Removing \(A\setminus B\) from \(A\cup B\) leaves all elements of (B). In such questions, view separate Venn diagram regions.
Step 2
Why this answer is correct
The correct answer is A. (B). Removing \(A\setminus B\) from \(A\cup B\) leaves all elements of (B). In such questions, view separate Venn diagram regions.
Step 3
Exam Tip
\(A\cup B\) से \(A\setminus B\) हटाने पर (B) के सभी तत्व बचते हैं। ऐसे प्रश्न में वेन आरेख के क्षेत्रों को अलग-अलग देखें।
When \(A\subseteq B\), \(A\cup B=B\) and \(A\cap B=A\), hence the answer is \(B\setminus A\). Apply the subset condition first.
Step 2
Why this answer is correct
The correct answer is A. \(B\setminus A\). When \(A\subseteq B\), \(A\cup B=B\) and \(A\cap B=A\), hence the answer is \(B\setminus A\). Apply the subset condition first.
Step 3
Exam Tip
जब \(A\subseteq B\), तब \(A\cup B=B\) और \(A\cap B=A\), इसलिए उत्तर \(B\setminus A\) है। उपसमुच्चय की शर्त पहले लगाएं।
\(A\cup B\) is made of three separate parts, so (54=17+21+n\(A\cap B\)). Hence (n\(A\cap B\)=16), add Venn regions carefully.
Step 2
Why this answer is correct
The correct answer is A. (16). \(A\cup B\) is made of three separate parts, so (54=17+21+n\(A\cap B\)). Hence (n\(A\cap B\)=16), add Venn regions carefully.
Step 3
Exam Tip
\(A\cup B\) तीन अलग भागों से बनता है, इसलिए (54=17+21+n\(A\cap B\))। अतः (n\(A\cap B\)=16), वेन आरेख में क्षेत्रों को जोड़ें।
Both differences being empty means each set is contained in the other, so (A=B). To prove equality, check both containments.
Step 2
Why this answer is correct
The correct answer is A. (A=B). Both differences being empty means each set is contained in the other, so (A=B). To prove equality, check both containments.
Step 3
Exam Tip
दोनों अंतर रिक्त होने का अर्थ है कि दोनों समुच्चय एक-दूसरे में समाहित हैं, इसलिए (A=B)। समानता सिद्ध करने में दोनों दिशाएं जांचें।
Since (B=(-1,6]) excludes (-1) and includes (6), removing it from (A) leaves ([-4,-1]\cup(6,9]). In differences, handle open and closed endpoints carefully.
Step 2
Why this answer is correct
The correct answer is A. ([-4,-1]\cup(6,9]). Since (B=(-1,6]) excludes (-1) and includes (6), removing it from (A) leaves ([-4,-1]\cup(6,9]). In differences, handle open and closed endpoints carefully.
Step 3
Exam Tip
(B=(-1,6]) में (-1) नहीं है और (6) शामिल है, इसलिए (A) से बीच का भाग हटकर ([-4,-1]\cup(6,9]) बचता है। अंतर में खुले-बंद सिरों पर विशेष ध्यान दें।
Since \(A\cap B\subseteq A\cup B\) always, equality here forces both sets to be equal. Element-wise thinking is useful in such questions.
Step 2
Why this answer is correct
The correct answer is A. (A=B). Since \(A\cap B\subseteq A\cup B\) always, equality here forces both sets to be equal. Element-wise thinking is useful in such questions.
Step 3
Exam Tip
क्योंकि \(A\cap B\subseteq A\cup B\) सदैव होता है, समानता तभी होगी जब दोनों समुच्चय समान हों। ऐसे प्रश्न में तत्व-आधारित सोच उपयोगी है।
Since (1) is included in (B), (1) and the common part after it are removed from (A). Watch open and closed endpoints in interval differences.
Step 2
Why this answer is correct
The correct answer is A. ((-2,1)). Since (1) is included in (B), (1) and the common part after it are removed from (A). Watch open and closed endpoints in interval differences.
Step 3
Exam Tip
(B) में (1) शामिल है, इसलिए (A) से (1) और उसके बाद का साझा भाग हटेगा। अंतर में सिरों की खुली-बंद स्थिति ध्यान से देखें।
\(A\cap B\) and \(A\setminus B\) are disjoint parts of (A), and their union is (A). View such identities as partitioning a set.
Step 2
Why this answer is correct
The correct answer is A. (A). \(A\cap B\) and \(A\setminus B\) are disjoint parts of (A), and their union is (A). View such identities as partitioning a set.
