NCERT Solution Class 12 Mathematics

PART 1

Chapter 1 Relations and Functions

Solutions of  Problems On  Exercise 1.1

Solution

R =  {(x,y):3x – y = 0}

(ie) R = {(x,y): y = 3x}

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

R is not reflexive as (x,x) ∉R ,∀ x∈A
R is not symmetric as (x,y)∈R⇏(y,x)∈R [Eg: (1,3) ∈R ,but(3,1)∉R]
R is not transitive as (1,3)∈R and(3,9)∈R,but (1,9)∉R

Solution

R = {(x,y) : y = x+5, x <4 }
(ie) R = {(1,6),(2,7),(3,8)}
R is not reflexive as (x,x) ∉R ,∀ x∈N
R is not symmetric as (x,y)∈R⇏(y,x)∈R [Eg: (1,6)∈R ,but(6,1)∉R]
R is transitive as transitivity is not violated .

Solution

R ={(x,y):y is divisible by x}
(ie) R = {(x,y):x is a factor of x}
(ie R = {(1,1),(1,2),(1,3),(1,4),(1,5),(1,6),(2,2),(2,4),(2,6),
(3,3),(3,6),(4,4),(5,5),(6,6)}
Sol : R is reflexive as (x,x) ∈R ,∀ x∈A
R is not symmetric as (x,y)∈R⇏(y,x)∈R [Eg: (1,6)∈R ,but(6,1)∉R]
R is transitive as (x,y) ∈R and (y,z)∈R ⇒(x,z)∈R ,∀x ,y ,z ∈A

Solution

R = {(x,y): x – y is an integer}
Sol: R is reflexive as (x,x)∈R,∀x∈Z [∵ x – x = 0 , an integer]
R is symmetric as x-y∈R ⇒(y,x)∈R ,∀x,y∈Z [If x-y is an integer ,
y-x is also an integer]
R is transitive as (x,y) ∈R and (y,z)∈ R⇒(x,z)∈R ,∀ x ,y ,z ∈Z
[If x – y and y – z are integers , then (x – y ) + (y – z ) = x – z is an integer} .

Solution

R is reflexive as (x,x) ∈R ,∀ x∈A
R is symmetric as (x,y)∈R⇒ (y,x)∈R, ∀ x,y∈A
R is transitive as (x,y) ∈R and (y,z)∈ R⇒(x,z)∈R ,∀ x ,y ,z ∈A

Solution

R is reflexive as (x,x) ∈R ,∀ x∈A
R is symmetric as (x,y)∈R⇒ (y,x)∈R, ∀ x,y∈A
R is transitive as (x,y) ∈R and (y,z)∈ R⇒(x,z)∈R ,∀ x ,y ,z ∈A

Solution

R is not reflexive as (x,x) ∉R ,∀ x∈A [ x is not exactly 7cm taller than x]
R is not symmetric as (x,y)∈R⇏ (y,x)∈R[If x is 7cm taller than y , then y
is 7cm shorter than x]
R is not transitive as (x,y) ∈R and (y,z)∈ R ⇏(x,z)∈R
[If x is 7cm taller than y and y is 7cm taller than z , then x is 14cm taller than z ]

Solution

R is not reflexive as (x,x) ∉R ,∀ x∈A
R is not symmetric as (x,y)∈R⇏ (y,x)∈R
R is not transitive as (x,y) ∈R and (y,z)∈ R ⇏(x,z)∈R

Solution

R is not reflexive as (x,x) ∉R
R is not symmetric as (x,y)∈R⇏ (y,x)∈R
R is not transitive as (x,y) ∈R and (y,z)∈ R ⇏(x,z)∈R

Solution

R is not reflexive as (a ,a) ∉R ,∀ a∈R [Eg: 1/2≰(1/2)^2
R is not symmetric as (a,b)∈R⇏ (b,a)∈R , ∀ a,b∈R
[Eg: 2 ≤6^2 ,but6≰2^2]
R is not transitive as (a,b) ∈R and (b,c)∈ R ⇏(a,c)∈R
[Eg: 5 ≤3^2 and 3≤2^2,but5≰2^2]

