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A. ( -3) पर बंद बिंदु बाईं ओर और (1) पर खुला बिंदु दाईं ओर/Closed dot at (-3) shaded left and open dot at (1) shaded right
Step 1
Concept
\(x\le -3\) includes (-3), while (x>1) does not include (1). In exams, remember closed dot for \(\le\) and open dot for (>).
Step 2
Why this answer is correct
The correct answer is A. ( -3) पर बंद बिंदु बाईं ओर और (1) पर खुला बिंदु दाईं ओर / Closed dot at (-3) shaded left and open dot at (1) shaded right. \(x\le -3\) includes (-3), while (x>1) does not include (1). In exams, remember closed dot for \(\le\) and open dot for (>).
Step 3
Exam Tip
\(x\le -3\) में (-3) शामिल है और (x>1) में (1) शामिल नहीं है। परीक्षा में \(\le\) के लिए बंद और (>) के लिए खुला बिंदु याद रखें।
A round bracket means open endpoint and a square bracket means closed endpoint. In exams, use this rule while converting interval notation to inequality.
Step 2
Why this answer is correct
The correct answer is B. \(-4<x\le 7\). A round bracket means open endpoint and a square bracket means closed endpoint. In exams, use this rule while converting interval notation to inequality.
Step 3
Exam Tip
गोल कोष्ठक खुला endpoint और वर्ग कोष्ठक बंद endpoint बताता है। परीक्षा में interval notation को inequality में बदलते समय यही नियम लगाएँ।
A. \(x\ge 5\) यानी (5) से दाईं ओर बंद बिंदु/\(x\ge 5\), closed dot at (5) shaded right
Step 1
Concept
Solving gives \(2x\ge 10\), so \(x\ge 5\). In exams, solve first and then mark direction and endpoint.
Step 2
Why this answer is correct
The correct answer is A. \(x\ge 5\) यानी (5) से दाईं ओर बंद बिंदु / \(x\ge 5\), closed dot at (5) shaded right. Solving gives \(2x\ge 10\), so \(x\ge 5\). In exams, solve first and then mark direction and endpoint.
Step 3
Exam Tip
हल \(2x\ge 10\) से \(x\ge 5\) मिलता है। परीक्षा में पहले हल करें फिर दिशा और endpoint लगाएँ।
B. (x>2), (2) से दाईं ओर/(x>2), shaded right of (2)
Step 1
Concept
From (-3x<-6), dividing by a negative reverses the sign. In exams, reverse the inequality when dividing by a negative.
Step 2
Why this answer is correct
The correct answer is B. (x>2), (2) से दाईं ओर / (x>2), shaded right of (2). From (-3x<-6), dividing by a negative reverses the sign. In exams, reverse the inequality when dividing by a negative.
Step 3
Exam Tip
(-3x<-6) में ऋणात्मक संख्या से भाग देने पर चिन्ह पलटता है। परीक्षा में negative से divide करते समय inequality reverse करें।
\(|x-2|\le 5\) gives \(-5\le x-2\le 5\), hence \(-3\le x\le 7\). In exams, absolute value with \(\le\) gives a closed middle interval.
Step 2
Why this answer is correct
The correct answer is A. ([-3,7]). \(|x-2|\le 5\) gives \(-5\le x-2\le 5\), hence \(-3\le x\le 7\). In exams, absolute value with \(\le\) gives a closed middle interval.
Step 3
Exam Tip
\(|x-2|\le 5\) से \(-5\le x-2\le 5\) और \(-3\le x\le 7\) मिलता है। परीक्षा में \(\le\) वाले absolute value में बीच का बंद interval लें।
(|x+1|>4) gives two outer regions, and endpoints are not included. In exams, absolute value with (>) is shaded outward.
Step 2
Why this answer is correct
The correct answer is C. (\(-\infty,-5\)\cup\(3,\infty\)). (|x+1|>4) gives two outer regions, and endpoints are not included. In exams, absolute value with (>) is shaded outward.
