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# NCERT Solutions class 12 Maths Exercise 1.1

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## NCERT Solutions class 12 Maths Relations and Functions

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 = .

(ii) Relation R in the set N of natural numbers defined as R =.

(iii) Relation R in the set A = {1, 2, 3, 4, 5, 6} as R =

(iv) Relation R in the set Z of all integers defined as R =

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

(a) R =

(b) R =

(c) R =

(d) R =

(e) R =

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.

#### 2. Show that the relation R in the set R of real numbers defined as R =  is neither reflexive nor symmetric 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.

#### 3. Check whether the relation R defined in the set {1, 2, 3, 4, 5, 6} as R =  is reflexive, symmetric or 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.

#### 4. Show that the relation R in R defined as R = , is reflexive and transitive but not symmetric.

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.

#### 5. Check whether the relation R in R defined by R =  is reflexive, symmetric or transitive.

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.

#### 6. Show that the relation in the set A = {1, 2, 3} given by R = {(1, 2), (2, 1)} is symmetric but neither reflexive 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.

#### 7. Show that the relation R in the set A of all the books in a library of a college, given by R =  is an equivalence relation.

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.

#### 8. Show that the relation R in the set A = {1, 2, 3, 4, 5} given by R =  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}.

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}.

#### 9. Show that each of the relation R in the set A =  given by:

(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}

#### 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.

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.

#### 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 points related to a point P  (0, 0) is the circle passing through P with origin as centre.

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

#### 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?

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.

#### 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?

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.

#### 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

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.

#### 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.

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.

#### 16. Let R be the relation in the set N given by R =  Choose the correct answer:

(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.

## NCERT Solutions class 12 Maths Exercise 1.1

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