A. ((0,0)), ((4,0)), (\left\(\frac{14}{5},\frac{18}{5}\right\)), ((0,5))
Step 1
Concept
The slant boundaries intersect at (\left\(\frac{14}{5},\frac{18}{5}\right\)). Use valid intercepts on the axes to list all corners.
Step 2
Why this answer is correct
The correct answer is A. ((0,0)), ((4,0)), (\left\(\frac{14}{5},\frac{18}{5}\right\)), ((0,5)). The slant boundaries intersect at (\left\(\frac{14}{5},\frac{18}{5}\right\)). Use valid intercepts on the axes to list all corners.
Step 3
Exam Tip
दोनों तिरछी सीमाओं का प्रतिच्छेद (\left\(\frac{14}{5},\frac{18}{5}\right\)) है। अक्षों पर वैध अवरोध लेकर सभी कोने चुनें।
The vertices are ((0,0)), ((6,0)), ((4,4)), and ((0,6)). The shoelace method gives area (24) square units.
Step 2
Why this answer is correct
The correct answer is C. (24) वर्ग इकाई / (24) square units. The vertices are ((0,0)), ((6,0)), ((4,4)), and ((0,6)). The shoelace method gives area (24) square units.
Step 3
Exam Tip
शीर्ष ((0,0)), ((6,0)), ((4,4)), ((0,6)) हैं। शूलेस विधि से क्षेत्रफल (24) वर्ग इकाई मिलता है।
C. \(\frac{25}{2}\) वर्ग इकाई/\(\frac{25}{2}\) square units
Step 1
Concept
The vertices are ((2,1)), ((7,1)), and ((2,6)). The base and height are (5), so the area is \(\frac{25}{2}\).
Step 2
Why this answer is correct
The correct answer is C. \(\frac{25}{2}\) वर्ग इकाई / \(\frac{25}{2}\) square units. The vertices are ((2,1)), ((7,1)), and ((2,6)). The base and height are (5), so the area is \(\frac{25}{2}\).
Step 3
Exam Tip
शीर्ष ((2,1)), ((7,1)), ((2,6)) हैं। आधार और ऊंचाई (5) हैं इसलिए क्षेत्रफल \(\frac{25}{2}\) है।
The \(\geq\) conditions give an 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 D. सीमा रहित और बंद / Unbounded and closed. The \(\geq\) conditions give an upper-side region in the first quadrant. Boundaries are included and the region extends infinitely.
Step 3
Exam Tip
\(\geq\) वाली शर्तें प्रथम चतुर्थांश में ऊपर की ओर क्षेत्र देती हैं। सीमाएं शामिल हैं और क्षेत्र अनंत तक जाता है।
A. दो समानांतर रेखाओं के बीच की बंद पट्टी/A closed strip between two parallel lines
Step 1
Concept
The condition \(-4\leq x-y\leq 2\) 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 \(-4\leq x-y\leq 2\) gives a strip between two parallel lines. Equality includes both boundaries.
Step 3
Exam Tip
शर्त \(-4\leq x-y\leq 2\) दो समानांतर रेखाओं के बीच की पट्टी देती है। बराबरी होने से दोनों सीमाएं शामिल हैं।
C. दो समानांतर टूटी रेखाओं के बीच खुली पट्टी/Open strip between two parallel dashed lines
Step 1
Concept
The two lines are parallel and give ( -x+2<y<-x+6). Because the inequalities are strict, neither boundary is included.
Step 2
Why this answer is correct
The correct answer is C. दो समानांतर टूटी रेखाओं के बीच खुली पट्टी / Open strip between two parallel dashed lines. The two lines are parallel and give ( -x+2<y<-x+6). Because the inequalities are strict, neither boundary is included.
Step 3
Exam Tip
दोनों रेखाएं समानांतर हैं और ( -x+2<y<-x+6) मिलता है। कठोर असमानताओं के कारण दोनों सीमाएं शामिल नहीं होंगी।
Together the two inequalities force equality (x+3y=9). Opposite inequalities with the same boundary often give only the line.
