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) R1
Reflexive:
Clearly, (a, a), (b, b) and (c, c)∈R1
So, R1 is reflexive.
Symmetric:
We see that the ordered pairs obtained by interchanging the components of R1 are also in R1.
So, R1 is symmetric.
Transitive:
Here,
a, b∈R1, b, c∈R1 and also a, c∈R1
So, R1 is transitive.
(ii) R2
Reflexive: Clearly a,a∈R2. So, R2 is reflexive.
Symmetric: Clearly a,a∈R⇒a,a∈R. So, R2 is symmetric.
Transitive: R2 is clearly a transitive relation, since there is only one element in it.
(iii) R3
Reflexive:
Here,
b, b∉R3 neither c, c∉R3
So, R3 is not reflexive.
Symmetric:
Here,
b, c∈R3, but c,b∉R3So, R3 is not symmetric.
Transitive:
Here, R3 has only two elements. Hence, R3 is transitive.
(iv) R4
Reflexive:
Here,
a, a∉R4, b, b∉ R4 c, c∉ R4So, R4 is not reflexive.
Symmetric:
Here,
a, b∈R4, but b,a∉R4.So, R4 is not symmetric.
Transitive:
Here,
a, b∈R4, b, c∈R4, but a, c∉R4So, R4 is not transitive.
Q3.
Answer :
(i) Reflexivity:
Let a be an arbitrary element of R1. Then,
a∈R1⇒a≠1a for all a∈Q0So, R1 is not reflexive.
Symmetry:
Let (a, b) ∈R1. Then,
a, b∈R1⇒a=1b⇒b=1a⇒b, a∈R1So, R1 is symmetric.
Transitivity:
Here,
a, b∈R1 and b, c∈R2⇒a=1b and b=1c⇒a=11c=c⇒a≠1c⇒a, c∉R1 So, R1 is not transitive.
(ii)
Reflexivity:
Let a be an arbitrary element of R2. Then,
a∈R2 ⇒a-a=0≤5So, R1 is reflexive.
Symmetry:
Let a, b∈R2⇒a-b≤5⇒b-a≤5 Since, a-b = b-a⇒b, a∈R2So, R2 is symmetric.
Transitivity:
Let 1, 3∈R2 and 3, 7∈R2⇒1-3≤5 and 3-7≤5But 1-7≰5 ⇒1,7∉R2So, R2 is not transitive.
(iii)
Reflexivity: Let a be an arbitrary element of R3. Then,
a∈R3⇒a2-4a×a+3a2=0 So, R3 is reflexive.
Symmetry:
Let a, b∈R3⇒a2-4ab+3b2=0But b2-4ba+3a2≠0 for all a, b ∈RSo, R3 is not symmetric.
Transitivity:
1, 2∈R3 and 2, 3∈R3⇒1-8+6=0 and 4-24+27=0But 1-12+9≠0So, R3 is not transitive.
Q4.
Answer :
1 R1
Reflexivity:
Here,
1, 1, 2, 2, 3, 3∈RSo, R1 is reflexive.
Symmetry:
Here,2, 1∈R1, but 1, 2∉R1 So, R1 is not symmetric.
Transitivity:
Here, 2, 1∈R1 and 1, 3∈R1, but 2, 3∉R1 So, R1
2 R2
Reflexivity:
Clearly, 1, 1 and 3, 3∉R2 So, R2 is not reflexive.
Symmetry:
Here, 1, 3∈R2 and 3, 1∈R2So, R2 is symmetric.
Transitivity:
Here, 1, 3∈R2 and 3, 1∈R2 But 3, 3∉R2So, R2 is not transitive.
3 R3
Reflexivity:
Clearly, 1, 1∉R3 So, R3 is not reflexive.
Symmetry:
Here, 1, 3∈R3, but 3, 1∉R3So, R3 is not symmetric.
Transitivity:
Here, 1, 3∈R3 and 3, 3∈R3 Also, 1, 3∈R3So, R3 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.
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) f1 = {(1, 3), (2, 5), (3, 7)} ; A = {1, 2, 3}, B = {3, 5, 7}
Injectivity:
f1 (1) = 3
f1 (2) = 5
f1 (3) = 7
⇒Every element of A has different images in B.
