ) 3, 4, 5 3 T 3
A T
L XY R L R = {( , ) : is parallel to }L1 L2 L1 L2 R
y = 2x+ 4
7/4/2018 https://community.toppr.com/content/questions/print/?show_answer=1&show_topic=1&show_solution=1&page=1&qid=421842%2C+4278…
https://community.toppr.com/content/questions/print/?show_answer=1&show_topic=1&show_solution=1&page=1&qid=421842%2C+427819%2C+4… 11/22
is reflexive as any line is parallel to itself i.e., .
Now, let .
is parallel to .
is parallel to .
is symmetric.
Now, let .
is parallel to . Also, is parallel to .
is parallel to .
is transitive.
Hence, is an equivalence relation.
The set of all lines related to the line is the set of all lines that are parallel to the line .
Slope of line is
It is known that parallel lines have the same slopes.
The line parallel to the given line is of the form , where .
Hence, the set of all lines related to the given line is given by , where .
#425308
Topic: Types of Relations
Let be the relation in the set given by .
Choose the correct answer.
A is reflexive and symmetric but not transitive.
B is reflexive and transitive but not symmetric.
C is symmetric and transitive but not reflexive.
D is an equivalence relation.
Solution
It is seen that , for every .
is reflexive.
It is seen that , but .
is not symmetric.
Also, it is observed that for all
is transitive.
Hence, is reflexive and transitive but not symmetric.
The correct answer is B.
#425309
Topic: Types of Relations
Let be the relation in the set given by . Choose the correct answer.
A
B
C
D
Solution
R = {( , ) : is parallel to }L1 L2 L1 L2
R L1 ( , ) ∈ RL1 L1
( , ) ∈ RL1 L2
⇒ L1 L2
⇒ L2 L1
⇒ ( , ) ∈ RL2 L1
∴ R
( , ), ( , ) ∈ RL1 L2 L2 L3
⇒ L1 L2 L2 L3
⇒ L1 L3
⇒ R
R
y = 2x+ 4 y = 2x+ 4
y = 2x+ 4 m = 2
y = 2x+ c c ∈ R
y = 2x+ c c ∈ R
R {1, 2, 3, 4} R = {(1, 2), (2, 2), (1, 1), (4, 4), (1, 3), (3, 3), (3, 2)}
R
R
R
R
R = {(1, 2), (2, 2), (1, 1), (4, 4), (1, 3), (3, 3), (3, 2)}
(a, a) ∈ R a ∈ {1, 2, 3, 4}
∴ R
(1, 2) ∈ R (2, 1) ∉ R
∴ R
(a, b), (b, c) ∈ R ⇒ (a, c) ∈ R a, b, c ∈ {1, 2, 3, 4}
∴ R
R
R N R = {(a, b) : a = b− 2, b > 6}
(2, 4) ∈ R
(3, 8) ∈ R
(6, 8) ∈ R
(8, 7) ∈ R
7/4/2018 https://community.toppr.com/content/questions/print/?show_answer=1&show_topic=1&show_solution=1&page=1&qid=421842%2C+4278…
https://community.toppr.com/content/questions/print/?show_answer=1&show_topic=1&show_solution=1&page=1&qid=421842%2C+427819%2C+4… 12/22
Now, since
Also, as
And, as
Now, consider .
We have and also, .
Hence the correct answer is C.
#425310
Topic: Types of Functions
Show that the function defined by is one-one and onto, where ' ' is the set of all non-zero real numbers. Is the result true, if the domain ' ' is replaced
by with co-domain being same as ' '?
Solution
R = {(a, b) : a = b− 2, b > 6}
b > 6, (2, 4) ∉ R
3 ≠ 8 − 2, (3, 8) ∉ R
8 ≠ 7 − 2
∴ (8, 7) ∉ R
(6, 8)
8 > 6 6 = 8 − 2
∴ (6, 8) ∈ R
f : R⋅ → R⋅ f(x) =1
xR⋅ R⋅
N R⋅
7/4/2018 https://community.toppr.com/content/questions/print/?show_answer=1&show_topic=1&show_solution=1&page=1&qid=421842%2C+4278…
https://community.toppr.com/content/questions/print/?show_answer=1&show_topic=1&show_solution=1&page=1&qid=421842%2C+427819%2C+4… 13/22
Given defined by
.....(1)
We have to check if is one-one and onto
(i) One-one:
Let for all
(by (1))
Hence, is one-one.
(ii) Onto :
For every , there is an such that
Hence, is onto.
