Symmetric relation

Step 1: From (a+b=2b), we get (a=b). Step 2: Thus (R) contains only diagonal pairs such as ((1,1),(2,2),(3,3),(4,4)). Each is its own reverse. Step 3: Sometimes a condition that looks different may still define only a diagonal relation.

For any relation, R⁻¹ is obtained by reversing ordered pairs. Since R is symmetric, reversing its pairs produces no new pairs and gives R⁻¹ = R. Substituting this into the expression yields R ∪ R⁻¹ = R ∪ R. The union of a set with itself is the set itself, so the expression equals R. It is not necessarily empty or universal.

The condition defining R is a+b=5. If (a,b) satisfies it, then reversing the components gives b+a=5 because addition is commutative; hence (b,a) also belongs to R, so R is symmetric. Reflexivity would require (a,a) for every a in A. But for a=1, 1+1≠5, and similarly the diagonal pairs are absent. Therefore R is symmetric but not reflexive.

For reflexivity on A, all diagonal pairs (1,1), (2,2), and (3,3) must be present; all three are listed, so R is reflexive. For symmetry, every non-diagonal pair must have its reverse. The pair (1,2) is accompanied by (2,1), and diagonal pairs reverse to themselves. Thus R satisfies both properties, making option A correct.

To test symmetry, assume (a,b) ∈ R. By definition, a+b≤5. Since addition is commutative, b+a has exactly the same value as a+b, so b+a≤5 as well. Therefore (b,a) ∈ R whenever (a,b) ∈ R, which proves that R is symmetric. The relation is not empty—for example, (1,1) belongs to it—and the question does not require testing reflexivity.

A relation R on A is reflexive when every diagonal pair (a,a) belongs to R, and it is symmetric when (a,b) in R implies (b,a) in R. Here (1,1), (2,2), and (3,3) are all present, so R is reflexive. The pair (1,2) occurs together with its reverse (2,1), and diagonal pairs reverse to themselves. Hence R is symmetric as well. Therefore option A is correct; options B and C contradict the listed pairs, while D does not state the required combined property.

The governing test for symmetry is: whenever (a,b) belongs to R, the reversed pair (b,a) must also belong to R. If |a-b|=1, then |b-a|=|-(a-b)|=|a-b|=1. Thus every pair satisfying the condition has its reverse satisfying the same condition. For example, (1,2) and (2,1), or (3,4) and (4,3), occur together. Therefore R is symmetric. It is not reflexive because |a-a|=0, and it is not universal because many pairs have distance other than 1.

The governing concept is symmetry: a relation S is symmetric when (a,b) ∈ S implies (b,a) ∈ S. If (a,b) belongs to R, then (b,a) belongs to R⁻¹; if (a,b) belongs to R⁻¹, then (b,a) belongs to R. Thus every pair in R ∪ R⁻¹ brings its reverse pair into the same union. Therefore option A is always true. Reflexivity and transitivity are not guaranteed, and the union need not be empty.

For a symmetric relation on an n-element set, each diagonal pair (a,a) can be selected independently, giving n choices of position. For each unordered pair of distinct elements, either both ordered pairs are included or both are excluded. There are n(n−1)/2 such unordered pairs, so the total number of independent choices is n + n(n−1)/2 = n(n+1)/2. For n = 3, this is 3 + 3 = 6, giving 2⁶ relations. Hence option B is correct.

A symmetric relation is determined by independent choices on the diagonal and on unordered pairs of distinct elements. For n elements, the number of such choices is n + n(n−1)/2 = n(n+1)/2. With n = 5, the exponent is 5×6/2 = 15. Each choice is binary: include the corresponding diagonal pair, or include both members of an off-diagonal pair; alternatively, omit it. Therefore the number of symmetric relations is 2¹⁵, so option A is correct. The exponents 20 and 25 do not count the required paired positions.

On a three-element set, a symmetric relation has 3 diagonal choices and 3 unordered off-diagonal choices, for a total of 6 independent binary choices. The condition that (1,1) is absent fixes one diagonal choice: it must be excluded rather than chosen. The remaining diagonal pairs (2,2) and (3,3), together with the three off-diagonal pair groups, remain independently selectable. Thus five binary choices remain, giving 2⁵ symmetric relations. Option A is correct; 2⁶ would allow (1,1), while the smaller powers omit valid choices.

Substitute \(x=a\) into the function: \(f(a)=4a-1\). Since \(f(a)=19\), we have \(4a-1=19\). Adding 1 to both sides gives \(4a=20\), and dividing by 4 gives \(a=5\). Option B is incorrect because \(a=4\) would give \(f(4)=16-1=15\), not 19. Exam tip: Verify the answer by substituting it back into the original function.