Step 3
Exam Tip
\(A\cap B\) और \(A\setminus B\), (A) के अलग-अलग भाग हैं और उनका संघ (A) है। ऐसे पहचान प्रश्न में भागों को विभाजन की तरह देखें।
The symmetric difference contains elements present in exactly one set, so it is ({1,3,5,6}). Do not include common elements.
Step 2
Why this answer is correct
The correct answer is A. ({1,3,5,6}). The symmetric difference contains elements present in exactly one set, so it is ({1,3,5,6}). Do not include common elements.
Step 3
Exam Tip
सममित अंतर में वे तत्व आते हैं जो केवल एक समुच्चय में हों, इसलिए ({1,3,5,6}) है। साझा तत्वों को हटाना न भूलें।
Adding (B) to (A) by union does not change (A), so every element of (B) is in (A). Detect subset relations from union identities.
Step 2
Why this answer is correct
The correct answer is A. \(B\subseteq A\). Adding (B) to (A) by union does not change (A), so every element of (B) is in (A). Detect subset relations from union identities.
Step 3
Exam Tip
संघ में (B) जोड़ने पर (A) नहीं बदलता, इसलिए (B) के सभी तत्व (A) में हैं। संघ से उपसमुच्चय संबंध पहचानें।
Being outside \(B\cup C\) means being outside both (B) and (C). De Morgan style thinking also helps with set difference.
Step 2
Why this answer is correct
The correct answer is A. (\(A\setminus B\)\cap\(A\setminus C\)). Being outside \(B\cup C\) means being outside both (B) and (C). De Morgan style thinking also helps with set difference.
Step 3
Exam Tip
\(B\cup C\) से बाहर रहने का अर्थ है (B) से भी बाहर और (C) से भी बाहर। डी मॉर्गन जैसी सोच अंतर में भी लागू होती है।
An element not in \(B\cap C\) is missing from at least one of (B) or (C). Hence union appears, not intersection.
Step 2
Why this answer is correct
The correct answer is A. (\(A\setminus B\)\cup\(A\setminus C\)). An element not in \(B\cap C\) is missing from at least one of (B) or (C). Hence union appears, not intersection.
Step 3
Exam Tip
जो तत्व \(B\cap C\) में नहीं है, वह कम-से-कम (B) या (C) में नहीं होगा। इसलिए संघ आता है, प्रतिच्छेद नहीं।
When both (A) and (B) lie in (C), \(A\cup B\subseteq C\), so nothing remains outside (C). Subset facts can determine differences quickly.
Step 2
Why this answer is correct
The correct answer is A. \(\varnothing\). When both (A) and (B) lie in (C), \(A\cup B\subseteq C\), so nothing remains outside (C). Subset facts can determine differences quickly.
Step 3
Exam Tip
जब (A) और (B) दोनों (C) में हैं, तब \(A\cup B\subseteq C\), इसलिए बाहर कुछ नहीं बचता। उपसमुच्चय से अंतर तुरंत तय हो सकता है।
Disjoint sets have no common elements, so the union size is (p+q). If the intersection is empty, no subtraction is needed.
Step 2
Why this answer is correct
The correct answer is A. (p+q). Disjoint sets have no common elements, so the union size is (p+q). If the intersection is empty, no subtraction is needed.
Step 3
Exam Tip
असंबद्ध समुच्चयों में कोई साझा तत्व नहीं होता, इसलिए संघ की संख्या (p+q) है। प्रतिच्छेद रिक्त हो तो घटाने की जरूरत नहीं।
\(A=\{-3,-2,-1,0,1,2,3\}\) and \(B=\{-2,-1,0,1,2\}\), so the difference is ({-3,3}). Check the integer condition separately.
Step 2
Why this answer is correct
The correct answer is A. ({-3,3}). \(A=\{-3,-2,-1,0,1,2,3\}\) and \(B=\{-2,-1,0,1,2\}\), so the difference is ({-3,3}). Check the integer condition separately.
Step 3
Exam Tip
\(A=\{-3,-2,-1,0,1,2,3\}\) और \(B=\{-2,-1,0,1,2\}\), इसलिए अंतर ({-3,3}) है। पूर्णांक शर्त अलग से देखें।
Every real number is either less than (2) or greater than (-1), so the union is \(\mathbb{R}\). In unions, visualize the whole number line.
