यदि (q(x)=(k-3)2x^{12}+\(k^2-9\)x-7+6x-2-5) और (k=3), तो (q(x)) की डिग्री क्या है?

If (q(x)=(k-3)2x^{12}+\(k^2-9\)x-7+6x-2-5) and (k=3), what is the degree of (q(x))?

Author: Muft Shiksha Editorial Team Published:
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Correct Answer

C. (2)

Step 1

Concept

Putting (k=3) removes both \(x^{12}\) and \(x^7\) terms. The remaining \(6x^2-5\) has degree (2).

Step 2

Why this answer is correct

The correct answer is C. (2). Putting (k=3) removes both \(x^{12}\) and \(x^7\) terms. The remaining \(6x^2-5\) has degree (2).

Step 3

Exam Tip

(k=3) रखने पर \(x^{12}\) और \(x^7\) दोनों पद हट जाते हैं। बचा \(6x^2-5\) है जिसकी डिग्री (2) है।

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Mathematics Answer, Explanation and Revision Hints

यदि (q(x)=(k-3)2x^{12}+\(k^2-9\)x-7+6x-2-5) और (k=3), तो (q(x)) की डिग्री क्या है? / If (q(x)=(k-3)2x^{12}+\(k^2-9\)x-7+6x-2-5) and (k=3), what is the degree of (q(x))?

Correct Answer: C. (2). Explanation: (k=3) रखने पर \(x^{12}\) और \(x^7\) दोनों पद हट जाते हैं। बचा \(6x^2-5\) है जिसकी डिग्री (2) है। / Putting (k=3) removes both \(x^{12}\) and \(x^7\) terms. The remaining \(6x^2-5\) has degree (2).

Which concept should I revise for this Mathematics MCQ?

Putting (k=3) removes both \(x^{12}\) and \(x^7\) terms. The remaining \(6x^2-5\) has degree (2).

What exam hint can help solve this Mathematics question?

(k=3) रखने पर \(x^{12}\) और \(x^7\) दोनों पद हट जाते हैं। बचा \(6x^2-5\) है जिसकी डिग्री (2) है।