CBSE Sample Paper 2024-25 · Section E · 4 marks

Q37: Case study: relations and functions on the finalists of a bike race

Sample 2024-25Ch 1 · Relations and Functions

The question

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 marks)

(iii) (b) If the track of the final race (for the biker $b_1$ ⁠) follows the curve $x^2 = 4y$ ⁠; (where $0 \le x \le 20\sqrt2$ & $0 \le y \le 200$ ⁠), then state whether the track represents a one-one and onto function or not. (Justify). (2 marks)

Exam answer

(i) Number of relations from B to G is equal to the number of subsets of $B \times G$ ⁠.

Why?

A relation from B to G is any set of ordered pairs $(b, g)$ with $b$ in B and $g$ in G, that is, any subset of $B \times G$ ⁠, including the empty set.

\begin{aligned} 2^{n(B \times G)} &= 2^{n(B) \times n(G)} \\ &= 2^{3 \times 2} = 2^6 \end{aligned}

Why?

$n(B \times G) = n(B) \times n(G) = 3 \times 2 = 6$ ⁠, and a set with $m$ elements has $2^m$ subsets: each element is either in or out.

$2^6 = 64$ ⁠.

\begin{aligned} 2^6 &= 2 \times 2 \times 2 \times 2 \times 2 \times 2 \\ &= 4 \times 4 \times 4 = 64 \end{aligned}

So 64 relations are possible.

(ii) The smallest equivalence relation on G is $\{(g_1, g_1), (g_2, g_2)\}$ ⁠.

Why?

An equivalence relation must be reflexive, so it must contain $(g_1, g_1)$ and $(g_2, g_2)$ ⁠. Any relation without one of them is not an equivalence relation.

These two pairs alone are also symmetric (each pair reversed is itself) and transitive (they only chain to themselves). So nothing more is needed.

(iii) (a)

(A) Reflexive but not symmetric:

$\{(b_1, b_2), (b_2, b_1),$ $(b_1, b_1), (b_2, b_2),$ $(b_3, b_3), (b_2, b_3)\}$

Why?

Reflexive on B needs all three pairs $(b_1, b_1)$ ⁠, $(b_2, b_2)$ ⁠, $(b_3, b_3)$ ⁠.

With only those added, the relation is still symmetric, since $R_1$ already holds both $(b_1, b_2)$ and $(b_2, b_1)$ ⁠. To break symmetry we must add one more pair whose reverse is missing, such as $(b_2, b_3)$ ⁠.

It is reflexive, and not symmetric as $(b_3, b_2)$ is not in it. So the minimum pairs to be added are $(b_1, b_1)$ ⁠, $(b_2, b_2)$ ⁠, $(b_3, b_3)$ ⁠, $(b_2, b_3)$ ⁠. (Any one of $(b_3, b_2)$ ⁠, $(b_1, b_3)$ ⁠, $(b_3, b_1)$ may be added in place of $(b_2, b_3)$ ⁠.)

(B) Reflexive and symmetric but not transitive:

$\{(b_1, b_2), (b_2, b_1),$ $(b_1, b_1), (b_2, b_2),$ $(b_3, b_3), (b_2, b_3),$ $(b_3, b_2)\}$

Why?

Again the three pairs $(b, b)$ are needed. $R_1$ with them is an equivalence relation, which is transitive, so at least one more pair is needed.

To stay symmetric, a new pair must come with its reverse, so we add $(b_2, b_3)$ and $(b_3, b_2)$ together.

It is reflexive and symmetric, and not transitive as $(b_1, b_2)$ and $(b_2, b_3)$ are in it but $(b_1, b_3)$ is not.

Why?

Transitive means: whenever $(a, b)$ and $(b, c)$ are in R, $(a, c)$ is in R too. One case where this fails is enough to show R is not transitive.

So the minimum pairs to be added are $(b_1, b_1)$ ⁠, $(b_2, b_2)$ ⁠, $(b_3, b_3)$ ⁠, $(b_2, b_3)$ ⁠, $(b_3, b_2)$ ⁠.

(iii) (b)

$x^2 = 4y$ ⁠. Let $y = f(x) = \dfrac{x^2}{4}$ ⁠, $f: [0, 20\sqrt2] \to [0, 200]$ ⁠.

Why?

Solving $x^2 = 4y$ for $y$ gives $y = \frac{x^2}{4}$ ⁠. The allowed values of $x$ form the domain, and the allowed values of $y$ the co-domain.

