Contents

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

Board |
CBSE |

Textbook |
NCERT |

Class |
Class 12 |

Subject |
Maths |

Chapter |
Chapter 1 |

Chapter Name |
Relations and Functions |

Exercise |
Ex 1.1, Ex 1.2, Ex 1.3, Ex 1.4 |

Number of Questions Solved |
55 |

Category |
NCERT Solutions |

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

### Chapter 1 Relations and Functions Exercise 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.

### Chapter 1 Relations and Functions Exercise 1.2

**Question 1.**

**Show that the function f: R —> R defined by f (x) = is one-one onto, where R is the set of all non-zero real numbers. Is the result true, if the domain R is replaced by N with co-domain being same as R ?**

**Solution:**

(a) We observe the following properties of f:

(i) f(x) = if f(x_{1}) = f(x_{2})

=> x_{1} = x_{2}

Each x ∈ R has a unique image in codomain

=> f is one-one.

(ii) For each y belonging codomain then

or there is a unique pre image of y.

=> f is onto.

(b) When domain R is replaced by N. codomain remaining the same, then f: N—> R If f(x_{1}) = f(x_{2})

=> => n1 = n2 where n1; n2 ∈ N

=> f is one-one.

But for every real number belonging to codomain may not have a pre-image in N.

eg:

∴ f is not onto.

**Question 2.**

**Check the injectivity and surjectivity of the following functions:**

**(i) f: N -> N given by f (x) = x²**

**(ii) f: Z -> Z given by f (x) = x²**

**(iii) f: R -> R given by f (x) = x²**

**(iv) f: N -> N given by f (x) = x³**

**(v) f: Z ->Z given by f (x) = x³**

**Solution:**

(i) f: N —> N given by f (x) = x²

(a) f(x_{1}) =>f(x_{2})

=>x_{1}^{2} = x_{2}^{2} =>x_{1} = x_{2}

∴ f is one-one i.e. it is injective.

(b) There are such member of codomain which have no image in domain N.

e.g. 3 ∈ codomain N. But there is no pre-image in domain of f.

=> f is not onto i.e. not surjective.

(ii) f: z —> z given by f(x) = x²

(a) f (-1) = f (1) = 1 => -1 and 1 have the same image.

∴ f is not one-one i.e. not injective.

(b) There are many such elements belonging to codomain have no pre-image in its codomain z.

e.g. 3 ∈ codomain z but √3 ∉ domain z of f,

∴ f is not onto i.e. not surjective

(iii) f: R->R, given by f(x) = x²

(a) f is not one-one since f(-1) = f(1) = 1

– 1 and 1 have the same image i.e., f is not injective

(b) – 2∈ codomain R off but √-2 does not belong to domain R of f.

=> f is not onto i.e. f is not surjective.

(iv) Injective but not surjective.

(v) Injective but not surjective.

**Question 3.**

**Prove that the Greatest Integer Function f: R->R given by f (x)=[x], is neither one-one nor onto, where [x] denotes the greatest integer less than or equal to x.**

**Solution:**

f: R—> R given by f (x) = [x]

(a) f(1. 2) = 1, f(1. 5) = 1 => f is not one-one

(b) All the images of x e R belonging to its domain have integers as the images in codomain. But no fraction proper or improper belonging to codomain of f has any pre-image in its domain.

=> f is not onto.

**Question 4.**

**Show that the Modulus Function f: R -> R given by f (x) = |x|, is neither one-one nor onto, where |x| is x, if x is positive or 0 and |x| is- x, if xis negative.**

**Solution:**

f: R->R given by f(x) = |x|

(a) f(-1) = |-1| = 1,f(1) = |1| = 1

=> -1 and 1 have the same image

∴ f is not one-one

(b) No negative value belonging to codomain of f has any pre-image in its domain

∴ f is not onto. Hence, f is neither one-one nor onto.

**Question 5.**

**Show that the Signum Function f: R–>R given by**

**f(x) = 1, if x > 0**

**f(x) = 0, if x = 0**

**f(x) = – 1, if x < 0**

**is neither one-one nor onto.**

**Solution:**

f: R–>R given by

f(x) = 1, if x > 0

f(x) = 0, if x = 0

f(x) = -1, if x < 0

(a) f(x_{1}) = f(x_{2}) = 1

∴ 1 and 2 have the same image i.e.

f(x_{1}) = f(x_{2}) = 1 for x>0

=> x_{1}≠x_{2}

Similarly f(x_{1}) = f(x_{2}) = – 1, for x<0 where x_{1} ≠ x_{2} => f is not one-one.

