यदि (f(x)=x-2+6x+9) और (g(x)=x+3) हों, तो (\left\(\frac{f}{g}\right\)(0)) क्या होगा, जहाँ \(x\ne -3\)?

If (f(x)=x-2+6x+9) and (g(x)=x+3), what is (\left\(\frac{f}{g}\right\)(0)), where \(x\ne -3\)?

Explanation opens after your attempt
Correct Answer

A. (3)

Step 1

Concept

(\frac{(x+3)2}{x+3}=x+3), so at (x=0) the value is (3). Keep the restriction \(x\ne -3\).

Step 2

Why this answer is correct

The correct answer is A. (3). (\frac{(x+3)2}{x+3}=x+3), so at (x=0) the value is (3). Keep the restriction \(x\ne -3\).

Step 3

Exam Tip

(\frac{(x+3)2}{x+3}=x+3), इसलिए (x=0) पर मान (3) है। प्रतिबंध \(x\ne -3\) साथ रखें।

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

यदि (f(x)=x-2+6x+9) और (g(x)=x+3) हों, तो (\left\(\frac{f}{g}\right\)(0)) क्या होगा, जहाँ \(x\ne -3\)? / If (f(x)=x-2+6x+9) and (g(x)=x+3), what is (\left\(\frac{f}{g}\right\)(0)), where \(x\ne -3\)?

Correct Answer: A. (3). Explanation: (\frac{(x+3)2}{x+3}=x+3), इसलिए (x=0) पर मान (3) है। प्रतिबंध \(x\ne -3\) साथ रखें। / (\frac{(x+3)2}{x+3}=x+3), so at (x=0) the value is (3). Keep the restriction \(x\ne -3\).

Which concept should I revise for this Mathematics MCQ?

(\frac{(x+3)2}{x+3}=x+3), so at (x=0) the value is (3). Keep the restriction \(x\ne -3\).

What exam hint can help solve this Mathematics question?

(\frac{(x+3)2}{x+3}=x+3), इसलिए (x=0) पर मान (3) है। प्रतिबंध \(x\ne -3\) साथ रखें।