NCERT Solutions for Class 12 Maths Chapter 1 Relations and Functions Ex 1.1 are part of NCERT Solutions for Class 12 Maths. Here we have given NCERT Solutions for Class 12 Maths Chapter 1 Relations and Functions Ex 1.1.

- Relations and Functions Class 12 Ex 1.2
- Relations and Functions Class 12 Ex 1.3
- Relations and Functions Class 12 Ex 1.4

Board |
CBSE |

Textbook |
NCERT |

Class |
Class 12 |

Subject |
Maths |

Chapter |
Chapter 1 |

Chapter Name |
Relations and Functions |

Exercise |
Ex 1.1 |

Number of Questions Solved |
16 |

Category |
NCERT Solutions |

## NCERT Solutions for Class 12 Maths Chapter 1 Relations and Functions Ex 1.1

**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}**

**(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 an 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) Relation R in the set A = {1, 2,….,14} defined as R= {(x,y): 3x -y = 0}

(a) Put y = x, 3x – x ≠ 0 => R is not reflexive.

(b) If 3x – y = 0, then 3y – x ≠ 0, R is not symmetric

(c) If 3x – y = 0, 3y – z = 0,then 3x – z ≠ 0,R is not transitive.

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

(a) Putting y = x, x ≠ x + 5, R is not reflexive

(b) Putting y = x + 5, then x ≠ y + 5,R is not symmetric.

(c) If y = x + 5, z = y + 5, then z ≠ x + 5 =>R is not transitive.

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

(a) Putting y = x, x is divisible by x => R is reflexive.

(b) If y is divisible by x, then x is not divisible by y when x ≠ y => R is not symmetric.

(c) If y is divisible by x and z is divisible by y then z is divisible by x e.g., 2 is divisible by 1,4 is divisible by 2.

=> 4 is divisible by 1 => R is transitive.

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

(a) x – x=0 is an integer => R is reflexive

(b) x – y is an integer so is y – x => R is transitive.

(c) x – y is an integer, y- z is an integer and x – z is also an integer => R is transitive.

(v) R is a set of human beings in a town at a particular time.

(a) R = {(x, y)} : x and y work at the same place. It is reflexive as x works at the same place. It is symmetric since x and y or y and x work at same place.

It is transitive since X, y work at the same place and if y, z work at the same place,u then x and z also work at the same place.

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

With similar reasoning as in part (a), R is reflexive, symmetrical and transitive.

(c) R: {(x, y)}: x is exactly 7 cm taller than y it is not reflexive: x cannot 7 cm taller than x. It is not symmetric: x is exactly 7 cm taller than y, y cannot be exactly 7 cm taller than x. It is not transitive: If x is exactly 7 cm taller than y and if y is exactly taller than z, then x is not exactly 7 cm taller than z.

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

R is not reflexive: x cannot be wife of x. R is not symmetric: x is wife of y but y is not wife of x.

R is not transitive: if x is a wife of y then y cannot be the wife of anybody else.

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

It is not reflexive: x cannot be father of himself. It is not symmetric: x is a father of y but y cannot be the father of x.

It is not transitive: x is a father of y and y is a father of z then x cannot be the father of z.

**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:**

(i) R is not reflexive, a is not less than or equal to a² for all a ∈ R, e.g., is not less than .

(ii) R is not symmetric since if a ≤ b² then b is not less than or equal to a² e.g. 2 < 5² but 5 is not less than 2².

(iii) R is not transitive: If a ≤ b², b ≤ c², then a is not less than c², e.g. 2 < (-2)², -2 < (-1)², but 2 is not less than (-1)².

**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:**

(i) R is not reflexive a ≠ a + 1.

(ii) R is not symmetric if b = a + 1, then a ≠ b+ 1

(iii) R is not transitive if b = a + 1, c = b + 1 then c ≠ a + 1.

