JEE Main 2025 — Mathematics Algebra

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

Let $(2 m+1, m),(2 \lambda-1, \lambda)$(2m+1,m),(2λ−1,λ) are such ordered pairs.

According to the condition

$$\mathrm{m}=2 \lambda-1 \Rightarrow \mathrm{~m}=\text { odd number }$$m=2λ−1⇒ m= odd number 

$\Rightarrow 1^{\text {st }}$⇒1st  element of ordered pair $(\mathrm{a}, \mathrm{b})$(a,b)

$$a=2(2 \lambda-1)+1=4 \lambda-1$$a=2(2λ−1)+1=4λ−1

Hence $a \in\{3,7, \ldots, 99\}$a∈{3,7,…,99}

$$\Rightarrow \lambda \in\{1,2, \ldots, 25\}$$⇒λ∈{1,2,…,25}

$\Rightarrow$⇒ set of sequence

$$\begin{aligned} & \left\{(4 \lambda-1,2 \lambda-1),(2 \lambda-1, \lambda-1),\left(\lambda-1, \frac{\lambda-2}{2}\right), \ldots \ldots .\right\} \\ & 2^{\text {nd }} \text { element of each ordered pair }=\frac{\lambda-2^{\mathrm{r}-2}}{2^{\mathrm{r}-2}}\end{aligned}$$​{(4λ−1,2λ−1),(2λ−1,λ−1),(λ−1,2λ−2​),…….}2nd  element of each ordered pair =2r−2λ−2r−2​​

For maximum number of ordered pairs in such sequence

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

$$\begin{aligned} & \text { Case-I : } \lambda=2 \mathrm{r}-1 \\ & \lambda=2,2^2, 2^3, 2^4 \\ & r=2,3,4,5\end{aligned}$$​ Case-I : λ=2r−1λ=2,22,23,24r=2,3,4,5​

Hence maximum value of $r$r is 5 when $\lambda=16$λ=16

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

Hence maximum value of $r$r is 5 when $\lambda=24$λ=24