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What is the domain of \(f(x)=\sqrt{x^2-9}\)?\(f(x)=\sqrt{x^2-9}\) का domain क्या है?
Answer and explanationउत्तर और व्याख्या
Correct answer:सही उत्तर: \(( -\infty,-3]\cup[3,\infty)\)\(( -\infty,-3]\cup[3,\infty)\)
For \(f(x)=\sqrt{x^2-9}\) to be real, we need \(x^2-9\ge 0\). That means \(x^2\ge 9\), so \(x\le -3\) or \(x\ge 3\). Therefore the domain is \(( -\infty,-3]\cup[3,\infty)\).\(f(x)=\sqrt{x^2-9}\) को वास्तविक (real) होने के लिए \(x^2-9\ge 0\) होना चाहिए। यानी \(x^2\ge 9\), इसलिए \(x\le -3\) या \(x\ge 3\)। अतः domain \(( -\infty,-3]\cup[3,\infty)\) है।
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