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Mathematics 2023 Paper I 50 marks Solve

Paper I — Q6

(a) Solve the differential equation: d³y/dx³ - 3d²y/dx² + 4dy/dx - 2y = eˣ + cos x. (15 marks) (b) When a particle is projected…

(a)

Solve the differential equation: d³y/dx³ - 3d²y/dx² + 4dy/dx - 2y = eˣ + cos x. 15 marks

(b)

When a particle is projected from a point O₁ on the sea level with a velocity v and angle of projection θ with the horizon in a vertical plane, its horizontal range is R₁. If it is further projected from a point O₂, which is vertically above O₁ at a height h in the same vertical plane, with the same velocity v and same angle θ with the horizon, its horizontal range is R₂. Prove that R₂ > R₁ and (R₂-R₁):R₁ is equal to (1/2){√(1 + 2gh/v²sin²θ) - 1}:1. 15 marks

(c)

Evaluate the integral ∬ₛ (3y²z²î + 4z²x²ĵ + z²y²k̂)·n̂ dS, where S is the upper part of the surface 4x² + 4y² + 4z² = 1 above the plane z = 0 and bounded by the xy-plane. Hence, verify Gauss-Divergence theorem. 20 marks

हिंदी में प्रश्न पढ़ें
(a)

अवकल समीकरण : d³y/dx³ - 3d²y/dx² + 4dy/dx - 2y = eˣ + cos x का हल कीजिए। (15 अंक)

(b)

एक कण को समुद्र तल पर बिन्दु O₁ से वेग v तथा क्षैतिज से प्रक्षेप कोण θ पर उद्वाधर तल में प्रक्षेपित किया जाता है तो क्षैतिज परास R₁ है। यदि इसको पुनः बिन्दु O₂, जो उसी उद्वाधर तल में O₁ के उद्वाधरतः h ऊँचाई पर है, से उसी वेग v तथा क्षैतिज से समान कोण θ पर प्रक्षेपित किया जाता है तो क्षैतिज परास R₂ है। सिद्ध कीजिए R₂ > R₁ तथा (R₂ - R₁) : R₁ = (1/2){√(1 + 2gh/v²sin²θ) - 1} : 1. (15 अंक)

(c)

समाकल ∬ₛ (3y²z²î + 4z²x²ĵ + z²y²k̂)·n̂ dS का मान ज्ञात कीजिए; जहाँ S समतल z = 0 के ऊपर पृष्ठ 4x² + 4y² + 4z² = 1 का ऊपरी भाग है और xy-समतल द्वारा परिबद्ध है। अतः गॉस-अपसरण प्रमेय को सत्यापित कीजिए। (20 अंक)

Q6 of the 2023 UPSC Mains Mathematics Paper I, as printed
The question as printed in the 2023 Mathematics paper
Evaluate your answer to this question → See all 2023 Mathematics questions

Model answer

Written by UPSC Answer Check against this question's marking rubric, to the expected length. UPSC does not publish answers for Mains — this is one way to score well, not an official key.

(a) Let L = d³/dx³ - 3d²/dx² + 4d/dx - 2. The characteristic equation is r³ - 3r² + 4r - 2 = 0, i.e. (r - 1)(r² - 2r + 2) = 0. Hence r = 1, 1 ± i. Therefore the complementary function is y_c = C₁eˣ + eˣ(C₂ cos x + C₃ sin x).

For eˣ, since r = 1 is a simple root, take y_p1 = A x eˣ. Here P(r) = r³ - 3r² + 4r - 2, so P(1) = 0 and P′(1) = 1. Hence L(A x eˣ) = A eˣ, giving A = 1. Thus y_p1 = x eˣ.

For cos x, take y_p2 = B cos x + C sin x. Substitution gives L(y_p2) = (B + 3C) cos x + (C - 3B) sin x. Equating with cos x gives B + 3C = 1 and C - 3B = 0. Thus B = 1/10, C = 3/10, so y_p2 = (1/10)(cos x + 3 sin x).

Therefore the general solution is y = C₁eˣ + eˣ(C₂ cos x + C₃ sin x) + x eˣ + (1/10)(cos x + 3 sin x).

(b) Let O₁ be the origin, x horizontal and y vertical upward. For projection from sea level, y = v sinθ t - (1/2)gt², x = v cosθ t. Time of flight is T₁ = 2v sinθ/g, hence R₁ = v cosθ T₁ = v² sin2θ/g = 2v² sinθ cosθ/g.

For projection from O₂ at height h above O₁, y = h + v sinθ t - (1/2)gt². At sea level y = 0, so (1/2)gt² - v sinθ t - h = 0, giving the positive root t = [v sinθ + √(v² sin²θ + 2gh)]/g. Therefore R₂ = v cosθ t = (v² sinθ cosθ/g)[1 + √(1 + 2gh/(v² sin²θ))]. Since v² sinθ cosθ/g = R₁/2, R₂ = (R₁/2)[1 + √(1 + 2gh/(v² sin²θ))]. As h > 0, the square root is greater than 1, so R₂ > R₁. Also R₂ - R₁ = (R₁/2)[√(1 + 2gh/(v² sin²θ)) - 1]. Hence (R₂ - R₁):R₁ = (1/2)[√(1 + 2gh/(v² sin²θ)) - 1]:1. This assumes 0 < θ < π/2 and neglects air resistance.

(c) Write F = 3y²z² î + 4z²x² ĵ + z²y² k̂. The surface is the upper hemisphere of 4x² + 4y² + 4z² = 1, i.e. x² + y² + z² = a² with a = 1/2.