(b) Except – 1,0,1 no other member of codomain of f has any pre-image in its domain.

∴ f is not onto.

=> f is neither into nor onto.

**Question 6.**

**Let A= {1,2,3}, B = {4,5,6,7} and let f = {(1,4), (2,5), (3,6)} be a function from A to B. Show that f is one-one.**

**Solution:**

A= {1,2,3},B= {4,5,6,7} f= {(1,4), (2,5), (3,6) }.

Every member of A has a unique image in B

∴ f is one – one.

**Question 7.**

**In each of die following cases, state whether the function is one-one, onto or bijective. Justify your answer.**

**(i) f: R–>R defined by f(x) = 3 – 4x**

**(ii) f: R–>R defined by f(x) = 1 + x²**

**Solution:**

(i) f: R —> R defined by 3 – 4x,

f (x_{1}) = 3 – 4x_{1}, f(x_{2}) = 3 – 4x_{2}

(a) f(x_{1}) = f(x_{2}) =>3 – 4x_{1} = 3 – 4x_{2}

=> x_{1} = x_{2}. This shows that f is one-one

(b) f(x) = y = 3 – 4x

For every value of y belonging to its codomain. There is a pre-image in its domain => f is onto.

Hence, f is one-one onto

(ii) f: R—>R given by f(x)= 1 + x²

(a) f(1) = 1 + 1 = 2,f(-1) = 1 +1 = 2

∴ f (-1) = f (1) = 2 i.e.-1 and 1 have the same image 2.

=> f is not one-one.

(b) No negative number belonging to its codomain has its pre-image in its domain

=> f is not onto. Thus f is neither one- one nor onto.

**Question 8.**

**Let A and B be sets. Show that f:A x B –>B x A such that f (a, b) = (b, a) is bijective function.**

**Solution:**

We have f: (A x B) —> B x A such that f (a, b) = b, a

(a) ∴ f(a1, b1)= (b1, a1) f(a2, b2) = (b2, a2) f(a1, b1) = f(a2, b2)

=>(b1, a1)

= (b2, a2)

=> b1 = b2 and a1 = a2 f is one-one

(b) Every member (p, q) belonging to its codomain has its pre-image in its domain as (q, p) f is onto. Thus, f is one-one and onto i.e. it is bijective.

**Question 9.**

**Let f: N —> N be defined by
f (n) = **

**,if n is odd**

f (n) = ,if n is even

for all n∈N

f (n) = ,if n is even

for all n∈N

**State whether the function f is bijective. Justify your answer.**

**Solution:**

f: N —> N, defined by

The elements 1, 2 belonging to domain of f have the same image 1 in its codomain

=> f is not one-one.

∴ it is not injective,

(b) Every member of codomain has pre-image in its domain e.g. 1 has two pre-images 1 and 2

=> f is onto. Thus f is not one-one but it is onto

=> f is not bijective.

**Question 10.**

**Let A = R-{3} and B = R-{1}. consider the function f: A -> B defined by f (x) = **

**Solution:**

Is f one-one and onto? Justify your answer.

f: A –> B where A = R – {3}, B = R – {1} f is defined by

**Question 11.**

**Let f: R -> R be defined as f (x)=x ^{4}. Choose the correct answer.**

**(a) f is one-one onto**

**(b) f is many-one onto**

**(c) f is one-one but not onto**

**(d) f is neither one-one nor onto**

**Solution:**

f(-1) = (-1)

^{4}= 1,f(1) = 1

^{4}= 1

∴ – 1, 1 have the same image 1 => f is not one- one

Further – 2 in the codomain of f has no pre-image in its domain.

∴ f is not onto i.e. f is neither one-one nor onto Option (d) is correct.

