An organization conducted bike race under 2 different

Case study from CBSE sample paper 2024 – 2025 class 12

Case Study-2:- 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 forms two sets B and G with these participants for his college project.

Let $B=\{ b_1, b_2, b_3\}, G = \{g_1, g_2\}$ 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:

(i) Ravi wishes to form all the relations possible from B to G. How many such relations are possible? [1 Mark]

(ii) Write the smallest equivalence relation on G. [1 mark]

(iii) (a) Ravi defines a relation from B to B as $R_1 = \{(b_1, b_2), (b_2, b_1)\}$ . Write the minimum ordered pairs to be added in $R_1$ so that it becomes (A) reflexive but not symmetric, (B) reflexive and symmetric but not transitive. [2 mark]

(iii) (b) If the track of the final race (for the biker b1) follows the curve 𝑥² = 4𝑦; (where0 ≤ 𝑥 ≤ 20√2 & 0 ≤ 𝑦 ≤ 200), then state whether the track represents a one-one and onto function or not. (Justify). [2 mark]

Solution:- Given, $B=\{ b_1, b_2, b_3\}, G = \{g_1, g_2\}$

Number of element in set B = 3

and nuber of element in set G = 2

(i) Total number of relation from B to g = $2^(3\times 2)= 2^6 = 64$

(ii) The smallest equivalence relation on G

$\{(g_1, g_1), (g_2, g_2)\}$

This relation is reflexive, symmetric and transitive. Hence this relation is a smallest equivalence relation.

(iii) (a) Ravi defines a relation from B to B as $R_1 = \{(b_1, b_2), (b_2, b_1)\}$

(A) $R_2 = \{(b_1, b_2), (b_2, b_1) (b_1, b_1), (b_2, b_2), (b_3, b_3) (b_2, b_3)\}$

$\{(b_1, b_1), (b_2, b_2) (b_3, b_3)\} \in R_2$ so this is reflexive

$(b_2, b_3) \in R_2 \Rightarrow (b_3, b_2) \notin R_$ thus this is not Symmetric

If we add $(b_1, b_1), (b_2, b_2), (b_3, b_3) (b_2, b_3)$ this 4 ordered pair in $R_1$ this becomes reflexive but not symmetric

(B) $R_3 = \{(b_1, b_2), (b_2, b_1) (b_1, b_1), (b_2, b_2), (b_3, b_3) (b_2, b_3),(b3, b_2)\}$

$\{(b_1, b_1), (b_2, b_2) (b_3, b_3)\} \in R_3$ so this is reflexive

$(b_1, b_2), (b_2, b_1), (b_2, b_3),(b3, b_2) \in R_3$ thus this is symmetric

Since, $(b_1, b_2) \in R_3 & (b_2, b_3) \in R_3 \Rightarrow (b_1, b_3) \notin R_3$

Thus thus the relation R3 is not a Transitive

If we add $(b_1, b_1), (b_2, b_2), (b_3, b_3) (b_2, b_3),(b3, b_2)$ this 5 ordered pair in $R_1$ then this relation become reflexive and symmetric but not transitive.

(iii) (b) Given curve, 𝑥² = 4𝑦

f(x) = x²/4

Wher x ∈ [0, 20√2] and y ∈ [0, 200]

For One-one:- Let, $x_1, x_2 \in [0, 20\sqrt{2}]$

Such that, $f(x_1) = f(x_2)$

$\dfrac{x_1^2}{4} = \dfrac{x_2^2}{4}$

⇒ $x_1^2 = x_2^2$

⇒ $x_1 = x_2$

Thus, the given curve is one-one

Onto:- Given curve, 𝑥² = 4𝑦

⇒ x = 2√y

Let $y = 100 \in [0, 200] = \Rightarrow x = 2\sqrt{100} = 20\in [0, 20\sqrt{2}]$

Every value of co-domain having preimage in domain so this function is onto function

Thus this curve is one-one and onto function.

Case Study-1:- Ramesh the owner of a sweet selling shop purchased some rectangular card board sheets of dimension 25 cm by 40 cm to make container packets without top. Let x cm be the length of the side of the square to be cut out from each corner to give that sheet the shape of the container by folding up the flaps.

Based on the above information answer the following questions

(i) Express the volume (V) of each container as function of x only. [1 Mark]

(ii) Find $\dfrac{dv}{dx}$ [1 Mark]

(iii) (a) For what value of x , the volume of each container is maximum ? [2 Marks]

(iii) (b) Check whether V has a point of inflection at $x = \dfrac{65}{6}$ or not? [2 Marks]

Solution:- See full solution

Case Study- 3 :-Arka bought two cages of birds: Cage-I contains 5 parrots and 1 owl and Cage –II contains 6 parrots. One day Arka forgot to lock both cages and two birds flew from Cage-I to Cage-II (simultaneously). Then two birds flew back from cage-II to cage-I(simultaneously). Assume that all the birds have equal chances of flying. On the basis of the above information, answer the following questions:

(i) When two birds flew from Cage-I to Cage-II and two birds flew back from Cage-II to Cage-I then
find the probability that the owl is still in Cage-I. [2 mark]

(ii) When two birds flew from Cage-I to Cage-II and two birds flew back from Cage-II to Cage-I, the owl is still seen in Cage-I, what is the probability that one parrot and the owl flew from Cage-I to Cage-II? [2 mark]

Solution:- See full solution

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