Step 2
Why this answer is correct
The correct answer is D. रेखा (x+3y=9) / Line (x+3y=9). Together the two inequalities force equality (x+3y=9). Opposite inequalities with the same boundary often give only the line.
Step 3
Exam Tip
दोनों असमानताएं मिलकर बराबरी (x+3y=9) को मजबूर करती हैं। विपरीत दिशाओं की समान सीमा अक्सर केवल रेखा देती है।
A. हल-क्षेत्र में और सीमा (x+y=5) पर/In the solution region and on the boundary (x+y=5)
Step 1
Concept
At ((2,3)), (x+y=5), and the remaining conditions also hold. Therefore it is a solution point on the boundary.
Step 2
Why this answer is correct
The correct answer is A. हल-क्षेत्र में और सीमा (x+y=5) पर / In the solution region and on the boundary (x+y=5). At ((2,3)), (x+y=5), and the remaining conditions also hold. Therefore it is a solution point on the boundary.
Step 3
Exam Tip
((2,3)) पर (x+y=5) है और बाकी शर्तें भी पूरी हैं। इसलिए यह सीमा पर स्थित हल-बिंदु है।
Solving the two equations gives \(x=\frac{27}{7}\) and \(y=\frac{32}{7}\). The intersection should then be checked in the inequalities.
Step 2
Why this answer is correct
The correct answer is D. (\left\(\frac{27}{7},\frac{32}{7}\right\)). Solving the two equations gives \(x=\frac{27}{7}\) and \(y=\frac{32}{7}\). The intersection should then be checked in the inequalities.
Step 3
Exam Tip
दोनों समीकरण हल करने पर \(x=\frac{27}{7}\) और \(y=\frac{32}{7}\) मिलता है। प्रतिच्छेद को बाद में असमानताओं में जांचना चाहिए।
B. टूटी सीमा और मूल-बिंदु वाली ओर/Dashed boundary and toward the origin
Step 1
Concept
Substituting the origin gives (0<10), which is true. Because the inequality is strict, the boundary is dashed and shading is toward the origin.
Step 2
Why this answer is correct
The correct answer is B. टूटी सीमा और मूल-बिंदु वाली ओर / Dashed boundary and toward the origin. Substituting the origin gives (0<10), which is true. Because the inequality is strict, the boundary is dashed and shading is toward the origin.
Step 3
Exam Tip
मूल-बिंदु रखने पर (0<10) सही है। कठोर असमानता के कारण सीमा टूटी होगी और छायांकन मूल-बिंदु वाली ओर होगा।
The same expression (x+y) cannot be at most (3) and at least (5) at the same time. Identify such contradictions before graphing.
Step 2
Why this answer is correct
The correct answer is A. कोई हल नहीं / No solution. The same expression (x+y) cannot be at most (3) and at least (5) at the same time. Identify such contradictions before graphing.
Step 3
Exam Tip
एक ही राशि (x+y) एक साथ (3) से छोटी या बराबर और (5) से बड़ी या बराबर नहीं हो सकती। ऐसे विरोध को ग्राफ से पहले पहचानें।
If \(a\geq 0\), at least ((0,0)) is in the solution and the region remains bounded. If (a<0), there is no solution in the first quadrant.
Step 2
Why this answer is correct
The correct answer is B. \(a\geq 0\). If \(a\geq 0\), at least ((0,0)) is in the solution and the region remains bounded. If (a<0), there is no solution in the first quadrant.
Step 3
Exam Tip
यदि \(a\geq 0\) है तो कम से कम ((0,0)) हल में आता है और क्षेत्र सीमित रहता है। (a<0) होने पर प्रथम चतुर्थांश में कोई हल नहीं मिलेगा।
The two slant boundaries meet at ((3,2)). The remaining corners come from valid parts of (x=1) and (y=0).
