JEE Advanced 2024 — Paper 2 — MATH-IV solution

JEE Advanced2024Paper 2MATH-IV

PARAGRAPH I

Let $S = {1, 2, 3, 4, 5, 6}$ and $X$ be the set of all relations $R$ from $S$ to $S$ that satisfy both the following properties:

i. $R$ has exactly 6 elements.

ii. For each $(a, b) \in R$ , we have $∣ a - b ∣ \geq 2$ .

Let $Y = {R \in X : The range of R has exactly one element}$ and $Z = {R \in X : R is a function from S to S}$ .

Let $n (A)$ denote the number of elements in a set $A$ .

(There are two questions based on PARAGRAPH “I”, the question given below is one of them)

Q2. If the value of $n (Y) + n (Z)$ is $k^{2}$ , then $∣ k ∣$ is ______.

Your answer

Solution

Computing $n (Y)$ : If all six pairs in $R$ share the same second coordinate $b$ , then $R = {(a_{1}, b), (a_{2}, b), \dots, (a_{6}, b)}$ with all $a_{i}$ distinct and each $∣ a_{i} - b ∣ \geq 2$ . Looking at the admissible- $b$ table from the previous part, for any fixed $b$ the number of admissible $a$ 's is at most 4 (achieved when $b \in {1, 6}$ ). Since we need 6 distinct admissible $a$ 's, this is impossible. Hence $n (Y) = 0$ .

$Computing n (Z) :$ For $R$ to be a function from $S$ to $S$ with exactly 6 elements, each element of $S$ must appear exactly once as a first coordinate. We then independently choose the image $f (a)$ for each $a$ from the admissible options:

$n (Z) = 4 \cdot 3 \cdot 3 \cdot 3 \cdot 3 \cdot 4 = 16 \cdot 81 = 1296 = 3 6^{2} .$

Hence $n (Y) + n (Z) = 0 + 1296 = 3 6^{2}$ , giving $∣ k ∣ = 36$ .

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