Mathematics

UPSC Mathematics 2025 — Paper I

All 8 questions from UPSC Civil Services Mains Mathematics 2025 Paper I (400 marks total). Every stem reproduced in full, with directive-word analysis, marks, word limits, and answer-approach pointers.

8Questions
400Total marks
2025Year
Paper IPaper

Topics covered

Linear algebra, calculus and 3D geometry (1)Linear transformation, mean value theorem and 3D geometry (1)Linear algebra, analytical geometry, multivariable calculus (1)Analytical geometry, partial derivatives, linear algebra (1)Differential equations, ellipses, orbital mechanics, catenary, vector calculus (1)Laplace transforms, convolution, integral equations, elastic string, directional derivative, Maxwell's equations (1)Mechanics, vector calculus and differential equations (1)Differential equations, vector calculus and particle dynamics (1)

A

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

(a) Can the set {(0, 0, 0, 3), (1, 1, 0, 0), (0, 1, –1, 0)} be extended to form a basis of the vector space ℝ⁴? Justify your answer. 10 marks (b) Find the range, rank, kernel and nullity of the linear transformation T : ℝ⁴ → ℝ³ given by T(x, y, z, w) = (x – w, y + z, z – w). 10 marks (c) A rectangular sheet of metal of length 6 meters and width 2 meters is given. Four equal squares are removed from the four corners. The sides of this sheet are now folded up to form an open rectangular box. Find approximately the height of the box, such that the volume of the box is maximum. 10 marks (d) Given that f(x + y) = f(x) f(y) for all real x, y, f(x) ≠ 0 for any real x and f'(0) = 2. Show that for all real x, f'(x) = 2f(x). Hence find f(x). 10 marks (e) Find the equation of the cone whose vertex is the point (1, 1, 0) and whose guiding curve is y = 0, x² + z² = 4. 10 marks

हिंदी में पढ़ें

(a) क्या समुच्चय {(0, 0, 0, 3), (1, 1, 0, 0), (0, 1, –1, 0)} को सदिश समष्टि ℝ⁴ का एक आधार बनाने के लिए विस्तारित किया जा सकता है? अपने उत्तर की पुष्टि कीजिए। 10 अंक (b) रैखिक रूपांतरण T : ℝ⁴ → ℝ³, जो T(x, y, z, w) = (x – w, y + z, z – w) द्वारा दिया गया है, का परिसर (रेंज), कोटि (रैंक), अश्टि (कर्नेल) और शून्यता ज्ञात कीजिए। 10 अंक (c) लम्बाई 6 मीटर और चौड़ाई 2 मीटर की एक आयताकार धातु की चादर दी गई है। चारों कोनों से चार बराबर वर्गों को हटाया गया है। इस चादर के फलकों को मोड़कर एक खुला आयताकार संदूक बनाना है। संदूक की ऐसी सन्निकट ऊँचाई ज्ञात कीजिए कि संदूक का आयतन अधिकतम हो। 10 अंक (d) दिया गया है कि f(x + y) = f(x) f(y), सभी वास्तविक x, y के लिए, f(x) ≠ 0 किसी भी वास्तविक x के लिए और f'(0) = 2 है। सभी वास्तविक x के लिए दर्शाइए कि f'(x) = 2f(x) है। अतः f(x) ज्ञात कीजिए। 10 अंक (e) उस शंकु का समीकरण ज्ञात कीजिए जिसका शीर्ष बिंदु (1, 1, 0) है तथा जिसका निर्देशक वक्र y = 0, x² + z² = 4 है। 10 अंक

Answer approach & key points

Solve each sub-part systematically with clear mathematical reasoning. For (a), verify linear independence and extend to basis; for (b), construct the matrix representation and apply rank-nullity; for (c), set up the volume function and optimize using calculus; for (d), use the functional equation to derive the differential equation; for (e), use the cone generator method. Allocate approximately 20% time per sub-part given equal marks distribution, ensuring each part receives complete treatment with proper justification.