Step 2
Why this answer is correct
The correct answer is C. ((1,0)), ((4,0)), ((3,2)), ((1,3)). The two slant boundaries meet at ((3,2)). The remaining corners come from valid parts of (x=1) and (y=0).
Step 3
Exam Tip
दो तिरछी सीमाएं ((3,2)) पर मिलती हैं। (x=1) और (y=0) की वैध सीमाओं से बाकी कोने मिलते हैं।
The vertices are ((1,0)), ((4,0)), ((3,2)), and ((1,3)). The shoelace method gives area (6) square units.
Step 2
Why this answer is correct
The correct answer is D. (6) वर्ग इकाई / (6) square units. The vertices are ((1,0)), ((4,0)), ((3,2)), and ((1,3)). The shoelace method gives area (6) square units.
Step 3
Exam Tip
शीर्ष ((1,0)), ((4,0)), ((3,2)), ((1,3)) हैं। शूलेस विधि से क्षेत्रफल (6) वर्ग इकाई मिलता है।
Checking (x+2y) at the corners gives the maximum (12) at ((2,5)). For a linear expression, checking corners is sufficient.
Step 2
Why this answer is correct
The correct answer is A. (12). Checking (x+2y) at the corners gives the maximum (12) at ((2,5)). For a linear expression, checking corners is sufficient.
Step 3
Exam Tip
कोनों पर (x+2y) जांचने से ((2,5)) पर अधिकतम (12) मिलता है। रैखिक व्यंजक के लिए कोनों की जांच पर्याप्त होती है।
The three strict inequalities form a bounded triangle, but no boundary is included. Therefore the region is open and bounded.
Step 2
Why this answer is correct
The correct answer is C. खुला सीमित त्रिभुज / Open bounded triangle. The three strict inequalities form a bounded triangle, but no boundary is included. Therefore the region is open and bounded.
Step 3
Exam Tip
तीनों कठोर असमानताएं एक सीमित त्रिभुज बनाती हैं लेकिन कोई सीमा शामिल नहीं होती। इसलिए क्षेत्र खुला और सीमित है।
A solid boundary includes equality. Since the origin gives (0<12), the side opposite the origin is \(3x+2y\geq 12\).
Step 2
Why this answer is correct
The correct answer is B. \(3x+2y\geq 12\). A solid boundary includes equality. Since the origin gives (0<12), the side opposite the origin is \(3x+2y\geq 12\).
Step 3
Exam Tip
ठोस सीमा के लिए बराबरी शामिल होगी। मूल-बिंदु पर (0<12) है इसलिए मूल-बिंदु के विपरीत ओर \(3x+2y\geq 12\) होगा।
Right side including the line means the value of (x) is greater than or equal to (-2). For vertical lines, read the (x)-value directly.
Step 2
Why this answer is correct
The correct answer is C. \(x\geq -2\). Right side including the line means the value of (x) is greater than or equal to (-2). For vertical lines, read the (x)-value directly.
Step 3
Exam Tip
रेखा सहित दाईं ओर का अर्थ (x) का मान (-2) से बड़ा या बराबर है। लंबवत रेखाओं में सीधे (x)-मान देखें।
The first two conditions give (x+y=6), and the next two give (x-y=2). Their intersection is ((4,2)).
Step 2
Why this answer is correct
The correct answer is D. केवल बिंदु ((4,2)) / Only the point ((4,2)). The first two conditions give (x+y=6), and the next two give (x-y=2). Their intersection is ((4,2)).
Step 3
Exam Tip
पहली दो शर्तें (x+y=6) और अगली दो शर्तें (x-y=2) देती हैं। दोनों रेखाओं का प्रतिच्छेद ((4,2)) है।
The square has area (25), and the triangle below (x+y=4) has area (8). Therefore the remaining area is (17).
Step 2
Why this answer is correct
The correct answer is C. (17) वर्ग इकाई / (17) square units. The square has area (25), and the triangle below (x+y=4) has area (8). Therefore the remaining area is (17).
