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D. \(x\ge6\), (6) पर बंद बिंदु और दाईं ओर/\(x\ge6\), closed dot at (6) shaded right
Step 1
Concept
\(3x\ge18\) gives \(x\ge6\). In exams, include the boundary point when the sign is \(\ge\).
Step 2
Why this answer is correct
The correct answer is D. \(x\ge6\), (6) पर बंद बिंदु और दाईं ओर / \(x\ge6\), closed dot at (6) shaded right. \(3x\ge18\) gives \(x\ge6\). In exams, include the boundary point when the sign is \(\ge\).
Step 3
Exam Tip
\(3x\ge18\) से \(x\ge6\) मिलता है। परीक्षा में \(\ge\) होने पर boundary point को शामिल करें।
The first part is open before (-4), and the second starts from (3) inclusive. In exams, read \(\cup\) as or.
Step 2
Why this answer is correct
The correct answer is C. (x<-4) या \(x\ge3\) / (x<-4) or \(x\ge3\). The first part is open before (-4), and the second starts from (3) inclusive. In exams, read \(\cup\) as or.
Step 3
Exam Tip
पहला भाग (-4) से पहले खुला है और दूसरा (3) सहित दाईं ओर है। परीक्षा में \(\cup\) को "या" की तरह पढ़ें।
Multiplying all parts by (3) gives \(-6\le x+5<12\), so \(-11\le x<7\). In exams, apply the same operation to every part of a compound inequality.
Step 2
Why this answer is correct
The correct answer is A. ([-11,7)). Multiplying all parts by (3) gives \(-6\le x+5<12\), so \(-11\le x<7\). In exams, apply the same operation to every part of a compound inequality.
Step 3
Exam Tip
सभी भागों को (3) से गुणा करने पर \(-6\le x+5<12\), इसलिए \(-11\le x<7\)। परीक्षा में compound inequality में हर भाग पर समान operation करें।
Integers greater than \(-\frac{5}{2}\) start at (-2), and the last integer up to \(\frac{13}{3}\) is (4). In exams, choose valid integers at fractional boundaries.
Step 2
Why this answer is correct
The correct answer is C. ({-2,-1,0,1,2,3,4}). Integers greater than \(-\frac{5}{2}\) start at (-2), and the last integer up to \(\frac{13}{3}\) is (4). In exams, choose valid integers at fractional boundaries.
Step 3
Exam Tip
\(-\frac{5}{2}\) से बड़े पूर्णांक (-2) से शुरू होते हैं और \(\frac{13}{3}\) तक (4) अंतिम पूर्णांक है। परीक्षा में fractional boundary पर valid integer चुनें।
The given interval includes (-6) and excludes (2), so the complement excludes (-6) and includes (2). In exams, check endpoints carefully while taking complements.
Step 2
Why this answer is correct
The correct answer is D. (\(-\infty,-6\)\cup[2,\infty)). The given interval includes (-6) and excludes (2), so the complement excludes (-6) and includes (2). In exams, check endpoints carefully while taking complements.
Step 3
Exam Tip
दिए गए interval में (-6) शामिल है और (2) शामिल नहीं है, इसलिए complement में (-6) हटेगा और (2) शामिल होगा। परीक्षा में complement लेते समय endpoints उलटकर जाँचें।
\(-9\le4x+1\le9\) gives \(-10\le4x\le8\), so \(-\frac{5}{2}\le x\le2\). In exams, modulus with \(\le\) gives a closed interval.
Step 2
Why this answer is correct
The correct answer is C. \([-\frac{5}{2},2]\). \(-9\le4x+1\le9\) gives \(-10\le4x\le8\), so \(-\frac{5}{2}\le x\le2\). In exams, modulus with \(\le\) gives a closed interval.
Step 3
Exam Tip
\(-9\le4x+1\le9\) से \(-10\le4x\le8\), इसलिए \(-\frac{5}{2}\le x\le2\)। परीक्षा में \(\le\) वाले modulus में closed interval लें।
The common part is from (-2) to (3), and both endpoints are excluded. In exams, take only the overlapping region for \(\cap\).
Step 2
Why this answer is correct
The correct answer is B. ((-2,3)). The common part is from (-2) to (3), and both endpoints are excluded. In exams, take only the overlapping region for \(\cap\).
