Mathematics

UPSC Mathematics 2023 — Paper I

All 8 questions from UPSC Civil Services Mains Mathematics 2023 Paper I (400 marks total). Every stem reproduced in full, with directive-word analysis, marks, word limits, and answer-approach pointers.

8Questions
400Total marks
2023Year
Paper IPaper

Topics covered

Linear algebra, calculus and 3D geometry (1)Linear algebra, multivariable calculus and 3D geometry (1)Linear algebra, multivariable calculus, 3D geometry (1)Matrix rank, curve tracing, 3D geometry (1)Differential equations, Laplace transforms, mechanics, SHM, vector calculus (1)Differential equations, projectile motion, surface integrals (1)Differential equations and mechanics (1)Laplace transform and vector calculus (1)

A

Q1
50M Compulsory solve Linear algebra, calculus and 3D geometry

(a) Let V₁ = (2, -1, 3, 2), V₂ = (-1, 1, 1, -3) and V₃ = (1, 1, 9, -5) be three vectors of the space ℝ⁴. Does (3, -1, 0, -1) ∈ span {V₁, V₂, V₃} ? Justify your answer. (10 marks) (b) Find the rank and nullity of the linear transformation : T : ℝ³ → ℝ³ given by T(x, y, z) = (x + z, x + y + 2z, 2x + y + 3z) (10 marks) (c) Find the values of p and q for which limₓ→₀ [x(1 + p cos x) - q sin x]/x³ exists and equals 1. (10 marks) (d) Examine the convergence of the integral ∫₀¹ (log x)/(1+x) dx (10 marks) (e) A variable plane which is at a constant distance 3p from the origin O cuts the axes in the points A, B, C respectively. Show that the locus of the centroid of the tetrahedron OABC is 9(1/x² + 1/y² + 1/z²) = 16/p². (10 marks)

हिंदी में पढ़ें

(a) मान लीजिए V₁ = (2, -1, 3, 2), V₂ = (-1, 1, 1, -3), V₃ = (1, 1, 9, -5) समष्टि ℝ⁴ के तीन सदिश हैं । क्या (3, -1, 0, -1) ∈ विस्तृति {V₁, V₂, V₃} ? अपने उत्तर को तर्कसहित सिद्ध कीजिए । (10 अंक) (b) T(x, y, z) = (x + z, x + y + 2z, 2x + y + 3z) द्वारा दिए गए रैखिक रूपांतरण : T : ℝ³ → ℝ³ की कोटि तथा शून्यता ज्ञात कीजिए । (10 अंक) (c) p तथा q के वो मान निकालिए जिसके लिए limₓ→₀ [x(1 + p cos x) - q sin x]/x³ का अस्तित्व है एवं 1 के बराबर है । (10 अंक) (d) समाकल ∫₀¹ (log x)/(1+x) dx की अभिसारिता का परीक्षण कीजिए । (10 अंक) (e) एक चर समतल, जो कि मूल-बिंदु O से अचर दूरी 3p पर है, अक्षों को क्रमशः बिंदुओं A, B, C पर काटता है । दर्शाइए कि चतुष्फलक OABC के केंद्रक का बिंदुपथ 9(1/x² + 1/y² + 1/z²) = 16/p² है । (10 अंक)

Answer approach & key points

Solve each sub-part systematically with equal time allocation (~20% per part) since all carry 10 marks each. For (a), set up and solve the linear system for span membership; for (b), construct the matrix of T and apply rank-nullity theorem; for (c), use Taylor series expansion or L'Hôpital's rule to find p and q; for (d), apply comparison tests for improper integrals; for (e), use the distance formula and centroid coordinates to derive the locus. Present solutions clearly with proper mathematical notation and concluding statements for each part.

  • Part (a): Set up the augmented matrix [V₁ V₂ V₃ | (3,-1,0,-1)] and determine consistency via row reduction; conclude whether the target vector lies in the span
  • Part (b): Construct the 3×3 matrix of T, compute its rank via determinant or row reduction, then apply rank-nullity theorem (rank + nullity = 3)
  • Part (c): Expand cos x and sin x using Taylor series up to x³ terms, equate coefficients to ensure limit exists and equals 1, solving for p = -5/2 and q = -3/2
  • Part (d): Identify the singularity at x=0, use limit comparison test with x^(-1/2) or direct evaluation showing |log x|/(1+x) is integrable, proving convergence
  • Part (e): Use the plane equation x/a + y/b + z/c = 1 with distance condition 1/√(1/a²+1/b²+1/c²) = 3p, find centroid (a/4, b/4, c/4), eliminate parameters to obtain 9(1/x²+1/y²+1/z²) = 16/p²
Q2
50M solve Linear algebra, multivariable calculus and 3D geometry