Step 3
Exam Tip
पूरे वर्ग का क्षेत्रफल (25) है और रेखा (x+y=4) के नीचे का त्रिभुज क्षेत्रफल (8) है। इसलिए बचा क्षेत्रफल (17) है।
The two half-planes intersect to form an infinite angular region. Since equality is included in both, the region is closed.
Step 2
Why this answer is correct
The correct answer is A. सीमा रहित और बंद / Unbounded and closed. The two half-planes intersect to form an infinite angular region. Since equality is included in both, the region is closed.
Step 3
Exam Tip
दो अर्द्ध-तल कटकर एक अनंत कोणीय क्षेत्र बनाते हैं। दोनों में बराबरी शामिल है इसलिए क्षेत्र बंद है।
The first condition gives (k<3), and the second gives \(k\geq 3\). They cannot hold together.
Step 2
Why this answer is correct
The correct answer is C. ऐसा कोई (k) नहीं है / No such (k) exists. The first condition gives (k<3), and the second gives \(k\geq 3\). They cannot hold together.
Step 3
Exam Tip
पहली शर्त से (k<3) और दूसरी से \(k\geq 3\) मिलता है। दोनों साथ संभव नहीं हैं।
If (p>0), the (x)-intercept \(\frac{6}{p}\) is finite. If \(p\leq 0\), the region is not bounded in the (x)-direction in the first quadrant.
Step 2
Why this answer is correct
The correct answer is C. (p>0). If (p>0), the (x)-intercept \(\frac{6}{p}\) is finite. If \(p\leq 0\), the region is not bounded in the (x)-direction in the first quadrant.
Step 3
Exam Tip
यदि (p>0), तो (x)-अवरोध \(\frac{6}{p}\) सीमित होता है। \(p\leq 0\) होने पर प्रथम चतुर्थांश में क्षेत्र (x) दिशा में सीमित नहीं रहता।
D. एक ओर बंद और दूसरी ओर खुली पट्टी/Strip closed on one side and open on the other
Step 1
Concept
The condition becomes \(4<x+2y\leq 8\). The boundary (x+2y=8) is included, but (x+2y=4) is excluded.
Step 2
Why this answer is correct
The correct answer is D. एक ओर बंद और दूसरी ओर खुली पट्टी / Strip closed on one side and open on the other. The condition becomes \(4<x+2y\leq 8\). The boundary (x+2y=8) is included, but (x+2y=4) is excluded.
Step 3
Exam Tip
शर्त \(4<x+2y\leq 8\) मिलती है। (x+2y=8) शामिल है लेकिन (x+2y=4) शामिल नहीं है।
The rectangle \([2,6]\times[1,5]\) is cut by the line (x+y=8). The corners are ((2,1)), ((6,1)), ((6,2)), ((3,5)), and ((2,5)).
Step 2
Why this answer is correct
The correct answer is C. (5). The rectangle \([2,6]\times[1,5]\) is cut by the line (x+y=8). The corners are ((2,1)), ((6,1)), ((6,2)), ((3,5)), and ((2,5)).
Step 3
Exam Tip
आयत \([2,6]\times[1,5]\) को रेखा (x+y=8) काटती है। कोने ((2,1)), ((6,1)), ((6,2)), ((3,5)), ((2,5)) हैं।
Among boundary corner-like points, (2x+y=6) at ((0,6)) is the smallest. Even in an unbounded region, a minimum often occurs at a boundary corner.
Step 2
Why this answer is correct
The correct answer is C. (6). Among boundary corner-like points, (2x+y=6) at ((0,6)) is the smallest. Even in an unbounded region, a minimum often occurs at a boundary corner.
Step 3
Exam Tip
कोनों जैसे सीमा-बिंदुओं में ((0,6)) पर (2x+y=6) सबसे छोटा है। सीमा रहित क्षेत्र में भी न्यूनतम अक्सर किसी सीमा-कोने पर मिलता है।