**Question 12.**

**Let f: R –> R be defined as f (x)=3x. Choose the correct answer.**

**(a) f is one-one onto**

**(b) f is many-one onto**

**(c) f is one-one but not onto**

**(d) f is neither one-one nor onto**

**Solution:**

f: R –> R is defined by f (x) = 3x

(a) f(x_{1}) = 3x_{1}, f(x_{2}) = 3x_{2}

=> f(x_{1}) = f(x_{2})

=> 3x_{1} = 3x_{2}

=> x_{1} = x_{2}

=> f is one-one

(b) for every member y belonging to co-domain has pre-image x in domain of f.

∵ y = 3x

=>

f is onto

f is one-one and onto. Option (a) is correct.

### Chapter 1 Relations and Functions Exercise 1.3

**Question 1.**

**Let f: {1,3,4} –> {1,2, 5} and g : {1, 2,5} –> {1,3} be given by f = {(1, 2), (3,5), (4,1) and g = {(1,3), (2,3), (5,1)}. Write down g of.**

**Solution:**

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

g= {(1,3),(2,3),(5,1)}

f(1) = 2, g(2) = 3 => gof(1) = 3

f(3) = 5, g(5)= 1 =>gof(3) = 1

f(4) = 1, g(1) = 3 => gof(4) = 3

=> gof= {(1,3), (3,1), (4,3)}

**Question 2.**

**Let f, g and h be functions from R to R. Show that (f + g) oh = foh + goh, (f • g) oh = (foh) • (goh)**

**Solution:**

f + R –> R, g: R –> R, h: R –> R

(i) (f+g)oh(x)=(f+g)[h(x)]

= f[h(x)]+g[h(x)]

={foh} (x)+ {goh} (x)

=>(f + g) oh = foh + goh

(ii) (f • g) oh (x) = (f • g) [h (x)]

= f[h (x)] • g [h (x)]

= {foh} (x) • {goh} (x)

=> (f • g) oh = (foh) • (goh)

**Question 3.**

**Find gof and fog, if**

**(i) f (x) = |x| and g (x) = |5x – 2|**

**(ii) f (x) = 8x³ and g (x) = .**

**Solution:**

(i) f(x) = |x|, g(x) = |5x – 2|

(a) gof(x) = g(f(x)) = g|x|= |5| x | – 2|

(b) fog(x) = f(g (x)) = f(|5x – 2|) = ||5 x – 2|| = |5x-2|

(ii) f(x) = 8x³ and g(x) =

(a) gof(x) = g(f(x)) = g(8x³) = = 2x

(b) fog (x) = f(g (x))=f() = 8.()³ = 8x

**Question 4.**

**If , show that fof (x) = x, for all . What is the inverse of f?**

**Solution:**

(a) fof (x) = f(f(x)) =

**Question 5.**

**State with reason whether following functions have inverse**

**(i) f: {1,2,3,4}–>{10} with f = {(1,10), (2,10), (3,10), (4,10)}**

**(ii) g: {5,6,7,8}–>{1,2,3,4} with g = {(5,4), (6,3), (7,4), (8,2)}**

**(iii) h: {1,2,3,4,5}–>{7,9,11,13} with h = {(2,7), (3,9), (4,11), (5,13)}**

**Solution:**

f: {1,2,3,4} –> {10} with f = {(1,10), (2,10), (3,10), (4,10)}

(i) f is not one-one since 1,2,3,4 have the same image 4.

=> f has no inverse.

(ii) g: {5,6,7,8} –> {1,2,3,4} with g = {(5,4), (6,3) , (7,4), (8,2)}

Here also 5 and 7 have the same image

∴ g is not one-one. Therefore g is not invertible.

(iii) f has an inverse

**Question 6.**

**Show that f: [-1,1] –> R, given by f(x) = is one-one. Find the inverse of the function f: [-1,1] –> Range f.**

**Hint – For y ∈ Range f, y = f (x) = for some x in [- 1,1], i.e., x = **

**Solution:**

**Question 7.**

**Consider f: R –> R given by f (x) = 4x + 3. Show that f is invertible. Find the inverse of f.**

**Solution:**

f: R—>R given by f(x) = 4x + 3

f (x_{1}) = 4x_{1} + 3, f (x_{2}) = 4x_{2} + 3

If f(x_{1}) = f(x_{2}), then 4x_{1} + 3 = 4x_{2} + 3

or 4x_{1} = 4x_{2} or x_{1} = x_{2}

f is one-one

Also let y = 4x + 3, or 4x = y – 3

∴

For each value of y ∈ R and belonging to co-domain of y has a pre-image in its domain.

