Mathematics - Sets and Relations Question with Solution | TestHub

$\begin{array}{rl}& \mathrm{a}=2\mathrm{b}+1\\ & 2\mathrm{b}=\mathrm{a}-1\\ & \mathrm{R}=\{(3,1),(5,2),\dots ,(99,49)\}\end{array}$
Let $(2m+1,m),(2\lambda -1,\lambda )$ 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 $a\in \{3,7,\dots ,99\}$
$\Rightarrow \lambda \in \{1,2,\dots ,25\}$
$\Rightarrow$ set of sequence
$\begin{array}{rl}& \left\{(4\lambda -1,2\lambda -1),(2\lambda -1,\lambda -1),\left(\lambda -1,\frac{\lambda -2}{2}\right),\dots \dots .\right\}\\ & 2^{\text{nd }}\text{ element of each ordered pair }=\frac{\lambda -2^{\mathrm{r}-2}}{2^{\mathrm{r}-2}}\end{array}$
For maximum number of ordered pairs in such sequence
$\begin{array}{rl}& \frac{\lambda -2^{\mathrm{r}-2}}{2^{\mathrm{r}-2}}=1\text{ or }2;1\le \lambda \le 25\\ & \lambda =2^{\mathrm{r}-1}\text{ or }\lambda =3.2^{\mathrm{r}-2}\end{array}$
$\begin{array}{rl}& \text{ Case-I : }\lambda =2\mathrm{r}-1\\ & \lambda =2,2^2,2^3,2^4\\ & r=2,3,4,5\end{array}$
Hence maximum value of $r$ is 5 when $\lambda =16$
$\begin{array}{rl}& \text{ Case-II }:\lambda =3.2^{\mathrm{r}-2}\\ & \lambda =3,6,12,24\\ & \mathrm{r}=2,3,4,5\end{array}$
Hence maximum value of $r$ is 5 when $\lambda =24$