(a) If the matrix of a linear transformation T : IR³→IR³ relative to the basis {(1, 0, 0), (0, 1, 0), (0, 0, 1)} is $$\begin{bmatrix} 1 & 1 & 2 \\ -1 & 2 & 1 \\ 0 & 1 & 3 \end{bmatrix},$$ then find the matrix of T relative to the basis {(1, 1, 1), (0, 1, 1), (0, 0, 1)}. (15 marks) (b) Evaluate the triple integral which gives the volume of the solid enclosed between the two paraboloids Z = 5(x² + y²) and Z = 6 – 7x² – y². (15 marks) (c)(i) Show that the equation 2x² + 3y² – 8x + 6y – 12z + 11 = 0 represents an elliptic paraboloid. Also find its principal axis and principal planes. (10 marks) (c)(ii) The plane $\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1$ meets the coordinate axes in A, B, C respectively. Prove that the equation of the cone generated by the lines drawn from the origin O to meet the circle ABC is $$yz\left(\frac{b}{c}+\frac{c}{b}\right)+zx\left(\frac{c}{a}+\frac{a}{c}\right)+xy\left(\frac{b}{a}+\frac{a}{b}\right)=0.$$ (10 marks)

हिंदी में पढ़ें

(a) यदि आधार {(1, 0, 0), (0, 1, 0), (0, 0, 1)} के सापेक्ष रैखिक रूपांतरण T : IR³→IR³ का आव्यूह $$\begin{bmatrix} 1 & 1 & 2 \\ -1 & 2 & 1 \\ 0 & 1 & 3 \end{bmatrix}$$ है, तब आधार {(1, 1, 1), (0, 1, 1), (0, 0, 1)} के सापेक्ष T का आव्यूह ज्ञात कीजिए। (15 अंक) (b) दो परवलयजों Z = 5(x² + y²) और Z = 6 – 7x² – y² के बीच घिरे ठोस के आयतन को दर्शाने वाले त्रिशः समाकल का मान निकालिए। (15 अंक) (c)(i) दर्शाइए कि समीकरण 2x² + 3y² – 8x + 6y – 12z + 11 = 0 एक दीर्घवृत्तीय परवलयज प्रदर्शित करता है। साथ ही मुख्य अक्ष और मुख्य समतलों को भी ज्ञात कीजिए। (10 अंक) (c)(ii) समतल $\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1$, निर्देशांक अक्षों को क्रमशः A, B, C में मिलता है। सिद्ध कीजिए कि मूल बिंदु O से वृत्त ABC को मिलाने वाली रेखाओं द्वारा जनित शंकु का समीकरण $$yz\left(\frac{b}{c}+\frac{c}{b}\right)+zx\left(\frac{c}{a}+\frac{a}{c}\right)+xy\left(\frac{b}{a}+\frac{a}{b}\right)=0$$ है। (10 अंक)

Answer approach & key points

Solve all three parts systematically, allocating approximately 30% time to part (a) on change of basis matrices, 30% to part (b) on triple integration using cylindrical coordinates, and 40% to part (c) covering both the elliptic paraboloid identification and the cone equation derivation. Begin each part with clear statement of the method, show all computational steps with proper justification, and conclude with boxed final answers.

  • Part (a): Correct construction of change of basis matrix P using new basis vectors, computation of P⁻¹, and application of similarity transformation P⁻¹AP to find the new matrix representation
  • Part (b): Identification of intersection curve x² + y² = 0.5, correct setup of cylindrical coordinate bounds (r: 0 to 1/√2, θ: 0 to 2π, z: 5r² to 6-7r²), and accurate evaluation of the triple integral
  • Part (c)(i): Completion of squares to standard form, verification of elliptic paraboloid by identifying one linear term and two squared terms with same sign, identification of principal axis (z-axis direction) and principal planes
  • Part (c)(ii): Correct determination of coordinates A(a,0,0), B(0,b,0), C(0,0,c), equation of sphere through OABC, intersection with plane to get circle ABC, and homogenization to derive the cone equation
  • Proper matrix notation and determinant calculations for part (a) with verification of non-singularity
  • Clear geometric interpretation in part (b) showing the paraboloid intersection forms a bounded solid
  • Systematic algebraic manipulation in (c)(ii) showing the homogenization process with λ = 0 substitution
Q3
50M solve Linear algebra, multivariable calculus, 3D geometry

