Q: Let A=[ Rectangles of the chess Board]. Define a relation R in A: Say (R1,R2)∈ Relation R, if rectangles R1 and R2 have a common side. Examine R.

Clearly, R is reflexive. Also, if (R1,R2)∈R,(R2,R1) will be in R. i.e., R is symmetric.However, if R1 and R2 have a common side and R2 and R3 have a common side, it does not imply R1 and R3 have a common side. In other words, R is not transitive.

Q: Given f(x)={x2,x+2,0<x<22≤x≤5 and f(x+5)=f(x) for all x. Compute f(11)+f(−11)f(11)−f(−11)

f(x+5)=f(x)⇒f(x) is periodic with period 5Now f(11)=f(10+1)=f(1)=1;f(−11)=f(−11+15)=f(4)=4+2=6∴f(11)+f(−11)f(11)−f(−11)=1+61−6=7−5

Question 1

A town has total population of 40000 out of which 400 people own cars, 12000 people own motorcycles and 350 people own both cars and motorcycles. How many in the town do not own either?

Accepted Answer

Let A= people owning cars; B= people owning motor cycles; S= people in the town

Question 2

Let A={1,2,3,4,…,20}. Find the number of subsets of A which contain 5,6,7,8,9 and 10.

Accepted Answer

Since 5,6,7,8,9 and 10 are to be there in all the subsets, the number of subsets is clearly the number of subsets that can be formed with the remaining 14 elements of A. The answer is 214.

Question 3

In an examination, 75% students passed in English, 65% passed in Hindi and 10% failed in both. Find the percentage of students who passed in both subjects.

Accepted Answer

Let A= Set of students who passed in English; B= Set of students who passed in HindiGiven n(A)=75,n(B)=65,n(A′∩B′)=10.⇒n(A∪B)=90We have, 90=75+65−n(A∩B) or n(A∩B)=50 or The required answer is 50%.

Question 4

In a survey of 100 students in a music school, the number of students learning different musical instruments was found to be: Guitar: 28, Veena: 30, Flute: 42, Guitar and Veena: 8, Guitar and Flute: 10, Veena and Flute: 5, All musical instruments: 3(i) How many students were learning none of these three musical instruments?(ii) How many students were learning only the flute?

Accepted Answer

(i) n(G)=28,n(V)=30,n(F)=42⇒n(G∪V∪F)=28+30+42−8−10−5+3=80Hence n(G′∩V′∩F′)=100−80=20.(ii) n( Flute only )=42−10−5+3=30

Question 5

Out of 500 students who appeared at a competitive examination from a centre, 140 failed in Mathematics, 155 failed in Physics and 142 failed in Chemistry. Those who failed in both Mathematics and Physics were 98, in Physics and Chemistry were 105, and in Mathematics and Chemistry 100. The number of students who failed in all the three subjects was 85 . Assuming that each student appeared in all the 3 subjects, find(i) the number of students who failed in at least one of the three subjects.(ii) the number of students who passed in all the subjects.(iii) the number of students who failed in Mathematics only.

Accepted Answer

Let M: failed in Mathematics; P : failed in Physics; C: failed in Chemistry

Question 6

Prove that if A,B,C are any three sets,
(i) A\(B∪C)=(A\B)∩(A\C)
(ii) A\(B∩C)=(A\B)∪(A\C)

Accepted Answer

(i) Let x be an element of A/(B∪C), i.e., x∈A\(B∪C)by definition of the set A\B⇒x∈A and x∈/(B∪C),⇒x∈A and (x∈/B and x∈/C)⇒(x∈A and x∈/B) and (x∈A and x∈/C)⇒(x∈A\B) and (x∈A\C)⇒x∈(A\B)∩(A\C)

Question 7

Let A=[ Rectangles of the chess Board]. Define a relation R in A: Say (R1​,R2​)∈ Relation R, if rectangles R1​ and R2​ have a common side. Examine R.

Accepted Answer

Clearly, R is reflexive. Also, if (R1​,R2​)∈R,(R2​,R1​) will be in R. i.e., R is symmetric.However, if R1​ and R2​ have a common side and R2​ and R3​ have a common side, it does not imply R1​ and R3​ have a common side. In other words, R is not transitive.

Question 8

A represents the set of real numbers. Examine the relation R in A : (x,y)∈R, if ∣x−y∣≤5 where, x,y∈A

Accepted Answer

x∈R. Since ∣x−x∣=0≤5,R is reflexive.Also, ∣x−y∣≤5 implies ∣y−x∣≤5 or R is symmetric.∣x−y∣≤5 and ∣y−z∣≤5 does not imply ∣x−z∣≤5. Therefore, R is not transitive.

Question 9

Given f(x)={x2,x+2,​0<x<22≤x≤5​ and f(x+5)=f(x) for all x. Compute f(11)+f(−11)f(11)−f(−11)​

Accepted Answer

f(x+5)=f(x)⇒f(x) is periodic with period 5Now f(11)=f(10+1)=f(1)=1;f(−11)=f(−11+15)=f(4)=4+2=6∴f(11)+f(−11)f(11)−f(−11)​=1+61−6​=7−5​

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Solutions for IIT-JEE Super Course in Mathematics - Calculus by Trishna Knowledge Systems

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