Step 2
Why this answer is correct
The correct answer is A. \(\mathbb{R}\). Every real number is either less than (2) or greater than (-1), so the union is \(\mathbb{R}\). In unions, visualize the whole number line.
Step 3
Exam Tip
हर वास्तविक संख्या या तो (2) से छोटी है या (-1) से बड़ी है, इसलिए संघ \(\mathbb{R}\) है। संघ में क्षेत्रों को मिलाकर पूरी रेखा देखें।
A. जब (B') सार्वत्रिक समुच्चय (U) के सापेक्ष पूरक हो/When (B') is complement relative to universal set (U)
Step 1
Concept
\(A\setminus B\) contains elements of (A) not in (B), which is \(A\cap B'\). Always understand complement relative to (U).
Step 2
Why this answer is correct
The correct answer is A. जब (B') सार्वत्रिक समुच्चय (U) के सापेक्ष पूरक हो / When (B') is complement relative to universal set (U). \(A\setminus B\) contains elements of (A) not in (B), which is \(A\cap B'\). Always understand complement relative to (U).
Step 3
Exam Tip
\(A\setminus B\) में (A) के वे तत्व हैं जो (B) में नहीं, यानी \(A\cap B'\)। पूरक हमेशा (U) के सापेक्ष समझें।
\(A\cup B={1,2,3,5,7,9,11}\), so remaining elements in (U) are ({4,6,8,10,12}). The universal set is essential for complements.
Step 2
Why this answer is correct
The correct answer is A. ({4,6,8,10,12}). \(A\cup B={1,2,3,5,7,9,11}\), so remaining elements in (U) are ({4,6,8,10,12}). The universal set is essential for complements.
Step 3
Exam Tip
\(A\cup B={1,2,3,5,7,9,11}\), इसलिए (U) में बचे तत्व ({4,6,8,10,12}) हैं। पूरक निकालते समय (U) जरूरी है।
Set difference is not commutative, so \(A\setminus B\) and \(B\setminus A\) are generally different. Order matters greatly in difference.
Step 2
Why this answer is correct
The correct answer is A. \(A\setminus B=B\setminus A\). Set difference is not commutative, so \(A\setminus B\) and \(B\setminus A\) are generally different. Order matters greatly in difference.
Step 3
Exam Tip
समुच्चय अंतर अदला-बदली योग्य नहीं है, इसलिए \(A\setminus B\) और \(B\setminus A\) सामान्यतः अलग होते हैं। अंतर में क्रम बहुत महत्वपूर्ण है।
Removing (B) from (A) changes nothing, so (A) and (B) have no common element. Difference identities reveal disjointness.
Step 2
Why this answer is correct
The correct answer is A. \(A\cap B=\varnothing\). Removing (B) from (A) changes nothing, so (A) and (B) have no common element. Difference identities reveal disjointness.
Step 3
Exam Tip
(B) हटाने पर (A) नहीं बदला, इसलिए (A) और (B) में कोई साझा तत्व नहीं है। अंतर पहचान से असंबद्धता समझें।
No element of (A) lies outside (B), so \(A\subseteq B\). An empty difference often indicates a subset relation.
Step 2
Why this answer is correct
The correct answer is A. \(A\subseteq B\). No element of (A) lies outside (B), so \(A\subseteq B\). An empty difference often indicates a subset relation.
Step 3
Exam Tip
(A) का कोई भी तत्व (B) के बाहर नहीं है, इसलिए \(A\subseteq B\)। रिक्त अंतर अक्सर उपसमुच्चय बताता है।
A number divisible by both (2) and (3) is a multiple of (6). In intersection, both conditions apply together.
Step 2
Why this answer is correct
The correct answer is A. (6) के गुणज / Multiples of (6). A number divisible by both (2) and (3) is a multiple of (6). In intersection, both conditions apply together.
Step 3
Exam Tip
जो संख्या (2) और (3) दोनों से विभाज्य है, वह (6) की गुणज है। प्रतिच्छेद में दोनों शर्तें साथ लगती हैं।
(A) has (7) elements, (B) has (3), and the common set is ({10}), so the count is (7+3-1=9). Do not count common elements twice.
Step 2
Why this answer is correct
The correct answer is A. (9). (A) has (7) elements, (B) has (3), and the common set is ({10}), so the count is (7+3-1=9). Do not count common elements twice.
Step 3
Exam Tip
(A) में (7), (B) में (3) और साझा ({10}) है, इसलिए संख्या (7+3-1=9) है। साझा तत्व दो बार न गिनें।
The union is made of three disjoint parts \(A\setminus B\), \(B\setminus A\) and \(A\cap B\), so (5+7+4=16). Add separate Venn regions.
