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The open dot excludes (-5), and the closed dot includes (2). In exams, write an open dot with a round bracket and a closed dot with a square bracket.
Step 2
Why this answer is correct
The correct answer is A. ((-5,2]). The open dot excludes (-5), and the closed dot includes (2). In exams, write an open dot with a round bracket and a closed dot with a square bracket.
Step 3
Exam Tip
खुला बिंदु (-5) को बाहर रखता है और बंद बिंदु (2) को शामिल करता है। परीक्षा में खुला बिंदु गोल कोष्ठक और बंद बिंदु वर्ग कोष्ठक से लिखें।
B. \(x\le -3\), (-3) पर बंद बिंदु और बाईं ओर/\(x\le -3\), closed dot at (-3) shaded left
Step 1
Concept
From \(-4x\ge 12\), dividing by a negative gives \(x\le -3\). In exams, reverse the inequality sign when dividing by a negative.
Step 2
Why this answer is correct
The correct answer is B. \(x\le -3\), (-3) पर बंद बिंदु और बाईं ओर / \(x\le -3\), closed dot at (-3) shaded left. From \(-4x\ge 12\), dividing by a negative gives \(x\le -3\). In exams, reverse the inequality sign when dividing by a negative.
Step 3
Exam Tip
\(-4x\ge 12\) में ऋणात्मक संख्या से भाग देने पर \(x\le -3\) मिलता है। परीक्षा में ऋणात्मक से भाग देने पर असमानता का चिन्ह पलटें।
The first part is to the left including (-6), and the second part is to the right after (4). In exams, read \(\cup\) as or.
Step 2
Why this answer is correct
The correct answer is C. \(x\le -6\) या (x>4) / \(x\le -6\) or (x>4). The first part is to the left including (-6), and the second part is to the right after (4). In exams, read \(\cup\) as or.
Step 3
Exam Tip
पहला भाग (-6) सहित बाईं ओर है और दूसरा भाग (4) के बाद दाईं ओर है। परीक्षा में \(\cup\) को या के रूप में पढ़ें।
Multiplying all parts by (4) gives \(12<x-2\le32\), so \(14<x\le34\). 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. ((14,34]). Multiplying all parts by (4) gives \(12<x-2\le32\), so \(14<x\le34\). In exams, apply the same operation to every part of a compound inequality.
Step 3
Exam Tip
सभी भागों को (4) से गुणा करने पर \(12<x-2\le32\), इसलिए \(14<x\le34\)। परीक्षा में संयुक्त असमानता में हर भाग पर समान क्रिया करें।
(-7) is included, and integers less than \(\frac{5}{2}\) go up to (2). In exams, with an integer restriction, mark separate points instead of a continuous line.
Step 2
Why this answer is correct
The correct answer is B. ({-7,-6,-5,-4,-3,-2,-1,0,1,2}). (-7) is included, and integers less than \(\frac{5}{2}\) go up to (2). In exams, with an integer restriction, mark separate points instead of a continuous line.
Step 3
Exam Tip
(-7) शामिल है और \(\frac{5}{2}\) से छोटे पूर्णांक (2) तक हैं। परीक्षा में पूर्णांक प्रतिबंध हो तो रेखा नहीं, अलग बिंदु बनाएं।
The given interval excludes (-3) and includes (5), so its complement includes (-3) and excludes (5). In exams, check endpoint inclusion carefully while taking complements.
Step 2
Why this answer is correct
The correct answer is B. ((-\infty,-3]\cup\(5,\infty\)). The given interval excludes (-3) and includes (5), so its complement includes (-3) and excludes (5). In exams, check endpoint inclusion carefully while taking complements.
Step 3
Exam Tip
दिए गए अंतराल में (-3) शामिल नहीं है और (5) शामिल है, इसलिए पूरक में (-3) शामिल और (5) बाहर होगा। परीक्षा में पूरक लेते समय सिरों की स्थिति बदलकर जाँचें।
\(-10\le3x-2\le10\) gives \(-8\le3x\le12\), so \(-\frac{8}{3}\le x\le4\). In exams, keep both endpoints closed for \(\le\).
