**RELATIONS**

**Page 1.11 Ex. 1.1**

**Q1.**

**Answer :**

**(i)** Reflexivity:

Let x be an arbitrary element of R. Then,x∈R ⇒x and x work at the same place is true since they are the same.⇒x, x∈RSo, R is a reflexive relation.

Symmetry:

Let x, y∈R⇒x and y work at the same place ⇒y and x work at the same place⇒y, x∈RSo, R is a symmetric relation.

Transitivity:

Let x, y∈R and y, z∈R. Then,x and y work at the same place.y and z also work at the same place.⇒x , y and z all work at the same place.⇒x and z work at the same place.⇒x, z∈RSo, R is a transitive relation.

**(ii)** Reflexivity:

Let x be an arbitrary element of R. Then,x∈R ⇒x and x live in the same locality is true since they are the same.So, R is a reflexive relation.

Symmetry:

Let x, y∈R⇒x and y live in the same locality⇒y and x live in the same locality⇒y, x∈R So, R is a symmetric relation.

Transitivity:

Let x, y∈R and y, z∈R. Then,x and y live in the same locality and y and z live in the same locality⇒x, y and z all live in the same locality⇒x and z live in the same locality ⇒x, z ∈RSo, R is a transitive relation.

**(iii)**

Reflexivity:

Let x be an element of R. Then,x is wife of x cannot be true.⇒x, x∉RSo, R is not a reflexive relation.

Symmetry:

Let x, y∈R⇒x is wife of y ⇒x is female and y is male⇒y cannot be wife of x as y is husband of x⇒y, x∉R So, R is not a symmetric relation.

Transitivity:

Let x, y∈R, but y, z∉RSince x is wife of y, but y cannot be the wife of z, y is husband of x.⇒x is not the wife of z⇒x, z∈RSo, R is a transitive relation.

**(iv)**

Reflexivity:

Let x be an arbitrary element of R. Then,x is father of x cannot be true since no one can be father of himself.So, R is not a reflexive relation.

Symmetry:

Let x, y∈R⇒x is father of y⇒y is son/daughter of x⇒y, x∉R So, R is not a symmetric relation.

Transitivity:

Let x, y∈R and y, z∈R. Then, x is father of y and y is father of z⇒x is grandfather of z⇒x, z∉RSo, R is not a transitive relation.

**Page 1.12 Ex. 1.1**

**Q2.**

**Answer :**

**(i)** R_{1}

Reflexive:

Clearly, (a, a), (b, b) and (c, c)∈R_{1}

So, R_{1} is reflexive.

Symmetric:

We see that the ordered pairs obtained by interchanging the components of R_{1} are also in R_{1}.

So, R_{1} is symmetric.

Transitive:

Here,

a, b∈R_{1}, b, c∈R_{1} and also a, c∈R_{1}

So, R1 is transitive.

**(ii)** R_{2}

Reflexive: Clearly a,a∈R_{2}. So, R_{2} is reflexive.

Symmetric: Clearly a,a∈R⇒a,a∈R. So, R_{2} is symmetric.

Transitive: R_{2} is clearly a transitive relation, since there is only one element in it.

**(iii)** R_{3}

Reflexive:

Here,

b, b∉R_{3} neither c, c∉R_{3}

So, R_{3} is not reflexive.

Symmetric:

Here,

b, c∈R_{3}, but c,b∉R_{3}So, R_{3} is not symmetric.

Transitive:

Here, R_{3} has only two elements. Hence, R_{3} is transitive.

**(iv)** R_{4}

Reflexive:

Here,

a, a∉R_{4}, b, b∉ R_{4} c, c∉ R_{4}So, R_{4} is not reflexive.

Symmetric:

Here,

a, b∈R_{4}, but b,a∉R_{4}.So, R_{4} is not symmetric.

Transitive:

Here,

a, b∈R_{4}, b, c∈R4, but a, c∉R_{4}So, R_{4} is not transitive.

**Q3.**

**Answer :**

(i) Reflexivity:

Let a be an arbitrary element of R_{1}. Then,

a∈R1⇒a≠1a for all a∈Q0So, R_{1} is not reflexive.

Symmetry:

Let (a, b) ∈R_{1}. Then,

a, b∈R_{1}⇒a=1b⇒b=1a⇒b, a∈R_{1}So, R_{1} is symmetric.

Transitivity:

Here,

a, b∈R_{1} and b, c∈R_{2}⇒a=1b and b=1c⇒a=11c=c⇒a≠1c⇒a, c∉R_{1} So, R_{1} is not transitive.

(ii)

Reflexivity:

Let a be an arbitrary element of R_{2}. Then,

a∈R_{2} ⇒a-a=0≤5So, R1 is reflexive.

Symmetry:

Let a, b∈R_{2}⇒a-b≤5⇒b-a≤5 Since, a-b = b-a⇒b, a∈R_{2}So, R_{2} is symmetric.

Transitivity:

Let 1, 3∈R_{2} and 3, 7∈R_{2}⇒1-3≤5 and 3-7≤5But 1-7≰5 ⇒1,7∉R_{2}So, R_{2} is not transitive.

(iii)

Reflexivity: Let a be an arbitrary element of R_{3}. Then,

a∈R_{3}⇒a2-4a×a+3a2=0 So, R_{3} is reflexive.

Symmetry:

Let a, b∈R_{3}⇒a2-4ab+3b2=0But b2-4ba+3a2≠0 for all a, b ∈RSo, R_{3} is not symmetric.

Transitivity:

1, 2∈R_{3} and 2, 3∈R_{3}⇒1-8+6=0 and 4-24+27=0But 1-12+9≠0So, R_{3} is not transitive.

**Q4.**

**Answer :**

1 R_{1}

Reflexivity:

Here,

1, 1, 2, 2, 3, 3∈RSo, R_{1} is reflexive.

Symmetry:

Here,2, 1∈R_{1}, but 1, 2∉R_{1} So, R_{1} is not symmetric.

Transitivity:

Here, 2, 1∈R_{1} and 1, 3∈R_{1}, but 2, 3∉R_{1} So, R_{1}

2 R_{2}

Reflexivity:

Clearly, 1, 1 and 3, 3∉R_{2} So, R_{2} is not reflexive.

Symmetry:

Here, 1, 3∈R_{2} and 3, 1∈R_{2}So, R_{2} is symmetric.

Transitivity:

Here, 1, 3∈R_{2} and 3, 1∈R_{2} But 3, 3∉R_{2}So, R_{2} is not transitive.

3 R3

Reflexivity:

Clearly, 1, 1∉R_{3} So, R_{3} is not reflexive.

Symmetry:

Here, 1, 3∈R_{3}, but 3, 1∉R_{3}So, R_{3} is not symmetric.

Transitivity:

Here, 1, 3∈R_{3} and 3, 3∈R_{3} Also, 1, 3∈R_{3}So, R_{3} is transitive.

**Q5.**

**Answer :**

**(i)**

Reflexivity: Let a be an arbitrary element of R. Then,

a∈RBut a-a = 0 ≯0So, this relation is not reflexive.

Symmetry:

Let a, b∈R⇒a-b>0⇒-(b-a)>0⇒b-a<0So, the given relation is not symmetric.

Transitivity:

Let a, b∈R and b, c∈R. Then,a-b>0 and b-c>0Adding the two, we geta-b+b-c>0⇒a-c>0 ⇒a, c∈R. So, the given relation is transitive.

**(ii)**

Reflexivity: Let a be an arbitrary element of R. Then,

a∈R⇒1+a×a>0i.e. 1+a2>0 Since, square of any number is positiveSo, the given relation is reflexive.

Symmetry:

Let a, b∈R⇒1+ab>0⇒1+ba>0⇒b, a∈RSo, the given relation is symmetric.

Transitivity:

Let a, b∈R and b, c∈R⇒1+ab>0 and 1+bc>0But 1+ac≯0⇒a, c∉RSo, the given relation is not transitive.

**(iii)**

Reflexivity: Let a be an arbitrary element of R. Then,

a∈R ⇒a≮a Since, a=aSo, R is not reflexive.

Symmetry:

Let a, b∈R⇒a≤b ⇒ b≰a for all a, b∈R⇒b, a∉R So, R is not symmetric.

Transitivity:

Let a, b∈R and b, c∈R⇒a≤b and b≤cMultiplying the corresponding sides, we get a b≤bc⇒a≤c⇒a, c∈RThus, R is transitive.

**Q6.**

**Answer :**

Reflexivity:

Letabeanarbitraryelementof R.Then,a=a+1 cannot be true for all a∈A.⇒a, a∉R So, R is not reflexive on A.

Symmetry:

Let a, b∈R⇒b=a+1⇒-a=-b+1⇒a=b-1Thus, b, a∉RSo, R is not symmetric on A.

Transitivity:

Let 1, 2 and 2, 3∈R⇒2=1+1 and 3 2+1 is true.But 3 ≠ 1+1⇒1, 3∉RSo, R is not transitive on A.

**Q7.**

**Answer :**

Reflexivity:

Since 12>123,12, 12∉RSo, R is not reflexive.

Symmetry:

Since 12, 2∈R,12<23But 2>123⇒2, 12∈RSo, R is not symmetric.

Transitivity:

Since 7, 3∈R and 3, 313∈R,7<33 and 3=3133But 7>3133⇒7, 313∉RSo, R is not transitive.

**Q8.**

**Answer :**

Let A be a set. Then,

Identity relation IA=IA is reflexive, since a, a∈A∀a

The converse of it need not be necessarily true.

Consider the set A = {1, 2, 3}

Here,

Relation R = {(1, 1), (2, 2) , (3, 3), (2, 1), (1, 3)} is reflexive on A.

However, R is not an identity relation.

**Q9.**

**Answer :**

(i) The relation on A having properties of being reflexive, transitive, but not symmetric is

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

Relation R satisfies reflexivity and transitivity.⇒1, 1, 2, 2, 3, 3∈R and 1, 1, 2, 1∈R ⇒1, 1∈RHowever, 2, 1∈R, but 1, 2∉R

(ii) The relation on A having properties of being symmetric, but neither reflexive nor transitive is

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

The relation R on A is neither reflexive nor transitive, but symmetric.

(iii) The relation on A having properties of being symmetric, reflexive and transitive is

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

The relation R is an equivalence relation on A.

**Q10.**

**Answer :**

Domain of R is the values of x and range of R is the values of y that together should satisfy 2x+y = 41.

So,

Domain of R = {1, 2, 3, 4, … , 20}

Range of R = {1, 3, 5, … , 37, 39}

Reflexivity: Let x be an arbitrary element of R. Then,

x∈R⇒2x+x=41 cannot be true.⇒x, x∉R So, R is not reflexive.

Symmetry:

Let x, y∈R. Then, 2x+y=41⇒ 2y+x = 41 ⇒y, x∉RSo, R is not symmetric.

Transitivity:

Let x, y and y, z∈R⇒2x+y=41 and 2y +z=41⇒2x+z=2x+41-2y 41-y-2y=41-3y⇒x, z∉RThus, R is not transitive.

**Q11.**

**Answer :**

No, it is not true.

Consider a set A = {1, 2, 3} and relation R on A such that R = {(1, 2), (2, 1), (2, 3), (1, 3)}

The relation R on A is symmetric and transitive. However, it is not reflexive.

1, 1, 2, 2 and 3, 3∉ R

Hence, R is not reflexive.

**Q12.**

**Answer :**

R=m, n : m, n∈Z, m=kn, where k∈NReflexivity:Let m be an arbitrary element of R. Then,m=km is true for k=1⇒m, m∈RThus, R is reflexive.Symmetry: Let m, n∈R⇒m=kn for some k∈N→n=1km⇒n, m∉R Thus, R is not symmetric.Transitivity: Let m, n and n, o∈R⇒m=kn and n=lo for some k, l ∈N⇒m=(kl) oHere, kl∈R⇒m, o∈RThus, R is transitive.

**Q13.**

**Answer :**

Let R be the set such that R = {(a, b) : a, b∈R; a≥b}

Reflexivity:

Let a be an arbitrary element of R. ⇒a∈R⇒a=a ⇒a≥a is true for a=a⇒a, a∈R Hence, R is reflexive.

Symmetry:

Let a, b∈R⇒a≥b is same as b≤a, but not b≥aThus, b, a∉R Hence, R is not symmetric.

Transitivity:

Let a, b and b, c∈R⇒a≥b and b≥c⇒a≥b≥c⇒a≥c⇒a, c∈RHence, R is transitive.

**Q14.**

**Answer :**

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

For R to be reflexive it must have (1, 1), (2, 2), (3, 3).For R to be symmetric, all the ordered pairs upon interchanging the elements must be present in R.Therefore, R must contain 2, 1 and 3, 2, 3, 1, 1, 3.Finally, for R to be transitive, it must contain 1,3.

Hence, the number of ordered pairs to be added to R is 7, i.e. (1, 1), (2, 2), (3, 3), (1, 3), (3, 1), (2, 1), (3, 2).

**Q15.**

**Answer :**

The relation R on A is such that

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

For relation R to be transitive, we must have1, 2∈R, 2, 3∈R⇒1, 3∈RTherefore, the minimun number of ordered pairs to be added to R is 1, i.e. (1, 3) to make it a transitive relation on A.

**Page 1.13 Ex. 1.1**

**Q16.**

**Answer :**

Suppose A be the set such that A = {1, 2, 3}

(i) Let R be the relation on A such that

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

Thus,

R is reflexive and symmetric, but not transitive.

(ii) Let R be the relation on A such that

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

Clearly, the relation R on A is reflexive and transitive, but not symmetric.

(iii) Let R be the relation on A such that

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

We see that the relation R on A is symmetric and transitive, but not reflexive.

(iv) Let R be the relation on A such that

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

The relation R on A is symmetric, but neither reflexive nor transitive.

(v) Let R be the relation on A such that

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

The relation R on A is transitive, but neither symmetric nor reflexive.

**Page 1.25 Ex. 1.2**

**Q1.**

**Answer :**

We observe the following relations of relation R.

Reflexivity:

Let a be an arbitrary element of R. Then,a-a=0=0 × 3⇒a-a is divisible by 3⇒a, a∈R for all a∈ZSo, R is reflexive on Z.

Symmetry:

Let a, b∈R⇒a-b is divisible by 3⇒a-b 3p for some p∈Z⇒b-a=3 -p Here, -p∈Z⇒b-a is divisible by 3⇒b, a∈R for all a, b∈ZSo, R is symmetric on Z.

Transitivity:

Let a, b and b, c∈R⇒a-b and b-c are divisible by 3⇒a-b=3p for some p∈Zand b-c=3q for some q∈ZAdding the above two, we get a-b+b-c=3p+3q⇒a-c=3 p+qHere, p+q∈Z⇒a-c is divisible by 3⇒a, c∈R for all a, c ∈ZSo, R is transitive on Z.

Hence, R is an equivalence relation on Z.

**Q2.**

**Answer :**

We observe the following properties of relation R.

Reflexivity:

Let a be an arbitrary element of the set Z. Then,a∈R⇒a-a=0=0 × 2⇒2 divides a-a⇒a, a∈R for all a∈ZSo, R is reflexive on Z.

Symmetry:

Let a, b∈R⇒2 divides a-b⇒a-b2=p for some p∈Z⇒b-a2=-p Here, -p∈Z⇒2 divides b-a⇒b, a∈R for all a, b ∈ZSo, R is symmetric on Z.

Transitivity:

Let a, b and b, c∈R⇒2 divides a-b and 2 divides b-c⇒a-b2=p and b-c2=q for some p, q∈ZAdding the above two, we geta-b2+b-c2=p+q⇒a-c2=p+q Here, p+q∈Z⇒2 divides a-c⇒a, c∈R for all a, c ∈ZSo, R is transitive on Z.

Hence, R is an equivalence relation on Z.

**Q3.**

**Answer :**

We observe the following properties of relation R.

Reflexivity:

Let a be an arbitrary element of R. Then,⇒a-a = 0 = 0 × 5⇒a-a is divisible by 5⇒a, a∈R for all a∈ZSo, R is reflexive on Z.

Symmetry:

Let a, b∈R⇒a-b is divisible by 5⇒a-b = 5p for some p∈Z⇒b-a = 5 -p Here, -p∈Z [Since p∈Z]⇒b-a is divisible by 5⇒b, a∈R for all a, b∈ZSo, R is symmetric on Z.

Transitivity:

Let a, b and b, c∈R⇒a-b is divisible by 5⇒a-b = 5p for some ZAlso, b-c is divisible by 5⇒b-c = 5q for some ZAdding the above two, we geta-b+b-c = 5p+5q⇒a-c = 5 (p+q)⇒a-c is divisible by 5Here, p+q∈Z⇒a, c∈R for all a, c∈ZSo, R is transitive on Z.

Hence, R is an equivalence relation on Z.

**Q4.**

**Answer :**

We observe the following properties of R. Then,

Reflexivity:

Let a∈NHere,a-a=0=0 × n⇒a-a is divisible by n⇒a, a∈R⇒a, a∈R for all a∈ZSo, R is reflexive on Z.

Symmetry:

Let a, b∈RHere,a-b is divisible by n⇒a-b=np for some p∈Z⇒b-a=n -p⇒b-a is divisible by n [p∈Z⇒-p∈Z]⇒b, a∈R So, R is symmetric on Z.

Transitivity:

Let a, b and b, c∈RHere, a-b is divisible by n and b-c is divisible by n.⇒a-b=np for some p∈Zand b-c=nq for some q∈ZAdding the above two, we geta-b+b-c=np+nq⇒a-c=n (p+q)Here, p+q∈Z⇒a, c∈R for all a, c∈ZSo, R is transitive on Z.

Hence, R is an equivalence relation on Z.

**Q5.**

**Answer :**

We observe the following properties of R.

Reflexivity:

Let a be an arbitrary element of Z. Then, a∈RClearly, a+a=2a is even for all a∈Z.⇒a, a∈R for all a∈ZSo, R is reflexive on Z.

Symmetry:

Let a, b∈R⇒a+b is even⇒b+a is even⇒b, a∈R for all a, b∈ZSo, R is symmetric on Z.

Transitivity:

Let a, b and b, c∈R⇒a+b and b+c are evenNow, let a+b=2x for some x∈Zand b+c=2y for some y∈ZAdding the above two, we get a+2b+c=2x+2y⇒a+c=2(x+y-b), which is even for all x, y, b∈ZThus, a, c∈RSo, R is transitive on Z.

Hence, R is an equivalence relation on Z.

**Q6.**

**Answer :**

We observe the following properties of relation R.

Let R={m, n : m, n∈Z : m-n is divisible by 13}Relexivity: Let m be an arbitrary element of Z. Then,m∈R⇒m-m=0=0 × 13⇒m-m is divisible by 13⇒m, m is reflexive on Z.Symmetry: Let m, n∈R. Then,m-n is divisible by 13⇒m-n=13pHere, p∈Z⇒n-m=13 -p Here, -p∈Z⇒n-m is divisible by 13⇒n, m∈R for all m, n∈ZSo, R is symmetric on Z.Transitivity: Let m, n and n, o∈R⇒m-n and n-o are divisible by 13⇒m-n=13p and n-o=13q for some p, q∈ZAdding the above two, we get m-n+n-o=13p+13q⇒m-o=13 p+qHere, p+q∈Z⇒m-o is divisible by 13⇒m, o∈R for all m, o∈ZSo, R is transitive on Z.

Hence, R is an equivalence relation on Z.

**Q7.**

**Answer :**

We observe the following properties of R.

Reflexivity: Let a, b be an arbitrary element of the set A. Then, a, b∈A⇒ab=ba ⇒a, b R a, bThus, R is reflexive on A.Symmetry: Let x, y and u, v∈A such that x, y R u, v. Then, xv=yu⇒vx=uy⇒uy=vx⇒u, v R x, ySo, R is symmetric on A.Transitivity: Let x, y, u, v and p, q∈R such that x, y R u, v and u, v R p, q.⇒xv=yu and uq=vpMultiplying the corresponding sides, we getxv × uq=yu × vp⇒xq=yp⇒x, y R p, qSo, R is transitive on A.

Hence, R is an equivalence relation on A.

**Q8.**

**Answer :**

We observe the following properties of R.

Reflexivity: Let a be an arbitrary element of A. Then,

a∈R⇒a=a Since, every element is equal to itself⇒a, a∈R for all a∈ASo, R is reflexive on A.Symmetry: Let a, b ∈R⇒a b⇒b=a⇒b, a∈R for all a, b∈ASo, R is symmetric on A.Transitivity: Let a, b and b, c∈R⇒a=b and b=c⇒a=b c⇒a=c⇒a, c∈RSo, R is transitive on A.

Hence, R is an equivalence relation on A.

The set of all elements related to 1 is {1}.

**Q9.**

**Answer :**

We observe the following properties of R.

Reflexivity: Let L1 be an arbitrary element of the set L. Then,L1∈L⇒L1 is parallel to L1 Every line is parallel to itself⇒L1, L1∈R for all L1∈LSo, R is reflexive on L.Symmetry: Let L1, L2∈R⇒L1 is parallel to L2⇒L2 is parallel to L1⇒L2, L1∈R for all L1 and L2∈LSo, R is symmetric on L.Transitivity: Let L1, L2 and L2, L3∈R⇒L1 is parallel to L2 and L2 is parallel to L3⇒L1, L2 and L3 are all parallel to each other⇒L1 is parallel to L3⇒L1, L3∈RSo, R is transitive on L.

Hence, R is an equivalence relation on L.

Set of all the lines related to y = 2x+4

= L’ = {(x, y) : y = 2x+c, where c∈R}

**Q10.**

**Answer :**

We observe the following properties on R.

Reflexivity: Let P1 be an arbitrary element of A.Then, polygon P1 and P1 have the same number of sides, since they are one and the same.⇒P1, P1∈R for all P1∈ASo, R is reflexive on A.Symmetry: Let P1, P2∈R⇒P1 and P2 have the same number of sides.⇒P2 and P1 have the same number of sides.⇒P2, P1∈R for all P1, P2∈ASo, R is symmetric on A.Transitivity: Let P1, P2, P2, P3∈R⇒P1 and P2 have the same number of sides and P2 and P3 have the same number of sides.⇒P1, P2 and P3 have the same number of sides.⇒P1 and P3 have the same number of sides.⇒P1, P3∈R for all P1, P3 ASo, R is transitive on A.

Hence, R is an equivalence relation on the set A.

Also, the set of all the triangles∈A is related to the right angle triangle T with the sides 3, 4, 5.

**Q11.**

**Answer :**

Let A be the set of all points in a plane such that

A={P : P is a point in the plane}Let R be the relation such that R=P, Q : P, Q∈A and OP=OQ, where O is the origin

We observe the following properties of R.

Reflexivity: Let P be an arbitrary element of R.

The distance of a point P will remain the same from the origin.

So, OP = OP

⇒P, P∈RSo, R is reflexive on A.Symmetry: Let P, Q∈R⇒OP=OQ⇒OQ=OP⇒Q, P∈RSo, R is symmetric on A.Transitivity: Let P, Q, Q, R∈R⇒OP=OQ and OQ=OR⇒OP=OQ=OR⇒OP=OR⇒P, R∈RSo, R is transitive on A.

**Q12.**

**Answer :**

We observe the following properties of R.

Reflexivity:

Let a be an arbitrary element of R. Then,a∈R⇒a, a∈R for all a∈ASo, R is reflexive on A.Symmetry: Let a, b∈R⇒Both a and b are either even or odd.⇒Both b and a are either even or odd.⇒b, a∈R for all a, b∈ASo, R is symmetric on A.Transitivity: Let a, b and b, c∈R⇒Both a and b are either even or odd and both b and c are either even or odd.⇒a, b and c are either even or odd.⇒a and c both are either even or odd.⇒a, c ∈R for all a, c∈ASo, R is transitive on A.

Thus, R is an equivalence relation on A.

We observe that all the elements of the subset {1, 3, 5, 7} are odd. Thus, they are related to each other.

This is because the relation R on A is an equivalence relation.

Similarly, the elements of the subset {2, 4, 6} are even. Thus, they are related to each other because every element is even.

Hence proved.

**Q13.**

**Answer :**

We observe the following properties of S.

Reflexivity:Let a be an arbitrary element of R. Then, a∈R⇒a2+a2≠1∀a∈R⇒a, a∉SSo, S is not reflexive on R.Symmetry: Let a, b∈R⇒a2+b2=1⇒b2+a2=1⇒b, a∈S for all a, b∈RSo, S is symmetric on R.Transitivity: Let a, b and b, c∈S⇒a2+b2=1 and b2+c2=1Adding the above two, we geta2+c2=2-2b2≠1 for all a, b, c∈RSo, S is not transitive on R.