∴ f is onto i.e. f is one-one and onto

f is invertible and f^{-1} (y) = g (y) =

**Question 8.**

**Consider f: R _{+} –> [4, ∞] given by f (x) = x² + 4. Show that f is invertible with the inverse f^{-1} of f given by f^{-1} (y) = √y-4 , where R_{+} is the set of all non-negative real numbers.**

**Solution:**

f(x

_{1}) = x

_{1}

^{2}+ 4 and f(x

_{2}) = x

_{2}

^{2}+ 4

f(x

_{1}) = f(x

_{2}) => x

_{1}

^{2}+ 4 = x

_{2}

^{2}+ 4

or x

_{1}

^{2}= x

_{2}

^{2}=> x

_{1}= x

_{2}As x ∈ R

∴ x>0, x

_{1}

^{2}= x

_{2}

^{2}=> x

_{1}= x

_{2}=>f is one-one

Let y = x² + 4 or x² = y – 4 or x = ±√y-4

x being > 0, -ve sign not to be taken

x = √y – 4

∴ f

^{-1}(y) = g(y) = √y-4 ,y ≥ 4

For every y ≥ 4, g (y) has real positive value.

∴ The inverse of f is f

^{-1}(y) = √y-4

**Question 9.**

**Consider f: R _{+} –> [- 5, ∞) given by f (x) = 9x² + 6x – 5. Show that f is invertible with**

**Solution:**

Let y be an arbitrary element in range of f.

Let y = 9x² + 6x – 5 = 9x² + 6x + 1 – 6

=> y = (3x + 1)² – 6

=> y + 6 = (3x + 1)²

=> 3x + 1 = √y + 6

**Question 10.**

**Let f: X –> Y be an invertible function. Show that f has unique inverse.**

**Hint – suppose g _{1} and g_{2} are two inverses of f. Then for all y∈Y, fog_{1}(y)=I_{y}(y)=fog_{2}(y).Use one-one ness of f.**

**Solution:**

If f is invertible gof (x) = I

_{x}and fog (y) = I

_{y}

∴ f is one-one and onto.

Let there be two inverse g

_{1}and g

_{2}

fog

_{1}(y) = I

_{y}, fog

_{2}(y) = I

_{y}

I

_{y}being unique for a given function f

=> g

_{1}(y) = g

_{2}(y)

f is one-one and onto

f has a unique inverse.

**Question 11.**

**Consider f: {1,2,3} –> {a, b, c} given by f (1) = a, f (2)=b and f (3)=c. Find f ^{-1} and show that (f^{-1})f^{-1}=f.**

**Solution:**

f: {1,2, 3,} –> {a,b,c} so that f(1) = a, f(2) = b, f(3) = c

Now let X = {1,2,3}, Y = {a,b,c}

∴ f: X –> Y

∴ f

^{-1}: Y –> X such that f

^{-1}(a)= 1, f

^{-1}(b) = 2; f

^{-1}(c) = 3

Inverse of this function may be written as

(f

^{-1})

^{-1}: X –> Y such that

(f

^{-1})

^{-1}(1) = a, (f

^{-1})

^{-1}(2) = b, (f

^{-1})

^{-1}(3) = c

We also have f: X –> Y such that

f(1) = a,f(2) = b,f(3) = c => (f

^{-1})

^{-1 }= f

**Question 12.**

**Let f: X –> Y be an invertible function. Show that the inverse of f ^{-1} is f, i.e., (f^{-1})^{-1} = f.**

**Solution:**

f: X —> Y is an invertible function

f is one-one and onto

=> g : Y –> X, where g is also one-one and onto such that

gof (x) = I

_{x}and fog (y) = I

_{y}=> g = f

^{-1}

Now f

^{-1}o (f

^{-1})

^{-1}= I

and fo[f

^{-1}o (f

^{-1})

^{-1}] =fol

or (fof

^{-1})

^{-1}o (f

^{-1})

^{-1}= f

=> Io (f

^{-1})

^{-1}= f

=> (f

^{-1})