Step 2
Why this answer is correct
The correct answer is B. \([- \frac{8}{3},4]\). \(-10\le3x-2\le10\) gives \(-8\le3x\le12\), so \(-\frac{8}{3}\le x\le4\). In exams, keep both endpoints closed for \(\le\).
Step 3
Exam Tip
\(-10\le3x-2\le10\) से \(-8\le3x\le12\), इसलिए \(-\frac{8}{3}\le x\le4\)। परीक्षा में \(\le\) होने पर दोनों सिरे बंद रखें।
The common part is between (-4) and (-1), and both endpoints are excluded. In exams, take only the commonly shaded part for \(\cap\).
Step 2
Why this answer is correct
The correct answer is B. ((-4,-1)). The common part is between (-4) and (-1), and both endpoints are excluded. In exams, take only the commonly shaded part for \(\cap\).
Step 3
Exam Tip
साझा भाग (-4) से (-1) के बीच है और दोनों सिरे बाहर हैं। परीक्षा में \(\cap\) के लिए केवल समान छायांकित भाग लें।
A. \(x<\frac{11}{3}\), \(\frac{11}{3}\) पर खुला बिंदु और बाईं ओर/\(x<\frac{11}{3}\), open dot at \(\frac{11}{3}\) shaded left
Step 1
Concept
(5-6x+2>-15) gives (7-6x>-15), so \(x<\frac{11}{3}\). In exams, expand brackets and reverse the sign for a negative coefficient.
Step 2
Why this answer is correct
The correct answer is A. \(x<\frac{11}{3}\), \(\frac{11}{3}\) पर खुला बिंदु और बाईं ओर / \(x<\frac{11}{3}\), open dot at \(\frac{11}{3}\) shaded left. (5-6x+2>-15) gives (7-6x>-15), so \(x<\frac{11}{3}\). In exams, expand brackets and reverse the sign for a negative coefficient.
Step 3
Exam Tip
(5-6x+2>-15) से (7-6x>-15), इसलिए \(x<\frac{11}{3}\)। परीक्षा में कोष्ठक खोलकर ऋणात्मक गुणांक पर चिन्ह पलटें।
The first condition gives (x<6), and the second gives \(x\ge -4\), so the common solution is ([-4,6)). In exams, take the common region for both conditions.
Step 2
Why this answer is correct
The correct answer is A. ([-4,6)). The first condition gives (x<6), and the second gives \(x\ge -4\), so the common solution is ([-4,6)). In exams, take the common region for both conditions.
Step 3
Exam Tip
पहली शर्त (x<6) देती है और दूसरी \(x\ge -4\), इसलिए साझा हल ([-4,6)) है। परीक्षा में दोनों शर्तों के लिए common भाग लें।
The first part is before (0), and the second includes (0) up to (7), so no gap remains. In exams, combine connected parts into one interval.
Step 2
Why this answer is correct
The correct answer is C. (\(-\infty,7\)). The first part is before (0), and the second includes (0) up to (7), so no gap remains. In exams, combine connected parts into one interval.
Step 3
Exam Tip
पहला भाग (0) से पहले है और दूसरा भाग (0) को शामिल करके (7) तक जाता है, इसलिए कोई अंतराल नहीं छूटता। परीक्षा में जुड़े हुए भागों को एक अंतराल में मिलाएं।
The closed dot includes (-1), and the open dot excludes (9). In exams, decide \(\le\) or (<) from the type of endpoint.
Step 2
Why this answer is correct
The correct answer is D. \(-1\le x<9\). The closed dot includes (-1), and the open dot excludes (9). In exams, decide \(\le\) or (<) from the type of endpoint.
Step 3
Exam Tip
बंद बिंदु (-1) को शामिल करता है और खुला बिंदु (9) को बाहर रखता है। परीक्षा में बिंदु के प्रकार से \(\le\) या (<) तय करें।
A. (x>11), (11) पर खुला बिंदु और दाईं ओर/(x>11), open dot at (11) shaded right
Step 1
Concept
(2x-7>15) gives (2x>22), so (x>11). In exams, removing a positive denominator does not change the sign.
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. (2x-7>15) gives (2x>22), so (x>11). In exams, removing a positive denominator does not change the sign.