Hence, S is not an equivalence relation on R.

**Q14.**

**Answer :**

We observe the following properties of R.

Reflexivity:

Let a, b be an arbitrary element of Z × Z0. Then,a, b∈Z × Z0⇒a, b∈Z, Z0⇒ab=ba⇒a, b∈R for all a, b∈Z × Z0So, R is reflexive on Z × Z0.

Symmetry:

Let a, b, c, d∈Z×Z0 such that a, b R c, d. Then,a, b R c, d⇒ad=bc⇒cb=da⇒c, d R a, bThus, a, b R c, d⇒c, d R a, b for all a, b, c, d∈Z×Z0So, R is symmetric on Z×Z0.

Transitivity:

Let a, b, c, d, e, f∈N×N0 such that a, b R c, d and c, d R e, f. Then,a, b R c, d⇒ad=bcc, d R e, f⇒cf=de⇒ad cf=bc de⇒af=be⇒a, b R e, fThus, a, b R c, d and c, d R e, f⇒a, b R e, f⇒a, b R e, f for all values a, b, c, d, e, f∈N×N0So, R is transitive on N×N0.

**Page 1.26 Ex. 1.2**

**Q15,**

**Answer :**

(i) R and S are symmetric relations on the set A.

⇒R⊂A×A and S⊂A×A⇒R∩S⊂A×AThus, R∩S is a relation on A.Let a, b∈A such that a, b∈R∩S. Then,a, b∈R∩S⇒a, b∈R and a, b∈S⇒b, a∈R and b, a∈S Since R and S are symmetric⇒b, a∈R∩SThus, a, b∈R∩S⇒b, a∈R∩S for all a, b∈ASo, R∩S is symmetric on A.

Also,

Let a, b∈A such that a, b∈R∪S⇒a, b∈R or a, b∈S⇒b, a∈R or b, a∈S Since R and S are symmetric⇒b, a∈R∪SSo, R∪S is symmetric on A.

(ii) R is reflexive and S is any relation.

Suppose a∈A. Then, a, a∈R Since R is reflexive⇒a, a∈R∪S⇒R∪S is reflexive on A.

**Q16.**

**Answer :**

Let A = {a, b, c} and R and S be two relations on A, given by

R = {(a, a), (a, b), (b, a), (b, b)} and

S = {(b, b), (b, c), (c, b), (c, c)}

Here, the relations R and S are transitive on A.

a, b∈R∪S and b, c∈R∪SBut a, c∉R∪S

Hence, R∪S is not a transitive relation on A.

**Page 1.26 (Very Short Answers)**

**Q1.**

**Answer :**

Domain of R is the set of values satisfying the relation R.

As a should be an integer, we get the given values of a:

0, ±3, ±4, ±5Thus,Domain of R=0, ±3, ±4, ±5

**Q2.**

**Answer :**

Domain of R is the set of values of x satisfying the relation R.

As x must be an integer, we get the given values of x:

0, ±1, ±2Thus, Domain of R=0, ±1, ±2

**Q3.**

**Answer :**

Identity set of A is

I = {(a, a), (b, b), (c, c)}

Every element of this relation is related to itself.

**Q4.**

**Answer :**

Here,

A = {1, 2, 3, 4}

Also, a relation is reflexive iff every element of the set is related to itself.

So, the smallest reflexive relation on the set A is

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

**Q5.**

**Answer :**

R = {(x, y) : x + 2y = 8, x, y∈N}

Then, the values of y can be 1, 2, 3 only.

Also, y = 4 cannot result in x = 0 because x is a natural number.

Therefore, range of R is {1, 2, 3}.

**Q6.**

**Answer :**

Here, R is symmetric on the set A.

Let a, b∈R⇒b, a∈R Since R is symmetric⇒a, b∈R-1 By definition of inverse relation⇒R⊂R-1Let x, y∈R-1⇒y, x∈R By definition of inverse relation⇒x, y∈R Since R is symmetric⇒ R-1⊂RThus, R=R-1

**Q7.**

**Answer :**

R is the set of ordered pairs satisfying the above relation. Also, no two different elements can satisfy the relation; only the same elements can satisfy the given relation.

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

**Q8.**

**Answer :**

Since R = {(x, y) : x ∈ A, y ∈ A and x < y},

R = {(2, 3), (2, 7), (3, 7), (4, 7)}

So, R-1 = {(3, 2), (7, 2), (7, 3), (7, 4)}

**Q9.**

**Answer :**

R = {(x, y) : x and y are relatively prime}

Then,

R = {(3, 2), (5, 2), (7, 2), (3, 10), (7, 10), (5, 6), (7, 6)}

So, R-1 = {(2, 3), (2, 5), (2, 7), (10, 3), (10, 7), (6, 5), (6, 7)}

**Q10.**

**Answer :**

A relation R on A is said to be reflexive iff every element of A is related to itself.

i.e. R is reflexive ⇔a, a∈R for all a∈A

**Q11.**

**Answer :**

A relation R on a set A is said to be symmetric iff

a, b∈R⇒b, a∈R for all a, b∈Ai.e. aRb⇒bRa for all a, b∈A

**Q12.**

**Answer :**

A relation R on a set A is said to be transitive iff

a, b∈R and b, c∈R⇒a, c∈R for all a, b, c∈Ri.e. aRb and bRc⇒aRc for all a, b, c∈R

**Q13.**

**Answer :**

A relation R on set A is said to be an equivalence relation iff

(i) it is reflexive,

(ii) it is symmetric and

(iii) it is transitive.

Relation R on set A satisfying all the above three properties is an equivalence relation.

**Q14.**

**Answer :**

Since, R=x, y : x, y∈N and x<y,R = {(3, 4), (3, 9), (5, 9), (7,9)}

**Q15.**

**Answer :**

Since R = {(x, y) : y is one half of x; x, y∈A}

So, R = {(2, 1), (4, 2), (6, 3), (8, 4)}

**Q16.**

**Answer :**

Since R = a, b : a, b∈N : a is a divisor of b

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

**Q17.**

**Answer :**

Since 1, 2∈R, 2, 1∈R but 1, 1∉R, R is not transitive on the set 1, 2, 3.For R to be in a transitive relation, we must have 1, 1∈R.

**Page 1.27 (Multiple Choice Questions)**

**Q1.**

**Answer :**

(c) (6, 8) ∈ R

6, 8∈R Then, a=b-2⇒6=8-2and b=8 > 6Hence, 6, 8∈R

**Q2.**

**Answer :**

(c) {0, ± 3, ± 4, ± 5}

R=a, b : a2+b2=25, a, b∈Z⇒a∈-5, -4, -3, -2, -1, 0, 1, 2, 3, 4, 5 and b∈-5, -4, -3, -2, -1, 0, 1, 2, 3, 4, 5

So, domain (R)=0, ± 3,± 4, ±5

**Q3.**

**Answer :**

(b) reflexive and symmetric

Reflexivity: Let x∈R. Then,x-x=0 < 1⇒x-x≤1⇒x, x∈R for all x∈ZSo, R is reflexive on Z.

Symmetry: Let x, y∈R. Then,x-y ≤ 0⇒-(y-x) ≤ 1⇒y-x ≤ 1 Since x-y=y-x⇒y, x∈R for all x, y∈ZSo, R is symmetric on Z.

Transitivity: Let x, y∈R and y, z∈R. Then,x-y ≤ 1 and y-z ≤ 1⇒It is not always true that x-y ≤ 1.⇒x, z∉RSo, R is not transitive on Z.

**Q4.**

**Answer :**

(d) none of these

R is given by {(1, 2), (2, 1), (2, 3), (3, 2), (3, 4), (4, 3), (4, 5), (5, 4), (1, 3), (3, 1), (1, 4), (4, 1) ,(2, 4), (4, 2)}, which is not mentioned in (a), (b) or (c).

**Q5.**

**Answer :**

(a) symmetric

A = Set of all straight lines in the plane

R=l1, l2 : l1, l2∈A : l1⊥l2Reflexivity: l1 is not ⊥ l1⇒l1, l1∉RSo, R is not reflexive on A.Symmetry: Let l1, l2∈R⇒l1⊥l2⇒l2⊥l1⇒l2, l1∈RSo, R is symmetric on A.Transitivity: Let l1, l2∈R, l2, l3∈R⇒l1⊥ l2 and l2⊥ l3But l1 is not ⊥ l3⇒l1, l3∉RSo, R is not transitive on A.

**Q6.**

**Answer :**

(c) transitive only

The relation R = {(b,c)} is neither reflexive nor symmetric because every element of A is not related to itself. Also, the ordered pair of R obtained by interchanging its elements is not contained in R.

We observe that R is transitive on A because there is only one pair.

**Q7.**

**Answer :**

(c) 6

The ordered pairs of the equivalence class of (3, 2) are {(3, 2), (6, 4), (9, 6), (12, 8), (15, 10), (18, 12)}.

We observe that these are 6 pairs.

**Q8.**

**Answer :**

(a) 1

The required relation is R.

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

Hence, there is only 1 such relation that is reflexive and symmetric, but not transitive.

**Q9.**

**Answer :**

(c) {−2, −1, 0, 1, 2}

Domain of R includes all values of x satisfying the relation.

**Q10.**

**Answer :**

(c) {1}

Here,

R=x, y : x∈A and y∈B : x > y⇒R=2, 1, 3, 1

Thus,

Range of R = {1}

**Q11.**

**Answer :**

(d) {2, 3, 4, 5}

The relation R is defined as

R = x, y : x∈2, 3, 4, 5, y∈3, 6, 7, 10 : x is relatively prime to y⇒R= 2, 3, 2, 7, 3, 7, 3, 10, 4, 7, 5, 3, 5, 7

Hence, the domain of R includes all the values of x, i.e. {2, 3, 4, 5}.

**Q12.**

**Answer :**

(d) i ϕ 1

∵ 2+3i=13≠13 3≠-3 1+i=2≠2and i =1So, i, 1∈ϕ

**Q13.**

**Answer :**

(c) {2,4,6}

The relation R is defined as

R=x, y : x, y∈N and x+2y = 8⇒R=x, y : x, y∈N and y = 8-x2

Domain of R is all values of x∈N satisfying the relation R. Also, there are only three values of x that result in y, which is a natural number. These are {2, 6, 4}.

**Page 1.28 (Multiple Choice Questions)**

**Q14.**

**Answer :**

(a) {(8, 11), (10, 13)}

The relation R is defined by

R=x, y : x∈11, 12, 13, y∈8, 10, 12 : y = x-3⇒R=11, 8, (13, 10)So, R-1=8, 11, 10, 13

**Q15.**

**Answer :**

(b) reflexive

Reflexivity: Since a, a∈R∀ a∈A, R is reflexive on A.Symmetry: Since a, b∈R but b, a∉R, R is not symmetric on A.⇒R is not antisymmetric on A.Also, R is not an identity relation on A.

**Q16.**

**Answer :**

(c) transitive

Reflexivity: Since (1, 1)∉B, B is not reflexive on A.Symmetry: Since 1, 2∈B but 2, 1∉B, B is not symmetric on A.Transitivity: Since 1, 2∈B, 2, 3∈B and 1, 3∈B, B is transitive on A.

**Q17.**

**Answer :**

(b) S ⊂ R

Since R is the largest equivalence relation on set A,

R ⊆ A × A

Since S is any relation on A,

S ⊂ A × A

So, S ⊂ R

**Q18.**

**Answer :**

(d) none of these

The relation R is defined as

R = x, y : x, y∈A : y = 3x⇒R = 1, 3, 2, 6, 3, 9

**Q19.**

**Answer :**

(d) all the three options

R=a, b : a=b and a, b∈AReflexivity: Let a∈A. Then,a=a⇒a, a∈R for all a∈ASo, R is reflexive on A.Symmetry: Let a, b∈A such that a, b∈R. Then,a, b∈R⇒a=b⇒b=a⇒b, a∈R for all a∈ASo, R is symmetric on A.Transitivity: Let a, b, c∈A such that a, b∈R and b, c∈R. Then,a, b∈R⇒a=band b, c∈R⇒b=c⇒a=c⇒a, c∈R for all a∈ASo, R is transitive on A.

Hence, R is an equivalence relation on A.

**Q20.**

**Answer :**

(a) symmetric and transitive only

Reflexivity: Since b, b∉R, R is not reflexive on A.Symmetry: Since a, b∈R and b, a∈R, R is symmetric on A.Transitivity: Since a, b∈R, b, a∈R and a, a∈R, R is transitive on A.

**Q21.**

**Answer :**

(c) transitive only

The relation R is not reflexive because every element of A is not related to itself. Also, R is not symmetric since on interchanging the elements, the ordered pair in R is not contained in it.

R is transitive by default because there is only one element in it.

**Q22.**

**Answer :**

(b) R is reflexive and transitive but not symmetric.

Reflexivity: Clearly, (a, a)∈R ∀ a∈ASo, R is reflexive on A.Symmetry: Since 1, 2∈R, but 2, 1∉R,R is not symmetric on A.Transitivity: Since, 1, 3, 3, 2∈R and 1, 2∈R,R is transitive on A.

**Q23.**

**Answer :**

(b) 2

There are 2 equivalence relations containing {1, 2}.

R = {(1, 2)}

S = {(1, 1), (2, 2), (3, 3), (1, 2), (2, 1), (2, 3), (3, 2), (1, 3), (3, 1)}

**Q24.**

**Answer :**

(b) reflexive and symmetric

Reflexivity: Let x be an arbitrary element of R. Then,x∈R⇒x-x = 0 < 1⇒x, x∈R for all x∈ZSo, R is reflexive on Z.Symmetry: Let x, y∈R⇒x-y ≤ 1⇒y-x ≤ 1 Since x-y = y-x⇒y,x∈R for all x, y∈ZSo, R is symmetric on Z.Transitivity: Let x, y∈R and y, z∈R⇒x-y ≤ 1 and y-z ≤ 1But x-z >1⇒x, z∉RSo, R is not transitive on Z.

**Q25.**

**Answer :**

(d) an equivalence relation

Reflexivity: Let a∈R

Then,

aa=a2>0⇒a, a∈R ∀ a∈R

So, S is reflexive on R.

Symmetry: Let (a, b)∈S

Then,

a, b∈S⇒ab≥0 ⇒ba≥0 ⇒b, a∈S∀ a, b∈R

So, S is symmetric on R.

Transitivity:

If a, b, b, c∈S⇒ab≥0 and bc≥0⇒ab×bc≥0⇒ac≥0 ∵ b2 ≥ 0⇒a, c∈S for all a, b, c∈set R

Hence, S is an equivalence relation on R.

**Q26.**

**Answer :**

(a) x R y : if x ≤ y

Clearly, R is not symmetric because x < y does not imply y < x.

Hence, (a) is not an equivalence relation.

**Q27.**

**Answer :**

(c) an equivalence relation

R=a, b : a=b and a, b∈AReflexivity: Let a∈A Here,a=a⇒a, a∈R for all a∈ASo, R is reflexive on A.Symmetry: Let a, b∈A such that a, b∈R. Then,a, b∈R⇒a=b⇒b=a⇒b, a∈R for all a∈ASo, R is symmetric on A.Transitive: Let a, b, c∈A such that a, b∈R and b, c∈R. Then, a, b∈R⇒a=band b, c∈R⇒b=c⇒a=c⇒a, c∈R for all a∈ASo, R is transitive on A.

Hence, R is an equivalence relation on A.

**Page 1.29 (Multiple Choice Questions)**

**Q28.**

**Answer :**

(c) an equivalence relation

We observe the following properties of relation R.

Reflexivity: Let (a, b)∈N×N⇒a, b∈N⇒a+b=b+a⇒a, b∈R So, R is reflexive on N×N.Symmetry: Let a, b, c, d∈N×N such that a, b R c, d⇒a+d=b+c⇒d+a=c+b⇒d, c, b, a∈R So, R is symmetric on N×N.Transitivity: Let a, b, c, d, e, f∈N×N such that a, b R c, d and c, d R e, f⇒a+d=b+c and c+f=d+e⇒a+d+c+f=b+c+d+e⇒a+f=b+e⇒a, b R e, fSo, R is transitive on N×N.

Hence, R is an equivalence relation on N.

**Q29.**

**Answer :**

(b) S ⊂ R

Since R is the largest equivalence relation on set A,

R ⊆ A × A

Also, since S is any relation on A,

S ⊂ A × A

So, S ⊂ R

**Q30.**

**Answer :**

(c) a R b ⇔ a < b

Clearly, R is not a symmetric relation. This is because if (a, b) is an element of relation R, then, a < b does not imply b < a ∀ a, b ∈ Z.

Hence, R is not an equivalence relation on Z.

**FUNCTIONS**

**Page 2.29 Ex. 2.1**

**Q1.**

**Answer :**

(i) which is one-one but not onto.

f: Z → Z given by f(x)=3x+2

Injectivity:

Let x and y be any two elements in the domain (Z), such that f(x) = f(y).

f(x)=f(y)

⇒3x + 2 =3y + 2

⇒3x = 3y

⇒x = y

⇒f(x) = f(y) ⇒x = y

So, f is one-one.

Surjectivity:

Let y be any element in the co-domain (Z), such that f(x) = y for some element x in Z (domain).

f(x) = y

⇒3x + 2 = y

⇒3x = y – 2

⇒x = y-23. It may not be in the domain (Z) because if we take y = 3,x = y-23 = 3-23 = 13∉ domain Z.

So, for every element in the co domain there need not be any element in the domain such that f(x) = y.

Thus, f is not onto.

(ii) which is not one-one but onto.

f: Z → N ∪ {0} given by f(x) = |x|

Injectivity:

Let x and y be any two elements in the domain (Z), such that f(x) = f(y).

⇒|x| = |y|

⇒x= ± y

So, different elements of domain f may give the same image.

So, f is not one-one.

Surjectivity:

Let y be any element in the co domain (Z), such that f(x) = y for some element x in Z (domain).

f(x) = y

⇒|x| = y

⇒x = ± y, which is an element in Z (domain).

So, for every element in the co-domain, there exists a pre-image in the domain.

Thus, f is onto.

(iii) which is neither one-one nor onto.

f: Z → Z given by f(x) = 2×2 + 1

Injectivity:

Let x and y be any two elements in the domain (Z), such that f(x) = f(y).

fx = fy⇒2×2+1 = 2y2+1⇒2×2 = 2y2⇒x2 = y2⇒x = ±y

So, different elements of domain f may give the same image.

Thus, f is not one-one.

Surjectivity:

Let y be any element in the co-domain (Z), such that f(x) = y for some element x in Z (domain).

f(x) = y

⇒2×2+1=y⇒2×2=y-1⇒x2=y-12⇒x=±y-12, ∉ Z always.For example, if we take, y = 4,x=±y-12=±4-12=±32, ∉ ZSo, x may not be in Z (domain).

Thus, f is not onto.

**Q2.**

**Answer :**

(i) f_{1} = {(1, 3), (2, 5), (3, 7)} ; A = {1, 2, 3}, B = {3, 5, 7}

Injectivity:

f_{1} (1) = 3

f_{1} (2) = 5

f_{1} (3) = 7

⇒Every element of A has different images in B.

So, f1 is one-one.

Surjectivity:

Co-domain of f_{1} = {3, 5, 7}

Range of f_{1} =set of images = {3, 5, 7}

⇒Co-domain = range

So, f_{1} is onto.

(ii) f_{2} = {(2, a), (3, b), (4, c)} ; A = {2, 3, 4}, B = {a, b, c}

Injectivity:

f_{2} (2) = a

f_{2} (3) = b

f_{2} (4) = c

⇒Every element of A has different images in B.

So, f_{2} is one-one.

Surjectivity:

Co-domain of f_{2} = {a, b, c}

Range of f_{2} = set of images = {a, b, c}

⇒Co-domain = range

So, f_{2} is onto.

(iii) f_{3} = {(a, x), (b, x), (c, z), (d, z)} ; A = {a, b, c, d,}, B = {x, y, z}

Injectivity:

f_{3} (a) = x

f_{3} (b) = x

f_{3} (c) = z

f_{3} (d) = z

⇒a and b have the same image x. (Also c and d have the same image z)

So, f_{3} is not one-one.

Surjectivity:

Co-domain of f_{1} ={x, y, z}

Range of f_{1} =set of images = {x, z}

So, the co-domain is not same as the range.

So, f_{3} is not onto.

**Q3.**

**Answer :**

f : N → N, defined by f(x) = x2 + x + 1

Injectivity:

Let x and y be any two elements in the domain (N), such that f(x) = f(y).

⇒x2+x+1=y2+y+1⇒x2-y2+x-y=0⇒x+yx-y+x-y=0⇒x-yx+y+1=0⇒x-y=0 x+y+1 cannot be zero because x and y are natural numbers⇒x=y

So, f is one-one.

Surjectivity:

The minimum number in N is 1.When x=1,x2+x+1=1+1+1=3⇒x2+x+1≥3, for every x in N.⇒fx will not assume the values 1 and 2.So, f is not onto.

**Q4.**

**Answer :**

A = {−1, 0, 1} and f = {(x, x^{2}) : x ∈ A}

Given, f(x) = x^{2}

Injectivity:

f(1) = 1^{2}=1 and

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

⇒1 and -1 have the same images.

So, f is not one-one.

Surjectivity:

Co-domain of f = {-1, 0, 1}

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

f(-1) = (-1)^{2} = 1 and

f(0) = 0

⇒Range of f = {0, 1}

So, both are not same.

Hence, f is not onto.

**Q5.**

**Answer :**

(i) f : N → N, given by f(x) = x^{2}

Injection test:

Let x and y be any two elements in the domain (N), such that f(x) = f(y).

f(x)=f(y)

x2=y2x=y(We do not get ± because x and y are in N)

So, f is an injection .

Surjection test:

Let y be any element in the co-domain (N), such that f(x) = y for some element x in N (domain).

f(x) = y

x2=yx=y, which may not be in N.For example, if y=3,x=3 is not in N.

So, f is not a surjection.

So, f is not a bijection.

(ii) f : Z → Z, given by f(x) = x^{2}

Injection test:

Let x and y be any two elements in the domain (Z), such that f(x) = f(y).

f(x)=f(y)

x2=y2x= ±y

So, f is not an injection .

Surjection test:

Let y be any element in the co-domain (Z), such that f(x) = y for some element x in Z (domain).

f(x) = y

x2=yx=±y which may not be in Z.For example, if y=3,x=±3 is not in Z.

So, f is not a surjection.

So, f is not a bijection.

(iii) f : N → N, given by f(x) = x^{3}

Injection test:

Let x and y be any two elements in the domain (N), such that f(x) = f(y).

f(x)=f(y)

x3=y3x=y

So, f is an injection .

Surjection test:

Let y be any element in the co-domain (N), such that f(x) = y for some element x in N (domain).

f(x) = y

x3=yx=y3which may not be in N.For example, if y=3,x=33 is not in N.

So, f is not a surjection and f is not a bijection.

(iv) f : Z → Z, given by f(x) = x^{3}

Injection test:

Let x and y be any two elements in the domain (Z), such that f(x) = f(y)

f(x)=f(y)

x3=y3x=y

So, f is an injection .

Surjection test:

Let y be any element in the co-domain (Z), such that f(x) = y for some element x in Z (domain).

f(x) = y

x3=yx=y3which may not be in Z.For example, if y=3,x=33 is not in Z.

So, f is not a surjection and f is not a bijection.

(v) f : R → R, defined by f(x) = | x |

Injection test:

Let x and y be any two elements in the domain (R), such that f(x) = f(y)

f(x)=f(y)

x=yx= ±y

So, f is not an injection .

Surjection test:

Let y be any element in the co-domain (R), such that f(x) = y for some element x in R (domain).

f(x) = y

x=yx=±y ∈ Z

So, f is a surjection and f is not a bijection.

(vi) f : Z → Z, defined by f(x) = x2 + x

Injection test:

Let x and y be any two elements in the domain (Z), such that f(x) = f(y).

f(x)=f(y)

x2+x=y2+yHere, we cannot say that x = y.For example, x = 2 and y = – 3Then, x2+x=22+2= 6y2+y=-32-3= 6So, we have two numbers 2 and -3 in the domain Z whose image is same as 6.

So, f is not an injection .

Surjection test:

Let y be any element in the co-domain (Z), such that f(x) = y for some element x in Z (domain).

f(x) = y

x2+x=yHere, we cannot say x ∈ Z.For example, y = -4.×2+x=-4×2+x+4=0x=-1±-152=-1±i152 which is not in Z.

