UPSC Mathematics 2026 — Paper II

(a) Prove a dichotomy for a group G of order pq (p, q primes): either G is abelian, or the centre Z(G) is trivial (no non-identity element commutes with every element of G). 'Show that' demands a complete proof using Lagrange's theorem and the G/Z(G) cyclic lemma. (b) Prove that in a Euclidean domain every non-zero prime ideal is maximal. Requires establishing (or quoting with justification) that a Euclidean domain is a PID and then arguing that a non-zero prime ideal (p) admits no proper intermediate ideal. (c) Evaluate the pointwise limit defining f on (0, 1), identify f explicitly as a sign function, and then list precisely the points of (0, 1) at which f is discontinuous. (d) For f(z) = sqrt(|xy|) (u = sqrt(|xy|), v = 0), verify that the Cauchy-Riemann equations hold at z = 0 by computing the four partial derivatives from first principles, and then show f is not differentiable (hence not analytic) at 0 by a path-dependent difference quotient. (e) Form the dual of the given 4-variable minimisation LPP (which has only 2 constraints, so its dual has only 2 variables), and solve that dual graphically; report the optimal point and optimal value, and (by duality) the optimal value of the given primal.

• |Z(G)| divides pq by Lagrange, so |Z(G)| ∈ {1, p, q, pq}
• Proof of the lemma: G/Z(G) cyclic ⇒ G abelian
• |Z(G)| = p or q gives a prime-index (cyclic) quotient, hence G abelian, a contradiction; so G is abelian or Z(G) = {e}
• Euclidean domain ⇒ PID (element of least d(a) generates the ideal, via the division algorithm)
• P = (p) ⊆ (a) ⇒ p = ab ⇒ a ∈ P or b ∈ P
• a ∈ P gives (a) = P; b = pc gives ac = 1 by cancellation (p ≠ 0, R a domain), so (a) = R; hence P is maximal
• Limits: t > 1 ⇒ 1, t = 1 ⇒ 0, 0 ≤ t < 1 ⇒ −1, so f(x) = sgn(sin(π/x))
• Discontinuities exactly at x = 1/n, n ≥ 2 (jump discontinuities; x = 1 excluded)
• Continuity at all other points (f locally constant)
• u = √|xy|, v = 0; first-principles partials at (0, 0) all zero, so the C–R equations hold
• Along y = mx the difference quotient tends to √|m|/(1 + im), which depends on m
• Hence f′(0) does not exist and f is not analytic at 0 (C–R necessary, not sufficient)
• Dual: maximise Z = 2x₁ + x₂ subject to x₁ + 2x₂ ≤ 10, x₁ + x₂ ≤ 6, x₁ − x₂ ≤ 2, x₁ − 2x₂ ≤ 1, x₁, x₂ ≥ 0
• Graph with corners (0, 0), (1, 0), (3, 1), (4, 2), (2, 4), (0, 5); optimum at (4, 2) with max Z = 10
• By duality min Y = 10

(a) Given a finite group G = <u, v> with u^2 = v^2 = e, prove G is isomorphic to the dihedral group D_{2n} of order 2n where n = o(uv). Requires producing the rotation element r = uv, the relation uru^{-1} = r^{-1}, the coset decomposition, the order count and an explicit isomorphism. (b) For u_{n+1} = sqrt(7 + u_n), u_1 = sqrt(7), prove convergence (monotone increasing and bounded above) and identify the limit as the positive root of x^2 - x - 7 = 0, i.e. (1 + sqrt(29))/2. (c) Prove, by contour integration on the unit circle, that the integral of sin^2(theta)/(a + b cos theta) over [0, 2pi] equals (2pi/b^2){a - sqrt(a^2 - b^2)} for a > b > 0. The method (contour integration) is prescribed, so a real-variable evaluation does not answer the demand.