Step 3
Exam Tip
साझा भाग (-2) से (3) तक है और दोनों endpoints बाहर हैं। परीक्षा में \(\cap\) के लिए केवल overlapping region लें।
A. (x<2), (2) पर खुला बिंदु और बाईं ओर/(x<2), open dot at (2) shaded left
Step 1
Concept
(8-6x-3>-7) gives (5-6x>-7), so (x<2). In exams, reverse the sign when dividing by a negative coefficient.
Step 2
Why this answer is correct
The correct answer is A. (x<2), (2) पर खुला बिंदु और बाईं ओर / (x<2), open dot at (2) shaded left. (8-6x-3>-7) gives (5-6x>-7), so (x<2). In exams, reverse the sign when dividing by a negative coefficient.
Step 3
Exam Tip
(8-6x-3>-7) से (5-6x>-7), इसलिए (x<2)। परीक्षा में negative coefficient से divide करते समय sign reverse करें।
The first condition gives (x<5), and the second gives \(x\ge-4\), so the common solution is ([-4,5)). In exams, and means the common shaded part.
Step 2
Why this answer is correct
The correct answer is B. ([-4,5)). The first condition gives (x<5), and the second gives \(x\ge-4\), so the common solution is ([-4,5)). In exams, and means the common shaded part.
Step 3
Exam Tip
पहली शर्त (x<5) देती है और दूसरी \(x\ge-4\), इसलिए साझा हल ([-4,5)) है। परीक्षा में "और" का अर्थ common shaded part होता है।
The first part includes (-3), and the second starts after (-3), so no gap remains. In exams, merge connected intervals.
Step 2
Why this answer is correct
The correct answer is C. (\(-\infty,4\)). The first part includes (-3), and the second starts after (-3), so no gap remains. In exams, merge connected intervals.
Step 3
Exam Tip
पहला भाग (-3) को शामिल करता है और दूसरा (-3) के बाद शुरू होता है, इसलिए gap नहीं बचता। परीक्षा में जुड़े intervals को merge करें।
The closed dot includes (5), and the open dot excludes (14). In exams, decide \(\le\) or (<) from the endpoint.
Step 2
Why this answer is correct
The correct answer is D. \(5\le x<14\). The closed dot includes (5), and the open dot excludes (14). In exams, decide \(\le\) or (<) from the endpoint.
Step 3
Exam Tip
बंद बिंदु (5) को शामिल करता है और खुला बिंदु (14) को बाहर रखता है। परीक्षा में endpoint देखकर \(\le\) या (<) तय करें।
A. (x>11), (11) पर खुला बिंदु और दाईं ओर/(x>11), open dot at (11) shaded right
Step 1
Concept
(4x-9>35) gives (4x>44), so (x>11). In exams, make an open endpoint for a strict inequality.
Step 2
Why this answer is correct
The correct answer is A. (x>11), (11) पर खुला बिंदु और दाईं ओर / (x>11), open dot at (11) shaded right. (4x-9>35) gives (4x>44), so (x>11). In exams, make an open endpoint for a strict inequality.
Step 3
Exam Tip
(4x-9>35) से (4x>44), इसलिए (x>11)। परीक्षा में strict inequality में open endpoint बनाएं।
The removed part does not remove (2), but it removes (7). In exams, pay attention to the endpoints removed in set difference.
Step 2
Why this answer is correct
The correct answer is A. ([-3,2]\cup(7,9]). The removed part does not remove (2), but it removes (7). In exams, pay attention to the endpoints removed in set difference.
Step 3
Exam Tip
हटाया गया भाग (2) को नहीं हटाता पर (7) को हटाता है। परीक्षा में set difference में हटने वाले endpoints पर ध्यान दें।
\(x^2\ge64\) gives \(|x|\ge8\), so two closed outer rays are obtained. In exams, include boundary points when the sign is \(\ge\).
Step 2
Why this answer is correct
The correct answer is C. (\(-\infty,-8]\cup[8,\infty\)). \(x^2\ge64\) gives \(|x|\ge8\), so two closed outer rays are obtained. In exams, include boundary points when the sign is \(\ge\).
Step 3
Exam Tip
\(x^2\ge64\) से \(|x|\ge8\), इसलिए बाहर के दो closed rays मिलते हैं। परीक्षा में \(\ge\) होने पर boundary points शामिल करें।
No number can be less than or equal to (7) and greater than (7) at the same time. In exams, contradictory conditions have an empty intersection.
Step 2
Why this answer is correct
The correct answer is A. \(\emptyset\). No number can be less than or equal to (7) and greater than (7) at the same time. In exams, contradictory conditions have an empty intersection.