Parametrize S by r(φ,θ) = (a sinφ cosθ, a sinφ sinθ, a cosφ), where 0 ≤ φ ≤ π/2, 0 ≤ θ ≤ 2π. Then r_φ × r_θ = a² sinφ (sinφ cosθ î + sinφ sinθ ĵ + cosφ k̂), giving the outward normal. Computing F · (r_φ × r_θ), the terms involving cosθ and sinθ vanish after θ-integration from 0 to 2π. What remains is ∬ₛ F·n dS = a⁶ (∫₀^π/2 sin³φ cos³φ dφ)(∫₀^2π sin²θ dθ). Now ∫₀^π/2 sin³φ cos³φ dφ = 1/12, ∫₀^2π sin²θ dθ = π. Thus ∬ₛ F·n dS = a⁶π/12 = (1/64)π/12 = π/768.

To verify Gauss-Divergence theorem, take V as the upper half-ball bounded by S and the disk D: z = 0, x² + y² ≤ a². Now ∇·F = ∂(3y²z²)/∂x + ∂(4z²x²)/∂y + ∂(z²y²)/∂z = 2y²z. In spherical coordinates, ∭ᵥ 2y²z dV = ∫₀^a∫₀^π/2∫₀^2π 2(ρ² sin²φ sin²θ)(ρ cosφ)(ρ² sinφ) dθ dφ dρ = 2(a⁶/6)(1/4)(π) = a⁶π/12 = π/768. On D, z = 0, so F = 0 and ∬ᴅ F·n dS = 0. Therefore ∬ₛ F·n dS + ∬ᴅ F·n dS = π/768 + 0 = ∭ᵥ ∇·F dV. Hence Gauss-Divergence theorem is verified.

What "Solve" is asking you to do

Choose the method, then carry it through to a final answer. Identifying what kind of problem this is and why that method applies is the first thing marked; a correct figure arrived at invisibly earns almost nothing.

Structure that answers it

Given data and what is required → method chosen, with the reason it applies → set-up (equation, circuit, free body, trial balance) → working, step by step → answer with units and any condition of validity

Where marks are lost

Doing the middle steps mentally and writing only the result. In mathematics papers, a further loss comes from giving a decimal where the exact value in surds or fractions was wanted, or from skipping the justification a part explicitly asks for.

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How this answer will be evaluated

Approach

(a) calculate: given > formula > substitution > result with units > interpretation | (b) justify: claim > 3-4 reasons > evidence > conclusion | (c) calculate: given > formula > substitution > result with units > interpretation Full marks: Flawless derivations, correct algebra, and clear verification of the theorem.

Key points expected

  • Characteristic equation m³ - 3m² + 4m - 2 = 0
  • Identification of root m=1 and quadratic factor m² - 2m + 2
  • Complementary function y_c = c₁eˣ + e²ˣ(c₂cosx + c₃sinx)
  • Particular integral for eˣ using x²eˣ/2
  • Equation of trajectory for projection from height h
  • Derivation of R₂ = (v²sin2θ + 2v²sin²θ√(1+2gh/v²sin²θ))/2g
  • Derivation of R₁ = v²sin2θ/2g
  • Algebraic simplification to the required ratio form

Evaluation rubric

Each sub-part is marked on its own, against the marks and word limit printed on the paper.

  1. (a) Complete solution of the third-order linear ODE with constant coefficients. 15 marks

    calculate— given → formula → substitution → result with units → interpretation

    Must cover

    • Characteristic equation m³ - 3m² + 4m - 2 = 0
    • Identification of root m=1 and quadratic factor m² - 2m + 2
    • Complementary function y_c = c₁eˣ + e²ˣ(c₂cosx + c₃sinx)
    • Particular integral for eˣ using x²eˣ/2

    Loses marks

    • Incorrect factorization of characteristic polynomial
    • Missing x² factor in PI for eˣ

    Earns more

    • Particular integral for cosx using (cosx - 2sinx)/5
    • Verification of final solution by substitution

    Extra mark

    • Alternative method for finding roots
  2. (b) Proof that R₂ > R₁ and derivation of the specific ratio formula. 15 marks

    justify— claim → 3-4 reasons → evidence → conclusion

    Must cover

    • Equation of trajectory for projection from height h
    • Derivation of R₂ = (v²sin2θ + 2v²sin²θ√(1+2gh/v²sin²θ))/2g
    • Derivation of R₁ = v²sin2θ/2g
    • Algebraic simplification to the required ratio form

    Loses marks

    • Incorrect time of flight calculation for height h
    • Failure to isolate the ratio (R₂-R₁)/R₁

    Earns more

    • Explicit statement that R₂ > R₁ due to positive h term
    • Clear definition of variables v, θ, h, g

    Extra mark

    • Neat diagram of the two trajectories
  3. (c) Evaluation of the surface integral and verification of Gauss-Divergence theorem. 20 marks

    calculate— given → formula → substitution → result with units → interpretation

    Must cover

    • Surface integral calculation using projection on xy-plane
    • Divergence calculation: ∇·F = 12x²z² + 8yz² + 2z²y²
    • Volume integral setup in spherical coordinates
    • Equating surface and volume integral results

    Loses marks

    • Incorrect divergence calculation
    • Failure to verify the theorem by equating results

    Earns more

    • Correct handling of the normal vector n̂
    • Explicit limits of integration for the volume integral

    Extra mark

    • Alternative method for surface integral

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