Let $A = \begin{bmatrix} 1 & 0 & 0 \\ 1 & 0 & 1 \\ 0 & 1 & 0 \end{bmatrix}$ (i) Verify the Cayley-Hamilton theorem for the matrix $A$. (ii) Show that $A^n = A^{n-2} + A^2 - I$ for $n \geq 3$, where $I$ is the identity matrix of order 3. Hence, find $A^{40}$. 10+10 (b) Justify whether $(0, 0)$ is an extreme point for the function $f(x, y) = 2x^4 - 3x^2y + y^2$. 15 (c) Find the equation of the sphere through the circle $x^2 + y^2 + z^2 - 4x - 6y + 2z - 16 = 0$; $3x + y + 3z - 4 = 0$ in the following two cases. (i) the point $(1, 0, -3)$ lies on the sphere. (ii) the given circle is a great circle of the sphere. 15

हिंदी में पढ़ें

दिया गया है $A=\begin{bmatrix} 1 & 0 & 0 \\ 1 & 0 & 1 \\ 0 & 1 & 0 \end{bmatrix}$ (i) आव्यूह A के लिये कैले-हैमिल्टन प्रमेय को सत्यापित कीजिए। (ii) दर्शाइए कि n ≥ 3 के लिये Aⁿ = Aⁿ⁻² + A² – I; जहाँ I कोटि 3 का तत्समक आव्यूह है। अतः A⁴⁰ ज्ञात कीजिए। 10+10 (b) तर्क सहित दर्शाइये कि $(0, 0)$, फलन $f(x, y) = 2x^4 - 3x^2y + y^2$ का चरम-बिन्दु है अथवा नहीं। 15 (c) वृत्त $x^2 + y^2 + z^2 - 4x - 6y + 2z - 16 = 0$; $3x + y + 3z - 4 = 0$ से होकर गुजरने वाले गोले का समीकरण निम्न दो स्थितियों में ज्ञात कीजिए। (i) बिन्दु $(1, 0, -3)$ गोले पर हो। (ii) दिया गया वृत्त गोले का एक बृहत् वृत्त हो। 15

Answer approach & key points

Solve this multi-part problem by allocating approximately 40% time to part (a) covering Cayley-Hamilton verification and recurrence relation (20 marks), 30% to part (b) on extreme point analysis using Hessian and higher-order tests (15 marks), and 30% to part (c) on sphere equations through given circle with two conditions (15 marks). Begin with clear statement of characteristic polynomial for (a), proceed to systematic matrix powers, apply second derivative test with discriminant analysis for (b), and use sphere family through circle intersection for (c).

  • For (a)(i): Correct computation of characteristic polynomial det(A - λI) = -λ³ + λ² + λ - 1 and verification that A³ = A² + A - I
  • For (a)(ii): Proof of recurrence Aⁿ = Aⁿ⁻² + A² - I using Cayley-Hamilton, and efficient computation of A⁴⁰ via pattern or binary exponentiation
  • For (b): Computation of first partials fx = 8x³ - 6xy, fy = -3x² + 2y, verification that (0,0) is critical point, and application of discriminant D = fxx·fyy - (fxy)² with higher-order analysis showing saddle point
  • For (c)(i): Formation of sphere family S: x²+y²+z²-4x-6y+2z-16 + λ(3x+y+3z-4) = 0 and substitution of (1,0,-3) to find λ
  • For (c)(ii): Condition that given circle is great circle requires sphere center to lie on plane 3x+y+3z-4=0, yielding center (-4+3λ)/2, (-6+λ)/2, (2+3λ)/2 satisfying plane equation
  • Explicit final answers: A⁴⁰ expression, conclusion that (0,0) is not extreme point (saddle), and two sphere equations with specific λ values
Q4
50M solve Matrix rank, curve tracing, 3D geometry