Step 3
Exam Tip
कोई संख्या एक साथ (7) से कम या बराबर और (7) से बड़ी नहीं हो सकती। परीक्षा में विरोधी शर्तों का intersection रिक्त होता है।
For \(x\ne9\), the whole number line is taken except (9). In exams, show the removed point with an open circle.
Step 2
Why this answer is correct
The correct answer is B. (\(-\infty,9\)\cup\(9,\infty\)). For \(x\ne9\), the whole number line is taken except (9). In exams, show the removed point with an open circle.
Step 3
Exam Tip
\(x\ne9\) में पूरी संख्या रेखा ली जाती है लेकिन (9) हटाया जाता है। परीक्षा में हटे हुए बिंदु को open circle से दिखाएँ।
The first inequality gives \(-2\le x\le4\), and the second gives (x>0), so the common part is ((0,4]). In exams, apply the extra condition as an intersection.
Step 2
Why this answer is correct
The correct answer is B. ((0,4]). The first inequality gives \(-2\le x\le4\), and the second gives (x>0), so the common part is ((0,4]). In exams, apply the extra condition as an intersection.
Step 3
Exam Tip
पहली असमानता से \(-2\le x\le4\) और दूसरी से (x>0), इसलिए common भाग ((0,4]) है। परीक्षा में extra condition लगाकर intersection लें।
The first interval includes (-5), and the second starts after (-5), so there is no gap. In exams, merge touching intervals.
Step 2
Why this answer is correct
The correct answer is A. ((-12,1]). The first interval includes (-5), and the second starts after (-5), so there is no gap. In exams, merge touching intervals.
Step 3
Exam Tip
पहला interval (-5) को शामिल करता है और दूसरा (-5) के बाद शुरू होता है, इसलिए कोई gap नहीं है। परीक्षा में touching intervals को merge करें।
In \(x\le-2\), (-2) is included, and in (4<x<10), both endpoints are excluded. In exams, write or using union.
Step 2
Why this answer is correct
The correct answer is B. (\(-\infty,-2]\cup(4,10)\). In \(x\le-2\), (-2) is included, and in (4<x<10), both endpoints are excluded. In exams, write or using union.
Step 3
Exam Tip
\(x\le-2\) में (-2) शामिल है और (4<x<10) में दोनों सिरे बाहर हैं। परीक्षा में "या" को union से लिखें।
A. \(x\le7\), (7) पर बंद बिंदु और बाईं ओर/\(x\le7\), closed dot at (7) shaded left
Step 1
Concept
\(11-3x\ge-10\) gives \(-3x\ge-21\), so \(x\le7\). In exams, reverse the inequality while dividing by a negative.
Step 2
Why this answer is correct
The correct answer is A. \(x\le7\), (7) पर बंद बिंदु और बाईं ओर / \(x\le7\), closed dot at (7) shaded left. \(11-3x\ge-10\) gives \(-3x\ge-21\), so \(x\le7\). In exams, reverse the inequality while dividing by a negative.
Step 3
Exam Tip
\(11-3x\ge-10\) से \(-3x\ge-21\), इसलिए \(x\le7\)। परीक्षा में ऋणात्मक से divide करते समय inequality reverse करें।
Because the endpoints are closed, (0) and (12) are included, and two outer rays are taken. In exams, read outer shading as a union of two intervals.
Step 2
Why this answer is correct
The correct answer is C. (\(-\infty,0]\cup[12,\infty\)). Because the endpoints are closed, (0) and (12) are included, and two outer rays are taken. In exams, read outer shading as a union of two intervals.
Step 3
Exam Tip
बंद endpoints के कारण (0) और (12) शामिल हैं और बाहर की दो rays ली जाती हैं। परीक्षा में outer shading को दो intervals की union समझें।
Multiplying by (6) gives (2x-8<2x+5), that is (-8<5), which is always true. In exams, when (x)-terms cancel, check the truth of the remaining statement.
Step 2
Why this answer is correct
The correct answer is B. सभी वास्तविक संख्याएँ / All real numbers. Multiplying by (6) gives (2x-8<2x+5), that is (-8<5), which is always true. In exams, when (x)-terms cancel, check the truth of the remaining statement.
Step 3
Exam Tip
(6) से गुणा करने पर (2x-8<2x+5) यानी (-8<5), जो हमेशा सत्य है। परीक्षा में (x) पद कटने पर शेष कथन की सत्यता जाँचें।
The common part is from (6) to (15), and both endpoints are included. In exams, keep closed boundaries in the intersection.
