Q: Which of the following is a singleton set?

(a) (a) {x:∣x∣<1,x∈I}={x:−1<x<1,x∈I}={0}(b) {x:∣x∣=5,x∈I}={x:x=±5}={±5}(c) {x:x2=1,x∈I}={x:x=±1}={±1}(d) {x:x2+x+1=0,x∈R}={x:x=2(1)−1±12−4×1,x∈R}={x:x=2−1±3i,x∈R}={ϕ}[∵x is real but here x is a complex number ] Hence, option (a) is correct.

Q: Set A has m elements and Set B has n elements. If the total number of subsets of A is 112 more than the total number of subsets of B, then the value of m⋅n is .....

It is given that n(A)=m and n(B)=nand 2m=2n+112.[∵ number of subsets of set A and B are 2m and 2n respectively]⇒2m−2n=24(7)⇒2n(2m−n−1)=24(23−1)On comparing n=4 and m−n=3∴m=7So, m⋅n=28

Q: Let S be the set of all real roots of the equation, 3x(3x−1)+2=∣3x−1∣+∣3x−2∣. Then, S

given equation 3x(3x−1)+2=∣3x−1∣+∣3x−2∣ Let 3x=t>0,∀x∈R, so equation is t2−t+2=∣t−1∣+∣t−2∣∵∣t−1∣+∣t−2∣={1−t+2−t,02={3−2t,02 Now, Case I If ∣t−1∣+∣t−2∣=3−2t, for 0 2 then t2−t+2=2t−3 ⇒t2−3t+5=0 have no real solution. Since, S be the set of all real roots of the given equation, then S is a singleton set.

Q: If A={x∈R:∣x∣<2} and B={x∈R:∣x−2∣≥3}; then

Given sets A={x∈R:∣x∣<2},and B={x∈R:∣x−2∣≥3}then, A={x∈R:−2<x<2}and B={x∈R:(x−2)∈(−∞,−3]∪[3,∞)}∴ B−A=R−(−2,5),

Q: In a town of 10000 families it was found that 40% families buy newspaper A,20% families buy newspaper B and 10% families buy newspaper C,5% buy A and B,3% buy B and C and 4% buy A and C. If 2% families buy all of three newspapers, then the number of families which buy A only, is

(b) n(A)=40% of 10000=4000,n(B)=2000,n(C)=1000,n(A∩B)=500,n(B∩C)=300,n(C∩A)=400,n(A∩B∩C)=200∴n(A∩Bˉ∩Cˉ) =n{A∩(B∪C)′}=n(A)−n{A∩(B∪C)} =n(A)−n(A∩B)−n(A∩C)+n(A∩B∩C)=4000−500−400+200=3300

Q: A survey shows that 63% of the people in a city read newspaper A whereas 76% read newspaper B. If x% of the people read both the newspapers, then a possible value of x can be

(a) Given, n(A)=63,n(B)=76 and n(A∩B)=xWe know that,n(A∪B)=n(A)+n(B)−n(A∩B)⇒100≥63+76−x⇒x≥139−100≥39and n(A∩B)≤n(A)⇒x≤63∴39≤x≤63

Question 1

Which of the following is a singleton set?

Accepted Answer

(a) (a) {x:∣x∣<1,x∈I}={x:−1<x<1,x∈I}={0}(b) {x:∣x∣=5,x∈I}={x:x=±5}={±5}(c) {x:x2=1,x∈I}={x:x=±1}={±1}(d) {x:x2+x+1=0,x∈R}={x:x=2(1)−1±12−4×1​​,x∈R}={x:x=2−1±3​i​,x∈R}={ϕ}[∵x is real but here x is a complex number ] Hence, option (a) is correct.

Question 2

Set A has m elements and Set B has n elements. If the total number of subsets of A is 112 more than the total number of subsets of B, then the value of m⋅n is .....

