A. शीर्ष ((2,0)), ((6,0)), ((4,2)) वाला बंद त्रिभुज/Closed triangle with vertices ((2,0)), ((6,0)), ((4,2))
Step 1
Concept
The three half-planes form a closed triangle. In exams, first find the intersection points of boundary lines.
Step 2
Why this answer is correct
The correct answer is A. शीर्ष ((2,0)), ((6,0)), ((4,2)) वाला बंद त्रिभुज / Closed triangle with vertices ((2,0)), ((6,0)), ((4,2)). The three half-planes form a closed triangle. In exams, first find the intersection points of boundary lines.
Step 3
Exam Tip
तीनों अर्द्ध-तल मिलकर बंद त्रिभुज बनाते हैं। परीक्षा में पहले रेखाओं के प्रतिच्छेद बिंदु निकालें।
Solving the two boundary equations gives \(x=\frac{16}{5}\) and \(y=\frac{27}{5}\). While finding corners, keep only feasible intersections.
Step 2
Why this answer is correct
The correct answer is A. (\left\(\frac{16}{5},\frac{27}{5}\right\)). Solving the two boundary equations gives \(x=\frac{16}{5}\) and \(y=\frac{27}{5}\). While finding corners, keep only feasible intersections.
Step 3
Exam Tip
दोनों सीमा समीकरण हल करने पर \(x=\frac{16}{5}\) और \(y=\frac{27}{5}\) मिलता है। कोना निकालते समय केवल वैध प्रतिच्छेद लें।
A. ((0,0)), ((6,0)), (\left\(\frac{21}{5},\frac{18}{5}\right\)), ((0,5))
Step 1
Concept
The intersection of the two slant lines is (\left\(\frac{21}{5},\frac{18}{5}\right\)). Use valid intercepts on the axes to choose all polygon corners.
Step 2
Why this answer is correct
The correct answer is A. ((0,0)), ((6,0)), (\left\(\frac{21}{5},\frac{18}{5}\right\)), ((0,5)). The intersection of the two slant lines is (\left\(\frac{21}{5},\frac{18}{5}\right\)). Use valid intercepts on the axes to choose all polygon corners.
Step 3
Exam Tip
दोनों तिरछी रेखाओं का प्रतिच्छेद (\left\(\frac{21}{5},\frac{18}{5}\right\)) है। अक्षों पर वैध अवरोध लेकर पूरे बहुभुज के कोने चुनें।
The vertices are ((2,0)), ((6,0)), ((3,6)), and ((2,7)). Using the shoelace method or splitting into parts gives area (20).
Step 2
Why this answer is correct
The correct answer is A. (20) वर्ग इकाई / (20) square units. The vertices are ((2,0)), ((6,0)), ((3,6)), and ((2,7)). Using the shoelace method or splitting into parts gives area (20).
Step 3
Exam Tip
शीर्ष ((2,0)), ((6,0)), ((3,6)), ((2,7)) हैं। शूलेस विधि या भागों में बांटकर क्षेत्रफल (20) मिलता है।
The vertices are ((2,1)), (\left\(\frac{15}{2},1\right\)), and (\(2,\frac{14}{3}\)). Check parallel distances carefully before using triangle area.
Step 2
Why this answer is correct
The correct answer is B. (15) वर्ग इकाई / (15) square units. The vertices are ((2,1)), (\left\(\frac{15}{2},1\right\)), and (\(2,\frac{14}{3}\)). Check parallel distances carefully before using triangle area.
Step 3
Exam Tip
शीर्ष ((2,1)), (\left\(\frac{15}{2},1\right\)), (\(2,\frac{14}{3}\)) हैं। आधार \(\frac{11}{2}\) और ऊंचाई \(\frac{11}{3}\) से क्षेत्रफल \(\frac{121}{12}\) नहीं बल्कि सही गणना के लिए अक्षों के समांतर दूरी जांचें।
Both inequalities select the upper sides of the lines, and in the first quadrant the region extends infinitely. Since \(\geq\) is used, boundaries are included.
