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।
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।
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।
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।
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।
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।
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 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 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।
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।
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।
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।
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 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।
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।