  • (a) Verify linear independence of the three vectors by checking no non-trivial combination equals zero; extend to basis by adding a fourth vector not in their span, such as (0,0,1,0) or (1,0,0,0)
  • (b) Construct the 3×4 matrix representation of T; find rank by row reduction or determinant of 3×3 minors; determine nullity using rank-nullity theorem (nullity = 4 - rank = 2)
  • (c) Express volume V = (6-2h)(2-2h)h = 4h(3-h)(1-h); find dV/dh = 0 yielding 3h² - 8h + 3 = 0; select valid root h = (4-√7)/3 ≈ 0.45m within domain 0 < h < 1
  • (d) Use f(x+y) = f(x)f(y) to show [f(x+h)-f(x)]/h = f(x)[f(h)-1]/h; as h→0, f'(x) = f(x)f'(0) = 2f(x); solve to get f(x) = e^(2x)
  • (e) For cone with vertex (1,1,0) and guiding curve y=0, x²+z²=4: parametrize generator lines (1+(x₀-1)t, 1-t, z₀t) where x₀²+z₀²=4; eliminate parameters to get (x-1)² + z² = 4(y-1)² with y ≤ 1
Q2
50M solve Linear transformation, mean value theorem and 3D geometry

(a) Let T : ℝ³ → ℝ² be a linear transformation such that T(1, 1, -1) = (1, 0), T(4, 1, 1) = (0, 1) and T(1, -1, 2) = (1, 1). Find T. 15 marks (b) Using Mean Value Theorem, prove that π/6 + √3/15 < sin⁻¹(3/5) < π/6 + 1/8 15 marks (c) (i) Find the equation of the cylinder whose generators are parallel to the line x/1 = y/2 = z/3 and that passes through the curve x² + y² = 16, z = 0. 10 marks (ii) Find the shortest distance between the straight lines (x-3)/3 = (y-8)/(-1) = (z-3)/1 and (x+3)/(-3) = (y+7)/2 = (z-6)/4. 10 marks

हिंदी में पढ़ें

(a) माना T : ℝ³ → ℝ² एक ऐसा रैखिक रूपांतरण है कि T(1, 1, -1) = (1, 0), T(4, 1, 1) = (0, 1) तथा T(1, -1, 2) = (1, 1) है। T ज्ञात कीजिए। 15 अंक (b) माध्यमान प्रमेय का प्रयोग करते हुए सिद्ध कीजिए कि π/6 + √3/15 < sin⁻¹(3/5) < π/6 + 1/8 15 अंक (c) (i) उस बेलन का समीकरण ज्ञात कीजिए जिसके जनक, रेखा x/1 = y/2 = z/3 के समांतर हैं और जो वक्र x² + y² = 16, z = 0 से होकर गुजरता है। 10 अंक (ii) सरल रेखाओं (x-3)/3 = (y-8)/(-1) = (z-3)/1 और (x+3)/(-3) = (y+7)/2 = (z-6)/4 के बीच की न्यूनतम दूरी ज्ञात कीजिए। 10 अंक

Answer approach & key points

Solve all four sub-parts systematically, allocating approximately 30% time to part (a) on linear transformation, 30% to part (b) on Mean Value Theorem proof, 20% to part (c)(i) on cylinder equation, and 20% to part (c)(ii) on shortest distance. Begin each part with clear statement of the approach, show complete working with proper mathematical notation, and conclude with boxed final answers.

  • For (a): Verify that {(1,1,-1), (4,1,1), (1,-1,2)} forms a basis for ℝ³, then express T as a 2×3 matrix by solving for images of standard basis vectors or using linearity directly
  • For (b): Apply MVT to f(x) = sin⁻¹x on [1/2, 3/5], showing f'(c) = 1/√(1-c²) lies between 4/5 and 5/√39, then manipulate to obtain the required bounds
  • For (c)(i): Use the condition that for any point (x,y,z) on cylinder, its distance from the axis line x/1=y/2=z/3 equals the radius 4, giving (2x-y)² + (3y-2z)² + (3x-z)² = 16×14 or equivalent simplified form
  • For (c)(ii): Verify the lines are skew, then apply SD formula |(a₂-a₁)·(b₁×b₂)|/|b₁×b₂| with correct identification of points and direction vectors, obtaining exact numerical value
  • Clear demonstration of linearity properties in (a), proper interval selection in (b), correct generator direction handling in (c)(i), and accurate cross product computation in (c)(ii)
Q3
50M solve Linear algebra, analytical geometry, multivariable calculus

