UPSC NDA 2 2017 Mathematics Solved Paper

Concept:Use the triangle inequality to relate $|z|$ and $\left| z - \frac{4}{z} \right|$, then solve a quadratic inequality.Explanation:We are given $\left| z - \frac{4}{z} \right| = 2$.Apply the triangle inequality: $|a + b| \le |a| + |b|$.Write $|z| = \left| \left(z - \frac{4}{z}\right) + \frac{4}{z} \right|$.Then $|z| \le \left| z - \frac{4}{z} \right| + \left| \frac{4}{z} \right| = 2 + \frac{4}{|z|}$.Multiply both sides by $|z| > 0$: $|z|^2 \le 2|z| + 4$.Rearrange: $|z|^2 - 2|z| - 4 \le 0$.Solve the quadratic equation $|z|^2 - 2|z| - 4 = 0$ using the quadratic formula.$|z| = \frac{2 \pm \sqrt{4 + 16}}{2} = \frac{2 \pm \sqrt{20}}{2} = 1 \pm \sqrt{5}$.Since $|z| \ge 0$, the inequality $|z|^2 - 2|z| - 4 \le 0$ holds for $|z|$ between $1 - \sqrt{5}$ (negative) and $1 + \sqrt{5}$.Thus the maximum possible value of $|z|$ is $1 + \sqrt{5}$.Answer:$1 + \sqrt{5}$ (Option B)