Step 3
Exam Tip
(|x+1|>4) में केंद्र से बाहर के दो भाग मिलते हैं और endpoint शामिल नहीं होते। परीक्षा में (>) वाले absolute value में बाहर की ओर shading करें।
A. \(x\le 7\), (7) पर बंद बिंदु और बाईं ओर छाया/\(x\le 7\), closed dot at (7) shaded left
Step 1
Concept
Multiplying by (2) gives \(x-1\le 6\), so \(x\le 7\). In exams, multiplying by a positive number does not reverse the sign.
Step 2
Why this answer is correct
The correct answer is A. \(x\le 7\), (7) पर बंद बिंदु और बाईं ओर छाया / \(x\le 7\), closed dot at (7) shaded left. Multiplying by (2) gives \(x-1\le 6\), so \(x\le 7\). In exams, multiplying by a positive number does not reverse the sign.
Step 3
Exam Tip
दोनों ओर (2) से गुणा करने पर \(x-1\le 6\), इसलिए \(x\le 7\)। परीक्षा में धनात्मक गुणा करने पर चिन्ह नहीं बदलता।
B. \(x\le -1\), (-1) से बाईं ओर बंद बिंदु/\(x\le -1\), closed dot at (-1) shaded left
Step 1
Concept
Solving gives \(5-x\ge 6\), then \(-x\ge 1\), so \(x\le -1\). In exams, remember to reverse the sign when removing (-x).
Step 2
Why this answer is correct
The correct answer is B. \(x\le -1\), (-1) से बाईं ओर बंद बिंदु / \(x\le -1\), closed dot at (-1) shaded left. Solving gives \(5-x\ge 6\), then \(-x\ge 1\), so \(x\le -1\). In exams, remember to reverse the sign when removing (-x).
Step 3
Exam Tip
हल \(5-x\ge 6\) से \(-x\ge 1\) और \(x\le -1\) आता है। परीक्षा में (-x) हटाते समय चिन्ह पलटना न भूलें।
In intersection, the common part of both intervals is ([-4,0]). In exams, look for the overlapping region for \(\cap\).
Step 2
Why this answer is correct
The correct answer is A. ([-4,0]). In intersection, the common part of both intervals is ([-4,0]). In exams, look for the overlapping region for \(\cap\).
Step 3
Exam Tip
Intersection में दोनों intervals का common भाग ([-4,0]) है। परीक्षा में \(\cap\) के लिए overlapping region देखें।
A. (x<-1) या \(4\le x\le 9\)/(x<-1) or \(4\le x\le 9\)
Step 1
Concept
The first part is before (-1), and the second part is closed from (4) to (9). In exams, read separate intervals separately.
Step 2
Why this answer is correct
The correct answer is A. (x<-1) या \(4\le x\le 9\) / (x<-1) or \(4\le x\le 9\). The first part is before (-1), and the second part is closed from (4) to (9). In exams, read separate intervals separately.
Step 3
Exam Tip
पहला भाग (-1) से पहले है और दूसरा भाग (4) से (9) तक बंद है। परीक्षा में अलग intervals को अलग-अलग पढ़ें।
The complement of ([2,6)) excludes (2) but includes (6). In exams, check endpoints carefully when taking complements.
Step 2
Why this answer is correct
The correct answer is A. (\(-\infty,2\)\cup[6,\infty)). The complement of ([2,6)) excludes (2) but includes (6). In exams, check endpoints carefully when taking complements.
Step 3
Exam Tip
([2,6)) के complement में (2) बाहर नहीं आता पर (6) शामिल होता है। परीक्षा में complement लेते समय endpoints उलटकर देखें।
\(x^2\ge 16\) means \(|x|\ge 4\), giving two closed outer regions. In exams, include (-4) and (4) for \(\ge\).
Step 2
Why this answer is correct
The correct answer is C. (\(-\infty,-4]\cup[4,\infty\)). \(x^2\ge 16\) means \(|x|\ge 4\), giving two closed outer regions. In exams, include (-4) and (4) for \(\ge\).