Step 2
Why this answer is correct
The correct answer is B. ([6,15]). The common part is from (6) to (15), and both endpoints are included. In exams, keep closed boundaries in the intersection.
Step 3
Exam Tip
साझा भाग (6) से (15) तक है और दोनों सिरे शामिल हैं। परीक्षा में closed boundaries को intersection में बनाए रखें।
The removed part removes (-2) but does not remove (1). In exams, check the remaining endpoints separately in set difference.
Step 2
Why this answer is correct
The correct answer is A. (\(-\infty,-2\)\cup[1,4]). The removed part removes (-2) but does not remove (1). In exams, check the remaining endpoints separately in set difference.
Step 3
Exam Tip
हटाया गया भाग (-2) को हटाता है पर (1) को नहीं हटाता। परीक्षा में set difference में बचने वाले endpoints अलग से जाँचें।
The intervals overlap and cover from (-7) to (6) without a gap. In exams, take the entire covered region for a union.
Step 2
Why this answer is correct
The correct answer is A. ((-7,6)). The intervals overlap and cover from (-7) to (6) without a gap. In exams, take the entire covered region for a union.
Step 3
Exam Tip
दोनों intervals overlap करते हैं और (-7) से (6) तक बिना gap के जाते हैं। परीक्षा में union में पूरा covered region लें।
The common part is from (-9) to (-6), and both endpoints are included. In exams, choose only the overlapping segment in an intersection.
Step 2
Why this answer is correct
The correct answer is B. ([-9,-6]). The common part is from (-9) to (-6), and both endpoints are included. In exams, choose only the overlapping segment in an intersection.
Step 3
Exam Tip
साझा भाग (-9) से (-6) तक है और दोनों सिरे शामिल हैं। परीक्षा में intersection में केवल overlapping segment चुनें।
A. \(x\ge -1\), (-1) पर बंद बिंदु और दाईं ओर/\(x\ge -1\), closed dot at (-1) shaded right
Step 1
Concept
\(15-3x\le2x+20\) gives \(-5\le5x\), so \(x\ge-1\). In exams, write the final inequality in the standard direction.
Step 2
Why this answer is correct
The correct answer is A. \(x\ge -1\), (-1) पर बंद बिंदु और दाईं ओर / \(x\ge -1\), closed dot at (-1) shaded right. \(15-3x\le2x+20\) gives \(-5\le5x\), so \(x\ge-1\). In exams, write the final inequality in the standard direction.
Step 3
Exam Tip
\(15-3x\le2x+20\) से \(-5\le5x\), इसलिए \(x\ge-1\)। परीक्षा में final inequality को standard direction में लिखें।
The given parts cover the whole number line except (2). In exams, identify the missing point while simplifying intervals.
Step 2
Why this answer is correct
The correct answer is B. (\(-\infty,2\)\cup\(2,\infty\)). The given parts cover the whole number line except (2). In exams, identify the missing point while simplifying intervals.
Step 3
Exam Tip
दिए गए भाग पूरी संख्या रेखा को ढकते हैं लेकिन (2) को नहीं लेते। परीक्षा में missing point पहचानकर interval सरल करें।
\(8\le3x-1\le20\) gives \(9\le3x\le21\), so \(3\le x\le7\). In exams, a closed compound inequality gives a closed interval.
Step 2
Why this answer is correct
The correct answer is A. ([3,7]). \(8\le3x-1\le20\) gives \(9\le3x\le21\), so \(3\le x\le7\). In exams, a closed compound inequality gives a closed interval.
Step 3
Exam Tip
\(8\le3x-1\le20\) से \(9\le3x\le21\), इसलिए \(3\le x\le7\)। परीक्षा में closed compound inequality से closed interval मिलता है।
The first part goes up to (-2), and the second starts after (-2), so there is no common point. In exams, decide intersection after checking endpoint inclusion.
Step 2
Why this answer is correct
The correct answer is C. \(\emptyset\). The first part goes up to (-2), and the second starts after (-2), so there is no common point. In exams, decide intersection after checking endpoint inclusion.
Step 3
Exam Tip
पहला भाग (-2) तक है और दूसरा (-2) के बाद शुरू होता है, इसलिए कोई common point नहीं है। परीक्षा में endpoint inclusion देखकर intersection तय करें।
B. \(x\ge-2\), (-2) पर बंद बिंदु और दाईं ओर/\(x\ge-2\), closed dot at (-2) shaded right
Step 1
Concept
Multiplying by (-4) reverses the sign to \(x-10\ge-12\), so \(x\ge-2\). In exams, reverse the sign when removing a negative denominator.