**Question 4.**

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

**R = {(a,b):a≤b}**

**Solution:**

(i) R is reflexive, replacing b by a, a ≤ a =>a = a is true.

(ii) R is not symmetric, a ≤ b, and b ≤ a which is not true 2 < 3, but 3 is not less than 2.

(iii) R is transitive, if a ≤ b and b ≤ c, then a ≤ c, e.g. 2 < 3, 3 < 4 => 2 < 4.

**Question 5.**

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

**Solution:**

(i) R is not reflexive.

(ii) R is not symmetric.

(iii) R is not 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:**

(i) (1, 1),(2, 2),(3, 3) do not belong to relation R

∴ R is not reflexive.

(ii) It is symmetric (1,2) and (2,1) belong to R.

(iii) there are only two element 1 and 2 in this relation and there is no third element c in it =>R is not 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:**

(i) The number of pages in a book remain the same

=> Relation R is reflexive.

(ii) The book x has the same number of pages as the book y.

=> Book y has the same number of pages as the book x.

=> The relation R is symmetric.

(iii) Book x and y have the same number of pages. Also book y and z have the same number of pages.

=> Books x and z also have the same number of pages.

R is transitive also Thus, 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 element of {1,3,5} is related to any element of {2,4}.**

**Solution:**

A= {1,2,3,4,5} and R= {(a,b): |a – b| is even}

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

(a) (i) Let us take any element of a set A. then |a – a| = 0 which is even.

=> R is reflexive.

(ii) If |a – b| is even, then |b – a| is also even, where,

R = {(a, b) : |a – b| is even} => R is symmetric.

(iii) Further a – c = a – b + b – c

If |a – b| and |b – c| are even, then their sum |a – b + b – c| is also even.

=> |a – c| is even,

∴ R is transitive. Hence R is an equivalence relation.

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

Since|1 – 3| = 2, |3 – 5| = 2, |1 – 5| = 4. 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 |2 – 4| = 2 an even number.

No element of set {1, 3, 5} is related to any element of {2,4}.

**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:**

The set A ∴ {x ∈ Z : 0 ≤ x ≤ 12} = {0, 1, 2,…..12}

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

|a – b| = 4k on b = a + 4k.

∴R ={(1,5),(1,9),(2, 6), (2, 10), (3, 7), (3, 11) , (4,8), (4,12), (5,9), (6,10), (7,11), (8.12) ,(0,0),(1,1), (2,2),…..(12,12)

(a) (a – a) = 0 = 4k,where k = 0=>(a,a)∈R

∴ R is reflexive.

(b) If |a – b| = 4k,then|b – a| = 4k i.e. (a,b) and (b, a) both belong to R. R is symmetric.

(c) a – c = a – b + b – c

when a – b and b – c are both multiples of 4 then a – c is also a multiple of 4. This shows if (a,b)(b,c) ∈ R then a – c also ∈ R

∴ R is an equivalence relation. The sets related 1 are {(1,5), (1,9)}.

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

{(0,0), (1,1), (2,2)…..(12,12)}

(a) a = a => (a, a) ∈R

R is reflexive.

(b) Again if (a, b) ∈R, then (b, a) also ∈R

Since a = b and (a, b) ∈ R => R is symmetric.

(c) If(a, b) ∈R,then (b, c) ∈R =>a = b = c

∴ a = c => (a, c) ∈R, Hence, R is transitive set related 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:**

Let A = set of straight lines in a plane

(i) R: {(a, b): a is perpendicular to b} let a, b be two perpendicular lines.

(ii) Let A = set of real numbers R= {(a,b):a>b}

(a) An element is not greater than itself

∴R is not reflexive.

(b) If a > b than b is not greater than a => R is not symmetric

(c) If a > b also b > c, then a > c thus R is transitive

Hence, R is transitive but neither reflexive nor symmetric.

