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Choose the range of (f(x)=\frac{x^2+2}{x^2+5}).फलन (f(x)=\frac{x^2+2}{x^2+5}) का परिसर चुनिए।
Answer and explanationउत्तर और व्याख्या
Correct answer:सही उत्तर: ([\frac{2}{5},1))([\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).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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