UPSC NDA I 2018 Mathematics Solved Paper

Concept:The modulus $|1-x|$ splits the equation into two cases based on whether $(1-x) \ge 0$ or $(1-x) < 0$.Each case gives a quadratic equation, solved using the quadratic formula $x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}$.Only roots satisfying the case condition are accepted.Explanation:Case I: $1-x \ge 0$ i.e. $x \le 1$.Then $|1-x| = 1-x$, so $1-x + x^2 = 5$.Simplify: $x^2 - x - 4 = 0$.Using quadratic formula: $x = \frac{1 \pm \sqrt{1 - 4(1)(-4)}}{2} = \frac{1 \pm \sqrt{17}}{2}$.Both roots are $\frac{1+\sqrt{17}}{2} \approx 2.56$ and $\frac{1-\sqrt{17}}{2} \approx -1.56$.Since $x \le 1$, only $x = \frac{1-\sqrt{17}}{2}$ is valid. This root is irrational.Case II: $1-x < 0$ i.e. $x > 1$.Then $|1-x| = -(1-x) = x-1$, so $-(1-x) + x^2 = 5$ becomes $x-1 + x^2 = 5$.Simplify: $x^2 + x - 6 = 0$.Using quadratic formula: $x = \frac{-1 \pm \sqrt{1 - 4(1)(-6)}}{2} = \frac{-1 \pm \sqrt{25}}{2} = \frac{-1 \pm 5}{2}$.Roots: $x = 2$ and $x = -3$.Since $x > 1$, only $x = 2$ is valid. This root is rational.Thus the equation has two real roots: one irrational $\left(\frac{1-\sqrt{17}}{2}\right)$ and one rational $(2)$.Answer:A. a rational root and an irrational root