Step 3
Exam Tip
\(x^2\ge 16\) में \(|x|\ge 4\), इसलिए बाहर के दो closed भाग मिलते हैं। परीक्षा में \(\ge\) होने पर (-4) और (4) शामिल करें।
\(\le -4\) gives a closed endpoint, and (>5) gives an open endpoint. In exams, always use round brackets on the infinite side.
Step 2
Why this answer is correct
The correct answer is A. ((-\infty,-4]\cup\(5,\infty\)). \(\le -4\) gives a closed endpoint, and (>5) gives an open endpoint. In exams, always use round brackets on the infinite side.
Step 3
Exam Tip
\(\le -4\) बंद endpoint देता है और (>5) खुला endpoint देता है। परीक्षा में infinite side पर हमेशा round bracket लगाएँ।
A. (x<2), (2) पर खुला बिंदु और बाईं ओर/(x<2), open dot at (2) shaded left
Step 1
Concept
(7-3x>1) gives (-3x>-6), hence (x<2). In exams, a negative coefficient reverses the direction.
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. (7-3x>1) gives (-3x>-6), hence (x<2). In exams, a negative coefficient reverses the direction.
Step 3
Exam Tip
(7-3x>1) से (-3x>-6) और (x<2) मिलता है। परीक्षा में negative coefficient हो तो दिशा पलटती है।
The common part is from (2) to (3), but (2) is excluded and (3) is included. In exams, check endpoint inclusion separately in intersections.
Step 2
Why this answer is correct
The correct answer is A. ((2,3]). The common part is from (2) to (3), but (2) is excluded and (3) is included. In exams, check endpoint inclusion separately in intersections.
Step 3
Exam Tip
Common भाग (2) से (3) तक है, पर (2) शामिल नहीं और (3) शामिल है। परीक्षा में intersection में endpoints की inclusion अलग से जाँचें।
A closed ray includes the endpoint, and a left ray means less than or equal. In exams, check both ray direction and endpoint.
Step 2
Why this answer is correct
The correct answer is B. \(x\le -7\). A closed ray includes the endpoint, and a left ray means less than or equal. In exams, check both ray direction and endpoint.
Step 3
Exam Tip
Closed ray में endpoint शामिल होता है और left ray का मतलब कम या बराबर है। परीक्षा में ray की दिशा और endpoint दोनों देखें।
A. (x>-4), (-4) पर खुला बिंदु और दाईं ओर/(x>-4), open dot at (-4) shaded right
Step 1
Concept
Multiplying by (-3) reverses the inequality to (2x+5>-3), so (x>-4). In exams, reverse the sign when removing a negative denominator.
Step 2
Why this answer is correct
The correct answer is A. (x>-4), (-4) पर खुला बिंदु और दाईं ओर / (x>-4), open dot at (-4) shaded right. Multiplying by (-3) reverses the inequality to (2x+5>-3), so (x>-4). In exams, reverse the sign when removing a negative denominator.
Step 3
Exam Tip
(-3) से गुणा करने पर inequality reverse होकर (2x+5>-3) बनती है, इसलिए (x>-4)। परीक्षा में negative denominator हटाते समय चिन्ह पलटें।
In (x<0), (0) is open; in \(2\le x<5\), (2) is closed and (5) is open. In exams, convert or into union.
Step 2
Why this answer is correct
The correct answer is A. (\(-\infty,0\)\cup[2,5)). In (x<0), (0) is open; in \(2\le x<5\), (2) is closed and (5) is open. In exams, convert or into union.
Step 3
Exam Tip
(x<0) में (0) खुला है और \(2\le x<5\) में (2) बंद तथा (5) खुला है। परीक्षा में or को union में बदलें।
Usually in Class (11), \(\mathbb{N}={1,2,3,\ldots}\), so points from (1) to (6) are marked. In exams, if the domain is natural numbers, mark only natural points.
Step 2
Why this answer is correct
The correct answer is B. ({1,2,3,4,5,6}). Usually in Class (11), \(\mathbb{N}={1,2,3,\ldots}\), so points from (1) to (6) are marked. In exams, if the domain is natural numbers, mark only natural points.