(a) Reduce the following matrix to echelon form: $$A = \begin{bmatrix} 2 & -2 & 2 & 1 \\ -3 & 6 & 0 & -1 \\ 1 & -7 & 10 & 2 \end{bmatrix}$$ (15 marks) (b) Find the equations of the spheres which pass through the circle $x^2 + y^2 + z^2 - 2x + 2y + 4z - 3 = 0$, $2x + y + z = 4$ and touch the plane $3x + 4y = 14$. (15 marks) (c) (i) Evaluate $\displaystyle\iint\limits_R y\, dx\, dy$, where R is the region bounded by $y = x$ and $y = 4x - x^2$. (10 marks) (ii) If $u(x,y) = x\, f\left(\dfrac{y}{x}\right) + g\left(\dfrac{y}{x}\right)$, where f and g are arbitrary functions, then show that I. $\quad x\,\dfrac{\partial u}{\partial x} + y\,\dfrac{\partial u}{\partial y} = x\, f\left(\dfrac{y}{x}\right)$, II. $\quad x^2\,\dfrac{\partial^2 u}{\partial x^2} + 2xy\,\dfrac{\partial^2 u}{\partial x\,\partial y} + y^2\,\dfrac{\partial^2 u}{\partial y^2} = 0$. (10 marks)

हिंदी में पढ़ें

(a) निम्नलिखित आव्यूह को सोपानक (एशेलोन) रूप में समानीत कीजिए : $$A = \begin{bmatrix} 2 & -2 & 2 & 1 \\ -3 & 6 & 0 & -1 \\ 1 & -7 & 10 & 2 \end{bmatrix}$$ (15 अंक) (b) उन गोलों के समीकरण ज्ञात कीजिए जो वृत्त $x^2 + y^2 + z^2 - 2x + 2y + 4z - 3 = 0$, $2x + y + z = 4$ से होकर गुजरते हैं और समतल $3x + 4y = 14$ को स्पर्श करते हैं। (15 अंक) (c) (i) $\displaystyle\iint\limits_R y\, dx\, dy$ का मान ज्ञात कीजिए, जहाँ R, $y = x$ तथा $y = 4x - x^2$ से परिवृत क्षेत्र है। (10 अंक) (ii) यदि $u(x,y) = x\, f\left(\dfrac{y}{x}\right) + g\left(\dfrac{y}{x}\right)$ है, जहाँ f और g स्वेच्छ फलन हैं, तो दर्शाइए कि I. $\quad x\,\dfrac{\partial u}{\partial x} + y\,\dfrac{\partial u}{\partial y} = x\, f\left(\dfrac{y}{x}\right)$ है, II. $\quad x^2\,\dfrac{\partial^2 u}{\partial x^2} + 2xy\,\dfrac{\partial^2 u}{\partial x\,\partial y} + y^2\,\dfrac{\partial^2 u}{\partial y^2} = 0$ है। (10 अंक)

Answer approach & key points

Solve all four sub-parts systematically, allocating approximately 30% time to part (a) matrix reduction, 30% to part (b) sphere equations, 25% to part (c)(i) double integration, and 15% to part (c)(ii) partial derivative proofs. Begin each sub-part with clear identification of the mathematical technique, show complete working with row operations for (a), sphere family parameterization for (b), region sketching and limits for (c)(i), and chain rule applications for (c)(ii). Conclude with boxed final answers for each part.

  • For (a): Correct application of elementary row operations to achieve row echelon form with leading entries 1, 2, 0 in successive rows and proper identification of pivot positions
  • For (b): Formation of sphere family equation S + λP = 0, correct application of tangency condition using distance from center to plane equals radius, yielding two valid sphere equations
  • For (c)(i): Accurate sketch of region R bounded by line y=x and parabola y=4x-x², correct intersection points (0,0) and (3,3), proper order of integration with limits 0 to 3 for x and x to 4x-x² for y
  • For (c)(ii): Verification of Euler's homogeneous function theorem for part I using substitution v=y/x and chain rule, and confirmation of part II as second-degree Euler equation for homogeneous functions
  • Clear presentation of all row operation steps in (a) with explicit notation (R₂ → R₂ + ³⁄₂R₁ etc.) to enable partial credit tracing
Q4
50M prove Analytical geometry, partial derivatives, linear algebra

(a) Show that there is no tangent plane to the sphere x² + y² + z² - 4x + 2y - 4z + 4 = 0 that can be passed through the straight line (x+6)/2 = y + 3 = z + 1. (15 marks) (b) If f(x, y) = { xy(x²-y²)/(x²+y²), when (x,y) ≠ (0,0) { 0, when (x,y) = (0,0), then find f_xy(0,0) and f_yx(0,0). (15 marks) (c) (i) Find the eigenvalues and the corresponding eigenvectors of the matrix A = [1 2 0] [2 1 -6] [2 -2 3] (12 marks) (ii) Let P_n denote the vector space of all polynomials of degree ≤ n over R. Verify that dim(P_4/P_2) = dim P_4 - dim P_2. (8 marks)