Solution

R= {(1,2),(2,3),(3,4) ,(4,5),(5,6)}
R is not reflexive as (a ,a) ∉R ,∀ a∈A Eg: (1,1)∈R
R is not symmetric as (a,b)∈R⇏ (b,a)∈R , ∀ a,b∈R Eg: (1,2)∈R ,but (2,1)∉R
R is not transitive as (a,b) ∈R and (b,c)∈ R ⇏(a,c)∈ R Eg: (1,2) ∈R and (2,3)∈R ,but (1,3)∉R

Solution

R is reflexive as (a,a)∈R ,∀a∈R (∵a≤a)
R is not symmetric as (a,b) ∈R ⇏(b,a)∈R , ∀ a,b∈A
R is transitive as (a,b) ∈R and (b,c) ∈R⇒(a,c)∈R,∀ a,b,c∈A [If a≤b
andb≤c , then a≤c]

Solution

R is not reflexive as (a,a)∉R ,∀a∈R [ Eg∶1/3≰(1/3)^3]
R is not symmetric as (a,b) ∈R ⇏(b,a)∈R, ,∀a,b∈R [Eg:1 ≤2^3,
but2≰1^3]
R is not transitive as (a,b) ∈R and (b,c) ∈R⇏(a,c)∈R,,∀a,b,c∈R
[Eg: 60≤4^3 and 4 ≤2^3 , but 60≰2^3]

Solution

R is not reflexive as (a,a)∉R ,∀a∈A [Eg: (1,1) ∉R]
R is symmetric as (a,b)∈R⇒ (b,a)∈R,∀a,b∈A
R is not transitive as (1,2) ∈R and (2,1) ∈R , but (1,1) ∉R

Solution

R is reflexive as (a,a)∈R ,∀a∈A
R is symmetric as (a,b)∈R⇒ (b,a)∈R,∀ a ,b ∈A
R is transitive as (a,b)∈Rand (b,c)∈R⇒ (a,c)∈R,∀ a ,b,c∈A
Since R is reflexive , symmetric and transitive , R is an equivalence relation .

Solution

R is reflexive as (a,a)∈R ,

∀a∈A [ |a – a | = 0 , an even number ]
R is symmetric as

(a,b)∈R⇒ (b,a)∈R,∀ a ,b ∈A [If |a – b|is even ,
then|b – a|is also even ]
R is transitive as (a,b)∈Rand (b,c)∈R⇒ (a,c)∈R,∀ a ,b,c∈A
[ If |a – b|is even and If |b – c|is even then |(a – b)+(b – c)| = |a – c | is even]
Since R is reflexive , symmetric and transitive , R is an equivalence relation .
All the elements of the set B = {1,3,5}are related to each other as |x – y| is even for all x , y ∈B.
All the elements of the set C ={2,4}are related to each other as |x – y| is even for all x , y ∈C.
No element of {1,3,5} is related to any element of {2,4} as |x-y| is odd for all
x ∈{1,3,5} and y ∈{2,4}or vice versa

Solution

A ={ 0,1,2,3,…..,12}
(i) R is reflexive as (a,a) ∈R ,∀ a∈A [ |a – a | = 0 , a multiple of 4 ]
R is symmetric as (a,b)∈R⇒ (b,a)∈R,∀ a ,b ∈A [If |a – b|is a multiple of 4 ,then|b – a|is also a multiple of 4
R is transitive as (a,b)∈Rand (b,c)∈R⇒ (a,c)∈R,∀ a ,b,c∈A
[ If |a – b| and If |b – c|are multiples of 4 , then |(a – b)+(b – c)| = |a – c | is also a multiple of 4 ]
Since R is reflexive , symmetric and transitive , R is an equivalence relation .
Set of elements related to 1 = { x ∈A∶(1,x)∈R} (ie){x∈A:|1-x| is a multiple of 4} = {1,5,9}

