असमानताओं \(x+y\geq 4\), (x+y<4), \(x\geq 0\) का हल-क्षेत्र कौन सा है?
Which is the solution region of \(x+y\geq 4\), (x+y<4), \(x\geq 0\)?
Explanation opens after your attempt
Correct Answer
B. खाली समुच्चयEmpty set
Concept
The first and second inequalities contradict each other. Even on the common boundary, the strict inequality removes the points.
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.
Exam Tip
पहली और दूसरी असमानता परस्पर विरोधी हैं। समान सीमा पर भी कठोर असमानता बिंदुओं को हटा देती है।
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