फलन (f(x)=\sqrt{x-1}+\frac{1}{x-5}) का प्रांत क्या है?

What is the domain of (f(x)=\sqrt{x-1}+\frac{1}{x-5})?

Explanation opens after your attempt
Correct Answer

A. \([1,\infty\)-{5})

Step 1

Concept

For the square root \(x\ge 1\), and for the denominator \(x\ne 5\) are needed. Hence the domain is \([1,\infty\)-{5}).

Step 2

Why this answer is correct

The correct answer is A. \([1,\infty\)-{5}). For the square root \(x\ge 1\), and for the denominator \(x\ne 5\) are needed. Hence the domain is \([1,\infty\)-{5}).

Step 3

Exam Tip

वर्गमूल के लिए \(x\ge 1\) और हर के लिए \(x\ne 5\) चाहिए। इसलिए प्रांत \([1,\infty\)-{5}) है।

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Mathematics Answer, Explanation and Revision Hints

फलन (f(x)=\sqrt{x-1}+\frac{1}{x-5}) का प्रांत क्या है? / What is the domain of (f(x)=\sqrt{x-1}+\frac{1}{x-5})?

Correct Answer: A. \([1,\infty\)-{5}). Explanation: वर्गमूल के लिए \(x\ge 1\) और हर के लिए \(x\ne 5\) चाहिए। इसलिए प्रांत \([1,\infty\)-{5}) है। / For the square root \(x\ge 1\), and for the denominator \(x\ne 5\) are needed. Hence the domain is \([1,\infty\)-{5}).

Which concept should I revise for this Mathematics MCQ?

For the square root \(x\ge 1\), and for the denominator \(x\ne 5\) are needed. Hence the domain is \([1,\infty\)-{5}).

What exam hint can help solve this Mathematics question?

वर्गमूल के लिए \(x\ge 1\) और हर के लिए \(x\ne 5\) चाहिए। इसलिए प्रांत \([1,\infty\)-{5}) है।