हिंदी में पढ़ें

(a) दर्शाइए कि गोले x² + y² + z² - 4x + 2y - 4z + 4 = 0 का कोई ऐसा स्पर्श समतल नहीं है, जो कि सरल रेखा (x+6)/2 = y + 3 = z + 1 से होकर गुजर सके। (15 अंक) (b) यदि f(x, y) = { xy(x²-y²)/(x²+y²), जब (x,y) ≠ (0,0) { 0, जब (x,y) = (0,0) है, तो f_xy(0,0) और f_yx(0,0) ज्ञात कीजिए। (15 अंक) (c) (i) आव्यूह A = [1 2 0] [2 1 -6] [2 -2 3] के अभिलक्षणिक मान और संगत अभिलक्षणिक सदिश ज्ञात कीजिए। (12 अंक) (ii) माना P_n, R पर घात ≤ n के सभी बहुपदों के सदिश समष्टि को दर्शाता है। सत्यापित कीजिए कि dim(P_4/P_2) = dim P_4 - dim P_2। (8 अंक)

Answer approach & key points

Prove the geometric impossibility in (a), calculate mixed partial derivatives in (b), and solve the eigenvalue problem with quotient space verification in (c). Allocate approximately 30% time to part (a) as it requires conceptual rigour, 30% to part (b) for careful limit analysis, 25% to (c)(i) for eigenvalue computation, and 15% to (c)(ii) for the dimension theorem verification. Begin with sphere-line analysis, proceed to partial derivative limits, then matrix characteristic polynomial, and conclude with basis construction for the quotient space.

  • For (a): Complete the square to find sphere centre (2,-1,2) and radius 3; verify the given line does not intersect the sphere or lies entirely outside; show the distance from centre to line exceeds radius, making tangent plane impossible
  • For (b): Compute f_x(0,y) using limit definition, then f_xy(0,0); compute f_y(x,0) then f_yx(0,0); demonstrate f_xy(0,0) = -1 and f_yx(0,0) = 1, showing inequality of mixed partials
  • For (c)(i): Find characteristic equation det(A-λI)=0 yielding eigenvalues λ=3,3,-2; find eigenvectors for λ=3 (geometric multiplicity 1) and λ=-2; handle the defective case for repeated eigenvalue appropriately
  • For (c)(ii): Identify basis {1,x,x²,x³,x⁴} for P₄ and {1,x,x²} for P₂; construct coset basis {x³+P₂, x⁴+P₂} for P₄/P₂; verify dim(P₄/P₂)=2=5-3=dim P₄-dim P₂
  • Cross-part rigour: Use ε-δ arguments or explicit limit calculations in (b); justify why tangent plane condition fails via distance formula or system inconsistency in (a)

B

Q5
50M Compulsory solve Differential equations, ellipses, orbital mechanics, catenary, vector calculus

(a) Solve $\left(1-y^{2}+\frac{y^{4}}{x^{2}}\right)\left(\frac{dy}{dx}\right)^{2}-2\frac{y}{x}\frac{dy}{dx}+\frac{y^{2}}{x^{2}}=0$. 10 marks (b) Form the differential equation of all ellipses whose axes coincide with coordinate axes. 10 marks (c) Prove that the time taken by the Earth to travel over half of its orbit, which is separated by the minor axis and is remote from the Sun, when the Sun is at the focus of the elliptic orbit, is two days more than half of the year. The eccentricity of the orbit is taken as $\frac{1}{60}$. 10 marks (d) Given that A and B are two points in the same horizontal line distant 2a apart. AO and BO are two equal heavy strings tied together at O and carrying their weight at O. If $l$ is length of each string and $d$ is depth of O below AB, then show that the parameter $c$ of this catenary, in which the strings hang, is given by $$l^{2}-d^{2}=2c^{2}\left[\cosh\left(\frac{a}{c}\right)-1\right].$$ 10 marks (e) If $u=x+y+z$, $v=x^{2}+y^{2}+z^{2}$ and $w=xy+yz+zx$, then show that grad u, grad v and grad w are coplanar. 10 marks

