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Choose the range of \(f(x)=\sqrt{x^2-16}\).\(f(x)=\sqrt{x^2-16}\) का परास (range) चुनिए।
Answer and explanationउत्तर और व्याख्या
Correct answer:सही उत्तर: \([0,\infty)\)\([0,\infty)\)
For \(f(x)=\sqrt{x^2-16}\) to be defined, we need \(x^2-16\ge 0\Rightarrow x^2\ge 16\Rightarrow |x|\ge 4\). The square root is always nonnegative, so \(f(x)\ge 0\). As \(|x|\) increases, \(x^2-16\) increases without bound, so \(\sqrt{x^2-16}\) also has no upper limit. Therefore the range is \([0,\infty)\).\(f(x)=\sqrt{x^2-16}\) परिभाषित होने के लिए \(x^2-16\ge 0\Rightarrow x^2\ge 16\Rightarrow |x|\ge 4\) होना चाहिए। वर्गमूल हमेशा अशून्य (nonnegative) होता है, इसलिए \(f(x)\ge 0\)। जब \(|x|\) बढ़ता है, \(x^2-16\) भी अनंत तक बढ़ता है, तो \(\sqrt{x^2-16}\) का कोई ऊपरी सीमा नहीं है। अतः परास \([0,\infty)\) है।
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