# NCERT Solutions Class 12 maths Chapter-1 (Relation And Functions)Exercise 1.1

**NCERT Solutions Class 12 Maths**from class

**12th**Students will get the answers of

**Chapter-1 (Relation And Functions) Exercise 1.1**. This chapter will help you to learn the basics and you should expect at least one question in your exam from this chapter.

We have given the answers of all the questions of

**NCERT Board Mathematics Textbook**in very easy language, which will be very easy for the students to understand and remember so that you can pass with good marks in your examination.

**NCERT Question-Answer**

**Class 12 ****Mathematics**

**Chapter-1 ****(****Relation And Functions****)**

**Questions and answers given in practice**

**Chapter****-1 ****(Relation And Functions)**

### Exercise 1.1

### Set 1

**Question 1. **

**Determine
whether each of the following relations are reflexive, symmetric and
transitive:**

**(i) Relation R in the set
A={ 1, 2,3, . . ., 13, 14} defined as R={ ( x , y):3x-y=0}**

**Solution:**

A={1 ,2 ,3,…,13, 14}

R={(x,y): 3x-y=0}

Therefore
R={(1,3),(2,6),(3,9),(4,12)}

R is not reflexive since
(1,1),(2,2),(3,3),…,(14,14)∉R

Also, R is not reflexive
since (1,3)∈R, but (3,1)∉R.[since 3(3)-1≠0]

Also, R is not
transitive as (1,3), (3,9)∈R,
but (1,9)∉R.[since 3(1)-9≠0]

Hence, R is not
reflexive, nor symmetric nor transitive.

**(ii) Relation R in the set N of natural numbers defined as R={ ( x, y) : y=x+5
and x<4}**

**Solution:**

R={(x, y): y=x+5 and
x<4}={(1,6), (2, 7), (3,8)}

It is seen that (1, 1)∉R. Therefore, R is not reflexive.

(1, 6)∈R. But, (6, 1)∉R so, R is not symmetric.

Now, since their is no
pair in R such that (x, y) and (y, z)∈R, so (x, z) can not belong to R. Therefore, R is
not transitive.

So, we can conclude that R is neither reflexive, nor symmetric,
nor transitive.

**(iii) Relation R in the
set A= {1, 2, 3, 4, 5, 6} as R={(x, y): y is divisible by x}**

**Solution:**

A={1, 2, 3, 4, 5, 6}

R={(x, y): y is
divisible by x}

We know that any number
is always divisible by itself.⇒ (x,
x)∈ R. Therefore, R is reflexive.

We see, (2, 4)∈R [4 is divisible by 2]. But, (4, 2)∉R [2 is not divisible by 4]. Therefore, R is not symmetric.

Let’s assume (x, y), (y, z) ∈ R. Then, y is divisible by x and z is divisible by
y. Therefore, z is divisible by x.⇒ (x, z) ∈ R. Therefore, R is
transitive.

**(iv) Relation R in the set Z of all integers defined as R={(x, y): x-y is an
integer}**

**Solution:**

R={(x, y): x-y is an
integer}

For every x ∈ **Z,
(x, x) **∈ R [x-x=0 which is an
integer]. Therefore, R is reflexive.

For every x, y ∈ **Z** if
(x, y) ∈ R, then x-y is an
integer. ⇒-(x-y) is also an
integer. ⇒ (y-x) is also an
integer. Therefore, R is
symmetric.

Let’s assume, (x, y) and (y, z) ∈ R, where x, y and z ∈ **Z.**
⇒ (x-y) and (y-z) are
integers. ⇒ x-z=(x-y)+(y-z) is an
integer. ⇒ (x, z) ∈ R. Therefore, R is
transitive.

**(v) Relation R in a set A
of human beings in a town at a particular time, given by:**

**(a)R={(x,y) : x and y work at the same place. **

**Solution:**

We can see (x,x) ∈ R .Therefore, R is
reflexive.

If (x,y) ∈ R, then x and y work at the same place. So, (y,x) ∈ R. Therefore, R is
symmetric.

Let, (x,y), (y,z) ∈ R.

⇒ x and y work mat the
same place and y and z work at the same place.

⇒ x and z work at the
same place.

Therefore, R is
transitive.

**(b) R={(x,y): x and y live in the same locality}. **

**Solution:**

We can see (x,x) ∈ R. Therefore, R is
reflexive.

