Case study Chapter 1 (Relation and Function)
Case study 4:- Read the following answer the question(Case study relation and function 4)
Student of gade 9,planned to plant saplings along straight lines, parallel to each other to one side of the play ground ensuring that they had enough play area. Let us assume that they planted one of the rows of the saplings along the line y = x – 4. Let L be the set of all lines which are parallel on the ground and R be a relation on L.
(i) Let relation R be defined by 
Then R is ______ Relation
(a) Equivalence (b) Only Reflexive
(c) Not Reflexive (d) Symmetric but not Transitive
(ii) Let 
which of the following is true ?
(a) R is Symmetric but neither reflexive nor transitive
(b) R is Reflexive and transitive but not Symmetric.
(c) R is Reflexive but neither Symmetric nor transitive.
(d) R is equivalence relation.
(iii) The function f:R → R defined by f(x) = x – 4 is
(a) Bijective (b) Surjective but not injective
(c) Injective but not Surjective
(d) Neither Surjective nor Injective
(iv) Let f:R → R be defined by f(x) = x – 4. then the range of f(x) is
(a) R (b) Z
(c) W (d) Q
(v) Let 
then which of the following can be taken as 
(a) 2x – 2y + 5 = 0 (b) 2x + y = 5
(c) 2x + 2y + 7 =0 (d) x + y = 7
Solution ;(i) Answer (a)
Given relation R defined by
Reflexive: Let 
It is reflexive
Symmetric: Let 
Such that
It is Symmetric
Transitive: Let 
It is transitive.
Hence R is an equivalence relation.
(ii) Answer (a)
Given relation R defined by
Reflexive: Since every line is not perpendicular to itself.
It is nor reflexive.
Symmetric: Let
It is symmetric.
Transitive: Let 
Let 
∴ It is not Transitive
Hence relation R is symmetric but neither reflexive nor transitive.
(iii) Answer (a)
Given function f:R → R defined by f(x) = x – 4
Injective: Let 
such that 
∴ It is injective.
Surjective: Let y = x – 4 ⇒ x = y + 4
For every y ∈ R there exists x = y + 4 ∈ R.
i.e. Co- domain = Range
∴ It is surjective.
Hence given function is bijective.
(iv) Answer (a)
Given function f:R → R defined by f(x) = x – 4
Let y = f(x) ⇒ y = x – 4 ⇒ x = y + 4
⇒ x ∈ R ⇒ y ∈ R
∴ Range of f(x) is R (Set of real numbers).
(v) Answer (a)
2x – 2y + 5 = 0 is parallel to y = x – 4.
Some other case study problem
Case study 1:- Read the following and answer the question:(Case study relation and function 1)
A general election of Lock sabha is a gigantic exercise. About 311 million people were eligible to vote and voter turnout was about 67%, the highest ever
Let I be the set of all citizens of India who were eligible to exercise their voting right in general electio held in 2019. A relation ‘R’ is defined on I as follows:
and both use their voting right in general election – 2019]
Solution: For solution click here
Case study 2:- Sherlin and Danju are playing Ludo at home during COVID- 19. While rolling the dice, Sherlin’s sister Raji cbserved and noted the possible outcomes of the throw every time belongs to set {1, 2, 3, 4, 5, 6}. Let A be the set of players while B be the set of all possible outcomes.(Case study relation and function 2)
A = {S, D}, B = {1, 2, 3, 4, 5, 6}
Based on the above information answer the question.
Solution: For solution click here
Case Study 3:- Read the following and answer the question:(Case study relation and function 3)
An organization conducted bike race under 2 different categories- boys and girls. In all, there were 250 participants. Among all of them finally three from category 1 and two from category 2 were selected for the final race. Ravi from two sets B and G with these participants for his college project.
Let 
and 
where B represents the set of boys selected and G the set of girls who were selected for the final race.
Ravi decides to explore these sets for various types of relations and functions
Solution : For solution click here
