The conditions \(0\le x\le6\) and \(0\le y\le5\) restrict the region inside a rectangle. The inequality \(x+y\ge4\) only cuts off one corner.
Step 2
Why this answer is correct
The correct answer is B. सीमित / Bounded. The conditions \(0\le x\le6\) and \(0\le y\le5\) restrict the region inside a rectangle. The inequality \(x+y\ge4\) only cuts off one corner.
Step 3
Exam Tip
\(0\le x\le6\) और \(0\le y\le5\) क्षेत्र को आयत में सीमित करते हैं। \(x+y\ge4\) केवल उसका एक कोना काटता है।
Writing the line as \(y=\frac{18-2x}{3}\), the upper side is \(y\ge\frac{18-2x}{3}\). This is equivalent to \(2x+3y\ge18\).
Step 2
Why this answer is correct
The correct answer is B. \(2x+3y\ge18\). Writing the line as \(y=\frac{18-2x}{3}\), the upper side is \(y\ge\frac{18-2x}{3}\). This is equivalent to \(2x+3y\ge18\).
Step 3
Exam Tip
रेखा को \(y=\frac{18-2x}{3}\) लिखने पर ऊपर का भाग \(y\ge\frac{18-2x}{3}\) है। यह \(2x+3y\ge18\) के बराबर है।
Putting (x=0) gives \(4y\le16\), so \(y\le4\). To find an axis intercept limit, put the other variable equal to zero.
Step 2
Why this answer is correct
The correct answer is C. ( (0,4) ). Putting (x=0) gives \(4y\le16\), so \(y\le4\). To find an axis intercept limit, put the other variable equal to zero.
Step 3
Exam Tip
(x=0) रखने पर \(4y\le16\), इसलिए \(y\le4\) है। अक्षों पर अधिकतम बिंदु निकालने के लिए दूसरे चर को शून्य रखें।
The conditions \(x\le5\) and \(y\le4\) with the axes form a bounded rectangular region. A bounded region needs closing boundaries in all directions.
Step 2
Why this answer is correct
The correct answer is C. \(x\ge0\), \(y\ge0\), \(x\le5\), \(y\le4\). The conditions \(x\le5\) and \(y\le4\) with the axes form a bounded rectangular region. A bounded region needs closing boundaries in all directions.
Step 3
Exam Tip
\(x\le5\) और \(y\le4\) अक्षों के साथ आयताकार सीमित क्षेत्र बनाते हैं। सीमित क्षेत्र के लिए सभी दिशाओं में बंद सीमा चाहिए।
C. हल क्षेत्र के विपरीत तरफ/On the opposite side of the solution region
Step 1
Concept
At the origin, (0>24) is false, so the side containing the origin is not the solution. If the test point fails, shade the other side.
Step 2
Why this answer is correct
The correct answer is C. हल क्षेत्र के विपरीत तरफ / On the opposite side of the solution region. At the origin, (0>24) is false, so the side containing the origin is not the solution. If the test point fails, shade the other side.
Step 3
Exam Tip
मूलबिंदु पर (0>24) असत्य है, इसलिए मूलबिंदु वाला भाग हल नहीं है। असत्य परीक्षण-बिंदु मिलने पर दूसरी तरफ छायांकन करें।
A. ( (10,0) ) और ( (0,4) )/( (10,0) ) and ( (0,4) )
Step 1
Concept
The boundary line is (2x+5y=20), giving intercepts (x=10) and (y=4). Drawing the line first is the base of graphical solution.
Step 2
Why this answer is correct
The correct answer is A. ( (10,0) ) और ( (0,4) ) / ( (10,0) ) and ( (0,4) ). The boundary line is (2x+5y=20), giving intercepts (x=10) and (y=4). Drawing the line first is the base of graphical solution.
Step 3
Exam Tip
सीमा रेखा (2x+5y=20) है, जिससे (x=10) और (y=4) प्रतिच्छेद मिलते हैं। पहले रेखा बनाना ग्राफीय हल का आधार है।
From (5x-2y<10), we get \(y>\frac{5x-10}{2}\). Do not forget to reverse the sign when removing a negative coefficient.
Step 2
Why this answer is correct
The correct answer is A. \(y>\frac{5x-10}{2}\). From (5x-2y<10), we get \(y>\frac{5x-10}{2}\). Do not forget to reverse the sign when removing a negative coefficient.
Step 3
Exam Tip
(5x-2y<10) से \(y>\frac{5x-10}{2}\) मिलता है। ऋणात्मक गुणांक हटाते समय चिन्ह पलटना न भूलें।
From \(x\le4\) and \(y\le3\), the maximum possible (x+y) is (7). Hence \(x+y\le8\) does not further reduce the region.
Step 2
Why this answer is correct
The correct answer is C. \(x+y\le8\). From \(x\le4\) and \(y\le3\), the maximum possible (x+y) is (7). Hence \(x+y\le8\) does not further reduce the region.
