UPSC CDS 2 2023 Math Paper

Concept:Use the Componendo-Dividendo rule: if $\frac{a+b}{a-b} = \frac{c+d}{c-d}$, then $\frac{a}{b} = \frac{c}{d}$. This helps simplify the given equation to isolate $x$.Explanation:Start with the given equation:$$\frac{(x-a)(x-b)}{(x-ma)(x-mb)} = \frac{(x+a)(x+b)}{(x+ma)(x+mb)}$$Cross-multiply to bring similar terms together:$$\frac{(x-a)(x-b)}{(x+a)(x+b)} = \frac{(x-ma)(x-mb)}{(x+ma)(x+mb)}$$Expand each numerator and denominator:$$\frac{x^2 - (a+b)x + ab}{x^2 + (a+b)x + ab} = \frac{x^2 - (ma+mb)x + m^2ab}{x^2 + (ma+mb)x + m^2ab}$$Apply Componendo-Dividendo (add 1 and subtract 1 on both sides, then take ratio):$$\frac{(x^2 + ab)}{(a+b)x} = \frac{(x^2 + m^2ab)}{(ma+mb)x}$$Cancel $x$ (assuming $x \neq 0$) and simplify the denominator on the right:$$\frac{x^2 + ab}{x^2 + m^2ab} = \frac{(a+b)}{(a+b)m} = \frac{1}{m}$$Cross-multiply:$$m(x^2 + ab) = x^2 + m^2ab$$Expand and collect like terms:$$mx^2 + mab = x^2 + m^2ab$$$$mx^2 - x^2 = m^2ab - mab$$$$x^2(m-1) = mab(m-1)$$If $m \neq 1$, cancel $(m-1)$:$$x^2 = mab$$Take square root:$$x = \pm \sqrt{mab}$$Answer:$x = \pm \sqrt{mab}$ (Option A).