फलन (f(x)=\sqrt{x-3}+\frac{1}{x-5}) का डोमेन क्या है?

What is the domain of (f(x)=\sqrt{x-3}+\frac{1}{x-5})?

Explanation opens after your attempt
Correct Answer

A. \([3,\infty\)\setminus{5})

Step 1

Concept

The square root gives \(x\ge 3\) and the denominator gives \(x\ne 5\). In exams take the intersection of all restrictions.

Step 2

Why this answer is correct

The correct answer is A. \([3,\infty\)\setminus{5}). The square root gives \(x\ge 3\) and the denominator gives \(x\ne 5\). In exams take the intersection of all restrictions.

Step 3

Exam Tip

वर्गमूल से \(x\ge 3\) और हर से \(x\ne 5\) चाहिए। परीक्षा में सभी प्रतिबंधों का प्रतिच्छेद लें।

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Mathematics Answer, Explanation and Revision Hints

फलन (f(x)=\sqrt{x-3}+\frac{1}{x-5}) का डोमेन क्या है? / What is the domain of (f(x)=\sqrt{x-3}+\frac{1}{x-5})?

Correct Answer: A. \([3,\infty\)\setminus{5}). Explanation: वर्गमूल से \(x\ge 3\) और हर से \(x\ne 5\) चाहिए। परीक्षा में सभी प्रतिबंधों का प्रतिच्छेद लें। / The square root gives \(x\ge 3\) and the denominator gives \(x\ne 5\). In exams take the intersection of all restrictions.

Which concept should I revise for this Mathematics MCQ?

The square root gives \(x\ge 3\) and the denominator gives \(x\ne 5\). In exams take the intersection of all restrictions.

What exam hint can help solve this Mathematics question?

वर्गमूल से \(x\ge 3\) और हर से \(x\ne 5\) चाहिए। परीक्षा में सभी प्रतिबंधों का प्रतिच्छेद लें।