हिंदी में पढ़ें

(a) $\left(1-y^{2}+\frac{y^{4}}{x^{2}}\right)\left(\frac{dy}{dx}\right)^{2}-2\frac{y}{x}\frac{dy}{dx}+\frac{y^{2}}{x^{2}}=0$ को हल कीजिए । 10 अंक (b) सभी दीर्घवृत्तों, जिनके अक्ष निर्देशांक अक्षों के संपाती हैं, का अवकल समीकरण बनाइए । 10 अंक (c) सिद्ध कीजिए कि पृथ्वी को अपनी कक्षा के आधे भाग, जो कि लघु अक्ष द्वारा अलग किया गया है और सूर्य से सुदूर है, जब सूर्य दीर्घवृत्तीय कक्षा की नाभि (फोकस) पर है, की यात्रा करने में लगने वाला समय आधे वर्ष से दो दिन अधिक है। कक्षा की उत्केन्द्रता $\frac{1}{60}$ ली गई है। 10 अंक (d) दिया गया है कि A और B एक ही क्षैतिज रेखा पर स्थित दो बिंदु हैं, जिनके बीच की दूरी 2a है। AO और BO दो समान भारी डोरी हैं जो O पर एक साथ बंधी हैं और जिनका भार O पर है। यदि प्रत्येक डोरी की लंबाई $l$ है तथा $d$, AB से नीचे O की गहराई है, तो दर्शाइए कि इस कैटनरी, जिसमें डोरी लटकी है, का प्राचल $c$ $$l^{2}-d^{2}=2c^{2}\left[\cosh\left(\frac{a}{c}\right)-1\right]$$ द्वारा दिया गया है। 10 अंक (e) यदि $u=x+y+z$, $v=x^{2}+y^{2}+z^{2}$ और $w=xy+yz+zx$ है, तो दर्शाइए कि grad u, grad v और grad w समतलीय हैं। 10 अंक

Answer approach & key points

Solve each sub-part systematically, allocating approximately equal time (~20%) to each 10-mark section. For (a), identify the substitution v = y/x to reduce to Clairaut's form; for (b), eliminate parameters from the standard ellipse equation; for (c), apply Kepler's second law with area integration; for (d), use catenary boundary conditions at symmetric points; for (e), compute gradients and verify scalar triple product vanishes. Present solutions with clear headings for each part.

  • Part (a): Recognize homogeneous structure, substitute v = y/x, transform to Clairaut's equation v = px + f(p), obtain complete primitive and singular solution
  • Part (b): Start with ellipse equation x²/a² + y²/b² = 1, eliminate two arbitrary constants a and b to get second-order differential equation xy(d²y/dx²) + x(dy/dx)² - y(dy/dx) = 0
  • Part (c): Apply Kepler's second law (equal areas in equal times), compute sector area from minor axis to aphelion using polar ellipse equation r = a(1-e²)/(1+e cos θ), integrate and compare with half-year
  • Part (d): Set up symmetric catenary y = c cosh(x/c) with origin at lowest point O, apply boundary conditions at x = ±a with y = c + d, eliminate parameter to derive required relation
  • Part (e): Compute ∇u, ∇v, ∇w explicitly, form scalar triple product [∇u ∇v ∇w] or show linear dependence via ∇v = x∇u + z∇w-type relation, verify coplanarity condition
Q6
50M prove Laplace transforms, convolution, integral equations, elastic string, directional derivative, Maxwell's equations

(a) If F(s) and G(s) are Laplace transforms of f(t) and g(t) respectively, then prove that $$\mathcal{L}\left\{\int_{0}^{t} f(x) g(t-x) dx\right\} = F(s) G(s).$$ Using this result, solve the equation $$y(t) = t + \int_{0}^{t} y(x) \sin(t-x) dx.$$ 15 marks (b) One end of an elastic string, having natural length a, is fixed at some point O and a heavy particle is attached to the other end of the string. The string is drawn vertically downward till it is four times its natural length at the point C and then released. If the modulus of elasticity of the string is equal to the weight of the particle, then show that the particle will return to the same point C in the time $$\sqrt{\frac{a}{g}}\left(2\sqrt{3} + \frac{4\pi}{3}\right).$$ 15 marks (c) (i) Find the absolute value of the directional derivative of $\phi(x, y, z) = x^2y^2z^2$ at the point $(1, 1, -1)$ in the direction of the tangent to the curve $x = e^t$, $y = 2\sin t + 1$, $z = t - \cos t$, at $t = 0$. 10 marks (ii) If $\nabla \cdot \overrightarrow{\mathrm{E}}=0$, $\nabla \cdot \overrightarrow{\mathrm{H}}=0$, $\nabla \times \overrightarrow{\mathrm{E}}=-\frac{\partial \overrightarrow{\mathrm{H}}}{\partial t}$ and $\nabla \times \overrightarrow{\mathrm{H}}=\frac{\partial \overrightarrow{\mathrm{E}}}{\partial t}$, then show that $\nabla^{2} \overrightarrow{\mathrm{H}}=\frac{\partial^{2} \overrightarrow{\mathrm{H}}}{\partial t^{2}}$ and $\nabla^{2} \overrightarrow{\mathrm{E}}=\frac{\partial^{2} \overrightarrow{\mathrm{E}}}{\partial t^{2}}$. 10 marks

