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What is the domain of the function \(f(x)=\frac{\sqrt{9-x^2}}{x+1}\)?फलन \(f(x)=\frac{\sqrt{9-x^2}}{x+1}\) का प्रांत क्या है?
Answer and explanationउत्तर और व्याख्या
Correct answer:सही उत्तर: \([-3,3]\setminus\{-1\}\)\([-3,3]\setminus\{-1\}\)
The radicand must satisfy \(9-x^2\ge 0\), giving \(-3\le x\le 3\). The denominator must also be nonzero, so exclude \(x=-1\). Thus the domain is \([-3,3]\setminus\{-1\}\).वर्गमूल के अंदर का व्यंजक \(9-x^2\ge 0\) होना चाहिए, जिससे \(-3\le x\le 3\) मिलता है। हर शून्य नहीं हो सकता, इसलिए \(x=-1\) को हटाएँ। अतः प्रांत \([-3,3]\setminus\{-1\}\) है।
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