• r = uv with o(r) = n, v = ur so G = ⟨r, u⟩, and the derived relation uru⁻¹ = r⁻¹
• Every element is rᵏ or rᵏu, and u ∉ ⟨r⟩, so |G| = 2n
• φ: a ↦ r, b ↦ u preserves aⁿ = b² = e, bab⁻¹ = a⁻¹ and is a bijective homomorphism, so G ≅ D₂ₙ
• Boundedness 0 < uₙ < L = (1 + √29)/2 by induction, and monotonicity from uₙ₊₁² − uₙ² = −(uₙ − L)(uₙ − L′) > 0
• Existence of the limit by the monotone convergence theorem before computing it
• l = √(7 + l) ⇒ l² − l − 7 = 0 ⇒ l = (1 + √29)/2 ≈ 3.193, negative root rejected
• z = e^{iθ}: I = (i/2)∮ (z² − 1)²/[z²(bz² + 2az + b)] dz over |z| = 1
• Poles inside: z = 0 (double, residue −2a/b²) and α = (−a + √(a² − b²))/b with |α| < 1 (simple, residue 2√(a² − b²)/b²)
• Residue theorem gives I = (2π/b²){a − √(a² − b²)}

(a) Prove that cosh(z + 1/z) has the stated symmetric Laurent expansion a_0 + sum a_n(z^n + z^{-n}), and that the coefficients are given by the real integral a_n = (1/2pi) times the integral over [0, 2pi] of cos(n theta) cosh(2 cos theta) d theta. (b) For f on [0,1] equal to x^2 at rationals and x^3 at irrationals, compute the lower Darboux integral, compute the upper Darboux integral, and decide (with reason) whether the Riemann integral exists. Three distinct answers are required. (c) Solve a 3-job x 4-machine unbalanced assignment problem (add a dummy job) by the Hungarian method to minimise total cost; then decide whether the optimal assignment is unique, justify the answer, and exhibit the alternate optimal assignment.

• f analytic in 0 < |z| < ∞ (essential singularity at 0), so a Laurent series with cₙ = (1/2πi)∮_{|z|=1} f(z) z^{−(n+1)} dz
• f(1/z) = f(z) ⇒ c₋ₙ = cₙ, giving the grouping a₀ + Σ aₙ(zⁿ + z⁻ⁿ)
• With z = e^{iθ}, z + 1/z = 2 cos θ and the sin nθ part vanishes, so aₙ = (1/2π)∫₀^{2π} cos nθ cosh(2 cos θ) dθ
• On every subinterval sup f = xᵢ² and inf f = xᵢ₋₁³, using density of rationals and irrationals and x³ ≤ x² on [0, 1]
• Upper integral = ∫₀¹ x² dx = 1/3; lower integral = ∫₀¹ x³ dx = 1/4
• 1/3 ≠ 1/4, so the Riemann integral does not exist
• Balance with a dummy job (zero row) and apply the Hungarian method: reduction, line covering, and the adjustment iterations until 4 lines are needed
• Optimal assignment J₁ → M₁, J₂ → M₂, J₃ → M₃, M₄ idle; minimum cost 40
• Not unique: alternate optimum J₁ → M₁, J₂ → M₃, J₃ → M₂, M₄ idle, also 40, justified by the alternative independent zeros

(a) Prove the equivalence of three statements about Z/nZ for a positive integer n: (i) it is an integral domain, (ii) it is a field, (iii) n is prime. A cyclic chain of implications (iii) => (ii) => (i) => (iii) is the expected structure. (b) Prove that the improper integral from 0 to infinity of dx/(1 + x^2 sin^2 x) diverges. The expected route is to bound the integral over each period [n pi, (n+1) pi] from below and compare with the harmonic series. (c) Solve the given 2-variable, 3-constraint maximisation LPP by the Simplex method (tableaux required, not a graph), state the optimal solution and value, and then obtain and explain the alternate optimal solution.

• (iii) ⇒ (ii): for p prime and a ≢ 0, gcd(a, p) = 1, Bézout au + pv = 1 gives [a][u] = [1]
• (ii) ⇒ (i): a field has no zero divisors
• (i) ⇒ (iii): n = ab with 1 < a, b < n gives [a][b] = [0], a zero divisor; n = 1 gives the zero ring
• Split into periods [nπ, (n+1)π] and use x ≤ (n+1)π to get Iₙ ≥ ∫₀^π dt/(1 + (n+1)²π² sin²t)
• ∫₀^π dt/(1 + k² sin²t) = π/√(1 + k²), giving Iₙ ≥ 1/(n+2)
• Positive integrand and divergent Σ 1/(n+2), so the integral diverges
• Standard form with slacks; x₂ enters, s₂ leaves (ratio 4); tableau gives x₁ = 0, x₂ = 4, Z = 40 with Z-row (0, 0, 0, 2, 0)
• Zero reduced cost of non-basic x₁ signals alternate optima; pivot x₁ in (s₁ leaves, ratio 3.75)
• Alternate optimum x₁ = 15/4, x₂ = 5/2, Z = 40; the whole segment between (0, 4) and (15/4, 5/2) is optimal because Z = 2(2x₁ + 5x₂)