If (x,y) ∈ R, then x and y live in same locality. So, (y,x) ∈ R. Therefore, R is
symmetric.

Let (x,y) ∈ R and (y,z) ∈ R. So, x, y and z live in the same locality. So,
(x,z) ∈ R. Therefore, R is transitive.

**(c) R={(x,y): x is exactly 7 cm taller than y}. **

**Solution:**

(x,x)∉R since, human being can not be taller than
himself. So, R is not reflexive.

Let (x,y) ∈ R ,then x is exactly 7 cm taller than y. Then, y
is not taller than x. Therefore, R is not
symmetric.

Let (x,y), (y,z) ∈ R, then x is exactly 7 cm taller than y and y is
exactly 7 cm taller than z which means x is 14 cm taller than z. So, (x,z)∉R . Therefore, R is not
transitive.

**(d) R={(x,y): x is wife of y}. **

**Solution:**

(x,x) ∉ R. Since, x can not be the wife of herself.
Therefore, R is not reflexive.

Let (x,y) ∈ R, then x is the wife of y. So, y is not the wife
of x ,i.e., (y,x) ∉ R. Therefore, R is not symmetric.

Let (x,y), (y,z) ∈ R, then x is the wife of y and y is the wife of z
which is not possible. So, R can not be
transitive.

**(e) R={(x,y): x is father of y}. **

**Solution:**

(x,x) ∉ R. Since, x can not be the father of himself.
Therefore, R is not reflexive.

Let (x,y) ∈ R, then x is the father of y. So, y can not be the
father of x. So, (y,x) ∉ R. Therefore, R is not symmetric.

Let (x,y), (y,z) ∈ R, then x is the father of y and y is the father
of z which means x is the grandfather of z. So, So, (x, z) ∉ R. Therefore, R is not
transitive.

**Question 2. **

**Show that the relation R in the set R of
real numbers, defined as R= {(a, b): a ≤ b ^{2}**

**} is neither reflexive nor symmetric nor transitive.**

**Solution:**

It can be observed that
(½, ½) ∉ R, since ½>(½)^{2} =¼. Therefore, R is not reflexive.

(1,4) ∈ R as 1<4^{2} .But, (4,1) ∉ R. Therefore, R is not symmetric.

(3,2), (2,1.5) ∈ R. But, 3> (1.5)^{2} =2.25 . So, (3,1.5) ∉ R. Therefore, R is not transitive.

Hence, R
is neither reflexive, nor symmetric, nor 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}. **

**Solution:**

Let the set {1, 2, 3, 4,
5, 6} be named A.

R={(1,2), (2, 3), (3,
4), (4, 5

), (5, 6)}

We can see (x, x) ∉ R. Since, x ≠ x+1. Therefore, R is not
reflexive.

It is observed that
(1,2) ∈ R but, (2,1) ∉ R. Therefore, R is not symmetric.

We can see, (1,2), (2,
3) ∈ R, but (1,3) ∉ R. Therefore, R is not transitive.

Hence, R
is neither reflexive, nor symmetric, nor transitive.

**Question 4. **

**Show that the relation R in R defined as R={ (a, b): a≤ b}, is reflexive and
transitive but not symmetric.**

**Solution:**

Clearly, (a,a) ∈ R as a=a. Therefore, R is reflexive.

(2,4) ∈ R (as 2<4) but (4,2) ∉ R as 4 is greater than 2. Therefore, R is
not symmetric.

Let (a,b), (b,c) ∈ R. Then, a≤ b and b≤ c.

⇒a ≤ c.

(a, c) ∈ R. Therefore, R is transitive.

Hence, R
is reflexive and transitive but not symmetric.

**Question 5. **

**Check whether the relation R in R defined as R ={ (a, b): a ≤ b ^{3}**

**} is reflexive, symmetric or transitive.**

**Solution:**

It is observed that (½,
½) ∉ R as ½ > (½)^{3} =(1/8). Therefore, R is not reflexive.

(1,2) ∈ R(as 1<8) but, (2,1) ∉ R. Therefore, R is not symmetric.

We have, (3, 3/2), (3/2,
6/5) ∈ R but, (3, 6/5) ∉ R. Therefore, R is not transitive.

Hence, R
is neither reflexive, nor symmetric, nor transitive.

**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. **

**Solution:**

Let the set {1, 2, 3} be
named A.

It is seen that, (1, 1),
(2,2), (3,3)∉ R. Therefore, R
is not reflexive.

