Practice focused topic-wise MCQs with answers and explanations for quick revision and exam preparation.
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C. (8)
(B) में (10) अवयव हैं और \(A\cap B\) में (35) के (2) गुणज हैं। इसलिए \(A'\cap B\) में (10-2=8) अवयव हैं। / (B) has (10) elements, and \(A\cap B\) has (2) multiples of (35). Therefore \(A'\cap B\) has (10-2=8) elements.
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If \(U={1,2,\ldots,24}\), (A'={1,5,7,11,13,17,19,23}), how many elements are in (A)?
C. (16)
(|U|=24) और (|A'|=8), इसलिए (|A|=24-8=16)। पूरक की संख्या कुल में से घटाकर मिलती है। / (|U|=24) and (|A'|=8), so (|A|=24-8=16). The count of a set is found by subtracting its complement from the total.
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If \(U=\mathbb{R}\), \(A={x:x^2-4x-12<0}\), what is (A')?
A. (\(-\infty,-2]\cup[6,\infty\))
\(x^2-4x-12<0\Rightarrow -2<x<6\)। इसका पूरक \(x\le -2\) या \(x\ge 6\), यानी (\(-\infty,-2]\cup[6,\infty\)) है।
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If \(U={1,2,\ldots,18}\), \(A=\{2,4,6,8,10,12\}\) and \(B=\{6,9,12,15,18\}\), what is ((B-A)')?
A. ({1,2,3,4,5,6,7,8,10,11,12,13,14,16,17})
(B-A={9,15,18}), इसलिए इसका पूरक (U-{9,15,18}) है। विकल्प (A) वही समुच्चय देता है। / (B-A={9,15,18}), so its complement is (U-{9,15,18}). Option (A) gives exactly that set.
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\(If (U={1,2,\ldots,144}) and (A={x:x\) is not divisible by 12}), what is (|A'|)?
C. (12)
(A') वे संख्याएँ हैं जो (12) से विभाज्य हैं। (144) तक (12) के \(\frac{144}{12}=12\) गुणज हैं। / (A') is the set of numbers divisible by (12). Up to (144), there are \(\frac{144}{12}=12\) multiples of (12).
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If \(A\triangle B\) is the symmetric difference, what does (\(A\triangle B\)') represent?
A. अवयव दोनों में हैं या दोनों में नहीं हैंElements are in both sets or in neither set
\(A\triangle B\) में वे अवयव होते हैं जो ठीक एक समुच्चय में हों। उसका पूरक वे अवयव देता है जो दोनों में हों या दोनों में न हों। / \(A\triangle B\) contains elements in exactly one set. Its complement gives elements that are in both or in neither.
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\(If (U={1,2,\ldots,30}), (A={x:x\) is even\(}) and (B={x:x\) is prime\(}), what is (|A'\cap B'|)\)?
A. (8)
(A') विषम संख्याएँ हैं और (B') अभाज्य नहीं संख्याएँ हैं। (1) से (30) तक विषम अभाज्य नहीं संख्याएँ ({1,9,15,21,25,27}) नहीं, साथ में (? ) नहीं; सही सूची ({1,9,15,21,25,27}) है, इसलिए संख्या (6) है। / (A') is the set of odd numbers and (B') is the set of non-prime numbers. The odd non-primes from (1) to (30) are ({1,9,15,21,25,27}), so the count is (6).
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If \(U=\mathbb{R}\), (A=(-\infty,0]\cup\(5,\infty\)), what is (A')?
A. ((0,5])
(0) (A) में है और (5) (A) में नहीं है। इसलिए पूरक ((0,5]) होगा। / (0) is in (A), while (5) is not in (A). Therefore the complement is ((0,5]).
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If \(A'\cap B'=A'\), which conclusion is always true?
A. \(A'\subseteq B'\)
यदि \(A'\cap B'=A'\), तो (A') का हर अवयव (B') में है। इसलिए \(A'\subseteq B'\) है। / If \(A'\cap B'=A'\), every element of (A') lies in (B'). Hence \(A'\subseteq B'\).
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If (\(A\cup B'\)\cap\(A'\cup B'\)=B'), which simplification shows it correctly?
A. (B'\cup\(A\cap A'\)=B')
वितरण नियम से (\(A\cup B'\)\cap\(A'\cup B'\)=B'\cup\(A\cap A'\))। चूंकि \(A\cap A'=\varnothing\), परिणाम (B') है। / By distributive law, (\(A\cup B'\)\cap\(A'\cup B'\)=B'\cup\(A\cap A'\)). Since \(A\cap A'=\varnothing\), the result is (B').
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