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Choose the range of (f(x)=\frac{x}{x^2+1}).फलन (f(x)=\frac{x}{x^2+1}) का परिसर चुनिए।
Answer and explanationउत्तर और व्याख्या
Correct answer:सही उत्तर: ([-\frac{1}{2},\frac{1}{2}])([-\frac{1}{2},\frac{1}{2}])
If (y=\frac{x}{x^2+1}), then the discriminant of (yx^2-x+y=0) must satisfy (1-4y^2\ge 0). Thus (-\frac{1}{2}\le y\le \frac{1}{2}).यदि (y=\frac{x}{x^2+1}), तो (yx^2-x+y=0) में विविक्तकर (1-4y^2\ge 0) चाहिए। अतः (-\frac{1}{2}\le y\le \frac{1}{2})।
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