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What is the range of (f(x)=\frac{1}{\sqrt{4-x^2}})?फलन (f(x)=\frac{1}{\sqrt{4-x^2}}) का परिसर क्या है?
Answer and explanationउत्तर और व्याख्या
Correct answer:सही उत्तर: ([\frac{1}{2},\infty))([\frac{1}{2},\infty))
The maximum value of (\sqrt{4-x^2}) is (2), so the minimum value of (f) is (\frac{1}{2}). Near the endpoints, the value grows without bound.(\sqrt{4-x^2}) का अधिकतम मान (2) है, इसलिए (f) का न्यूनतम मान (\frac{1}{2}) है। किनारों के पास मान असीमित बढ़ता है।
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