Let A = {-3, -2, -1, 0, 1, 2, 3}. Let R be a relation on A defined by xRy if and only if $0\le x^2+2y\le 4$. Let $l$ be the number of elements in R and $m$ be the minimum number of elements required to be added in R to make it a reflexive relation. Then $l + m$ is equal to
18
Given set A = {-3, -2, -1, 0, 1, 2, 3} and relation R defined by $xRy \iff 0 \le x^2 + 2y \le 4$. We need to find $l$, the number of elements in R, and $m$, the minimum elements to add for R to be reflexive, then calculate $l + m$.
We check the condition $0 \le x^2 + 2y \le 4$ for all pairs $(x, y)$ where $x, y \in A$. This involves finding the range for $y$ for each $x$ in A that satisfies the inequalities.
| $x$ | $x^2$ | Condition $0 \le x^2+2y \le 4$ | Range for $y$ | Valid $y$ in A | Count | Pairs in R |
|---|---|---|---|---|---|---|
| -3 | 9 | $0 \le 9+2y \le 4$ | $y \le -2.5$ | {-3} | 1 | (-3, -3) |
| -2 | 4 | $0 \le 4+2y \le 4$ | $-2 \le y \le 0$ | {-2, -1, 0} | 3 | (-2, -2), (-2, -1), (-2, 0) |
| -1 | 1 | $0 \le 1+2y \le 4$ | $-0.5 \le y \le 1.5$ | {0, 1} | 2 | (-1, 0), (-1, 1) |
| 0 | 0 | $0 \le 0+2y \le 4$ | $0 \le y \le 2$ | {0, 1, 2} | 3 | (0, 0), (0, 1), (0, 2) |
| 1 | 1 | $0 \le 1+2y \le 4$ | $-0.5 \le y \le 1.5$ | {0, 1} | 2 | (1, 0), (1, 1) |
| 2 | 4 | $0 \le 4+2y \le 4$ | $-2 \le y \le 0$ | {-2, -1, 0} | 3 | (2, -2), (2, -1), (2, 0) |
| 3 | 9 | $0 \le 9+2y \le 4$ | $y \le -2.5$ | {-3} | 1 | (3, -3) |
Summing the counts for each $x$: $l = 1 + 3 + 2 + 3 + 2 + 3 + 1 = 15$.
A relation R is reflexive if $(x, x) \in R$ for every $x \in A$. We test the condition $0 \le x^2 + 2x \le 4$ for each $x \in A$ to see if the diagonal pairs are present.
The pairs needed for reflexivity that are missing from R are $(-1, -1)$, $(2, 2)$, and $(3, 3)$.
Therefore, the minimum number of elements to add to make R reflexive is $m = 3$.
We have calculated $l = 15$ (the number of elements in R) and $m = 3$ (the minimum elements needed for reflexivity).
The sum $l + m$ is $15 + 3 = 18$.
Let f and g be functions satisfying $f(x+y) = f(x)f(y), f(1) = 7$ and $g(x+y) = g(xy), g(1) = 1$, for all $x, y \in \mathbb{N}$. If $\sum_{x=1}^{n} \left(\frac{f(x)}{g(x)}\right) = 19607$, then n is equal to :
Let S be the set of the first 11 natural numbers. Then the number of elements in $A = \{B \subseteq S : n(B) \geq 2$ and the product of all elements of B is even is ________.