UPSC NDA I 2018 Mathematics Solved Paper

Concept:Greatest integer function $\lfloor x \rfloor$ and quadratic equation.Explanation:Rewrite the equation $x^{2} - 4x + \lfloor x \rfloor = 0$ as $\lfloor x \rfloor = -x^{2} + 4x$.Let $f(x) = -x^{2} + 4x$ and $g(x) = \lfloor x \rfloor$.In the interval $[0, 2]$, $\lfloor x \rfloor$ takes these values:• For $x \text{ in } [0, 1)$, $\lfloor x \rfloor = 0$• For $x \text{ in } [1, 2)$, $\lfloor x \rfloor = 1$• At $x = 2$, $\lfloor x \rfloor = 2$Consider each subinterval.Case 1: $x \text{ in } [0, 1)$Equation becomes $x^{2} - 4x + 0 = 0 \Rightarrow x(x - 4) = 0$.Solutions are $x = 0$ and $x = 4$. Only $x = 0$ lies in $[0, 1)$.So one solution $x = 0$.Case 2: $x \text{ in } [1, 2)$Equation becomes $x^{2} - 4x + 1 = 0$.Using quadratic formula: $x = \frac{4 \pm \sqrt{16 - 4}}{2} = \frac{4 \pm \sqrt{12}}{2} = 2 \pm \sqrt{3}$.$\sqrt{3} \approx 1.732$, so $2 - \sqrt{3} \approx 0.268$ (not in $[1,2)$) and $2 + \sqrt{3} \approx 3.732$ (outside interval).No solution in this case.Case 3: $x = 2$Substitute: $2^{2} - 4 \cdot 2 + 2 = 4 - 8 + 2 = -2 \neq 0$. No solution.Thus the only solution in $[0,2]$ is $x = 0$.Answer:One (Option B).