Find the rank of the matrix $A = \begin{bmatrix} 1 & 2 & -1 & 0 \\ -1 & 3 & 0 & -4 \\ 2 & 1 & 3 & -2 \\ 1 & 1 & 1 & -1 \end{bmatrix}$ by reducing it to row-reduced echelon form. 15 (b) Trace the curve $y^2(x^2 - 1) = 2x - 1$. 20 (c) Prove that the locus of a line which meets the lines y = mx, z = c; y = -mx, z = -c and the circle x² + y² = a², z = 0 is c²m²(cy - mzx)² + c²(yz - cmx)² = a²m²(z² - c²)². 15

हिंदी में पढ़ें

आव्यूह $A = \begin{bmatrix} 1 & 2 & -1 & 0 \\ -1 & 3 & 0 & -4 \\ 2 & 1 & 3 & -2 \\ 1 & 1 & 1 & -1 \end{bmatrix}$ का पंक्ति समानतित सोपानक रूप में समान्यन करके उसकी कोटि ज्ञात कीजिए। 15 (b) वक्र $y^2(x^2 - 1) = 2x - 1$ को अनुरेखित कीजिए। 20 (c) सिद्ध कीजिए कि रेखाओं y = mx, z = c; y = -mx, z = -c और वृत्त x² + y² = a², z = 0 से मिलने वाली रेखा का विद्यु-पथ c²m²(cy - mzx)² + c²(yz - cmx)² = a²m²(z² - c²)² है। 15

Answer approach & key points

Solve all three parts systematically, allocating approximately 30% time to part (a) matrix rank reduction, 40% to part (b) curve tracing as it carries highest marks, and 30% to part (c) 3D geometry proof. Begin with clear identification of each part, show complete row operations for (a), detailed curve analysis with asymptotes and intercepts for (b), and rigorous parametric derivation for (c).

  • Part (a): Correct reduction of 4×4 matrix to row-reduced echelon form using elementary row operations, identification of pivot positions, and accurate rank determination
  • Part (b): Complete curve tracing of y²(x²-1)=2x-1 including symmetry analysis, asymptotes (vertical at x=±1), intercepts, domain restrictions, and sketch of two branches
  • Part (c): Proper parameterization of lines meeting given skew lines and circle, elimination of parameters to derive the stated quartic locus surface
  • Verification of rank by checking linear dependence of rows/columns or using determinant test for part (a)
  • Analysis of singular points and behavior near asymptotes for part (b) curve

B

Q5
50M Compulsory solve Differential equations, Laplace transforms, mechanics, SHM, vector calculus

(a) Obtain the solution of the initial-value problem dy/dx - 2xy = 2, y(0) = 1 in the form y = eˣ²[1 + √π erf(x)]. (10 marks) (b) Given that L{f(t); p} = F(p). Show that ∫₀^∞ f(t)/t dt = ∫₀^∞ F(x)dx. Hence evaluate the integral ∫₀^∞ (e⁻ᵗ - e⁻³ᵗ)/t dt. (10 marks) (c) A cylinder of radius 'a' touches a vertical wall along a generating line. Axis of the cylinder is fixed horizontally. A uniform flat beam of length 'l' and weight 'W' rests with its extremities in contact with the wall and the cylinder, making an angle of 45° with the vertical. If frictional forces are neglected, then show that a/l = (√5 + 5)/(4√2). Also, find the reactions of the cylinder and wall. (10 marks) (d) A particle is moving under Simple Harmonic Motion of period T about a centre O. It passes through the point P with velocity v along the direction OP and OP = p. Find the time that elapses before the particle returns to the point P. What will be the value of p when the elapsed time is T/2 ? (10 marks) (e) If ā = sinθî + cosθĵ + θk̂, b̄ = cosθî - sinθĵ - 3k̂, c̄ = 2î + 3ĵ - 3k̂, then find the values of the derivative of the vector function ā × (b̄ × c̄) w.r.t. θ at θ = π/2 and θ = π. (10 marks)

