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MAH-CET 2026 Mathematics PYQ — Let and be a relation on such that . Let be a sequence of element… | Mathem Solvex

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MAH-CET 2026 Mathematics PYQ — Let and be a relation on such that . Let be a sequence of element… | Mathem Solvex | Mathem Solvex

Explanation

Let us analyze the problem step-by-step to understand the relation and find the maximum length of the chain $k$ .

Step 1: Understand the Relation Condition

The relation is given by:

$a = 2 b + 1$

Every ordered pair $(x, y) \in R$ must satisfy the condition where the first element is $2$ times the second element plus $1$ .

For our given chain of elements:

$(a_{1}, a_{2}) \in R ⟹ a_{1} = 2 a_{2} + 1$
$(a_{2}, a_{3}) \in R ⟹ a_{2} = 2 a_{3} + 1$
$(a_{3}, a_{4}) \in R ⟹ a_{3} = 2 a_{4} + 1$
$\dots$
$(a_{k}, a_{k + 1}) \in R ⟹ a_{k} = 2 a_{k + 1} + 1$

Notice that as we move down the chain from $a_{1}$ to $a_{k + 1}$ , the values must strictly decrease because each subsequent term is roughly half of the previous term. All elements $a_{1}, a_{2}, \dots, a_{k + 1}$ must belong to the set $A = {1, 2, 3, \dots, 100}$ .

Step 2: Work Backward from the Smallest Possible Element

To maximize the number of links ( $k$ ) in our sequence, we should make the final elements as small as possible.

Let the smallest positive integer in the set $A$ be the last element of our sequence, $a_{k + 1}$ .

Let us assume the smallest valid positive integer for the final entry is:

$a_{k + 1} = 1$

Now, let's build the sequence backward using the relation rule $a_{n} = 2 a_{n + 1} + 1$ :

For the $k$ -th pair $(a_{k}, a_{k + 1})$ :
$a_{k} = 2 (1) + 1 = 3$
(Pair: $(3, 1) \in R$ )
For the previous pair:
$a_{k - 1} = 2 (3) + 1 = 7$
(Pair: $(7, 3) \in R$ )
Moving backward again:
$a_{k - 2} = 2 (7) + 1 = 15$
(Pair: $(15, 7) \in R$ )
Moving backward again:
$a_{k - 3} = 2 (15) + 1 = 31$
(Pair: $(31, 15) \in R$ )
Moving backward again:
$a_{k - 4} = 2 (31) + 1 = 63$
(Pair: $(63, 31) \in R$ )
If we try one more step backward:
$a_{k - 5} = 2 (63) + 1 = 127$
However, $127 \in / A$ because the set $A$ only contains numbers up to $100$ . Therefore, $63$ must be our starting element ( $a_{1}$ ).

Step 3: List the Complete Sequence and Count $k$

Let's align our calculated values with the sequence terms starting from $a_{1} = 63$ :

$a_{1} = 63$
$a_{2} = 31$
$a_{3} = 15$
$a_{4} = 7$
$a_{5} = 3$
$a_{6} = 1$

Now, let us write out the resulting ordered pairs in the sequence:

$(a_{1}, a_{2}) = (63, 31)$
$(a_{2}, a_{3}) = (31, 15)$
$(a_{3}, a_{4}) = (15, 7)$
$(a_{4}, a_{5}) = (7, 3)$
$(a_{5}, a_{6}) = (3, 1)$

Counting the total number of ordered pairs in this chain, we get exactly $5$ pairs.

Thus, the maximum value of $k$ is $5$ .

Conclusion

The largest integer $k$ for which such a sequence exists is $5$ .

Hence, the correct option is (a) 5.

Explanation

Let us analyze the problem step-by-step to understand the relation and find the maximum length of the chain $k$ .

Step 1: Understand the Relation Condition

The relation is given by:

$a = 2 b + 1$

Every ordered pair $(x, y) \in R$ must satisfy the condition where the first element is $2$ times the second element plus $1$ .

For our given chain of elements:

$(a_{1}, a_{2}) \in R ⟹ a_{1} = 2 a_{2} + 1$
$(a_{2}, a_{3}) \in R ⟹ a_{2} = 2 a_{3} + 1$
$(a_{3}, a_{4}) \in R ⟹ a_{3} = 2 a_{4} + 1$
$\dots$
$(a_{k}, a_{k + 1}) \in R ⟹ a_{k} = 2 a_{k + 1} + 1$

Step 2: Work Backward from the Smallest Possible Element

To maximize the number of links ( $k$ ) in our sequence, we should make the final elements as small as possible.

Let the smallest positive integer in the set $A$ be the last element of our sequence, $a_{k + 1}$ .

Let us assume the smallest valid positive integer for the final entry is:

$a_{k + 1} = 1$

Now, let's build the sequence backward using the relation rule $a_{n} = 2 a_{n + 1} + 1$ :

For the $k$ -th pair $(a_{k}, a_{k + 1})$ :
$a_{k} = 2 (1) + 1 = 3$
(Pair: $(3, 1) \in R$ )
For the previous pair:
$a_{k - 1} = 2 (3) + 1 = 7$
(Pair: $(7, 3) \in R$ )
Moving backward again:
$a_{k - 2} = 2 (7) + 1 = 15$
(Pair: $(15, 7) \in R$ )
Moving backward again:
$a_{k - 3} = 2 (15) + 1 = 31$
(Pair: $(31, 15) \in R$ )
Moving backward again:
$a_{k - 4} = 2 (31) + 1 = 63$
(Pair: $(63, 31) \in R$ )
If we try one more step backward:
$a_{k - 5} = 2 (63) + 1 = 127$
However, $127 \in / A$ because the set $A$ only contains numbers up to $100$ . Therefore, $63$ must be our starting element ( $a_{1}$ ).

Step 3: List the Complete Sequence and Count $k$

Let's align our calculated values with the sequence terms starting from $a_{1} = 63$ :

$a_{1} = 63$
$a_{2} = 31$
$a_{3} = 15$
$a_{4} = 7$
$a_{5} = 3$
$a_{6} = 1$

Now, let us write out the resulting ordered pairs in the sequence:

$(a_{1}, a_{2}) = (63, 31)$
$(a_{2}, a_{3}) = (31, 15)$
$(a_{3}, a_{4}) = (15, 7)$
$(a_{4}, a_{5}) = (7, 3)$
$(a_{5}, a_{6}) = (3, 1)$

Counting the total number of ordered pairs in this chain, we get exactly $5$ pairs.

Thus, the maximum value of $k$ is $5$ .

Conclusion

The largest integer $k$ for which such a sequence exists is $5$ .

Hence, the correct option is (a) 5.

Mathem Solvex

MAH-CET 2026 — Mathematics PYQ

Choose the correct answer:

Explanation

Step 1: Understand the Relation Condition

Step 2: Work Backward from the Smallest Possible Element

Step 3: List the Complete Sequence and Count $k$

Conclusion

Explanation

Step 1: Understand the Relation Condition

Step 2: Work Backward from the Smallest Possible Element

Step 3: List the Complete Sequence and Count $k$

Conclusion