As (1, 2) ∈ R and (2, 1) ∈ R. Therefore, R is symmetric.

However, (1, 1)∉ R. Therefore, R is not transitive.

Hence, R
is symmetric but neither reflexive nor 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 the same number of pages} is a equivalence relation.**

**Solution:**

Set A is the set of all
books in the library of a college.

R={(x,y):x and y have
the same number of pages}

R is reflexive since
(x,x) ∈ R as x and x have the
same number of pages.

Let (x,y) ∈ R

⇒x and y have the
same number of pages

⇒y and x have the same
number of pages.

⇒(y,x)∈ R

Therefore , R is
symmetric.

Let (x,y) ∈ R and (y,z)∈ R.

⇒x and y have the same
number of pages and y and z have the same number of pages.

⇒x and z have the same
number of pages.

⇒(x,z) ∈ R

Therefore, R is
transitive.

Hence, R
is an equivalence relation.

**Question 8. **

**Show that the relation R in the set A
={1,2,3,4,5} given by R={(a,b):|a-b| is even} , is an equivalence relation .
Show that all the elements of { 1,3,5} are related to each other and all the
elements of {2, 4} are related to each other . But no elements of {1,3,5} is
related to any element of {2,4}.**

**Solution:**

A={ 1,2,3,4,5}

R={(a,b):|a-b| is even }

It is clear that for any
element a∈ A, we have |a-a|=0
(which is even).

Therefore, R is
reflexive.

Let (a,b) ∈ R.

⇒|a-b| is even.

⇒|-(a-b)|=|b-a| is also
even.

⇒(b,a)∈ R

Therefore, R is
symmetric.

Now , let (a,b)∈ R and (b,c)∈ R.

⇒|a-b| is even and |b-c|
is even.

⇒(a-b) is even and (b-c)
is even.

⇒(a-c)=(a-b)+(b-c) is
even. [Sum of two even integers is even]

⇒|a-c| is even.

Therefore, R is
transitive.

Hence, R is an equivalence relation.

All elements of the set {1,2,3} 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 {2,4} are related to each other as all the elements of this subset are even.

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 .** Thus,
the modulus of the difference between the two elements (from each of these two
subsets) will not be even.**

**Question 9: **

**Show that
each of the relation R in the set A={x∈ Z:0<=x<=12} , given by**

**(i) R={(a,b):|a-b| is a multiple of 4}**

**(ii) R={(a,b):a=b}**

**is an equivalence relation. Find the set of all elements related to 1 in
each case.**

**Solution:**

A={x∈ Z:0<=x<=12}={0,1,2,3,4,5,6,7,8,9,10,11,12}

** (i) R={(a,b):|a-b| is a multiple of 4}**

For any element a ∈A , we have (a,a)∈R as |a-a|=0 is a multiple of 4.

Therefore, R is
reflexive.

Now , let (a,b)∈R ⇒|a-b|
is a multiple of 4.

⇒|-(a-b)|=|b-a| is a
multiple of 4.

⇒(b,a)∈R

Therefore, R is
symmetric.

Let (a,b) ,(b,c) ∈ R.

⇒|a-b| is a multiple of 4
and |b-c| is a multiple of 4.

⇒(a-b) is a multiple of 4
and (b-c) is a multiple of 4.

⇒(a-c)=(a-b)+(b-c) is a
multiple of 4.

⇒|a-c| is a multiple of
4.

⇒(a,c)∈R

Therefore, R is
transitive.

Hence, R is an
equivalence relation .

The set of elements
related to 1 is {1,5,9} since |1-1|=0 is a multiple of 4,

|5-1|=4 is a multiple of
4,and

|9-1|=8 is a multiple of
4.

**(ii) R={(a,b):a=b}**

For any element a∈A, we have (a,a) ∈ R , since a=a.

Therefore, R is
reflexive.

Now , let (a,b)∈R.

⇒a=b

⇒b=a

⇒(b,a)∈R

Therefore, R is
symmetric.

Now, let (a,b)∈R and (b,c)∈R.

⇒a=b and b=c

⇒a=c

⇒(a,c)∈R

*Therefore, R is
transitive.*

*Hence, R is an
equivalence relation.*

*The elements in R that
are related to 1 will be those elements from set A which are equal to 1.*

*Hence, the set of elements related to 1 is {1].*

**Question 10: **

**Give 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 A ={5,6,7}.**

Define a relation R on A
as R ={(5,6),(6,5)}.