Step 3
Exam Tip
Class (11) में सामान्यतः \(\mathbb{N}={1,2,3,\ldots}\), इसलिए (1) से (6) तक बिंदु होंगे। परीक्षा में domain natural हो तो only natural points mark करें।
The two intervals meet at (-2), and (5) is not included. In exams, merge touching intervals when the common endpoint is included.
Step 2
Why this answer is correct
The correct answer is A. (\(-\infty,5\)). The two intervals meet at (-2), and (5) is not included. In exams, merge touching intervals when the common endpoint is included.
Step 3
Exam Tip
दोनों intervals (-2) पर जुड़ते हैं और (5) शामिल नहीं है। परीक्षा में touching intervals को merge करें जब common endpoint included हो।
(x<1) and \(x\ge 1\) together give all real numbers. In exams, the union of complementary rays is the whole number line.
Step 2
Why this answer is correct
The correct answer is A. (\(-\infty,\infty\)). (x<1) and \(x\ge 1\) together give all real numbers. In exams, the union of complementary rays is the whole number line.
Step 3
Exam Tip
(x<1) और \(x\ge 1\) मिलकर सभी real numbers देते हैं। परीक्षा में complementary rays का union पूरी number line होता है।
No (x) can be greater than (3) and less than or equal to (3) at the same time. In exams, the intersection of contradictory conditions is empty.
Step 2
Why this answer is correct
The correct answer is A. \(\emptyset\). No (x) can be greater than (3) and less than or equal to (3) at the same time. In exams, the intersection of contradictory conditions is empty.
Step 3
Exam Tip
कोई (x) एक साथ (3) से बड़ा और (3) से कम या बराबर नहीं हो सकता। परीक्षा में contradictory conditions का intersection empty होता है।
A. \(x\le 16\), (16) पर बंद बिंदु और बाईं ओर/\(x\le 16\), closed dot at (16) shaded left
Step 1
Concept
\(3x-6\le 2x+10\) gives \(x\le 16\). In exams, expand brackets and move like terms carefully.
Step 2
Why this answer is correct
The correct answer is A. \(x\le 16\), (16) पर बंद बिंदु और बाईं ओर / \(x\le 16\), closed dot at (16) shaded left. \(3x-6\le 2x+10\) gives \(x\le 16\). In exams, expand brackets and move like terms carefully.
Step 3
Exam Tip
\(3x-6\le 2x+10\) से \(x\le 16\) मिलता है। परीक्षा में brackets खोलकर like terms को सही side पर रखें।
A. (x>-2), (-2) पर खुला बिंदु और दाईं ओर/(x>-2), open dot at (-2) shaded right
Step 1
Concept
(2-2x<x+8) gives (-3x<6), so (x>-2). In exams, change the sign while dividing by a negative coefficient.
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. (2-2x<x+8) gives (-3x<6), so (x>-2). In exams, change the sign while dividing by a negative coefficient.
Step 3
Exam Tip
(2-2x<x+8) से (-3x<6), इसलिए (x>-2)। परीक्षा में negative coefficient से divide करते समय sign बदलें।
The common part is from (-5) to (1), including (-5) and excluding (1). In exams, both conditions must be satisfied in an intersection.
Step 2
Why this answer is correct
The correct answer is A. ([-5,1)). The common part is from (-5) to (1), including (-5) and excluding (1). In exams, both conditions must be satisfied in an intersection.
Step 3
Exam Tip
Common भाग (-5) से (1) तक है, जिसमें (-5) शामिल और (1) बाहर है। परीक्षा में intersection में दोनों शर्तें साथ पूरी होनी चाहिए।
Open dots exclude the endpoints, and outer shading gives two rays. In exams, read outer shading as an or inequality.
Step 2
Why this answer is correct
The correct answer is A. (x<-1) या (x>4) / (x<-1) or (x>4). Open dots exclude the endpoints, and outer shading gives two rays. In exams, read outer shading as an or inequality.
Step 3
Exam Tip
Open dots endpoints को exclude करते हैं और बाहर shading दो rays देती है। परीक्षा में outer shading को or inequality समझें।
The intervals overlap and together extend from (-4) to (3). In exams, the union of overlapping intervals forms one larger interval.
