फलन (f(x)=\frac{x-2+2}{x-2+5}) का परिसर चुनिए।
Choose the range of (f(x)=\frac{x-2+2}{x-2+5}).
Explanation opens after your attempt
Correct Answer
A. \([\frac{2}{5},1\))
Concept
Let \(t=x^2\ge 0\), then \(f=\frac{t+2}{t+5}\). It is \(\frac{2}{5}\) at (t=0) and approaches (1) but never equals (1).
Why this answer is correct
The correct answer is A. \([\frac{2}{5},1\)). Let \(t=x^2\ge 0\), then \(f=\frac{t+2}{t+5}\). It is \(\frac{2}{5}\) at (t=0) and approaches (1) but never equals (1).
Exam Tip
मान लें \(t=x^2\ge 0\), तब \(f=\frac{t+2}{t+5}\)। यह (t=0) पर \(\frac{2}{5}\) है और (1) तक पहुंचता है पर (1) नहीं होता।
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