UPSC CDS II 2025 Mathematics Paper

Concept:Factorisation of quadratic expressions and algebraic simplification using common denominators.Explanation:First, factor the second denominator: $c^{2}-bc-a^{2}-ab = (c+a)(c-a-b)$.Rewrite the first two fractions with common denominator $(c-a)(c+a+b)$:$\frac{(a+b)^{2}}{(c-a)(c+a+b)}+\frac{(a+b)c}{(c+a)(c-a-b)}$.Note that $(c+a)(c-a-b) = (c-a)(c+a+b)$ (since $(c+a)(c-a-b) = (c-a)(c+a+b)$ after factoring). So the common denominator is $(c-a)(c+a+b)$.Combine numerators: $(a+b)^{2}+(a+b)c = (a+b)(a+b+c)$.Thus, the sum of first two terms simplifies to $\frac{(a+b)(a+b+c)}{(c-a)(c+a+b)} = \frac{a+b}{c-a}$.Now subtract the third term: $\frac{a+b}{c-a}-\frac{a(a+2b+c)}{2(c-a)}$.Take LCM $2(c-a)$: numerator = $2(a+b)-a(a+2b+c) = 2a+2b-a^{2}-2ab-ac$.Simplify: $-a^{2}-2ab-ac+2a+2b = -a(a+2b+c)+2(a+b)$.But note $a+2b+c = (a+b)+(b+c)$. However, a simpler approach: set $a=0,\ b=1,\ c=2$ (random values satisfying conditions) to verify the expression becomes $\frac{1}{2}$.Algebraically, the numerator simplifies to $(c-a)$: actually compute $2(a+b)-a(a+2b+c) = 2a+2b-a^{2}-2ab-ac$. Factor: $2a+2b-a(a+2b+c) = (c-a)(?)$? Let's do step:$2(a+b)-a(a+2b+c) = 2a+2b-a^{2}-2ab-ac$.Rewrite as $-a^{2}-2ab-ac+2a+2b = -a(a+2b+c)+2(a+b)$. Not obviously factorable. Instead, use identity: $(a+b+c)(c-a) = (c^{2}-a^{2})+(bc-ab)$? Not needed. The given solution states the numerator simplifies to $c-a$. Let's check: if numerator = $c-a$, then fraction = $\frac{c-a}{2(c-a)} = \frac{1}{2}$. So we need to verify that $2(a+b)-a(a+2b+c) = c-a$.Expand $c-a$: no. Let's compute directly: $2a+2b-a^{2}-2ab-ac$. Set equal to $c-a$? That would imply $2a+2b-a^{2}-2ab-ac-c+a = 0$ => $3a+2b-a^{2}-2ab-ac-c = 0$, not generally true. Wait, the original solution might have a mistake? Let's re-evaluate the expression carefully.