(a) Find the integral surface (general solution) of the Lagrange auxiliary system dx/(xz - y) = dy/(yz - x) = dz/(1 - z^2), i.e. obtain two independent first integrals and write the surface as an arbitrary relation between them. (b) Two tasks: (a) derive the cube-root iteration x_{i+1} = (1/3)(2x_i + N/x_i^2) from Newton-Raphson applied to f(x) = x^3 - N; (b) use it with x_0 = 3.9 to compute the cube root of 63 correct to four decimal places, showing the successive iterates. (c) Read the printed three-AND-plus-OR logic circuit L and answer three sub-parts: (i) write Y as a Boolean expression, (ii) write the standard 8-bit distinct input sequences for A, B and C, (iii) build the truth table of L from those sequences. (d) Three tasks for a three-dimensional simple source of strength m at the origin in a uniform stream U i: find the velocity potential phi at a point P with OP = r and polar angle theta; find the differential equation of the streamlines; and show the streamlines lie on U r^2 sin^2(theta) - 2 m cos(theta) = constant. (e) A uniform solid hemisphere of radius a rests with its base vertical and its curved surface on a rough horizontal plane; it is released and rolls without slipping. Derive, by energy conservation, that (d(theta)/dt)^2 = 15 g cos(theta)/[a(28 - 15 cos theta)], where theta is the angle of the base with the horizontal.

• Multipliers (1, 1, 0): d(x + y)/(x + y) = −dz/(1 + z) ⇒ (x + y)(1 + z) = c₁
• Multipliers (1, −1, 0): d(x − y)/(x − y) = dz/(1 − z) ⇒ (x − y)(1 − z) = c₂
• Integral surface F((x + y)(1 + z), (x − y)(1 − z)) = 0, F arbitrary
• f(x) = x³ − N, f′(x) = 3x² in xᵢ₊₁ = xᵢ − f(xᵢ)/f′(xᵢ) gives xᵢ₊₁ = (1/3)(2xᵢ + N/xᵢ²)
• Iterates from 3.9: 3.9806706, 3.9790579, 3.9790572
• ∛63 = 3.9791 (successive iterates agree to 4 decimals)
• Y = ABC + AB′C + A′B, simplified to Y = AC + A′B
• A = 00001111, B = 00110011, C = 01010101
• AND outputs 00000001, 00000100, 00110000 and Y = 00110101 (truth table Y = 0, 0, 1, 1, 0, 1, 0, 1 for ABC = 000 … 111)
• φ = m/r − Ur cos θ (q = −∇φ), u_r = U cos θ + m/r², u_θ = −U sin θ
• Streamline equation dr/u_r = r dθ/u_θ
• Integration to (1/2)Ur² sin²θ − m cos θ = constant, i.e. Ur² sin²θ − 2m cos θ = constant
• OG = 3a/8; I_G = (2/5)ma² − m(3a/8)² = (83/320)ma²
• I_P = I_G + m·GP² with GP² = a² + 9a²/64 − (3a²/4) cos θ, giving I_P = ma²(28 − 15 cos θ)/20
• Energy from rest at θ = π/2 (friction does no work): (1/2)I_P θ̇² = mg(3a/8) cos θ ⇒ θ̇² = 15g cos θ/[a(28 − 15 cos θ)]

(a) Classify and reduce y z_xx + (x + y) z_xy + x z_yy = 0 to canonical form via its characteristic curves, and then integrate the canonical equation to obtain the general solution. (b(i)) Convert two 10-bit binary numbers to decimal, where the last 5 bits are the fractional part (i.e. the binary point sits after the fifth bit, giving a 5-bit integer part and a 5-bit fraction). (b(ii)) Add three numbers given in different bases — (11001.1011) with the printed subscript 10, (33.24) octal and (101101) binary — by converting to a common base, and then express the final result in hexadecimal. (c) Four tasks for a circular cylinder of radius a in a uniform stream -V i: apply the Milne-Thomson circle theorem to obtain the complex potential; add a circulation k; locate the stagnation points; and decide, with explanation, whether a vertical lifting tendency results.