Step 2
Why this answer is correct
The correct answer is A. सीमा रहित और बंद / Unbounded and closed. Both inequalities select the upper sides of the lines, and in the first quadrant the region extends infinitely. Since \(\geq\) is used, boundaries are included.
Step 3
Exam Tip
दोनों असमानताएं रेखाओं के ऊपर वाले भाग को चुनती हैं और प्रथम चतुर्थांश में क्षेत्र ऊपर की ओर अनंत है। \(\geq\) होने से सीमाएं शामिल हैं।
The same expression (2x-y) cannot be at most (4) and greater than (8) at the same time. Check contradictions before drawing the graph.
Step 2
Why this answer is correct
The correct answer is C. खाली क्षेत्र / Empty region. The same expression (2x-y) cannot be at most (4) and greater than (8) at the same time. Check contradictions before drawing the graph.
Step 3
Exam Tip
एक ही राशि (2x-y) एक साथ (4) से कम या बराबर और (8) से अधिक नहीं हो सकती। विरोधी शर्तें दिखें तो ग्राफ बनाने से पहले जांच लें।
A. मूल-बिंदु वाला अर्द्ध-तल/Half-plane containing the origin
Step 1
Concept
The origin satisfies the inequality, so shading is on that side. Choose a test point not lying on the line.
Step 2
Why this answer is correct
The correct answer is A. मूल-बिंदु वाला अर्द्ध-तल / Half-plane containing the origin. The origin satisfies the inequality, so shading is on that side. Choose a test point not lying on the line.
Step 3
Exam Tip
मूल-बिंदु असमानता को संतुष्ट करता है इसलिए उसी ओर छायांकन होगा। जांच बिंदु रेखा पर न हो।
B. टूटी रेखा और मूल-बिंदु वाली ओर/Dashed line and toward the origin
Step 1
Concept
In a strict inequality, the boundary is excluded, so the line is dashed. The origin satisfies the condition, so shade its side.
Step 2
Why this answer is correct
The correct answer is B. टूटी रेखा और मूल-बिंदु वाली ओर / Dashed line and toward the origin. In a strict inequality, the boundary is excluded, so the line is dashed. The origin satisfies the condition, so shade its side.
Step 3
Exam Tip
कठोर असमानता में सीमा शामिल नहीं होती इसलिए रेखा टूटी होगी। मूल-बिंदु शर्त पूरी करता है इसलिए उसी ओर छायांकन होगा।
Checking the corners gives (10) at ((0,5)), (12) at ((4,0)), and (16) at (\left\(\frac{14}{5},\frac{18}{5}\right\)). A linear expression attains its maximum at a corner.
Step 2
Why this answer is correct
The correct answer is A. (16). Checking the corners gives (10) at ((0,5)), (12) at ((4,0)), and (16) at (\left\(\frac{14}{5},\frac{18}{5}\right\)). A linear expression attains its maximum at a corner.
Step 3
Exam Tip
कोनों पर जांचने से ((0,5)) पर (10), ((4,0)) पर (12), और (\left\(\frac{14}{5},\frac{18}{5}\right\)) पर (16) मिलता है। रैखिक व्यंजक का अधिकतम कोनों पर मिलता है।
For parallel lines with the same slope, (2x+1) is always greater than (2x-3). Hence no (y) can satisfy both conditions.
Step 2
Why this answer is correct
The correct answer is C. कोई हल नहीं / No solution. For parallel lines with the same slope, (2x+1) is always greater than (2x-3). Hence no (y) can satisfy both conditions.
Step 3
Exam Tip
एक ही ढाल वाली रेखाओं के लिए (2x+1) हमेशा (2x-3) से बड़ा है। इसलिए कोई (y) दोनों शर्तें पूरी नहीं कर सकता।
The \(\geq\) inequalities give the upper-side region in the first quadrant. Boundaries are included and the region extends infinitely.
