व्यंजक (\(w^2-1\)x-4+(w-1)x-2+6) नियतांक बहुपद कब बनेगा?

When will (\(w^2-1\)x-4+(w-1)x-2+6) become a constant polynomial?

Explanation opens after your attempt
Correct Answer

A. जब (w=1)When (w=1)

Step 1

Concept

To become constant, both \(w^2-1=0\) and (w-1=0) are needed. Both conditions are satisfied together at (w=1).

Step 2

Why this answer is correct

The correct answer is A. जब (w=1) / When (w=1). To become constant, both \(w^2-1=0\) and (w-1=0) are needed. Both conditions are satisfied together at (w=1).

Step 3

Exam Tip

नियतांक बनने के लिए \(w^2-1=0\) और (w-1=0) दोनों चाहिए। दोनों शर्तें साथ में (w=1) पर पूरी होती हैं।

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Mathematics Answer, Explanation and Revision Hints

व्यंजक (\(w^2-1\)x-4+(w-1)x-2+6) नियतांक बहुपद कब बनेगा? / When will (\(w^2-1\)x-4+(w-1)x-2+6) become a constant polynomial?

Correct Answer: A. जब (w=1) / When (w=1). Explanation: नियतांक बनने के लिए \(w^2-1=0\) और (w-1=0) दोनों चाहिए। दोनों शर्तें साथ में (w=1) पर पूरी होती हैं। / To become constant, both \(w^2-1=0\) and (w-1=0) are needed. Both conditions are satisfied together at (w=1).

Which concept should I revise for this Mathematics MCQ?

To become constant, both \(w^2-1=0\) and (w-1=0) are needed. Both conditions are satisfied together at (w=1).

What exam hint can help solve this Mathematics question?

नियतांक बनने के लिए \(w^2-1=0\) और (w-1=0) दोनों चाहिए। दोनों शर्तें साथ में (w=1) पर पूरी होती हैं।