UPSC CDS I 2019 Math Paper

Concept:Use algebraic manipulation and componendo-dividendo to solve the given equation for $x$.Explanation:Start with the given equation:$\frac{1 + px}{1 - px} \sqrt{\frac{1 - qx}{1 + qx}} = 1$Isolate the square root term:$\sqrt{\frac{1 - qx}{1 + qx}} = \frac{1 - px}{1 + px}$Square both sides:$\frac{1 - qx}{1 + qx} = \left(\frac{1 - px}{1 + px}\right)^2$Expand the right side:$\frac{1 - qx}{1 + qx} = \frac{1 + p^2 x^2 - 2px}{1 + p^2 x^2 + 2px}$Apply componendo and dividendo to both sides:$\frac{(1 - qx) + (1 + qx)}{(1 - qx) - (1 + qx)} = \frac{(1 + p^2 x^2 - 2px) + (1 + p^2 x^2 + 2px)}{(1 + p^2 x^2 - 2px) - (1 + p^2 x^2 + 2px)}$Simplify numerators and denominators:$\frac{2}{-2qx} = \frac{2(1 + p^2 x^2)}{-4px}$Cancel common factors:$\frac{1}{-qx} = \frac{1 + p^2 x^2}{-2px}$Multiply both sides by $-1$:$\frac{1}{qx} = \frac{1 + p^2 x^2}{2px}$Cross multiply (assuming $x \neq 0$):$2px = qx(1 + p^2 x^2)$Cancel $x$ (since $x \neq 0$):$2p = q(1 + p^2 x^2)$Rearrange:$q p^2 x^2 = 2p - q$Solve for $x^2$:$x^2 = \frac{2p - q}{p^2 q}$Take square root to get non-zero solutions:$x = \pm \frac{1}{p} \sqrt{\frac{2p - q}{q}}$The condition $2p \neq q$ ensures the expression is defined.Answer:$x = \pm \frac{1}{p} \sqrt{\frac{2p - q}{q}}$ (Option A).