Sets, Relations and Functions — JEE Main practice

Let the set be

A = {𝑎, 𝑏, 𝑐, 𝑑},

so |A| = 4.

For a reflexive relation, all diagonal pairs

(𝑎,𝑎), (𝑏,𝑏), (𝑐,𝑐), (𝑑,𝑑)

must be included.

There are

4C2 = 6

unordered pairs of distinct elements.

For each unordered pair {𝑢, 𝑣}, symmetry gives two possibilities:

• Include both (𝑢, 𝑣) and (𝑣, 𝑢), or • Exclude both.

Thus, each of the 6 unordered pairs has 2 independent choices.

Hence, the total number of reflexive and symmetric relations is

2⁶ = 64.

List all ordered pairs satisfying

2x ≤ 3y.

For x = 2:

(2,2), (2,3), (2,5), (2,7), (2,9) ⇒ 5 pairs.

For x = 3:

(3,2), (3,3), (3,5), (3,7), (3,9) ⇒ 5 pairs.

For x = 5:

(5,5), (5,7), (5,9) ⇒ 3 pairs.

For x = 7:

(7,5), (7,7), (7,9) ⇒ 3 pairs.

For x = 9:

(9,7), (9,9) ⇒ 2 pairs.

Hence,

l = 5 + 5 + 3 + 3 + 2 = 18.

To make the relation symmetric, add the reverse pair whenever it is missing.

The missing reverse pairs are:

(5,2), (7,2), (9,2), (5,3), (7,3), (9,3), (9,5).

Thus,

m = 7.

Therefore,

l + m = 18 + 7 = 25.

First, determine set A.

From

|x² − 10| ≤ 6,

we get

−6 ≤ x² − 10 ≤ 6,

which simplifies to

4 ≤ x² ≤ 16.

Hence,

2 ≤ |x| ≤ 4,

so

A = [−4, −2] ∪ [2, 4].

Now determine set B.

From

|x − 2| > 1,

we obtain

x − 2 < −1 or x − 2 > 1,

which gives

x < 1 or x > 3.

Thus,

B = (−∞, 1) ∪ (3, ∞).

Now check each option.

A ∪ B = (−∞, 1) ∪ [2, ∞), so Option A is incorrect.

B − A:

Remove [−4, −2] from (−∞, 1), giving

(−∞, −4) ∪ (−2, 1).

Remove [3, 4] from (3, ∞), leaving

(4, ∞).

Therefore,

B − A = (−∞, −4) ∪ (−2, 1) ∪ (4, ∞),

which matches Option B.

Also,

A − B = [2, 3],

not [2, 3),

and

A ∩ B = [−4, −2] ∪ (3, 4],

not [−4, −2] ∪ [3, 4].

Hence, Option B is correct.