हिंदी में पढ़ें

(a) प्रारंभिक-मान समस्या : dy/dx - 2xy = 2, y(0) = 1 का हल y = eˣ²[1 + √π erf(x)] के रूप में प्राप्त कीजिए। (10 अंक) (b) दिया गया है L{f(t); p} = F(p). दर्शाइए कि ∫₀^∞ f(t)/t dt = ∫₀^∞ F(x)dx. अतः समाकल ∫₀^∞ (e⁻ᵗ - e⁻³ᵗ)/t dt का मान ज्ञात कीजिए। (10 अंक) (c) अर्ध्व्यास 'a' का एक बेलन (सिलिंडर) एक जनक रेखा के अनुदिश एक उद्वाधर दीवार को स्पर्श किया हुआ है। बेलन का अक्ष क्षैतिजतः स्थिर है। लम्बाई 'l' तथा भार 'W' का एक एक समान समतल दंड उद्वाधर से 45° का कोण बनाते हुए अपने सिरों को दीवार के सहारे तथा बेलन पर टिकाए है। अगर घर्षण बल नगण्य हैं, तब दर्शाइए कि a/l = (√5 + 5)/(4√2). दीवार और बेलन की प्रतिक्रियाएँ भी ज्ञात कीजिए। (10 अंक) (d) कोई कण केन्द्र 'O' के सापेक्ष आवर्त काल T के साथ सरल आवर्त गति में गतिशील है। कण बिन्दु P से OP के अनुदिश दिशा में v वेग से गुजरता है तथा OP = p है। कण का बिन्दु P पर पुनः लौटने में लगा समय ज्ञात कीजिए। यदि लगा समय T/2 हो, तो p का मान क्या होगा ? (10 अंक) (e) यदि ā = sinθî + cosθĵ + θk̂, b̄ = cosθî - sinθĵ - 3k̂, c̄ = 2î + 3ĵ - 3k̂, तो सदिश फलन ā × (b̄ × c̄) के θ के सापेक्ष अवकलज के मान, θ = π/2 और θ = π पर ज्ञात कीजिए। (10 अंक)

Answer approach & key points

Solve all five sub-parts systematically, allocating approximately 20% time to each part given equal 10-mark weighting. For (a), apply integrating factor method for linear ODE; for (b), use Laplace transform properties with Fubini's theorem; for (c), draw free-body diagram and apply equilibrium conditions; for (d), use SHM equations with phase analysis; for (e), apply vector triple product identity before differentiation. Present solutions with clear intermediate steps and boxed final answers.

  • Part (a): Identify integrating factor e^(-x²), solve using standard formula, apply initial condition y(0)=1, and manipulate to express using error function erf(x)
  • Part (b): Prove the identity using Laplace definition and changing order of integration (Fubini/Tonelli), then apply to f(t)=e^(-t)-e^(-3t) with F(p)=1/(p+1)-1/(p+3) to get ln(3)
  • Part (c): Construct geometry with 45° beam, locate contact points, write three equilibrium equations (ΣFx=0, ΣFy=0, ΣM=0), eliminate reactions to find a/l ratio, then back-substitute for reaction magnitudes
  • Part (d): Use SHM equation x = Asin(ωt+φ), apply conditions at point P to find phase, determine return time using periodicity and symmetry, evaluate p when elapsed time is T/2
  • Part (e): Apply vector identity ā×(b̄×c̄) = (ā·c̄)b̄ - (ā·b̄)c̄, compute scalar products, differentiate resultant vector component-wise, evaluate at θ=π/2 and θ=π
Q6
50M solve Differential equations, projectile motion, surface integrals

(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 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)

Answer approach & key points

Solve this multi-part numerical problem by allocating approximately 30% time to part (a) on third-order linear ODE, 30% to part (b) on projectile motion derivation, and 40% to part (c) on surface integral evaluation and Gauss-Divergence theorem verification. Begin each part with clear problem identification, show complete working with intermediate steps, and conclude with boxed final answers. For part (c), explicitly compute both the surface integral directly and the volume integral via divergence theorem to demonstrate verification.

  • Part (a): Correctly find complementary function by solving characteristic equation m³ - 3m² + 4m - 2 = 0 with roots m = 1, 1±i, and construct particular integrals for both eˣ and cos x terms using appropriate methods
  • Part (a): Apply correct operator methods or undetermined coefficients to obtain PI for eˣ (resonance case) and PI for cos x, then combine for complete general solution
  • Part (b): Set up projectile equations from sea level O₁: derive range R₁ = v²sin(2θ)/g using standard trajectory equations y = xtanθ - gx²/(2v²cos²θ)
  • Part (b): Set up modified equations from height h at O₂, find new range by solving when projectile hits sea level (y = -h), derive R₂ = (v²cosθ/g)[sinθ + √(sin²θ + 2gh/v²)], prove R₂ > R₁ and establish required ratio
  • Part (c): Identify S as upper hemisphere 4x² + 4y² + 4z² = 1, z ≥ 0 with radius 1/2, correctly compute divergence ∇·F = 2yz² + 2z²x + 2zy² for verification
  • Part (c): Evaluate surface integral directly using spherical coordinates or projection method, then compute volume integral ∭(∇·F)dV over hemisphere, show equality to verify Gauss-Divergence theorem
Q7
50M solve Differential equations and mechanics

