(A) has (18) elements, (B) has (8), and \(A\cap B\) has (2) multiples of (36), so \(|A\cup B|=24\). Hence (|\(A\cup B\)'|=72-24=48).
Step 2
Why this answer is correct
The correct answer is A. (48). (A) has (18) elements, (B) has (8), and \(A\cap B\) has (2) multiples of (36), so \(|A\cup B|=24\). Hence (|\(A\cup B\)'|=72-24=48).
Step 3
Exam Tip
(A) में (18), (B) में (8) और \(A\cap B\) में (36) के (2) गुणज हैं, इसलिए \(|A\cup B|=18+8-2=24\)। अतः (|\(A\cup B\)'|=72-24=48)।
\(x^2-4x\le 0\Rightarrow 0\le x\le 4\), so \(A=\{0,1,2,3,4\}\). The remaining integers of (U) form (A').
Step 2
Why this answer is correct
The correct answer is A. ({-8,-7,-6,-5,-4,-3,-2,-1,5,6,7,8}). \(x^2-4x\le 0\Rightarrow 0\le x\le 4\), so \(A=\{0,1,2,3,4\}\). The remaining integers of (U) form (A').
Step 3
Exam Tip
\(x^2-4x\le 0\Rightarrow 0\le x\le 4\), इसलिए \(A=\{0,1,2,3,4\}\)। (U) के बाकी पूर्णांक (A') बनाते हैं।
(A) includes (-3) and (10), but excludes (4) and (7). Therefore the complement is (\(-\infty,-3\)\cup[4,7]\cup\(10,\infty\)).
Step 2
Why this answer is correct
The correct answer is A. (\(-\infty,-3\)\cup[4,7]\cup\(10,\infty\)). (A) includes (-3) and (10), but excludes (4) and (7). Therefore the complement is (\(-\infty,-3\)\cup[4,7]\cup\(10,\infty\)).
Step 3
Exam Tip
(A) में (-3) और (10) शामिल हैं, पर (4) और (7) शामिल नहीं हैं। इसलिए पूरक (\(-\infty,-3\)\cup[4,7]\cup\(10,\infty\)) है।
\(A\cap B=[-1,2]\), so its complement is (\(-\infty,-1\)\cup\(2,\infty\)). Watch the endpoints in interval questions.
Step 2
Why this answer is correct
The correct answer is A. (\(-\infty,-1\)\cup\(2,\infty\)). \(A\cap B=[-1,2]\), so its complement is (\(-\infty,-1\)\cup\(2,\infty\)). Watch the endpoints in interval questions.
Step 3
Exam Tip
\(A\cap B=[-1,2]\), इसलिए उसका पूरक (\(-\infty,-1\)\cup\(2,\infty\)) है। अंतरालों में सीमा बिंदुओं पर ध्यान दें।
By De Morgan's law, (\(A'\cap B\)'=(A')'\cup B'=A\cup B'). When taking complement, \(\cap\) changes to \(\cup\).
Step 2
Why this answer is correct
The correct answer is A. \(A\cup B'\). By De Morgan's law, (\(A'\cap B\)'=(A')'\cup B'=A\cup B'). When taking complement, \(\cap\) changes to \(\cup\).
Step 3
Exam Tip
डी मॉर्गन नियम से (\(A'\cap B\)'=(A')'\cup B'=A\cup B')। पूरक लेते समय \(\cap\) बदलकर \(\cup\) होता है।
\(|x-3|<5\Rightarrow -2<x<8\), so the complement has \(x\le -2\) or \(x\ge 8\). The complement of a strict inequality includes equality.
Step 2
Why this answer is correct
The correct answer is A. (\(-\infty,-2]\cup[8,\infty\)). \(|x-3|<5\Rightarrow -2<x<8\), so the complement has \(x\le -2\) or \(x\ge 8\). The complement of a strict inequality includes equality.
Step 3
Exam Tip
\(|x-3|<5\Rightarrow -2<x<8\), इसलिए पूरक में \(x\le -2\) या \(x\ge 8\) होगा। सख्त असमानता का पूरक बराबरी जोड़ता है।
(\(A\cup B\)'=A'\cap B'), and \({b,e,i}\cap{a,d,j}=\varnothing\). Apply De Morgan directly to the given complements.