Accepted Answer

It is given that n(A)=m and n(B)=nand 2m=2n+112.[∵ number of subsets of set A and B are 2m and 2n respectively]⇒2m−2n=24(7)⇒2n(2m−n−1)=24(23−1)On comparing n=4 and m−n=3∴m=7So, m⋅n=28

Question 3

Let S be the set of all real roots of the equation, 3x(3x−1)+2=∣3x−1∣+∣3x−2∣. Then, S

Accepted Answer

given equation 3x(3x−1)+2=∣3x−1∣+∣3x−2∣ Let 3x=t>0,∀x∈R, so equation is t2−t+2=∣t−1∣+∣t−2∣∵∣t−1∣+∣t−2∣={1−t+2−t,02={3−2t,02 Now, Case I If ∣t−1∣+∣t−2∣=3−2t, for 02 then t2−t+2=2t−3 ⇒t2−3t+5=0 have no real solution. Since, S be the set of all real roots of the given equation, then S is a singleton set.

Question 4

If set A={1,3,5}, then number of elements in P{P(A)} is

Accepted Answer

Given, A={1,3,5}∴n{P(A)}=23=8∴n[P{P(A)}]=28=256

Question 5

If A={ϕ,{ϕ}}, then the power set of A is

Accepted Answer

We have,A={ϕ,{ϕ}}Subset of set A are ϕ,{ϕ},{{ϕ}},{ϕ,{ϕ}}∴ Power set of A i.e. P(A)={ϕ,{ϕ},{{ϕ}},{ϕ,{ϕ}}} ⇒P(A)={ϕ,{ϕ},{{ϕ}},A}

Question 6

If X={4n−3n−1:n∈N} and Y={9(n−1):n∈N}, where N is the set of natural numbers, then X∪Y is equal to

Accepted Answer

(d) We have, X={4n−3n−1:n∈N}⇒X=0,9,54,243,.... put n=1,2,3,…Y=9(n−1):n∈N  put n=1,2,3, …⇒ Y ={0,9,18,27, …}It is clear that X⊂Y . ∴ X∪Y=Y

Question 7

Consider the two sets A={m∈R : both the roots of x2−(m+1)x+m+4=0 are real } and B=[−3,5).Which of the following is not true?

Accepted Answer

(a) Given sets A={m∈R : both the roots of x2−(m+1)x+m+4=0 are real } and B=[−3,5)∵ Roots of x2−(m+1)x+m+4=0 are real, m∈R∴D≥0⇒(m+1)2−4(m+4)≥0⇒m2−2m−15≥0⇒m2−5m+3m−15≥0⇒m(m−5)+3(m−5)≥0⇒(m+3)(m−5)≥0⇒m∈(−∞,−3]∪[5,∞)[∵A=(−∞,−3]∪[5,∞)]∴A−B=(−∞,−3)∪[5,∞)A∩B={−3},B−A=(−3,5) and A∪B=RHence, option (a) is correct.

Question 8

If A={x∈R:∣x∣<2} and B={x∈R:∣x−2∣≥3}; then

Accepted Answer

Given sets A={x∈R:∣x∣<2},and B={x∈R:∣x−2∣≥3}then, A={x∈R:−2<x<2}and B={x∈R:(x−2)∈(−∞,−3]∪[3,∞)}∴ B−A=R−(−2,5),

Question 9

In a town of 10000 families it was found that 40% families buy newspaper A,20% families buy newspaper B and 10% families buy newspaper C,5% buy A and B,3% buy B and C and 4% buy A and C. If 2% families buy all of three newspapers, then the number of families which buy A only, is

Accepted Answer

(b) n(A)=40% of 10000=4000,n(B)=2000,n(C)=1000,n(A∩B)=500,n(B∩C)=300,n(C∩A)=400,n(A∩B∩C)=200∴n(A∩Bˉ∩Cˉ) =n{A∩(B∪C)′}=n(A)−n{A∩(B∪C)} =n(A)−n(A∩B)−n(A∩C)+n(A∩B∩C)=4000−500−400+200=3300

Question 10

A survey shows that 63% of the people in a city read newspaper A whereas 76% read newspaper B. If x% of the people read both the newspapers, then a possible value of x can be