हिंदी में पढ़ें

(a) यदि f(t) और g(t) के लाप्लास रूपान्तर क्रमशः F(s) और G(s) हैं, तो सिद्ध कीजिए कि $$\mathcal{L}\left\{\int_{0}^{t} f(x) g(t-x) dx\right\} = F(s) G(s)$$ है। इस परिणाम का प्रयोग करते हुए, समीकरण $$y(t) = t + \int_{0}^{t} y(x) \sin(t-x) dx$$ को हल कीजिए। 15 अंक (b) एक प्रत्यास्थ डोरी, जिसकी प्राकृतिक लंबाई a है, का एक छोर किसि बिंदु O पर स्थिर है और डोरी के दूसरे छोर पर एक भारी कण जुड़ा हुआ है। डोरी को उर्ध्वाधर नीचे की ओर बिंदु C तक तब तक खींचा जाता है जब तक वह अपनी प्राकृतिक लंबाई से चार गुना न हो जाए तथा फिर छोड़ दिया जाता है। यदि डोरी का प्रत्यास्थता गुणांक कण के भार के बराबर है, तो दर्शाइए कि कण $$\sqrt{\frac{a}{g}}\left(2\sqrt{3} + \frac{4\pi}{3}\right)$$ समय में उसी बिंदु C पर वापस आ जाएगा। 15 अंक (c) (i) $\phi(x, y, z) = x^2y^2z^2$ का बिंदु $(1, 1, -1)$ पर, वक्र $x = e^t$, $y = 2\sin t + 1$, $z = t - \cos t$, के बिंदु $t = 0$ पर स्पर्श-रेखा की दिशा में दिक्-अवकलज का निरपेक्ष मान ज्ञात कीजिए। 10 अंक (ii) यदि $\nabla \cdot \overrightarrow{\mathrm{E}}=0$, $\nabla \cdot \overrightarrow{\mathrm{H}}=0$, $\nabla \times \overrightarrow{\mathrm{E}}=-\frac{\partial \overrightarrow{\mathrm{H}}}{\partial t}$ और $\nabla \times \overrightarrow{\mathrm{H}}=\frac{\partial \overrightarrow{\mathrm{E}}}{\partial t}$ है, तो दर्शाइए कि $\nabla^{2} \overrightarrow{\mathrm{H}}=\frac{\partial^{2} \overrightarrow{\mathrm{H}}}{\partial t^{2}}$ और $\nabla^{2} \overrightarrow{\mathrm{E}}=\frac{\partial^{2} \overrightarrow{\mathrm{E}}}{\partial t^{2}}$ है। 10 अंक

Answer approach & key points

Begin with the proof of the convolution theorem in part (a) using Fubini's theorem and the definition of Laplace transform, then apply it to solve the integral equation via algebraic manipulation in s-domain and partial fractions. For part (b), establish the equation of motion in two phases (stretched string SHM and free fall), carefully handling the transition conditions at natural length. Part (c)(i) requires computing the gradient and unit tangent vector before taking the dot product, while (c)(ii) demands vector calculus identities to derive the wave equations. Allocate approximately 30% time to (a), 35% to (b), 20% to (c)(i), and 15% to (c)(ii) based on mark distribution and computational complexity.