Relation R is not
reflexive as (5,5) , (6,6),(7,7)∉R.

Now, as (5,6)∈R and also (6,5)∈R, R is symmetric.

⇒(5,6),(6,5)∈R , but (5,5)∉R

Therefore , R is not
transitive.

Hence, relation R is symmetric but not
reflexive or 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 itself . In
fact a=a.

Therefore, R is not
reflexive. Now,

(1,2)∈R (as 1<2)

But, 2 is not less than
1.

Therefore, (2,1)∉R

Therefore, R is not
symmetric.

Now, let (a,b),(b,c)∈R.

⇒a<b and b<c

⇒a<c

⇒ (a,c)∈R

Therefore, R is
transitive.

Hence, relation R is transitive but not
reflexive and symmetric.

**(iii) ****Let A={4,6,8}.**

Define a relation R on A
as:

A={(4,4),(6,6),(8,8),(4,6),(6,4),(6,8),(8,6)}

Relation R is reflexive
since for every a∈A , (a,a)∈R i.e.,(4,4),(6,6),(8,8)∈R.

Relation R is symmetric
since (a,b)∈R⇒(b,a)∈R for all a,b∈R.

Relation R is not
transitive since (4,6),(6,8)∈R ,
but (4,8)∉R.

Hence, relation R is reflexive and symmetric
but not transitive.

**(iv)**** Define a relation R in R as:**

R={ (a,b): a^{3} ≥ b^{3} }

Clearly (a,a)∈R as a^3=a^3

Therefore, R is
reflexive.

Now, (2,1)∈R(as 2^{3}>=1^{3})

But, (1,2)∉ R (as 1^{3}< 2^{3})

Therefore, R is not
symmetric

Let (a,b), (b,c) ∈ R.

⇒a^{3 }>= b^{3} and b^{3} >=c^{3}

⇒a^{3}>=c^{3}

⇒(a, c)∈R

Therefore, R is
transitive.

Hence, relation R is reflexive and transitive
but not symmetric.

**(v)**** Let A={-5,-6},**

Define a relation R on A
as:

R={(-5,-6), (-6,-5),
(-5,-5)}

Relation R is not
reflexive as (-6,-6)∉ R.

Relation R is symmetric
as (-5,-6)∈ R and (-6,-5)∈ R.

It is seen that
(-5,-6),(-6,-5)∈R . Also , (-5,-5)∈R.

Therefore, the relation
R is transitive .

Hence, relation
R is symmetric and transitive but not reflexive.

**Exercise 1.1**

**Set 2**

**Question 11. **

**Show that
the relation R in the set A of points in a plane given by R ={ (P,Q): distance
of the point P from the origin is the same as the distance of the point Q from
the origin}, is an equivalence relation. Further, show that the set of all
points related to a point P ≠ (0, 0) is the circle passing through P with
origin as centre.**

**Solution:**

We can see (P, P) ∈ R since, the distance of point P from the origin
is always the same as the distance of the same point P from the origin.
Therefore, R is reflexive.

Let (P,Q)∈ R.

⇒The distance of point P
from the origin is the same as the distance of point Q from the origin.

⇒The distance of point Q
from the origin is the same as the distance of point P from the origin.

So, (Q,P) ∈ R. Therefore, R is symmetric.

Let (P,Q), (Q,S) ∈ R.

⇒The distance of point P
from the origin is the same as the distance of point Q from the origin and
also, the distance of point P from the origin is the same.

⇒ The distance of points
P and S from the origin is the same.

⇒(P, S) ∈ R. Therefore, R is transitive.

Therefore, R is an
equivalence relation.

The set of all points
related to P ≠ (0, 0) will be those points whose distance from the origin is
the same as the distance of point P from the origin.

In other words, if O(0 0) is the origin and OP = k,
then the set of all points related to P is at the same as k from the origin.
Hence, this set of points forms a circle with the centre as the origin and the
circle passes through point P.

**Question 12. **

**Show that the relation R defined in the
set A of all triangles as R={(T _{1} ,T_{2} ): T_{1} is similar to T_{2} }, is equivalence relation. Consider three right angle triangles T_{1} with sides 3, 4, 5, T_{2} with sides 5, 12,
13 and T_{3} with sides 6, 8, 10. Which triangles among T_{1}, T_{2}, and T_{3}**

**are related?**

**Solution:**

R is reflexive since,
every triangle is similar to itself.

