ग्राफ \(y=\sqrt{16-x^2}\) की रेंज क्या है?

What is the range of the graph \(y=\sqrt{16-x^2}\)?

Explanation opens after your attempt
Correct Answer

A. ([0,4])

Step 1

Concept

It is the upper semicircle of \(x^2+y^2=16\). Hence (y) ranges from (0) to (4).

Step 2

Why this answer is correct

The correct answer is A. ([0,4]). It is the upper semicircle of \(x^2+y^2=16\). Hence (y) ranges from (0) to (4).

Step 3

Exam Tip

यह वृत्त \(x^2+y^2=16\) का ऊपरी अर्धभाग है। इसलिए (y) का मान (0) से (4) तक है।

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ग्राफ \(y=\sqrt{16-x^2}\) की रेंज क्या है? / What is the range of the graph \(y=\sqrt{16-x^2}\)?

Correct Answer: A. ([0,4]). Explanation: यह वृत्त \(x^2+y^2=16\) का ऊपरी अर्धभाग है। इसलिए (y) का मान (0) से (4) तक है। / It is the upper semicircle of \(x^2+y^2=16\). Hence (y) ranges from (0) to (4).

Which concept should I revise for this Mathematics MCQ?

It is the upper semicircle of \(x^2+y^2=16\). Hence (y) ranges from (0) to (4).

What exam hint can help solve this Mathematics question?

यह वृत्त \(x^2+y^2=16\) का ऊपरी अर्धभाग है। इसलिए (y) का मान (0) से (4) तक है।