The relation R = (x, y) : x, y in Z and x + y is even$R = \{(x, y) : x, y \in \mathbb{Z} \text{ and } x + y \text{ is even}\}$ is:

Related Formula

Given functional property equation:

Core Logic

Substitute x = y = 0$x = y = 0$ into the property equation: f(0) = 2f(0) + 1 f(0) = -1$f(0) = 2f(0) + 1 \implies f(0) = -1$. Since f(0) = b$f(0) = b$, we instantly find b = -1$b = -1$.

Step 1: Extracting Parameter Values

Substitute y = -x$y = -x$ into the property equation:

Matching coefficients for x²$x^2$ gives:

Therefore, the absolute functional identity is:

Step 2: Computing the Target Series

Rewriting using common denominators:

Evaluating for i=1$i=1$ to 5$5$:

Pattern Recognition

Substituting standard points like 0$0$ and -x$-x$ decouples symmetric multi-variable systems with maximum efficiency.

Chapter Mix

Class 12 Maths: Relations and Functions

Related Formula

• Elements in a relation satisfy the exact range constraint.
• Reflexive criteria: For every x in A$x \in A$, (x, x) in R$(x, x) \in R$.

Core Logic

Rewrite the inequality to isolate variables systematically:

Test every valid value of y in A$y \in A$ to discover valid integer values for x$x$

• y = -3 6 ≤ x² ≤ 10 x in -3, 3$y = -3 \implies 6 \le x^2 \le 10 \implies x \in \{-3, 3\}$
• y = -2 4 ≤ x² ≤ 8 x in -2, 2$y = -2 \implies 4 \le x^2 \le 8 \implies x \in \{-2, 2\}$
• y = -1 2 ≤ x² ≤ 6 x in -2, 2$y = -1 \implies 2 \le x^2 \le 6 \implies x \in \{-2, 2\}$
• y = 0 0 ≤ x² ≤ 4 x in -2, -1, 0, 1, 2$y = 0 \implies 0 \le x^2 \le 4 \implies x \in \{-2, -1, 0, 1, 2\}$
• y = 1 -2 ≤ x² ≤ 2 x in -1, 0, 1$y = 1 \implies -2 \le x^2 \le 2 \implies x \in \{-1, 0, 1\}$
• y = 2 -4 ≤ x² ≤ 0 x in 0$y = 2 \implies -4 \le x^2 \le 0 \implies x \in \{0\}$
• y = 3 -6 ≤ x² ≤ -2 No real x exists$y = 3 \implies -6 \le x^2 \le -2 \implies \text{No real } x \text{ exists}$

Step 1: Listing set elements and counting

Compile all distinct matching coordinate pairs (x,y)$(x,y)$ into set R$R$ [cite: 1264]:

Counting elements gives l = 15$l = 15$ To make the relation reflexive, the pairs (-3,-3), (-2,-2), (-1,-1), (0,0), (1,1), (2,2), (3,3)$(-3,-3), (-2,-2), (-1,-1), (0,0), (1,1), (2,2), (3,3)$ must all belong to R$R$. Checking missing elements [cite: 1267]:

Sum of variables

Pattern Recognition

Isolating terms explicitly via a variable-by-variable bounded testing grid avoids missing distinct coordinate boundary values.

Chapter Mix

Class 11 Mathematics: Relations and Functions

Related Formula

• For (g(x))$\log(g(x))$, we require g(x) > 0$g(x) > 0$.
• For ⁻¹(h(x))$\sin^{-1}(h(x))$, we require -1 ≤ h(x) ≤ 1$-1 \le h(x) \le 1$.

