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 $1/2$.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 = $\frac{c-a}{2(c-a)} = 1/2$. 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.Original given expression: $\frac{(a+b)^{2}}{(c-a)(c+a+b)}+\frac{(a+b)c}{c^{2}-bc-a^{2}-ab}-\frac{a(a+2b+c)}{2(c-a)}$.We already did first two: sum = $(a+b)/(c-a)$. Then subtract third term: $\frac{a+b}{c-a} - \frac{a(a+2b+c)}{2(c-a)} = \frac{2(a+b) - a(a+2b+c)}{2(c-a)}$. Now simplify numerator: $2a+2b - a^{2} - 2ab - ac = -a^{2} - 2ab - ac + 2a + 2b$. Factor by grouping? Write as $-(a^{2}+2ab+ac) + 2(a+b) = -a(a+2b+c) + 2(a+b)$. Can we write $2(a+b) - a(a+2b+c) = (c-a)$? Let's test with numbers: choose a=1, b=2, c=3. Then numerator = $2(1+2) - 1(1+4+3) = 2*3 - 1*8 = 6-8 = -2$. Denominator = $2(c-a)=2(2)=4$, so fraction = -1/2. But that gives negative? Check original expression with a=1,b=2,c=3: first term: $(1+2)^2/((3-1)(3+1+2)) = 9/(2*6)=9/12=3/4$. Second term: $(1+2)*3/(9-6-1-2)=3*3/(0)$ denominator zero? Wait $c^{2}-bc-a^{2}-ab=9-6-1-2=0$, so second term undefined. So my test values invalid because they make denominator zero. The condition a≠b,b≠c,c≠a but also denominators must not be zero. So need to pick values where all denominators non-zero. Let's choose a=1, b=2, c=4. Then compute: first term: $(3)^2/((3)(4+1+2))? Actually (c-a)=3, (c+a+b)=4+1+2=7 => 9/(3*7)=9/21=3/7. Second term:$(3)*4/(16-8-1-2)=12/(5)=12/5=2.4. Sum = 3/7+12/5 = (15+84)/35=99/35≈2.8286. Third term: a(a+2b+c)/2(c-a)=1*(1+4+4)/2*3=9/6=1.5. So expression = 99/35 - 1.5 = 99/35 - 3/2 = (198-105)/70=93/70≈1.3286. Not 1/2. Something off. Let's recompute second term denominator: $c^{2}-bc-a^{2}-ab = 16-8-1-2=5$ correct. So the sum is not 1/2. Maybe I mis-factored? Actually the original solution says $c^{2} - bc - a^{2} - ab = (c+a)(c-a-b)$. Check: (c+a)(c-a-b) = (c+a)(c - (a+b)) = c^2 - c(a+b) + ac - a(a+b) = c^2 -ac -bc + ac - a^2 -ab = c^2 - bc - a^2 - ab. Yes correct. So second term denominator = (c+a)(c-a-b). First term denominator = (c-a)(c+a+b). Are these equivalent? No, they are different. So the common denominator is not simply (c-a)(c+a+b) because the second denominator has (c+a) instead of (c-a). We need to find a common denominator for the first two fractions. Multiply numerator and denominator of second fraction to get denominator (c-a)(c+a+b). Note that (c+a)(c-a-b) and (c-a)(c+a+b) are related: multiply numerator and denominator of second fraction by (c-a)/(c-a)? Actually to make denominator (c-a)(c+a+b), we need to multiply second fraction by $\frac{c-a}{c-a} \cdot \frac{c+a+b}{c+a+b}$? That would be messy. Alternatively, observe that $(c+a)(c-a-b) = (c-a)(c+a+b) - 2a(b+c)? Not obvious. Let's compute product: (c-a)(c+a+b) = (c-a)(c+a) + (c-a)b = (c^2 - a^2) + b(c-a). (c+a)(c-a-b) = (c+a)(c-a) - (c+a)b = (c^2 - a^2) - b(c+a). So they differ by sign of b terms. Actually (c-a)(c+a+b) = (c^2 - a^2) + b(c-a) and (c+a)(c-a-b) = (c^2 - a^2) - b(c+a). So not equal. Therefore the solution's step "common denominator is (c-a)(c+a+b)" is incorrect unless we adjust numerators accordingly. The existing solution seems flawed. Let's re-derive correctly. Given expression: Term1: T1 = (a+b)^2 / [(c-a)(c+a+b)] Term2: T2 = (a+b)c / [(c+a)(c-a-b)] Term3: T3 = a(a+2b+c) / [2(c-a)] We need to simplify T1+T2-T3. Let me denote S = T1+T2. Find a common denominator. Perhaps factor further. Note that (c-a-b) = (c - (a+b)). Also note that (c+a+b) = (c + (a+b)). So we have terms with (c - (a+b)) and (c + (a+b)). Also denominators have (c-a) and (c+a). So there is symmetry. Maybe combine using partial fractions? Alternatively, set x = a+b, y = c? But a and b separate. Let's try algebraic manipulation. Write T1 = (a+b)^2 / [(c-a)(c+x)] where x = a+b. T2 = (a+b)c / [(c+a)(c-x)]. T3 = a(a+2b+c) / [2(c-a)]. We need to find a common denominator for T1 and T2. The LCM of (c-a)(c+x) and (c+a)(c-x) is (c-a)(c+a)(c+x)(c-x)? That would be degree 4, but maybe simplifications occur. Alternatively, observe that (c+a)(c-x) = (c+a)(c - (a+b)) = c^2 - c(a+b) + ac - a(a+b) = c^2 - bc - a^2 - ab (as before). And (c-a)(c+x) = (c-a)(c+a+b) = c^2 - a^2 + b(c-a). Not symmetric. Maybe a better approach: rewrite T2 using the identity we have. Let's compute T1+T2 as a single fraction over a common denominator, but perhaps there is a known simplification. Let's attempt to combine step by step with algebra. Let D1 = (c-a)(c+a+b) and D2 = (c+a)(c-a-b). Find a common denominator D = (c-a)(c+a)(c+a+b)(c-a-b)? That's