Maths · Relations and functions · chapter 1 of 13

Class 12 Maths Chapter 1: Relations and Functions case study question with answers

One case-based question of the kind the board asks: a passage, then 3 parts worth 4 marks in all, with the answer and the marking scheme for each. Written from the NCERT chapter.

Case study · 4 marks
A school in Nagpur places students in three houses: Red, Blue and Green. For a group S of five students, the allotment is a function f : S → H, where H = {Red, Blue, Green}. This year the allotment was: Aarav → Red, Bhavna → Blue, Chetan → Red, Diya → Green, Eshan → Blue. In S, the relation R = {(x, y) : f(x) = f(y)}, that is, x and y are in the same house, is used to seat students at house tables.
Part 1 · 1 mark
The number of different ways in which the five students could have been allotted to the three houses is
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Answer: (a) 243

Each of the 5 students has 3 choices of house: 35 = 243. 125 = 53 reverses the roles; 15 = 3 × 5.

Part 2 · 1 mark
The relation R in S is
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Answer: (a) an equivalence relation

R = {(x, y) : f(x) = f(y)} is an equivalence relation for any function f: f(x) = f(x); f(x) = f(y) gives f(y) = f(x); and f(x) = f(y), f(y) = f(z) give f(x) = f(z). It is not universal, since Aarav and Diya are in different houses.

Part 3 · 2 marks
Write the equivalence classes of R, and state with reason whether this year's allotment f is one-one and whether it is onto.
Answer and marking scheme

The classes are the house groups: [Aarav] = {Aarav, Chetan}, [Bhavna] = {Bhavna, Eshan}, [Diya] = {Diya}. f(Aarav) = f(Chetan) = Red, so f is not one-one. Each of Red, Blue, Green is the image of some student, so f is onto.

  • three classes written (1)
  • not one-one and onto, with reasons (1)
Or
Find the number of allotments of the five students to the three houses in which every house gets at least one student.
Answer and marking scheme

These are the onto functions S → H. All allotments: 35 = 243. Missing a given house: 25 = 32, for each of 3 houses; missing two given houses: 1, for each of 3 pairs. Onto allotments = 243 − 3 × 32 + 3 × 1 = 150.

  • counts allotments missing one or two houses (1)
  • 243 − 96 + 3 = 150 (1)
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Chapter 2: Inverse Trigonometric Functions →