Step 2
Why this answer is correct
The correct answer is A. \(\varnothing\). (\(A\cup B\)'=A'\cap B'), and \({b,e,i}\cap{a,d,j}=\varnothing\). Apply De Morgan directly to the given complements.
Step 3
Exam Tip
(\(A\cup B\)'=A'\cap B') और \({b,e,i}\cap{a,d,j}=\varnothing\)। दिए गए पूरकों पर सीधे डी मॉर्गन लगाएं।
(0) and (10) are in (A), while (4) and (7) are not in (A). So the complement is (\(-\infty,0\)\cup[4,7]\cup\(10,\infty\)).
Step 2
Why this answer is correct
The correct answer is A. (\(-\infty,0\)\cup[4,7]\cup\(10,\infty\)). (0) and (10) are in (A), while (4) and (7) are not in (A). So the complement is (\(-\infty,0\)\cup[4,7]\cup\(10,\infty\)).
Step 3
Exam Tip
(0) और (10) (A) में हैं, जबकि (4) और (7) (A) में नहीं हैं। इसलिए पूरक (\(-\infty,0\)\cup[4,7]\cup\(10,\infty\)) है।
\(A\cap A'\) is always \(\varnothing\), so it cannot be (U) when \(U\neq\varnothing\). A set and its complement are disjoint.
Step 2
Why this answer is correct
The correct answer is D. \(A\cap A'=U\). \(A\cap A'\) is always \(\varnothing\), so it cannot be (U) when \(U\neq\varnothing\). A set and its complement are disjoint.
Step 3
Exam Tip
\(A\cap A'\) हमेशा \(\varnothing\) होता है, इसलिए यह (U) नहीं हो सकता जब \(U\neq\varnothing\)। पूरक और मूल समुच्चय असंयुक्त होते हैं।
(A') is the set of numbers divisible by (7). Up to (100), there are \(\left\lfloor \frac{100}{7}\right\rfloor=14\) multiples of (7).
Step 2
Why this answer is correct
The correct answer is B. (14). (A') is the set of numbers divisible by (7). Up to (100), there are \(\left\lfloor \frac{100}{7}\right\rfloor=14\) multiples of (7).
Step 3
Exam Tip
(A') वे संख्याएँ हैं जो (7) से विभाज्य हैं। (100) तक (7) के \(\left\lfloor \frac{100}{7}\right\rfloor=14\) गुणज हैं।
The symmetric difference contains elements in exactly one set. Its complement contains elements in both or in neither, that is (\(A\cap B\)\cup\(A'\cap B'\)).
Step 2
Why this answer is correct
The correct answer is A. \(\(A\cap B\)\cup\(A'\cap B'\)). The symmetric difference contains elements in exactly one set. Its complement contains elements in both or in neither, that is (\(A\cap B\)\cup\(A'\cap B'\)).
Step 3
Exam Tip
सममित अंतर में वे अवयव हैं जो केवल एक समुच्चय में हों। उसका पूरक वे अवयव हैं जो दोनों में हों या दोनों में न हों, यानी (\(A\cap B\)\cup\(A'\cap B'\))।
(A') is the set of odd numbers and (B') is the set of non-prime numbers. The correct odd non-primes in (U) are ({1,9,15}), so the count is (3).
Step 2
Why this answer is correct
The correct answer is B. (6). (A') is the set of odd numbers and (B') is the set of non-prime numbers. The correct odd non-primes in (U) are ({1,9,15}), so the count is (3).
Step 3
Exam Tip
(A') विषम संख्याएँ हैं और (B') अभाज्य नहीं संख्याएँ हैं। इनमें ({1,9,15,21}) नहीं बल्कि (U) में ({1,9,15}) के साथ ({? }) सोचने की गलती न करें; सही समुच्चय ({1,9,15}) और (? ) नहीं, बल्कि ({1,9,15}) केवल (3) हैं।
Factoring (B') gives (\(A\cap B'\)\cup\(A'\cap B'\)=B'\cap\(A\cup A'\)). Since \(A\cup A'=U\), the result is (B').
