NCERT Solutions class 12 Maths Exercise 1.1 Ch 1 Relations and Functions

Ans. (i) R =  in A = {1, 2, 3, 4, 5, 6, ……13, 14}

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

Since,  R,  R is not reflexive.

Again  R but  R R is not symmetric.

Also (1, 3) R and (3, 9)  R but (1, 9)  R,  R is not transitive.

Therefore, R is neither reflexive, nor symmetric and nor transitive.

(ii) R =  in set N of natural numbers.

Clearly R = {(1, 6), (2, 7), (3, 8)}

Now  R, R is not reflexive.

Again  R but  R R is not symmetric.

Also (1, 6) R and (2, 7)  R but (1, 7)  R,  R is not transitive.

Therefore, R is neither reflexive, nor symmetric and nor transitive.

(iii) R =  in A = {1, 2, 3, 4, 5, 6}

Clearly 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)

Now  i.e., (1, 1), (2, 2) and (3, 3)  R  R is reflexive.

Again  i.e., (1, 2) R but  R R is not symmetric.

Also (1, 4) R and (4, 4)  R and (1, 4)  R, R is transitive.

Therefore, R is reflexive and transitive but not symmetric.

(iv) R =  in set Z of all integers.

Now  i.e., (1, 1) = 1 – 1 = 0  Z  R is reflexive.

Again   R and   R, i.e.,  and  are an integerR is symmetric.

Also   Z and   Z and

  R,  R is transitive.

Therefore, R is reflexive, symmetric and transitive.

(v) Relation R in the set A of human being in a town at a particular time.

(a) R = 

Since  R, because  and  work at the same place. R is reflexive.

Now, if  R and  R, since  and  work at the same place and  and  work at the same place.  R is symmetric.

Now, if  R and  R   R. R is transitive

Therefore, R is reflexive, symmetric and transitive.

(b) R = 

Since  R, because  and  live in the same locality. R is reflexive.

Also  R  R because  and  live in same locality and  and  also live in same locality. R is symmetric.

Again R and  R  R  R is transitive.

Therefore, R is reflexive, symmetric and transitive.

(c) R = 

 is not exactly 7 cm taller than , so  R R is not reflexive.

Also  is exactly 7 cm taller than  but  is not 7 cm taller than , so  R but  R R is not symmetric.

Now  is exactly 7 cm taller than  and  is exactly 7 cm taller than  then it does not  imply that  is exactly 7 cm taller than  R is not transitive.

Therefore, R is neither reflexive, nor symmetric and nor transitive.

(d) R = 

 is not wife of , so  R  R is not reflexive.

Also  is wife of  but  is not wife of , so  R but  R  R is not symmetric.

Also  R and  R then  R  R is not transitive.

Therefore, R is neither reflexive, nor symmetric and nor transitive.

(e) R = 

 is not father of , so  R R is not reflexive.

Also  is father of  but  is not father of ,

so  R but  R  R is not symmetric.

Also  R and  R then  R  R is not transitive.

Therefore, R is neither reflexive, nor symmetric and nor transitive.

Ans. R = , Relation R is defined as the set of real numbers.

(i) Whether  R, then  which is false.  R is not reflexive.

(ii) Whether , then  and , it is false. R is not symmetric.

(iii) Now ,   , which is false. R is not transitive

Therefore, R is neither reflexive, nor symmetric and nor transitive.

Ans. R =  R

Now R = {(1, 2), (2, 3), (3, 4), (4, 5), (5, 6)} and 

(i) When    which is false, so  R, R is not reflexive.

(ii) Whether , then  and , false R is not symmetric.

(iii) Now if  R,  R   R

(iv) Then  and    which is false. R is not transitive.

Therefore, R is neither reflexive, nor symmetric and nor transitive.

Ans. (i)  which is true, so  R,  R is reflexive.

(ii)  but  R is not symmetric.

(iii)  and    which is true. R is transitive.

Therefore, R is reflexive and transitive but not symmetric.

Ans. (i) For   which is false.  R is not reflexive.

(ii) For  and  which is false.  R is not symmetric.

(iii) For  and ,  which is false.  R is not transitive.

Therefore, R is neither reflexive, nor symmetric and nor transitive.

Ans. R = {(1, 2), (2, 1)}, so for , (1, 1)  R.  R is not reflexive.

Also if  then  RR is symmetric.

Now  R and  then does not imply  R  R is not transitive..

Therefore, R is symmetric but neither reflexive nor transitive.

Ans. Books  and  have same number of pages   R  R is reflexive.

If  R   R, so  R is symmetric.

Now if   R,  R   R  R is transitive.

Therefore, R is an equivalence relation.

Ans.  A = {1, 2, 3, 4, 5} and R = , then R = {(1, 3), (1, 5), (3, 5), (2, 4)}

(a) For   which is even. R is reflexive.

If  is even, then  is also even.   R is symmetric.

Now, if  and  is even then   is also even.  R is transitive.

Therefore, R is an equivalence relation.

(b) Elements of {1, 3, 5} are related to each other.

Since  all are even numbers

 Elements of {1, 3, 5} are related to each other.

Similarly elements of (2, 4) are related to each other.

Since  an even number, then no element of the set {1, 3, 5} is related to any element of (2, 4).

Hence no element of {1, 3, 5} is related to any element of {2, 4}.

