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

(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.

(46)

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

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

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

(10)