JEE Mathematics 2023 Question with Solution

Given: $A = \{1, 2, 3, \ldots, 10\}$ and $B = \{0, 1, 2, 3, 4\}$. The relation is

Find: The number of ordered pairs $(a,b)$ in $R$.

Let $a-b = k$. Then the condition becomes

Since $a,b \in A$, the integer $k=a-b$ can take values from $-9$ to $9$.

So we need

Check the relevant integer values of $k$ shown in the solution:

• If $k=0$, then

So $a=b$. This gives $10$ ordered pairs.

If $k=-2$, then

So $a=b-2$. For $1 \le a,b \le 10$, this gives

so there are $8$ ordered pairs.

Also, from the extracted solution:

• If $k=-1$, then

• If $k=1$, then

Therefore, the valid cases are only $a=b$ and $a=b-2$. Hence the total number of elements is

Therefore, the required number of elements in R is $18$.

Given: The relation depends only on the difference $a-b$. Find: Count all ordered pairs $(a,b)$ satisfying the condition.

Set $k=a-b$. Then the relation condition is

Because $a,b \in \{1,2,\ldots,10\}$, we have

From the extracted explanation, the admissible values are obtained by checking integer values of $k$. The valid ones are:

The values $k=-1$ and $k=1$ fail because they give $-1$ and $5$ respectively, which are not in $B$.

For $k=0$:

Thus the pairs are

so the count is $10$.

For $k=-2$:

Hence the pairs are

so the count is $8$.

Adding both counts,

Therefore, the answer is $18$.