Step 2
Why this answer is correct
The correct answer is C. सीमा रहित और बंद / Unbounded and closed. The \(\geq\) inequalities give the upper-side region in the first quadrant. Boundaries are included and the region extends infinitely.
Step 3
Exam Tip
\(\geq\) वाली असमानताएं प्रथम चतुर्थांश में ऊपर की दिशा का क्षेत्र देती हैं। सीमाएं शामिल हैं और क्षेत्र अनंत तक जाता है।
Solving the two boundary equations gives \(x=\frac{16}{5}\) and \(y=\frac{27}{5}\). It is a good method to test the intersection in all inequalities.
Step 2
Why this answer is correct
The correct answer is A. (\left\(\frac{16}{5},\frac{27}{5}\right\)). Solving the two boundary equations gives \(x=\frac{16}{5}\) and \(y=\frac{27}{5}\). It is a good method to test the intersection in all inequalities.
Step 3
Exam Tip
दोनों सीमा समीकरण हल करने पर \(x=\frac{16}{5}\) और \(y=\frac{27}{5}\) मिलता है। प्रतिच्छेद को सभी असमानताओं से जांचना अच्छा तरीका है।
The first and second inequalities contradict each other. Even on the common boundary, the strict inequality removes the points.
Step 2
Why this answer is correct
The correct answer is B. खाली समुच्चय / Empty set. The first and second inequalities contradict each other. Even on the common boundary, the strict inequality removes the points.
Step 3
Exam Tip
पहली और दूसरी असमानता परस्पर विरोधी हैं। समान सीमा पर भी कठोर असमानता बिंदुओं को हटा देती है।
The vertices are ((1,2)), ((5,2)), and ((1,6)). Base and height are (4), so the area is (8).
Step 2
Why this answer is correct
The correct answer is B. (8) वर्ग इकाई / (8) square units. The vertices are ((1,2)), ((5,2)), and ((1,6)). Base and height are (4), so the area is (8).
Step 3
Exam Tip
शीर्ष ((1,2)), ((5,2)), ((1,6)) मिलते हैं। आधार और ऊंचाई (4) हैं इसलिए क्षेत्रफल (8) है।
Solving the two equations gives \(x=\frac{18}{5}\) and \(y=\frac{14}{5}\). This is the inner corner on the graph.
Step 2
Why this answer is correct
The correct answer is A. (\left\(\frac{18}{5},\frac{14}{5}\right\)). Solving the two equations gives \(x=\frac{18}{5}\) and \(y=\frac{14}{5}\). This is the inner corner on the graph.
Step 3
Exam Tip
दोनों समीकरण हल करने पर \(x=\frac{18}{5}\) और \(y=\frac{14}{5}\) मिलता है। ग्राफ में यही आंतरिक कोना है।
At ((2,3)), (2x+5y=19) and (x-y=-1), so it cannot be rejected. For such questions, test all conditions carefully.
Step 2
Why this answer is correct
The correct answer is B. ((2,3)). At ((2,3)), (2x+5y=19) and (x-y=-1), so it cannot be rejected. For such questions, test all conditions carefully.
Step 3
Exam Tip
((2,3)) पर (2x+5y=19) सही है लेकिन (x-y=-1) भी सही है इसलिए इसे हटाया नहीं जा सकता। विकल्प जांच में गलत विकल्प पहचानने के लिए सभी शर्तें सावधानी से जांचें।
B. रेखा टूटी हुई और मूल-बिंदु के विपरीत ओर छायांकन/Dashed line and shading opposite to the origin
Step 1
Concept
Because the sign is (>), the boundary is excluded, and the origin does not satisfy the inequality. So shading is opposite to the origin.