Step 3
Exam Tip
\(x\le4\) और \(y\le3\) से अधिकतम (x+y=7) हो सकता है। इसलिए \(x+y\le8\) अलग से क्षेत्र नहीं घटाती।
In the first quadrant, the region above (x+y=5) extends without bound. In such questions, check both direction and axis restrictions.
Step 2
Why this answer is correct
The correct answer is C. असीमित क्षेत्र / Unbounded region. In the first quadrant, the region above (x+y=5) extends without bound. In such questions, check both direction and axis restrictions.
Step 3
Exam Tip
पहले चतुर्थांश में रेखा (x+y=5) के ऊपर का भाग असीमित फैलता है। ऐसे प्रश्न में दिशा और अक्षों की शर्त साथ देखें।
D. किसी भी असमानता को संतुष्ट नहीं करता/It satisfies neither inequality
Step 1
Concept
At ( (2,2) ), both left sides equal (6), which is less than (10). In a combined solution, all inequalities must be true together.
Step 2
Why this answer is correct
The correct answer is D. किसी भी असमानता को संतुष्ट नहीं करता / It satisfies neither inequality. At ( (2,2) ), both left sides equal (6), which is less than (10). In a combined solution, all inequalities must be true together.
Step 3
Exam Tip
( (2,2) ) पर दोनों पक्षों में मान (6) आता है, जो (10) से कम है। संयुक्त हल में सभी असमानताएं एक साथ सत्य होनी चाहिए।
B. मूलबिंदु वाला भाग स्वीकार होगा/The side containing origin is accepted
Step 1
Concept
Since \(0\ge-5\) is true, the side containing the origin is the solution region. A simple test point quickly decides the shading.
Step 2
Why this answer is correct
The correct answer is B. मूलबिंदु वाला भाग स्वीकार होगा / The side containing origin is accepted. Since \(0\ge-5\) is true, the side containing the origin is the solution region. A simple test point quickly decides the shading.
Step 3
Exam Tip
\(0\ge-5\) सत्य है, इसलिए मूलबिंदु वाला भाग हल क्षेत्र है। सरल परीक्षण-बिंदु से छायांकन जल्दी तय होता है।
At ( (3,2) ), (8<10) and (9<12), so it is strictly inside. A boundary point must make at least one inequality an equality.
Step 2
Why this answer is correct
The correct answer is C. ( (3,2) ). At ( (3,2) ), (8<10) and (9<12), so it is strictly inside. A boundary point must make at least one inequality an equality.
Step 3
Exam Tip
( (3,2) ) पर (8<10) और (9<12), इसलिए यह अंदर है। सीमा पर होने के लिए कम से कम एक असमानता बराबरी बननी चाहिए।
B. रेखा (4x+y=10) बिंदुदार होगी/The line (4x+y=10) is dotted
Step 1
Concept
The sign (<) is strict, so the boundary line is not included. Such a boundary is drawn dotted in the graph.
Step 2
Why this answer is correct
The correct answer is B. रेखा (4x+y=10) बिंदुदार होगी / The line (4x+y=10) is dotted. The sign (<) is strict, so the boundary line is not included. Such a boundary is drawn dotted in the graph.
Step 3
Exam Tip
(<) सख्त असमानता है, इसलिए सीमा रेखा शामिल नहीं होती। ग्राफ में ऐसी सीमा रेखा बिंदुदार बनाते हैं।
From \(3x-2y \ge 6\), we get \(y \le \frac{3x-6}{2}\). When dividing by a negative number, the inequality sign reverses.
Step 2
Why this answer is correct
The correct answer is B. \(y \le \frac{3x-6}{2}\). From \(3x-2y \ge 6\), we get \(y \le \frac{3x-6}{2}\). When dividing by a negative number, the inequality sign reverses.
Step 3
Exam Tip
\(3x-2y \ge 6\) से \(y \le \frac{3x-6}{2}\) मिलता है। ऋणात्मक संख्या से भाग देते समय असमानता का चिन्ह बदलता है।
A. सीमा रेखा ठोस होगी और मूलबिंदु वाला भाग छायांकित होगा/Boundary is solid and the side containing origin is shaded
Step 1
Concept
Since \(0 \le 12\) is true, the half-plane containing the origin is taken. In exams, check the boundary line first and then use a test point.
Step 2
Why this answer is correct
The correct answer is A. सीमा रेखा ठोस होगी और मूलबिंदु वाला भाग छायांकित होगा / Boundary is solid and the side containing origin is shaded. Since \(0 \le 12\) is true, the half-plane containing the origin is taken. In exams, check the boundary line first and then use a test point.
