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NCERT Solutions Class 12 maths Chapter-1 (Relation And Functions)Exercise 1.1

NCERT Solutions Class 12 maths Chapter-1 (Relation And Functions)Exercise 1.1

NCERT Solutions Class 12 Maths from class 12th Students will get the answers of Chapter-1 (Relation And Functions) Exercise 1.1. This chapter will help you to learn the basics and you should expect at least one question in your exam from this chapter.
We have given the answers of all the questions of NCERT Board Mathematics Textbook in very easy language, which will be very easy for the students to understand and remember so that you can pass with good marks in your examination.
Solutions Class 12 maths Chapter-1 (Relation And Functions)Exercise 1.1
NCERT Question-Answer

Class 12 Mathematics

Chapter-1 (Relation And Functions)

Questions and answers given in practice

Chapter-1 (Relation And Functions)

Exercise 1.1

Set 1

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):3x-y=0}

Solution:

A={1 ,2 ,3,…,13, 14}

R={(x,y): 3x-y=0}

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

R is not reflexive since (1,1),(2,2),(3,3),…,(14,14)R

Also, R is not reflexive since (1,3)R, but (3,1)R.[since 3(3)-1≠0]

Also, R is not transitive as (1,3), (3,9)R, but (1,9)R.[since 3(1)-9≠0]

Hence, R is not reflexive, nor symmetric nor transitive.

(ii) Relation R in the set N of natural numbers defined as R={ ( x, y) : y=x+5 and x<4}

Solution:

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

It is seen that (1, 1)R. Therefore, R is not reflexive.

(1, 6)R. But, (6, 1)R so, R is not symmetric.

Now, since their is no pair in R such that (x, y) and (y, z)R, so (x, z) can not belong to R. Therefore, R is not transitive.

So, we can conclude that R is neither reflexive, nor symmetric, nor transitive.

(iii) Relation R in the set A= {1, 2, 3, 4, 5, 6} as R={(x, y): y is divisible by x}

Solution:

A={1, 2, 3, 4, 5, 6}

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

We know that any number is always divisible by itself. (x, x) R. Therefore, R is reflexive.

We see, (2, 4)R [4 is divisible by 2]. But, (4, 2)R [2 is not divisible by 4]. Therefore, R is not symmetric.

Let’s assume (x, y), (y, z) R. Then, y is divisible by x and z is divisible by y. Therefore, z is divisible by x. (x, z) R. Therefore, R is transitive.

(iv) Relation R in the set Z of all integers defined as R={(x, y): x-y is an integer}

Solution:

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

For every x  Z, (x, x)  R [x-x=0 which is an integer]. Therefore, R is reflexive.

For every x, y  Z if (x, y) R, then x-y is an integer. -(x-y) is also an integer. (y-x) is also an integer. Therefore, R is symmetric.

Let’s assume, (x, y) and (y, z) R, where x, y and z  Z. (x-y) and (y-z) are integers. x-z=(x-y)+(y-z) is an integer. (x, z) R. Therefore, R is transitive.

(v) Relation R in a 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. 

Solution:

We can see (x,x) R .Therefore, R is reflexive.

If (x,y) R, then x and y work at the same place. So, (y,x) R. Therefore, R is symmetric.

Let, (x,y), (y,z) R. 

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

x and z work at the same place.

Therefore, R is transitive.

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

Solution:

We can see (x,x) R. Therefore, R is reflexive.

If (x,y) R, then x and y live in same locality. So, (y,x) R. Therefore, R is symmetric.

Let (x,y) R and (y,z) R. So, x, y and z live in the same locality. So, (x,z) R. Therefore, R is transitive.

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

Solution:

(x,x)R since, human being can not be taller than himself. So, R is not reflexive.

Let (x,y) R ,then x is exactly 7 cm taller than y. Then, y is not taller than x. Therefore, R is not symmetric.

Let (x,y), (y,z) R, then x is exactly 7 cm taller than y and y is exactly 7 cm taller than z which means x is 14 cm taller than z. So, (x,z)R . Therefore, R is not transitive.

(d) R={(x,y): x is wife of y}. 

Solution:

(x,x) R. Since, x can not be the wife of herself. Therefore, R is not reflexive.