Step 2
Why this answer is correct
The correct answer is A. (16). The union is made of three disjoint parts \(A\setminus B\), \(B\setminus A\) and \(A\cap B\), so (5+7+4=16). Add separate Venn regions.
Step 3
Exam Tip
संघ तीन असंबद्ध भागों \(A\setminus B\), \(B\setminus A\) और \(A\cap B\) से बनता है, इसलिए (5+7+4=16)। वेन आरेख में अलग क्षेत्रों को जोड़ें।
Every element of (A) is in the complement of (B), so (A) and (B) have no common element. Being in a complement means being outside the set.
Step 2
Why this answer is correct
The correct answer is A. \(A\cap B=\varnothing\). Every element of (A) is in the complement of (B), so (A) and (B) have no common element. Being in a complement means being outside the set.
Step 3
Exam Tip
(A) का हर तत्व (B) के पूरक में है, इसलिए (A) और (B) में साझा तत्व नहीं है। पूरक में होने का अर्थ बाहर होना है।
Removing \(B\setminus C\) means removing elements in (B) and outside (C), so elements of (A) remain if they are outside (B) or in (C). Verify identities by element method.
Step 2
Why this answer is correct
The correct answer is A. (\(A\setminus B\)\cup\(A\cap C\)). Removing \(B\setminus C\) means removing elements in (B) and outside (C), so elements of (A) remain if they are outside (B) or in (C). Verify identities by element method.
Step 3
Exam Tip
\(B\setminus C\) से हटाने का अर्थ है (B) में और (C) से बाहर तत्व हटाना, इसलिए (A) में या तो (B) से बाहर या (C) में मौजूद तत्व बचते हैं। पहचान को तत्व विधि से जांचें।
Choosing (2) elements for \(A\setminus B\) is like choosing (2) elements from (A), so \(\binom{4}{2}=6\). Difference size is linked to complementary selection.
Step 2
Why this answer is correct
The correct answer is A. (6). Choosing (2) elements for \(A\setminus B\) is like choosing (2) elements from (A), so \(\binom{4}{2}=6\). Difference size is linked to complementary selection.
Step 3
Exam Tip
\(A\setminus B\) के (2) तत्व चुनना (A) से (2) तत्व चुनने जैसा है, इसलिए \(\binom{4}{2}=6\)। अंतर का आकार पूरक उपचयन से जुड़ा है।
\(A\cup B\) is formed by the disjoint parts \(A\setminus B\) and all of (B), so (12+20=32). Sometimes adding regions directly is easier.
Step 2
Why this answer is correct
The correct answer is A. (32). \(A\cup B\) is formed by the disjoint parts \(A\setminus B\) and all of (B), so (12+20=32). Sometimes adding regions directly is easier.
Step 3
Exam Tip
\(A\cup B\) में \(A\setminus B\) और पूरा (B) असंबद्ध रूप से जुड़ते हैं, इसलिए (12+20=32)। कभी-कभी सीधे क्षेत्रों को जोड़ना आसान है।
A. \(B\subseteq A\) और \(A\subseteq C\)/\(B\subseteq A\) and \(A\subseteq C\)
Step 1
Concept
The left side contains (A), and the right side is contained in (A), so both must equal (A). Hence \(B\subseteq A\) and \(A\subseteq C\).
Step 2
Why this answer is correct
The correct answer is A. \(B\subseteq A\) और \(A\subseteq C\) / \(B\subseteq A\) and \(A\subseteq C\). The left side contains (A), and the right side is contained in (A), so both must equal (A). Hence \(B\subseteq A\) and \(A\subseteq C\).
Step 3
Exam Tip
बायां पक्ष (A) को समाहित करता है और दायां पक्ष (A) में समाहित है, इसलिए दोनों (A) के बराबर हैं। इससे \(B\subseteq A\) और \(A\subseteq C\) मिलता है।
\(A\cap B\) and \(A\setminus B\) are always disjoint, so they can be equal only when both are empty. Then (A) is empty too.
Step 2
Why this answer is correct
The correct answer is A. \(A=\varnothing\). \(A\cap B\) and \(A\setminus B\) are always disjoint, so they can be equal only when both are empty. Then (A) is empty too.