Step 3
Exam Tip
(2x-7>15) से (2x>22), इसलिए (x>11)। परीक्षा में धनात्मक हर हटाने पर चिन्ह नहीं बदलता।
The removed part does not remove (1), but it removes (6). In exams, check the endpoints of the removed interval carefully.
Step 2
Why this answer is correct
The correct answer is A. ([-2,1]\cup(6,8]). The removed part does not remove (1), but it removes (6). In exams, check the endpoints of the removed interval carefully.
Step 3
Exam Tip
हटाया गया भाग (1) को नहीं हटाता पर (6) को हटाता है। परीक्षा में set difference में हटने वाले सिरे ध्यान से देखें।
\(x^2>49\) gives (|x|>7), so two open outer regions are obtained. In exams, do not include boundary points for strict (>).
Step 2
Why this answer is correct
The correct answer is D. (\(-\infty,-7\)\cup\(7,\infty\)). \(x^2>49\) gives (|x|>7), so two open outer regions are obtained. In exams, do not include boundary points for strict (>).
Step 3
Exam Tip
\(x^2>49\) से (|x|>7), इसलिए बाहर के दो खुले भाग मिलते हैं। परीक्षा में strict (>) होने पर boundary points शामिल न करें।
The natural numbers are (4,5,6,7,8,9), so there are (6) points. In exams, count only the separate valid natural number values.
Step 2
Why this answer is correct
The correct answer is B. (6). The natural numbers are (4,5,6,7,8,9), so there are (6) points. In exams, count only the separate valid natural number values.
Step 3
Exam Tip
प्राकृतिक संख्याएँ (4,5,6,7,8,9) हैं, इसलिए कुल (6) बिंदु होंगे। परीक्षा में प्राकृतिक संख्या के लिए केवल अलग-अलग मान गिनें।
Together, the two parts cover every real number. In exams, the union of complementary rays gives the whole number line.
Step 2
Why this answer is correct
The correct answer is C. (\(-\infty,\infty\)). Together, the two parts cover every real number. In exams, the union of complementary rays gives the whole number line.
Step 3
Exam Tip
दोनों भाग मिलकर हर वास्तविक संख्या को ढक लेते हैं। परीक्षा में पूरक किरणों का union पूरी संख्या रेखा देता है।
\(-6\le5-x<12\) gives \(-11\le -x<7\), so \(-7<x\le11\). In exams, reverse order and signs when changing (-x) to (x).
Step 2
Why this answer is correct
The correct answer is A. ((-7,11]). \(-6\le5-x<12\) gives \(-11\le -x<7\), so \(-7<x\le11\). In exams, reverse order and signs when changing (-x) to (x).
Step 3
Exam Tip
\(-6\le5-x<12\) से \(-11\le -x<7\), इसलिए \(-7<x\le11\)। परीक्षा में (-x) को (x) बनाते समय क्रम और चिन्ह पलटें।
For \(x\ne -4\), the whole number line is taken except (-4). In exams, show the removed point as an open dot.
Step 2
Why this answer is correct
The correct answer is B. (\(-\infty,-4\)\cup\(-4,\infty\)). For \(x\ne -4\), the whole number line is taken except (-4). In exams, show the removed point as an open dot.
Step 3
Exam Tip
\(x\ne -4\) में पूरी संख्या रेखा ली जाती है लेकिन (-4) हटाया जाता है। परीक्षा में हटे हुए बिंदु को खुले बिंदु से दिखाएँ।
The first part gives \(x\le6\) and the second gives (x>-3), so the correct common solution is ((-3,6]). In exams, split a compound inequality into two conditions and check carefully.
Step 2
Why this answer is correct
The correct answer is A. \([-6,\infty\)). The first part gives \(x\le6\) and the second gives (x>-3), so the correct common solution is ((-3,6]). In exams, split a compound inequality into two conditions and check carefully.
Step 3
Exam Tip
पहले भाग से \(x\le6\) नहीं बल्कि \(x\le6\) और दूसरे से (x>-3) मिलता है, इसलिए सही साझा हल ((-3,6]) नहीं है। परीक्षा में संयुक्त असमानता को दो अलग शर्तों में बाँटकर जाँचें।
The first interval includes (-2), and the second starts after (-2), so no gap remains. In exams, merge connected intervals.
