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SubjectsMathematics
समुच्चयों का कार्तीय गुणनफल
In Class 11 Mathematics, under Relations and Functions, students learn how to form the Cartesian product of two sets as the set of all possible ordered pairs (a, b), where a belongs to the first set and b to the second. The topic explains why the order of elements matters, how to represent products using roster form and diagrams, and how to find their number of elements. This foundation helps students describe relations as subsets of a Cartesian product and understand domain and codomain.
Correct answer: B. ({(r,p),(r,q),(s,p),(s,q)})
Explanation: In (B\times A), the first component is from (B) and the second from (A). Do not confuse (A\times B) with (B\times A).
Correct answer: A. Order matters in ordered pairs
Explanation: In ordered pairs, ((1,3)) and ((3,1)) are not the same. Hence (A\times B) is generally not equal to (B\times A).
Correct answer: B. (6)
Explanation: (|A\times B|=2\cdot1=2), then (|(A\times B)\times C|=2\cdot3=6). Cardinalities multiply in nested products too.
Correct answer: B. 3
Explanation: Check all three values of \(x\): \((-1)^2=1\) gives \((-1,1)\), \(0^2=0\) gives \((0,0)\), and \(1^2=1\) gives \((1,1)\). Thus, there are 3 ordered pairs. Option 6 is the total number of pairs in \(A\times B\), but not all of them satisfy \(x^2=y\). Exam tip: For this type of question, compute \(x^2\) for each \(x\) and verify whether the resulting \(y\) belongs to \(B\).
Correct answer: B. (6)
Explanation: For (a=2,3,4), the counts are (1,2,3), giving (6). For inequalities, count using the first component.
Correct answer: B. 4
Explanation: For the Cartesian product, |A×B|=|A|×|B|. Here, |A|=3 and |A×B|=12, so 3×|B|=12 and hence |B|=12÷3=4. The value 12 is the cardinality of the product, not of B. In an exam, first find |A| and divide the product cardinality by it.
Correct answer: C. (2^{mn})
Explanation: (|A\times B|=mn), so the number of subsets is (2^{mn}). First find the cardinality of the base set.
Correct answer: D. (64)
Explanation: A relation is any subset of (A\times B), and (|A\times B|=6). Hence the number of relations is (2^6=64).
Correct answer: B. (63)
Explanation: (|A\times B|=3\cdot2=6), so non-empty subsets are (2^6-1=63). Do not forget to subtract the empty set.
Correct answer: B. ({a,b})
Explanation: Direct answer: Option B, \\(B=\\{a,b\\}\\). A Cartesian product contains ordered pairs \\( (u,v) \\) where the first component comes from the first set and the second component comes from the second set. Since the first components are 1 and 2, they match A. The second components appearing are a and b, so B must be \\(\\{a,b\\}\\). Option A repeats A, not B. Option B correctly collects the second entries. Option C lists pairs, so it is a subset of the product rather than the set B. Option D mixes first and second components into one set and does not describe the second factor. The order matters: \\(A\\times B\\) is not generally the same as \\(B\\times A\\). Memory cue: in every ordered pair, first coordinate belongs to A and second coordinate belongs to B.
Correct answer: A. (B=\varnothing)
Explanation: If (A) is non-empty and (A\times B=\varnothing), then (B=\varnothing). A Cartesian product is empty if at least one set is empty.
Correct answer: A. (\varnothing)
Explanation: The first set is empty, so no ordered pair can be formed. Treat the empty set as a set with (0) elements.
Correct answer: B. (20)
Explanation: Since (R) can be at most the whole (A\times B), and (|A\times B|=20). If maximum elements are asked, do not write (2^{20}).
Correct answer: C. (24)
Explanation: For Cartesian product of three sets, the cardinality is (2\cdot3\cdot4=24). For ordered triples, multiply all three sizes.
Correct answer: C. (6)
Explanation: (B\cup C={3,4,5}), so (|A\times(B\cup C)|=2\cdot3=6). First evaluate the set inside the brackets.