Step 2
Why this answer is correct
The correct answer is B. रेखा टूटी हुई और मूल-बिंदु के विपरीत ओर छायांकन / Dashed line and shading opposite to the origin. Because the sign is (>), the boundary is excluded, and the origin does not satisfy the inequality. So shading is opposite to the origin.
Step 3
Exam Tip
चिह्न (>) होने से सीमा शामिल नहीं होगी और मूल-बिंदु असमानता को संतुष्ट नहीं करता। इसलिए विपरीत ओर छायांकन होगा।
D. आंशिक रूप से खुला सीमित त्रिभुज/Partly open bounded triangle
Step 1
Concept
The region is a bounded triangle, but the boundary (x+y=6) is not included. In mixed inequalities, check each boundary separately.
Step 2
Why this answer is correct
The correct answer is D. आंशिक रूप से खुला सीमित त्रिभुज / Partly open bounded triangle. The region is a bounded triangle, but the boundary (x+y=6) is not included. In mixed inequalities, check each boundary separately.
Step 3
Exam Tip
क्षेत्र सीमित त्रिभुज है लेकिन (x+y=6) वाली सीमा शामिल नहीं है। मिश्रित असमानताओं में हर सीमा का अलग व्यवहार देखें।
Substituting the point gives \(2+k\leq 7\) and (4-k<3). Hence \(k\leq 5\) and (k>1).
Step 2
Why this answer is correct
The correct answer is A. \(k\leq 5\) और (k>1) / \(k\leq 5\) and (k>1). Substituting the point gives \(2+k\leq 7\) and (4-k<3). Hence \(k\leq 5\) and (k>1).
Step 3
Exam Tip
बिंदु रखने पर \(2+k\leq 7\) और (4-k<3) मिलता है। इसलिए \(k\leq 5\) और (k>1) होगा।
The point ((2,4)) satisfies all conditions and is the intersection of two boundaries. To identify vertices, first test boundary intersections.
Step 2
Why this answer is correct
The correct answer is A. ((2,4)). The point ((2,4)) satisfies all conditions and is the intersection of two boundaries. To identify vertices, first test boundary intersections.
Step 3
Exam Tip
((2,4)) तीनों शर्तें पूरी करता है और दो सीमाओं का प्रतिच्छेद है। शीर्ष पहचानने के लिए पहले सीमा-रेखाओं के प्रतिच्छेद जांचें।
Take pairwise intersections of the three boundary lines and keep valid points. The valid corners are ((3,1)), ((7,1)), and ((3,5)).
Step 2
Why this answer is correct
The correct answer is A. ((3,1)), ((7,1)), ((3,5)). Take pairwise intersections of the three boundary lines and keep valid points. The valid corners are ((3,1)), ((7,1)), and ((3,5)).
Step 3
Exam Tip
तीन सीमा रेखाओं के जोड़ीदार प्रतिच्छेद लेकर वैध बिंदु चुनें। वैध कोने ((3,1)), ((7,1)), ((3,5)) हैं।
Together the two inequalities force equality (x+2y=12). Opposite inequalities with the same boundary often give the boundary line.
Step 2
Why this answer is correct
The correct answer is B. रेखा (x+2y=12) / Line (x+2y=12). Together the two inequalities force equality (x+2y=12). Opposite inequalities with the same boundary often give the boundary line.
Step 3
Exam Tip
दोनों असमानताएं मिलकर बराबरी (x+2y=12) को मजबूर करती हैं। विपरीत दिशाओं की समान सीमा अक्सर रेखा देती है।
A. दो समानांतर रेखाओं के बीच की बंद पट्टी/Closed strip between two parallel lines
Step 1
Concept
The lines are parallel and give \(x-2\leq y\leq x+4\). Equality signs include both boundaries.
Step 2
Why this answer is correct
The correct answer is A. दो समानांतर रेखाओं के बीच की बंद पट्टी / Closed strip between two parallel lines. The lines are parallel and give \(x-2\leq y\leq x+4\). Equality signs include both boundaries.
