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What is the range of (f(x)=\frac{\sqrt{x+1}}{\sqrt{x+1}+2})?फलन (f(x)=\frac{\sqrt{x+1}}{\sqrt{x+1}+2}) का परिसर क्या है?
Answer and explanationउत्तर और व्याख्या
Correct answer:सही उत्तर: ( [0,1) )( [0,1) )
Put (t=\sqrt{x+1}\ge 0), then (f=\frac{t}{t+2}). At (t=0), the value is (0), and as (t\to\infty), it approaches (1) but never equals (1).(t=\sqrt{x+1}\ge 0) रखने पर (f=\frac{t}{t+2})। (t=0) पर मान (0) और (t\to\infty) पर मान (1) के पास जाता है पर (1) नहीं होता।
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