फलन (f(x)=\frac{x}{x-2+1}) का परिसर चुनिए।
Choose the range of (f(x)=\frac{x}{x-2+1}).
Explanation opens after your attempt
Correct Answer
A. \([-\frac{1}{2},\frac{1}{2}]\)
Concept
If \(y=\frac{x}{x^2+1}\), then the discriminant of \(yx^2-x+y=0\) must satisfy \(1-4y^2\ge 0\). Thus \(-\frac{1}{2}\le y\le \frac{1}{2}\).
Why this answer is correct
The correct answer is A. \([-\frac{1}{2},\frac{1}{2}]\). If \(y=\frac{x}{x^2+1}\), then the discriminant of \(yx^2-x+y=0\) must satisfy \(1-4y^2\ge 0\). Thus \(-\frac{1}{2}\le y\le \frac{1}{2}\).
Exam Tip
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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