So, f1 is one-one.
Surjectivity:
Co-domain of f1 = {3, 5, 7}
Range of f1 =set of images = {3, 5, 7}
⇒Co-domain = range
So, f1 is onto.
(ii) f2 = {(2, a), (3, b), (4, c)} ; A = {2, 3, 4}, B = {a, b, c}
Injectivity:
f2 (2) = a
f2 (3) = b
f2 (4) = c
⇒Every element of A has different images in B.
So, f2 is one-one.
Surjectivity:
Co-domain of f2 = {a, b, c}
Range of f2 = set of images = {a, b, c}
⇒Co-domain = range
So, f2 is onto.
(iii) f3 = {(a, x), (b, x), (c, z), (d, z)} ; A = {a, b, c, d,}, B = {x, y, z}
Injectivity:
f3 (a) = x
f3 (b) = x
f3 (c) = z
f3 (d) = z
⇒a and b have the same image x. (Also c and d have the same image z)
So, f3 is not one-one.
Surjectivity:
Co-domain of f1 ={x, y, z}
Range of f1 =set of images = {x, z}
So, the co-domain is not same as the range.
So, f3 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, x2) : x ∈ A}
Given, f(x) = x2
Injectivity:
f(1) = 12=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) = 12 = 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) = x2
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) = x2
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) = x3
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) = x3
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) = ex
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 ex 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 f1: R → R, given by f1(x)=x, and f2(x)=-x are one-one.
Proving f1 is one-one:
Let f1x=f1y⇒x=y
So, f1 is one-one.
Proving f2 is one-one:
Let f2x=f2y⇒-x=-y⇒x=y
So, f2 is one-one.
Proving (f1 + f2) is not one-one:
Given:
(f1 + f2) (x) = f1 (x) + f2 (x)= x + (-x) =0
So, for every real number x, (f1 + f2) (x)=0
So, the image of ever number in the domain is same as 0.
Thus, (f1 + f2) is not one-one.
Q18.
Answer :
We know that f1: R → R, given by f1(x) = x, and f2(x) = -x are surjective functions.
Proving f1 is surjective :
Let y be an element in the co-domain (R), such that f1(x) = y.
f1(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)=f1(y)x=y
So, f1is surjective .
Proving f2 is surjective :Let f2(x)=f2(y)−x=−yx=y
Let y be an element in the co domain (R) such that f2(x) = y.
f2(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)=f1(y)x=y
So, f2 is surjective .
Proving (f1 + f2) is not surjective :
Given:
(f1 + f2) (x) = f1 (x) + f2 (x)= x + (-x) =0
So, for every real number x, (f1 + f2) (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, f1 + f2 is not surjective.
Q19.
Answer :
We know that f1: R → R, given by f1(x) = x, and f2(x) = x are one-one.
Proving f1 is one-one:
Let x and y be two elements in the domain R, such that
f1(x) = f1(y)
⇒x = yet f1(x)=f1(y)x=y
So, f1 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 f1(x)=f1(y)x=y
So, f2 is one-one.
Proving f1 × f2 is not one-one:
Given:
f1 × f2x=f1 x × f2 x=x × x=x2Let x and y be two elements in the domain R, such thatf1 × f2x=f1 × f2y⇒x2 = y2⇒x=± ySo, f1 × f2 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) = x2 + 5
Now, (gof) (x)
= g (f (x))
= g (2x +3)
= (2x + 3)2 + 5
= 4x2+ 9 + 12x +5
=4x2+ 12x + 14
(fog) (x)
=f (g (x))
= f (x2 + 5)
= 2 (x2 + 5) +3
= 2 x2+ 10 + 3
= 2x2 + 13
(ii) f(x) = 2x + x2 and g(x) = x3
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) = x2 + 8 and g(x) = 3x3 + 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) = 8x3 and g(x) = x1/3
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 = 23 = 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).
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 + 3b2
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 = a2 + b2
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 = a2 − b2 + 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) 412 + 32
Given: a * b = a2 + b2
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 = a2 + b2 + 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) 105
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 *.
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