Step 2
Why this answer is correct
The correct answer is A. ([-4,3)). The intervals overlap and together extend from (-4) to (3). In exams, the union of overlapping intervals forms one larger interval.
Step 3
Exam Tip
दोनों intervals overlap करते हैं और मिलकर (-4) से (3) तक जाते हैं। परीक्षा में overlapping intervals का union एक बड़ा interval बनता है।
Removing ([-1,2)) removes (-1) but keeps (2). In exams, check endpoints carefully in set difference.
Step 2
Why this answer is correct
The correct answer is A. (\(-\infty,-1\)\cup[2,4]). Removing ([-1,2)) removes (-1) but keeps (2). In exams, check endpoints carefully in set difference.
Step 3
Exam Tip
([-1,2)) हटाने पर (-1) भी हटता है पर (2) बचता है। परीक्षा में set difference में हटाए गए interval के endpoints को ध्यान से जाँचें।
\(4\le -2x<10\) gives \(-5<x\le -2\) after negative division. In exams, endpoints also reverse when dividing a compound inequality by a negative.
Step 2
Why this answer is correct
The correct answer is A. ((-5,-2]). \(4\le -2x<10\) gives \(-5<x\le -2\) after negative division. In exams, endpoints also reverse when dividing a compound inequality by a negative.
Step 3
Exam Tip
\(4\le -2x<10\) से negative division के बाद \(-5<x\le -2\) मिलता है। परीक्षा में compound inequality को negative से divide करते समय endpoints भी उलटते हैं।
The common part is from (2) to (5), including (2) and excluding (5). In exams, read intersection as a simultaneous condition.
Step 2
Why this answer is correct
The correct answer is A. \(2\le x<5\). The common part is from (2) to (5), including (2) and excluding (5). In exams, read intersection as a simultaneous condition.
Step 3
Exam Tip
Common भाग (2) से (5) तक है, जिसमें (2) शामिल और (5) बाहर है। परीक्षा में intersection को simultaneous condition की तरह पढ़ें।
B. \(x\le -13\), (-13) पर बंद बिंदु और बाईं ओर/\(x\le -13\), closed dot at (-13) shaded left
Step 1
Concept
Multiplying by (15) gives \(3x+6\ge 5x-20\), so \(x\le 13\), not \(x\ge 13\). In exams, recheck the final sign and arithmetic.
Step 2
Why this answer is correct
The correct answer is B. \(x\le -13\), (-13) पर बंद बिंदु और बाईं ओर / \(x\le -13\), closed dot at (-13) shaded left. Multiplying by (15) gives \(3x+6\ge 5x-20\), so \(x\le 13\), not \(x\ge 13\). In exams, recheck the final sign and arithmetic.
Step 3
Exam Tip
(15) से गुणा करने पर \(3x+6\ge 5x-20\), इसलिए \(x\le 13\) नहीं बल्कि \(x\le 13\) आता है। परीक्षा में final sign और arithmetic दोबारा जाँचें।
A. (x<11), (11) पर खुला बिंदु और बाईं ओर/(x<11), open dot at (11) shaded left
Step 1
Concept
Multiplying by (4) gives (3x-1<2x+10), so (x<11). In exams, after clearing denominators, a strict sign gives an open endpoint.
Step 2
Why this answer is correct
The correct answer is A. (x<11), (11) पर खुला बिंदु और बाईं ओर / (x<11), open dot at (11) shaded left. Multiplying by (4) gives (3x-1<2x+10), so (x<11). In exams, after clearing denominators, a strict sign gives an open endpoint.
Step 3
Exam Tip
(4) से गुणा करने पर (3x-1<2x+10), इसलिए (x<11)। परीक्षा में denominators हटाने के बाद strict sign खुला endpoint देता है।
Multiplying all parts by (2) gives \(14\le 3x+2<32\), so \(4\le x<10\). In exams, apply the same operation to all parts of a compound inequality.