So, f is not a surjection and f is not a bijection.

(vii) f : Z → Z, defined by f(x) = x − 5

Injection test:

Let x and y be any two elements in the domain (Z), such that f(x) = f(y).

f(x)=f(y)

x – 5 = y – 5

x = y

So, f is an injection .

Surjection test:

Let y be any element in the co domain (Z), such that f(x) = y for some element x in Z (domain).

f(x) = y

x – 5 = y

x = y + 5, which is in Z.

So, f is a surjection and f is a bijection.

(viii) f : R → R, defined by f(x) = sin x

Injection test:

Let x and y be any two elements in the domain (R), such that f(x) = f(y).

f(x)=f(y)

sin x = sin yHere, x may not be equal to y because sin 0= sin π.So, 0 and π have the same image 0.

So, f is not an injection .

Surjection test:

Range of f = [-1, 1]

Co-domain of f = R

Both are not same.

So, f is not a surjection and f is not a bijection.

(ix) f : R → R, defined by f(x) = x3 + 1

Injection test:

Let x and y be any two elements in the domain (R), such that f(x) = f(y).

f(x)=f(y)

x3+1=y3+1×3=y3x=y

So, f is an injection .

Surjection test:

Let y be any element in the co-domain (R), such that f(x) = y for some element x in R (domain).

f(x) = y

x3+1=yx=y-13∈R

So, f is a surjection.

So, f is a bijection.

(x) f : R → R, defined by f(x) = x3 − x

Injection test:

Let x and y be any two elements in the domain (R), such that f(x) = f(y).

f(x)=f(y)

x3-x=y3-yHere, we cannot say x=y.For example,x=1 and y=-1×3-x=1-1= 0y3-y=-13–1-1+1=0So, 1 and -1 have the same image 0.

So, f is not an injection.

Surjection test:

Let y be any element in the co-domain (R), such that f(x) = y for some element x in R (domain).

f(x) = y

x3-x=yBy observation we can say that there exist some x in R, such that x3-x=y.

So, f is a surjection and f is not a bijection.

(xi) f : R → R, defined by f(x) = sin2 x + cos2 x

f(x) = sin2 x + cos2 x=1

So, f(x) =1 for every x in R.

So, for all elements in the domain, the image is 1.

So, f is not an injection.

Range of f = {1}

Co-domain of f =R

Both are not same.

So, f is not a surjection and f is not a bijection.

(xii) f : Q − {3} → Q, defined by fx=2x+3x-3

Injection test:

Let x and y be any two elements in the domain (Q-{3}), such that f(x) = f(y).

f(x)=f(y)

2x+3x-3=2y+3y-32x+3y-3=2y+3x-32xy-6x+3y-9=2xy-6y+3x-99x=9yx=y

So, f is an injection.

Surjection test:

Let y be any element in the co-domain ((Q-{3}), such that f(x) = y for some element x in Q (domain).

f(x) = y

2x+3x-3=y2x+3=xy-3y2x-xy=-3y-3×2-y=-3y+1x=3y+1y-2, which is not defined at y=2.

So, f is not a surjection and f is not a bijection.

(xiii) f : Q → Q, defined by f(x) = x3 + 1

Injection test:

Let x and y be any two elements in the domain (Q), such that f(x) = f(y).

f(x)=f(y)

x3+1=y3+1×3=y3x=y

So, f is an injection .

Surjection test:

Let y be any element in the co domain (Q), such that f(x) = y for some element x in Q (domain).

f(x) = y

x3+1=yx=y-1,3 which may not be in Q.For example, if y= 8,×3+1= 8×3=7x=73 , which is not in Q.

So, f is not a surjection and f is not a bijection.

So, f is a surjection and f is a bijection.

(xiv) f : R → R, defined by f(x) = 5×3 + 4

Injection test:

Let x and y be any two elements in the domain (R), such that f(x) = f(y).

f(x)=f(y)

5×3+4 = 5y3+45×3= 5y3x3= y3x=y

So, f is an injection .

Surjection test:

Let y be any element in the co-domain (R), such that f(x) = y for some element x in R (domain).

f(x) = y

5×3+4=y5x3=y-4×3=y-45x=y-453∈R

So, f is a surjection and f is a bijection.

(xv) f : R → R, defined by f(x) = 3 − 4x

Injection test:

Let x and y be any two elements in the domain (R), such that f(x) = f(y).

f(x)=f(y)

3-4x=3-4y-4x=-4yx= y

So, f is an injection .

Surjection test:

Let y be any element in the co-domain (R), such that f(x) = y for some element x in R (domain).

f(x) = y

3-4x=y4x=3-yx=3-y4∈R

So, f is a surjection and f is a bijection.

(xvi) f : R → R, defined by f(x) = 1 + x2

Injection test:

Let x and y be any two elements in the domain (R), such that f(x) = f(y).

f(x)=f(y)

1+x2=1+y2x2=y2x= ±y

So, f is not an injection .

Surjection test:

Let y be any element in the co-domain (R), such that f(x) = y for some element x in R (domain).

f(x) = y

1+x2=yx2=y-1x=±y-1 which may not be in RFor example, if y=0,x=±-1=±i is not in R.

So, f is not a surjection and f is not a bijection.

**Page 2.30 Ex. 2.1**

**Q6.**

**Answer :**

Range of f = {a}

So, the number of images of f = 1

Since, f is an injection, there will be exactly one image for each element of f .

So, number of elements in A = 1.

**Q7.**

**Answer :**

f : R − {3} → R − {2} given by

fx=x-2x-3

Injectivity:

Let x and y be any two elements in the domain (R − {3}), such that f(x) = f(y).

f(x) = f(y)

⇒x-2x-3=y-2y-3⇒x-2y-3=y-2x-3⇒xy-3x-2y+6=xy-3y-2x+6⇒x=y

So, f is one-one.

Surjectivity:

Let y be any element in the co-domain (R − {2}), such that f(x) = y for some element x in R − {3} (domain).

f(x) = y

⇒x-2x-3=y⇒x-2=xy-3y⇒xy-x=3y-2⇒xy-1=3y-2⇒x=3y-2y-1, which is in R-{3}

So, for every element in the co-domain, there exists some pre-image in the domain.

⇒f is onto.

Since, f is both one-one and onto, it is a bijection.

**Q8.**

**Answer :**

f : R → R, given by f(x) = x − [x]

Injectivity:

As fx=0∀x∈Z,

f is not one-one.

Surjectivity:

Range of f = [0, 1)

Co-domain of f = R

Both are not same.

So, f is not onto.

**Q9.**

**Answer :**

Injectivity:

Let x and y be any two elements in the domain (N).

Case-1: Let both x and y be even and

fx=fy⇒x-1=y-1⇒x=y

Case-2: Let both x and y be odd and

fx=fy⇒x+1=y+1⇒x=y

Case-3: Let x be even and y be odd then, x ≠y.

Then, x + 1is odd and y – 1 is even.

⇒x +1 ≠y-1⇒fx≠fySo, x≠y ⇒fx≠fy

In all the 3 cases, f is one-one.

Surjectivity:

Co-domain of f = N={1, 2, 3, 4, …}

Range of f = {1+1, 2-1, 3+1, 4-1, …} = {2, 1, 4, 3, …} = {1, 2, 3, 4, …}

Both are same.

⇒f is onto.

So, f is a bijection.

**Q10.**

**Answer :**

A ={1, 2, 3}

Number of elements in A = 3

Number of one-one functions = number of ways of arranging 3 elements = 3! = 6

(i) {(1, 1), (2, 2), (3, 3)}

(ii) {(1, 1), (2, 3), (3, 2)}

(iii) {(1, 2 ), (2, 2), (3, 3 )}

(iv) {(1, 2), (2, 1), (3, 3)}

(v) {(1, 3), (2, 2), (3, 1)}

(vi) {(1, 3), (2, 1), (3,2 )}

**Q11.**

**Answer :**

Injectivity:

Let x and y be any two elements in the domain (R), such that f(x) = f(y)

⇒4×3+7=4y3+7⇒4×3=4y3⇒x3=y3⇒x=y

So, f is one-one.

Surjectivity:

Let y be any element in the co-domain (R), such that f(x) = y for some element x in R (domain).

f(x) = y

⇒4×3+7=y⇒4×3=y-7⇒x3=y-74⇒x=y-743∈R

So, for every element in the co-domain, there exists some pre-image in the domain.

⇒f is onto.

Since, f is both one-to-one and onto, it is a bijection.

**Q12.**

**Answer :**

f : R → R, given by f(x) = e^{x}

Injectivity:

Let x and y be any two elements in the domain (R), such that f(x) = f(y)

f(x)=f(y)

⇒ex=ey⇒x=y

So, f is one-one.

Surjectivity:

We know that range of e^{x} is (0, ∞) = R^{+}

Co-domain = R

Both are not same.

So, f is not onto.

If the co-domain is replaced by R^{+}, then the co-domain and range become the same and in that case, f is onto and hence, it is a bijection.

**Q13.**

**Answer :**

f:R^{+}→R given by fx= loga x, a>0

Injectivity:

Let x and y be any two elements in the domain (N), such that f(x) = f(y).

f(x) = f(y)

loga x=loga y⇒x=y

So, f is one-one.

Surjectivity:

Let y be any element in the co-domain (R), such that f(x) = y for some element x in R^{+} (domain).

f(x) = y

loga x=y⇒x=ay ∈R^{+}

So, for every element in the co-domain, there exists some pre-image in the domain.

⇒f is onto.

Since f is one-one and onto, it is a bijection.

**Q14.**

**Answer :**

A ={1, 2, 3}

Number of elements in A = 3

Number of one – one functions = number of ways of arranging 3 elements = 3! = 6

So, the possible one -one functions can be the following:

(i) {(1, 1), (2, 2), (3, 3)}

(ii) {(1, 1), (2, 3), (3, 2)}

(iii) {(1, 2 ), (2, 2), (3, 3 )}

(iv) {(1, 2), (2, 1), (3, 3)}

(v) {(1, 3), (2, 2), (3, 1)}

(vi) {(1, 3), (2, 1), (3,2 )}

Here, in each function, range = {1, 2, 3}, which is same as the co-domain.

So, all the functions are onto.

**Q15.**

**Answer :**

A ={1, 2, 3}

Possible onto functions from A to A can be the following:

(i) {(1, 1), (2, 2), (3, 3)}

(ii) {(1, 1), (2, 3), (3, 2)}

(iii) {(1, 2 ), (2, 2), (3, 3 )}

(iv) {(1, 2), (2, 1), (3, 3)}

(v) {(1, 3), (2, 2), (3, 1)}

(vi) {(1, 3), (2, 1), (3,2 )}

Here, in each function, different elements of the domain have different images.

So, all the functions are one-one.

**Q16.**

**Answer :**

We know that every onto function from A to itself is one-one.

So, the number of one-one functions = number of bijections = n!

**Q17.**

**Answer :**

We know that f_{1}: R → R, given by f_{1}(x)=x, and f_{2}(x)=-x are one-one.

Proving f_{1} is one-one:

Let f_{1}x=f_{1}y⇒x=y

So, f_{1} is one-one.

Proving f_{2} is one-one:

Let f_{2}x=f_{2}y⇒-x=-y⇒x=y

So, f_{2} is one-one.

Proving (f_{1} + f_{2}) is not one-one:

Given:

(f_{1} + f_{2}) (x) = f_{1} (x) + f_{2} (x)= x + (-x) =0

So, for every real number x, (f_{1} + f_{2}) (x)=0

So, the image of ever number in the domain is same as 0.

Thus, (f_{1} + f_{2}) is not one-one.

**Q18.**

**Answer :**

We know that f_{1}: R → R, given by f_{1}(x) = x, and f_{2}(x) = -x are surjective functions.

Proving f_{1} is surjective :

Let y be an element in the co-domain (R), such that f_{1}(x) = y.

f_{1}(x) = y

⇒x = y, which is in R.

So, for every element in the co-domain, there exists some pre-image in the domain.1(x)=f_{1}(y)x=y

So, f_{1}is surjective .

Proving f_{2} is surjective :Let f_{2}(x)=f_{2}(y)−x=−yx=y

Let y be an element in the co domain (R) such that f_{2}(x) = y.

f_{2}(x) = y

⇒x = y, which is in R.

So, for every element in the co-domain, there exists some pre-image in the domain.1(x)=f_{1}(y)x=y

So, f_{2} is surjective .

Proving (f_{1} + f_{2}) is not surjective :

Given:

(f_{1} + f_{2}) (x) = f_{1} (x) + f_{2} (x)= x + (-x) =0

So, for every real number x, (f_{1} + f_{2}) (x)=0

So, the image of every number in the domain is same as 0.

⇒Range = {0}

Co-domain = R

So, both are not same.

So, f_{1} + f_{2} is not surjective.

**Q19.**

**Answer :**

We know that f_{1}: R → R, given by f1(x) = x, and f_{2}(x) = x are one-one.

Proving f_{1} is one-one:

Let x and y be two elements in the domain R, such that

f_{1}(x) = f_{1}(y)

⇒x = yet f_{1}(x)=f_{1}(y)x=y

So, f_{1} is one-one.

Proving f2 is one-one:

Let x and y be two elements in the domain R, such that

f2(x) = f2(y)

⇒x = yet f_{1}(x)=f_{1}(y)x=y

So, f_{2} is one-one.

Proving f_{1} × f_{2} is not one-one:

Given:

f_{1} × f_{2}x=f_{1} x × f_{2} x=x × x=x2Let x and y be two elements in the domain R, such thatf_{1} × f2x=f_{1} × f_{2}y⇒x2 = y2⇒x=± ySo, f_{1} × f_{2} is not one-one.

**Q20.**

**Answer :**

We know that f1: R → R, given by f1(x)=x3 and f2(x)=x are one-one.

Injectivity of f1:

Let x and y be two elements in the domain R, such that

f1x=f2y⇒x3=y⇒x=y3∈RLet f1(x)=f1(y)x=y

So, f1 is one-one.

Injectivity of f2:

Let x and y be two elements in the domain R, such that

f2x=f2y⇒x=y ⇒x∈R.Let f2(x)=f2(y)−x=−yx=y

So, f2 is one-one.

Proving f1f2is not one-one:

Given that f1f2x=f1xf2x=x3x=x2

Let x and y be two elements in the domain R, such that

f1f2x=f1f2y⇒x2=y2⇒x=±y

So, f1f2 is not one-one.

**Page 2.45 Ex. 2.2**

**Q1.**

**Answer :**

Given, f : R → R and g : R → R

So, gof : R → R and fog : R → R

(i) f(x) = 2x + 3 and g(x) = x^{2} + 5

Now, (gof) (x)

= g (f (x))

= g (2x +3)

= (2x + 3)^{2} + 5

= 4x^{2}+ 9 + 12x +5

=4x^{2}+ 12x + 14

(fog) (x)

=f (g (x))

= f (x^{2} + 5)

= 2 (x^{2} + 5) +3

= 2 x^{2}+ 10 + 3

= 2x^{2} + 13

(ii) f(x) = 2x + x^{2} and g(x) = x^{3}

gof x=g f x=g 2x+x2=2x+x23fog x=f g x=f x3=2 x3+x32=2×3+x6

(iii) f(x) = x^{2} + 8 and g(x) = 3x^{3} + 1

gof x=g fx=g x2+8=3 x2+83+1fog x=f g x=f 3×3+1=3×3+12+8=9×6+6×3+1+8=9×6+6×3+9

(iv) f(x) = x and g(x) = |x|

gof x=g fx=g x=xfog x=f g x=f x=x

(v) f(x) = x2 + 2x − 3 and g(x) = 3x − 4

gof x=g fx=g x2+2x-3=3 x2+2x-3-4=3×2+6x-9-4=3×2+6x-13fog x=f g x=f 3x-4=3x-42+2 3x-4-3=9×2+16-24x+6x-8-3=9×2-18x+5

(vi) f(x) = 8x^{3} and g(x) = x^{1/3}

gof x=g f x=g 8×3=8×313=2×313=2xfog x=f g x=f x13=8 x133=8x

**Q2.**

**Answer :**

f = {(3, 1), (9, 3), (12, 4)} and g = {(1, 3), (3, 3) (4, 9) (5, 9)}

f : {3, 9, 12} → {1, 3,4} and g : {1, 3, 4, 5} → {3, 9}

Co-domain of f is a subset of the domain of g.

So, gof exists and gof : {3, 9, 12} → {3, 9}

gof 3=g f 3=g 1=3gof 9=g f 9=g 3=3gof 12=g f 12=g 4=9⇒gof =3, 3, 9, 3, 12, 9

Co-domain of g is a subset of the domain of f.

So, fog exists and fog : {1, 3, 4, 5} → {3, 9, 12}

fog 1=f g 1=f 3=1fog 3=f g 3=f 3=1fog 4=f g 4=f 9=3fog 5=f g 5=f 9=3⇒fog=1, 1, 3, 1, 4, 3, 5, 3

**Q3.**

**Answer :**

f = {(1, −1), (4, −2), (9, −3), (16, 4)} and g = {(−1, −2), (−2, −4), (−3, −6), (4, 8)}

f : {1, 4, 9, 16} → {-1, -2, -3, 4} and g : {-1, -2, -3, 4} → {-2, -4, -6, 8}

Co-domain of f = domain of g

So, gof exists and gof : {1, 4, 9, 16} → {-2, -4, -6, 8}

gof 1=g f 1=g -1=-2gof 4=g f 4=g -2=-4gof 9=g f 9=g -3=-6gof 16=g f 16=g 4=8So, gof=1, -2, 4, -4, 9, -6, 16, 8

But the co-domain of g is not same as the domain of f.

So, fog does not exist.

**Q4.**

**Answer :**

Proving f is a bijection:

f = {(a, v), (b, u), (c, w)} and f : A → B

Injectivity of f: No two elements of A have the same image in B.

So, f is one-one.

Surjectivity of f: Co-domain of f = {u v, w}

Range of f = {u v, w}

Both are same.

So, f is onto.

Hence, f is a bijection.

Proving g is a bijection:

g = {(u, b), (v, a), (w, c)} and g : B → A

Injectivity of g: No two elements of B have the same image in A.

So, g is one-one.

Surjectivity of g: Co-domain of g = {a, b, c}

Range of g = {a, b, c}

Both are the same.

So, g is onto.

Hence, g is a bijection.

Finding fog:

Co-domain of g is same as the domain of f.

So, fog exists and fog : {u v, w} → {u v, w}

fog u=f g u=f b=ufog v=f g v=f a=vfog w=f g w=f c=wSo, fog =u, u, v, v, w, w

Finding gof:

Co-domain of f is same as the domain of g.

So, fog exists and gof : {a, b, c} → {a, b, c}

gof a=g f a=g v=agof b=g f b=g u=bgof c=g f c=g w=cSo, gof=a, a, b, b, c, c

**Q5.**

**Answer :**

fog 2=f g 2=f3×23+1=f25=252+8=633gof 1=g f 1=g 12+8=g 9=3×93+1=2188

**Q6.**

**Answer :**

Given, f : R^{+} → R^{+} and g : R^{+} → R^{+}

So, fog : R^{+} → R+ and gof : R^{+} → R^{+}

Domains of fog and gof are the same.

fog x=f g x=f x=x2=xgof x=g f x=g x2=x2=xSo, fog x=gof x,∀x∈R^{+}

Hence, fog = gof

**Q7.**

**Answer :**

Given, f : R → R and g : R → R.

So, the domains of f and g are the same.

fog x=f g x=f x+1=x+12=x2+1+2xgof x=g f x=g x2=x2+1

So, fog ≠ gof

**Q8.**

**Answer :**

Given, f : R → R and g : R → R

⇒fog : R → R and gof : R → R (Also, we know that IR : R → R)

So, the domains of all fog, gof and IR are the same.

fog x=f g x=f x-1=x-1+1=x=IR x … 1gof x=g f x=g x+1=x+1-1=x=IR x … 2From 1 and 2, fog x=gof x=IR x, ∀x∈RHence, fog=gof=IR

**Q9.**

**Answer :**

Given that f : N → Z0 , g : Z0 → Q and h : Q → R .

gof : N → Q and hog : Z0 → R

⇒h o (gof ) : N → R and (hog) o f: N → R

So, both have the same domains.

gof x=g f x=g 2x=12x …1hog x=h g x=h 1x=e1x …2Now,h ogof x=hgof x=h 12x=e12x [from 1]hog o fx=hog f x= hog 2x=e12x [from 2]⇒h ogof x=hog o fx, ∀x∈NSo, h ogof=hog o f

Hence, the associative property has been verified.

**Page 2.46 Ex. 2.2**

**Q10.**

**Answer :**

Given, f : N → N, g : N → N and h : N → R

⇒gof : N → N and hog : N → R

⇒ho (gof) : N → R and (hog) of : N → R

So, both have the same domains.

gof x=g f x=g 2x=3 2x+4=6x+4 …1hog x=hg x=h 3x+4=sin 3x+4 … 2Now,h o gof x=h gof x=h6x+4 = sin 6x+4 [from 1]hog o f x=hog f x=hog 2x=sin 6x+4 [from 2]So, h o gof x=hog o f x, ∀x∈NHence, h o gof=hog o f

**Q11.**

**Answer :**

Let us consider a function f : N → N given by f(x) = x +1 , which is not onto.

[This not onto because if we take 0 in N (co-domain), then,

0=x+1

⇒x=-1∉N]

Let us consider g : N → N given by

g x=x-1, if x>11, if x=1Now, let us find gof xCase 1: x>1gof x=g f x=g x+1=x+1-1=xCase 2: x=1gof x=g f x=g x+1=1From case-1 and case-2, gof x=x, ∀x∈N, which is an identity function and, hence, it is onto.

**Q12.**

**Answer :**

Let f : N → Z be given by f (x) = x, which is injective.

(If we take f(x) = f(y), then it gives x = y)

Let g : Z → Z be given by g (x) = |x|, which is not injective.

If we take f(x) = f(y), we get:

|x| = |y|

⇒x = ± y

Now, gof : N → Z.

gof x=g f x=g x=x

Let us take two elements x and y in the domain of gof , such that

gof x=gof y⇒x=y⇒x=y We don’t get ± here because x, y ∈N

So, gof is injective.

**Q13.**

**Answer :**

Given, f : A → B and g : B → C are one – one.

Then, gof : A → B

Let us take two elements x and y from A, such that

gof x=gof y⇒g f x=g f y⇒f x=f y As, g is one-one⇒x=y As, f is one-one

Hence, gof is one-one.

**Q14.**

**Answer :**

Given, f : A → B and g : B → C are onto.

Then, gof : A → C

Let us take an element z in the co-domain (C).

Now, z is in C and g : B → C is onto.

So, there exists some element y in B, such that g (y) = z … (1)

Now, y is in B and f : A → B is onto.

So, there exists some x in A, such that f (x) = y … (2)

From (1) and (2),

z = g (y) = g (f (x)) = (gof) (x)

So, z = (gof) (x), where x is in A.

Hence, gof is onto.