^{-1}= f

**Question 13.**

**If f: R –> R be given by f(x) = , then fof (x) is**

**(a) **

**(b) x³**

**(c) x**

**(d) (3 – x³)**

**Solution:**

f: R-> R defined by f(x) =

fof (x) = f[f(x)] =

=

=

=

= x

**Question 14.**

**Let f: be a function defined as f (x) = . The inverse of f is the map g: Range f–> given by**

**(a) **

**(b) **

**(c) **

**(d) **

**Solution:**

(b)

### Chapter 1 Relations and Functions Exercise 1.4

**Question 1.**

**Determine whether or not each of the definition of * given below gives a binary operation. In the event that * is not a binary operation, give justification for this.**

**(i) On Z ^{+},define * by a * b = a – b**

**(ii) On Z**

^{+}, define * by a * b = ab**(iii) On R, define * by a * b = ab²**

**(iv) On Z**

^{+},define * by a * b = |a – b|**(v) On Z**

^{+}, define * by a * b = a**Solution:**

(i) If a > b, a * b = a – b > 0, which belongs to Z

^{+}.

But if a < b, a * b = a – b < 0, which does not belong to Z

^{+}

=> * given operation is not a binary operation.

(ii) For all a and b belonging to Z

^{-1}, ab also belongs to Z

^{+}.

∴ The operation *, defined by a * b = ab is a binary operation.

(iii) For all a and b belonging to R, ab² also belongs to R.

∴ The operation * defined by a * b = ab² is binary operation.

(iv) For all a and b belonging to Z

^{+}, |a – b| also belongs to Z

^{+1}

∴ The operation a * b = |a – b| is a binary operation.

(v) On Z

^{+}defined by a * b = a

a, b ∈ Z

^{+}= a ∈ Z

^{+}

∴ The operation * is a binary operation.

**Question 2.**

**For each binary operation * defined below, determine whether * is commutative or associative.**

**(i) OnZ, define a * b = a – b**

**(ii) OnQ, define a * b = ab + 1**

**(iii) On Q, define a * b = **

**(iv) On Z ^{+}, define a * b = 2ab**

**(v) On Z**

^{+}, define a * b = ab**(vi) On R- {-1}, define a * b =**

**Solution:**

(i) On Z, operation * is defined as

(a) a * b = a – b => b * a = b – a

But a – b ≠ b – a ==> a * b ≠ b * a

Defined operation is not commutative

(b) a – (b – c) ≠ (a – b) – c

Binary operation * as defined is not associative.

(ii) On Q, Operation * is defined as a * b = ab + 1

(a) ab + 1 = ba + 1, a * b = b * a

Defined binary operation is commutative.

(b) (a*b)*c = (ab + 1)*c = (ab + 1)c + 1 = abc + c + 1

a * (b * c) = a * (bc + 1) = a(bc + 1)+ 1 = abc + a+ 1

=> a * (b * c) ≠ (a * b) * c

∴ Binary operation defined is not associative.

(iii) (a) On Q, operation * is defined as a * b =

∴ a * b = b * a

∴ Operation binary defined is commutative.

be abc

(b) a * (b * c) = a * = and

(a * b) * c = * c

=> (a * b) * c = * c

Defined binary operation is associative.

(iv) On Z^{+} operation * is defined as a * b = 2^{ab}

(a) a * b = 2^{ab}, b * a = 2^{ba} = 2^{ab}

=> a * b = b * a .

∴ Binary operation defined as commutative.

(b) a * (b * c) = a * 2^{ba} = 2^{a.2bc}

(a * b) * c = 2^{ab} * c = 2^{2ab}

Thus (a * b) * c ≠ a * (b * c)

∴ Binary operation * as defined as is not associative.

(v) On Z^{+}, a * b = ab

(a) b * a = b^{a} .

∴ a^{b} ≠ b^{a} = a * b ≠ b * a.

* is not commutative.

(b) (a*b)*c = a^{b}*c = (a^{b})^{c} = a^{bc} a*(b* c)

= a*b^{c} = a^{bc}

Thus (a * b) * c ≠ (a * b * c)

∴ Operation * is not associative.

(vi) Neither commutative nor associative.

**Question 3.**

**Consider the binary operation ^ on the set {1, 2, 3, 4, 5} defined by a ^ b=min {a, b}. Write the operation table of the operation ^.**

**Solution:**

Operation ^ table on the set {1, 2, 3, 4, 5} is as follows.