(a)(i) Find the solution of the differential equation : $\dfrac{dy}{dx}=-\dfrac{2xy^3+2}{3x^2y^2+8e^{4y}}$ 10 (a)(ii) Reduce the equation $x^2p^2+y(2x+y)p+y^2=0$ to Clairaut's form by the substitution $y=u$ and $xy=v$. Hence solve the equation and show that $y+4x=0$ is a singular solution of the differential equation. 10 (b) A solid hemisphere is supported by a string fixed to a point on its rim and to a point on a smooth vertical wall with which the curved surface is in contact. If $\theta$ is the angle of inclination of the string with vertical and $\phi$ is the angle of inclination of the plane base of the hemisphere to the vertical, then find the value of $(\tan\phi-\tan\theta)$. 15 (c) If the tangent to a curve makes a constant angle θ with a fixed line, then prove that the ratio of radius of torsion to radius of curvature is proportional to tanθ. Further prove that if this ratio is constant, then the tangent makes a constant angle with a fixed direction. 15

हिंदी में पढ़ें

(a)(i) अवकल समीकरण : $\dfrac{dy}{dx}=-\dfrac{2xy^3+2}{3x^2y^2+8e^{4y}}$ का हल ज्ञात कीजिए। 10 (a)(ii) समीकरण $x^2p^2+y(2x+y)p+y^2=0$ का प्रतिस्थापन $y=u$ और $xy=v$ द्वारा क्लेरो रूप में समान्यन कीजिए। अतः समीकरण का हल निकालिए और दर्शाइए कि $y+4x=0$ अवकल समीकरण का एक विचित्र हल है। 10 (b) एक ठोस अर्ध-गोलक एक डोरी द्वारा, जिसका एक सिरा एक चिकनी उच्चाधर दीवार पर एक बिंदु से और दूसरा सिरा अर्धगोलक के किनारे (रिम) पर स्थित एक बिंदु से बंधा है, उच्चाधर दीवार के सहारे टिका है। ठोस अर्धगोलक का वक्रित पृष्ठ दीवार को स्पर्श करता है। अगर उच्चाधर के साथ डोरी का आनति कोण $\theta$ है और अर्धगोलक के समतल आधार (बेस) का आनति कोण $\phi$ है तो $(\tan\phi-\tan\theta)$ का मान ज्ञात कीजिए। 15 (c) अगर एक वक्र की स्पर्श रेखा एक नियत रेखा के साथ एक स्थिर कोण θ बनाती है तो सिद्ध कीजिए कि वक्रता की त्रिज्या के साथ व्यवर्तन त्रिज्या का अनुपात tanθ के समानुपाती है। और आगे सिद्ध कीजिए कि अगर यह अनुपात एक स्थिरांक है, तो स्पर्श रेखा एक नियत दिशा के साथ एक स्थिर कोण बनाती है। 15

Answer approach & key points

Solve this multi-part problem by allocating approximately 40% time to part (a) covering both differential equations (20 marks), 30% to the mechanics problem (b) involving equilibrium of a hemisphere (15 marks), and 30% to the differential geometry proof in part (c) on curvature and torsion (15 marks). Begin with clear identification of the solution method for each sub-part, execute calculations systematically with proper justification, and conclude with verification of results including the singular solution in (a)(ii) and the constant angle property in (c).

  • For (a)(i): Recognize the equation as exact after rewriting in differential form Mdx + Ndy = 0, verify ∂M/∂y = ∂N/∂x, and find the potential function ψ(x,y) = x²y³ + 2x + 2e^(4y) = c
  • For (a)(ii): Apply substitution y = u, xy = v to transform to Clairaut's form v = pu + f(p) where p = dv/du, obtain general solution v = cu + c²/(c+1), and verify y + 4x = 0 as singular solution via c-discriminant or envelope method
  • For (b): Draw correct free-body diagram with three forces (weight W at center of mass, tension T along string, normal reaction R perpendicular to wall), apply equilibrium conditions ΣFx = 0, ΣFy = 0, ΣM = 0 about appropriate point, use geometry relating θ and φ through hemisphere radius a, and derive tanφ - tanθ = 1/2
  • For (c): Use Frenet-Serret formulas with fixed direction making angle θ with tangent, express d𝐭/ds = κ𝐧 and d𝐛/ds = -τ𝐧, show that 𝐭·𝐚 = cosθ implies 𝐧·𝐚 = 0 and 𝐛·𝐚 = sinθ, derive τ/κ = tanθ, and prove converse by showing constant τ/κ implies 𝐭·𝐚 is constant
  • Cross-cutting: Maintain dimensional consistency, state all assumptions explicitly (smooth wall, uniform hemisphere, inextensible string), and verify boundary conditions or special cases where applicable
Q8
50M solve Laplace transform and vector calculus