**Page 2.55 Ex. 2.3**

**Q1.**

**Answer :**

i) f x=ex, gx=loge xf:R→0,∞; g:0,∞→RComputing fog:Clearly, the range of g is a subset of the domain of f.fog : 0,∞→Rfog x=f g x=f loge x=loge ex=xComputing gof:Clearly, the range of f is a subset of the domain of g.⇒fog : R→Rgof x=g f x=g ex=loge ex=x

ii) f x=x2, gx=cos xf:R→[0, ∞) ; g:R→-1, 1Computing fog:Clearly, the range of g is not a subset of the domain of f.⇒Domain fog=x: x∈domain of g and gx∈domain of f⇒Domain fog=x: x∈R and cos x ∈R}⇒Domain of fog=Rfog: R→Rfog x=f g x=f cos x=cos2xComputing gof:Clearly, the range of f is a subset of the domain of g.⇒fog : R→Rgof x=g f x=g x2=cos x2

iii) f x=x, gx=sin xf:R→0, ∞; g:R→-1, 1Computing fog:Clearly, the range of g is a subset of the domain of f.⇒fog : R→Rfog x=f g x=f sin x=sin xComputing gof:Clearly, the range of f is a subset of the domain of g.⇒fog : R→Rgof x=g f x=g x=sin x

iv) f x=x+1, gx=exf:R→R; g:R→[1, ∞)Computing fog:Clearly, range of g is a subset of domain of f.⇒fog : R→Rfog x=f g x=f ex=ex+1Computing gof:Clearly, range of f is a subset of domain of g.⇒fog : R→Rgof x=g f x=g x+1=ex+1

v) f x=sin-1x, gx=x2f:-1,1→-π2,π2 ; g:R→[0, ∞)Computing fog:Clearly, the range of g is not a subset of the domain of f.Domain fog=x: x∈domain of g and gx∈domain of fDomain fog=x: x∈R and x2∈-1,1Domain fog=x: x∈R and x∈-1,1Domain of fog=-1,1fog: -1,1→Rfog x=f g x=f x2=sin-1 x2Computing gof:Clearly, the range of f is a subset of the domain of g.fog : -1,1→Rgof x=g f x=g sin-1x=sin-1 x2

vi) fx=x+1, gx=sin xf:R→R ; g:R→-1, 1Computing fog:Clearly, the range of g is a subset of the domain of f.⇒fog: R→Rfog x=f g x=f sin x=sin x+1Computing gof:Clearly, the range of f is a subset of the domain of g.⇒fog : R→Rgof x=g f x=g x+1=sin x+1

vii) f x=x+1, gx=2x+3f:R→R ; g:R→RComputing fog:Clearly, the range of g is a subset of the domain of f.⇒fog: R→Rfog x=f g x=f 2x+3=2x+3+1=2x+4Computing gof:Clearly, the range of f is a subset of the domain of g.⇒fog : R→Rgof x=g f x=g x+1=2 x+1+3=2x+5

viii) f x=c, gx=sin x2f:R→c ; g:R→0, 1Computing fog:Clearly, the range of g is a subset of the domain of f.fog: R→Rfog x=f g x=f sin x2=cComputing gof:Clearly, the range of f is a subset of the domain of g.⇒fog : R→Rgof x=g f x=g c=sin c2

ix) fx=x2+2f:R→[2,∞) gx=1-11-xFor domain of g: 1-x≠0 ⇒x≠1⇒Domain of g=R-1gx=1-11-x=1-x-11-x=-x1-xFor range of g:y=-x1-x⇒y-xy=-x⇒y=xy-x⇒y=xy-1⇒x=yy-1Range of g =R-1So, g: R-1→R-1Computing fog:Clearly, the range of g is a subset of the domain of f.⇒fog: R-1→Rfog x=f g x=f -xx-1=-xx-12+2=x2+2×2+2-4×1-x2=3×2-4x+21-x2Computing gof:Clearly, the range of f is a subset of the domain of g.⇒gof : R→Rgof x=g f x=g x2+2=1-11-x2+2=1-1-x2+1=x2+2×2+1

**Q2.**

**Answer :**

fog x=f g x=fsin x=sin2x+sin x+1and gof x=g f x=g x2+x+1= sin x2+x+1So, fog≠gof.

**Q3.**

**Answer :**

Domains of f and fof are same as R.

fof x=f f x=f x= x =x=f xSo,fof x=f x, ∀x∈RHence, fof=f

**Q4.**

**Answer :**

f(x) and g(x) are polynomials.

⇒f : R → R and g : R → R.

So, fog : R → R and gof : R → R.

i fog x=f g x=f x2+1=2 x2+1+5=2×2+2+5=2×2+7

ii gof x=g f x=g 2x+5=2x+52+1=4×2+20x+26

iii fof x=f f x=f 2x+5=2 2x+5+5=4x+10+5=4x+15

iv f2 x=fx×fx=2x+52x+5=2x+52=4×2+20x+25

**Q5.**

**Answer :**

fx=1-xFor domain, 1-x≥0⇒x≤1⇒domain of f =(-∞, 1]⇒f:(-∞, 1]→0,∞ gx=loge xClearly, g : 0, ∞→RComputation of fog:Clearly, the range of g is not a subset of the domain of f.So,we need to compute the domain of fog.⇒Domain fog=x : x∈Domain g and gx∈Domain of f⇒Domain fog=x: x∈0, ∞ and loge x ∈ (-∞, 1]⇒Domain fog=x:x∈0, ∞ and x∈ (0, e]⇒Domain fog=x: x ∈(0, e]⇒Domain fog=(0, e]⇒fog: 0, e→RSo, fog x=f g x=f loge x=1-loge x Computation of gof:Clearly, the range of f is a subset of the domain of g.⇒gof:(-∞,1]→R⇒gof x=g f x=g 1-x=loge1-x=loge 1-x12=12loge 1-x

**Q6.**

**Answer :**

g x=1-x2⇒x2≥0, ∀x∈-1, 1⇒-x2≤0, ∀x∈-1, 1⇒1-x2≤1, ∀x∈-1, 1We know that 1-x2≥0⇒0≤1-x2≤1⇒Range of gx=0, 1So, f:-π2, π2→R and g:-1, 1→ 0, 1Computation of fog:Clearly, the range of g is a subset of the domain of f.So, fog: -1, 1→Rfog x=f g x=f 1-x2=tan 1-x2Computation of gof:Clearly, the range of f is not a subset of the domain of g.⇒Domain gof=x∈domain of f and fx∈domain of g⇒Domain gof=x∈-π2, π2 and tan x ∈-1,1⇒Domain gof=x∈-π2, π2 and x ∈-π4, π4⇒Domain gof=x∈-π4, π4Now, gof:-π4, π4→RSo, gof x=g f x=g tan x=1-tan2x

**Page 2.56 Ex. 2.3**

**Q7.**

**Answer :**

We know thatf:R→-1, 1 and g: R→Ri gofClearly, the range of f is a subset of the domain of g.gof:R→Rgof x=g f x=g sin x=2 sin x

ii fogClearly, the range of g is a subset of the domain of f.fog:R→RSo, fog x=f g x=f 2x=sin 2x

Clearly, fog≠gof

**Q8.**

**Answer :**

fx=x+3For domain,x+3≥0⇒x≥-3Domain of f =[-3, ∞)Since f is a square root function, range of f =[0, ∞)f: [-3, ∞)→[0, ∞)g(x)=x2+1 is a polynomial.⇒g:R→RComputation of fog:Range of g is not a subset of the domain of f.and domain fog=x: x∈domain of g and gx∈domain of fx⇒Domain fog=x:x∈R and x2+1∈[-3, ∞)⇒Domain fog=x:x∈R and x2+1≥-3⇒Domain fog=x:x∈R and x2+4≥0⇒Domain fog=x:x∈R and x∈R⇒Domain fog=Rfog:R→Rfog x=fg x=f x2+1=x2+1+3=x2+4Computation of gof:Range of f is a subset of the domain of g.gof: [-3, ∞)→R⇒gof x=g f x=g x+3=x+32+1=x+3+1=x+4

**Q9.**

**Answer :**

We know that f:R→-1, 1 and g:R→RClearly, the range of g is a subset of the domain of f.fog:R→RNow, fh x=fxhx=sin x cos x=12 sin 2xDomain of fh is R.Since range of sin x is [-1,1],-1≤sin 2x≤1⇒-12≤sin x2≤12Range of fh =-12, 12So, fh:R→-12, 12Clearly, range of fh is a subset of g.⇒gofh:R→R⇒domains of fog and gofh are the same.So, fog x=f g x=f 2x=sin 2xand gofhx= g fh x=g sinx cos x=2sin x cos x=sin 2x⇒fog x= gofhx, ∀x∈RHence, fog = gofh

**Q10.**

**Answer :**

fx=x-2For domain,x-2≥0⇒x≥2Domain of f=[2,∞)Since fis a square-root function, range of f=0,∞So, f:[2,∞)→0,∞i fofRange of f is not a subset of the domain of f.⇒Domainfof=x: x ∈domain of fand fx∈domain of f⇒Domainfof=x: x ∈[2,∞) and x-2∈[2,∞)⇒Domainfof=x: x ∈[2,∞) and x-2≥2⇒Domainfof=x: x ∈[2,∞) and x-2≥4⇒Domainfof=x: x ∈[2,∞) and x≥6⇒Domainfof=x: x≥6⇒Domainfof=[6, ∞)fof :[6, ∞)→Rfof x=f f x=f x-2=x-2-2

ii fofof= (fof) ofWe have, f:[2,∞)→0,∞ and fof : [6, ∞)→R⇒Range of f is not a subset of the domain of fof.Then, domainfofof=x: x ∈domain of fand fx∈domain of fof⇒Domainfofof=x: x ∈[2,∞) and x-2∈[6,∞)⇒Domainfofof=x: x ∈[2,∞) and x-2≥6⇒Domainfofof=x: x ∈[2,∞) and x-2≥36⇒Domainfofof=x: x ∈[2,∞) and x≥38⇒Domainfofof=x: x≥38⇒Domainfofof=[38, ∞)fof :[38,∞)→RSo, fofof x=fof f x=fof x-2=x-2-2-2

iii We have, fofof x=x-2-2-2So, fofof 38=38-2-2-2=36-2-2=6-2-2=2-2=0

iv We have, fof=x-2-2⇒f2x=fx×fx=x-2×x-2=x-2So, fof ≠ f2

**Q11.**

**Answer :**

Given, f:R→RSince gx=2x is a polynomial, g:R→RClearly, gof:R→R and f+f:R→RSo, domains of gof and f+f are the same.gof x=g f x=2 fxf+f x=fx+fx=2 fx⇒gof x=f+f x, ∀x∈RHence, gof =f+f

**Q12.**

**Answer :**

fof x=f f x=f a-xn1n=a-a-xn1nn1n=a-a-xn1n=xn1n=x

**Q13.**

**Answer :**

fx=1+x,0≤x≤23-x,2<x≤3It can be written as,fx= 1+x,0≤x≤11+x,1<x≤23-x,2<x≤3 When,0≤x≤1Then, f(x)=1+xNow when ,0≤x≤1 then ,1≤x+1≤2Then, f(f(x))=1+1+x=2+x ∵1≤f(x)<2When ,1<x≤2Then, f(x)=1+xNow when ,1<x≤2 then,2<x+1≤3Then, f(f(x))=3-1+x=2-x ∵2≤f(x)<3When ,2<x≤3Then, f(x)=3-xNow when ,2<x≤3 then ,0≤3-x<1Then, f(f(x))=1+3-x=4-x ∵0≤f(x)<1ffx= 2+x,0≤x≤12-x,1<x≤24-x,2<x≤3

**Page 2.71 Ex. 2.3**

**Q1.**

**Answer :**

(i) f : {1, 2, 3, 4} → {10} with f = {(1, 10), (2, 10), (3, 10), (4, 10)}

We have:

f (1) = f (2) = f (3) = f (4) = 10

⇒f is not one-one.

⇒f is not a bijection.

So, f does not have an inverse.

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

g (5) = g (7) = 4

⇒f is not one-one.

⇒f is not a bijection.

So, f does not have an inverse.

(iii) h : {2, 3, 4, 5} → {7, 9, 11, 13} with h = {(2, 7), (3, 9), (4, 11), (5, 13)}

Here, different elements of the domain have different images in the co-domain.

⇒h is one-one.

Also, each element in the co-domain has a pre-image in the domain.

⇒h is onto.

⇒h is a bijection.

⇒h has an inverse and it is given by

h-1={(7, 2), (9, 3), (11, 4), (13, 5)}

**Q2.**

**Answer :**

(i) A = {0, −1, −3, 2}; B = {−9, −3, 0, 6} and f(x) = 3 x.

Given: f(x) = 3 x

So, f = {(0, 0), (-1, -3), (-3, -9), (2, 6)}

Clearly, this is one-one.

Range of f = Range of f =B

So, f is a bijection and, thus, f -1 exists.

Hence, f -1= {(0, 0), (-3, -1), (-9, -3), (6, 2)}

(ii) A = {1, 3, 5, 7, 9}; B = {0, 1, 9, 25, 49, 81} and f(x) = x2

Given: f(x) = x2

So, f = {(1, 1), (3, 9), (5, 25), (7,49), (9, 81)}

Clearly, f is one-one.

But this is not onto because the element 0 in the co-domain (B) has no pre-image in the domain (A) .

⇒f is not a bijection.

So, f -1does not exist.

**Q3.**

**Answer :**

f=1, a, 2, b, 3, c and g=a, apple, b, ball, c, catClearly, f and g are bijections.So, f and g are invertible.Now,f-1=a, 1, b, 2, c, 3 and g-1=apple, a, ball, b, cat, cSo, f-1o g-1=apple, 1, ball, 2, cat, 3 …1f:1, 2, 3→a, b, c and g:a, b, c→apple, ball, catSo, gof:1, 2, 3→apple, ball, cat⇒gof 1=g f 1=g a=applegof 2=g f 2=g b=ball,and gof 3=g f 3=g c=cat∴gof =1, apple, 2, ball, 3, catClearly, gofis a bijection.So, gof is invertible.gof-1=apple, 1, ball, 2, cat, 3 …2From 1 and 2, we get:gof-1=f-1o g-1

**Q4.**

**Answer :**

fx=2x+1⇒f=1, 21+1, 2, 22+1, 3, 23+1, 4, 24+1=1, 3, 2, 5, 3, 7, 4, 9gx=x2-2⇒g=3, 32-2, 5, 52-2, 7, 72-2, 9, 92-2=3, 7, 5, 23, 7, 47, 9, 79Clearly f and g are bijections and, hence, f-1:B→A and g-1: C→B exist.So, f-1=3, 1, 5, 2, 7, 3, 9, 4 and g-1=7, 3, 23, 5, 47, 7, 79, 9Now, f-1 o g-1:C→Af-1 o g-1=7, 1, 23, 2, 47, 3, 79, 4 …1Also, f:A→B and g:B→C,⇒gof:A→C, gof-1:C→ASo, f-1 o g-1and gof-1 have same domains.gofx=g f x=g 2x+1=2x+12-2⇒ gofx=4×2+4x+1-2⇒ gofx=4×2+4x-1Then, gof1=g f 1=4+4-1=7,gof2=g f 2=4+4-1=23,gof3=g f 3=4+4-1=47 and gof4=g f 4=4+4-1=79So, gof=1, 7, 2, 23, 3, 47, 4, 79⇒gof-1=7, 1, 23, 2, 47, 3, 79, 4 …2From 1 and 2, we get: gof-1=f-1 o g-1

**Q5.**

**Answer :**

Injectivity of f:

Let x and y be two elements of the domain (Q), such that

f(x)=f(y)

⇒3x + 5 =3y + 5

⇒3x = 3y

⇒x = y

So, f is one-one.

Surjectivity of f:

Let y be in the co-domain (Q), such that f(x) = y

⇒3x+5=y⇒3x=y-5⇒x=y-53∈Q domain

⇒f is onto.

So, f is a bijection and, hence, it is invertible.

Finding f -1:

Let f-1x=y …1⇒x=fy⇒x=3y+5⇒x-5=3y⇒y=x-53So, f-1x=x-53 [from 1]

**Q6.**

**Answer :**

Injectivity of f :

Let x and y be two elements of domain (R), such that

f(x) = f(y)

⇒4x + 3 = 4y + 3

⇒4x = 4y

⇒x = y

So, f is one-one.

Surjectivity of f :

Let y be in the co-domain (R), such that f(x) = y.

⇒4x+3=y⇒4x=y- 3⇒x=y- 34∈RDomain

⇒f is onto.

So, f is a bijection and, hence, is invertible.

Finding f -1:

Let f-1x=y …1⇒x=fy⇒x=4y+3⇒x-3=4y⇒y=x-34So, f-1x=x-34 [from 1]

**Q7.**

**Answer :**

Injectivity of f :

Let x and y be two elements of the domain (Q), such that

f(x)=f(y)

⇒x2+4=y2+4⇒x2=y2⇒x=y as co-domain as R+

So, f is one-one.

Surjectivity of f :

Let y be in the co-domain (Q), such that f(x) = y

⇒x2+4=y⇒x2=y-4⇒x=y-4∈R

⇒f is onto.

So, f is a bijection and, hence, it is invertible.

Finding f -1:

Let f-1x=y …1⇒x=fy⇒x=y2+4⇒x-4=y2⇒y=x-4So, f-1x=x-4 [from 1]

**Q8.**

**Answer :**

fofx=ffx=f 4x+36x-4=44x+36x-4+364x+36x-4-4=16x+12+18x-1224x+18-24x+16=34×34=x⇒fofx=x=IX, where I is an identity function.So, f=f-1 Hence, f-1=4x+36x-4

**Page 2.72 Ex. 2.5**

**Q9.**

**Answer :**

Injectivity of f :

Let x and y be two elements of domain (R+), such that

f(x)=f(y)

⇒9×2+6x-5=9y2+6y-5⇒9×2+6x=9y2+6y⇒x=y As, x, y∈R+

So, f is one-one.

Surjectivity of f:

Let y is in the co domain (Q) such that f(x) = y

⇒9×2+6x-5=y⇒9×2+6x=y+5⇒9×2+6x+1=y+6 Adding 1 on both sides⇒3x+12=y+6⇒3x+1=y+6⇒3x=y+6-1⇒x=y+6-13∈R+domain

⇒f is onto.

So, f is a bijection and hence, it is invertible.

Finding f -1:

Let f-1x=y …1⇒x=fy⇒x=9y2+6y-5⇒x+5=9y2+6y⇒x+6=9y2+6y+1 adding 1 on both sides⇒x+6=3y+12⇒3y+1=x+6⇒3y=x+6-1⇒y=x+6-13So, f-1x=x+6-13 [from 1]

**Q10.**

**Answer :**

Injectivity of f :

Let x and y be two elements in domain (R),

such that, x3-3=y3-3 ⇒x3=y3 ⇒x=y

So, f is one-one.

Surjectivity of f :

Let y be in the co-domain (R) such that f(x) = y

⇒x3-3=y⇒x3=y+3⇒x=y+33∈R

⇒f is onto.

So, f is a bijection and, hence, it is invertible.

Finding f -1:

Let f-1x=y …1⇒x=fy⇒x=y3-3⇒x+3=y3⇒y=x+33 = f-1x [from 1]So, f-1x=x+33 Now, f-124=24+33=273=333=3 and f-15=5+33=83=233=2

**Q11.**

**Answer :**

Injectivity of f:

Let x and y be two elements of domain (R), such that

fx=fy⇒x3+4=y3+4⇒x3=y3⇒x=y

So, f is one-one.

Surjectivity of f:

Let y be in the co-domain (R), such that f(x) = y.

⇒x3+4=y⇒x3=y-4⇒x=y-43∈R domain

⇒ f is onto.

So, f is a bijection and, hence, is invertible.

Finding f -1:

Let f-1x=y …1⇒x=fy⇒x=y3+4⇒x-4=y3⇒y=x-43So, f-1x=x-43 [from 1]f-13=3-43 =-13=-1

**Q12.**

**Answer :**

Injectivity of f:

Let x and y be two elements of domain (Q), such that

f(x) = f(y)

⇒2x = 2y

⇒x = y

So, f is one-one.

Surjectivity of f:

Let y be in the co-domain (Q), such that f(x) = y.

⇒2x= y⇒x= y2∈Q domain

⇒ f is onto.

So, f is a bijection and, hence, it is invertible.

Finding f -1:

Let f-1x=y …1⇒x=fy⇒x=2y⇒y=x2 So, f-1x=x2 from 1

Injectivity of g:

Let x and y be two elements of domain (Q), such that

g(x) = g(y)

⇒x + 2 = y + 2

⇒x = y

So, g is one-one.

Surjectivity of g:

Let y be in the co domain (Q), such that g(x) = y.

⇒x+2=y⇒x= 2-y∈Q domain

⇒ g is onto.

So, g is a bijection and, hence, it is invertible.

Finding g -1:

Let g-1x=y …2⇒x=gy⇒x=y+2⇒y=x-2So, g-1x=x-2 From 2

Verification of (gof)−1 = f−1 og −1:

fx=2x; gx=x+2and f-1x=x2; g-1x=x-2Now, f-1o g-1x=f-1g-1x⇒f-1o g-1x=f-1x-2⇒f-1o g-1x=x-22 …3gofx=g f x=g 2x=2x+2Let gof-1x=y …. 4x=gofy⇒x=2y+2⇒2y=x-2⇒y=x-22 ⇒gof-1x=x-22 [from 4 … 5]From 3 and 5, gof-1=f-1o g-1

**Q13.**

**Answer :**

Injectivity of f:

Let x and y be two elements of domain (R), such that

fx=fy⇒10x-10-x10x-10-x=10y-10-y10y-10-y⇒10-x102x-110-x102x+1=10-y102y-110-y102y+1⇒102x-1102x+1=102y-1102y+1⇒102x-1102y+1=102x+1102y-1⇒102x+2y+102x-102y-1=102x+2y-102x+102y-1⇒2×102x=2×102y⇒102x=102y⇒2x=2y⇒x=y

So, f is one-one.

Surjectivity of f:

Let y is in the co domain (R), such that f(x) = y

⇒10x-10-x10x+10-x=y⇒10-x102x-110-x102x+1=y⇒102x-1=y×102x+y⇒102×1-y=1+y⇒102x=1+y1-y⇒2x=log 1+y1-y⇒x=12log 1+y1-y∈R domain

⇒ f is onto.

So, f is a bijection and hence, it is invertible.

Finding f -1:

Let f-1x=y …1⇒fy=x⇒10y-10-y10y+10-y=x⇒10-y102y-110-y102y+1=x⇒102y-1=x×102y+x⇒102y1-x=1+x⇒102y=1+x1-x⇒2y=log 1+x1-x⇒y=12log 1+x1-xSo, f-1x=12log 1+x1-x [from 1]

**Q14.**

**Answer :**

Injectivity of f :

Let x and y be two elements of domain (R), such that

fx=fy⇒ex-e-xex-e-x+1=ey-e-ye-e-y+1⇒ex-e-xex-e-x=ey-e-ye-e-y⇒e-xe2x-1e-xe2x+1=e-ye2y-1e-ye2y+1⇒e2x-1e2x+1=e2y-1e2y+1⇒e2x-1e2y+1=e2x+1e2y-1⇒e2x+2y+e2x-e2y-1=e2x+2y-e2x+e2y-1⇒2×e2x=2×e2y⇒e2x=e2y⇒2x=2y⇒x=y

So, f is one-one.

Surjectivity of f:

Let y be in the co-domain 0, 2, such that f(x) = y.

ex-e-xex+e-x+1=y⇒e-xe2x-1e-xe2x+1+1=y⇒e-xe2x-1e-xe2x+1=y-1⇒e2x-1=y-1e2x+1⇒e2x-1=y×e2x+y-e2x-1⇒e2x=y×e2x+y-e2x⇒e2x2-y=y⇒e2x=y2-y⇒2x=loge y2-y⇒x=12loge y2-y∈R domain

So, f is onto.

∴ f is a bijection and, hence, it is invertible.

Finding f -1:

Let f-1x=y …1⇒fy=x⇒ey-e-yey+e-y+1=x⇒e-ye2y-1e-ye2y+1+1=x⇒e-ye2y-1e-ye2y+1=x-1⇒e2y-1=x-1e2y+1⇒e2y-1=x×e2y+x-e2y-1⇒e2y=x×e2y+x-e2y⇒e2y2-x=x⇒e2y=x2-x⇒2y=loge x2-x⇒y=12loge x2-

**Q15.**

**Answer :**

Injectivity: Let x and y ∈[-1, ∞), such that fx=fy⇒x+12-1=y+12-1⇒x+12=y+12⇒x+1=y+1⇒x=ySo, f is a injection.Surjectivity: Let y ∈[-1, ∞). Then, fx=y⇒x+12-1=y⇒x+1=y+1⇒x=y+1-1Clearly, x=y+1-1 is real for all y≥-1.Thus, every element y ∈[-1, ∞) has its pre-image x∈[-1, ∞) given by x=y+1-1.⇒f is a surjection.So, f is a bijection.Hence, f is invertible.Let f-1x=y …(1)⇒fy=x⇒y+12-1=x⇒y+12=x+1⇒y+1=x+1⇒y=±x+1-1⇒f-1x=±x+1-1 [from 1]fx=f-1x⇒x+12-1=±x+1-1⇒x+12=±x+1⇒x+14=x+1⇒x+1x+13-1=0⇒x+1=0 or x+13-=0⇒x=-1 or x+13=1⇒x=-1 or x+1=1⇒x=-1 or x=0⇒S=0, -1

**Q16.**

**Answer :**

f is not one-one because

f-1=-12=1and f1=12=1

⇒ -1 and 1 have the same image under f.

⇒ f is not a bijection.

So, f -1 does not exist.

Injectivity of g:

Let x and y be any two elements in the domain (A), such that

gx=gy⇒sin πx2=sin πy2⇒πx2=πy2⇒x=y

So, g is one-one.

