Sets & Relations
Counting ordered relation pairs
nta_pyq_2025_apr
Grade 12
Question:
Let$A = {1$, 2, 3,$\ldots$, 10} and R be a relation$o_n$A such that$R = {(a$, b) :$a = 2$$b + 1}.$Let (a , a ), 1 2 ($a_{2}$,$a_{3}$) , ($a_{3}$,$a_{4}$) ,$\ldots$. , (ak ,$ak+1$) be a sequence of k elements of R such that the second entry of an ordered pair is equal to the first entry of the next ordered pair. Then the largest integer k , for which such a sequence exists, is equal to :
Step-by-Step Solution
Key Concept: Translate the finite relation rule into explicit admissible ordered pairs and count the required set.
$a = 2$$b + 1$2$b = a - 1$(3)$R = {(3$, 1), (5, 2),$\ldots$, (99, 49)} Let$(2m + 1$, m), (2$\lambda$- 1,$\lambda$) are such ordered pairs. According to the condition m = 2$\lambda$- 1 $\Rightarrow$$m = odd$number $\Rightarrow$ 1 st element of ordered pair (a, b)$a = 2($2$\lambda$-$1) + 1$= 4$\lambda$- 1 Hence a$\ in ${3, 7,$\ldots$, 99} $\Rightarrow$$\lambda$$\ in ${1, 2,$\ldots$, 25} $\Rightarrow$ set of sequence$\lambda$- 2 {(4$\lambda$- 1, 2$\lambda$- 1), (2$\lambda$- 1,$\lambda$- 1), ($\lambda$- 1, ),$\ldots$$\ldots$.} 2$r-2$nd$\lambda$- 2 2 element of each ordered$pair = r-2$2 For maximum number of ordered pairs in such sequence$r-2$$\lambda$-$2 = 1$or 2; 1$\le$$\lambda$$\le$25$r-2$2$r-1$$r-2$$\lambda$= 2 or$\lambda$= 3.2$Case-I$:$\lambda$=$2r - 1$2 3 4$\lambda$= 2, 2 , 2 , 2$r = 2$, 3, 4, 5 Hence maximum value of r is 5 when$\lambda$= 16$r-2$$Case-II$:$\lambda$= 3.2$\lambda$= 3, 6, 12, 24$r = 2$, 3, 4, 5 Hence maximum value of r is 5 when$\lambda$= 24
Correct Answer: 3