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What is the range of (f(x)=\frac{x^2}{x^2+4x+8})?फलन (f(x)=\frac{x^2}{x^2+4x+8}) का परिसर क्या है?
Answer and explanationउत्तर और व्याख्या
Correct answer:सही उत्तर: ( [0,\infty) )( [0,\infty) )
From (y=\frac{x^2}{x^2+4x+8}), real (x) is needed in ((y-1)x^2+4yx+8y=0). The discriminant gives (0\le y\le 2), and (y=1) is also possible, so the range is ([0,2]).(y=\frac{x^2}{x^2+4x+8}) से ((y-1)x^2+4yx+8y=0) में वास्तविक (x) चाहिए। विविक्तकर (16y^2-32y(y-1)=16y(2-y)) से (0\le y\le 2), पर (y=1) भी संभव है, इसलिए विस्तृत जांच से ([0,2]) मिलता है।
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