यदि (f(x)=x-2+4x+4) और (g(x)=x+2) हैं, तो \(\frac{f}{g}\) का सरलीकृत रूप और प्रांत क्या है?

If (f(x)=x-2+4x+4) and (g(x)=x+2), what are the simplified form and domain of \(\frac{f}{g}\)?

Explanation opens after your attempt
Correct Answer

C. (x+2), \(x\ne -2\)

Step 1

Concept

(f=(x+2)2), so \(\frac{f}{g}=x+2\), but (x=-2) is excluded. A cancelled denominator still gives a domain restriction.

Step 2

Why this answer is correct

The correct answer is C. (x+2), \(x\ne -2\). (f=(x+2)2), so \(\frac{f}{g}=x+2\), but (x=-2) is excluded. A cancelled denominator still gives a domain restriction.

Step 3

Exam Tip

(f=(x+2)2), इसलिए \(\frac{f}{g}=x+2\), पर (x=-2) हटेगा। रद्द हुआ हर भी प्रांत में प्रतिबंध देता है।

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Mathematics Answer, Explanation and Revision Hints

यदि (f(x)=x-2+4x+4) और (g(x)=x+2) हैं, तो \(\frac{f}{g}\) का सरलीकृत रूप और प्रांत क्या है? / If (f(x)=x-2+4x+4) and (g(x)=x+2), what are the simplified form and domain of \(\frac{f}{g}\)?

Correct Answer: C. (x+2), \(x\ne -2\). Explanation: (f=(x+2)2), इसलिए \(\frac{f}{g}=x+2\), पर (x=-2) हटेगा। रद्द हुआ हर भी प्रांत में प्रतिबंध देता है। / (f=(x+2)2), so \(\frac{f}{g}=x+2\), but (x=-2) is excluded. A cancelled denominator still gives a domain restriction.

Which concept should I revise for this Mathematics MCQ?

(f=(x+2)2), so \(\frac{f}{g}=x+2\), but (x=-2) is excluded. A cancelled denominator still gives a domain restriction.

What exam hint can help solve this Mathematics question?

(f=(x+2)2), इसलिए \(\frac{f}{g}=x+2\), पर (x=-2) हटेगा। रद्द हुआ हर भी प्रांत में प्रतिबंध देता है।