Let A=\ 1,2,3, , 10\ and R be a relation on A such that R=\ (a, b): a=2 ~b+1\ . Let (a_1, a_2) , (a_2, a_3), (a_3, a_4) , .,(a_k, a_ k+1 ) be a sequence of k elements of R such that the second entry of an ordered pair

Correct answer: C. 5 aligned & a=2 ~b+1 & 2 ~b=a-1 & R=\ (3,1),(5,2), ,(99,49)\ aligned Let (2 m+1, m),(2 λ-1, λ) are such ordered pairs. According to the condition m=2 λ-1 ~m= odd number 1^ st element of ordered pair (a, b) a=2(2 λ-1)+1=4 λ-1 Hence a ∈\ 3,7, , 99\ λ ∈\ 1,2, , 25\ set of sequence aligned & \ (4 λ-1,2 λ-1),(2 λ-1, λ-1),(λ-1, (λ-2)/(2)), .\ & 2^ nd element of each ordered pair = λ-2^ r-2 2^ r-2 aligned For maximum number of ordered pairs in such sequence aligned & λ-2^ r-2 2^ r-2 =1 or 2 ; 1 ≤ λ ≤ 25 & λ=2^ r-1 or λ=3.2^ r-2 aligned aligned & Case-I : λ=2 r-1 & λ=2,2^2, 2^3, 2^4 & r=2,3,4,5 aligned Hence maximum value of r is 5 when λ=16 aligned & Case-II : λ=3.2^ r-2 & λ=3,6,12,24 & r=2, 3, 4,5 aligned Hence maximum value of r is 5 when λ=24

JEE Main 20252 AprilEvening Shift

Exam: JEE Main
Year: 2025
Date: 2 April
Shift: Evening Shift
Subject: Mathematics
Chapter: Sets and Relations

JEE Main 2025 · 2 April · Evening Shift · Mathematics · Sets and Relations

Let A=\ 1,2,3, , 10\ and R be a relation on A such that R=\ (a, b):...

Let

\mathrm{A}=\{1,2,3, \ldots, 10\} and R be a relation on A such that

\mathrm{R}=\{(\mathrm{a}, \mathrm{b}): \mathrm{a}=2 \mathrm{~b}+1\}

. Let

\left(\mathrm{a}_1, \mathrm{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 :

Options

A $6$
B $7$
C $5$
D $8$

Correct answer

C. $5$

Step-by-step solution

\begin{aligned}<br/>& \mathrm{a}=2 \mathrm{~b}+1 \\ & 2 \mathrm{~b}=\mathrm{a}-1 \\ & \mathrm{R}=\{(3,1),(5,2), \ldots,(99,49)\}<br/>\end{aligned}

Let

(2 m+1, m),(2 \lambda-1, \lambda)

are such ordered pairs.
According to the condition

\mathrm{m}=2 \lambda-1 \Rightarrow \mathrm{~m}=\text{ odd number }

\Rightarrow 1^{\text{st }}

element of ordered pair

(\mathrm{a}, \mathrm{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

\begin{aligned} & \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 }} \text{ element of each ordered pair }=\frac{\lambda-2^{\mathrm{r}-2}}{2^{\mathrm{r}-2}}\end{aligned}

For maximum number of ordered pairs in such sequence

\begin{aligned} & \frac{\lambda-2^{\mathrm{r}-2}}{2^{\mathrm{r}-2}}=1 \text{ or } 2 ; 1 \leq \lambda \leq 25 \\ & \lambda=2^{\mathrm{r}-1} \text{ or } \lambda=3.2^{\mathrm{r}-2}\end{aligned}

\begin{aligned} & \text{ Case-I : } \lambda=2 \mathrm{r}-1 \\ & \lambda=2,2^2, 2^3, 2^4 \\ & r=2,3,4,5\end{aligned}

Hence maximum value of

r

is 5 when

\lambda=16

\begin{aligned} & \text{ Case-II }: \lambda=3.2^{\mathrm{r}-2} \\ & \lambda=3,6,12,24 \\ & \mathrm{r}=2, \quad 3, \quad 4,5\end{aligned}

Hence maximum value of

r

is 5 when

\lambda=24

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