Step 2
Why this answer is correct
The correct answer is A. ((-8,3]). The first interval includes (-2), and the second starts after (-2), so no gap remains. In exams, merge connected intervals.
Step 3
Exam Tip
पहला अंतराल (-2) को शामिल करता है और दूसरा (-2) के बाद शुरू होता है, इसलिए कोई gap नहीं बचता। परीक्षा में जुड़े हुए intervals को मिलाकर लिखें।
A. \(x\le\frac{19}{2}\), \(\frac{19}{2}\) पर बंद बिंदु और बाईं ओर/\(x\le\frac{19}{2}\), closed dot at \(\frac{19}{2}\) shaded left
Step 1
Concept
\(7-2x\ge -12\) gives \(-2x\ge -19\), so \(x\le\frac{19}{2}\). In exams, reverse the sign when dividing by a negative.
Step 2
Why this answer is correct
The correct answer is A. \(x\le\frac{19}{2}\), \(\frac{19}{2}\) पर बंद बिंदु और बाईं ओर / \(x\le\frac{19}{2}\), closed dot at \(\frac{19}{2}\) shaded left. \(7-2x\ge -12\) gives \(-2x\ge -19\), so \(x\le\frac{19}{2}\). In exams, reverse the sign when dividing by a negative.
Step 3
Exam Tip
\(7-2x\ge -12\) से \(-2x\ge -19\), इसलिए \(x\le\frac{19}{2}\)। परीक्षा में ऋणात्मक से भाग देने पर चिन्ह पलटें।
Because the endpoints are open, (-10) and (-3) are excluded, and two outer parts are taken. In exams, outer shading means two separate rays.
Step 2
Why this answer is correct
The correct answer is B. (\(-\infty,-10\)\cup\(-3,\infty\)). Because the endpoints are open, (-10) and (-3) are excluded, and two outer parts are taken. In exams, outer shading means two separate rays.
Step 3
Exam Tip
खुले सिरों के कारण (-10) और (-3) शामिल नहीं हैं और बाहर के दो भाग लिए जाते हैं। परीक्षा में बाहर छायांकन का अर्थ दो अलग किरणें होता है।
B. (x>4), (4) पर खुला बिंदु और दाईं ओर/(x>4), open dot at (4) shaded right
Step 1
Concept
Multiplying by (6) gives (3x+3<3x-5), which is impossible. In exams, when (x)-terms cancel, compare the constants.
Step 2
Why this answer is correct
The correct answer is B. (x>4), (4) पर खुला बिंदु और दाईं ओर / (x>4), open dot at (4) shaded right. Multiplying by (6) gives (3x+3<3x-5), which is impossible. In exams, when (x)-terms cancel, compare the constants.
Step 3
Exam Tip
(6) से गुणा करने पर (3x+3<3x-5) बनता है, जो असंभव है। परीक्षा में समान (x) पद कटने पर स्थिर संख्याओं की तुलना करें।
The common part is from (3) to (12), and both endpoints are included. In exams, two included boundaries give a closed interval in the common region.
Step 2
Why this answer is correct
The correct answer is B. ([3,12]). The common part is from (3) to (12), and both endpoints are included. In exams, two included boundaries give a closed interval in the common region.
Step 3
Exam Tip
साझा भाग (3) से (12) तक है और दोनों सिरे शामिल हैं। परीक्षा में दो बंद सीमाएँ common भाग में closed interval देती हैं।
The removed part does not remove (-4), but it removes (0). In exams, think carefully about the open and closed endpoints of the removed interval.
Step 2
Why this answer is correct
The correct answer is A. ((-\infty,-4]\cup(0,2]). The removed part does not remove (-4), but it removes (0). In exams, think carefully about the open and closed endpoints of the removed interval.
Step 3
Exam Tip
हटाया गया भाग (-4) को नहीं हटाता पर (0) को हटाता है। परीक्षा में हटाए गए interval के खुले और बंद सिरे उलटकर सोचें।
The two intervals overlap and cover from (-2) to (9) 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. ((-2,9)). The two intervals overlap and cover from (-2) to (9) without a gap. In exams, take the entire covered region for a union.
