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What is the range of (f(x)=\frac{1}{2+\sqrt{x^2+1}})?फलन (f(x)=\frac{1}{2+\sqrt{x^2+1}}) का परिसर क्या है?
Answer and explanationउत्तर और व्याख्या
Correct answer:सही उत्तर: ( (0,\frac{1}{3}] )( (0,\frac{1}{3}] )
Since (\sqrt{x^2+1}\ge 1), the denominator is at least (3). The maximum is (\frac{1}{3}), and the value approaches (0).(\sqrt{x^2+1}\ge 1), इसलिए हर कम से कम (3) है। अधिकतम (\frac{1}{3}) है और मान (0) के पास जाता है।
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