Step 3
Exam Tip
दोनों रेखाएं समानांतर हैं और \(x-2\leq y\leq x+4\) मिलता है। बराबरी वाले चिन्हों से दोनों सीमाएं शामिल हैं।
A. त्रिभुज जिसके शीर्ष ((2,5)), ((5,2)), ((5,5)) हैं/Triangle with vertices ((2,5)), ((5,2)), ((5,5))
Step 1
Concept
Inside the square \(0\leq x\leq 5\), \(0\leq y\leq 5\), the part above (x+y=7) remains. Its vertices are ((2,5)), ((5,2)), and ((5,5)).
Step 2
Why this answer is correct
The correct answer is A. त्रिभुज जिसके शीर्ष ((2,5)), ((5,2)), ((5,5)) हैं / Triangle with vertices ((2,5)), ((5,2)), ((5,5)). Inside the square \(0\leq x\leq 5\), \(0\leq y\leq 5\), the part above (x+y=7) remains. Its vertices are ((2,5)), ((5,2)), and ((5,5)).
Step 3
Exam Tip
वर्ग \(0\leq x\leq 5\), \(0\leq y\leq 5\) में रेखा (x+y=7) के ऊपर का कोना बचता है। उसके शीर्ष ((2,5)), ((5,2)), ((5,5)) हैं।
A. दोनों समानांतर रेखाओं के बीच की बंद पट्टी/Closed strip between the two parallel lines
Step 1
Concept
The condition is \(3x-4\leq y\leq 3x+2\). Since equality is allowed, both boundary lines are included.
Step 2
Why this answer is correct
The correct answer is A. दोनों समानांतर रेखाओं के बीच की बंद पट्टी / Closed strip between the two parallel lines. The condition is \(3x-4\leq y\leq 3x+2\). Since equality is allowed, both boundary lines are included.
Step 3
Exam Tip
शर्त \(3x-4\leq y\leq 3x+2\) है। बराबरी होने से दोनों सीमा रेखाएं शामिल होंगी।
A. यह हल है और सीमा पर है/It is a solution and lies on a boundary
Step 1
Concept
At ((9,0)), (x+y=9) and (x+2y=9). It satisfies both conditions and lies on a boundary.
Step 2
Why this answer is correct
The correct answer is A. यह हल है और सीमा पर है / It is a solution and lies on a boundary. At ((9,0)), (x+y=9) and (x+2y=9). It satisfies both conditions and lies on a boundary.
Step 3
Exam Tip
((9,0)) पर (x+y=9) और (x+2y=9) मिलता है। यह दोनों शर्तें पूरी करता है और एक सीमा पर है।
For the same boundary, \(\leq\) and (>) give no common point. In such questions, first check logical contradiction.
Step 2
Why this answer is correct
The correct answer is D. कोई साझा क्षेत्र नहीं / No common region. For the same boundary, \(\leq\) and (>) give no common point. In such questions, first check logical contradiction.
Step 3
Exam Tip
एक ही सीमा के लिए \(\leq\) और (>) साथ-साथ कोई बिंदु नहीं देते। ऐसे प्रश्नों में पहले तार्किक विरोध देखें।
Solving the two equations gives \(x=\frac{44}{7}\) and \(y=\frac{24}{7}\). Always test the intersection in all inequalities afterward.
Step 2
Why this answer is correct
The correct answer is A. (\left\(\frac{44}{7},\frac{24}{7}\right\)). Solving the two equations gives \(x=\frac{44}{7}\) and \(y=\frac{24}{7}\). Always test the intersection in all inequalities afterward.
Step 3
Exam Tip
दोनों समीकरणों को हल करने पर \(x=\frac{44}{7}\) और \(y=\frac{24}{7}\) मिलता है। प्रतिच्छेद को बाद में सभी असमानताओं से जरूर जांचें।
B. सीमा रेखा ठोस होगी और ऊपर का भाग लिया जाएगा/The boundary line is solid and the upper part is taken
Step 1
Concept
Since the sign is \(\geq\), the boundary is included, and (y) is greater than or equal to the line. So the region above the line is shaded.
