Mathematics 2024 Paper II 50 marks Prove

Q4

(a) Consider the polynomial ring Z[x] over the ring Z of integers. Let S be an ideal of Z[x] generated by x. Show that S is prime but not a maximal ideal of Z[x]. (15 marks) (b) Find the upper and lower Riemann integrals for the function f defined on [0, 1] as follows: f(x) = (1-x²)^½, if x is rational; f(x) = (1-x), if x is irrational. Hence, show that f is not Riemann integrable on [0, 1]. (15 marks) (c) The personnel manager of a company wants to assign officers A, B and C to the regional offices at Delhi, Mumbai, Kolkata and Chennai. The cost of relocation (in thousand Rupees) of the three officers at the four regional offices are given below: | Officer | Delhi | Mumbai | Kolkata | Chennai | |---------|-------|--------|---------|---------| | A | 16 | 22 | 24 | 20 | | B | 10 | 32 | 26 | 16 | | C | 10 | 20 | 46 | 30 | Find the assignment which minimizes the total cost of relocation and also determine the minimum cost. (20 marks)

हिंदी में प्रश्न पढ़ें

(a) पूर्णांकों के वलय Z पर बहुपद वलय Z[x] का विचार कीजिए। मान लीजिए x द्वारा जनित Z[x] की एक गुणजावली S है। दर्शाइए कि S, Z[x] की एक अभाज्य गुणजावली है लेकिन उच्चिष्ट गुणजावली नहीं है। (15 अंक) (b) [0, 1] पर परिभाषित निम्नलिखित फलन f के लिए ऊपर तथा निम्न रीमान समाकल ज्ञात कीजिए: f(x) = (1-x²)^½, यदि x परिमेय है; f(x) = (1-x), यदि x अपरिमेय है। अतः दर्शाइए कि [0, 1] पर f रीमान समाकलनीय नहीं है। (15 अंक) (c) एक कंपनी का कर्मिक प्रबंधक, अधिकारियों A, B और C को क्षेत्रीय कार्यालयों दिल्ली, मुंबई, कोलकाता और चेन्नई में नियुक्त करना चाहता है। चार क्षेत्रीय कार्यालयों में इन तीन अधिकारियों के स्थानांतरण की लागत (हजार रुपयों में) नीचे दी गई है: [तालिका दी गई है] वह नियतन (असाइनमेंट) ज्ञात कीजिए, जो स्थानांतरण की कुल लागत को न्यूनतम करता है और न्यूनतम लागत भी निर्धारित कीजिए। (20 अंक)

Directive word: Prove

This question asks you to prove. The directive word signals the depth of analysis expected, the structure of your answer, and the weight of evidence you must bring.

See our UPSC directive words guide for a full breakdown of how to respond to each command word.

How this answer will be evaluated

Approach

Prove the three mathematical claims systematically: for (a) establish primality via quotient ring isomorphism to Z and non-maximality by embedding in (x,2); for (b) compute upper and lower Darboux sums using density of rationals and irrationals, showing unequal integrals; for (c) solve the unbalanced assignment problem by adding a dummy officer with zero costs, then apply Hungarian algorithm. Allocate approximately 30% time to (a), 35% to (b), and 35% to (c) reflecting their computational demands.

Key points expected

  • For (a): Prove S=(x) is prime by showing Z[x]/(x) ≅ Z is an integral domain; prove not maximal by showing (x) ⊂ (x,2) ⊂ Z[x] or noting Z is not a field
  • For (a): Correctly identify that maximality would require Z[x]/(x) to be a field, which Z is not
  • For (b): Establish that on any subinterval, sup f = √(1-x²) (achieved at rationals dense in [0,1]) and inf f = (1-x) (achieved at irrationals), using density arguments
  • For (b): Compute upper integral ∫₀¹ √(1-x²)dx = π/4 and lower integral ∫₀¹ (1-x)dx = 1/2, showing π/4 ≠ 1/2 hence non-integrability
  • For (c): Convert unbalanced 3×4 problem to balanced 4×4 by adding dummy officer D with zero costs; apply Hungarian algorithm steps correctly
  • For (c): Identify optimal assignment (e.g., A-Chennai, B-Delhi, C-Mumbai or equivalent) with minimum cost calculation in thousand Rupees

Evaluation rubric

DimensionWeightMax marksExcellentAveragePoor
Setup correctness20%10Correctly defines S=(x) in Z[x], properly sets up quotient ring isomorphism for (a); correctly identifies sup/inf on arbitrary partitions for (b); properly constructs 4×4 cost matrix with dummy for (c)Basic setup correct but misses subtlety like explicit isomorphism in (a) or density justification in (b); adds dummy but with errors in cost matrixMisidentifies S as maximal or confuses Z[x] with Q[x]; confuses function definitions or uses wrong supremum/infimum; fails to balance the assignment problem
Method choice20%10Uses isomorphism theorem for prime ideals, proper containment for non-maximality; applies Darboux criterion with density arguments; employs Hungarian algorithm with row/column reductions systematicallyCorrect general methods but inefficient approach or missing key theorems; uses Riemann sum definition without Darboux; solves by enumeration rather than algorithmAttempts element-wise verification of prime ideal; uses Lebesgue criterion incorrectly; brute force without systematic optimization
Computation accuracy20%10Accurate integration yielding π/4 and 1/2 with exact inequality; correct Hungarian algorithm iterations with proper covering lines and zero assignments; arithmetic error-free throughoutMinor arithmetic errors in integration constants or final cost calculation; correct algorithm but slips in matrix updatesMajor errors in evaluating ∫√(1-x²)dx; incorrect minimum cost due to wrong assignment; computational breakdown in algorithm
Step justification20%10Rigorous justification of why sup/inf are achieved on dense sets; explicit appeal to First Isomorphism Theorem; clear proof that π/4 ≠ 1/2; step-by-step Hungarian algorithm justification with optimality proofStates key steps but skips density justification or assumes without proof; mentions theorems without full hypotheses; algorithm steps shown but optimality briefly justifiedUnjustified claims about function behavior; missing crucial logical steps; algorithm applied without explaining why solution is optimal
Final answer & units20%10Clear conclusion: S prime not maximal with explicit counterexample; explicit statement that f not Riemann integrable due to unequal upper/lower integrals; optimal assignment stated with officer-office mapping and minimum cost in thousand RupeesCorrect conclusions but poorly organized or missing explicit non-integrability statement; assignment correct but cost unit unclearMissing final conclusions; wrong assignment or cost; no units specified

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