JEE Main 2025 Apr 2 Shift 2, Mathematics Q4: Relations

JEE Main2025Apr 2, Shift 2Mathematics

Let $A = {1, 2, 3, \dots, 100}$ and $R$ be a relation on $A$ such that $R = {(a, b) : a = 2 b + 1}$ . Let $(a_{1}, a_{2}), (a_{2}, a_{3}), (a_{3}, a_{4}), \dots, (a_{k}, a_{k + 1})$ 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 :

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

Solution

Given,

$R = {(a, b) : a = 2 b + 1}$

Since $a \leq 100$ ,

$2 b + 1 \leq 100 \Rightarrow b \leq 49$

Hence,

$R = {(3, 1), (5, 2), (7, 3), \dots, (99, 49)}$

If two consecutive ordered pairs are

$(2 m + 1, m)$ and $(2 n + 1, n)$ ,

then the chaining condition gives

$m = 2 n + 1$ .

Thus, the second entries decrease as

$m, \frac{m - 1}{2}, \frac{m - 3}{4}, \frac{m - 7}{8}, \dots$

or, after $(r - 1)$ steps,

$b_{r} = \frac{m - ( 2 ^{r - 1} - 1 )}{2 ^{r - 1}}$

For the chain to continue, $b_{r}$ must be a positive integer.

The largest possible starting value is $m = 49$ .

Now,

$49 \to 24 \to 11 \to 5 \to 2 \to stop$

Hence, the longest chain is

$(99, 49) \to (49, 24) \to (24, 11) \to (11, 5) \to (5, 2)$

which contains $5$ ordered pairs.

$k = 5$

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JEE Main 2025 Apr 2 Shift 2, Mathematics Q4: Relations

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