Now, if co-domain of is replaced by N, then the function defined by
We have to check if is one -one and onto.
Let for
Hence, is one-one.
Onto :
is not onto as for every negative real number there is no in such that
Also, for but there is no such that belongs to .
Hence, the result is not true if domain is replaced by
f : R− {0} → R− {0}
f(x) =1
x
f
f(x) = f(y) x, y ∈ R− {0}
⇒ =1
x
1
y
⇒ x = y
f
y ∈ R− {0} x = ∈ R− {0}1
y
f(x) =1
1
y
⇒ f(x) = y
f
f g : N → R− {0}
g(x) =1
x
g
g(x) = g(y)
⇒ =1
x
1
y
⇒ x = y
g
g x N f(x) = y
1.6 ∈ R− {0} x ∈ N1
1.6N
R− {0} N
7/4/2018 https://community.toppr.com/content/questions/print/?show_answer=1&show_topic=1&show_solution=1&page=1&qid=421842%2C+4278…
https://community.toppr.com/content/questions/print/?show_answer=1&show_topic=1&show_solution=1&page=1&qid=421842%2C+427819%2C+4… 14/22
It is given that is defined by .
One-one.
is one-one.
Onto:
It is clear that for , there exists (Exists as ) such that
is onto.
Thus, the given function is one-one and onto.
Now, consider function $$g: N \rightarrow R\cdot $$ defined by
we have,
is one-one.
Further, it is clear that g is not onto as for there does not exit any in such that
Hence, function g is one-one but onto.
#427806
Topic: Types of Functions
Check the injectivity and surjectivity of the following functions:
(i) given by
(ii) given by
(iii) given by
(iv) given by
(v) given by
Solution
(i)
It is seen that for ,
is injective
Now, , But, there does not exist any
in such that .
is not surjective.
Hence, function is injective but not surjective.
(ii)
It is seen that , but
is not injective.
Now, . But, there does not exist any
element such that
f : R⋅ → R⋅ f(x) =1
x
f(x) = f(y)
⇒ =1
x
1
y
⇒ x = y
∴ f
y ∈ R⋅ x = ∈ R⋅1
yy ≠= 0 f(x) = = y1
1
y
∴ f
(f)
g(x) =1
x
g( ) = g( ) ⇒ = ⇒ =x1 x21
x1
1
x2
x1 x2
⇒ g
1.2 ∈ R⋅ x N g(x) =1
1.2
f : N → N f(x) = x2
f : Z → Z f(x) = x2
f : R → R f(x) = x2
f : N → N f(x) = x3
f : Z → Z f(x) = x3
f(x) = x2
x, y ∈ N
f(x) = f(y) ⇒ = ⇒ x = yx2 y2
∴ f
2 ∈ N x
N f(x) = = 2x2
∴ f
f
f(x) = x2
f(−1) = f(1) = 1 −1 ≠ 1
∴ f
−2 ∈ Z
x ∈ Z f(x) = = −2x2
7/4/2018 https://community.toppr.com/content/questions/print/?show_answer=1&show_topic=1&show_solution=1&page=1&qid=421842%2C+4278…
https://community.toppr.com/content/questions/print/?show_answer=1&show_topic=1&show_solution=1&page=1&qid=421842%2C+427819%2C+4… 15/22
is not surjective.
Hence, function is neither injective nor surjective.
(iii)
It is seen that , but .
is not injective.
Now, . But, there does not exist any element such that
is not surjective.
Hence, function is neither injective nor surjective.
(iv)
It is seen that for
is injective.
Now, . But, there does not exist any element
in domain such that
is not surjective.
Hence, function is injective but not surjective.
(v)
It is seen that for
is injective.
Now, . But, there does not exist any element
in domain such that
is not surjective.
Hence, function is injective but not surjective.
#427811
Topic: Types of Functions
Prove that the Greatest Integer Function given by , is neither one-one nor onto, where denotes the greatest integer less that or equal to
Solution
It is seen that .
, but
is not one-one
Now, consider
It is known that is always an integer. Thus, there does not exist any element such that
is not onto.
Hence, the greatest integer function is neither one-one nor onto.
