UPSC NDA I 2019 Mathematics Solved Paper

Concept:The absolute value equation $|f(x)| = g(x)$ is solved by considering two cases: when $f(x) \ge 0$ we have $f(x) = g(x)$, and when $f(x) < 0$ we have $-f(x) = g(x)$. Additionally, we must check that the solutions lie in the domain of each case.Explanation:Given $|x^{2} - x - 6| = x + 2$.Factor the quadratic: $x^{2} - x - 6 = (x-3)(x+2)$.So the equation becomes $|(x-3)(x+2)| = x+2$.We consider two cases based on the sign of $(x-3)(x+2)$.Case 1: $(x-3)(x+2) \ge 0$ i.e. $x \le -2$ or $x \ge 3$.Then $|(x-3)(x+2)| = (x-3)(x+2) = x^{2} - x - 6$.Equation: $x^{2} - x - 6 = x + 2$ → $x^{2} - 2x - 8 = 0$ → $(x-4)(x+2)=0$.So $x = 4$ or $x = -2$. Both belong to $(-\infty, -2] \cup [3, \infty)$.Thus valid roots from this case: $x = -2, 4$.Case 2: $(x-3)(x+2) < 0$ i.e. $-2 < x < 3$.Then $|(x-3)(x+2)| = -(x-3)(x+2) = -x^{2} + x + 6$.Equation: $-x^{2} + x + 6 = x + 2$ → $-x^{2} + 6 = 2$ → $-x^{2} + 4 = 0$ → $x^{2} = 4$.So $x = 2$ or $x = -2$. Only $x = 2$ lies in $(-2, 3)$; $x = -2$ is not in this open interval.Thus valid root from this case: $x = 2$.Combining both cases, the roots are $x = -2$, $x = 2$, and $x = 4$.These correspond to option D: $-2, 2, 4$.Answer:$-2, 2, 4$ (Option D)