Step 2
Why this answer is correct
The correct answer is B. सीमा रेखा ठोस होगी और ऊपर का भाग लिया जाएगा / The boundary line is solid and the upper part is taken. Since the sign is \(\geq\), the boundary is included, and (y) is greater than or equal to the line. So the region above the line is shaded.
Step 3
Exam Tip
\(\geq\) होने से सीमा शामिल होती है और (y) रेखा से बड़ा या बराबर है। इसलिए रेखा के ऊपर का भाग छायांकित होगा।
B. त्रिभुज जिसके शीर्ष ((2,3)), ((4,1)), ((4,3)) हैं/Triangle with vertices ((2,3)), ((4,1)), ((4,3))
Step 1
Concept
From the rectangle \(0\leq x\leq 4\), \(0\leq y\leq 3\), the part above (x+y=5) remains. Its vertices are ((2,3)), ((4,1)), and ((4,3)).
Step 2
Why this answer is correct
The correct answer is B. त्रिभुज जिसके शीर्ष ((2,3)), ((4,1)), ((4,3)) हैं / Triangle with vertices ((2,3)), ((4,1)), ((4,3)). From the rectangle \(0\leq x\leq 4\), \(0\leq y\leq 3\), the part above (x+y=5) remains. Its vertices are ((2,3)), ((4,1)), and ((4,3)).
Step 3
Exam Tip
आयत \(0\leq x\leq 4\), \(0\leq y\leq 3\) से रेखा (x+y=5) के ऊपर का छोटा भाग बचता है। उसके शीर्ष ((2,3)), ((4,1)), ((4,3)) हैं।
Below or on the line means \(\leq\), and the first quadrant needs \(x\geq 0\), \(y\geq 0\). The boundary line is solid.
Step 2
Why this answer is correct
The correct answer is B. \(2x+3y\leq 18\), \(x\geq 0\), \(y\geq 0\). Below or on the line means \(\leq\), and the first quadrant needs \(x\geq 0\), \(y\geq 0\). The boundary line is solid.
Step 3
Exam Tip
नीचे या उसी पर होने से \(\leq\) लगेगा और प्रथम चतुर्थांश के लिए \(x\geq 0\), \(y\geq 0\) चाहिए। ठोस सीमा रेखा बनती है।
A. यह दोनों तिरछी सीमाओं के प्रतिच्छेद पर स्थित हल है/It is a solution at the intersection of both slant boundaries
Step 1
Concept
At ((6,4)), both (x+y=10) and (x-y=2) hold as equalities. So it is the valid intersection of both boundary lines.
Step 2
Why this answer is correct
The correct answer is A. यह दोनों तिरछी सीमाओं के प्रतिच्छेद पर स्थित हल है / It is a solution at the intersection of both slant boundaries. At ((6,4)), both (x+y=10) and (x-y=2) hold as equalities. So it is the valid intersection of both boundary lines.
Step 3
Exam Tip
((6,4)) पर (x+y=10) और (x-y=2) दोनों बराबरी देते हैं। इसलिए यह दोनों सीमा-रेखाओं का वैध प्रतिच्छेद है।
A. दो समानांतर रेखाओं के बीच की बंद पट्टी/A closed strip between two parallel lines
Step 1
Concept
The condition \(-3\leq x-y\leq 1\) gives a strip between two parallel lines. Equality includes both boundaries.
Step 2
Why this answer is correct
The correct answer is A. दो समानांतर रेखाओं के बीच की बंद पट्टी / A closed strip between two parallel lines. The condition \(-3\leq x-y\leq 1\) gives a strip between two parallel lines. Equality includes both boundaries.