∴ f
f
f(x) = x2
f(−1) = f(1) = 1 −1 ≠ 1
∴ f
−2 ∈ R x ∈ R f(x) = = −2x2
∴ f
f
f(x) = x3
x, y ∈ N, f(x) = f(y) ⇒ = ⇒ x = yx3 y3
∴ f
2 ∈ N
x N f(x) = = 2x3
∴ f
f
f(x) = x3
x, y ∈ Z, f(x) = f(y) ⇒ = ⇒ x = yx3 y3
∴ f
2 ∈ Z
x Z f(x) = = 2x3
∴ f
f
f : R → R f(x) = [x] [x] x
f(x) = [x]
f(1.2) = [1.2] = 1, f(1.9) = [1.9] = 1
∴ f(1.2) = f(1.9) 1.2 ≠ 1.9
∴ f
0.7 ∈ R
f(x) = [x] x ∈ R f(x) = 0.7
∴ f
7/4/2018 https://community.toppr.com/content/questions/print/?show_answer=1&show_topic=1&show_solution=1&page=1&qid=421842%2C+4278…
https://community.toppr.com/content/questions/print/?show_answer=1&show_topic=1&show_solution=1&page=1&qid=421842%2C+427819%2C+4… 16/22
#427812
Topic: Types of Functions
Show that the modulus function given by , is neither one-one nor onto, where is , if is positive or and is , if is negative.
Solution
It is seen that , but
is not one-one
Now, consider .
but it is known that is always non-negative.
Thus, there does not exist any element in domain such that
is not onto.
Hence, the modulus function is neither one-one nor onto.
#427813
Topic: Types of Functions
Show that the Signum function , given by
.
is neither one-one nor onto.
Solution
.
It is seen that , but .
is not one-one.
Now, as takes only values or for the element in co-domain ,
there does not exist any in domain such that
is not onto
Hence, the signum function is neither one-one nor onto.
#427814
Topic: Types of Functions
Let { }, { } and let be a function from to . Show that is one-one.
Solution
It is given that { }, { }.
is defined as .
It is seen that the images of distinct elements of
under are distinct
Hence, function is one-one
#427817
Topic: Types of Functions
f : R → R f(x) = |x| |x| x x 0 |x| −x x
f(x) = |x| = { x if x > 0−x if x < 0
f(−1) = |−1| = 1, f(1) = |1| = 1∴ f(−1) = f(1) −1 ≠ 1
∴ f
−1 ∈ R
f(x) = |x|
x R f(x) = |x| = −1
∴ f
f : R → R
f(x) =⎧⎩⎨
1, if x > 00, if x = 0−1, if x < 0
f(x) =
⎧
⎩⎨⎪
⎪
1
0
−1
ifx > 0
ifx = 0
ifx < 0
⎫
⎭⎬⎪
⎪
f(1) = f(2) = 1 1 ≠ 2
∴ f
f(x) 3 (1, 0, −1) −2 R
x R f(x) = −2
∴ f
A = 1, 2, 3 B = 4, 5, 6, 7 f = (1, 4), (2, 5), (3, 6) A B f
A = 1, 2, 3 B = 4, 5, 6, 7
f : A → B f = (1, 4), (2, 5), (3, 6)
∴ f(1) = 4, f(2) = 5, f(3) = 6
A
f
f
7/4/2018 https://community.toppr.com/content/questions/print/?show_answer=1&show_topic=1&show_solution=1&page=1&qid=421842%2C+4278…
https://community.toppr.com/content/questions/print/?show_answer=1&show_topic=1&show_solution=1&page=1&qid=421842%2C+427819%2C+4… 17/22
In each of the following cases, state whether the function is one-one, onto or bijective. Justify your answer.
(i) defined by
(ii) defined by
Solution
(i)
Let such that
Hence f is one-one
Also the range of f is R = codomain of f,
since it is first order polynomial
Therefore f is both one-one and onto function
(ii)
is defined as
Let such that
does not imply that .
For instance,
is not one-one.
Consider an element in co-domain .
It is seen that is positive for all
Thus, there does not exist any in domain such that
is not onto.
Hence, is neither one-one nor onto
#427818
Topic: Types of Functions
Let and be sets. Show that such that is bijective function.
Solution
is defined as .
Let such that .
and
is one-one.
Now, let be any element.
Then, there exists such that
. By definition of
is onto.
