यदि (f(x)=\frac{1}{\sqrt{x-4}}) और (g(x)=x+1) हैं, तो ((f+g)(x)) का प्रांत क्या है?

If (f(x)=\frac{1}{\sqrt{x-4}}) and (g(x)=x+1), what is the domain of ((f+g)(x))?

Explanation opens after your attempt
Correct Answer

B. (\(4,\infty\))

Step 1

Concept

The denominator contains \(\sqrt{x-4}\), so (x-4>0) is required. Thus the domain is (\(4,\infty\)).

Step 2

Why this answer is correct

The correct answer is B. (\(4,\infty\)). The denominator contains \(\sqrt{x-4}\), so (x-4>0) is required. Thus the domain is (\(4,\infty\)).

Step 3

Exam Tip

हर में \(\sqrt{x-4}\) है, इसलिए (x-4>0) चाहिए। अतः प्रांत (\(4,\infty\)) है।

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Mathematics Answer, Explanation and Revision Hints

यदि (f(x)=\frac{1}{\sqrt{x-4}}) और (g(x)=x+1) हैं, तो ((f+g)(x)) का प्रांत क्या है? / If (f(x)=\frac{1}{\sqrt{x-4}}) and (g(x)=x+1), what is the domain of ((f+g)(x))?

Correct Answer: B. (\(4,\infty\)). Explanation: हर में \(\sqrt{x-4}\) है, इसलिए (x-4>0) चाहिए। अतः प्रांत (\(4,\infty\)) है। / The denominator contains \(\sqrt{x-4}\), so (x-4>0) is required. Thus the domain is (\(4,\infty\)).

Which concept should I revise for this Mathematics MCQ?

The denominator contains \(\sqrt{x-4}\), so (x-4>0) is required. Thus the domain is (\(4,\infty\)).

What exam hint can help solve this Mathematics question?

हर में \(\sqrt{x-4}\) है, इसलिए (x-4>0) चाहिए। अतः प्रांत (\(4,\infty\)) है।