**Question 4.**

**Consider a binary operation * on the set {1, 2, 3, 4, 5} given by the following multiplication table.**

**(i) Compute (2 * 3) * 4 and 2 * (3 * 4)**

**(ii) Is * commutative?**

**(iii) Compute (2 * 3) * (4 * 5).**

**Hint – use the following table)**

**Solution:**

(i) From the given table, we find

2*3 = 1, 1*4 = 1

(a) (2*3)*4 = 1 * 4 = 1

(b) 2*(3*4) = 2 * 1 = 1

(ii) Let a, b ∈ {1,2,3,4,5} From the given table, we find

a*a = a

a*b = b*a = 1 when a or b or are odd and a b.

2 * 4 = 4 * 2 = 2, when a and b are even and a ≠ b

Thus a * b = b * c

∴Binary operation * given is commutative.

(iii) Binary operation * given is commutative (2 * 3) * (4 * 5) = 1 * 1 = 1.

**Question 5.**

**Let *’ be the binary operation on the set {1,2,3,4,5} defined by a *’ b=H.C.F. of a and b. Is the operation *’ same as the operation * defined in the exp no. 4 above? Justify your answer.**

**Solution:**

The set is {1,2,3,4, 5} and a * b = HCF of a and b.

Let us prepare the table of operation *.

**Question 6.**

**Let * be the binary operation on N given by a * b = L.C.M. of a and b.Find**

**(i) 5 * 7, 20 * 16**

**(ii) Is * commutative?**

**(iii) Is * associative?**

**(iv) Find the identity of * in N**

**Solution:**

Binary operation * defined as a * b = 1 cm. of a and b.

(i) 5 * 7 = 1 cm of 5 and 7 = 35

20 * 16= 1 cm of 20 and 16 = 80

(ii) a * b= 1 cm of a and b b * a = 1 cm of b and a

=> a * b = b * a, 1 cm of a, b and b, a are equal

∴ Binary operation * is commutative.

(iii) a * (b * c) = 1 cm of a, b, c and (a * b) * c = 1 cm of a, b, c

=> a * (b * c) = (a * b) * c

=> Binary operation * given is associative.

(iv) Identity of * in N is 1

1 * a = a * 1 = a = 1 cm of 1 and a.

(v) Let * : N x N—> N defined as a * b = 1.com of (a, b)

For a = 1, b = 1, a * b = l b * a

Otherwise a * b ≠ 1

∴ Binary operation * is not invertible

=> 1 is invertible for operation *

**Question 7.**

**Is * defined on the set {1, 2, 3, 4, 5} by a * b = L.C.M. of a and b a binary operation? Justify your answer.**

**Solution:**

The given set = {1,2,3,4,5} Binary operation is defined as a * b = 1 cm of a and b. 4 * 5 = 20 which does not belong to the given set {1, 2, 3, 4, 5}.

It is not a binary operation.

**Question 8.**

**Let * be the binary operation on N defined by a * b = H.C.F. of a and b. Is * commutative? Is * associative? Does there exist identity for this binary operation on N?**

**Solution:**

Binary operation on set N is defined as a * b = HCF of a and b

(a) We know HCF of a, b = HCF of b, a

∴ a * b = b * a

∴ Binary operation * is commutative.

(b) a*(b*c) = a* (HCF of b, c) = HCF of a and (HCF of b, c) = HCF of a, b and c

Similarly (a * b) * c = HCF of a, b, and c

=> (a * b) * c = a * (b * c)

Binary operation * as defined above is associative.

(c) 1 * a = a * 1 = 1 ≠ a

∴ There does not exists any identity element.

**Question 9.**

**Let * be a binary operation on the set Q of rational numbers as follows:**

**(i) a * b = a – b**

**(ii) a * b = a² + b²**

**(iii) a * b = a + ab**

**(iv) a * b = (a – b)²**

**(v) a * b = **

**(vi) a * b = ab²**

Find which of the binary operations are commutative and which are associative.

**Solution:**

Operation is on the set Q.,,

(i) defined as a * b = a – b

(a) Now b * a = b – a

But a – b ≠ b – a

∴ a * b ≠ b * a

∴ Operation * is not commutative.

(b) a * (b * c) = a * (b – c) = a – (b – c) = a – b + c (a * b) * c = (a – b) * c = a – b – c

Thus a * (b * c)¹ (a * b) * c = (a² + b²)² + c²

=> a * (b * c) ≠ (a * b) * c

∴ The operation * as defined is not associative.