Hence, is bijective.
f : R → R f(x) = 3 − 4x
f : R → R f(x) = 1 + x2
, ∈ Rx1 x2 f( ) = f( )x1 x2
⇒ 3 − 4 = 3 − 4 ⇒ =x1 x2 x1 x2
f : R → R f(x) = 1 + x2
, ∈ Rx1 x2 f( ) = f( )x1 x2
⇒ 1 + = 1 +x21 x2
2
⇒ =x21 x2
2
⇒ = ±x1 x2
∴ f( ) = f( )x1 x2 =x1 x2
f(1) = f(−1) = 2
∴ f
−2 R
f(x) = 1 + x22 x ∈ R
x R f(x) = −2
∴ f
f
A B f : A×B → B×A f(a, b) = (b, a)
f : A×B → B×A f(a, b) = (b, a)
( , ), ( , ) ∈ A×Ba1 b1 a2 b2 f( , ) = f( , )a1 b1 a2 b2
⇒ ( , ) = ( , )b1 a1 b2 a2
⇒ = )b1 b2 ( = )a1 a2
⇒ ( , ) = ( , )a1 b1 a2 b2
∴ f
(b, a) ∈ B×A
(a, b) ∈ A×B
f(a, b) = (b, a) [ f]
∴ f
f
7/4/2018 https://community.toppr.com/content/questions/print/?show_answer=1&show_topic=1&show_solution=1&page=1&qid=421842%2C+4278…
https://community.toppr.com/content/questions/print/?show_answer=1&show_topic=1&show_solution=1&page=1&qid=421842%2C+427819%2C+4… 18/22
#427819
Topic: Types of Functions
Let be defined by for all .
State whether the function is bijective. Justify your answer.
Solution
for all
is defined as
It can be observed that:
and
, where
is not one-one.
Consider a natural number in co-domain Case : is odd
for some .
Then, there exists such that
Case : is even
for some .
Then, there exists such that
is onto.
Hence, is not a bijective function
#427820
Topic: Types of Functions
Let and . Consider the function defined by
. Is one-one and onto? Justify your answer
Solution
f : N → N f(n) = { }, if n is oddn+1
2
, if n is evenn
2
n ∈ N
f
f(n) = { }, if n is oddn+1
2
, if n is evenn
2
n ∈ N
f : N → N
f(1) = = 11 + 1
2f(2) = = 1
2
2
∴ f(1) = f(2) 1 ≠ 2
∴ f
(n) N I n
∴ n = 2r+ 1 r ∈ N
4r+ 1 ∈ N f(4r+ 1) = = 2r+ 14r+ 1 + 1
2
II n
∴ n = 2r r ∈ N
4r ∈ N f(4r) = = 2r4r
2
∴ f
f
A = R− 3 B = R− 1 f : A → B
f(x) = ( )x− 2
x− 3f
7/4/2018 https://community.toppr.com/content/questions/print/?show_answer=1&show_topic=1&show_solution=1&page=1&qid=421842%2C+4278…
https://community.toppr.com/content/questions/print/?show_answer=1&show_topic=1&show_solution=1&page=1&qid=421842%2C+427819%2C+4… 19/22
is defined as
Let, such that
is one-one.
Let . Then,
The function is onto if there exists
such that
Now,
Thus, for any , there exists
such that
is onto.
Hence, function is one-one and onto.
#427821
Topic: Types of Functions
Let be defined as .
Choose the correct answer.
A is one-one onto
B is many-one onto
C is one-one but not onto
D is neither one-one nor onto
Solution
A = R− 3,B = R− 1
f(x) = ( )x− 2
x− 3
f : A → B
x, y ∈ A f(x) = f(y)
⇒ =x− 2
x− 3
y − 2
y − 3
⇒ x− 2y − 3 = y − 2x− 3
⇒ xy − 3x− 2y + 6 = xy − 3y − 2x+ 6
⇒ −3x− 2y = −3y − 2x
⇒ 3x− 2x = 3y − 2y
⇒ x = y
∴ f
y ∈ B = R− 1 y ≠ 1
f x ∈ A
f(x) = y
f(x) = y
⇒ = yx− 2
x− 3
⇒ x− 2 = xy − 3y
⇒ x(1 − y) = −3y + 2
⇒ x = ∈ A2 − 3y
1 − y[y ≠ 1]
y ∈ B
∈ A2 − 3y
1 − y
f ( ) =2 − 3y
1 − y
( )− 22 − 3y
1 − y
( )− 32 − 3y
1 − y
∴ f
f
f : R → R f(x) = x4
f
f
f
f
7/4/2018 https://community.toppr.com/content/questions/print/?show_answer=1&show_topic=1&show_solution=1&page=1&qid=421842%2C+4278…
https://community.toppr.com/content/questions/print/?show_answer=1&show_topic=1&show_solution=1&page=1&qid=421842%2C+427819%2C+4… 20/22
is defined as .
Let such that
does not imply that
For instance
is not one-one.