Accepted Answer

(a) Given, n(A)=63,n(B)=76 and n(A∩B)=xWe know that,n(A∪B)=n(A)+n(B)−n(A∩B)⇒100≥63+76−x⇒x≥139−100≥39and n(A∩B)≤n(A)⇒x≤63∴39≤x≤63

Question 11

Two newspapers A and B are published in a city. It is known that 25% of the city population reads A and 20% reads B while 8% reads both A and B. Further, 30% of those who read A but not B look into advertisements and 40% of those who read B but not A also look into advertisements, while 50% of those who read both A and B look into advertisements. Then, the percentage of the population who look into advertisements is

Accepted Answer

Sol. (d) Let the population of city is 100 .Then, n(A)=25,n(B)=20 and n(A∩B)=8So, n(A∩Bˉ)=17 and n(Aˉ∩B)=12According to the question,Percentage of the population who look into advertisement is=[10030​×n(A∩Bˉ)]+[10040​×n(Aˉ∩B)] +[10050​×n(A∩B)]=(10030​×17)+(10040​×12)+(10050​×8)=5.1+4.8+4=13.9

Question 12

If Xn​={z=x+iy:∣z∣2≤n1​} for all integers n≥1 Then, ⋂n=1∞​Xn​ is

Accepted Answer

Given, Xn​={z=x+iy:∣z∣2≤n1​}={x2+y2≤n1​}∴X1​={x2+y2≤1}, X2​={x2+y2≤21​}X3​={x2+y2≤31​} ⇒X∞​={x2+y2≤0}∴⋂n=1∞​Xn​=[X1​∩X2​∩X3​∩⋯∩X∞​={x2+y2}]={0}Hence, ⋂n=1∞​Xn​ is a singleton set.

Question 13

The set A={x:∣2x+3∣<7} is equal to

Accepted Answer

Given, set A={x:∣2x+3∣<7}Now, ∣2x+3∣<7⇒−7<2x+3<7⇒−7−3<2x<7−3⇒−10<2x<4⇒−5<x<2⇒0<(x+5)<7

Question 14

Let A={1,{2,3}}. Then, the number of subsets of A, is

Accepted Answer

We have, A={1,{2,3}}Number of element in A=2Number of subsets of A=22=4

Question 15

Let n(A)=m and n(B)=n, if the number of subsets of A is 56 more than of subsets of B, then m+n is equal to

Accepted Answer

Since, total possible subsets of sets A and B are 2m and 2n, respectively.According to given condition,2m−2n=56 ⇒2n(2m−n−1)=23×(23−1)On comparing both sides, we get2n=23 and 2m−n=23⇒n=3 and m−n=3 ⇒m=6 and n=3Now, m+n=6+3=9

Question 16

Let A be a set represented by the squares of natural numbers and x,y are any two elements of A, then

Accepted Answer

Let x,y∈A. Then,
x=m2,y=n2 for some m,n∈N
⇒xy=(mn)2∈A

Question 17

If A and B are two non-empty subsets of a set X such that A is not a subset of B. Then,

Accepted Answer

Since, A⊂B therefore it is not necessary that B is subset of A.

And also, it is not necessary that A and B are disjoint. Obviously, A and B′ are not disjoint.

Question 18

If P={θ:sinθ−cosθ=2​cosθ} and Q={θ:sinθ+cosθ=2​sinθ} are two sets. Then,

Accepted Answer

Since, cosθ(2​+1)=sinθ ⇒tanθ=2​+1 and sinθ(2​−1)=cosθ∴tanθ=2​−11​×2​+12​+1​=(2​+1)∴P=Q

Question 19

If A and B are two given sets, then A∩(A∩B)c is equal to

Accepted Answer

A∩(A∩B)c​=A∩(Ac∪Bc)(∵ by De-Morgan’s law )=(A∩Ac)∪(A∩Bc)=ϕ∪(A∩Bc)=A∩Bc​

Question 20

If A and B are two given sets, then A∩(A∩B)c is equal to

Accepted Answer

A∩(A∩B)c​=A∩(Ac∪Bc)(∵ by De-Morgan’s law )=(A∩Ac)∪(A∩Bc)=ϕ∪(A∩Bc)=A∩Bc​

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