UPSC NDA I and II 2020 Math Solved Paper

Concept:For $\tan A$ to be real, $\tan^2 A$ must be non-negative.Hence, the given expression $\frac{k-3}{3k-1}$ must satisfy $\frac{k-3}{3k-1} \ge 0$.Explanation:We are given $\tan^2 A = \frac{k-3}{3k-1}$.For real $\tan A$, we require $\frac{k-3}{3k-1} \ge 0$.This inequality holds when both numerator and denominator have the same sign or numerator is zero.Solving, we find that the expression is positive outside the interval $\left(\frac{1}{3}, 3\right)$.At $k = \frac{1}{3}$ and $k = 3$, the fraction becomes negative (or undefined).Therefore, for real $\tan A$, $k$ cannot lie strictly between $\frac{1}{3}$ and $3$.Answer:Option B: $\frac{1}{3}$ and $3$