Step 2
Why this answer is correct
The correct answer is A. (B'\cap\(A\cup A'\)=B'). Factoring (B') gives (\(A\cap B'\)\cup\(A'\cap B'\)=B'\cap\(A\cup A'\)). Since \(A\cup A'=U\), the result is (B').
Step 3
Exam Tip
साझा (B') लेने पर (\(A\cap B'\)\cup\(A'\cap B'\)=B'\cap\(A\cup A'\))। चूंकि \(A\cup A'=U\), परिणाम (B') है।
\(A\cap B\) contains multiples of (\operatorname{lcm}(6,10)=30), namely (30,60). Hence the complement has (60-2=58) elements.
Step 2
Why this answer is correct
The correct answer is C. (58). \(A\cap B\) contains multiples of (\operatorname{lcm}(6,10)=30), namely (30,60). Hence the complement has (60-2=58) elements.
Step 3
Exam Tip
\(A\cap B\) में (\operatorname{lcm}(6,10)=30) के गुणज हैं, यानी (30,60)। इसलिए पूरक में (60-2=58) अवयव हैं।
In natural numbers, (1) is neither prime nor composite, so it belongs to the complement of composite numbers. (A') contains (1) and all prime numbers.
Step 2
Why this answer is correct
The correct answer is A. (1) और अभाज्य संख्याएँ / (1) and prime numbers. In natural numbers, (1) is neither prime nor composite, so it belongs to the complement of composite numbers. (A') contains (1) and all prime numbers.
Step 3
Exam Tip
प्राकृतिक संख्याओं में (1) न अभाज्य है न संयुक्त, इसलिए वह संयुक्त संख्याओं के पूरक में आएगा। (A') में (1) और सभी अभाज्य संख्याएँ होंगी।
(A'\cup B'=\(A\cap B\)'), and \(A\cap B=(2,5)\). Therefore the complement is (\(-\infty,2]\cup[5,\infty\)).
Step 2
Why this answer is correct
The correct answer is A. (\(-\infty,2]\cup[5,\infty\)). (A'\cup B'=\(A\cap B\)'), and \(A\cap B=(2,5)\). Therefore the complement is (\(-\infty,2]\cup[5,\infty\)).
Step 3
Exam Tip
(A'\cup B'=\(A\cap B\)') और \(A\cap B=(2,5)\) है। इसलिए पूरक (\(-\infty,2]\cup[5,\infty\)) होगा।
(A') has (27-9=18) elements and \(B\subseteq A\), so \(A'\cap B=\varnothing\). (B) has (3) elements, hence the total is (18+3=21).
Step 2
Why this answer is correct
The correct answer is D. (21). (A') has (27-9=18) elements and \(B\subseteq A\), so \(A'\cap B=\varnothing\). (B) has (3) elements, hence the total is (18+3=21).
Step 3
Exam Tip
(A') में (27-9=18) अवयव हैं और \(B\subseteq A\), इसलिए \(A'\cap B=\varnothing\)। (B) में (3) अवयव हैं, अतः कुल (18+3=21) है।
\(A\cup B=(1,7)\) because the intervals join at (4), and (7) is not included. Hence the complement is (\(-\infty,1]\cup[7,\infty\)).
Step 2
Why this answer is correct
The correct answer is A. (\(-\infty,1]\cup[7,\infty\)). \(A\cup B=(1,7)\) because the intervals join at (4), and (7) is not included. Hence the complement is (\(-\infty,1]\cup[7,\infty\)).
Step 3
Exam Tip
\(A\cup B=(1,7)\) क्योंकि (4) दोनों ओर से जुड़ा है और (7) शामिल नहीं है। इसलिए पूरक (\(-\infty,1]\cup[7,\infty\)) है।
By De Morgan's law, (\(A\cup B\)'=A'\cap B') is true in all cases. The condition \(A\cap B=\varnothing\) is extra information here.
Step 2
Why this answer is correct
The correct answer is A. \(A'\cap B'\). By De Morgan's law, (\(A\cup B\)'=A'\cap B') is true in all cases. The condition \(A\cap B=\varnothing\) is extra information here.
