हल-क्षेत्र \(x\geq 0\), \(y\geq 0\), \(px+y\leq 6\) सीमित होने के लिए (p) की सही शर्त क्या है?

For the region \(x\geq 0\), \(y\geq 0\), \(px+y\leq 6\) to be bounded, what is the correct condition on (p)?

Explanation opens after your attempt
Correct Answer

C. (p>0)

Step 1

Concept

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।

Question me issue ya doubt hai?

Answer, explanation, typing mistake ya suggestion directly hamari team ko bhejein. 📱Helpline (Call / WhatsApp): +91 7272824365

Related Mathematics Questions

FAQs

Mathematics Answer, Explanation and Revision Hints

हल-क्षेत्र \(x\geq 0\), \(y\geq 0\), \(px+y\leq 6\) सीमित होने के लिए (p) की सही शर्त क्या है? / For the region \(x\geq 0\), \(y\geq 0\), \(px+y\leq 6\) to be bounded, what is the correct condition on (p)?

Correct Answer: C. (p>0). Explanation: 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। / 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.

Which concept should I revise for this Mathematics MCQ?

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.

What exam hint can help solve this Mathematics question?

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।