Let $x_1, x_2 \in [0, 20\sqrt2]$ such that $f(x_1) = f(x_2)$ ⁠.

\begin{aligned} &\Rightarrow \frac{x_1^2}{4} = \frac{x_2^2}{4} \Rightarrow x_1^2 = x_2^2 \\ &\Rightarrow (x_1 - x_2)(x_1 + x_2) = 0 \\ &\Rightarrow x_1 = x_2 \end{aligned}

as $x_1, x_2 \in [0, 20\sqrt2]$ (if $x_1 + x_2 = 0$ ⁠, both are 0). $\therefore f$ is a one-one function.

Why?

One-one means equal outputs come only from equal inputs. Start from $f(x_1) = f(x_2)$ and show $x_1 = x_2$ ⁠.

\begin{aligned} \frac{x_1^2}{4} &= \frac{x_2^2}{4} \\ x_1^2 &= x_2^2 \quad (\times\,4) \\ x_1^2 - x_2^2 &= 0 \\ (x_1 - x_2)(x_1 + x_2) &= 0 \end{aligned}

Use $a^2 - b^2 = (a - b)(a + b)$ ⁠. Both numbers are at least 0, so $x_1 + x_2 = 0$ only when both are 0. Either way $x_1 = x_2$ ⁠.

Watch out: on all real numbers $x^2$ is not one-one, since $(-2)^2 = 2^2$ ⁠. It is the domain $[0, 20\sqrt2]$ that makes it one-one here.

Let $y \in [0, 200]$ and $f(x) = y$ ⁠. Then $x^2 = 4y$ ⁠, so $x = 2\sqrt y$ (as $x \ge 0$ ⁠).

Why?

Onto means every element of the co-domain is an output. So take any $y$ in $[0, 200]$ and solve for an $x$ in the domain.

Now $0 \le y \le 200$ ⁠, hence $y$ is non-negative, $2\sqrt y \in [0, 20\sqrt2]$ and $f(2\sqrt y) = y$ ⁠.

Why?

$\sqrt y$ grows with $y$ ⁠, so $2\sqrt y$ runs from $0$ up to

2\sqrt{200} = 2 \times 10\sqrt2 = 20\sqrt2

( $\sqrt{200} = \sqrt{100 \times 2} = 10\sqrt2$ ⁠. At $y = 0$ ⁠, $2\sqrt0 = 0$ ⁠.)

And $f(2\sqrt y) = \frac{(2\sqrt y)^2}{4} = \frac{4y}{4} = y$ ⁠.

$\therefore$ For any arbitrary $y \in [0, 200]$ ⁠, the pre-image of $y$ exists in $[0, 20\sqrt2]$ ⁠. Hence $f$ is an onto function.

$\therefore$ The track represents a one-one and onto function.

Where this comes from

Chapter 1, Relations and Functions: Relations: counting, equivalence; one-one and onto.

Question type: (i) Counting questions; (iii)(a) A relation given as a list of pairs; (iii)(b) Prove one-one and onto

How to spot it: click to see

(i) Type 8. Counting questions: list the possibilities

Spot it: "the number of relations (or functions) …" on a small set like $\{1, 2, 3\}$ ⁠, usually an MCQ.

Here: Each of the $3 \times 2 = 6$ pairs of $B \times G$ is in or out of a relation, so there are $2^6 = 64$ relations.

(iii)(a) Type 1. A relation given as a list of pairs: check each property

Spot it: $R$ is written out pair by pair, on a small set like $\{1, 2, 3\}$ ⁠.

Here: $R_1$ is listed pair by pair. Reflexive needs $(b_1, b_1), (b_2, b_2), (b_3, b_3)$ ⁠; then add $(b_2, b_3)$ alone to break symmetry, or with $(b_3, b_2)$ to break transitivity.

(iii)(b) Type 5. Prove one-one and onto: the bijective function

Spot it: "show that $f$ is one-one and onto", or "is $f$ bijective?", with a formula like $f(x) = 3 - 4x$ ⁠.

Here: Write the track as $f(x) = \frac{x^2}{4}$ on $[0, 20\sqrt2]$ ⁠. $x_1^2 = x_2^2$ gives $x_1 = x_2$ ⁠, as both are at least 0, and $x = 2\sqrt y$ lies in the domain.

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