Step 3
Exam Tip
दोनों intervals overlap करते हैं और (-2) से (9) तक बिना gap के जाते हैं। परीक्षा में union में पूरा covered भाग लें।
The common part is from (-7) to (-5), and both endpoints are included. In exams, choose only the overlapping segment for intersection.
Step 2
Why this answer is correct
The correct answer is B. ([-7,-5]). The common part is from (-7) to (-5), and both endpoints are included. In exams, choose only the overlapping segment for intersection.
Step 3
Exam Tip
साझा भाग (-7) से (-5) तक है और दोनों सिरे शामिल हैं। परीक्षा में intersection में केवल overlapping segment चुनें।
B. \(x\le3\), (3) पर बंद बिंदु और बाईं ओर/\(x\le3\), closed dot at (3) shaded left
Step 1
Concept
\(8-2x\ge3x-7\) gives \(15\ge5x\), so \(x\le3\). In exams, keep variables on one side and constants on the other.
Step 2
Why this answer is correct
The correct answer is B. \(x\le3\), (3) पर बंद बिंदु और बाईं ओर / \(x\le3\), closed dot at (3) shaded left. \(8-2x\ge3x-7\) gives \(15\ge5x\), so \(x\le3\). In exams, keep variables on one side and constants on the other.
Step 3
Exam Tip
\(8-2x\ge3x-7\) से \(15\ge5x\), इसलिए \(x\le3\)। परीक्षा में चर को एक ओर और संख्याओं को दूसरी ओर रखें।
The given parts cover the whole number line except (-1). In exams, identify the missing point separately.
Step 2
Why this answer is correct
The correct answer is B. (\(-\infty,-1\)\cup\(-1,\infty\)). The given parts cover the whole number line except (-1). In exams, identify the missing point separately.
Step 3
Exam Tip
दिए गए भाग पूरी संख्या रेखा को ढकते हैं लेकिन (-1) नहीं लेते। परीक्षा में missing point को अलग से पहचानें।
\(5\le2x+3\le15\) gives \(2\le2x\le12\), so \(1\le x\le6\). In exams, a closed compound inequality gives a closed interval.
Step 2
Why this answer is correct
The correct answer is A. ([1,6]). \(5\le2x+3\le15\) gives \(2\le2x\le12\), so \(1\le x\le6\). In exams, a closed compound inequality gives a closed interval.
Step 3
Exam Tip
\(5\le2x+3\le15\) से \(2\le2x\le12\), इसलिए \(1\le x\le6\)। परीक्षा में बंद संयुक्त असमानता से बंद interval मिलता है।
The first part excludes (5), and the second starts at (5), so there is no common point. In exams, decide intersection only after checking endpoint inclusion.
Step 2
Why this answer is correct
The correct answer is C. \(\emptyset\). The first part excludes (5), and the second starts at (5), so there is no common point. In exams, decide intersection only after checking endpoint inclusion.
Step 3
Exam Tip
पहला भाग (5) को शामिल नहीं करता और दूसरा (5) से शुरू होता है, इसलिए कोई common point नहीं है। परीक्षा में endpoint inclusion देखकर ही intersection तय करें।
The closed dot includes (-8), and right shading means \(x\ge -8\). In exams, read a right ray as the greater side.
Step 2
Why this answer is correct
The correct answer is C. \([-8,\infty\)). The closed dot includes (-8), and right shading means \(x\ge -8\). In exams, read a right ray as the greater side.
Step 3
Exam Tip
बंद बिंदु (-8) को शामिल करता है और दाईं ओर shading \(x\ge -8\) बताती है। परीक्षा में right ray को greater side समझें।
B. (x>1), (1) पर खुला बिंदु और दाईं ओर/(x>1), open dot at (1) shaded right
Step 1
Concept
Multiplying by (-2) reverses the sign to (9-x>-8), so (x<17), not (x>17). In exams, reverse the sign when removing a negative denominator.