Surjectivity of g:

Range of g = sin π-12, sin π12 = sin -π2, sin π2 = -1, 1 = A (co-domain of g)

⇒g is onto.

⇒g is a bijection.

So, g-1 exists.

Also,

let g-1x=y …1⇒gy=x⇒sinπy2=x⇒πy2=sin-1 x⇒y=2πsin-1 x⇒g-1x=2πsin-1 x [from 1]

**Q17.**

**Answer :**

Injectivity:

Let x and y be two elements in the domain (R), such that

fx=fy⇒cosx+2=cosy+2⇒x+2=y+2 or x+2=2π-y+2⇒x=y or x+2=2π-y-2⇒x=y or x=2π-y-4So, we cannot say that x=yFor example,cosπ2=cos 3π2=0So,π2 and 3π2 have the same image 0.

⇒ f is not one-one.

⇒ f is not a bijection.

Thus, f is not invertible.

**Q18.**

**Answer :**

f1=1, a, 2, b, 3, c, 4, d⇒f1-1=a, 1, b, 2, c, 3, d, 4f2=1, b, 2, a, 3, c, 4, d⇒f2-1=b, 1, a, 2, c, 3, d, 4f3=1, a, 2, b, 4, c, 3, d⇒f3-1=a, 1, b, 2, c, 4, d, 3f4=1, b, 2, a, 4, c, 3, d⇒f4-1=b, 1, a, 2, c, 4, d, 3

Clearly, all these are bijections because they are one-one and onto.

**Q19**

**Answer :**

A and B are two non empty sets. Let f be a function from A to B .It is given that there is injective map from A to B. That means f is one-one function .It is also given that there is injective map from B to A .That means every element of set B has its image in set A.⇒f is onto function or surjective.∴ f is bijective.If a function is both injective and surjective, then the function is bijective.

**Q20.**

**Answer :**

Given: A → A, g : A → A are two bijections.

Then, fog : A → A

(i) Injectivity of fog:

Let x and y be two elements of the domain (A), such that

fogx=fogy⇒fgx=fgy⇒gx=gy As, f is one-one⇒x=y As, g is one-one

So, fog is an injection.

(ii) Surjectivity of fog:

Let z be an element in the co-domain of fog (A).

Now, z∈A co-domain of f and f is a surjection.So, z=fy, where y∈A domain of f …1Now, y∈A co-domain of g and g is a surjection.So, y=gx, where x∈A domain of g …2From 1 and 2,z=fy=fgx=fogx, where x∈Adomain of fog

So, fog is a surjection.

**Page 2.73 (Very Short Answers)**

**Q1.**

**Answer :**

In graph (b), 0 has more than one image, whereas every value of x in graph (a) has a unique image.

Thus, graph (a) represents a function.

So, the answer is (a).

**Q2.**

**Answer :**

In the graph of (b), different elements on the x-axis have different images on the y-axis.

But in (a), the graph cuts the x-axis at 3 points, which means that 3 points on the x-axis have the same image as 0 and hence, it is not one-one.

**Q3.**

** Answer :**

Formula:

If set A has m elements and set B has n elements, then the number of functions from A to B is nm.

Given:

A = {1, 2, 3} and B = {a, b}

⇒nA = 3 and nB = 2

∴ Number of functions from A to B = 2^{3} = 8

**Q4.**

**Answer :**

Let f:A→B be a one-one function.

Then, fa can take 5 values, fb can take 4 values and fc can take 3 values.

Then, the number of one-one functions = 5 × 4 × 3 = 60

**Q5.**

**Answer :**

A has 4 elements and B has 3 elements.

Also, one-one function is only possible from A to B if nA≤nB.

But, here nA>nB.

So, the number of one-one functions from A to B is 0.

**Q6.**

**Answer :**

Let f-125=x … 1⇒fx=25⇒x2=25⇒x2-25=0⇒x-5x+5=0⇒x=±5⇒f-125=-5, 5 [from 1]

**Q7.**

**Answer :**

Let f-1-4=x … 1⇒fx=-4⇒x2=-4⇒x2+4=0⇒x+2ix-2i=0 using the identity: a2+b2=a-iba+ib⇒x=±2i as x∈C⇒f-125=-2i, 2i from 1

**Q8.**

**Answer :**

Let f-11= x … 1⇒fx= 1⇒x3= 1⇒x3-1= 0⇒x-1×2+x+1= 0 using the identity:a3-b3=a-ba2+ab+b2⇒x=1 ( as x∈R) ⇒f-11= 1 [from 1]

**Q9.**

**Answer :**

Let f-11=x … 1⇒fx=1⇒x3=1⇒x3-1=0⇒x-1×2+x+1=0 Using identity: a3-b3=a-ba2+ab+b2⇒x-1x-ωx-ω2=0, where ω=1±i32⇒x=1, ω or ω2 as x∈C⇒f-11=1, ω, ω2 [from 1]

**Page 2.74 (Very Short Answers)**

**Q10.**

**Answer :**

Let f-1-1=x … 1⇒fx=-1⇒x3=-1⇒x3+1=0⇒x+1×2-x+1=0 using the identity: a3+b3=a+ba2-ab+b2⇒x+1x+ωx+ω2=0, where ω= 1±i32 ⇒x=-1, -ω, -ω2 as x∈C⇒f-1-1=-1, -ω, -ω2 [from 1]

**Q11.**

**Answer :**

Let f-11=x … 1⇒fx=1⇒x4=1⇒x4-1=0⇒x2-1×2+1=0 using identity: a2-b2=a-ba+b⇒x-1x+1×2+1=0 using identity: a2-b2=a-ba+b⇒x=±1 as x∈R⇒f-11=-1, 1 [ from 1]

**Q12.**

**Answer :**

Let f-11=x … 1⇒fx=1⇒x4=1⇒x4-1=0⇒x2-1×2+1=0 using identity: a2-b2=a-ba+b⇒x-1x+1x-ix+i=0, where i=-1 using identity: a2-b2=a-ba+b⇒x=±1, ±i ⇒f-11=-1, 1, i,-i [from 1]

**Q13.**

**Answer :**

Let f-1-25=x⇒fx=-25⇒x2=-25We cannot find x∈R, such that x2=-25 as x2≥0 for all x∈RSo, f-1-25=ϕ

**Q14.**

**Answer :**

Let f-1-1=x … 1⇒fx=-1⇒x-23=-1⇒x-2=-1 or -ω or -ω2 as the roots of -113are -1, -ω and -ω2, where ω=1±i32⇒x=-1+2 or 2-ω or 2-ω2=1, 2-ω, 2-ω⇒f-1-1=1, 2-ω, 2-ω2 [from 1]

**Q15.**

**Answer :**

Let f-1x=y … 1⇒fy=x⇒10y-7=x⇒10y=x+7⇒y=x+710⇒f-1x=x+710 From 1

**Q16.**

**Answer :**

Domain =-π2, π2=-1.57, 1.57 (as π=227)So, cos x=cos -2=cos 2 ∀x∈-1.57, 0Also, cos 0=1 for x=0And cos x=cos 1 ∀x∈0, 1.57∴Range=1, cos 1, cos 2

**Q17.**

**Answer :**

Let f-1x=y … 1⇒fy=x⇒3y-4=x⇒3y=x+4⇒y=x+43⇒f-1x=x+43 [from 1]

**Q18.**

**Answer :**

fog-3=f g -3=f-32+1=f10=10+12=121

**Q19.**

**Answer :**

∵fx=xx=±xx=±1 ∀x∈A, range of f=-1, 1.

**Q20.**

**Answer :**

∵f is a bijection,

co-domain of f = range of f

As -1≤sin x≤1,

-1≤y≤1

So, A = [-1, 1]

**Q21.**

**Answer :**

Let f-1x=y … 1⇒fy=x⇒ay=x⇒y=loga x⇒f-1x=log a x [from 1]

**Q22.**

**Answer :**

Let f-1x=y … 1⇒fy=x⇒yy+1=x⇒y=xy+x⇒y-xy=x⇒y1-x=x⇒y=x1-x⇒f-1x=x1-x [from 1]

**Q23.**

**Answer :**

Let f-1x=y … 1⇒fy=x⇒2y5y+3=x⇒2y=5xy+3x⇒2y-5xy=3x⇒y2-5x=3x⇒y=3×2-5x⇒f-1x=3×2-5x [from 1]

**Q24.**

**Answer :**

fog-2=f g -2=f1–22=f-3=-32+-3+1=9-3+1=7

**Q25.**

**Answer :**

Let f-1x=y …1⇒fy=x⇒2y-34=x⇒2y-3=4x⇒2y=4x+3⇒y=4x+32⇒f-1x=4x+32 [from 1]⇒f-1x=4x+32∴fof-11=f41+32=f72=272-34=7-34=44=1

**Q26.**

**Answer :**

Given that f is an invertible real function.

f-1o f=I,where I is an identity function.So,f-1o f1+f-1o f2+…+f-1o f100=I1+I2+… +I100=1+2+…+100 As Ix=x, ∀x∈R=100100+12[Sum of first n natural numbers=nn+12]=5050

**Q27.**

**Answer :**

Formula:

When two sets A and B have m and n elements respectively, then the number of onto functions from A to B is

∑r=1n -1r nCr rm, if m≥no, if m<n

Here, number of elements in A = 4 = m

Number of elements in B = 2 = n

So, m > n

Number of onto functions

=∑r=12 -1r 2Cr r4=-11 2C1 14+-12 2C2 24=-2+16=14

**Q28.**

**Answer :**

[x] is the greatest integral function.

So, 0≤x-x<1⇒x-x exists for every x∈R.⇒Domain =R

**Q29.**

**Answer :**

[x] is the greatest integer function.

x≤x, ∀x∈R⇒x-x≤0,∀x∈R⇒x-x does not exist for any x∈R.Domain =ϕ

**Q30.**

**Answer :**

Case-1: When x>0x=x⇒1x-x=1x-x=10=∞Case-2: When x<0x=-x⇒1x-x=1-x-x=1-2x exists because when x<0, -2x>0⇒fx is defined when x<0So, domain =-∞,0

**Q31.**

**Answer :**

fx=x+x2=x±x=0 or 2xSo, each element x in the domain may contain 2 images.For example,f0=0+02=0f-1=-1+-12=-1+1=-1+1=0Here, the image of 0 and -1 is 0.

Hence, f is may-one.

**Q32.**

**Answer :**

fog7=f g7=f7-7=f 0=0+7=7

**Q33.**

**Answer :**

fx=x-1x-1=±x-1x-1=±1Range of f=-1, 1

**Q34.**

**Answer :**

fof x=f f x=f 3-x313=3-3-x313313=3-3-x313=x313=x

**Page 2.75 (Very Short Answers)**

**Q35.**

**Answer :**

ff x=f 3x+2=3 3x+2+2=9x+6+2=9x+8

**Q36.**

**Answer :**

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

Here, different elements of the domain have different images in the co-domain.

So, f is one-one.

**Page 2.75 (Multiple Choice Questions)**

**Q1.**

**Answer :**

(a) S defines a function from A to B

Let x∈A⇒-1≤x≤1Now, x2+y2=1⇒y2=1-x2⇒y=±1-x2⇒-1≤y≤1∴ y∈BThus, S defines a function from A to B.

**Q2.**

**Answer :**

fx=x+x2=x±x=0 or 2x⇒ Each element of the domain has 2 images.

⇒f is not a function.

So, the answer is (d).

**Q3.**

**Answer :**

(d) None of these

f:A→B3f(x)+2-x=4 ⇒3f(x)=4-2-xTaking log on both the sides , f(x) log 3=log 4-2-x⇒f(x)=log 4-2-xlog 3Logaritmic function will only be defined if 4-2-x>0⇒4>2-x⇒22>2-x⇒2>-x⇒-2<x⇒x∈ -2,∞That means A=x∈R:-2<x<∞As we know that, f(x)=log 4-2-xlog 3We take x=0∈ -2,∞⇒f(x)=1 which does not belong to any of the options .

**Q4.**

**Answer :**

(d) many-one and into

Graph for the given function is as follows.

A line parallel to X axis is cutting the graph at two different values.

Therefore, for two different values of x we are getting the same value of y.

That means it is many one function.

From the given graph we can see that the range is [2,∞)

and R is the co-domain of the given function.

Hence, Co-domain≠Range

Therefore, the given function is into.

**Q5.**

**Answer :**

(c) f is both one-one and onto

Injectivity:

Let x and y be two elements in the domain R- {-b}, such that

fx=fy⇒x+ax+b=y+ay+b⇒x+ay+b=x+by+a⇒xy+bx+ay+ab=xy+ax+by+ab⇒bx+ay=ax+by⇒a-bx=a-by⇒x=y

So, f is one-one.

Surjectivity:

Let y be an element in the co-domain of f, i.e. R-{1}, such that f (x)=y

fx=y⇒x+ax+b=y⇒x+a=yx+yb⇒x-yx=yb-a⇒x1-y=yb-a⇒x=yb-a1-y∈R–b

So, f is onto.

**Q6.**

**Answer :**

(a) A=(-∞, 3] and B=(-∞, 1]

fx=-x2+6x-8 , is a polynomial functionAnd the domain of polynomial function is real number.∴x∈R

f(x) =-x2+6x-8 =-x2-6x+8 =-x2-6x+9-1 =-x-32+1Maximum value of -x-32 woud be 0∴Maximum value of -x-32+1 woud be 1∴ f(x) ∈(-∞,1]

We can see from the given graph that function is symmetrical about x=3& the given function is bijective .So, x would be either (-∞,3] or [3,∞)The correct option which satisfy A and B both is:

A=(-∞, 3] and B=(-∞, 1]

**Q7.**

**Answer :**

Injectivity:

Let x and y be any two elements in the domain A.

Case-1: Let x and y be two positive numbers, such that

fx=fy⇒xx=yy⇒xx=yy⇒x2=y2⇒x=y

Case-2: Let x and y be two negative numbers, such that

fx=fy⇒xx=yy⇒x-x=y-y⇒-x2=-y2⇒x2=y2⇒x=y

Case-3: Let x be positive and y be negative.

Then, x≠y⇒fx=xx is positive and fy=yy is negative⇒fx≠fySo, x≠y⇒fx≠fy

From the 3 cases, we can conclude that f is one-one.

Surjectivity:

Let y be an element in the co-domain, such that y = f (x)

Case-1: Let y>0. Then, 0<y≤1⇒y=fx=xx>0⇒x>0⇒x=xfx=y⇒xx=y⇒xx=y⇒x2=y⇒x=y ∈ A We do not get ± because x>0Case-2: Let y<0. Then, -1≤y<0⇒y=fx=xx<0⇒x<0⇒x=-xfx=y⇒xx=y⇒x-x=y⇒-x2=y⇒x2=-y⇒x=–y ∈ A We do not get ± because x>0

⇒f is onto.

⇒f is a bijection.

So, the answer is (c).

**Q8.**

**Answer :**

(b) many-one and into

f : R→R

fx=x2+x+1-3

It is many one function because in this case for two different values of x

we would get the same value of f(x) .

For x=1.1, 1.2 ∈Rf(1.1)=1.12 +1.1+1-3 =1.21+2.1-3 =1+2-3 =0f(1.1)=1.22 +1.2+1-3 =1.44+2.2-3 =1+2-3 =0

It is into function because for the given domain we would only get the integral values of

f(x).

but R is the codomain of the given function.

That means , Codomain≠Range

Hence, the given function is into function.

Therefore, f(x) is many one and into

**Page 2.76 (Multiple Choice Questions)**

**Q9.**

**Answer :**

M=A=abcd: a, b, c, d∈R f: M→R is given by fA=A

Injectivity:

f0000=0000=0and f1000=1000=0⇒f0000=f1000=0

So, f is not one-one.

Surjectivity:

Let y be an element of the co-domain, such that

fA=-y, A=abcd⇒abcd=y⇒ad-bc=y⇒a, b, c, d∈R ⇒A=abcd∈M

⇒f is onto.

So, the answer is (d).

**Q10.**

**Answer :**

Injectivity:

Let x and y be two elements in the domain, such that

fx=fy⇒xx+1=yy+1⇒xy+x=xy+y⇒x=y

So, f is one-one.

Surjectivity:

Let y be an element in the co domain R, such that

y=fx⇒y=xx+1⇒xy+y=x⇒xy-1=-y⇒x=-yy-1Range of f=R-1≠co domain (R)

⇒f is not onto.

So, the answer is (b).

**Q11.**

**Answer :**

We know that

7-x>0; x-3 ≥0 and 7-x≥x-3⇒x<7; x≥3 and 2x≤10⇒x<7; x≥3 and x≤5So, x=3, 4, 5Range of f=P3-37-3, P4-37-4 , P7-55-3=4P0, 3P1, 2P2=1, 3, 2=1, 2, 3

So, the answer is (d).

**Q12.**

**Answer :**

(d) one-one and onto both

Injectivity:

Let x and y be any two elements in the domain (N).

Case-1: Both x and y are even.Let fx=fy⇒-x2=-y2⇒-x=-y⇒x=yCase-2: Both x and y are odd.Let fx=fy⇒x-12=y-12⇒x-1=y-1⇒x=yCase-3: Let x be even and y be odd. Then, fx=-x2and fy=y-12Then, clearly x≠y ⇒fx≠fyFrom all the cases, f is one-one.

Surjectivity:

Co-domain of f=Z=…,-3, -2, -1, 0, 1, 2, 3, ….Range of f=…, -3-12, –22,-1-12, 02, 1-12, -22, 3-12, …⇒Range of f=…,-2, 1, -1, 0, 0, -1, 1,…⇒Range of f=…, -2, -1, 0, 1, 2, ….⇒Co-domain of f=Range of f

⇒ f is onto.

**Q13.**

**Answer :**

Case-1: Let fx=1 be true.Then, fy≠1 and fz≠2 are false.So, f(y)=1 and fz=2⇒ fx=1, fy=1⇒x and y have the same images.This contradicts the fact that fis one-one.Case-2: Let fy≠1 be true.Then, fx=1 and fz≠2 are false.So, fx≠1 and fz=2⇒ fx≠1, fy≠1 and fz=2⇒There is no pre-image for 1.This contradicts the fact that range is 1, 2, 3.Case-3: Let fz≠2 be true.Then, fx=1 and fy≠1 are false.So, fx≠1 and fy=1⇒fx=2, fy=1 and fz=3⇒fy=1⇒f-11=y

So, the answer is (b).

**Q14.**

**Answer :**

a f is not onto because for y = 3∈Co-domain(Z), there is no value of x∈Domain(Z)x3=3⇒x=33∉Z⇒f is not onto.So, fis not a bijection.

(b) Injectivity:

Let x and y be two elements of the domain (Z), such that

x+2=y+2⇒x=y

So, f is one-one.

Surjectivity:

Let y be an element in the co-domain (Z), such that

y=fx⇒y=x+2⇒x=y-2∈Z Domain

⇒f is onto.

So, f is a bijection.

c fx=2x+1 is not onto because if we take 4 ∈ Zco domain, then 4=fx⇒4=2x+1⇒2x=3⇒x=32∉ZSo, f is not a bijection.d f0=02+0=0and f-1=-12+-1=1-1=0⇒0 and -1 have the same image.⇒f is not one-one.So, f is not a bijection.

So, the answer is (b).

**Q15.**

**Answer :**

a Range of f =-12, 12 ≠ ASo, f is not a bijection.b Range =sin-π2, sinπ2=-1,1=ASo, g is a bijection.c h-1=-1=1and h1=1=1⇒-1 and 1 have the same imagesSo, h is not a bijection.d k-1=-12=1and k1=12=1⇒-1 and 1 have the same imagesSo, k is not a bijection.

So, the answer is (b).

**Q16.**

**Answer :**

Injectivity:

Let x and y be any two elements in the domain A.

Case-1: Let x and y be two positive numbers, such that

fx=fy⇒xx=yy⇒xx=yy⇒x2=y2⇒x=y

Case-2: Let x and y be two negative numbers, such that

fx=fy⇒xx=yy⇒x-x=y-y⇒-x2=-y2⇒x2=y2⇒x=y

Case-3: Let x be positive and y be negative.

Then, x≠y⇒fx=xx is positive and fy=yy is negative⇒fx≠fySo, x≠y⇒fx≠fy

So, f is one-one.

Surjectivity:

Let y be an element in the co-domain, such that y = f (x)

Case-1: Let y>0. Then, 0<y≤1y=fx=xx>0⇒x>0⇒x=x⇒fx=y⇒xx=y⇒xx=y⇒x2=y⇒x=y ∈ A We do not get ±, as x>0Case-2: Let y<0. Then, -1≤y<0y=fx=xx<0⇒x<0⇒x=-x⇒fx=y⇒xx=y⇒x-x=y⇒-x2=y⇒x2=-y⇒x=–y ∈ A We do not get ±, as x>0

⇒f is onto

⇒f is a bijection.

So, the answer is (a).

**Q17.**

**Answer :**

As f is surjective, range of f=co-domain of f⇒A= range of f ∵fx= x2x2+1, y=x2x2+1⇒yx2+1= x2⇒y-1×2+y= 0⇒x2= -yy-1⇒x=y1-y⇒y1-y≥0⇒y∈[0, 1)⇒Range of f= [0, 1)⇒A= [0, 1)

So, the answer is (d).

**Q18.**

**Answer :**

Since f is a bijection, co-domain of f = range of f

⇒B = range of f

Given: fx=x2-4x+5Let fx=y⇒y=x2-4x+5⇒x2-4x+5-y=0∵Discrimant, D=b2-4ac≥0,-42-4×1×5-y≥0⇒16-20+4y≥0⇒4y≥4⇒y≥1⇒y∈[1, ∞)⇒Range of f=[1, ∞)⇒B=[1, ∞)

So, the answer is (b).

**Q19.**

**Answer :**

fx=x-1x-2x-3

Injectivity:

f1=1-11-21-3=0f2=2-12-22-3=0f3=3-13-23-3=0⇒ f1=f2=f3=0

So, f is not one-one.

Surjectivity:

Let y be an element in the co domain R, such that

y=fx⇒y=x-1x-2x-3Since y∈R and x∈R, f is onto.

So, the answer is (b).

**Page 2.77 (Multiple Choice Questions)**

**Q20.**

**Answer :**

fx=sin-13x-4×3⇒fx=3sin-1x

Injectivity:

Let x and y be two elements in the domain -12, 12 , such that

fx=fy⇒3sin-1x=3sin-1y⇒sin-1x=sin-1y⇒x=y

So, f is one-one.

Surjectivity:

Let y be any element in the co-domain, such that

fx=y⇒3sin-1x=y⇒sin-1x=y3⇒x=siny3∈-12, 12

⇒f is onto.

⇒f is a bijection.

So, the answer is (a).

**Q21.**

**Answer :**

(d) f is neither an injection nor a surjection

f : R→R

fx=e|x|-e-xex+e-xFor x=-2 and -3∈ R f(-2) =e-2-e2e-2+e2 =e2-e2e-2+e2 = 0& f(-3) =e-3-e3e-3+e3 =e3-e3e-3+e3 = 0Hence, for different values of x we are getting same values of f(x)That means , the given function is many one .

Therefore, this function is not injective.

For x<0f(x) =0For x>0f(x) =ex-e-xex+e-x =ex+e-xex+e-x-2e-xex+e-x =1-2e-xex+e-xThe value of 2e-xex+e-x is always positive.Therefore, the value of f(x) is always less than 1Numbers more than 1 are not included in the range but they are included in codomain.As the codomain is R.∴ Codomain≠RangeHence, the given function is not onto .

Therefore, this function is not surjective .

**Q22.**

**Answer :**

Injectivity:

Let x and y be two elements in the domain R-{n}, such that

fx=fy⇒x-mx-n=y-my-n⇒x-my-n=x-ny-m⇒xy-nx-my+mn=xy-mx-ny+mn⇒m-nx=m-ny⇒x=y

So, f is one-one.

Surjectivity:

Let y be an element in the co domain R, such that

fx=y⇒x-mx-n=y⇒x-m=xy-ny⇒ny-m=xy-x⇒ny-m=xy-1⇒x=ny-my-1, which is not defined for y =1So, 1 ∈ R co domain has no pre image in R-n

⇒f is not onto.

Thus, the answer is (b).

**Q23.**

**Answer :**

Injectivity:

Let x and y be two elements in the domain (R), such that

fx=fy⇒x2-8×2+2=y2-8y2+2⇒x2-8y2+2=x2+2y2-8⇒x2y2+2×2-8y2-16=x2y2-8×2+2y2-16⇒10×2=10y2⇒x2=y2⇒x=±y

So, f is not one-one.

