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If \(f:\mathbb{R}\to\mathbb{R}\) is given by \(f(x)=\frac{x}{1+x^2}\), what is the range of (f)?यदि \(f:\mathbb{R}\to\mathbb{R}\) को \(f(x)=\frac{x}{1+x^2}\) से दिया गया है, तो (f) का परास क्या है?
Answer and explanationउत्तर और व्याख्या
Correct answer:सही उत्तर: \(\left[-\frac{1}{2},\frac{1}{2}\right]\)\(\left[-\frac{1}{2},\frac{1}{2}\right]\)
Step 1: Put \(y=\frac{x}{1+x^2}\). Step 2: Treat \(yx^2-x+y=0\) as a quadratic in (x) and require its discriminant to be non-negative. Step 3: From \(1-4y^2\geq0\), we get \(-\frac{1}{2}\leq y\leq\frac{1}{2}\).चरण 1: \(y=\frac{x}{1+x^2}\) मानें। चरण 2: \(yx^2-x+y=0\) को (x) में द्विघात मानकर विविक्तकर \(\geq0\) रखें। चरण 3: \(1-4y^2\geq0\) से \(-\frac{1}{2}\leq y\leq\frac{1}{2}\) मिलता है।
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