Let (x,y) R, then x is the wife of y. So, y is not the wife of x ,i.e., (y,x) R. Therefore, R is not symmetric.

Let (x,y), (y,z) R, then x is the wife of y and y is the wife of z which is not possible. So, R can not be transitive.

(e) R={(x,y): x is father of y}. 

Solution:

(x,x) R. Since, x can not be the father of himself. Therefore, R is not reflexive.

Let (x,y) R, then x is the father of y. So, y can not be the father of x. So, (y,x) R. Therefore, R is not symmetric.

Let (x,y), (y,z) R, then x is the father of y and y is the father of z which means x is the grandfather of z. So, So, (x, z) R. Therefore, R is not 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:

It can be observed that (½, ½) R, since ½>(½)2 =¼. Therefore, R is not reflexive.

(1,4) R as 1<42 .But, (4,1)  R. Therefore, R is not symmetric.

(3,2), (2,1.5) R. But, 3> (1.5)2 =2.25 . So, (3,1.5)  R. Therefore, 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 the set {1, 2, 3, 4, 5, 6} be named A.

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

We can see (x, x)  R. Since, x ≠ x+1. Therefore, R is not reflexive.

It is observed that (1,2) R but, (2,1)  R. Therefore, R is not symmetric.

We can see, (1,2), (2, 3) R, but (1,3)  R. Therefore, R is not transitive.

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

Question 4. 

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

Solution:

Clearly, (a,a) R as a=a. Therefore, R is reflexive.

(2,4) R (as 2<4) but (4,2)  R as 4 is greater than 2. Therefore, R is not symmetric.

Let (a,b), (b,c) R. Then, a≤ b and b≤ c.

a ≤ c. 

(a, c) R. Therefore, R is transitive.

Hence, R is reflexive and transitive but not symmetric.

Question 5. 

Check whether the relation R in  defined as R ={ (a, b): a ≤ b3 } is reflexive, symmetric or transitive. 

Solution:

It is observed that (½, ½)  R as ½ > (½)3 =(1/8). Therefore, R is not reflexive.

(1,2) R(as 1<8) but, (2,1)  R. Therefore, R is not symmetric.

We have, (3, 3/2), (3/2, 6/5) R but, (3, 6/5)  R. Therefore, R is not transitive.

Hence, 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 the set {1, 2, 3} be named A.

It is seen that, (1, 1), (2,2), (3,3)  R. Therefore, R is not reflexive.

As (1, 2) R and (2, 1) R. Therefore, R is symmetric.

However, (1, 1)  R. Therefore, 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 the same number of pages} is a 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} 

R is reflexive since (x,x) R as x and x have the same number of pages.

Let (x,y)

x and  y have the same number of pages

y and x have the same number of pages.

(y,x) R

Therefore , R is symmetric.

Let (x,y) R and (y,z) R.

x and y 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

Therefore, 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 R={(a,b):|a-b| is even} , 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 elements of {1,3,5} is related to any element of {2,4}.

Solution:

A={ 1,2,3,4,5}

R={(a,b):|a-b| is even }

It is clear that for any element a A, we have |a-a|=0 (which is even).

Therefore, R is reflexive.

Let (a,b) R.

|a-b| is even.

|-(a-b)|=|b-a| is also even.

(b,a) R

Therefore, R is symmetric.

Now , let (a,b) R and (b,c) R.

|a-b| is even and |b-c| is even.

(a-b) is even and (b-c) is even.

(a-c)=(a-b)+(b-c) is even.   [Sum of two even integers is even]

|a-c| is even.

Therefore, R is transitive.

Hence, 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 all elements of {1,3,5} are odd and all elements of {2,4} are even . Thus, the modulus of the difference between the two elements (from each of these two subsets) will not be even.

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}

is an equivalence relation. Find the set of all elements related to 1 in each case.

Solution:

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

 (i) R={(a,b):|a-b| is a multiple of 4}

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

Therefore, R is reflexive.

Now , let (a,b)R |a-b| is a multiple of 4.

|-(a-b)|=|b-a| is a multiple of 4.

(b,a)R

Therefore, R is symmetric.

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.

Hence, R is an equivalence relation .

