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What is the range of (f(x)=\frac{2}{3+\sqrt{x^2+4}})?फलन (f(x)=\frac{2}{3+\sqrt{x^2+4}}) का परिसर क्या है?
Answer and explanationउत्तर और व्याख्या
Correct answer:सही उत्तर: ( \left(0,\frac{2}{5}\right] )( \left(0,\frac{2}{5}\right] )
Since (\sqrt{x^2+4}\ge 2), the denominator is at least (5), so the maximum value is (\frac{2}{5}). As (|x|\to\infty), the value approaches (0) but never becomes (0).(\sqrt{x^2+4}\ge 2), इसलिए हर कम से कम (5) है और अधिकतम मान (\frac{2}{5}) है। (|x|\to\infty) पर मान (0) के पास जाता है पर (0) नहीं होता।
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