JEE Main 2025 (Online) 2nd April Evening ShiftQuestion 1
Let $A=\{1,2,3,\ldots,100\}$ and $R$ be a relation on $A$ such that $R=\{(a,b):a=2b+1\}$. Let $(a_1,a_2),(a_2,a_3),(a_3,a_4),\ldots,(a_k,a_{k+1})$ 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 :
Sets, Relations and Function
Solution
$a_i=2a_{i+1}+1\Rightarrow a_i+1=2(a_{i+1}+1)$, so $a_1+1=2^k(a_{k+1}+1)$. Since $a_{k+1}\ge1$: $a_1+1\ge 2^{k+1}$, and $a_1\le100\Rightarrow 2^{k+1}\le101\Rightarrow k+1\le6$. So $k_{max}=5$ (e.g. $63,31,15,7,3,1$). Option A.