UPSC NDA I 2018 Mathematics Solved Paper

Concept:The modulus of a complex number $z = a + bi$ is $|z| = \sqrt{a^2 + b^2}$. When both sides have the same base, equate the exponents. We seek solutions where $x$ is a non‑zero integer.Explanation:First, compute $\lvert 1 - 2i \rvert$. Here $a = 1$, $b = -2$. So $\lvert 1 - 2i \rvert = \sqrt{1^2 + (-2)^2} = \sqrt{1 + 4} = \sqrt{5}$. The given equation becomes $(\sqrt{5})^{x} = 5^{x}$. Rewrite $\sqrt{5}$ as $5^{1/2}$. Thus $(5^{1/2})^{x} = 5^{x} \quad\Rightarrow\quad 5^{\frac{x}{2}} = 5^{x}$. Since the bases are equal and non‑zero, the exponents must be equal: $\frac{x}{2} = x \quad\Rightarrow\quad x = 2x \quad\Rightarrow\quad x = 0$. The only solution is $x = 0$, which is not a non‑zero integer. Therefore, there are no non‑zero integral solutions.Answer:Zero (No solution) – Option A.