Question

An organization conducted a 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 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 and functions.

On the basis of the above information, answer the following questions:

1. Ravi wishes to form all the relations possible from B to G. How many such relations are possible?  (1)
2. Write the smallest equivalence relation on G.  (1)
3.    
   1. Ravi defines a relation from B to B as R₁ = {(b₁, b₂), (b₂, b₁)}. Write the minimum ordered pairs to be added in R₁ so that it becomes 
      1. reflexive but not symmetric,
      2. reflexive and symmetric but not transitive.  (2)
                               OR
   2. If the track of the final race (for the biker b₁) follows the curve
      `x^2 = 4y; ("where"  0 ≤ x ≤ 20sqrt2   "and"  0 ≤ y ≤ 200)`, then state whether the track represents a one-one and onto function or not. (Justify).  (2)

Answer 1

i. The number of relations is equal to the number of subsets of the set B × G = `2^(n(B × G))`

= 2^n(B) × n^(G)

= 2^{3 × 2}

= 2⁶
Where n(A) denotes the number of the elements in the finite set A.

ii. The smallest equivalence relation on G is {(g₁, g₁), (g₂, g₂)}.

iii. a. A. Reflexive but not symmetric = {(b₁, b₂), (b₂, b₁), (b₁, b₁), (b₂, b₂), (b₃, b₃), (b₂, b₃)}.

So the minimum number of elements to be added are (b₁, b₁), (b₂, b₂), (b₃, b₃), (b₂, b₃).

{Note: It can be any one of the pair from (b₃, b₂), (b₁, b₃), (b₃, b₁) in place of (b₂, b₃) also.}

B. Reflexive and symmetric but not transitive = {(b₁, b₂), (b₂, b₁), (b₁, b₁), (b₂, b₂), (b₃, b₃), (b₂, b₃), (b₃, b₂)}

So the minimum number of elements to be added are (b₁, b₁), (b₂, b₂), (b₃, b₃), (b₂, b₃), (b3, b₂).

OR

b. One-one and onto function.

x² = 4y. let `y = f(x) = x^2/4`

Let `x_1, x_2 in [0, 20sqrt2]` such that f(x₁) = f(x₂) ⇒ `x_1^2/4 = x_1^2/4`

⇒ `x_1^2 = x_2^2 ⇒ (x_1 - x_2)(x_1 +x_2) = 0 ⇒ x_1 = x_2    "as"  x_1, x_2 in [0, 20sqrt2]`

∴ f is a one-one function.

Now 0 ≤ y ≤ 200, hence the value of y is non-negative and `f(2sqrty) = y`.

∴ For any arbitrary y ∈ [0, 200], the pre-image of y exists in `[0, 20sqrt2]`, hence f is an onto function.