  • Part (a): Correct proof of convolution theorem using change of order of integration and identification of Laplace kernel; proper setup of subsidiary equation ȳ(s) = 1/s² + ȳ(s)/(s²+1) and its solution yielding y(t) = t + t³/6
  • Part (b): Correct derivation of SHM equation ẍ = -g(x-a)/a for x > a with solution x = a + 3a cos(√(g/a)t); determination of time to reach natural length t₁ = √(a/g)·π/3 and velocity v₁ = 3√(ag)/2
  • Part (b) continued: Analysis of free motion phase with initial conditions, time to reach maximum height and return, and total time calculation showing the required expression √(a/g)(2√3 + 4π/3)
  • Part (c)(i): Computation of ∇φ = (2xy²z², 2x²yz², 2x²y²z) = (2, 2, -2) at (1,1,-1); tangent vector (1, 2, 1) at t=0; unit vector and directional derivative magnitude |8/√6| = 4√(2/3) or equivalent simplified form
  • Part (c)(ii): Application of curl to Faraday's and Ampère's equations, use of vector identity ∇×(∇×H) = ∇(∇·H) - ∇²H with divergence-free condition to obtain wave equations for both E and H fields
Q7
50M prove Mechanics, vector calculus and differential equations

(a) A solid sphere rests inside a fixed rough and hemispherical bowl of twice its radius. If a large amount of weight, whatsoever, is attached to the highest point of the sphere, then show that the equilibrium is stable. (15 marks) (b) Verify Green's theorem in the plane for $\oint\limits_{\mathrm{C}}\left[\left(x y+y^{2}\right) d x+x^{2} d y\right]$, where C is the boundary of the region bounded by the curves $y=x$ and $y=x^{2}$. (15 marks) (c) (i) Find the general solution and singular solution of the differential equation $\left(1+\frac{d y}{d x}\right)^{3}=\frac{27}{8 a}(x+y)\left(1-\frac{d y}{d x}\right)^{3}$. (10 marks) (ii) Find the complete solution of $x^{3} \frac{d^{3} y}{d x^{3}}+3 x^{2} \frac{d^{2} y}{d x^{2}}+x \frac{d y}{d x}+y=x \log x$. (10 marks)

हिंदी में पढ़ें

(a) एक ठोस गोला अपनी त्रिज्या से दुगुनी त्रिज्या के स्थिर रूक्ष अर्धगोलीय कटोरे में रखा हुआ है। यदि एक बड़ा भार, कितना भी हो, गोले के सबसे ऊँचे बिंदु पर जुड़ा है, तो दर्शाइए कि संतुलन स्थिर है। (15 अंक) (b) $\oint\limits_{\mathrm{C}}\left[\left(x y+y^{2}\right) d x+x^{2} d y\right]$, जहाँ C, वक्रों $y=x$ और $y=x^{2}$ द्वारा परिबद्ध क्षेत्र की परिसीमा है, के लिए समतल में ग्रीन का प्रमेय सत्यापित कीजिए। (15 अंक) (c) (i) अवकल समीकरण $\left(1+\frac{d y}{d x}\right)^{3}=\frac{27}{8 a}(x+y)\left(1-\frac{d y}{d x}\right)^{3}$ के व्यापक हल और विचित्र हल ज्ञात कीजिए। (10 अंक) (ii) $x^{3} \frac{d^{3} y}{d x^{3}}+3 x^{2} \frac{d^{2} y}{d x^{2}}+x \frac{d y}{d x}+y=x \log x$ का पूर्ण हल ज्ञात कीजिए। (10 अंक)

Answer approach & key points

Prove the stability condition in part (a) by analyzing the potential energy and center of mass displacement; verify Green's theorem in part (b) by computing both line integral and double integral separately; solve the Clairaut-type equation in (c)(i) for general and singular solutions; and apply the substitution x = e^t to reduce (c)(ii) to a linear equation with constant coefficients. Allocate approximately 30% time to (a), 25% to (b), 25% to (c)(i), and 20% to (c)(ii) based on mark distribution.

  • Part (a): Correct geometric setup with sphere radius r and bowl radius 2r, identification of contact point and angle θ, calculation of new center of mass position after adding weight W, and proof that potential energy minimum exists for any W
  • Part (b): Proper identification of region bounded by y=x and y=x² with intersection points (0,0) and (1,1), correct application of Green's theorem with P=xy+y² and Q=x², accurate computation of both ∮(Pdx+Qdy) and ∬(∂Q/∂x-∂P/∂y)dA
  • Part (c)(i): Substitution u=x+y to transform equation, recognition of Clairaut's equation form, derivation of general solution (x+y-c)³(1+8a/27c³)=0 and singular solution envelope
  • Part (c)(ii): Substitution x=e^t to convert to Cauchy-Euler form, reduction to linear ODE with constant coefficients, finding complementary function and particular integral for RHS e^t·t
  • Correct handling of rough constraint in (a) ensuring no slipping condition is satisfied
Q8
50M solve Differential equations, vector calculus and particle dynamics