(iii) The relation R in the set {1,2,3} is given by R= {(a, b) :a + b≤4}

R= {(1,1), (1,2), (2,1), (1,3), (3,1), (2,2)} (1,1), (2,2) ∈ R => R is reflexive

(1,2), (2,1), (1,3), (3,1) => R is symmetric

But it is not transitive, since (2,1) ∈R, (1,3) ∈R but (2,3)∉R.

(iv) The relation R in the set {1,2,3} given by R = {(a, b): a≤b} = {(1,2), (2,2), (3,3), (2,3), 0,3)}

(a) (1,1), (2,2), (3,3) ≤R => R is reflexive

(b) (1, 2) ∈R, but (2, 1) ∉R => R is not symmetric

(c) (1,2) ∈R, (2,3) ∈R,Also (1,3) ∈R=>R is transitive.

(v) The relation R in the set {1,2,3} given by R= {(a, b): 0 < |a – b| ≤ 2} = {(1,2), (2,1), (1, 3), (3,1), (2,3), (3,2)}

(a) R is not reflexive

∵ (1,1),(2,2), (3,3) do not belong to R

(b) R is symmetric

∵ (1,2), (2,1), (1,3), (3,1), (2,3), (3,2) ∈R

(c) R is transitive (1,2) ∈R, (2,3) ∈R, Also (1,3) ∈R

**Question 11.**

**Show that the relation R in the set A of punts in a plane given by R = {(P, Q) : distance of the point P from the origin is same as the distance of the punt Q from the origin}, is an equivalence relation. Further, show that the set of all punts related to a point P ≠ (0,0) is the circle passing through P with origin as centre.**

**Solution:**

Let O be the origin then the relation

R={(P,Q):OP=OQ}

(i) R is reflexive. Take any distance OP,

OP = OP => R is reflexive.

(ii) R is symmetric, if OP = OQ then OQ = OP

(iii) R is transitive, let OP = OQ and OQ = OR =>OP=OR

Hence, R is an equivalence relation.

Since OP = K (constant) => P lies on a circle with centre at the origin.

**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:**

(i) In a set of triangles R = {(T1, T2) : T1 is similar T2}

(a) Since A triangle T is similar to itself. Therefore (T, T) ∈ R for all T ∈ A.

∴ Since R is reflexive

(b) If triangle T1 is similar to triangle T2 then T2 is similar triangle T1

∴ R is symmetric.

(c) Let T1 is similar to triangle T2 and T2 to T3 then triangle T1 is similar to triangle T3,

∴ R is transitive.

Hence, R is an equivalence relation.

(ii) Two triangles are similar if their sides are proportional now sides 3,4,5 of triangle T1 are proportional to the sides 6, 8, 10 of triangle T3.

∴ 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:**

Let n be the number of sides of polygon P1.

R= {(P1, P2): P1 and P2 are n sides polygons}

(i) (a) Any polygon P1 has n sides => R is reflexive

(b) If P1 has n sides, P2 also has n sides then if P2 has n sides P1 also has n sides.

=> R is symmetric.

(c) Let P1, P2; P2, P3 are n sided polygons. P1 and P3 are also n sided polygons.

=> R is transitive. Hence R is an equivalence relation.

(ii) The set A = set of all the triangles in a plane.

**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:**

L = set of all the lines in XY plane, R= {(L1,L2) : L1 is parallel to L2}

(i) (a) L1 is parallel to itself => R is reflexive.

(b) L1 is parallel to L2 => L2 is parallel to L1 R is symmetric.

(c) Let L1 is parallel to L2 and L2 is parallel to L3 and L1 is parallel to L3 => R is transitive.

Hence, R is an equivalence relation.

(ii) Set of parallel lines related to y = 2x + 4 is y = 2x + c, where c is an arbitrary constant.

**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 an equivalence relation.**

**Solution:**

(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)∈R**

**(d)(8,7)∈R**

**Solution:**

Option (c) satisfies the condition that a = b – 2

i. e. 6 = 8 – 2 and b > 6, i.e. b = 8

=> option (c) is correct.

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