फलन (f(x)=\frac{\sqrt{x+1}}{\sqrt{x+1}+2}) का परिसर क्या है?
What is the range of (f(x)=\frac{\sqrt{x+1}}{\sqrt{x+1}+2})?
Explanation opens after your attempt
Correct Answer
A. ( [0,1) )
Concept
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).
Why this answer is correct
The correct answer is A. ( [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).
Exam Tip
\(t=\sqrt{x+1}\ge 0\) 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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