Surjectivity:

f-1=-12-8-12+2=1-81+2=-73 and f1=12-812+2=1-81+2=-73⇒f-1=f1=-73

⇒f is not onto.

The correct answer is (d).

**Q24.**

**Answer :**

(d) neither one-one nor onto

We have,fx=ex2-e-x2ex2+e-x2Here, -2, 2∈RNow, 2≠-2But, f2=f-2Therefore, function is not one-one.And,The minimum value of the function is 0 and maximum value is 1That is range of the function is 0, 1 but the co-domain of the function is given R.Therefore, function is not onto.∴function is neither one-one nor onto.

**Q25.**

**Answer :**

Injectivity:

Let x and y be any two elements in the domain (R), such that f(x) = f(y). Then,

x2=y2⇒x=±y

So, f is not one-one.

Surjectivity:

As f-1=-12=1and f1=12=1, f-1=f1

So, both -1 and 1 have the same images.

⇒f is not onto.

So, the answer is (d).

**Q26.**

**Answer :**

Injectivity:

Let x and y be any two elements in the domain (N).

Case-1: Both x and y are even.Let fx=fy⇒-x2=-y2⇒-x=-y⇒x=yCase-2: Both x and y are odd.Let fx=fy⇒x-12=y-12⇒x-1=y-1⇒x=yCase-3: Let x be even and y be odd. Then, fx=-x2and fy=y-12Then, clearly x≠y ⇒fx≠fyFrom all the cases, f is one-one.

Surjectivity:

Co-domain of f=Z=…,-3, -2, -1, 0, 1, 2, 3, ….Range of f=…, -3-12, –22,-1-12, 02, 1-12, -22, 3-12, …⇒Range of f=…,-2, 1, -1, 0, 0, -1, 1,…⇒Range of f=…, -2, -1, 0, 1, 2, ….⇒Co-domain of f=Range of f

⇒ f is onto.

So, the answer is (d).

**Q27.**

**Answer :**

(b) fx=sinπ x2

It is clear that f(x) is one-one.

Range of f=sinπ-12, sinπ12=sin -π2, sinπ2=-1,1= A=Co domain of f

⇒ f is onto.

So, f is a bijection.

**Q28.**

**Answer :**

Injectivity:

Let x and y be two elements in the domain (Z), such that

fx=fyCase-1: Let both x and y be even.Then,fx=fy⇒x2=y2⇒x=yCase-2: Let both x and y be odd.Then,fx=fy⇒0=0Here, we cannot determine whether x=y.

So, f is not one-one.

Surjectivity:

Let y be an element in the co-domain (Z), such that

Co-domain of f =Z={0, ±1, ±2, ±3, ±4, …}Range of f=0, 0, ±22, 0, ±42 ,…=0, ±1, ±2, …⇒Co-domain of f=Range of f

⇒ f is onto.

So, the answer is (a).

**Page 2.78 (Multiple Choice Questions)**

**Q29.**

**Answer :**

(d) many one and into

Graph of the given function is as follows :

A line parallel to X axis is cutting the graph at two different values.

Therefore, for two different values of x we are getting the same value of y .

That means it is many one function .

From the given graph we can see that the range is [2,∞)

and R is the codomain of the given function .

Hence, Codomain≠Range

Therefore, the given function is into .

**Q30.**

**Answer :**

Since fogx=gofx, fgx=gfx⇒f2x=gx2⇒2×2=2×2⇒22x=2×2⇒x2=2x⇒x2-2x=0⇒xx-2=0⇒x=0, 2⇒x∈0, 2

So, the answer is (c).

**Q31.**

**Answer :**

Clearly, f is a bijection.

So, f -1 exists.

Let f-1x=y …1⇒fy=x⇒3y-5=x⇒3y=x+5⇒y=x+53⇒f-1x=x+53 [from 1]

So, the answer is (b).

**Q32.**

**Answer :**

If we solve it by the trial-and-error method, we can see that (a) satisfies the given condition.

From (a):

fx=sin2x and gx=x⇒fgx=fx=sin2 x=sin x2

So, the answer is (a).

**Q33.**

**Answer :**

Let f-1x=y …1⇒fy=x⇒ey-e-yey+e-y=x⇒e-ye2y-1e-ye2y+1=x⇒e2y-1=xe2y+1⇒e2y-1=xe2y+x⇒e2y1-x=x+1⇒e2y=1+x1-x⇒2y=loge 1+x1-x⇒y=12loge 1+x1-x⇒f-1x=12loge 1+x1-x [from 1]

So, the answer is (a).

**Q34.**

**Answer :**

Let f-1x=y…1⇒fy=x⇒2yy-1=x⇒2y2-y=x⇒y2-y=log2 x⇒y2-y+14=log2 x+14⇒y-122=4log2 x+14⇒y-12=±4log2 x+12⇒y=12±4log2 x+12⇒y=12+4log2 x+12 ∵ y ≥1So, f-1x=12(1+1+4log2 x ) [from 1]

So, the answer is (b).

**Q35.**

**Answer :**

Let y be the element in the codomain R such that f-1x=y …1⇒fy=x and y ≤1⇒y2-y=x⇒2y-y2=x⇒y2-2y+x=0⇒y2-2y=-x⇒y2-2y+1=1-x⇒y-12=1-x⇒y-1=±1-x⇒y=1±1-x⇒y=1-1-x ∵ y ≤1

The correct answer is (b).

**Q36.**

**Answer :**

Domain of f:1-x≠0⇒x≠1Domain of f=R-1Range of f:y=11-x⇒1-x=1y⇒x=1-1y⇒x=y-1y⇒y≠0Range of f=R-0So, f:R-1→R-0 and f:R-1→R-0 Range of f is not a subset of the domain of f.Domain fof=x: x∈domain of f and fx∈domain of fDomain fof=x: x∈R-1 and 11-x∈R-1 Domain fof=x: x ≠1 and 11-x≠1Domain fof=x: x ≠1 and 1-x≠1Domain fof=x: x ≠1 and x≠0Domain fof=R-0, 1fofx=ffx=f11-x=11-11-x=1-x1-x-1=1-x-x=x-1xFor range of fof, x≠0Now, fof:R-0, 1→R -0 and f:R-1→R-0Range of fof is not a subset of domain of f.Domainf o fof=x: x∈domain of fof and fofx∈domain of fDomain f o fof=x: x∈R-0, 1 and x-1x∈R-1 Domainf o fof=x: x ≠0, 1 and x-1x≠1Domain f o fof=x: x ≠0, 1 and x-1≠xDomain f o fof=x: x ≠0, 1 and x∈RDomain f o fof=R-0, 1fofofx=ffofx=fx-1x=11-x-1x=xx-x+1=xSo, fofofx=x, where x≠0,1

So, the answer is (c).

**Q37.**

**Answer :**

f(x) = x – [x]

We know that the range of f is [0, 1).

Co-domain of f = R

As range of f ≠ Co-domain of f, f is not onto.

⇒ f is not a bijective function.

So, f -1 does not exist.

Thus, the answer is (c).

**Q38.**

**Answer :**

Let f-1x = y⇒fy=x⇒y+1y=x⇒y2+1=xy⇒y2-xy+1=0⇒y2-2×y×x2+x22-x22+1=0⇒y2-2×y×x2+x22=x2-14⇒y-x22=x2-14⇒y-x2=x2-42⇒y=x2+x2-42⇒y=x+x2-42⇒f-1x=x+x2-42

So, the answer is (a).

**Page 2.79 (Multiple Choice Questions)**

**Q39.**

**Answer :**

(b) 1

When, -1<x<0Then, g(x)=1+x-x =1+x–1=2+x∴fg(x)=1 When, x=0Then, g(x)=1+x-x =1+x-0=1+x∴fg(x)=1When, x>1Then, g(x)=1+x-x =1+x-1=x∴fg(x)=1

Therefore, for each interval f(g(x))=1

**Q40.**

**Answer :**

(d) −1

ffx=x⇒ fαxx+1=x⇒ααxx+1αxx+1+1=x⇒α2xαx+x+1=x⇒α2x=αx2+x2+x⇒α2x-αx2-x2-x=0⇒α2x-αx2-x2+x=0Solving for the α we get,α=–x2±-x22-4×x×-x2+x2x =x2±x4+4×3+4x22x =x+1, -1Here, -1 is independent of x,∴for, α=-1, ffx=x

**Q41.**

**Answer :**

Let us substitute the end-points of the intervals in the given functions. Here, domain = [-1, 1] and range =[0, 2]

By substituting -1 or 1 in each option, we get:

Option (a):

f-1=-1+1=0 and f1=1+1=2g-1=1+1=2 and g1=-1+1=0

So, option (a) is correct.

Option (b):

f-1=-1-1=-2 and f1=1-1=0g-1=-1+1=0 and g1=1+1=2

Here, f(-1) gives -2∉0, 2

So, (b) is not correct.

Similarly, we can see that (c) is also not correct.

**Q42.**

**Answer :**

Since f is invertible, range of f = co domain of f = X

So, we need to find the range of f to find X.

For finding the range, let

fx=y⇒4x-x2=y⇒x2-4x=-y⇒x2-4x+4=4-y⇒x-22=4-y⇒x-2=±4-y⇒x=2±4-yThis is defined only when 4-y≥0⇒y≤4X=Range of f=(-∞,4]

So, the answer is (c).

**Q43.**

**Answer :**

(b) -Sgn x x1-x

We have, fx=-x|x|1+x2 x∈-1, 1Case-IWhen, x<0,Then, x=-xAnd fx>0Now,fx=-x-x1+x2⇒y=x21+x2⇒y1=x21+x2⇒y+1y-1=x2+1+x2x2-1-x2 Using Componendo and dividendo⇒y+1y-1=2×2+1-1⇒-y+1y-1=2×2+1⇒2y1-y=2×2⇒y1-y=x2⇒x=-y1-y As x<0⇒x=-y1-y As y>0To find the inverse interchanging x and y we get,f-1x=-x1-x …iCase-IIWhen, x>0,Then, x=xAnd fx<0Now,fx=-xx1+x2⇒y=-x21+x2⇒y1=-x21+x2⇒y+1y-1=-x2+1+x2-x2-1-x2 Using Componendo and dividendo⇒y+1y-1=1-2×2-1⇒1+y1-y=12×2+1⇒1-y1+y=2×2+1⇒-2y1+y=2×2⇒x2=-y1+y⇒x=-y1+y As x>0⇒x=y1-y As y<0To find the inverse interchanging x and y we get,f-1x=x1-x …iiCase-IIIWhen, x=0,Then, fx=0Hence, f-1x=0 …iiiCombinig equation i , ii and iii we get,f-1x=-Sgnxx1-x

**Q44.**

**Answer :**

(c) hofog=hogof

We have,gx=x2 =0 As 12≤x≤ 12∴ 14≤x2≤ 12fogx=fgx=sin-10 =0hofogx=hfgx=2×0=0

Andfx=sin-1xNow,for, x∈12, 12fx∈π6, π4fx∈0.52, 0.78gofx=0 As, fx∈0.52, 0.78 =0hogofx=hgfx=2×0=0

∴ hofog=hogof=0

**Q45.**

**Answer :**

We will solve this problem by the trial-and-error method.

Let us check option (a) first.

If fx = 2x-312gofx = gfx= 12g2x-3= 122x-32+2x-3-2= 124×2+9-12x+2x-3-2= 124×2-10x+4= 2×2-5x+2

The given condition is satisfied by (a).

So, the answer is (a).

**Q46.**

**Answer :**

(b)

If we take gx = x, thengfx = gsin2x = sin2x = ±sin x = sin x

So, the answer is (b).

**Q47.**

**Answer :**

(c)

Let f-1x = yfy = x⇒y3+3 = x⇒y3 = x-3⇒y = x-33 ⇒y = x-313

So, the answer is (c).

**Q48.**

**Answer :**

(c) {0, 1, 8, 27}

fx=x3Domain = 0, 1, 2, 3Range = 03, 13, 23, 33 = 0, 1, 8, 27So, f = 0, 0, 1, 1, 2, 8, 3, 27f-1 = 0, 0, 1, 1, 8, 2, 27, 3Domain of f-1 = 0, 1, 8, 27

**Q49.**

**Answer :**

(d)

Let f-1x=yfy=xy2-3=xy2=x+3y=±x+3

So, the answer is (d).

**BINARY OPERATIONS**

**Page 3.5 Ex. 3.1**

**Q2.**

**Answer :**

(i) If a = 1 and b = 2 in Z+, then

a * b= a-b =1-2 =-1∉Z+ ∵ Z+ is the set of non-negative integers⇒For a=1 and b=2, a * b∉Z+Thus, * is not a binary operation on Z+.

ii a, b∈Z+⇒ab∈Z+⇒a * b∈Z+Therefore,a * b∈Z+, ∀ a, b∈Z+Thus, * is a binary operation on Z+.

iii a, b∈R⇒a, b2∈R⇒ab2∈R⇒a * b∈RThus, * is a binary operation on R.

iv a, b∈Z+⇒a-b∈Z+ ∵ a-b is a positive integer⇒a * b∈Z+Therefore,a * b∈Z+, ∀ a, b∈Z+Thus, * is a binary operation on Z+.

v a, b∈Z+⇒a∈Z+ ⇒a * b∈Z+Therefore,a * b∈Z+, ∀ a, b∈Z+Thus, * is a binary operation on Z+.

vi a, b∈R⇒a, 4b2∈R⇒a+4b2∈R⇒a * b∈RTherefore,a * b∈R,∀ a, b∈RThus, * is a binary operation on R.

**Q3.**

**Answer :**

Given: a * b = 2a + b − 3

3 * 4 = 2 (3) + 4 – 3

= 6 + 4 – 3

= 7

**Q4.**

**Answer :**

LCM | 1 | 2 | 3 | 4 | 5 |

1 | 1 | 2 | 3 | 4 | 5 |

2 | 2 | 2 | 6 | 4 | 10 |

3 | 3 | 5 | 3 | 12 | 15 |

4 | 4 | 4 | 12 | 4 | 20 |

5 | 5 | 10 | 15 | 20 | 5 |

In the given composition table, all the elements are not in the set {1, 2, 3, 4, 5}.

If we consider *a *= 2 and *b* = 3,* a * b = *LCM of *a* and *b* = 6

∉

{1, 2, 3, 4, 5}.

Thus, * is not a binary operation on {1, 2, 3, 4, 5}.

**Q5.**

**Answer :**

Number of binary operations on a set with n elements is nn2.

Here, S = {a, b, c}

Number of elements in S = 3

Number of binary operations on a set with 3 elements is 332=39

**Q6.**

**Answer :**

Number of binary operations on a set with n elements is nn2.

Here, S = {a, b}

Number of elements in S = 2

Number of binary operations on a set with 2 elements =222 =24 =16

**Q7.**

**Answer :**

S=a=mn : m∈Z, n∈1, 2, 3Let a=13, b=53∈Sa * b =ab =13×53 =59∉S ∵ 9 ∉1, 2, 3 Therefore, ∃ a, b ∈ S, such that a * b ∉ S

Thus, * is not a binary operation.

**Q8.**

**Answer :**

Let A=a100b1, B=a200b2∈MA * B =AB =a100b1a200b2 =a1a200b1b2∈M, ∵ a1a2 and b1b2∈R-0Therefore,A * B∈M,∀ A, B∈M

Thus, * is a binary operation on M.

**Page 3.13 Ex. 3.2**

**Q1.**

**Answer :**

a * b = 1.c.m. (a, b)

(i) 2 * 4 = 1.c.m. (2, 4)

= 4

3 * 5 = 1.c.m. (3, 5)

= 15

1 * 6 = 1.c.m. (1, 6)

= 6

(ii) Commutativity:

Let a, b ∈Na * b=1.c.m. a, b =1.c.m. b,a =b * aTherefore,a * b=b * a,∀ a, b∈N

Thus, * is commutative on N.

Associativity:

Let a, b, c∈Na * b * c=a * 1.c.m. b, c = 1.c.m. a, b, c = 1.c.m. a, b, ca * b * c=1.c.m. a, b * c = 1.c.m. a, b, c = 1.c.m. a, b, cTherefore,a * b * c=a * b * c,∀ a, b, c∈N

Thus, * is associative on N.

**Q2.**

**Answer :**

(i) Commutativity:

Let a, b∈N. Then,a * b =1 b * a =1Therefore,a * b=b * a,∀ a, b∈N

Thus, * is commutative on N.

Associativity:

Let a, b, c∈N. Then,a * b * c=a * 1 =1a * b * c=1 * c =1Therefore,a * b * c=a * b * c,∀ a, b, c∈N

Thus, * is associative on N.

(ii) Commutativity:

Let a, b∈N. Then,a * b =a+b2 =b+a2 =b * aTherefore,a * b=b * a,∀ a, b∈N

Thus, * is commutative on N.

Associativity:

Let a, b, c∈N. Then,a * b * c=a * b+c2 =a+b+c22 =2a+b+c4a * b * c=a+b2 * c =a+b2+c2 =a+b+2c4Thus, a * b * c≠a * b * cIf a=1, b=2, c=31 * 2 * 3=1 * 2+32 =1 * 52 =1+522 =741 * 2 * 3=1+22 * 3 =32 * 3 =32+32 =94Therefore, ∃ a=1, b=2, c=3∈N such that a * b * c≠a * b * c

**Q3.**

**Answer :**

Commutativity:

Let a, b∈A. Then,a * b=b b * a=aTherefore, a * b≠b * a

Thus, * is not commutative on A.

Associativity:

Let a, b, c∈A. Then,a * b * c=a * c =ca * b * c=b * c =cTherefore,a * b * c=a * b * c,∀ a, b, c∈A

Thus, * is associative on A.

**Page 3.14 Ex. 3.2**

**Q4.**

**Answer :**

(i) Commutativity:

Let a, b∈Z. Then,a * b=a+b+ab =b+a+ba = b * a Therefore,a * b=b * a,∀ a, b∈Z

Thus, * is commutative on Z.

Associativity:

Let a, b, c∈Z. Then, a * b * c=a * b+c+bc =a+b+c+bc+ab+c+bc =a+b+c+bc+ab+ac+abca * b * c=a+b+ab * c =a+b+ab+c+a+b+abc =a+b+ab+c+ac+bc+abcTherefore,a * b * c=a * b * c,∀ a, b, c∈Z

Thus, * is associative on Z.

(ii) Commutativity:

Let a, b∈N. Then, a * b=2ab =2ba = b * aTherefore,a * b = b * a,∀ a, b∈N

Thus, * is commutative on N.

Associativity:

Let a, b, c∈N. Then, a * b * c=a * 2bc =2a*2bca * b * c=2ab * c =2ab*2cTherefore,a * b * c≠a * b * c

Thus, * is not associative on N.

(iii) Commutativity:

Let a, b∈Q. Then, a * b=a-bb * a=b-aTherefore, a * b≠b * a

Thus, * is not commutative on Q.

Associativity:

Let a, b, c∈Q. Then, a * b * c=a * b-c =a-b-c =a-b+ca * b * c=a-b * c =a-b-cTherefore, a * b * c≠a * b * c

Thus, * is not associative on Q.

(iv) Commutativity:

Let a, b∈Q. Then, a⊙b=a2+b2 =b2+a2 =b⊙a Therefore, a⊙b=b⊙a,∀ a, b∈Q

Thus, ⊙ is commutative on Q.

Associativity:

Let a, b, c∈Q. Then, a⊙b⊙c=a⊙b2+c2 =a2+ b2+c22 =a2+b4+c4+2b2c2a⊙b⊙c=a2+b2⊙c =a2+b22+c2 =a4+b4+2a2b2+c2Therefore,a⊙b⊙c≠a⊙b⊙c

Thus, ⊙ is not associative on Q.

(v) Commutativity:

Let a, b∈Q. Then, a o b=ab2 =ba2 =b o a Therefore,a o b=b o a,∀ a, b∈Q

Thus, o is commutative on Q.

Associativity:

Let a, b, c∈Q. Then, a o b o c =a o bc2 =a bc22 =abc4a o b o c=ab2 o c =ab2c2 =abc4Therefore,a o b o c =a o b o c,∀ a, b, c∈Q

Thus, o is associative on Q.

(vi) Commutativity:

Let a, b∈Q. Then, a * b=ab2b * a=ba2Therefore,a * b≠b * a

Thus, * is not commutative on Q.

Associativity:

Let a, b, c∈Q. Then, a * b * c=a * bc2 =abc22 =ab2c4a * b * c=ab2 * c =ab2c2Therefore,a * b * c≠a * b * c

Thus, * is not associative on Q.

(vii) Commutativity:

Let a, b∈Q. Then, a * b=a+abb * a=b+ba =b+abTherefore,a * b≠b * a

Thus, * is not commutative on Q.

Associativity:

Let a, b, c∈Q. Then, a * b * c=a * b+bc =a+ab+bc =a+ab+abca * b * c=a+ab * c =a+ab +a+ab c =a+ab+ac+abcTherefore,a * b * c≠a * b * c

Thus, * is not associative on Q.

(viii) Commutativity:

Let a, b∈R. Then, a * b=a+b-7 =b+a-7 =b * a Therefore,a * b=b * a,∀ a, b∈R

Thus, * is commutative on R.

Associativity:

Let a, b, c∈R. Then, a * b * c=a * b+c-7 =a+b+c-7-7 =a+b+c-14a * b * c=a+b-7 * c =a+b-7 +c-7 =a+b+c-14Therefore,a * b * c=a * b * c,∀ a, b, c∈R

Thus, * is associative on R.

(ix) Commutativity:

Let a, b∈R–1. Then,a * b=ab+1b * a=ba+1Therefore,a * b≠b * a

Thus, * is not commutative on R – {-1}.

Associativity:

Let a, b, c∈R–1. Then, a * b * c=a * bc+1 =abc+1+1 =ac+1b+c+1a * b * c=ab+1 * c =ab+1c+1 =ab+1c+1Therefore,a * b * c≠a * b * c

Thus, * is not associative on R – {-1}.

(x) Commutativity:

Let a, b∈Q. Then, a * b=ab+1 =ba+1 =b * a Therefore,a * b, b * a,∀ a, b∈Q

Thus, * is commutative on Q.

Associativity:

Let a, b, c∈Q. Then, a * b * c=a * bc+1 =abc+1+1 =abc+a+1a * b * c=ab+1 * c =ab+1c+1 =abc+c+1Therefore,a * b * c≠a * b * c

Thus, * is not associative on Q.

(xi) Commutativity:

Let a, b∈N. Then, a * b=abb * a=baTherefore,a * b≠b * a

Thus, * is not commutative on N.

Associativity:

Let a, b, c∈N. Then,a * b * c=a * bc =abca * b * c=ab * c =abc =abcTherefore,a * b * c≠a * b * c

Thus, * is not associative on N.

(xii) Commutativity:

Let a, b∈Z. Then, a * b=a-bb * a=b-aTherefore,a * b≠b * a

Thus, * not is commutative on Z.

Associativity:

Let a, b, c∈Z. Then,a * b * c=a * b-c =a-b-c =a-b+ca * b * c=a-b-c =a-b-cTherefore,a * b * c≠a * b * c

Thus, * is not associative on Z.

(xiii) Commutativity:

Let a, b∈Q. Then, a * b=ab4 =ba4 =b * a Therefore,a * b=b * a,∀ a, b∈Q

Thus, * is commutative on Q.

Associativity:

Let a, b, c∈Q. Then,a * b * c=a * bc4 =abc44 =abc16a * b * c=ab4 * c =ab4c4 =abc16Therefore,a * b * c=a * b * c,∀ a, b, c∈Q

Thus, * is associative on Q.

(xiv) Commutativity:

Let a, b∈Q. Then, a * b=a-b2 =b-a2 =b * a Therefore,a * b=b * a,∀ a, b∈Q

Thus, * is commutative on Q.

Associativity:

Let a, b, c∈Q. Then, a * b * c=a * b-c2 =a * b2+c2-2bc =a-b2-c2+2bc2a * b * c=a-b2 * c =a2+b2-2ab * c =a2+b2-2ab-c2Therefore, a * b * c≠a * b * c

Thus, * is not associative on Q.

**Q5.**

**Answer :**

Let a, b∈Q–1. Then, a o b= a+b-ab =b+a-ba =b o aTherefore,a o b=b o a,∀ a, b∈Q–1

Thus, o is commutative on Q – {1}.