Consider an element in co-domain .
It is clear that there does not exist any in domain such that
is not onto.
Hence, function is neither one-one nor onto.
The correct answer is .
#427822
Topic: Types of Functions
Let be defined as . Choose the correct answer.
A is one-one onto
B is many-one onto
C is one-one but not onto
D is neither one-one nor onto
Solution
is defined as .
Let such that
is one-one
Also, for any real number in co-domian ,
there exists in such that
is onto.
Hence, function is one-one and onto
The correct answer is .
#427823
Topic: Composite and Inverse Functions
Let and be given by and . Write down .
Solution
f : R → R f(x) = x4
x, y ∈ R f(x) = f(y)
⇒ =x4 y4
⇒ x = ±y
∴ f( ) = f( )x1 x2 =x1 x2
f(1) = f(−1) = 1
∴ f
−2 R
x R f(x) = −2
∴ f
f
D
f : R → R f(x) = 3x
f
f
f
f
f : R → R f(x) = 3x
x, y ∈ R f(x) = f(y)
⇒ 3x = 3y
⇒ x = y
∴ f
(y) Ry
3R
f ( ) = 3 ( ) = yy
3
y
3
∴ f
f
A
f : 1, 3, 4 → 1, 2, 5 g : 1, 2, 5 → 1, 3 f = (1, 2), (3, 5), (4, 1) g = (1, 3). (2, 3), (5, 1) gof
7/4/2018 https://community.toppr.com/content/questions/print/?show_answer=1&show_topic=1&show_solution=1&page=1&qid=421842%2C+4278…
https://community.toppr.com/content/questions/print/?show_answer=1&show_topic=1&show_solution=1&page=1&qid=421842%2C+427819%2C+4… 21/22
The function and are defined as
and
as and
as and
as and
#427825
Topic: Composite and Inverse Functions
Let and be functions from to . Show that
(i)
(ii)
Solution
(i)
To prove:
Consider:
for all
Hence,
(ii)
To prove:
Consider:
for all
Hence,
#427827
Topic: Composite and Inverse Functions
Find and , if and
ii) and
Solution
f : 1, 3, 5 → 1, 2, 5 g : 1, 2, 5 → 1, 3
f = (1, 2), (3, 5), (4, 1) g = (1, 3). (2, 3), (5, 1)
gof(1) = g(f(1)) = g(2) = 3 [f(1) = 2 g(2) = 3]
gof(3) = g(f(3)) = g(5) = 1 [f(3) = 5 g(5) = 1]
gof(4) = g(f(4)) = g(1) = 3 [f(4) = 1 g(1) = 3]
∴ gof = (1, 3), (3, 1), (4, 3)
f, g h R R
(f + g)oh = foh+ goh
(f. g)oh = (foh). (goh)
(f + g)oh = foh+ goh
((f + g)oh)(x)
= (f + g)(h(x))
= f(h(x)) + g(h(x))
= (foh)(x) + (goh)(x)
= (foh) + (goh)(x)
∴ ((f + g)oh)(x) = (foh) + (goh)(x) x ∈ R
(f + g)oh = foh+ goh
(f. g)oh = (foh). (goh)
((f. g)oh)(x)
= (f. g)(h(x))
= f(h(x)). g(h(x))
= (foh)(x). (goh)(x)
= (foh). (goh)(x)
∴ ((f. g)oh)(x) = (foh). (goh)(x) x ∈ R
(f. g)oh = (foh). (goh)
gof fog (i) f(x) = |x| g(x) = |5x− 2|
f(x) = 8x3 g(x) = x1
3
7/4/2018 https://community.toppr.com/content/questions/print/?show_answer=1&show_topic=1&show_solution=1&page=1&qid=421842%2C+4278…
https://community.toppr.com/content/questions/print/?show_answer=1&show_topic=1&show_solution=1&page=1&qid=421842%2C+427819%2C+4… 22/22
and
And
and
And
(i) f(x) = |x| g(x) = |5x− 2|
∴ (gof)(x) = g(f(x)) = g(|x|) = |5|x| − 2|
(fog)(x) = f(g(x)) = f(|5x− 2|) = ||5x− 2|| = |5x− 2|
(ii)f(x) = 8x3 g(x) = x1
3
∴ (gof)(x) = g(f(x)) = g(8 ) = (8 = 2xx3 x3)
1
3
(fog)(x) = f(g(x)) = f ( ) = 8 = 8xx1
3 ( )x1
3
3
top related