Step 2
Why this answer is correct
The correct answer is B. \(x\ge-2\), (-2) पर बंद बिंदु और दाईं ओर / \(x\ge-2\), closed dot at (-2) shaded right. Multiplying by (-4) reverses the sign to \(x-10\ge-12\), so \(x\ge-2\). In exams, reverse the sign when removing a negative denominator.
Step 3
Exam Tip
(-4) से गुणा करने पर चिन्ह पलटकर \(x-10\ge-12\), इसलिए \(x\ge-2\)। परीक्षा में negative denominator हटाते समय sign reverse करें।
Only (-5) is removed, so both sides are shaded with an open point at (-5). In exams, identify it as \(x\ne-5\).
Step 2
Why this answer is correct
The correct answer is B. (\(-\infty,-5\)\cup\(-5,\infty\)). Only (-5) is removed, so both sides are shaded with an open point at (-5). In exams, identify it as \(x\ne-5\).
Step 3
Exam Tip
सिर्फ (-5) हटाया गया है, इसलिए दोनों ओर shading और (-5) पर open point होगा। परीक्षा में इसे \(x\ne-5\) के रूप में पहचानें।
The first interval includes (-3), and the second starts after (-3), so no gap remains. In exams, change connected intervals into one interval.
Step 2
Why this answer is correct
The correct answer is A. ([-8,2)). The first interval includes (-3), and the second starts after (-3), so no gap remains. In exams, change connected intervals into one interval.
Step 3
Exam Tip
पहला interval (-3) को शामिल करता है और दूसरा (-3) के बाद शुरू होता है, इसलिए gap नहीं बचता। परीक्षा में connected intervals को एक interval में बदलें।
A. (x>-2), (-2) पर खुला बिंदु और दाईं ओर/(x>-2), open dot at (-2) shaded right
Step 1
Concept
(7-5x<2x+21) gives (-14<7x), so (x>-2). In exams, match the final result with the number line direction.
Step 2
Why this answer is correct
The correct answer is A. (x>-2), (-2) पर खुला बिंदु और दाईं ओर / (x>-2), open dot at (-2) shaded right. (7-5x<2x+21) gives (-14<7x), so (x>-2). In exams, match the final result with the number line direction.
Step 3
Exam Tip
(7-5x<2x+21) से (-14<7x), इसलिए (x>-2)। परीक्षा में final result को number line की direction से मिलाएँ।
The common part is greater than (6) and less than (11), so both endpoints are open. In exams, read both conditions together in an intersection.
Step 2
Why this answer is correct
The correct answer is B. ((6,11)). The common part is greater than (6) and less than (11), so both endpoints are open. In exams, read both conditions together in an intersection.
Step 3
Exam Tip
Common भाग (6) से बड़ा और (11) से छोटा है, इसलिए दोनों endpoints खुले हैं। परीक्षा में intersection में दोनों conditions साथ पढ़ें।
C. सबसे छोटा (13), सबसे बड़ा (26)/Smallest (13), greatest (26)
Step 1
Concept
\(12\le x-1<27\) gives \(13\le x<28\), so integers go from (13) to (27). In exams, take the integer before the strict upper boundary.
Step 2
Why this answer is correct
The correct answer is C. सबसे छोटा (13), सबसे बड़ा (26) / Smallest (13), greatest (26). \(12\le x-1<27\) gives \(13\le x<28\), so integers go from (13) to (27). In exams, take the integer before the strict upper boundary.
Step 3
Exam Tip
\(12\le x-1<27\) से \(13\le x<28\), इसलिए पूर्णांक (13) से (27) तक हैं। परीक्षा में upper strict सीमा से पहले वाला integer लें।
A. \(x\le19\), (19) पर बंद बिंदु और बाईं ओर/\(x\le19\), closed dot at (19) shaded left
Step 1
Concept
Multiplying by (6) gives \(4x+14\le3x+33\), so \(x\le19\). In exams, multiply by the LCM to clear fractions.
Step 2
Why this answer is correct
The correct answer is A. \(x\le19\), (19) पर बंद बिंदु और बाईं ओर / \(x\le19\), closed dot at (19) shaded left. Multiplying by (6) gives \(4x+14\le3x+33\), so \(x\le19\). In exams, multiply by the LCM to clear fractions.