**Q6.**

**Answer :**

Let a, b∈Z. Then, a * b=3a+7bb * a=3b+7aThus, a * b≠b * aLet a=1 and b=2 1 * 2=3×1+7×2 =3+14 =172 * 1=3×2+7×1 =6+7 =13Therefore, ∃ a=1; b=2 ∈Z such that a * b≠b * a

Thus, * is not commutative on Z.

**Q7.**

**Answer :**

Let a, b, c∈Za * b * c=a * bc+1 =abc+1+1 =abc+a+1a * b * c=ab+1 * c =ab+1c+1 =abc+c+1Thus, a * b * c≠a * b * c

Thus, * is not associative on Z.

**Q8.**

**Answer :**

Checking for binary operation:

Let a, b∈S. Then,a, b∈R and a≠-1, b≠-1a * b=a+b+abWe need to prove that a+b+ab∈S. For this we have to prove that a+b+ab∈R and a+b+ab≠-1Since a, b∈R, a+b+ab∈R, let us assume that a+b+ab=-1.a+b+ab+1=0a+ab+b+1=0a1+b+11+b=0a+1b+1=0a=-1, b=-1 which is falseHence, a+b+ab≠-1Therefore,a+b+ab∈S

Thus, * is a binary operation on S.

Commutativity:

Let a, b∈S. Then,a * b=a+b+ab =b+a+ba = b * a Therefore, a * b=b * a,∀ a, b∈S

Thus, * is commutative on N.

Associativity:

Let a, b, c∈Sa * b * c =a * b+c+bc =a+b+c+bc+ab+c+bc =a+b+c+bc+ab+ac+abca * b * c =a+b+ab * c =a+b+ab+c+a+b+abc =a+b+ab+c+ac+bc+abcTherefore,a * b * c=a * b * c,∀ a, b, c∈S

Thus, * is associative on S.

Now,

Given: 2 * x* 3=7⇒2+x+2x * 3=7⇒2+3x * 3=7⇒2+3x+3+2+3×3=7⇒5+3x+6+9x=7⇒12x+11=7⇒12x=-4⇒x=-412⇒x=-13

**Q9.**

**Answer :**

Let a, b, c∈Q. Then, a * b * c=a * b-c2 =a-b-c22 =2a-b+c4a * b * c=a-b2 * c =a-b2-c2 =a-b-2c4Thus, a * b * c≠a * b * cIf a=1, b=2, c=3 1 * 2 * 3=1 * 2-32 =1 * -12 =1+122 =341 * 2 * 3=1-22 * 3 =-12* 3 =-12-32 =-74Therefore, ∃ a=1, b=2, c=3 ∈R such that a * b * c≠a * b * c

Thus, * is not associative on Q.

**Q10.**

**Answer :**

Commutativity:

Let a, b∈Z. Then, a * b =a+3b-4b * a=b+3a-4a * b ≠b * aLet a=1, b=21 * 2=1+ 6-4 = 32 * 1=2+3-4 =1Therefore, ∃ a=1, b=2∈Z such that a * b ≠ b * a

Thus, * is not commutative on Z.

Associativity:

Let a, b, c∈Z. Then, a * b * c=a * b+3c-4 =a+3b+3c-4-4 =a+3b+9c-12-4 =a+3b+9c-16a * b * c=a+3b-4 * c =a+3b-4+3c-4 =a+3b+3c-8Thus, a * b * c≠a * b * cIf a=1, b=2, c=31 * 2 * 3=1 * 2+9-4 =1 * 7 =1+21-4 =181 * 2 * 3=1+6-4 * 3 =3 * 3 =3+9-4 =8Therefore, ∃ a=1, b=2, c=3∈Z such that a * b * c≠a * b * c

Thus, * is not associative on Z.

**Q11.**

**Answer :**

Let a, b, c∈Q. Then, a * b * c=a * bc5 =abc55 =abc25a * b * c=ab5 * c =ab5c5 =abc25Therefore,a * b * c=a * b * c,∀ a, b, c∈Q.

Thus, * is associative on Q.

**Q12.**

**Answer :**

Let a, b, c∈Q. Then, a * b * c=a * bc7 =abc77 =abc49a * b * c=ab7 * c =ab7c7 =abc49Therefore,a * b * c=a * b * c,∀ a, b, c∈Q

Thus, * is associative on Q.

**Q13.**

**Answer :**

Let a, b, c∈Q. Then, a * b * c=a * b+c2 =a+b+c22 =2a+b+c4a * b * c=a+b2 * c =a+b2+c2 =a+b+2c4Thus, a * b * c≠a * b * cIf a=1, b=2, c=3 1 * 2 * 3=1 * 2+32 =1 * 52 =1+522 =741 * 2 * 3=1+22 * 3 =32 * 3 =32+32 =94Therefore, ∃ a=1, b=2, c=3∈Q such that a * b * c≠a * b * c

Thus, * is not associative on Q.

**Page 3.17 Ex. 3.3**

**Q1.**

**Answer :**

Let e be the identity element in I+ with respect to * such that

a * e=a=e * a,∀ a∈I+a * e=a and e * a=a,∀ a∈I+a+e=a and e+a=a,∀ a∈I+e=0 ,∀ a∈I+

Thus, 0 is the identity element in I+ with respect to *.

**Q2.**

**Answer :**

Let e be the identity element in Q- {-1} with respect to * such that

a * e=a=e * a,∀ a∈Q-1a * e=a and e * a=a,∀ a∈Q-1a+e+ae=a and e+a+ea=a,∀ a∈Q-1e+ae=0 and e+ea=0,∀ a∈Q-1e1+a=0 and e1+a=0,∀ a∈Q-1e=0,∀ a∈Q–1 ∵ a≠-1

Thus, 0 is the identity element in Q – {-1} with respect to *.

**Q3.**

**Answer :**

Let e be the identity element in Z with respect to * such that

a * e=a=e * a,∀ a∈Za * e=a and e * a=a,∀ a∈Za+e-5=a and e+a-5=a,∀ a∈Ze=5,∀ a∈Z

Thus, 5 is the identity element in Z with respect to *.

**Q4.**

**Answer :**

Let e be the identity element in Z with respect to * such that

a * e=a=e * a,∀ a∈Za * e=a and e * a=a,∀ a∈Za+e+2=a and e+a+2=a,∀ a∈Ze=-2 ,∀ a∈Z

Thus, -2 is the identity element in Z with respect to *.

**Page 3.28 Ex. 3.4**

**Q1.**

**Answer :**

(i) Commutativity:

Let a, b∈Z. Then, a * b=a+b-4 =b+a-4 = b * aTherefore, a * b=b * a,∀ a, b∈Z

Thus, * is commutative on Z.

Associativity:

Let a, b, c∈Z. Then,a * b * c=a * b+c-4 =a+b+c-4-4 =a+b+c-8a * b * c=a+b-4 * c =a+b-4+c-4 =a+b+c-8Therefore,a * b * c=a * b * c,∀ a, b, c∈Z

Thus, * is associative on Z.

(ii) Let e be the identity element in Z with respect to * such that

a * e=a=e * a,∀ a∈Za * e=a and e * a=a,∀ a∈Za+e-4=a and e+a-4=a,∀ a∈Ze=4 ,∀ a∈Z

Thus, 4 is the identity element in Z with respect to *.

iii Let a∈Z and b∈Z be the inverse of a. Then, a * b=e=b * aa * b=e and b * a=ea+b-4=4 and b+a-4=4b=8-a ∈ZThus, 8-a is the inverse of a∈Z.

**Q2.**

**Answer :**

Commutativity:

Let a, b∈Q0a * b=ab5 =ba5 =b * a Therefore, a * b=b * a,∀ a, b∈Q0

Thus, * is commutative on Qo.

Associativity:

Let a, b, c∈Q0a * b * c=a * bc5 =abc55 =abc25a * b * c=ab5 * c =ab5c5 =abc25Therefore, a * b * c=a * b * c,∀ a, b, c∈Q0

Thus, * is associative on Qo.

Finding identity element:

Let e be the identity element in Z with respect to * such that

a * e=a=e * a,∀ a∈Q0a * e=a and e * a=a,∀ a∈Q0⇒ae5=a and ea5=a,∀ a∈Q0⇒e=5 ,∀ a∈Q0 ∵ a≠0

Thus, 5 is the identity element in Qo with respect to *.

**Q3.**

**Answer :**

(i) Commutativity:

Let a, b∈Q–1. Then, a * b=a+b+ab =b+a+ba = b * aTherefore, a * b=b * a,∀ a, b∈Q–1

Thus, * is commutative on Q -{-1}.

Associativity:

Let a, b, c∈Q–1. Then,a * b * c=a * b+c+bc =a+b+c+bc+a b+c+bc =a+b+c+bc+ab+ac+abca * b * c=a+b+ab * c =a+b+ab+c+a+b+abc =a+b+c+ab+ac+bc+abcTherefore, a * b * c=a * b * c,∀ a, b, c∈Q–1.

Thus, * is associative on Q – {-1}.

(ii) Let e be the identity element in Q- {-1} with respect to * such that

a * e=a=e * a,∀ a∈Q–1⇒a * e=a and e * a=a,∀ a∈Q–1⇒a+e+ae=a and e+a+ea=a, ∀a∈Q–1⇒e1+a=0,∀ a∈Q–1⇒e=0,∀ a∈Q–1 ∵ a≠-1

Thus, 0 is the identity element in Q – {-1} with respect to *.

iii Let a∈Q–1 and b∈Q–1 be the inverse of a. Then, a * b=e=b * a⇒a * b=e and b * a=e⇒a+b+ab=0 and b+a+ba=0⇒b1+a=-a ∈Q–1⇒b=-a1+a ∈Q–1 ∵a≠-1Thus, -a1+a is the inverse of a∈Q–1.

**Q4.**

**Answer :**

(i) Commutativity:

Let X=a, b and Y=c, d∈A,∀ a, c∈R0 & b, d∈R. Then, X⊙Y=ac, bc+d& Y⊙X=ca, da+bTherefore, X⊙Y=Y⊙X,∀ X,Y∈A

Thus, ⊙ is commutative on A.

Associativity:

Let X=(a, b), Y=(c, d) and Z=( e, f),∀ a, c, e∈R0 & b, d, f∈RX⊙Y⊙Z=(a, b)⊙ce, de+f =ace, bce+de+fX⊙Y⊙Z=ac, bc+d⊙e,f = ace, bc+de+f =ace, bce+de+f∴ X⊙Y⊙Z=X⊙Y⊙Z,∀ X, Y, Z∈AThus,⊙ is associative on A.

(ii) Let E=(x, y) be the identity element in A with respect to ⊙,∀ x∈ R0& y∈R such that X⊙E=X=E⊙X,∀ X∈A⇒X⊙E=X and E⊙X=X⇒ax, bx+y=a, b and xa, ya+b=a, bConsidering ax, bx+y=a, b⇒ax=a ⇒x=1 & bx+y=b ⇒y=0 ∵ x=1Considering xa, ya+b=a, b⇒xa=a⇒x=1& ya+b=b⇒y=0 ∵ x=1∴ 1, 0 is the identity element in A with respect to ⊙.

iii Let F=(m, n) be the inverse in A∀ m∈R0 & n∈RX⊙F=E and F⊙X=E⇒am, bm+n=1, 0 and ma, na+b=1, 0Considering am, bm+n=1, 0⇒am=1⇒m=1a& bm+n=0⇒n=-ba ∵ m=1aConsidering ma, na+b=1, 0⇒ma=1⇒m=1a& na+b=0⇒n=-ba∴ The inverse of a, b∈A with respect to ⊙ is 1a,-ba .

**Q5.**

**Answer :**

(i) Commutativity:

Let a, b∈Q0. Then, a o b=ab2 =ba2 = b o aTherefore,a o b=b o a,∀ a, b∈Q0

Thus, o is commutative on Qo.

Associativity:

Let a, b, c∈Q0. Then,a o b o c=a o bc2 =abc22 =abc4a o b o c=ab2 o c =ab2c2 =abc4Therefore,a o b o c=a o b o c,∀ a, b, c∈Q0

Thus, o is associative on Qo.

(ii) Let e be the identity element in Qo with respect to * such that

a o e=a=e o a,∀ a∈Q0a o e=a and e o a=a,∀ a∈Q0⇒ae2=a and ea2=a,∀ a∈Q0e=2 ∈Q0 ,∀ a∈Q0

Thus, 2 is the identity element in Qo with respect to o.

iii Let a∈Q0 and b∈Q0 be the inverse of a. Then, a o b=e=b o a⇒a o b=e and b o a=e⇒ab2=2 and ba2=2⇒b=4a ∈Q0Thus, 4a is the inverse of a∈Q0.

**Q6.**

**Answer :**

Commutativity:

Let a, b∈R-1. Then, a * b=a+b-ab =b+a-ba = b * aTherefore,a * b = b * a,∀ a, b∈R-1

Thus, * is commutative on R – {1}.

Associativity:

Let a, b, c∈R-1. Then, a * b * c=a * b+c-bc =a+b+c-bc-ab+c-bc =a+b+c-bc-ab-ac+abca * b * c=a+b-ab * c =a+b-ab+c-a+b-abc =a+b+c-ab-ac-bc+abcTherefore, a * b * c=a * b * c,∀ a, b, c∈R-1

Thus, * is associative on R – {1}.

Finding identity element:

Let e be the identity element in R – {1} with respect to * such that

a * e=a=e * a,∀ a∈R-1a * e=a and e * a=a,∀ a∈R-1⇒a+e-ae=a and e+a-ea=a,∀ a∈R-1e1-a=0,∀ a∈R-1e=0∈∀ a∈R-1,∀ a∈R-1 ∵ a≠1

Thus, 0 is the identity element in R -{1} with respect to *.

Finding inverse:

Let a∈R-1 and b∈R-1 be the inverse of a. Then, a * b=e=b * aa * b=e and b * a=e⇒a+b-ab=0 and b+a-ba=0⇒a=ab-b⇒a=ba-1 ⇒b=aa-1Thus, aa-1 is the inverse of a∈R-1.

**Q7.**

**Answer :**

(i) Commutativity: Let a, b & c, d∈A∀ a, b, c, d∈R0. Then,a, b* c, d=ac, bd =ca, db =c, d*a, b∴ a, b* c, d=c, d*a, bThus, * is commutaive on A.Associativity: Let a, b, c, d & e, f∈A∀ a, b, c, d, e, f∈ R0,. Then, a, b*c, d* e, f=a, b*ce, df =ace, bdf a, b*c, d* e, f=ac, bd*e, f =ace, bdf∴ a, b*c, d* e, f= a, b*c, d* e, fThus, * is associative on A.

(ii) Let x, y be the identity element in A∀ x, y∈A. Then,a, b*x, y=a, b=x, y*a, b ⇒a, b*x, y=a, b and x, y*a, b =a, b⇒ax, by=a, b and xa, yb=a, b⇒x=1 and y=1 Thus, 1, 1 is the identity element of A.

(iii) Let m, n be the inverse of a, b∀ a, b∈A. Then,a, b*m, n=1,1⇒ am, bn=1,1⇒am=1 & bn=1⇒m=1a& n=1bThus, 1a, 1b is the inverse of a, b∀ a, b∈A.

### Page 3.29 Ex. 3.4

**Q8.**

**Answer :**

Let e be the identity element. Then,

a * e=a=e * a,∀ a∈NHCF a, e=a=HCF e, a,∀ a∈N⇒HCF a, e=a,∀ a∈N

We cannot find e that satisfies this condition.

So, the identity element with respect to * does not exist in N.

### Page 3.37 Ex. 3.5

**Q1**

**Answer :**

Here,

1 ×4 1 = Remainder obtained by dividing 1 ×1 by 4

= 1

0 ×4 1 = Remainder obtained by dividing 0 × 1 by 4

= 0

2 ×4 3 = Remainder obtained by dividing 2 × 3 by 4

= 2

3 ×4 3 = Remainder obtained by dividing 3 × 3 by 4

= 1

So, the composition table is as follows:

×4 | 0 | 1 | 2 | 3 |

0 | 0 | 0 | 0 | 0 |

1 | 0 | 1 | 2 | 3 |

2 | 0 | 2 | 0 | 2 |

3 | 0 | 3 | 2 | 1 |

**Q2.**

**Answer :**

Here,

1 +5 1 = Remainder obtained by dividing 1 + 1 by 5

= 2

3 +5 4 = Remainder obtained by dividing 3 + 4 by 5

= 2

4 +5 4 = Remainder obtained by dividing 4 + 4 by 5

= 3

So, the composition table is as follows:

+5 | 0 | 1 | 2 | 3 | 4 |

0 | 0 | 1 | 2 | 3 | 4 |

1 | 1 | 2 | 3 | 4 | 5 |

2 | 2 | 3 | 4 | 0 | 1 |

3 | 3 | 4 | 0 | 1 | 2 |

4 | 4 | 0 | 1 | 2 | 3 |

**Q3.**

**Answer :**

Here,

1 ×6 1 = Remainder obtained by dividing 1 × 1 by 6

= 1

3 ×6 4 = Remainder obtained by dividing 3 × 4 by 6

= 0

4 ×6 5 = Remainder obtained by dividing 4× 5 by 6

= 2

So, the composition table is as follows:

×6 | 0 | 1 | 2 | 3 | 4 | 5 |

0 | 0 | 0 | 0 | 0 | 0 | 0 |

1 | 0 | 1 | 2 | 3 | 4 | 5 |

2 | 0 | 2 | 4 | 0 | 2 | 4 |

3 | 0 | 3 | 0 | 3 | 0 | 3 |

4 | 0 | 4 | 2 | 0 | 4 | 2 |

5 | 0 | 5 | 4 | 3 | 2 | 1 |

**Q4.**

**Answer :**

Here,

1 ×5 1 = Remainder obtained by dividing 1 × 1 by 5

= 1

3 ×5 4 = Remainder obtained by dividing 3 × 4 by 5

= 2

4 ×5 4 = Remainder obtained by dividing 4 × 4 by 5

= 1

So, the composition table is as follows:

×5 | 0 | 1 | 2 | 3 | 4 |

0 | 0 | 0 | 0 | 0 | 0 |

1 | 0 | 1 | 2 | 3 | 4 |

2 | 0 | 2 | 4 | 1 | 3 |

3 | 0 | 3 | 1 | 4 | 2 |

4 | 0 | 4 | 3 | 2 | 1 |

**Q5.**

**Answer :**

Here,

1 ×10 1 = Remainder obtained by dividing 1 × 1 by 10

=1

3 ×10 7 = Remainder obtained by dividing 3 × 7 by 10

=1

7 ×10 9 = Remainder obtained by dividing 7 × 9 by 10

= 3

So, the composition table is as follows:

×10 | 1 | 3 | 7 | 9 |

1 | 1 | 3 | 7 | 9 |

3 | 3 | 9 | 1 | 7 |

7 | 7 | 1 | 9 | 3 |

9 | 9 | 7 | 3 | 1 |

We observe that the elements of the first row are same as the top-most row.

So, 1∈S is the identity element with respect to ×10.

Finding inverse of 3:

From the above table we observe,

3 ×10 7 = 1

So, the inverse of 3 is 7.

**Q6.**

**Answer :**

Finding identity element:

Here,

1 ×7 1 = Remainder obtained by dividing 1 × 1 by 7

= 1

3 ×7 4 = Remainder obtained by dividing 3 × 4 by 7

= 5

4 ×7 5 = Remainder obtained by dividing 4× 5 by 7

= 6

So, the composition table is as follows:

×7 | 1 | 2 | 3 | 4 | 5 | 6 |

1 | 1 | 2 | 3 | 4 | 5 | 6 |

2 | 2 | 4 | 6 | 1 | 3 | 5 |

3 | 3 | 6 | 2 | 5 | 1 | 4 |

4 | 4 | 1 | 5 | 2 | 6 | 3 |

5 | 5 | 3 | 1 | 6 | 4 | 2 |

6 | 6 | 5 | 4 | 3 | 2 | 1 |

We observe that all the elements of the first row of the composition table are same as the top-most row.

So, the identity element is 1.

Also, 3 ×7 5=1

So, 3-1 = 5

Now,3-1×7 4=5 ×7 4=6

**Q7.**

**Answer :**

Z11=0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10Multiplication modulo 11 is defined as follows:For a, b∈Z11, a ×11 b is the remainder when a×b is divided by 11.

Here,

1 ×11 1 = Remainder obtained by dividing 1 × 1 by 11

= 1

3 ×11 4 = Remainder obtained by dividing 3 × 4 by 11

= 1

4 ×11 5 = Remainder obtained by dividing 4 × 5 by 11

= 9

So, the composition table is as follows:

×11 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 |

1 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 |

2 | 2 | 4 | 6 | 8 | 10 | 1 | 3 | 5 | 7 | 9 |

3 | 3 | 6 | 9 | 1 | 4 | 7 | 10 | 2 | 5 | 8 |

4 | 4 | 8 | 1 | 5 | 9 | 2 | 6 | 10 | 3 | 7 |

5 | 5 | 10 | 4 | 9 | 3 | 8 | 2 | 7 | 1 | 6 |

6 | 6 | 1 | 7 | 2 | 8 | 3 | 9 | 4 | 10 | 5 |

7 | 7 | 3 | 10 | 6 | 2 | 9 | 5 | 1 | 8 | 4 |

8 | 8 | 5 | 2 | 10 | 7 | 4 | 1 | 9 | 6 | 3 |

9 | 9 | 7 | 5 | 3 | 1 | 10 | 8 | 6 | 4 | 2 |

10 | 10 | 9 | 8 | 7 | 6 | 5 | 4 | 3 | 2 | 1 |

We observe that the first row of the composition table is same as the top-most row.

So, the identity element is 1.

Also,

5 ×11 9=1Hence, 5-1=9

**Q8.**

**Answer :**

Here,

1 ×5×5 1 = Remainder obtained by dividing 1× 1 by 5

= 1

3 ×5 4 = Remainder obtained by dividing 3 ×4 by 5

= 2

4 ×54 = Remainder obtained by dividing 4 × 4 by 5

= 1

So, the composition table is as follows:

×5 | 0 | 1 | 2 | 3 | 4 |

0 | 0 | 0 | 0 | 0 | 0 |

1 | 0 | 1 | 2 | 3 | 4 |

2 | 0 | 2 | 4 | 1 | 3 |

3 | 0 | 3 | 1 | 4 | 2 |

4 | 0 | 4 | 3 | 2 | 1 |

**Q9.**

**Answer :**

(i) Commutativity:

The table is symmetrical about the leading element. It means * is commutative on S.

Associativity:

a * b * c=a * d =da * b * c=b * c =dTherefore,a * b * c=a * b * c∀ a, b, c∈S

So, * is associative on S.

Finding identity element:

We observe that the first row of the composition table coincides with the top-most row and the first column coincides with the left-most column.

These two intersect at a.

⇒x * a=a * x=x, ∀x∈S

So, a is the identity element.

Finding inverse elements:

a * a =a⇒a-1=ab * b=a⇒b-1=bc * c=a⇒c-1=cd * d=a⇒d-1=d

(ii) Commutativity:

The table is symmetrical about the leading element. It means that o is commutative on S.

Associativity:

a o b o c=a o c =aa o b o c=a o c =aThus,a o b o c=a o b o c∀ a, b, c∈S

So, o is associative on S.

Finding identity element:

We observe that the second row of the composition table coincides with the top-most row and the first column coincides with the left-most column.

These two intersect at b.

⇒x o b=b o x =x,∀ x∈S

So, b is the identity element.

Finding inverse elements:

In the first row, we don’t have b, i.e. there does not exist an element x such that a o x=x o a=b.So, a-1 does not exist.b o b=b⇒b-1=bc o d=b⇒c-1=dd o c=b⇒d-1=c

### Page 3.38 Ex. 3.5

**Q10.**

**Answer :**

Here,

1 * 1 =1+1 (∵ 1+1 <6 )

= 2

3 * 4 = 3 + 4 -6 (∵ 3 + 4 >6 )

= 7 – 6

= 1

4 * 5 = 4 + 5-6 (∵ 4 + 5>6 )

= 9- 6

= 3 etc.

So, the composition table is as follows:

* | 0 | 1 | 2 | 3 | 4 | 5 |

0 | 0 | 1 | 2 | 3 | 4 | 5 |

1 | 1 | 2 | 3 | 4 | 5 | 0 |

2 | 2 | 3 | 4 | 5 | 0 | 1 |

3 | 3 | 4 | 5 | 0 | 1 | 2 |

4 | 4 | 5 | 0 | 1 | 2 | 3 |

5 | 5 | 0 | 1 | 2 | 3 | 4 |

We observe that the first row of the composition table coincides with the top-most row and the first column coincides with the left-most column.

