Question 3 - Case Based Questions (MCQ) - Chapter 1 Class 12 Relation and Functions
Last updated at April 16, 2024 by Teachoo
Case Based Questions (MCQ)
Last updated at April 16, 2024 by Teachoo
Question An organization conducted bike race under 2 different categories-boys and girls. Totally 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 forms two sets B and G with these participants for his college project. Let B = {b1, b2, b3} G = {g1, g2} 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 & functions Question 1 Ravi wishes to form all the relations possible from B to G. How many such relations are possible? (a) 26 (b) 25 (c) 0 (d) 23 Given B = {b1, b2, b3} G = {g1, g2} Numbers of Relation from B to G = 2Numbers of elements of B × Number of elements of G = 23 × 2 = 26 So, the correct answer is (a) Question 2 Let R: B → B be defined by R = {(𝑥, 𝑦): 𝑥 and y are students of same sex}. Then this relation R is_______ (a) Equivalence (b) Reflexive only (c) Reflexive and symmetric but not transitive (d) Reflexive and transitive but not symmetric R = {(𝑥, 𝑦): 𝑥 and y are students of same sex}. Check Reflexive Since x and x are of the same sex So, (x, x) ∈ R for all x ∴ R is reflexive Check symmetric If x and y are of the same sex Then, y and x are of the same sex Thus (x, y) ∈ R , and (y, x) ∈ R ∴ R is symmetric Check transitive To check whether transitive or not, If (x, y) ∈ R & (y, z) ∈ R , then (x, z) ∈ R If x and y are of the same sex, And y and z is of the same sex Then, x and z will be of the same sex Thus, for all values of x, y, z (x, y) ∈ R & (y, z) ∈ R , then (x, z) ∈ R ∴ R is transitive Since R is reflexive, symmetric and transitive ∴ R is an Equivalence relation So, the correct answer is (a) Question 3 Ravi wants to know among those relations, how many functions can be formed from B to G? (a) 22 (b) 212 (c) 32 (d) 23 Given B = {b1, b2, b3} G = {g1, g2} So, B has 3 elements, G has 2 elements Numbers of functions from B to G = 2 × 2 × 2 = 23 So, the correct answer is (d) Question 4 Let 𝑅: 𝐵 → 𝐺 be defined by R = {(b1, g1), (b2, g2), (b3, g1)}, then R is__________ (a) Injective (b) Surjective (c) Neither Surjective nor Injective (d) Surjective and Injective Given B = {b1, b2, b3}, G = {g1, g2} And, R = {(b1, g1), (b2, g2), (b3, g1)} So, our relation looks like Injective Injective means one-one Since b1 and b3 have the same image g1 ∴ R is not injective Surjective Surjective means onto Since all elements G has a pre-image ∴ R is surjective So, the correct answer is (b) Question 5 Ravi wants to find the number of injective functions from B to G. How many numbers of injective functions are possible? (a) 0 (b) 2! (c) 3! (d) 0! In a function, Every element of set B will have an image. Every element of set B will only one image in set G For injective functions All elements of set G should have a unique pre-image Which is not possible ∴ Number of possible injective functions = 0 So, the correct answer is (a)