UPSC CDS 2 2023 Math Paper

Concept:Componendo-Dividendo rule is used to simplify ratios.Explanation:Given: $\frac{(x-a)(x-b)}{(x-ma)(x-mb)} = \frac{(x+a)(x+b)}{(x+ma)(x+mb)}$ Cross-multiply terms to combine: $\frac{(x-a)(x-b)}{(x+a)(x+b)} = \frac{(x-ma)(x-mb)}{(x+ma)(x+mb)}$ Expand each product: Numerator left: $x^2 - (a+b)x + ab$ Denominator left: $x^2 + (a+b)x + ab$ Numerator right: $x^2 - (ma+mb)x + m^2ab$ Denominator right: $x^2 + (ma+mb)x + m^2ab$ Apply componendo-dividendo: If $\frac{P}{Q} = \frac{R}{S}$, then $\frac{P+Q}{P-Q} = \frac{R+S}{R-S}$ Here, $\frac{x^2 - (a+b)x + ab}{x^2 + (a+b)x + ab} = \frac{x^2 - (ma+mb)x + m^2ab}{x^2 + (ma+mb)x + m^2ab}$ Using the rule: $\frac{(x^2 + ab)}{(a+b)x} = \frac{(x^2 + m^2ab)}{(ma+mb)x}$ Cancel $x$ (assuming $x \neq 0$): $\frac{x^2 + ab}{a+b} = \frac{x^2 + m^2ab}{m(a+b)}$ Multiply both sides by $(a+b)$: $x^2 + ab = \frac{x^2 + m^2ab}{m}$ Thus, $\frac{x^2 + ab}{x^2 + m^2ab} = \frac{1}{m}$Answer:$\frac{x^{2}+ab}{x^{2}+m^{2}ab} = \frac{1}{m}$ Hence, the correct answer is $\frac{1}{m}$ (Option D).