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**Class XII CBSE Mathematics
NCERT Solutions **

**Chapter 1, relations and functions**

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__CBSE
Class 12 NCERT Mathematics Exercise 1.1 Solved__

__CBSE Class 12 NCERT Mathematics Exercise 1.1 Solved__

(

__Cbse Class 12 Ncert Maths Textbook Page 5, 6, 7__)

__For solutions of previous questions 1-8 please visit__–#### Class12 Maths NCERT Solutions | Chapter 1, Relations and Functions Exercise 1.1 |CBSE 12 NCERT Maths Exercise 1.1 Solved

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}

__Solution__: (i) Given that

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

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

Or,
|– (a – b)| = |b – a| is a multiple of 4.

=>
(b, a) ∈
R

∴R is symmetric.

Now,
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 and hence, R is an equivalence relation.

The
set of elements related to 1 is {1, 5, 9} as,

|1
– 1| = 0 is a multiple of 4,

|5
– 1| = 4 is a multiple of 4,

|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.

∴ R is reflexive.

Now,
let (a, b) ∈
R.

⇒ a = b

⇒ b = a ⇒ (b, a) ∈ R

∴ R is symmetric.

Now,
let (a, b) ∈
R and (b, c) ∈
R.

⇒ a = b and b = c

⇒ a = c

⇒ (a, c) ∈ R

∴ 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: 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

*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

∴ 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*a*itself.
In
fact, a = a.

∴ R is not reflexive.

Now,
(1, 2) ∈
R (as 1 < 2)

But,
2 is not less than 1.

Therefore,
(2, 1) ∉ R

∴ 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)
Consider

*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}.
∴ R is
reflexive.

Now, (2, 1) ∈ R (as 2

^{3}≥ 1^{3})
But, (1, 2) ∉ R (as 1

^{3 }< 2^{3})
∴ R is not
symmetric.

Let (

*a*,*b*), (*b*,*c*) ∈ R.
⇒ a

^{3}≥ b^{3}and b^{3}≥ c^{3}
⇒ (

*a*,*c*) ∈ R
∴ 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.

∴The relation R is transitive.

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

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 same
as the distance of the point Q from the origin}, is an equivalence relation.
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.

__Solution__: To show that

R
= {(P, Q): distance of point P from the origin is the same as the distance of
point Q from the origin}

Clearly, (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.

∴ R is reflexive.

Say
(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.

⇒ (Q, P) ∈ R

∴ R is symmetric.

Suppose
(P, Q), (Q, S) ∈
R.

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

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

⇒
(P, S) ∈
R

So, R is transitive and in 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.

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. Thus, it is clear that the set of points forms a circle with the
centre as the origin and this circle passes through point P.

####
__CBSE XII Maths NCERT Solutions of Relations and Functions Exercise
1.1__

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.
Consider three right angle triangles T1 with sides 3, 4, 5, T2 with sides 5,
12, 13 and T3 with sides 6, 8, 10. Which triangles among T1, T2 and T3 are
related?

__Solution__: R = {(T1, T2): T1 is similar to T2}

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

Also,
if (T1, T2) ∈
R, then T1 is similar to T2.

⇒ T2 is similar to T1.

⇒ (T2, T1) ∈R ∴R is symmetric.

Suppose,
(T1, T2), (T2, T3) ∈ R.

⇒ T1 is similar to T2
and T2 is similar to T3.

⇒ T1 is similar to T3.

⇒ (T1, T3) ∈ R

∴ R is transitive.

Thus,
R is an equivalence relation.

Now,
we can observe that: 3/6 = 4/8 = 5/10 = 1/2

Since
corresponding sides of triangles T1
and T3 are in the same ratio
so, triangle T1 is similar to
triangle T3.

Hence,
T1 is related to T3.

Question 13: Show that the relation R defined in the set A of
all polygons as R = {(P1, P2): P1 and P2 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 = {(P1, P2): P1 and P2 have same the number of sides}

R
is reflexive since (

*P*1,*P*1) ∈ R as the same polygon has the same number of sides with itself.
Let
(

*P*1,*P*2) ∈ R.
⇒ P1 and P2 have
the same number of sides.

⇒ P2 and P1 have
the same number of sides.

⇒ (P2, P1) ∈ R

∴ R is symmetric.

Suppose
(P1, P2), (P2, P3) ∈ R.

⇒ P1 and P2 have
the same number of sides. Also, P2
and P3 have the same number of
sides.

⇒ P1 and P3 have
the same number of sides.

⇒ (P1, P3) ∈ R

Hence
R is transitive and so, 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 =
{(L1, L2): L1 is parallel to L2}. Show that R is an equivalence relation. Find
the set of all lines related to the line y = 2x + 4.

__Solution__: R = {(

*L*1,

*L*2): L1 is parallel to

*L*2}

R is reflexive as any
line

*L*1 is parallel to itself i.e., (*L*1,*L*1) ∈ R.
Let (

*L*1,*L*2) ∈ R.
⇒

*L*1 is parallel to*L*2. ⇒*L*2 is parallel to*L*1.
⇒ (

*L*2,*L*1) ∈ R
∴ R is symmetric.

Say, (

*L*1,*L*2), (*L*2,*L*3) ∈R.
⇒

*L*1 is parallel to*L*2. Also,*L*2 is parallel to*L*3.
⇒

*L*1 is parallel to*L*3.
∴ R is transitive.

Hence, R is an
equivalence relation.

The set of all lines
related to the line

*y*= 2*x*+ 4 is the set of all lines that are parallel to the line*y*= 2*x*+ 4.
Slope of line

*y*= 2*x*+ 4 is*m*= 2
The line parallel to the
given line is of the form

*y*= 2*x*+*c*, where*c*∈ R.
(Parallel lines have the
same slopes).

Hence, the set of all
lines related to the given line is given by

*y*= 2*x*+*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 an equivalence relation.

__Solution__: (B)

[Hint: R = {(1, 2), (2, 2),
(1, 1), (4, 4), (1, 3), (3, 3), (3, 2)}

It is seen that (a, a) ∈

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

∴ R
is not symmetric.

(a, b), (b, c) ∈ R ⇒ (a,
c) ∈ R for all a, b, c ∈ {1,
2, 3, 4}.

∴ R
is transitive.

=> R is reflexive and
transitive but not symmetric.]

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) ∈ R (D) (8, 7) ∈ R

__Solution__: (c)

[Hint:
R = {(a, b): a = b − 2, b > 6}

Since
b > 6, (2, 4) ∉ R

Similarly,
3 ≠ 8 − 2, (3, 8) ∉ R

Also
as 8 ≠ 7 – 2 (8, 7) ∉ R

Now
considering (6, 8).

Since
8 > 6 and also, 6 = 8 − 2.

So,
(6, 8) ∈
R]

*For solutions of previous questions 1-8 please visit**–*

**Class12 Maths NCERT Solutions | Chapter 1, Relations and Functions Exercise 1.1 |CBSE 12 NCERT Maths Exercise 1.1 Solved**