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What is the range of (f(x)=\frac{1}{\sqrt{16-x^2}})?फलन (f(x)=\frac{1}{\sqrt{16-x^2}}) का परिसर क्या है?
Answer and explanationउत्तर और व्याख्या
Correct answer:सही उत्तर: ( \left[\frac{1}{4},\infty\right) )( \left[\frac{1}{4},\infty\right) )
The maximum of (\sqrt{16-x^2}) is (4), so the minimum fraction is (\frac{1}{4}). Near the endpoints, the denominator tends to (0).(\sqrt{16-x^2}) का अधिकतम (4) है, इसलिए भिन्न का न्यूनतम (\frac{1}{4}) है। किनारों के पास हर (0) की ओर जाता है।
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