(i) R = 

(ii) R =  is an equivalence relation. Find the set of all elements related to 1 in each case.

Ans. (a) (i) A = A = {0, 1, 2, 3, ……….., 12}

Now R = 

 R = {(4, 0), (0, 4), (5, 1), (1, 5), (6, 2), (2, 6), ….. (12, 9), (9, 12), …. (8, 0), (0, 8), …..  (8, 4), (4, 8), …… (12, 12)}

Here,  is a multiple of 4.

  is a multiple of 4. R is reflexive.

Also we observe that R is symmetric.

And   is the multiple of 4.  R is transitive.

Hence R is an equivalence relation.

(ii) R =  and A = {0, 1, 2, 3, ……….., 12}

 R = {(0, 0), (1, 1), (2, 2), …….. (12, 12)}

For  R is reflexive.

As  then R is symmetric.

Also   then R is transitive.

Hence R is an equivalence relation.

(b) Now set of all elements related to 1 in each case.

(i) Required set = {1, 5, 9}  (ii) Required set = {1}

(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.

Ans. (i) The relation “is perpendicular to”  is not perpendicular to 

If  then , however if  and  then  is not perpendicular to 

So it is clear that R “is perpendicular to” is a symmetric but neither reflexive nor transitive.

(ii) Relation R = 

We know that  is false. Also  but  is false and if ,  this implies 

Therefore, R is transitive, but neither reflexive nor symmetric.

(iii) “is friend of” R = 

It is clear that  is friend of R is reflexive.

Also  is friend of  and  is friend of  R is symmetric.

Also if   is friend of  and  is friend of  then

 cannot be friend of R is not transitive.

Therefore, R is reflexive and symmetric but not transitive.

(iv) “is greater or equal to” R = 

It is clear that R is reflexive.

And  does not imply R is not symmetric.

But ,    R is transitive.

Therefore, R is reflexive and transitive but not symmetric.

(v) “is brother of” R = 

It is clear that  is not the brother of R is not reflexive.

Also  is brother of  and  is brother of R is symmetric.

Also if   is brother of  and  is brother of  then

 can be brother of R is transitive.

Therefore, R is symmetric and transitive but not reflexive.

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

Let P and Q and O (0, 0).

OP = OQ   =  = 

Now, For (P, P), OP = OP R is reflexive.

Also OP = OQ and OQ = OP  (P, Q) = (Q, P) R R is symmetric.

Also OP = OQ and OQ = OR  OP = OQ  R is transitive.

Therefore, R is an equivalent relation.

Part II: As  =  =  (let)   which represents a circle with centre (0, 0) and radius 

Ans. Part I: R = {(T₁, T₂): T₁ is similar to T₂} and T₁, T₂ are triangle.

We know that each triangle similar to itself and thus (T₁, T₂) R R is reflexive.

Also two triangles are similar, then T₁ T₂  T₁ T₂ R is symmetric.

Again, if then T₁ T₂  and then T₂ T₃ then T₁ T₃ R is transitive.

Therefore, R is an equivalent relation.

Part II: It is given that T₁, T₂ and T₃ are right angled triangles.

T₁ with sides 3, 4, 5  T₂ with sides 5, 12, 13 and

T₃ with sides 6, 8, 10

Since, two triangles are similar if corresponding sides are proportional.

Therefore, 

Therefore, T₁ and T₃ are related.

Ans. Part I: R = {(P₁, P₂): P₁ and P₂ have same number of sides}

(i) Consider the element (P₁, P₂), it shows P₁ and P₂ have same number of sides. Therefore, R is reflexive.

(ii) If (P₁, P₂) R then also (P₂, P₁) R

 (P₁, P₂) = (P₂, P₁) as P₁ and P₂ have same number of sides, therefore, R is symmetric.

(iii) If (P₁, P₂) R and (P₂, P₃) R then also (P₁, P₃) R as P₁, P₂ and P₃ have same number of sides, therefore, R is transitive.

Therefore, R is an equivalent relation.

Part II: we know that if 3, 4, 5 are the sides of a triangle, then the triangle is right angled triangle. Therefore, the set A is the set of right angled triangle.

Ans. Part I: R = {(L₁, L₂): L₁ is parallel to L₂}

(i) It is clear that L₁  L₁ i.e., (L₁, L₁) R R is reflexive.

(ii) If L₁  L₂ and L₂  L₁ then (L₁, L₂) R  R is symmetric.

(iii) If L₁  L₂ and L₂  L₃  L₁  L₃ R is transitive.

Therefore, R is an equivalent relation.

Part II: All the lines related to the line  and  where  is a real number.

(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.

Ans. Let R be the relation in the set {1, 2, 3, 4} is given by

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

(a) (1, 1), (2, 2), (3, 3), (4, 4)  R  R is reflexive.

(b) (1, 2)  R but (2, 1)  R  R is not symmetric.

(c) If (1, 3)  R and (3, 2)  R then (1, 2)  R  R is transitive.

Therefore, option (B) is correct.

(A) (2, 4)  R

(B) (3, 8)  R

(C) (6, 8)  R

(D) (8, 7)  R

Ans. Given: 

(A) , Here  is not true, therefore, this option is incorrect.

(B) and  3 = 6, which is false.

Therefore, this option is incorrect.

(C) and  6 = 6, which is true.

Therefore, option (C) is correct.

(D) and  8 = 5, which is false.