(a) Solve the differential equation $(x + 2)\frac{d^2y}{dx^2} - (2x + 5)\frac{dy}{dx} + 2y = (1 + x) e^x$ by the method of variation of parameters. (15 marks) (b) Verify Gauss's divergence theorem for $\vec{F} = [(x^2 - yz)\hat{i} + (y^2 - zx)\hat{j} + (z^2 - xy)\hat{k}]$, taken over the rectangular parallelopiped $0 \leq x \leq a$, $0 \leq y \leq b$, $0 \leq z \leq c$. (15 marks) (c) A particle is projected inside a fixed smooth cylinder with circular cross-section in a vertical plane from the lowest point with initial horizontal velocity u. Show that for (i) $(u^2 \leq 2ag)$; the particle oscillates about the mean position in the lower half, (ii) $(u^2 \geq 5ag)$; the particle executes complete circular motion, and (iii) $(2ag < u^2 < 5ag)$; the particle will leave the curve in a tangential direction, making an angle $\alpha$ with the horizontal such that $\cos \alpha = \frac{u^2 - 2ag}{3ag}$. (20 marks)

हिंदी में पढ़ें

(a) अवकल समीकरण $(x + 2)\frac{d^2y}{dx^2} - (2x + 5)\frac{dy}{dx} + 2y = (1 + x) e^x$ को प्राचल विचरण विधि द्वारा हल कीजिए। (15 अंक) (b) समकोणिक समांतरपटलक $0 \leq x \leq a$, $0 \leq y \leq b$, $0 \leq z \leq c$ पर $\vec{F} = [(x^2 - yz)\hat{i} + (y^2 - zx)\hat{j} + (z^2 - xy)\hat{k}]$ के लिए गॉस अपसरण प्रमेय सत्यापित कीजिए। (15 अंक) (c) एक कण को उच्चाधर तल में वृत्ताकर अनुप्रस्थ-परिच्छेद वाले स्थिर चिकने बेलन के अंदर प्रारंभिक क्षैतिज वेग u के साथ सबसे निचले बिंदु से प्रक्षेपित किया जाता है। दर्शाइए कि (i) $(u^2 \leq 2ag)$ के लिए; कण निचले आधे भाग में माध्य स्थिति के आसपास (about) दोलन करता है, (ii) $(u^2 \geq 5ag)$ के लिए; कण पूर्णतः वृत्तीय गति करता है, और (iii) $(2ag < u^2 < 5ag)$ के लिए; कण, वक्र को एक स्पर्श की दिशा में, जो क्षैतिज के साथ कोण $\alpha$ बनाती है, छोड़ देगा, जबकि $\cos \alpha = \frac{u^2 - 2ag}{3ag}$ है। (20 अंक)

Answer approach & key points

Solve all three parts systematically, allocating approximately 30% time to part (a) on variation of parameters, 30% to part (b) on divergence theorem verification, and 40% to part (c) on particle dynamics which carries the highest marks. Begin each part with clear identification of the method/theorem being applied, show complete working with intermediate steps, and conclude with explicit verification or derived conditions. For part (c), clearly distinguish the three energy regimes with proper energy conservation and normal reaction analysis.

  • Part (a): Reduce to standard form, find complementary function by solving characteristic equation, apply variation of parameters correctly with Wronskian calculation, obtain particular integral and general solution
  • Part (b): Compute divergence of F correctly as 2(x+y+z), evaluate volume integral over rectangular parallelopiped, calculate surface integral over all six faces with proper orientation, show equality of both integrals
  • Part (c): Apply energy conservation between lowest point and arbitrary position, derive expression for normal reaction R in terms of angle and velocity, analyze R=0 condition for leaving the curve, establish critical velocity thresholds at u²=2ag and u²=5ag
  • Part (c)(i): Show particle cannot reach horizontal diameter when u²≤2ag, prove oscillatory motion in lower half with amplitude determined by initial energy
  • Part (c)(ii): Demonstrate complete circular motion requires R≥0 throughout, show u²≥5ag ensures positive normal reaction at topmost point
  • Part (c)(iii): Derive leaving condition R=0, obtain cos α = (u²-2ag)/(3ag) by simultaneous solution of energy and force equations

Practice Mathematics 2025 Paper I answer writing

Pick any question above, write your answer, and get a detailed AI evaluation against UPSC's standard rubric.

Start free evaluation →