Step 2
Why this answer is correct
The correct answer is A. ([4,10)). Multiplying all parts by (2) gives \(14\le 3x+2<32\), so \(4\le x<10\). In exams, apply the same operation to all parts of a compound inequality.
Step 3
Exam Tip
दोनों ओर (2) से गुणा करने पर \(14\le 3x+2<32\), इसलिए \(4\le x<10\)। परीक्षा में compound inequality में सभी हिस्सों पर समान operation करें।
Integers greater than \(-\frac{9}{2}\) start at (-4), and (2) is the last integer up to \(\frac{5}{2}\). In exams, choose the nearest valid integer at fractional boundaries.
Step 2
Why this answer is correct
The correct answer is A. ({-4,-3,-2,-1,0,1,2}). Integers greater than \(-\frac{9}{2}\) start at (-4), and (2) is the last integer up to \(\frac{5}{2}\). In exams, choose the nearest valid integer at fractional boundaries.
Step 3
Exam Tip
\(-\frac{9}{2}\) से बड़े पूर्णांक (-4) से शुरू होते हैं और \(\frac{5}{2}\) तक (2) अंतिम पूर्णांक है। परीक्षा में fractional boundary पर nearest valid integer चुनें।
The common part of the two intervals is from (-3) to (1), excluding (-3) and including (1). In exams, check endpoint inclusion separately in intersections.
Step 2
Why this answer is correct
The correct answer is A. ((-3,1]). The common part of the two intervals is from (-3) to (1), excluding (-3) and including (1). In exams, check endpoint inclusion separately in intersections.
Step 3
Exam Tip
दोनों intervals का common भाग (-3) से (1) तक है, जिसमें (-3) बाहर और (1) शामिल है। परीक्षा में intersection में endpoint inclusion अलग से जाँचें।
\(|2x-1|\ge 7\) gives \(2x-1\le -7\) or \(2x-1\ge 7\), so \(x\le -3\) or \(x\ge 4\). In exams, absolute value with \(\ge\) forms closed outer rays.
Step 2
Why this answer is correct
The correct answer is A. (\(-\infty,-3]\cup[4,\infty\)). \(|2x-1|\ge 7\) gives \(2x-1\le -7\) or \(2x-1\ge 7\), so \(x\le -3\) or \(x\ge 4\). In exams, absolute value with \(\ge\) forms closed outer rays.
Step 3
Exam Tip
\(|2x-1|\ge 7\) से \(2x-1\le -7\) या \(2x-1\ge 7\), इसलिए \(x\le -3\) या \(x\ge 4\)। परीक्षा में \(\ge\) वाले absolute value में बाहर के closed rays बनते हैं।
A. \(x\ge 3\), (3) पर बंद बिंदु और दाईं ओर/\(x\ge 3\), closed dot at (3) shaded right
Step 1
Concept
\(4-3x\le -5\) gives \(-3x\le -9\), so \(x\ge 3\). In exams, reverse the inequality when dividing by a negative coefficient.
Step 2
Why this answer is correct
The correct answer is A. \(x\ge 3\), (3) पर बंद बिंदु और दाईं ओर / \(x\ge 3\), closed dot at (3) shaded right. \(4-3x\le -5\) gives \(-3x\le -9\), so \(x\ge 3\). In exams, reverse the inequality when dividing by a negative coefficient.
Step 3
Exam Tip
\(4-3x\le -5\) से \(-3x\le -9\), इसलिए \(x\ge 3\)। परीक्षा में negative coefficient से divide करते समय inequality reverse करें।
Removing ([-2,0)) also removes (-2) but keeps (0). In exams, check endpoint inclusion carefully in set difference.
Step 2
Why this answer is correct
The correct answer is C. \((-6,-2)\cup[0,1]\). Removing ([-2,0)) also removes (-2) but keeps (0). In exams, check endpoint inclusion carefully in set difference.
Step 3
Exam Tip
([-2,0)) हटाने पर (-2) भी हटता है और (0) बचता है। परीक्षा में set difference में हटाए गए endpoint की inclusion जाँचें।
