sets relations and functions jee mains questions | JEE Main Mathematics Sets Relations And Functions

Q 11 :

The number of elements in the set $S = {(x, y, z) : x, y, z \in Z, x + 2 y + 3 z = 42, x, y, z \geq 0}$ equals _______ . [2024]

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(169)

$x + 2 y + 3 z = 42$

At $z = 0, x + 2 y = 42$

$y$ can take all value from 0 to 21

Hence 22 solutions.

Similarly,

at $z = 1, x + 2 y = 39$ $\to$ $20 solutions$

at $z = 2, x + 2 y = 36$ $\to$ $19 solutions$

at $z = 3, x + 2 y = 33$ $\to$ $17 solutions$

at $z = 4, x + 2 y = 30$ $\to$ $16 solutions$

at $z = 5, x + 2 y = 27$ $\to$ $14 solutions$

at $z = 6, x + 2 y = 24$ $\to$ $13 solutions$

at $z = 7, x + 2 y = 21$ $\to$ $11 solutions$

at $z = 8, x + 2 y = 18$ $\to$ $10 solutions$

at $z = 9, x + 2 y = 15$ $\to$ $8 solutions$

at $z = 10, x + 2 y = 12$ $\to$ $7 solutions$

at $z = 11, x + 2 y = 9$ $\to$ $5 solutions$

at $z = 12, x + 2 y = 6$ $\to$ $4 solutions$

at $z = 13, x + 2 y = 3$ $\to$ $2 solutions$

at $z = 14, x + 2 y = 0$ $\to$ $1 solutions$

Total cases = 169 solutions.

Q 12 :

Let the set $C = {(x, y) ∣ x^{2} - 2^{y} = 2023, x, y \in N}$ . Then $\sum_{(x, y) \in C} (x + y)$ is equal to ________ . [2024]

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(46)

$We have, x^{2} - 2^{y} = 2023 \Rightarrow 2^{y} = x^{2} - 2023$

$∴$ $Possible values of (x, y) are (45, 1)$

$S o, \sum_{(x, y) \in C} (x + y) = (45 + 1) = 46$

Q 13 :

A group of 40 students appeared in an examination of 3 subjects - Mathematics, Physics, and Chemistry. It was found that all students passed in at least one of the subjects, 20 students passed in Mathematics, 25 students passed in Physics, 16 students passed in Chemistry, at most 11 students passed in both Mathematics and Physics, atmost 15 students passed in both Physics and Chemistry, atmost 15 students passed in both Mathematics and Chemistry. The maximum number of students passed in all the three subjects is __________ . [2024]

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(10)

Q 14 :

Let $A = {(α, β) \in R \times R : | α - 1 | \leq 4 and | β - 5 | \leq 6}$ and $B = {(α, β) \in R \times R : 16 {(α - 2)}^{2} + 9 {(β - 6)}^{2} \leq 144}$ . Then [2025]

$A \subset B$
neither $A \subset B$ nor $B \subset A$
$B \subset A$
$A \cup B = {(x, y) : - 4 \leq x \leq 4, - 1 \leq y \leq 11}$

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(3)

$A : | α - 1 | \leq 4 and | β - 5 | \leq 6$

$\Rightarrow - 4 \leq α - 1 \leq 4 and - 6 \leq β - 5 \leq 6$

$\Rightarrow - 3 \leq α \leq 5 and - 1 \leq β \leq 11$

$B : 16 {(α - 2)}^{2} + 9 {(β - 6)}^{2} \leq 144$

$B : \frac{{(α - 2)}^{2}}{9} + \frac{{(β - 6)}^{2}}{16} \leq 1$

From Diagram, $B \subset A$ .

Q 15 :

Let A = {1, 2, 3, ..., 10} and $B = {\frac{m}{n} : m, n \in A, m < n and g c d (m, n) = 1}$ .

Then n(B) is equal to : [2025]

36
31
37
29

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(2)

We have, A = {1, 2, 3, ..., 10}

$B = {\frac{m}{n} : m, n \in A, m < n and g c d (m, n) = 1}$

For m = 1, n = 2, 3, ..., 10 $\Rightarrow$ 9 cases

m = 2, n = 3, 5, 7, 9 $\Rightarrow$ 4 cases

m = 3, n = 4, 5, 7, 8, 10 $\Rightarrow$ 5 cases

m = 4, n = 5, 7, 9 $\Rightarrow$ 3 cases

m = 5, n = 6, 7, 8, 9 $\Rightarrow$ 4 cases

m = 6, n = 7 $\Rightarrow$ 1 cases

m = 7, n = 8, 9, 10 $\Rightarrow$ 3 cases

m = 8, n = 9 $\Rightarrow$ 1 cases

m = 9, n = 10 $\Rightarrow$ 1 cases

$∴$ n(B) = 31.

Q 16 :

Let $S = {(m, n) : m, n \in {1, 2, 3, . . . . ., 50}}$ . If the number of elements $(m, n)$ in S such that $6^{m} + 9^{n}$ is a multiple of 5 is p and the number of elements $(m, n)$ in S such that $m + n$ is a square of a prime number is $q$ , then $p + q$ is equal to _______. [2026]

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(1333)

$S = {1, 2, 3, \dots, 50}$

$p = (6^{m} + 9^{n}) is divisible by 5$

$No. of ways$

$6^{m} = {(5 λ + 1)}^{m} = 5 k + 1$

$9^{n} = {(10 - 1)}^{n} =$ $10 μ - 1$ if $n$ is odd

$\Rightarrow n must be odd$

$10 μ + 1$ if $n$ is even

$\Rightarrow No. of ways = 50 \times 25 = 1250$

$q \Rightarrow (m + n) is square of a prime$

	$m + n = 4$	$m + n = 9$	$m + n = 25$	$m + n = 49$
No. of	$↓$	$↓$	$↓$	$↓$
ways	3	8	24	48

$q = 3 + 8 + 24 + 48 = 83$

$= p + q = 1250 + 83 = 1333$

Q 17 :

The number of elements in the set

$S = {x : x \in [0, 100] and \int_{0}^{x} t^{2} \sin (x - t) d t = x^{2}}$ is_____. [2026]

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16

Q 18 :

Let $A = {x : | x^{2} - 10 | \leq 6} and B = {x : | x - 2 | > 1} .$ Then [2026]

$A \cup B = (- \infty, 1] \cup (2, \infty)$
$A \cap B = [- 4, - 2] \cup [3, 4]$
$A - B = [2, 3)$
$B - A = (- \infty, - 4) \cup (- 2, 1) \cup (4, \infty)$

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