Step 3
Exam Tip
\(A\cap B\) और \(A\setminus B\) हमेशा असंबद्ध होते हैं, इसलिए वे समान तभी हो सकते हैं जब दोनों रिक्त हों। तब (A) भी रिक्त है।
\(A\setminus B\) and \(B\setminus A\) are always disjoint, so they cannot be the same non-empty set. Check disjointness in impossibility questions.
Step 2
Why this answer is correct
The correct answer is A. ऐसा संभव नहीं है / This is impossible. \(A\setminus B\) and \(B\setminus A\) are always disjoint, so they cannot be the same non-empty set. Check disjointness in impossibility questions.
Step 3
Exam Tip
\(A\setminus B\) और \(B\setminus A\) हमेशा असंबद्ध होते हैं, इसलिए दोनों समान गैर-रिक्त समुच्चय नहीं हो सकते। असंभवता में असंबद्धता जांचें।
Numbers divisible by both are multiples of (\operatorname{lcm}(4,6)=12), and up to (50) they are (12,24,36,48). Use LCM for intersections of multiples.
Step 2
Why this answer is correct
The correct answer is A. (4). Numbers divisible by both are multiples of (\operatorname{lcm}(4,6)=12), and up to (50) they are (12,24,36,48). Use LCM for intersections of multiples.
Step 3
Exam Tip
दोनों से विभाज्य संख्याएं (\operatorname{lcm}(4,6)=12) की गुणज हैं, और (50) तक (12,24,36,48) हैं। प्रतिच्छेद में लघुत्तम समापवर्त्य उपयोग करें।
\(A=\{-3,-2,-1,0,1,2,3\}\) and \(B=\{-1,1\}\), so removing them leaves ({-3,-2,0,2,3}). List both sets first.
Step 2
Why this answer is correct
The correct answer is A. ({-3,-2,0,2,3}). \(A=\{-3,-2,-1,0,1,2,3\}\) and \(B=\{-1,1\}\), so removing them leaves ({-3,-2,0,2,3}). List both sets first.
Step 3
Exam Tip
\(A=\{-3,-2,-1,0,1,2,3\}\) और \(B=\{-1,1\}\), इसलिए इन्हें हटाने पर ({-3,-2,0,2,3}) बचता है। पहले दोनों समुच्चय सूचीबद्ध करें।
No real number can satisfy \(x\le0\) and (x>2) together, so the intersection is empty. Recognize incompatible conditions quickly.
Step 2
Why this answer is correct
The correct answer is A. \(\varnothing\). No real number can satisfy \(x\le0\) and (x>2) together, so the intersection is empty. Recognize incompatible conditions quickly.
Step 3
Exam Tip
कोई वास्तविक संख्या एक साथ \(x\le0\) और (x>2) नहीं हो सकती, इसलिए प्रतिच्छेद रिक्त है। असंगत शर्तों को जल्दी पहचानें।
This is distribution of union over intersection. Repetition of (A) in two brackets signals distribution.
Step 2
Why this answer is correct
The correct answer is A. वितरण नियम / Distributive law. This is distribution of union over intersection. Repetition of (A) in two brackets signals distribution.
Step 3
Exam Tip
यह संघ का प्रतिच्छेद पर वितरण है। दो कोष्ठकों में (A) दोहरना वितरण का संकेत है।
\(A\setminus B\) is always a subset of (A), so \(C\subseteq A\) is true. A difference cannot contain elements outside the original set.
Step 2
Why this answer is correct
The correct answer is A. यह सदैव सत्य है / It is always true. \(A\setminus B\) is always a subset of (A), so \(C\subseteq A\) is true. A difference cannot contain elements outside the original set.
Step 3
Exam Tip
\(A\setminus B\) हमेशा (A) का उपसमुच्चय होता है, इसलिए \(C\subseteq A\) सत्य है। अंतर का परिणाम मूल समुच्चय से बाहर नहीं जा सकता।
Inside (A), the parts of (B) and (C) are the same, so no symmetric difference remains within (A). Do not conclude full (B=C) from equal intersections.
Step 2
Why this answer is correct
The correct answer is A. (A\cap\(B\triangle C\)=\varnothing). Inside (A), the parts of (B) and (C) are the same, so no symmetric difference remains within (A). Do not conclude full (B=C) from equal intersections.
Step 3
Exam Tip
(A) के अंदर (B) और (C) का हिस्सा समान है, इसलिए (A) में उनका सममित अंतर नहीं बचेगा। समान प्रतिच्छेद से पूरे (B=C) निष्कर्ष न निकालें।