Step 3
Exam Tip
(6) से गुणा करने पर \(4x+14\le3x+33\), इसलिए \(x\le19\)। परीक्षा में भिन्न हटाने के लिए LCM से गुणा करें।
\(18\le13-2x<45\) gives \(5\le-2x<32\), so \(-16<x\le-\frac{5}{2}\). In exams, reverse order and signs when dividing by a negative.
Step 2
Why this answer is correct
The correct answer is C. (\(-16,-\frac{5}{2}]\). \(18\le13-2x<45\) gives \(5\le-2x<32\), so \(-16<x\le-\frac{5}{2}\). In exams, reverse order and signs when dividing by a negative.
Step 3
Exam Tip
\(18\le13-2x<45\) से \(5\le-2x<32\), इसलिए \(-16<x\le-\frac{5}{2}\)। परीक्षा में ऋणात्मक से भाग देने पर क्रम और चिन्ह बदलें।
The first integer greater than \(-\frac{17}{4}\) is (-4), and the last integer up to \(\frac{7}{3}\) is (2). In exams, choose valid integers carefully at fractional boundaries.
Step 2
Why this answer is correct
The correct answer is A. ({-4,-3,-2,-1,0,1,2}). The first integer greater than \(-\frac{17}{4}\) is (-4), and the last integer up to \(\frac{7}{3}\) is (2). In exams, choose valid integers carefully at fractional boundaries.
Step 3
Exam Tip
\(-\frac{17}{4}\) से बड़ा पहला पूर्णांक (-4) है और \(\frac{7}{3}\) तक अंतिम पूर्णांक (2) है। परीक्षा में fractional boundary पर valid integer ध्यान से चुनें।
The first common part is ([-5,-3]), and the second is ((2,4]). In exams, find the intersection of each union part separately.
Step 2
Why this answer is correct
The correct answer is D. ([-5,-3]\cup(2,4]). The first common part is ([-5,-3]), and the second is ((2,4]). In exams, find the intersection of each union part separately.
Step 3
Exam Tip
पहला common भाग ([-5,-3]) है और दूसरा ((2,4]) है। परीक्षा में union के हर भाग का intersection अलग-अलग निकालें।
\(|3x-4|\ge11\) gives \(3x-4\le-11\) or \(3x-4\ge11\), so \(x\le-\frac{7}{3}\) or \(x\ge5\). In exams, modulus with \(\ge\) gives closed outer rays.
Step 2
Why this answer is correct
The correct answer is B. (\(-\infty,-\frac{7}{3}]\cup[5,\infty\)). \(|3x-4|\ge11\) gives \(3x-4\le-11\) or \(3x-4\ge11\), so \(x\le-\frac{7}{3}\) or \(x\ge5\). In exams, modulus with \(\ge\) gives closed outer rays.
Step 3
Exam Tip
\(|3x-4|\ge11\) से \(3x-4\le-11\) या \(3x-4\ge11\), इसलिए \(x\le-\frac{7}{3}\) या \(x\ge5\)। परीक्षा में \(\ge\) वाले modulus में बाहर के closed rays बनते हैं।
The removed part removes (-2) but does not remove (4). In exams, check separately which endpoint remains in set difference.
Step 2
Why this answer is correct
The correct answer is A. (\(-\infty,-2\)\cup[4,6]). The removed part removes (-2) but does not remove (4). In exams, check separately which endpoint remains in set difference.
Step 3
Exam Tip
हटाया गया भाग (-2) को हटाता है पर (4) को नहीं हटाता। परीक्षा में set difference में कौन-सा endpoint बचता है, यह अलग से जाँचें।
B. \(x>\frac{11}{4}\), \(\frac{11}{4}\) पर खुला बिंदु और दाईं ओर/\(x>\frac{11}{4}\), open dot at \(\frac{11}{4}\) shaded right
Step 1
Concept
Multiplying by (-2) reverses the sign to (5-4x<-6), so \(x>\frac{11}{4}\). In exams, reverse the inequality when removing a negative denominator.
Step 2
Why this answer is correct
The correct answer is B. \(x>\frac{11}{4}\), \(\frac{11}{4}\) पर खुला बिंदु और दाईं ओर / \(x>\frac{11}{4}\), open dot at \(\frac{11}{4}\) shaded right. Multiplying by (-2) reverses the sign to (5-4x<-6), so \(x>\frac{11}{4}\). In exams, reverse the inequality when removing a negative denominator.
Step 3
Exam Tip
(-2) से गुणा करने पर चिन्ह पलटकर (5-4x<-6), इसलिए \(x>\frac{11}{4}\)। परीक्षा में negative denominator हटाते समय inequality reverse करें।
