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For \(f(x)=\sqrt{9-x^2}\), which condition is correct for the real domain?\(f(x)=\sqrt{9-x^2}\) के लिए वास्तविक डोमेन के लिए कौन-सी शर्त सही है?
Answer and explanationउत्तर और व्याख्या
Correct answer:सही उत्तर: \(9-x^2\ge 0\)\(9-x^2\ge 0\)
A square root \(\sqrt{A}\) is real when \(A\ge 0\). Here \(A=9-x^2\). So the correct condition for the real domain is \(9-x^2\ge 0\).वर्गमूल \(\sqrt{A}\) वास्तविक होता है जब \(A\ge 0\) हो। यहाँ \(A=9-x^2\) है। इसलिए वास्तविक डोमेन के लिए सही शर्त \(9-x^2\ge 0\) है।
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