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**CBSE Class 12 Mathematics
NCERT Solutions **

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**Chapter
1, relations and functions**

**CBSE Class XII NCERT Mathematics Exercise 1.1 Solved***(*

__Cbse Class 12 Ncert Maths Textbook Page 5, 6, 7__)
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*): 3*x*−*y*= 0}
(ii) Relation R in the set N of natural numbers defined as

R = {(

*x*,*y*):*y*=*x*+ 5 and*x*< 4}
(iii) Relation R in the set

*A*= {1, 2, 3, 4, 5, 6} as
R = {(

*x*,*y*):*y*is divisible by*x*}
(iv) Relation R in the set Z of all integers defined as

R = {(

*x*,*y*):*x*−*y*is as integer}
(v) Relation R in the 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}

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

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

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

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

__Solution__:

(i)
A = {1, 2, 3 … 13, 14}

R
= {(x, y): 3x − y = 0}

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

We
can see that R is not reflexive since (1, 1), (2, 2) … (14, 14) ∉
R.

Also,
R is not symmetric as (1, 3) ∈R,
but (3, 1) ∉ R. [3(3) − 1 ≠ 0]

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

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

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

It can be seen that (1, 1) ∉
R.

∴ R is not reflexive.

(1, 6) ∈ R But, (1, 6) ∉
R.

∴ R is not symmetric.

Since there is no pair in R
such that (

*x*,*y*) and (*y*,*z*) ∈ R, then (*x*,*z*) cannot belong to R.
Therefore R is not
transitive.

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

(iii)
A = {1, 2, 3, 4, 5, 6}

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

We
know that any number (x) is divisible by itself.

So,
(x, x) ∈
R

∴ R is reflexive.

Now,
(2, 4) ∈
R [because 4 is divisible by 2]

But,
(4, 2) ∉ R. [Because 2 is not divisible by 4]

Therefore,
R is not symmetric.

Let
(x, y), (y, z) ∈
R. Then, y is divisible by x and z is divisible by y.

∴ z is divisible by x.

⇒ (x, z) ∈ R

∴ R is transitive.

Hence,
R is reflexive and transitive but not symmetric.

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

Now,
for every x ∈
Z, (x, x) ∈R
as x − x = 0 is an integer.

∴ R is reflexive.

Now,
for every x, y ∈
Z if (x, y) ∈
R, then x − y is an integer.

⇒ −(x − y) is also an
integer.

⇒ (y − x) is an integer.

∴ (y, x) ∈ R

∴ R is symmetric.

Let
(x, y) and (y, z) ∈R,
where x, y, z ∈
Z.

⇒ (x − y) and (y − z) are
integers.

⇒ x − z = (x − y) + (y − z)
is an integer.

∴ (x, z) ∈ R

∴ R is transitive.

Hence,
R is reflexive, symmetric, and transitive.

(v)

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

(x,
x) ∈
R

∴ R is reflexive.

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

⇒ y and x work at the same
place.

⇒ (y, x) ∈ R.

∴ R is symmetric.

Now,
let (x, y), (y, z) ∈
R

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

⇒ x and z work at the same
place.

⇒ (x, z) ∈R

∴ R is transitive.

Hence,
R is reflexive, symmetric, and transitive.

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

Clearly
(x, x) ∈
R as x and x is the same human being.

∴ R is reflexive.

If
(x, y) ∈R,
then x and y live in the same locality.

⇒ y and x live in the same
locality.

⇒ (y, x) ∈ R

∴ R is symmetric.

Now,
let (x, y) ∈
R and (y, z) ∈
R.

⇒ x and y live in the same
locality and y and z live in the same locality.

⇒ x and z live in the same
locality.

⇒ (x, z) ∈ R

∴ R is transitive.

Hence,
R is reflexive, symmetric, and transitive.

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

Now,
(x, x) ∉ R

Since
human being x cannot be taller than himself.

∴ R is not reflexive.

Now,
let (x, y) ∈
R.

⇒ x is exactly 7 cm taller
than y.

Then,
y is not taller than x.

∴ (y, x) ∉
R

Indeed
if x is exactly 7 cm taller than y, then y is exactly 7 cm shorter than x.

∴ R is not symmetric.

Now,
Let (x, y), (y, z) ∈
R.

⇒ x is exactly 7 cm taller
than y and y is exactly 7 cm taller than z.

⇒ x is exactly 14 cm taller
than z .