Step 3
Exam Tip
डी मॉर्गन नियम के अनुसार (\(A\cup B\)'=A'\cap B') हर स्थिति में सत्य है। \(A\cap B=\varnothing\) यहां अतिरिक्त जानकारी है।
(A') is the set of even numbers, and (B') contains numbers not in the prime list. Their intersection is ({4,6,8,10,12,14}).
Step 2
Why this answer is correct
The correct answer is A. ({4,6,8,10,12,14}). (A') is the set of even numbers, and (B') contains numbers not in the prime list. Their intersection is ({4,6,8,10,12,14}).
Step 3
Exam Tip
(A') सम संख्याएँ हैं और (B') वे संख्याएँ हैं जो अभाज्य सूची में नहीं हैं। उनका प्रतिच्छेद ({4,6,8,10,12,14}) है।
The complement changes the endpoint status of (0), (2), (5), and (9). The correct complement is (\(-\infty,0]\cup[2,5\)\cup[9,\infty)).
Step 2
Why this answer is correct
The correct answer is A. (\(-\infty,0]\cup[2,5\)\cup[9,\infty)). The complement changes the endpoint status of (0), (2), (5), and (9). The correct complement is (\(-\infty,0]\cup[2,5\)\cup[9,\infty)).
Step 3
Exam Tip
पहले भाग के पूरक में (0) और (2) की स्थिति बदलती है, और दूसरे भाग में (5) बाहर नहीं आएगा पर (9) आएगा। सही पूरक (\(-\infty,0]\cup[2,5\)\cup[9,\infty)) है।
(A'\cup B'=\(A\cap B\)'). If this is (U), then the complement of \(A\cap B\) is (U), so \(A\cap B=\varnothing\).
Step 2
Why this answer is correct
The correct answer is A. \(A\cap B=\varnothing\). (A'\cup B'=\(A\cap B\)'). If this is (U), then the complement of \(A\cap B\) is (U), so \(A\cap B=\varnothing\).
Step 3
Exam Tip
(A'\cup B'=\(A\cap B\)')। यदि यह (U) है, तो \(A\cap B\) का पूरक (U) है, इसलिए \(A\cap B=\varnothing\)।
From (1) to (21), there are (11) odd numbers, and (3,9,15,21) are divisible by (3). So odd elements in (A') are (11-4=7).
Step 2
Why this answer is correct
The correct answer is B. (7). From (1) to (21), there are (11) odd numbers, and (3,9,15,21) are divisible by (3). So odd elements in (A') are (11-4=7).
Step 3
Exam Tip
(1) से (21) तक (11) विषम संख्याएँ हैं और उनमें (3,9,15,21) (3) से विभाज्य हैं। इसलिए (A') में विषम अवयव (11-4=7) हैं।
\(x^2-2x-8\le 0\Rightarrow -2\le x\le 4\). Its complement is (x<-2) or (x>4), that is (\(-\infty,-2\)\cup\(4,\infty\)).
Step 2
Why this answer is correct
The correct answer is A. (\(-\infty,-2\)\cup\(4,\infty\)). \(x^2-2x-8\le 0\Rightarrow -2\le x\le 4\). Its complement is (x<-2) or (x>4), that is (\(-\infty,-2\)\cup\(4,\infty\)).
Step 3
Exam Tip
\(x^2-2x-8\le 0\Rightarrow -2\le x\le 4\)। इसका पूरक (x<-2) या (x>4), यानी (\(-\infty,-2\)\cup\(4,\infty\)) है।
एक कक्षा के (80) विद्यार्थियों में (46) क्रिकेट खेलते हैं, (38) फुटबॉल खेलते हैं और (20) दोनों खेलते हैं। कोई भी खेल न खेलने वाले विद्यार्थियों की संख्या कितनी है?
\(|C\cup F|=46+38-20=64\), so those playing neither are (|\(C\cup F\)'|=80-64=16). Read neither as the complement of the union.
Step 2
Why this answer is correct
The correct answer is B. (16). \(|C\cup F|=46+38-20=64\), so those playing neither are (|\(C\cup F\)'|=80-64=16). Read neither as the complement of the union.
Step 3
Exam Tip
\(|C\cup F|=46+38-20=64\), इसलिए न कोई खेल खेलने वाले (|\(C\cup F\)'|=80-64=16) हैं। neither को संघ के पूरक के रूप में पढ़ें।