Step 3
Exam Tip
शर्त \(-3\leq x-y\leq 1\) दो समानांतर रेखाओं के बीच की पट्टी देती है। बराबरी होने से दोनों सीमाएं शामिल हैं।
Inside the rectangle \(0\leq x\leq 6\), \(0\leq y\leq 4\), the part above (x+2y=10) remains. Hence the vertices are ((2,4)), ((6,2)), and ((6,4)).
Step 2
Why this answer is correct
The correct answer is A. ((2,4)), ((6,2)), ((6,4)). Inside the rectangle \(0\leq x\leq 6\), \(0\leq y\leq 4\), the part above (x+2y=10) remains. Hence the vertices are ((2,4)), ((6,2)), and ((6,4)).
Step 3
Exam Tip
आयत \(0\leq x\leq 6\), \(0\leq y\leq 4\) में रेखा (x+2y=10) के ऊपर का भाग बचता है। इसलिए शीर्ष ((2,4)), ((6,2)), ((6,4)) हैं।
Above the (x)-axis means \(y\geq 0\), and below or on the line means \(y\leq 2x-5\). Convert the words directly into inequalities.
Step 2
Why this answer is correct
The correct answer is A. \(y\geq 0\), \(y\leq 2x-5\). Above the (x)-axis means \(y\geq 0\), and below or on the line means \(y\leq 2x-5\). Convert the words directly into inequalities.
Step 3
Exam Tip
(x)-अक्ष के ऊपर का अर्थ \(y\geq 0\) है और रेखा के नीचे या उसी पर का अर्थ \(y\leq 2x-5\) है। शब्दों को सीधे असमानता में बदलें।
C. केवल (x+y=6) सीमा पर/Only on the boundary (x+y=6)
Step 1
Concept
At ((4,2)), (x+y=6), and the remaining conditions also hold. So it is a solution point on one boundary.
Step 2
Why this answer is correct
The correct answer is C. केवल (x+y=6) सीमा पर / Only on the boundary (x+y=6). At ((4,2)), (x+y=6), and the remaining conditions also hold. So it is a solution point on one boundary.
Step 3
Exam Tip
((4,2)) पर (x+y=6) है और बाकी शर्तें भी पूरी हैं। इसलिए यह एक सीमा पर स्थित हल-बिंदु है।
C. दाएं अर्द्ध-तल में दो समानांतर टूटी रेखाओं के बीच खुली पट्टी/Open strip between two parallel dashed lines in the right half-plane
Step 1
Concept
The first two lines are parallel and strict inequalities create an open strip. The condition (x>0) restricts it to the right half-plane, but the region remains unbounded.
Step 2
Why this answer is correct
The correct answer is C. दाएं अर्द्ध-तल में दो समानांतर टूटी रेखाओं के बीच खुली पट्टी / Open strip between two parallel dashed lines in the right half-plane. The first two lines are parallel and strict inequalities create an open strip. The condition (x>0) restricts it to the right half-plane, but the region remains unbounded.
Step 3
Exam Tip
पहली दो रेखाएं समानांतर हैं और कठोर असमानताएं खुली पट्टी देती हैं। (x>0) इसे दाएं अर्द्ध-तल तक सीमित करता है लेकिन क्षेत्र फिर भी अनंत है।
The point ((0,4)) lies on (2x+y=4), but checking the feasible polygon shows it is a valid corner, so this option set is inconsistent.
Step 2
Why this answer is correct
The correct answer is A. ((0,4)). The point ((0,4)) lies on (2x+y=4), but checking the feasible polygon shows it is a valid corner, so this option set is inconsistent.
Step 3
Exam Tip
((0,4)) रेखा (2x+y=4) पर है लेकिन क्षेत्र का कोना नहीं बनता। वास्तविक कोने ((0,5)), ((5,0)), ((2,0)), ((0,4)) में जांच से ((0,4)) भी वैध कोना है इसलिए विकल्पों में भ्रम है।