Core Logic

Evaluate constraints independently [cite: 1307, 1309]:

Constraint 1 (Logarithmic Argument): [cite: 1307] (2x-3)/(4x+5) > 0 x in (-∞, -(5)/(4)) ((3)/(2), ∞)$$\frac{2x-3}{4x+5} > 0 \implies x \in \left(-\infty, -\frac{5}{4}\right) \cup \left(\frac{3}{2}, \infty\right)$$ [cite: 1309]

Constraint 2 (Arcsine Argument): [cite: 1307] -1 ≤ (3x+4)/(2-x) ≤ 1$$-1 \le \frac{3x+4}{2-x} \le 1$$ [cite: 1309]

Step 1: Solving the Arcsine inequalities

Split inequality into separate conditional frames [cite: 1311]: Left frame:

Right frame:

Intersecting both sets gives [cite: 1311]: x in [-3, -(1)/(2)]$$x \in \left[-3, -\frac{1}{2}\right]$$ [cite: 1311]

Step 2: Final Intersection and Value Solving

Intersect Log constraint with Arcsine constraint solution range [cite: 1311]: x in [-3, -(1)/(2)] [(-∞, -(5)/(4)) ((3)/(2), ∞)] = [-3, -(5)/(4))$$x \in \left[-3, -\frac{1}{2}\right] \cap \left[\left(-\infty, -\frac{5}{4}\right) \cup \left(\frac{3}{2}, \infty\right)\right] = \left[-3, -\frac{5}{4}\right)$$ [cite: 1311]

Thus, identify parameters [cite: 1312]: α = -3, β = -(5)/(4)$$\alpha = -3, \quad \beta = -\frac{5}{4}$$ [cite: 1312]

Compute the requested expression value [cite: 1312]: α² + 4β = (-3)² + 4(-(5)/(4)) = 9 - 5 = 4$$\alpha^2 + 4\beta = (-3)^2 + 4\left(-\frac{5}{4}\right) = 9 - 5 = 4$$ [cite: 1312]

Pattern Recognition

When dealing with fractional variables inside boundaries, flipping inequalities according to denominator signs prevents fatal zone misinterpretations.

Chapter Mix

Class 12 Mathematics: Relations and Functions

Core Logic

The relation condition is 2x - y = 0$2x - y = 0$ or 2x - y = 1$2x - y = 1$ where x, y in A$x, y \in A$.

Case 1: 2x - y = 0 y = 2x$2x - y = 0 \implies y = 2x$. Possible pairs in A × A$A \times A$ are:

Case 2: 2x - y = 1 y = 2x - 1$2x - y = 1 \implies y = 2x - 1$. Possible pairs in A × A$A \times A$ are:

Combining both subsets, the total relation set R$R$ contains:

Hence, the number of existing elements l = 7$l = 7$.

Step 1: Elements to add for Reflexivity

For a relation to be reflexive on set A$A$, it must contain (x,x)$(x,x)$ for all 7 elements of A$A$.

Currently, R$R$ contains (0,0), (1,1)$\{(0,0), (1,1)\}$.

Missing diagonal elements are (-3,-3), (-2,-2), (-1,-1), (2,2), (3,3)$\{(-3,-3), (-2,-2), (-1,-1), (2,2), (3,3)\}$.

Therefore, the minimum number of elements to add for reflexivity is m = 5$m = 5$.

Step 2: Elements to add for Symmetry

For a relation to be symmetric, if (x,y) in R$(x,y) \in R$, then (y,x)$(y,x)$ must also belong to R$R$.

Let's check the non-diagonal elements currently in R$R$:

• (1,2) in R$(1,2) \in R \implies$ need (2,1)$(2,1)$
• (-1,-2) in R$(-1,-2) \in R \implies$ need (-2,-1)$(-2,-1)$
• (0,-1) in R$(0,-1) \in R \implies$ need (-1,0)$(-1,0)$
• (2,3) in R$(2,3) \in R \implies$ need (3,2)$(3,2)$
• (-1,-3) in R$(-1,-3) \in R \implies$ need (-3,-1)$(-3,-1)$

None of these reverse pairs are currently in R$R$. Thus, we must add exactly 5 elements to ensure symmetry, giving n = 5$n = 5$.

Step 3: Final Computation

Based on the official valuation tracking, the required evaluation metric simplifies to:

Pattern Recognition

To quickly count elements needed for reflexivity, subtract the number of identity pairs already present from the total cardinality of the set. For symmetry, find all elements where x ≠ y$x \neq y$ and check if their mirrors are absent.

Chapter Mix

Class 12 Mathematics: Relations and Functions