(ii) (a) a * b = a² + b * a = b² + a² = a² + b².

a * b = b * a

This binary operation is commutative,

(b) a*(b*c) = a*(b² + c²) = a² + (b²)² + c²)²

=> (a*b)*c = (a² + b²)*c = (a² + b²) + c²

Thus a * (b*c) (a*b) * c

The operation * given is not associative.

(iii) Operation * is defined as a * b = a + ab

(a) b * a = b + ba

a * b ≠ b * a

The operation is not commutative.

(b) a*(b*c) = a*(b + bc)

= a + a(b + bc)

= a + ab + abc (a * b) * c

= (a + ab) * c

= (a + ab) + (a + ab) • c

= a + ab + ac + abc

=> a * (b * c) ≠ (a * b) * c

=> The binary operation is not associative.

(iv) The binary operation is defined as a * b = (a – b)²

(a) b*a = (b – a)² = (a – b)² => a*b = b*a

.’. This binary operation * is commutative.

(b) a*(b*c) = a*(b – c)² = [a – (b – c)²]² (a * b) * c

= (a – b)² * c

= [(a – b)² – c]²

=> a * (b * c) ≠ (a * b) * c

the operation * is not associative.

(v) Commutative and associative.

(vi) Neither commutative nor associative.

**Question 10.**

**Show that none of the operations given above has identity.**

**Solution:**

The binary operation * on set Q is

(i) defined as a*b = a – b

For identity element e, a*e = e*a = a

But a*e = a – e≠a and e*a = e – a≠a

There is no identity element for this operation

(ii) Binary operation * is defined as a * b = a² + b² ≠ a

This operation * has no identity.

(iii) The binary operation is defined as a*b = a+ab

Putting b = e, a + e = a + eb ≠ a

There is no identity element.

(iv) The binary operation is defined as a * b = (a – b)²

Put b = e, a * e = (a – e)² ≠ a for any value of

e∈Q

=> there is no Identity Element.

(v) The operation is a * b =

∴ a * e = ≠ a for any value of e ∈ Q

∴ Operation * has no identity

(vi) The operation * is a * b = ab² Put b = e, a

*e = ae² and e * a = ea² ≠ a for any value of e∈Q

=> There is no Identity Element. Thus, these operations have no Identity.

**Question 11.**

**Let A=N x N and * be the binary operation on A defined by (a,b)*(c,d)=(a+c,b+d)**

**Show that * is commutative and associative. Find the identity element for * on A, if any.**

**Solution:**

A = N x N Binary operation * is defined as (a, b) * (c, d) = (a + c, b + d)

(a) Now (c, d) * (a,b) = (c+a, d+b) = (a+c,b+d)

=> (a, b) * (c, d) = (c, d) * (a, b)

∴ This operation * is commutative

(b) Next(a,b)* [(c,d)*(e,f)]=(a,b)*(c+e,d+f) = ((a + c + e), (b + d + f))

and [(a, b) * (c, d)] * (e, f)=(a+c, b+d) * (e, f) = ((a + c + e, b+d + f))

=> (a, b) * [(c, d) * (e, f)] = [(a, b) * (c, d)] * (e,f)

∴ The binary operation given is associative

(c) Identity element does not exists.

**Question 12.**

**State whether the following statements are true or false. Justify.**

**(i) For an arbitrary binary operation * on a set N,**

**a*a=a∀a∈N.**

**(ii) If * is a commutative binary operation on N, then**

**a * (b * c) = (c * b) * a**

**Solution:**

(i) A binary operation on N is defined as

a*a=a∀a∈N.

Here operation * is not defined.

∴ Given statement is false.

(ii) * is a binary commutative operation on N. c

* b = b * c

∵ * is commutative

∵ (c * b) * a = (b * c) * a = a * (b * c)

∴ Thus a * (b * c) = (c * b) * a

This statement is true.

**Question 13.**

**Consider a binary operation * on N defined as a * b = a³ + b³. Choose the correct answer.**

**(a) Is * both associative and commutative ?**

**(b) Is * commutative but not associative?**

**(c) Is * associative but not commutative?**

**(d) Is * neither commutative nor associative?**

**Solution:**

(b)

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