Correct answer: A. ({(1,3),(2,3)})
Explanation: ((A\times B)\cap(A\times C)=A\times(B\cap C)), and (B\cap C={3}). Thus the pairs are ((1,3)) and ((2,3)).
Correct answer: B. (15)
Explanation: (|A\times B|=6), and choosing (2) elements gives (\binom{6}{2}=15). Exactly (2) elements means a combination.
Correct answer: B. (6)
Explanation: (|A\times B|=6), so there are (6) ways to choose a one-element relation. A singleton relation is just one ordered pair.
Correct answer: B. 8
Explanation: The sum of two numbers is even when they have the same parity—both must be even or both must be odd. Each set contains 2 even elements, \(\{2,4\}\), and 2 odd elements, \(\{1,3\}\). Hence, the number of favourable ordered pairs is \(2\times2+2\times2=8\). Exam tip: remember that an even sum requires equal parity, while an odd sum requires opposite parity. The value 16 would count every pair in \(A\times B\), including pairs with one even and one odd component.
Correct answer: B. (5)
Explanation: The pairs are ((1,5),(2,4),(3,3),(4,2),(5,1)), so there are (5). Reversed pairs count separately in ordered pairs.
Correct answer: B. (8)
Explanation: For (a=1,2,3,4), the counts are (4,2,1,1), totaling (8). In divisibility, check each (a) separately.
Correct answer: C. (3)
Explanation: The direct answer is option C, 3 pairs. A pair (a,b) belongs to A × B only when a comes from A and b comes from B. We must test the rule b = a^2 for each a in A. For a = 1, b = 1^2 = 1, and 1 is in B, giving (1,1). For a = 2, b = 2^2 = 4, and 4 is in B, giving (2,4). For a = 3, b = 3^2 = 9, and 9 is in B, giving (3,9). Thus there are three valid pairs. Option A, 1, misses two valid cases. Option B, 2, also stops before checking all three values of a. Option C, 3, correctly counts all successful checks. Option D, 6, is the total number of pairs in A × B, but not every pair satisfies b = a^2. A useful exam method is to calculate the required b for every a and then check whether that b belongs to B.
Correct answer: B. (6)
Explanation: The direct answer is option B: 6 pairs. We count by fixing a and then listing b values that make a + b < 3. If a = 0, then b < 3, so b = 0, 1, 2: 3 choices. If a = 1, then b < 2, so b = 0, 1: 2 choices. If a = 2, then b < 1, so b = 0: 1 choice. The total is 3 + 2 + 1 = 6. The valid pairs are (0,0), (0,1), (0,2), (1,0), (1,1), and (2,0). Option A, 5, misses one pair. Option B, 6, matches the complete count. Option C, 7, usually results from including a boundary value such as a + b = 3, but the condition is strictly less than 3, not less than or equal to 3. Option D, 8, overcounts still more pairs. Memory cue: for “less than,” do not include equality; count each row carefully.
Correct answer: B. (3)
Explanation: The diagonal pairs are ((1,1),(2,2),(3,3)), so the count is (3). For equal sets, the diagonal count is usually (|A|).
Correct answer: C. (3)
Explanation: The direct answer is option C, 3 pairs. The condition a/b = 2 can be rewritten by multiplying both sides by b: a = 2b. Now test each b in B. For b = 1, a = 2, which is in A, so (2,1) works. For b = 2, a = 4, which is in A, so (4,2) works. For b = 3, a = 6, which is in A, so (6,3) works. Therefore there are three valid ordered pairs. Option A, 1, counts only one pair and ignores the other two. Option B, 2, also misses one valid pair. Option C, 3, includes every allowed value of b and is correct. Option D, 4, has no fourth possible value because B contains only 1, 2, and 3; alternatively, checking a = 2, 4, 6 gives exactly the same three pairs. The important idea is that ordered pairs must satisfy both membership and the equation. Converting a fraction condition into a = 2b makes the counting easy.