Solution

(ii) R is reflexive as (a,a) ∈R ,∀ a∈A
R is symmetric as (a,b)∈R⇒ (b,a)∈R,∀ a ,b ∈A
R is transitive as (a,b)∈Rand (b,c)∈R⇒ (a,c)∈R,∀ a ,b,c∈A
Since R is reflexive , symmetric and transitive , R is an equivalence relation .
Set of elements related to 1 = { x ∈A∶(1,x)∈R} ={1}

Solution

Consider the relation R in the set A = {1,2,3} defined as R = {(1,3),(3,1)}.Here R is symmetric but neither reflexive nor transitive

Solution

The relation R in the set A = {1,2,3} defined as R = {(1,3),(3,2),(1,2)}.
Here R is transitive but neither reflexive nor symmetric.

Solution

Consider the relation R in the set A = {1,2,3} defined as
R ={(1,1),(2,2),(3,3),(2,3),(3,2),(1,3),(3,1)} . Here R is reflexive and symmetric but not transitive [R is not transitive as (2,3)∈R and (3,1)∈R,but (2,1)∉R]

Solution

Solution

Consider the relation R in the set A = {1,2,3} defined as R =
{ (1,1),(1,2),(2,1),(2,2)} . Here R is symmetric and transitive but not reflexive .

Solution

R is reflexive as (P,P) ∈R ,∀ P∈A
R is symmetric as (P,Q) ∈R ⇒(Q,P)∈R , ∀ P ,Q∈A
R is transitive as (P,Q) ∈R and (Q,R)∈R⇒(P,R)∈R , ∀ P,Q,R∈A
Since R is reflexive , symmetric and transitive , R is an equivalence relation
Set of points related to P ≠(0,0) ={Q∈A ∶OP=OQ }
Thus , the set of points related to P ≠(0,0) is the circle passing through P with origin as centre .

Solution

R is reflexive as (T1, T(1 ))∈R ,∀ T1∈A
R is symmetric as (T1, T(2 )) ∈R ⇒(T2,T1) ∈R , ∀ T1,T2∈A
R is transitive as (T1, T(2 )) ∈R and (T2,T(3 ) )∈R ⇒ (T1, T(3 )) ∈R, ∀ T1,T2,T3∈A
Since R is reflexive , symmetric and transitive , R is an equivalence relation .
Since the sides of two similar triangles are proportional , triangles T1 and T3 are similar and so related .

Solution

R is reflexive as (P_1, P_(1 ))∈R ,∀ P_1∈A
R is symmetric as (P_1, P_(2 )) ∈R ⇒(P_2,P_1) ∈R , ∀ P_1,P_2∈A
R is transitive as (P_1, P_(2 )) ∈R and (P_2,P_(3 ) )∈R ⇒ (P_1, P_(3 )) ∈R,
∀ P_1,P_2,P_3∈A
Since R is reflexive , symmetric and transitive , R is an equivalence relation .
The set of all elements in A related to the right triangle T with sides 3,4 and 5 is the set of all triangles in A .

Solution

R is reflexive as (L1, L(1 ))∈R ,∀ L1∈L
R is symmetric as (L, L(2 )) ∈R ⇒(L2,L) ∈R , ∀ L1,L2∈L
R is transitive as (L1, L) ∈R and (L2,L(3 ) )∈R ⇒ (L, L) ∈R,
∀ L1,L2,L3∈A
Since R is reflexive , symmetric and transitive , R is an equivalence relation.
The set of all lines related to the line y = 2x + 4 is the set of all lines with equation y = 2x + k where k is some constant .

Solution

R is reflexive as (x,x) ∈A ,∀x∈A
R is not symmetric as (1,2) ∈R ,but (2,1)∉R
R is transitive as (x,y)∈R and (y,z)∈R ⇒(x,z)∈R,∀x,y,z ∈A
Thus , B is the correct option .

Solution

R = { (a,b): a = b – 2 , b >6} (ie) R = { (a,b): b = a+2 , b >6}
Thus , C is the correct option.