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

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

Explanation opens after your attempt
Correct Answer

A. (x>1)

Step 1

Concept

The denominator has \(\sqrt{x-1}\), so (x-1>0) is required. Hence the domain is (x>1).

Step 2

Why this answer is correct

The correct answer is A. (x>1). The denominator has \(\sqrt{x-1}\), so (x-1>0) is required. Hence the domain is (x>1).

Step 3

Exam Tip

हर में \(\sqrt{x-1}\) है, इसलिए (x-1>0) होना चाहिए। अतः प्रांत (x>1) है।

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Mathematics Answer, Explanation and Revision Hints

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

Correct Answer: A. (x>1). Explanation: हर में \(\sqrt{x-1}\) है, इसलिए (x-1>0) होना चाहिए। अतः प्रांत (x>1) है। / The denominator has \(\sqrt{x-1}\), so (x-1>0) is required. Hence the domain is (x>1).

Which concept should I revise for this Mathematics MCQ?

The denominator has \(\sqrt{x-1}\), so (x-1>0) is required. Hence the domain is (x>1).

What exam hint can help solve this Mathematics question?

हर में \(\sqrt{x-1}\) है, इसलिए (x-1>0) होना चाहिए। अतः प्रांत (x>1) है।