फलन (f(x)=\sqrt{x}+\frac{1}{\sqrt{4-x}}) का प्रांत क्या है?

What is the domain of (f(x)=\sqrt{x}+\frac{1}{\sqrt{4-x}})?

Explanation opens after your attempt
Correct Answer

A. ( [0,4) )

Step 1

Concept

For \(\sqrt{x}\), \(x\ge 0\), and for the denominator square root, (4-x>0) is needed. Together, \(0\le x<4\).

Step 2

Why this answer is correct

The correct answer is A. ( [0,4) ). For \(\sqrt{x}\), \(x\ge 0\), and for the denominator square root, (4-x>0) is needed. Together, \(0\le x<4\).

Step 3

Exam Tip

\(\sqrt{x}\) के लिए \(x\ge 0\) और हर वाले वर्गमूल के लिए (4-x>0) चाहिए। दोनों मिलाकर \(0\le x<4\) मिलता है।

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Mathematics Answer, Explanation and Revision Hints

फलन (f(x)=\sqrt{x}+\frac{1}{\sqrt{4-x}}) का प्रांत क्या है? / What is the domain of (f(x)=\sqrt{x}+\frac{1}{\sqrt{4-x}})?

Correct Answer: A. ( [0,4) ). Explanation: \(\sqrt{x}\) के लिए \(x\ge 0\) और हर वाले वर्गमूल के लिए (4-x>0) चाहिए। दोनों मिलाकर \(0\le x<4\) मिलता है। / For \(\sqrt{x}\), \(x\ge 0\), and for the denominator square root, (4-x>0) is needed. Together, \(0\le x<4\).

Which concept should I revise for this Mathematics MCQ?

For \(\sqrt{x}\), \(x\ge 0\), and for the denominator square root, (4-x>0) is needed. Together, \(0\le x<4\).

What exam hint can help solve this Mathematics question?

\(\sqrt{x}\) के लिए \(x\ge 0\) और हर वाले वर्गमूल के लिए (4-x>0) चाहिए। दोनों मिलाकर \(0\le x<4\) मिलता है।