∴ (x, z) ∉
R

∴ R is not transitive.

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

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

Now,
(x, x) ∉ R

Since
x cannot be the wife of herself.

∴ R is not reflexive.

Let (x, y) ∈
R

⇒ x is the wife of y.

Clearly
y is not the wife of x.

∴ (y, x) ∉
R

If
x is the wife of y, then y is the husband of x.

∴ R is not transitive.

Say (x, y), (y, z) ∈
R

⇒ x is the wife of y and y is
the wife of z which is not possible. Also, it does not imply that x is the wife
of z.

∴ (x, z) ∉
R

∴ R is not transitive.

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

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

(x,
x) ∉
R

As
x cannot be the father of himself.

∴ R is not reflexive.

Now,
let (x, y) ∈
R.

⇒ x is the father of y.

⇒ y cannot be the father of
y.

Indeed,
y is the son or the daughter of y.

∴ (y, x) ∉
R

∴ R is not symmetric.

Now,
let (x, y) ∈
R and (y, z) ∈
R.

⇒ x is the father of y and y
is the father of z.

⇒ x is not the father of z.

Indeed
x is the grandfather of z.

∴ (x, z) ∉
R

∴ R is not transitive.

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

Question 2: Show that the relation R
in the set R of real numbers, defined as R = {(a, b): a ≤ b2} is neither
reflexive nor symmetric nor transitive.

__Solution__: R = {(a, b): a ≤ b

^{2}}

It
can be observed that

∴

R is not reflexive.

Now,
(1, 4) ∈
R as 1 < 4

^{2}But, 4 is not less than 12.
∴ (4, 1) ∉
R

∴ R is not symmetric.

Now,

(3,
2), (2, 1.5) ∈
R

(since
3 < 22 = 4 and 2 < (1.5)

^{2}= 2.25)
But,
3 > (1.5)

^{2}= 2.25
∴ (3, 1.5) ∉
R

∴ 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 A = {1, 2, 3, 4, 5, 6}.

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.

For
instance,

(1,
1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6) ∉ R ∴R is not reflexive.

It
can be observed that (1, 2) ∈
R, but (2, 1) ∉ R.

∴ R is not symmetric.

Now,
(1, 2), (2, 3) ∈
R

But,
(1, 3) ∉ R

∴ R is not transitive

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

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**CBSE XII Maths NCERT Solutions - ****Relations and Functions Exercise 1.1**

Question 4: Show that the relation R in R defined as R = {(a, b): a ≤ b}, is reflexive and transitive but not symmetric.

__Solution__: R = {(a, b); a ≤ b} Clearly (a, a) ∈ R as a = a.

∴ R is reflexive.

Let
us take, (2, 4) ∈
R (as 2 < 4)

But,
(4, 2) ∉ R as 4 is greater than 2.

∴ R is not symmetric.

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

⇒ a ≤ c

⇒ (a, c) ∈ R

∴ 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__: R = {(a, b): a ≤ b

^{3}}

It
can be observed that

∴ R is not reflexive.

For
example, (1, 2) ∈
R (as 1 < 2

^{3}= 8)
Since,
R is not symmetric.

Since,
R is not transitive so, 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 A = {1, 2, 3}.

A relation R on A is
defined as R = {(1, 2), (2, 1)}.

It is seen that (1, 1),
(2, 2), (3, 3) ∉ R.

∴ R is not reflexive.

Now, as (1, 2) ∈ R and (2, 1) ∈ R, then R is symmetric.

Now, (1, 2) and (2, 1) ∈ R

However,

(1, 1) ∉ R

∴ 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 same number
of pages} is an 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}

Now,
R is reflexive since (x, x) ∈
R as x and x has 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

∴ R is symmetric.

Now,
let (x, y) ∈R
and (y, z) ∈
R.

⇒ x and y and 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

∴ 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

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 element of {1, 3, 5} is related to any element of 2, 4}.

__Solution__: Clearly for any element a ∈ A, we have |a – a| = 0 (which is even).

∴ R is reflexive.

Let
(a, b) ∈
R.

⇒ (a, c) ∈ R

Since
R is transitive so, 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 elements
of {1, 3, 5} are odd and elements of {2, 4} are even. Hence, the modulus of the
difference between the two elements from each of these two subsets will not be
even.

__To view solutions of Question 9-16 please visit__-**Class XII CBSE Maths Relations and Functions - NCERT Solutions of Class 12 Mathematics Exercise 1.1**

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