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1.Relation and Function
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Let $A =\{1,2,3, \ldots, 10\}$ and $R$ be a relation on $A$ such that $R =\{( a , b ): a =2 b+1\}$. Let $\left( a _1, a _2\right)$, $\left(a_2, a_3\right),\left(a_3, a_4\right), \ldots .,\left(a_k, a_{k+1}\right)$ 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 :
A$6$
B$7$
C$3$
D$8$
(JEE MAIN-2025)
Solution
$a=2 b+1$
$2 b=a-1$
$R=\{(3,1),(5,2), \ldots,(99,49)\}$
Let $(2 m+1, m),(2 \lambda-1, \lambda)$ are such ordered pairs.
According to the condition
$m =2 \lambda-1 \Rightarrow m=$ odd number
$\Rightarrow 1^{\text {st }}$ 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
$\left\{(4 \lambda-1,2 \lambda-1),(2 \lambda-1, \lambda-1),\left(\lambda-1, \frac{\lambda-2}{2}\right), \ldots \ldots\right\}$
$2^{\text {nd }}$ element of each ordered pair $=\frac{\lambda-2^{ r -2}}{2^{ r -2}}$
For maximum number of ordered pairs in such sequence
$\frac{\lambda-2^{r-2}}{2^{r-2}}=1 \text { or } 2 ; 1 \leq \lambda \leq 25$
$\lambda=2^{r-1} \text { or } \lambda=3.2^{r-2}$
$\text { Case-I }: \lambda=2 r-1$
$\lambda=2,2^2, 2^3, 2^4$
$r=2,3,4,5$
Hence maximum value of $r$ is $5$ when $\lambda=16$
Case-II : $\lambda =3.2^{ r -2}$
$\lambda=3, 6, 12,$
$r=24 3, 4,$
Hence maximum value of $r$ is $5$ when $\lambda=24$
$2 b=a-1$
$R=\{(3,1),(5,2), \ldots,(99,49)\}$
Let $(2 m+1, m),(2 \lambda-1, \lambda)$ are such ordered pairs.
According to the condition
$m =2 \lambda-1 \Rightarrow m=$ odd number
$\Rightarrow 1^{\text {st }}$ 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
$\left\{(4 \lambda-1,2 \lambda-1),(2 \lambda-1, \lambda-1),\left(\lambda-1, \frac{\lambda-2}{2}\right), \ldots \ldots\right\}$
$2^{\text {nd }}$ element of each ordered pair $=\frac{\lambda-2^{ r -2}}{2^{ r -2}}$
For maximum number of ordered pairs in such sequence
$\frac{\lambda-2^{r-2}}{2^{r-2}}=1 \text { or } 2 ; 1 \leq \lambda \leq 25$
$\lambda=2^{r-1} \text { or } \lambda=3.2^{r-2}$
$\text { Case-I }: \lambda=2 r-1$
$\lambda=2,2^2, 2^3, 2^4$
$r=2,3,4,5$
Hence maximum value of $r$ is $5$ when $\lambda=16$
Case-II : $\lambda =3.2^{ r -2}$
$\lambda=3, 6, 12,$
$r=24 3, 4,$
Hence maximum value of $r$ is $5$ when $\lambda=24$
Standard 12
Mathematics