The set of elements related to 1 is {1,5,9} since |1-1|=0 is a multiple of 4,

|5-1|=4 is a multiple of 4,and

|9-1|=8 is a multiple of 4.

(ii) R={(a,b):a=b}

For any element aA, we have (a,a) R , since a=a.

Therefore, R is reflexive.

Now , let (a,b)R.

a=b

b=a

(b,a)R

Therefore, R is symmetric.

Now, let (a,b)R and (b,c)R.

a=b and b=c

a=c

(a,c)R

Therefore, 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: 

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.

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

Therefore , 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 aR, we have (a,a)R since a cannot be strictly less than itself . In fact a=a.

Therefore, R is not reflexive. Now,

(1,2)R (as 1<2)

But, 2 is not less than 1.

Therefore, (2,1)R

Therefore, 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) Let 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 aA , (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,bR.

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): a3 ≥ b3 }

Clearly (a,a)R as a^3=a^3

Therefore, R is reflexive.

Now, (2,1)R(as 23>=13)

But, (1,2) R (as 13< 23)

Therefore, R is not symmetric

Let (a,b), (b,c) R.

a>= b3 and b3 >=c3

a3>=c3

(a, c)R

Therefore, 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.

Therefore, the relation R is transitive .

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

Exercise 1.1

Set 2

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 the 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.

Solution:

We can see (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. Therefore, R is reflexive.

Let (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.

So, (Q,P) R. Therefore, R is symmetric.

Let (P,Q), (Q,S) R.

The distance of point P from the origin is the same as the distance of point Q from the origin and also, the distance of point P from the origin is the same.

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

(P, S) R. Therefore, R is transitive.

Therefore, R is 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.

In other words, if O(0 0) is the origin and OP = k, then the set of all points related to P is at the same as k from the origin. Hence, this set of points forms a circle with the centre as the origin and the circle passes through point P.

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 is reflexive since, every triangle is similar to itself.

If (T1, T2) R, then T1 is similar to T2. So, (T2, T1) R. Therefore, R is symmetric.

Let (T1, T2), (T2, T3) R, then T1 is similar to T2 and T2 is similar to T3. So, T1 is also similar to T3. Therefore, (T1, T3) R so, R is transitive.

Thus, R is an equivalence relation.

We observe,

(3/6)=(4/8)=(5/10)=1/2

Therefore, the corresponding sides of triangles T1 and T3 are in the same ratio. Then, 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 is reflexive since(P1, P2) R as the same polygon has the same number of sides with itself.

Let (P1, P2) R, then P1 and P2 have the same number of sides. So, (P2, P1) R. Therefore, R is symmetric.

Let (P1, P2), (P2, P3) R, then P1 and P2 have the same number of sides. Also, P2 and P3 have the same number of sides. So, P1 and P3 have the same number of sides ,i.e., (P1, P3) R. Therefore, R is transitive.

Hence, 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 is reflexive as any line L1 is parallel to itself i.e., (L1, L2) R.

Let (L1, L2) R, then L1 is parallel to L2. So, (L2, L1) R. Therefore, R is symmetric.

Let (L1, L2), (L2, L3) R, then L1 is parallel to L2 and L2 is parallel to L3. So, L1 is parallel to L3. Therefore, R is transitive.

Hence, R is an equivalence relation.

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

It is known that parallel lines have the same slopes. The line parallel to the given line is of the form y=2x +c, where c R.

Hence, the set of all lines related to the given line is given by y=2x +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 equivalence relation.

Solution:

It is seen that (a,a) R, for every a {1, 2, 3, 4}. Therefore, R is reflexive.

It is seen that (1,2) R but (2,1) R. Therefore, R is not symmetric.

Also, it is observed that (a,b), (b,c) R (a,c) R for all a, b, c {1, 2, 3, 4}. Therefore, R is transitive.

Hence, R is reflexive and transitive but not symmetric. The correct answer is B.

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)

(D) (8,7) R

Solution:

Since b>a, (2,4) R also, as 3≠8-2, (3,8)R and as 8≠7-2. Therefore, (8,7)R

Consider (6,8). We have 8>6 and also, 6=8-2. Therefore, (6,8) R.

The correct answer is C.

 Chapter-1 (Relation And Functions)

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