Step 3
Exam Tip
\(0 \le 12\) सत्य है, इसलिए मूलबिंदु वाला अर्ध-तल लिया जाता है। परीक्षा में पहले सीमा रेखा और फिर परीक्षण-बिंदु जांचें।
For ((3,1)), (2x+y=7), so it fails the strict inequality. This item needs a point strictly inside the region.
Step 2
Why this answer is correct
The correct answer is B. बिंदु ((3,1)) / Point ((3,1)). For ((3,1)), (2x+y=7), so it fails the strict inequality. This item needs a point strictly inside the region.
Step 3
Exam Tip
((3,1)) रखने पर (7<7) नहीं बल्कि (2x+y=7) होता है, इसलिए यह गलत है। सही जांच से कोई विकल्प नहीं?
Boundary intersections give ((2,3)), ((4,1)), and ((4,3)). On a graph, keep only intersections satisfying all inequalities.
Step 2
Why this answer is correct
The correct answer is B. ((2,3)), ((4,1)), ((4,3)). Boundary intersections give ((2,3)), ((4,1)), and ((4,3)). On a graph, keep only intersections satisfying all inequalities.
Step 3
Exam Tip
सीमाओं के प्रतिच्छेद से ((2,3)), ((4,1)), ((4,3)) मिलते हैं। ग्राफ में केवल वे प्रतिच्छेद लें जो सभी असमानताओं को संतुष्ट करें।
A. वह आधा तल जिसमें ((0,0)) है/The half-plane containing ((0,0))
Step 1
Concept
Substituting ((0,0)) gives \(0\geq 6\), which is false, so the half-plane containing the origin is rejected. In exams, choose a test point not lying on the boundary.
Step 2
Why this answer is correct
The correct answer is A. वह आधा तल जिसमें ((0,0)) है / The half-plane containing ((0,0)). Substituting ((0,0)) gives \(0\geq 6\), which is false, so the half-plane containing the origin is rejected. In exams, choose a test point not lying on the boundary.
Step 3
Exam Tip
((0,0)) रखने पर \(0\geq 6\) गलत है, इसलिए मूल वाला आधा तल हटेगा। परीक्षा में जांच-बिंदु सीमा रेखा पर नहीं होना चाहिए।
D. (3x+y=6) टूटी और (x-y=2) ठोस होगी/(3x+y=6) dashed and (x-y=2) solid
Step 1
Concept
A strict inequality (>) excludes the boundary, so its line is dashed. The inequality \(\leq\) includes the boundary, so its line is solid.
Step 2
Why this answer is correct
The correct answer is D. (3x+y=6) टूटी और (x-y=2) ठोस होगी / (3x+y=6) dashed and (x-y=2) solid. A strict inequality (>) excludes the boundary, so its line is dashed. The inequality \(\leq\) includes the boundary, so its line is solid.
Step 3
Exam Tip
कड़ी असमानता (>) के लिए सीमा शामिल नहीं होती, इसलिए रेखा टूटी होती है। \(\leq\) में सीमा शामिल होती है, इसलिए रेखा ठोस होती है।
B. बिंदु (\left\(\frac{8}{3},\frac{8}{3}\right\))/Point (\left\(\frac{8}{3},\frac{8}{3}\right\))
Step 1
Concept
Solving the two boundary lines together gives \(x=y=\frac{8}{3}\). In exams, find vertices by treating boundary lines as equations.
Step 2
Why this answer is correct
The correct answer is B. बिंदु (\left\(\frac{8}{3},\frac{8}{3}\right\)) / Point (\left\(\frac{8}{3},\frac{8}{3}\right\)). Solving the two boundary lines together gives \(x=y=\frac{8}{3}\). In exams, find vertices by treating boundary lines as equations.
Step 3
Exam Tip
दोनों रेखाओं को साथ हल करने पर \(x=y=\frac{8}{3}\) मिलता है। परीक्षा में शीर्ष निकालने के लिए सीमा रेखाओं को समीकरण मानें।
C. \(\frac{15}{2}\) वर्ग इकाई/\(\frac{15}{2}\) square units
Step 1
Concept
The vertices are ((1,0)), ((4,0)), ((4,1)), and ((1,4)). With parallel sides (4) and (1) and distance (3), the area is (\frac{1}{2}(4+1)3=\frac{15}{2}).
Step 2
Why this answer is correct
The correct answer is C. \(\frac{15}{2}\) वर्ग इकाई / \(\frac{15}{2}\) square units. The vertices are ((1,0)), ((4,0)), ((4,1)), and ((1,4)). With parallel sides (4) and (1) and distance (3), the area is (\frac{1}{2}(4+1)3=\frac{15}{2}).
Step 3
Exam Tip
क्षेत्र के शीर्ष ((1,0)), ((4,0)), ((4,1)), ((1,4)) हैं। समांतर भुजाओं (4) और (1) तथा दूरी (3) से क्षेत्रफल (\frac{1}{2}(4+1)3=\frac{15}{2}) है।
