UPSC NDA I 2016 Mathematics Solved Paper

Concept:For a quadratic equation $a x^{2} + b x + c = 0$ with real coefficients, the condition for real $x$ is discriminant $D = b^{2} - 4ac \geq 0$.Explanation:Given $y = \frac{x^{2}}{1 + x^{2}}$ for all $x \in \mathbb{R}$.Rearrange to bring all terms to one side:$y(1 + x^{2}) = x^{2} \Rightarrow y + yx^{2} = x^{2} \Rightarrow (y-1)x^{2} + y = 0$.This is a quadratic in $x$ with coefficients $a = y-1$, $b = 0$, $c = y$.For $x$ to be real, the quadratic must have real roots, so $D \geq 0$.$D = b^{2} - 4ac = 0 - 4(y-1)(y) = -4y(y-1) \geq 0$.Divide by $-4$ (reverse inequality): $y(y-1) \leq 0$.Solving gives $0 \leq y \leq 1$.However, note that $a = y-1 \neq 0$ for the quadratic to exist; $y=1$ would make $a=0$ and the equation becomes $0 \cdot x^{2} + 1 = 0$, impossible.Also, $y=0$ is attained at $x=0$.Thus $0 \leq y < 1$.Answer:$[0, 1)$ (Option A).