These two intersect at 0.

So, 0 is the identity element .

⇒a * 0=0 * a=a, ∀ a∈0, 1, 2, 3, 4, 5

Finding inverse:

Let a∈0, 1, 2, 3, 4, 5 and b∈0, 1, 2, 3, 4, 5 such thata * b=b * a=ea * b=e and b * a =eCase 1: Let us assume that a+b<6Then,a * b=e and b * a =ea+b=0 and b+a=0a=-b, which is not possible because all the elements of the given set are non-negative.Case 2: Let us assume that a+b≥6Then,a * b=e and b * a =ea+b-6=0 and b+a-6=0b=6-a (from the table we can observe that this is true for all a≠0)Thus, 6-a is the inverse of a.

### Page 3.39 (Very Short Answers)

**Q1.**

**Answer :**

Let e be the identity element in R0 with respect to * such that

a * e=a=e * a, ∀ a∈R0a * e=a and e * a=a, ∀ a∈R0Then , ae2=a and ea2=a, ∀ a∈R0ae=2a, ∀ a∈R0ae-2=0, ∀ a∈R0e-2=0, ∀a∈R0 (∵ a≠0)e=2∈R0

Thus, 2 is the identity element in R0 with respect to *.

**Q2.**

**Answer :**

To find the identity element, let e be the identity element in Z with respect to * such that

a * e=a=e * a,∀a∈Za * e=a and e * a=a,∀a∈ZThen,a+e+2=a and e+a+2=a,∀a∈Ze=-2∈Z,∀a∈Z

Thus,-2 is the identity element in Z with respect to *.

Now,

Let b∈Z be the inverse of 4.Here,4 * b=e=b * 44 * b=e and b * 4=eThen,4+b+2=-2 and b+4+2=-2b=-8 ∈ZThus, -8 is the inverse of 4.

**Q3.**

**Answer :**

Let A be a non-empty set. An operation * is called a binary operation on A, if and only if

a * b∈A,∀ a, b∈A

**Q4.**

**Answer :**

An operation * on a set A is called a commutative binary operation if and only if it is a binary operation as well as commutative, i.e. it must satisfy the following two conditions.

i a * b∈A,∀a, b∈A (Binary operation)ii a * b=b * a,∀ a, b∈A (Commutaive)

**Q5.**

**Answer :**

An operation * on a set A is called an associative binary operation if and only if it is a binary operation as well as associative, i.e. it must satisfy the following two conditions:

i a * b∈A,∀ a, b∈A (Binary operation)ii a * b * c=a * b * c,∀ a, b, c∈A (Associative)

**Q6.**

**Answer :**

Number of binary operations on a set with n elements = nn2

Here,Number of binary operations on a set with 2 elements=222 = 24 =16

**Q7.**

**Answer :**

Let e be the identity element in R with respect to * such that

a * e=a=e * a,∀ a∈Ra * e=a and e * a=a,∀ a∈RThen, 3ae7=a and 3ea7=a,∀ a∈Re=73 , ∀ a∈R

Thus, 73 is the identity element in R with respect to *.

**Q8.**

**Answer :**

Given: 2 * x * 5=10 Here,2 * 5×5=10⇒2 * x=10⇒2×5=10⇒x=10×52⇒x=25

**Q9.**

**Answer :**

Here,

3 ×11 4= Remainder obtained by dividing 3 x 4 by 11

= 0

4 ×11 5 = Remainder obtained by dividing 4 x 5 by 11

= 9

So,

×11 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 |

1 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 |

2 | 2 | 4 | 6 | 8 | 10 | 1 | 3 | 5 | 7 | 9 |

3 | 3 | 6 | 9 | 1 | 4 | 7 | 10 | 2 | 5 | 8 |

4 | 4 | 8 | 1 | 5 | 9 | 2 | 6 | 10 | 3 | 7 |

5 | 5 | 10 | 4 | 9 | 3 | 8 | 2 | 7 | 1 | 6 |

6 | 6 | 1 | 7 | 2 | 8 | 3 | 9 | 4 | 10 | 5 |

7 | 7 | 3 | 10 | 6 | 2 | 9 | 5 | 1 | 8 | 4 |

8 | 8 | 5 | 2 | 10 | 7 | 4 | 1 | 9 | 6 | 3 |

9 | 9 | 7 | 5 | 3 | 1 | 10 | 8 | 6 | 4 | 2 |

10 | 10 | 9 | 8 | 7 | 6 | 5 | 4 | 3 | 2 | 1 |

We observe that the first row of the composition table coincides with the top-most row and the first column coincides with the left-most column.

These two intersect at 1.

So, the identity element is 1.

Also,

5 ×11 9=1Thus, 5-1=9

**Q10.**

**Answer :**

Let * be a binary operation on a set A.

An element e is called an identity element in A with respect to * if and only if

a * e=e * a=a,∀ a∈A

**Q11.**

**Answer :**

Here,

2 ×10 4 = Remainder obtained by dividing 2 × 4 by 10

= 8

4 ×10 6 = Remainder obtained by dividing 4 × 6 by 10

= 4

2 ×10 8 = Remainder obtained by dividing 2 × 8 by 10

= 6

3 ×10 4 = Remainder obtained by dividing 3 × 4 by 10

= 2

So, the composition table is as follows:

×10 | 2 | 4 | 6 | 8 |

2 | 4 | 8 | 2 | 6 |

4 | 8 | 6 | 4 | 2 |

6 | 2 | 4 | 6 | 8 |

8 | 6 | 2 | 8 | 4 |

**Q12.**

**Answer :**

Here,

1 ×10 1 = Remainder obtained by dividing 1 × 1 by 10

= 1

3 ×10 1 = Remainder obtained by dividing 3 × 1 by 10

= 3

7 ×10 3 = Remainder obtained by dividing 7 × 3 by 10

= 1

3 ×10 3 = Remainder obtained by dividing 3× 3 by 10

= 9

So, the composition table is as follows:

×10 | 1 | 3 | 7 | 9 |

1 | 1 | 3 | 7 | 9 |

3 | 3 | 9 | 1 | 7 |

7 | 7 | 1 | 9 | 3 |

9 | 9 | 7 | 3 | 1 |

We observe that the first row of the composition table coincides with the top-most row and the first column coincides with the left-most column.

These two intersect at 1.

⇒a * 1=1 * a=a, ∀a∈S

So, the identity element is 1.

Also,

3 ×10 7 = 1

3-1 = 7

**Q13.**

**Answer :**

Here,

1 ×51 = Remainder obtained by dividing 1 × 1 by 5

= 1

3 ×5 4 = Remainder obtained by dividing 3 × 4 by 5

= 2

4 ×5 4 = Remainder obtained by dividing 4 × 4 by 5

= 1

So, the composition table is as follows:

×5 × | 1 | 2 | 3 | 4 |

1 | 1 | 2 | 3 | 4 |

2 | 2 | 4 | 1 | 3 |

3 | 3 | 1 | 4 | 2 |

4 | 4 | 3 | 2 | 1 |

We observe that the first row of the composition table coincides with the top-most row and the first column coincides with the left-most column.

These two intersect at 1.

⇒a ×5 1=1 ×5 a=a, ∀a∈S

Thus, 1 is the identity element.

Now,3 ×5 4-1-1=3 ×5 4 -1 ∵ 4 ×5 4 = 1=2-1=3 ∵ 2 ×5 3 = 1

**Q14.**

**Answer :**

To find the identity element, let e be the identity element in Z with respect to * such that

a * e=a=e * a,∀ a∈Za * e=a and e * a=a,∀ a∈ZThen, a+e+2=a and e+a+2=a,∀ a∈Z⇒e=-2 ,∀ a∈Z

Thus, -2 is the identity element in Z with respect to *.

Let b be the inverse of 4.

Then, 4 * b=e=b * 44 * b=e and b *4=e⇒4+b+2=-2 and b+4+2=-2b=-8 ∈ZThus, -8 is the inverse of 4.

**Q15.**

**Answer :**

Here,

1×51 = Remainder obtained by dividing 1 × 1 by 5

= 1

3×54 = Remainder obtained by dividing 3 × 4 by 5

= 2

4×54 = Remainder obtained by dividing 4 × 4 by 5

= 1

So, the composition table is as follows:

×5 × | 0 | 1 | 2 | 3 | 4 |

0 | 0 | 0 | 0 | 0 | 0 |

1 | 0 | 1 | 2 | 3 | 4 |

2 | 0 | 2 | 4 | 1 | 3 |

3 | 0 | 3 | 1 | 4 | 2 |

4 | 0 | 4 | 3 | 2 | 1 |

**Q16.**

**Answer :**

Let e be the identity element in R with respect to * such that

a * e=a=e * a,∀ a∈Ra * e=a and e * a=a,∀ a∈RThen, a2+e2=a and e2+a2=a,∀ a∈R⇒a2+e=a and e+a2=a,∀ a∈R ∵ e2=e⇒a2+e=a2 and e+a2=a2,∀ a∈R⇒e=0∈R, ∀ a∈R

Thus, 0 is the identity element in R with respect to *.

**Q17.**

**Answer :**

Here,

1 +6 1 = Remainder obtained by dividing 1 + 1 by 6

= 2

3 +6 4 = Remainder obtained by dividing 3 + 4 by 6

= 1

4 +6 5 = Remainder obtained by dividing 4 + 5 by 6

= 3

So, the composition table is as follows:

+6 | 0 | 1 | 2 | 3 | 4 | 5 |

0 | 0 | 1 | 2 | 3 | 4 | 5 |

1 | 1 | 2 | 3 | 4 | 5 | 0 |

2 | 2 | 3 | 4 | 5 | 0 | 1 |

3 | 3 | 4 | 5 | 0 | 1 | 2 |

4 | 4 | 5 | 0 | 1 | 2 | 3 |

5 | 5 | 0 | 1 | 2 | 3 | 4 |

We observe that the first row of the composition table coincides with the the top-most row and the first column coincides with the left-most column.

These two intersect at 0.

⇒a+6 0=0+6 a=a,∀ a∈S

So, 0 is the identity element.

From the table,

4+6 2=0 ⇒4-1=23+6 3=0 ⇒3-1=3Now,2+6 4-1+6 3-1= 2+6 2+6 3 =4+6 3 =1

**Q18.**

**Answer :**

Given: a * b = 3a + 4b − 2

Here,

4 * 5 = 3 (4) + 4 (5) – 2

= 12 + 20 – 2

= 30

**Q19.**

**Answer :**

Given: a * b = a + 3b^{2}

Here,

2 * 4 = 2 + 3 (4)^{2}

= 2 + 3 (16)

= 2 + 48

= 50

**Q20.**

**Answer :**

Given: a * b = HCF (a, b)

Here,

22 * 4 = HCF (22, 4)

= 2 [because highest common factor of 22 and 4 is 2]

**Q21.**

**Answer :**

Given: a * b = 2a + b − 3

Here,

3 * 4 = 2 (3) + 4 -3

= 6 + 4 – 3

= 7

### Page 3.40 (Multiple Choice Questions)

**Q1.**

**Answer :**

(c) 412+32

Given: a * b = a^{2} + b^{2}

4 * 5 * 3=42+52 * 3 =42+522+32 =412+32

**Q2.**

**Answer :**

(c) 10

4.7 = (4 * 7) + 3

= 7 + 3

=10

**Q3.**

**Answer :**

(d) associative and commutative with an identity

Commutativity:X∆Y=X∩Y∪X∩Y =Y∩X∪Y∩X =Y∆XThus, X∆Y=Y∆XHence, ∆ is commutative on A.

Let ϕ be the identity element for ∆ on P.

A∆ϕ=A∩ϕ∪A∩ϕ =ϕ∪A =Aand, ϕ∆A=ϕ∩A∪ϕ∩A =A∪ϕ =A

**Q4.**

**Answer :**

(b) 33

Given: a * b = a^{2} − b^{2} + ab + 4

2 * 3=22-32+2×3+4 =4-9+6+4 =52 * 3 * 4=5 * 4 =52-42+5×4+4 =25-16+20+4 =33

**Q5.**

**Answer :**

(c) 41^{2} + 3^{2}

Given: a * b = a^{2} + b^{2}

4 * 5 * 3=42+52 * 3 =41 * 3 =412+32

**Q6.**

**Answer :**

(d) 5

Given: a * b = 3a − b

2 * 3 = 3 (2) – 3

= 6 – 3

= 3

(2 * 3) * 4 = 3 * 4

= 3 (3) – 4

= 9 – 4

= 5

**Q7.**

**Answer :**

(a) 43

Let e be the identity element in Q+ with respect to ⊙ such that

a * e=a=e * a,∀ a∈Q+a * e=a and e * a=a,∀ a∈Q+ae2=a and ea2=a,∀ a∈Q+e=2 ,∀ a∈Q+

Thus, 2 is the identity element in Q+ with respect to ⊙.

Let b∈Q+ be the inverse of 3. Then, 3 * b =e=b * 33 * b =e and b * 3=e3b2=2 and b32=2b=43Thus, 43 is the inverse of 3.

### Page 3.41 (Multiple Choice Questions)

**Q8.**

**Answer :**

Let xxxx∈G and eeee∈G such thatxxxx eeee==xxxx= eeeexxxxxxxx eeee=xxxx2ex2ex2ex2ex=xxxx2ex=xe=12∈R-0Thus, 12121212∈G, is the identity element in G.

*Disclaimer: The question in the book has some error, so, none of the options are matching with the solution. The solution is created according to the question given in the book.*

**Q9.**

**Answer :**

(d) 4a

Let e be the identity element in Q+ with respect to * such that

a * e=a=e * a,∀ a∈Q+a * e=a and e * a=a,∀ a∈Q+ae2=a and ea2=a,∀ a∈Q+e=2∈Q+,∀ a∈Q+

Thus, 2 is the identity element in Q+ with respect to *.

Let a∈Q+ and b∈Q+be the inverse of a.Then, a * b=e=b * aa * b=e and b * a=eab2=2 and ba2=2b=4a ∈Q+Thus, 4a is the inverse of a∈Q+.

**Q10.**

**Answer :**

(a) 3160

Given: a⊙b=ab4

3⊙15⊙12=3⊙15124 =3⊙140 =31404 =3160

**Q11.**

**Answer :**

(d) 0

Let e be the identity element in Q – {1} with respect to * such that

a * e=a=e * a,∀ a∈Q-1a * e=a and e * a=a,∀ a∈Q-1a+e-ae=a and e+a-ea=a,∀ a∈Q-1e1-a=0, ∀a∈Q-1e=0,∀ a∈Q-1 ∵ a≠1

Thus, 0 is the identity element in Q – {1} with respect to *.

**Q12.**

**Answer :**

(b) * defined by a*b=a+b2 is a binary operation on Q.

Let us check each option one by one.

(a)

If a=1 and b=2,a * b=a+b2 =1+22 =32∉Z

Hence, (a) is false.

(b)

a * b=a+b2∈Q,∀ a, b∈QFor example: Let a=32, b=56∈Qa*b=32+562 =9+512 =1412 =76∈Q

Hence, (b) is true.

(c)

Commutativity:

Let a, b∈N. Then, a * b=2ab =2ba = b * aTherefore, a * b=b * a,∀ a, b∈N

Thus, * is commutative on N.

Associativity:

Let a, b, c∈N. Then, a * b * c=a * 2bc =2a*2bca * b * c=2ab * c =2ab*2cTherefore,a * b * c≠a * b * c

Thus, * is not associative on N.

Therefore, all binary commutative operations are not associative.

Hence, (c) is false.

(d) Subtraction is not a binary operation on N because subtraction of any two natural numbers is not always a natural number.

For example: 2 and 4 are natural numbers.

2-4 = -2 which is not a natural number.

Hence, (d) is false.

**Q13.**

**Answer :**

(c) commutative and associative both

Commutativity:

Let a, b∈NThen, a * b=a+b+ab =b+a+ba = b * aThus, a * b=b * a,∀ a, b∈N

Thus, * is commutative on N.

Associativity:

Let a, b, c∈Na * b * c=a * b+c+bc =a+b+c+bc+a b+c+bc =a+b+c+bc+ab+ac+abca * b * c=a+b+ab * c =a+b+ab+c+a+b+ab c =a+b+c+ab+ac+bc+abcTherefore, a * b * c=a * b * c, ∀ a, b, c∈N

Thus, * is associative on N.

**Q14.**

**Answer :**

(d) 445

Given: a * b = a^{2} + b^{2} + ab + 1

2 * 3=22+32+2×3+1 =4+9+6+1 =202 * 3 * 2=20 * 2 =202+22+20×2+1 =400+4+40+1 =445

**Q15.**

**Answer :**

(a) commutative but not associative

Commutativity:

Let a, b∈Ra * b=ab+1 =ba+1 = b * aTherefore,a * b=b * a,∀ a, b∈R

Therefore, * is commutative on R.

Associativity:

Let a, b, c∈Ra * b * c=a * bc+1 =abc+1+1 =abc+a+1a * b * c=ab+1 * c =ab+1c+1 =abc+c+1∴ a * b * c≠a * b * cFor example: a=1, b=2 and c=3 which belong to RNow,1 * 2 * 3=1 * 6+1 =1 * 7 =7+1 =81 * 2 * 3=2+1 * 3 =3 * 3 =9+1 =10⇒1 * 2 * 3≠1 * 2 * 3Therefore, ∃ a=1, b=2 and c=3 which belong to R such that a * b * c≠a * b * c

**Q16.**

**Answer :**

(d) neither commutative nor associative

Subtraction of integers is not commutative

For example: If a = 1 and b = 2, then both are integers

1-2=-12-1=1

⇒-1≠1

∴ a-b≠b-a∀ a, b∈Z

Subtraction of integers is not associative.

For example: If a = 1, b = 2, c = 3, then all are integers

1-2-3=1+1 =21-2-3 =-1-3 =-4⇒2≠-4∴ a-b-c≠a-b-c ,∀ a, b, c∈Z

### Page 3.42 (Multiple Choice Questions)

**Q17.**

**Answer :**

(c) commutative law

The law a + b = b + a is commutative.

**Q18.**

**Answer :**

(d) none of these

* is not closure because when a = 1 and b = 2,

a * b=ab=12∉Z

* is not commutative because when a = 1 and b = 2,

1 * 2=122 * 1=211 * 2≠2 * 1

* is not associative because when a = 1, b = 2 and c = 3,

1 * 2 * 3=1 * 23 =12 3 =321 * 2 * 3=12 * 3 =123 =16Thus,1 * 2 * 3≠1 * 2 * 3

**Q19.**

**Answer :**

(c) not associative

Commutativity:

Let a, b∈Z. Then, a * b=a2+b2 =b2+a2 = b * aTherefore,a * b b * a,∀ a, b∈Z

Thus, * is commutative on Z.

Associativity:

Let a, b, c∈Za * b * c=a * b2+c2 =a2+b2+c22 =a2+b4+c4+2b2c2a * b * c=a2+b2 * c =a2+b22+c2 =a4+b4+2a2b2+c2Therefore,a * b * c≠a * b * c

Thus, * is not associative on Z.

**Q20.**

**Answer :**

(c) not commutative

Commutativity:

Let a, b∈Za * b=3a+bb * a=3b+aThus, a * b≠b * aIf a=1 and b=2,1 * 2=31+2 =52 * 1=32+1 =71 * 2≠2 * 1

Thus, * is not commutative on Z.

**Q21.**

**Answer :**

(a) 10^{5}

Let e be the identity element in Q+with respect to * such that

a * e=a=e * a,∀ a∈Q+a * e=a & e * a=a,∀ a∈Q+ae100=a &ea100=a,∀ a∈Q+e=100 ,∀ a∈Q+

Thus, 100 is the identity element in Q+ with respect to *.

Let b∈Q+ be the inverse of 0.1. Then,0.1 * b=e=b * 0.10.1* b=e and b * 0.1=e0.1b100=100 and b0.1100=100b=100×1000.1=105∈Q+Thus, 105 is the inverse of 0.1.

**Q22.**

**Answer :**

(d) non-existent

Let e be the identity element in N with respect to * such that

a * e=a=e * a,∀ a∈Na * e=a and e * a=a,∀ a∈NThen,a+e+10=a and e+a+10=a,∀ a∈Ne=-10∉N

So, the identity element with respect to * does not exist in N.

**Q23.**

**Answer :**

(a) 0

Let e be the identity element in Q – {1} with respect to * such that

a * e=a=e * a,∀ a∈Q-1a * e=a and e * a=a,∀ a∈Q-1Then, a+e-ae=a and e+a-ea=a,∀ a∈Q-1e1-a=0 ,∀ a∈Q-1e=0 ∈Q-1 ∵ a≠1

Thus, 0 is the identity element in Q – {1} with respect to *.

**Q24.**

**Answer :**

(b) -aa+1

Let e be the identity element in R – {1} with respect to * such that

a * e=a=e * a,∀ a∈R-1a * e=a and e * a=a,∀ a∈R-1Then, a+e+ae=a and e+a+ea=a,∀ a∈R-1e1+a=0 ,∀ a∈R-1e=0∈R-1

Thus, 0 is the identity element in R – {1}with respect to *.

Let a∈R-1 and b∈R-1 be the inverse of a. Then,a * b =e=b * aa * b =e and b * a=e⇒a+b+ab=0 and b+a+ba=0⇒b1+a=-a ∈R-1⇒b=-a1+a ∈R-1Thus,-a1+ais the inverse of a∈R-1.

**Q25.**

**Answer :**

(c) 2/13-3/133/132/13

To find the identity element,

Let A=ab-ba and I=xy-yx such that A.I=I.A=AA. I=Aab-baxy-yx=xy-yxax-byay+bx-ay+bxax-by=xy-yx⇒ax-by=x …1⇒ay+bx=y …2Solving these two equations, we getx=1 and y=0Thus, I=xy-yx=1001 (which is usually an identity matrix)Let mn-nm be the inverse of 23-32.∴ 23-32 mn-nm=1001⇒2m-3n2n+3m-3m-2n-3n+2m=1001⇒2m-3n=1 …(3) 2n+3m=0 …(4) -3m-2n=0 …(5) -3n+2m=1 …(6) From eq. (4) n=-3m2 …(7) Substituting the value of n in eq. (3) 2m-3-3m2=1⇒2m+9m2=1⇒13m2=1⇒m=213Substituting the value of m in eq. (7)⇒n=-32×213=-313Hence, the inverse of 23-32 is 213-313313213.

So, the answer is (c).

**Q26.**

**Answer :**

(b) 12

Let e be the identity element in Q+ with respect to * such that

a * e=a=e * a,∀ a∈Q+a * e=a and e * a=a,∀ a∈Q+Then,ae2=a and ea2=a,∀ a∈Q+e=2,∀ a∈Q+

Thus, 2 is the identity element in Q+ with respect to *.

Let b∈Q+ be the inverse of 8. Then,8 * b=e=b * 88 * b=e and b * 8=e8b2=2 and b82=2 ∵e=2b=12 Thus, 12 is the inverse of 8.

**Page 3.43 (Multiple Choice Questions)**

**Q27.**

**Answer :**

(a) 98

Let e be the identity element in Q+ with respect to * such that

a * e=a=e * a,∀ a∈Q+ a * e=a and e * a=a,∀ a∈Q+Then,ae3=a and ea3=a,∀ a∈Q+⇒e=3 ,∀ a∈Q+

Thus, 3 is the identity element in Q+ with respect to *.

Let a∈Q+ and b∈Q+ be the inverse of a. Then,a * b=e=b * aa * b=e and b * a=e∴ ab3=3 and ba3=3b=9a ∈Q+Thus, 9a is the inverse of a∈Q+.

Given: a * b=ab34 * 6=4×63 =8Now,a-1=9a4 * 6-1=8-1 =98

**Q28.**

**Answer :**

(c) 16

We know that the number of binary operations on a set of n elements is nn2.

So, the number of binary operations on a set of 2 elements is 222 (24), i.e. 16.

**Q29.**

**Answer :**

(d) 2

The number of commutative binary operations on a set of n elements is nnn-12.

Therefore,

Number of commutative binary operations on a set of 2 elements = 222-12 =21=2

**Q30.**

**Answer :**

(b) −1

Let e be the identity element in Z with respect to * such that

a * e=a=e * a,∀ a∈Za * e=a and e * a=a,∀ a∈ZThen, a+e+1=a and e+a+1=a,∀ a∈Ze=-1∈Z,∀ a∈Z

Thus, −1 is the identity element in Z with respect to *.