Step 2
Why this answer is correct
The correct answer is B. (x>1), (1) पर खुला बिंदु और दाईं ओर / (x>1), open dot at (1) shaded right. Multiplying by (-2) reverses the sign to (9-x>-8), so (x<17), not (x>17). In exams, reverse the sign when removing a negative denominator.
Step 3
Exam Tip
(-2) से गुणा करने पर चिन्ह पलटकर (9-x>-8), इसलिए (x<17) नहीं बल्कि (x<17) मिलता है। परीक्षा में negative denominator हटाते समय चिन्ह पलटें।
The first part is left including (0), and the second is open between (2) and (7). In exams, connect separate conditions using union.
Step 2
Why this answer is correct
The correct answer is A. (\(-\infty,0]\cup(2,7)\). The first part is left including (0), and the second is open between (2) and (7). In exams, connect separate conditions using union.
Step 3
Exam Tip
पहला भाग (0) सहित बाईं ओर है और दूसरा (2) तथा (7) के बीच खुला है। परीक्षा में अलग-अलग conditions को union से जोड़ें।
Only (6) is removed, so there is an open dot at (6) and shading on both sides. In exams, identify this as \(x\ne6\).
Step 2
Why this answer is correct
The correct answer is B. (\(-\infty,6\)\cup\(6,\infty\)). Only (6) is removed, so there is an open dot at (6) and shading on both sides. In exams, identify this as \(x\ne6\).
Step 3
Exam Tip
सिर्फ (6) हटाया गया है, इसलिए (6) पर खुला बिंदु और दोनों ओर shading होगी। परीक्षा में इसे \(x\ne6\) की तरह पहचानें।
The first interval includes (2), and the second starts after (2), so there is no gap. In exams, merge connected intervals into one.
Step 2
Why this answer is correct
The correct answer is A. ([-4,5]). The first interval includes (2), and the second starts after (2), so there is no gap. In exams, merge connected intervals into one.
Step 3
Exam Tip
पहला अंतराल (2) को शामिल करता है और दूसरा (2) के बाद शुरू होता है, इसलिए कोई gap नहीं है। परीक्षा में connected intervals को एक में मिलाएँ।
A. (x>-3), (-3) पर खुला बिंदु और दाईं ओर/(x>-3), open dot at (-3) shaded right
Step 1
Concept
(6-4x<2x+24) gives (-18<6x), so (x>-3). In exams, rewrite the final inequality in the standard direction.
Step 2
Why this answer is correct
The correct answer is A. (x>-3), (-3) पर खुला बिंदु और दाईं ओर / (x>-3), open dot at (-3) shaded right. (6-4x<2x+24) gives (-18<6x), so (x>-3). In exams, rewrite the final inequality in the standard direction.
Step 3
Exam Tip
(6-4x<2x+24) से (-18<6x), इसलिए (x>-3)। परीक्षा में अंतिम inequality को मानक दिशा में लिखें।
The common part is greater than (4) and less than (10), so both endpoints are open. In exams, read both inequalities together in an intersection.
Step 2
Why this answer is correct
The correct answer is B. ((4,10)). The common part is greater than (4) and less than (10), so both endpoints are open. In exams, read both inequalities together in an intersection.
Step 3
Exam Tip
Common भाग (4) से बड़ा और (10) से छोटा है, इसलिए दोनों सिरे खुले हैं। परीक्षा में intersection में दोनों असमानताओं को साथ पढ़ें।
A. सबसे छोटा (4), सबसे बड़ा (11)/Smallest (4), greatest (11)
Step 1
Concept
The solution \(6\le x+2<14\) gives \(4\le x<12\), so integers run from (4) to (11). In exams, take the integer just before a strict upper boundary.
Step 2
Why this answer is correct
The correct answer is A. सबसे छोटा (4), सबसे बड़ा (11) / Smallest (4), greatest (11). The solution \(6\le x+2<14\) gives \(4\le x<12\), so integers run from (4) to (11). In exams, take the integer just before a strict upper boundary.
