Reflexive relation

Step 1: The four diagonal pairs are compulsory. Step 2: Exactly (2) non-diagonal pairs are needed and ((1,2)) is already fixed. Step 3: The second non-diagonal pair can be chosen from the remaining (11), so there are (11) relations.

Step 1: On a five-element set, the number of reflexive relations is (2^{25-5}=2^{20}). Step 2: This includes the identity relation and the universal relation. Step 3: Removing both gives (2^{20}-2).

Step 1: Four diagonal pairs are compulsory. Step 2: To have (11) total pairs, choose (7) non-diagonal pairs. Step 3: The number of ways is (\binom{12}{7}).

Step 1: On a five-element set, (5) diagonal pairs are compulsory. Step 2: The relation has (18) total pairs. Step 3: The additional non-diagonal pairs are (18-5=13).

Step 1: (A=\mathcal{P}(S)) has (2^4=16) elements. Step 2: The number of reflexive relations on (A) is (2^{16^2-16}). Step 3: Since (16^2-16=240), the answer is (2^{240}).

Step 1: On the diagonal, (X\subset X) is false. Step 2: But (X\cap X=X) is true. Step 3: In the or condition, the second part includes every ((X,X)), so the relation is reflexive.

Step 1: Reflexivity requires ((1,1)) too. Step 2: The number (1) has no prime factor, so ((1,1)) will not belong to the relation. Step 3: Missing even one diagonal pair makes a relation non-reflexive.

Step 1: Every element of (A) is greater than (1). Step 2: Each such element has at least one prime factor, which it shares with itself. Step 3: Therefore all (6) diagonal pairs are in the relation.

Step 1: On the diagonal, put (b=a). Step 2: Then (\frac{b}{a}=\frac{a}{a}=1=2^0), which is an integral power of (2). Step 3: Hence every ((a,a)) belongs to the relation.

Step 1: For reflexivity, take ((a,a)). Step 2: For every (a), (a\mid a) is true, so the or condition is satisfied. Step 3: Self-divisibility directly proves reflexivity.

Step 1: On the diagonal, (a+b=2a) and (ab=a^2). Step 2: The condition becomes (2a=a^2), i.e. (a(a-2)=0). Step 3: In the given set, only (a=2) works, so there is one diagonal pair.

Step 1: A reflexive relation on this set needs five diagonal pairs. Step 2: The diagonal condition holds only for (a=2). Step 3: Therefore the remaining four diagonal pairs must be added.

Step 1: For ((a,a)), the left side is (a^2-a) and the right side is also (a^2-a). Step 2: A quantity is always equal to itself. Step 3: Therefore all diagonal pairs belong to the relation.

Step 1: Rewrite (a^2+b^2=2ab) as (a^2-2ab+b^2=0). Step 2: This is ((a-b)^2=0), which is true when (a=b). Step 3: Therefore every ((a,a)) belongs to the relation.

Step 1: On the diagonal, put (a=b), giving (a^2+a^2<2a^2). Step 2: This becomes (2a^2<2a^2), which is false. Step 3: No diagonal pair satisfies the condition, so the relation is not reflexive.

Step 1: To test reflexivity, take ((a,a)). Step 2: Then (a-a=0), and (0) is divisible by any non-zero number. Step 3: Therefore all diagonal pairs belong to the relation.

Step 1: For a composition evaluate the inner function first: \((f\circ g)(3)=f(g(3))\). Step 2: \(g(3)=3-1=2\). Step 3: \(f(2)=2^2+1=4+1=5\). Thus the correct value is 5. The option 4 is a common mistake if you forget the "+1" in \(f(x)\); other wrong choices arise from similar arithmetic slip-ups. Exam tip: always compute the inner function value first, then substitute into the outer function.

Substituting \(a\) for \(x\) in \(f(x)=2x+3\) gives \(f(a)=2a+3\). Thus, \(2a+3=11\), so \(2a=8\) and \(a=4\). Option C, 5, is incorrect because it would give \(f(5)=13\), not 11. Exam tip: substitute the given input into the function rule, solve the resulting equation, and verify the answer in the original function.

Substituting \(x=4\) into \(f(x)=kx-3\) gives \(f(4)=4k-3\). Since \(f(4)=17\), we get \(4k-3=17\), so \(4k=20\) and hence \(k=5\). Option C is the intermediate value of \(4k\), not the value of \(k\); dividing by 4 is still required. Exam tip: after substitution, isolate the unknown completely before selecting the final answer.

Step 1: (f(x)=(x+1)^2), so its range is ([0,\infty)). Step 2: The codomain is also ([0,\infty)), so every codomain element is attained. Step 3: In exams, first find the range and then compare it with the codomain.