huge. But note that D1 and D2 are both products of two linear factors. Maybe we can write T1+T2 = (a+b)/(c-a) * something? From the existing solution, they claimed T1+T2 = (a+b)/(c-a). Let's test with numbers: a=1,b=2,c=5 (to avoid zeros). Then T1 = 9/(4*(5+3)=9/(4*8)=9/32=0.28125. T2 = 3*5/((6)*(5-3)=15/(6*2)=15/12=1.25. Sum=1.53125. (a+b)/(c-a)=3/4=0.75. Not equal. So the claim is false. The existing solution is erroneous. Given the context, the original solution likely made a mistake. But as a tutor, I must provide the correct solution. However, the user asked to rewrite the given solution, not to correct it. But the instruction says "Verify the result internally before presenting the final solution." So I must check and provide correct answer if the given solution is wrong. Also the options are -1/2, 0, 1/2, 1. For the test values a=1,b=2,c=5, we got about 1.53, not any of these. Maybe I miscomputed T3? T3 = a(a+2b+c)/[2(c-a)] = 1*(1+4+5)/(2*4) = 10/8=1.25. Then T1+T2 - T3 = 1.53125 - 1.25 = 0.28125 = 9/32. Not constant. Maybe there is a condition a≠b,b≠c,c≠a but also something about the expression being independent of a,b,c? It seems not constant from this test. Let me try another set: a=2,b=3,c=7. Then T1: (5^2)/((5)*(7+2+3? Actually c-a=5, c+a+b=7+2+3=12) => 25/(5*12)=25/60=5/12≈0.4167. T2: (5*7)/((9)*(7-5)=35/(9*2)=35/18≈1.9444. Sum≈2.3611. T3: 2*(2+6+7)/(2*5)=2*15/10=30/10=3. So expression = 2.3611-3 = -0.6389 = -23/36? Not constant. Thus the given expression is not constant. Perhaps the original question had a different expression? The existing solution might have been for a different problem. Given the instruction to rewrite the existing solution, I must assume the solution is correct and just rewrite it as is, even if it's mathematically wrong? The rules say "Verify the result internally before presenting the final solution." So I should correct if necessary. But the user provided the existing solution and expects a rewrite. I think the safest is to present the rewrite exactly as per the given solution, but also note that the explanation might be flawed. However, the user is the one who provided the solution; they likely expect the rewritten version consistent with that solution. As an expert tutor, I should produce a clear, step-by-step rewrite that matches the logic of the existing solution, even if it's incorrect. But that would be misleading. The instruction says "strict rules: output ONLY the rewritten solution." and "Match the exact language of the question." It also says "Verify the result internally before presenting the final solution." This implies I should ensure the solution is correct. Given the discrepancy, I need to re-evaluate the original expression more carefully. Perhaps I mis-read the expression. The question:$\frac{(a+b)^{2}}{(c-a)(c+a+b)}+\frac{(a+b)c}{c^{2}-bc-a^{2}-ab}-\frac{(a+2b+c)}{2(c-a)}a≠b,b≠c,c≠a$? Wait, the question as provided has a minus sign before the third fraction, and the third fraction is$\frac{(a+2b+c)}{2(c-a)}$? Actually the user wrote: "-\frac{(a+2b+c)}{2(c-a)}a≠b,b≠c,c≠a". But in the existing solution, they wrote "-\frac{a(a+2b+c)}{2(c-a)}". There is an extra 'a' in the numerator. The question has no 'a' factor: it's just (a+2b+c) in numerator. Let me check the original: In the user's message, the question is: "What is$\frac{(a+b)^{2}}{(c-a)(c+a+b)}+\frac{(a+b)c}{c^{2}-bc-a^{2}-ab}-\frac{(a+2b+c)}{2(c-a)}a≠b,b≠c,c≠a$ equal to?" There is no 'a' multiplying (a+2b+c). But the existing solution includes 'a' factor: "-\frac{a(a+2b+c)}{2(c-a)}". So there is a discrepancy. The existing solution might be for a different version. The user might have copied the question incorrectly? In the "Existing Solution to Rewrite" they have: "-\frac{a(a+2b+c)}{2(c-a)}". So the solution uses 'a' factor. I'll assume the correct expression is with that 'a'. Actually the question text in the user's input has: "-\frac{(a+2b+c)}{2(c-a)}". But then the existing solution says "-\frac{a(a+2b+c)}{2(c-a)}". Possibly a typo. As the tutor, I should use the expression as given in the solution, because that's what is being solved. The user wants to rewrite that solution. So I will proceed with the expression that has an 'a' factor in the third term.Given that, let's test with the same numbers: a=1,b=2,c=5, T3 becomes 1*(1+4+5)/(2*4)=10/8=1.25 (same as before, because we already had a factor of a=1). For a=2,b=3,c=7, T3 = 2*(2+6+7)/(2*5)=2*15/10=3 (same). So the test remains. The sum is not constant. Could there be