Step 3
Exam Tip
हल \(6\le x+2<14\) से \(4\le x<12\), इसलिए पूर्णांक (4) से (11) तक हैं। परीक्षा में ऊपरी strict सीमा से ठीक पहले वाला पूर्णांक लें।
A. \(x\le\frac{32}{7}\), \(\frac{32}{7}\) पर बंद बिंदु और बाईं ओर/\(x\le\frac{32}{7}\), closed dot at \(\frac{32}{7}\) shaded left
Step 1
Concept
Multiplying by (6) gives \(10x-8\le3x+24\), so \(x\le\frac{32}{7}\). In exams, multiply by the LCM to clear fractions.
Step 2
Why this answer is correct
The correct answer is A. \(x\le\frac{32}{7}\), \(\frac{32}{7}\) पर बंद बिंदु और बाईं ओर / \(x\le\frac{32}{7}\), closed dot at \(\frac{32}{7}\) shaded left. Multiplying by (6) gives \(10x-8\le3x+24\), so \(x\le\frac{32}{7}\). In exams, multiply by the LCM to clear fractions.
Step 3
Exam Tip
(6) से गुणा करने पर \(10x-8\le3x+24\), इसलिए \(x\le\frac{32}{7}\)। परीक्षा में भिन्न हटाने के लिए लघुत्तम समापवर्त्य से गुणा करें।
\(4\le7-3x<22\) gives \(-3\le-3x<15\), so \(-5<x\le1\). In exams, dividing by a negative changes both order and signs.
Step 2
Why this answer is correct
The correct answer is A. ((-5,1]). \(4\le7-3x<22\) gives \(-3\le-3x<15\), so \(-5<x\le1\). In exams, dividing by a negative changes both order and signs.
Step 3
Exam Tip
\(4\le7-3x<22\) से \(-3\le-3x<15\), इसलिए \(-5<x\le1\)। परीक्षा में ऋणात्मक से भाग देने पर क्रम और चिन्ह दोनों बदलते हैं।
The first integer greater than or equal to \(-\frac{11}{3}\) is (-3), and the last integer less than \(\frac{9}{2}\) is (4). In exams, choose valid integers separately at fractional boundaries.
Step 2
Why this answer is correct
The correct answer is B. ({-3,-2,-1,0,1,2,3,4}). The first integer greater than or equal to \(-\frac{11}{3}\) is (-3), and the last integer less than \(\frac{9}{2}\) is (4). In exams, choose valid integers separately at fractional boundaries.
Step 3
Exam Tip
\(-\frac{11}{3}\) से बड़ा या बराबर पहला पूर्णांक (-3) है और \(\frac{9}{2}\) से छोटा अंतिम पूर्णांक (4) है। परीक्षा में fractional boundary पर valid integer अलग से चुनें।
The removed part does not remove (-1), but it removes (3). In exams, check the open and closed endpoints of the removed interval carefully.
Step 2
Why this answer is correct
The correct answer is C. \([-6,-1]\cup(3,4)\). The removed part does not remove (-1), but it removes (3). In exams, check the open and closed endpoints of the removed interval carefully.
Step 3
Exam Tip
हटाया गया भाग (-1) को नहीं हटाता पर (3) को हटाता है। परीक्षा में set difference में हटाए गए अंतराल के खुले और बंद सिरों को ध्यान से देखें।
(|2x+5|>9) gives (2x+5<-9) or (2x+5>9), so (x<-7) or (x>2). In exams, modulus with (>) gives open outer parts.
Step 2
Why this answer is correct
The correct answer is D. (\(-\infty,-7\)\cup\(2,\infty\)). (|2x+5|>9) gives (2x+5<-9) or (2x+5>9), so (x<-7) or (x>2). In exams, modulus with (>) gives open outer parts.
Step 3
Exam Tip
(|2x+5|>9) से (2x+5<-9) या (2x+5>9), इसलिए (x<-7) या (x>2)। परीक्षा में (>) वाले modulus में बाहर के खुले भाग बनते हैं।
The common part is from (-2) to (3), and both endpoints are included. In exams, take only the common shaded 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 included. In exams, take only the common shaded region for \(\cap\).
Step 3
Exam Tip
साझा भाग (-2) से (3) तक है और दोनों सिरे शामिल हैं। परीक्षा में \(\cap\) के लिए केवल common shaded region लें।