(a) Solve the following initial value problem by using Laplace transform technique : $$\frac{d^2y}{dt^2} - 4\frac{dy}{dt} + 3y(t) = f(t),$$ $y(0) = 1$, $y'(0) = 0$ and $f(t)$ is a given function of $t$. 15 (b) A particle is projected from an apse at a distance $\sqrt{c}$ from the centre of force with a velocity $\sqrt{\frac{2\lambda}{3}c^3}$ and is moving with central acceleration $\lambda(r^5 - c^2r)$. Find the path of motion of this particle. Will that be the curve $x^4 + y^4 = c^2$ ? 20 (c) For a scalar point function $\phi$ and vector point function $\vec{f}$, prove the identity $\nabla \cdot (\phi\vec{f}) = \nabla\phi \cdot \vec{f} + \phi(\nabla \cdot \vec{f})$. Also find the value of $\nabla \cdot \left(\frac{f(r)}{r}\vec{r}\right)$ and then verify stated identity. 15

हिंदी में पढ़ें

(a) लाप्लास रूपांतर प्रविधि का उपयोग कर निम्नलिखित प्रारंभिक मान समस्या को हल कीजिए। $$\frac{d^2y}{dt^2} - 4\frac{dy}{dt} + 3y(t) = f(t),$$ $y(0) = 1$, $y'(0) = 0$ और $f(t)$, $t$ का एक दिया गया फलन है। 15 (b) एक कण, बल-केंद्र से $\sqrt{c}$ दूरी पर स्थित एक स्तब्धिका से $\sqrt{\frac{2\lambda}{3}c^3}$ वेग से प्रक्षेपित किया जाता है और यह केंद्रीय त्वरण $\lambda(r^5 - c^2r)$ से गतिशील है। इस कण की गति का पथ ज्ञात कीजिए। क्या यह वक्र $x^4 + y^4 = c^2$ होगा ? 20 (c) एक अदिश बिंदु फलन $\phi$ और सदिश बिंदु फलन $\vec{f}$ के लिये निम्नलिखित सर्वसमिका सिद्ध कीजिए $$\nabla \cdot (\phi\vec{f}) = \nabla\phi \cdot \vec{f} + \phi(\nabla \cdot \vec{f})$$ $$\nabla \cdot \left(\frac{f(r)}{r}\vec{r}\right)$$ का मान भी ज्ञात कीजिए और तब उल्लेखित सर्वसमिका का सत्यापन कीजिए। 15

Answer approach & key points

Solve this three-part numerical problem by allocating approximately 30% time to part (a) Laplace transform IVP, 40% to part (b) central force motion and orbit determination, and 30% to part (c) vector calculus identity proof and verification. Begin with clear statement of given conditions, apply appropriate mathematical techniques systematically, show all computational steps explicitly, and conclude with precise final answers for each sub-part.

  • Part (a): Correct application of Laplace transform to second-order ODE with proper handling of initial conditions y(0)=1, y'(0)=0 and symbolic f(t)
  • Part (a): Accurate partial fraction decomposition and inverse Laplace transform to obtain complete solution y(t)
  • Part (b): Correct formulation of central force problem using Binet's equation or energy equation, identifying the given acceleration λ(r⁵ - c²r)
  • Part (b): Derivation of orbital equation and verification whether the path satisfies x⁴ + y⁴ = c² through polar to Cartesian conversion
  • Part (c): Rigorous proof of vector identity ∇·(φf⃗) = ∇φ·f⃗ + φ(∇·f⃗) using component-wise expansion or index notation
  • Part (c): Correct computation of ∇·(f(r)/r · r⃗) using spherical/polar coordinate formulas and explicit verification of the identity

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