If (T_{1}, T_{2}) ∈ R,
then T_{1} is similar to T_{2}. So, (T_{2}, T_{1}) ∈ R. Therefore, R is symmetric.

Let (T_{1}, T_{2}), (T_{2}, T_{3}) ∈ R, then T_{1} is similar to T_{2} and T_{2} is similar to T_{3}. So, T_{1} is also similar to T_{3}. Therefore, (T_{1}, T_{3}) ∈ R
so, R is transitive.

Thus, R is an
equivalence relation.

We observe,

(3/6)=(4/8)=(5/10)=1/2

Therefore, the
corresponding sides of triangles T_{1} and T_{3} are in the same ratio. Then, triangle T_{1} is similar to triangle T_{3}.

Hence, T_{1} is related to T_{3}.

**Question 13. **

**Show that the relation R defined in the
set A of all polygons as R={(P _{1}, P_{2}): P_{1} and P_{2}**

**have same number of sides}, is an equivalence relation. What is the set of all elements in A related to the right angle triangle T with sides 3, 4 and 5?**

**Solution:**

R is reflexive since(P_{1}, P_{2}) ∈ R
as the same polygon has the same number of sides with itself.

Let (P_{1}, P_{2}) ∈ R,
then P_{1} and P_{2} have the same
number of sides. So, (P_{2}, P_{1}) ∈ R. Therefore, R is symmetric.

Let (P_{1}, P_{2}), (P_{2}, P_{3}) ∈ R, then P_{1} and P_{2} have the same number of sides. Also, P_{2} and P_{3} have the same number of sides. So, P_{1} and P_{3} have the same number of sides ,i.e., (P_{1}, P_{3}) ∈ R.
Therefore, R is transitive.

Hence, R is an
equivalence relation.

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).

Hence, the set of all elements in A related to
triangle T is the set of all triangles.

**Question 14. **

**Let L be the set of all lines in XY plane and R be the relation
in L defined as R={(L _{1}, L_{2}): L_{1} is parallel to L_{2}**

**}. Show that R is an equivalence relation. Find the set of all lines related to the line y=2x+4.**

**Solution:**

R is reflexive as any
line L_{1} is parallel to itself i.e., (L_{1}, L_{2}) ∈ R.

Let (L_{1}, L_{2}) ∈ R,
then L_{1} is parallel to L_{2}. So, (L_{2}, L_{1}) ∈ R. Therefore, R is symmetric.

Let (L_{1}, L_{2}), (L_{2}, L_{3}) ∈ R, then L_{1} is parallel to L_{2} and L_{2} is parallel to L_{3}. So, L_{1} is parallel to L_{3}. Therefore, R is transitive.

Hence, R is an
equivalence relation.

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.Slope of the line is m=2.

It is known that
parallel lines have the same slopes. The line parallel to the given line is of
the form y=2x +c, where c ∈ R.

Hence, the set of all lines related to the given
line is given by y=2x +c, where c ∈ R.

**Question 15. **

**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)}. Choose the correct answer.**

**(A) R is reflexive and symmetric but not transitive.**

**(B) R is reflexive and transitive but not symmetric.**

**(C) R is symmetric and transitive but not reflexive.**

**(D) R is equivalence relation****.**

**Solution:**

It is seen that (a,a) ∈ R, for every a ∈ {1, 2, 3, 4}. Therefore, R is reflexive.

It is seen that (1,2) ∈ R but (2,1) ∉ R. Therefore, R is not symmetric.

Also, it is observed
that (a,b), (b,c) ∈ R⇒ (a,c) ∈ R for all a, b, c ∈ {1, 2, 3, 4}. Therefore, R is transitive.

Hence, R is reflexive and transitive but not
symmetric. The correct answer is** B.**

**Question 16. **

**Let R be the
relation in the set N given by R={(a,b):a=b-2; b>6}. Choose the
correct answer.**

**(A) (2,4) ∈ R**

**(B) (3,8) ∈ R**

**(C) (6,8)**

**(D) (8,7) ∈ R**

**Solution:**

Since b>a, (2,4) ∉ R also, as 3≠8-2, (3,8)∉R and as 8≠7-2. Therefore, (8,7)∉R

Consider (6,8). We have
8>6 and also, 6=8-2. Therefore, (6,8) ∈ R.

The correct answer is** C.**

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