• Δ = (x − y)² > 0 (hyperbolic for x ≠ y); characteristics (dy − dx)(y dy − x dx) = 0 ⇒ ξ = y − x, η = y² − x²
• Transformation giving −2ξ² z_ξη − 2ξ z_η = 0, i.e. canonical form z_ξη + (1/ξ) z_η = 0
• (ξ z_η)_ξ = 0 ⇒ z = G(y² − x²)/(y − x) + H(y − x)
• Binary point after the fifth bit: 10101.11011 and 01001.01010
• 10101.11011₂ = 21 + 0.84375 = 21.84375
• 01001.01010₂ = 9 + 0.3125 = 9.3125
• (11001.1011)₂ = 25.6875 (subscript 10 read as a misprint, stated), (33.24)₈ = 27.3125, (101101)₂ = 45
• Sum = 98
• 98 = 6 × 16 + 2 = (62)₁₆
• Circle theorem w = f(z) + f̄(a²/z) with f = −Vz gives w = −V(z + a²/z); with circulation w = −V(z + a²/z) − (ik/2π) ln z
• dw/dz = 0 ⇒ Vz² + (ik/2π)z − Va² = 0 ⇒ z = −ik/(4πV) ± √(a² − k²/(16π²V²)), on the cylinder with sin θ = −k/(4πVa) when k < 4πVa
• Yes: Blasius/Kutta–Joukowski gives a transverse force ρVk per unit length (upward for this sense of k) and zero drag, from the velocity asymmetry and Bernoulli

(a) Solve the linear PDE with constant coefficients (2D^2 - 5DD' + 2D'^2)z = 24(y - x) + sin(y - x): find the complementary function by factorising the operator and the particular integral for both terms on the right-hand side. (b) Evaluate the definite integral from 0 to 0.8 of the given quintic by a SINGLE application of Simpson's 3/8 rule (n = 3 subintervals, h = 0.8/3), computing all four ordinates and the resulting estimate. (c) For a rod OA (length 2a, mass m) pivoted at O with spherical angles (theta from the downward vertical, phi azimuth) carrying a bead of mass (lambda)m on an elastic string of modulus nmg and natural length a: derive the kinetic energy expression 2T and then obtain the Lagrange equations of motion in the coordinates theta, phi, x.

• (D − 2D′)(2D − D′), CF = φ₁(y + 2x) + φ₂(2y + x)
• PI of 24(y − x): z = k(y − x)³, 54k = 24 ⇒ (4/9)(y − x)³
• PI of sin(y − x): F = −9 ⇒ −(1/9) sin(y − x); z = CF + PI
• n = 3, h = 0.8/3 = 0.266667, weights (1, 3, 3, 1) with factor 3h/8
• Ordinates 0.2, 1.432724, 3.487177, 0.232
• I ≈ 1.5192
• I_O = (4/3)ma², rod angular velocity components θ̇ and φ̇ sin θ, bead v² = ẋ² + x²θ̇² + x²φ̇² sin²θ, giving 2T = (4/3)ma²(θ̇² + φ̇² sin²θ) + λm(ẋ² + x²θ̇² + x²φ̇² sin²θ)
• V = −mga cos θ − λmgx cos θ + nmg(x − a)²/(2a)
• Equations: [(4/3)ma² + λmx²] sin²θ φ̇ = const; d/dt{[(4/3)ma² + λmx²]θ̇} = [(4/3)ma² + λmx²]φ̇² sin θ cos θ − mg(a + λx) sin θ; λmẍ = λmx(θ̇² + φ̇² sin²θ) + λmg cos θ − nmg(x − a)/a

(a) Two tasks: deduce by separation of variables that the 2-D Laplace equation in polar form has solutions (A r^n + B r^-n) e^(+/- i n theta); then determine V in the disc 0 <= r <= a subject to finiteness at r = 0 and the boundary value V = sum c_n cos(n theta) on r = a. (b) For dy/dx = 1 + y/x with y(1) = 1 and step size h = 1: (i) apply the classical fourth-order Runge-Kutta method for one step to approximate y(2); (ii) apply Euler's method to obtain y(2) and then y(3). (c) For steady viscous incompressible flow between concentric cylinders of radii a and b (b > a) with no body forces, formulate the governing equation for the circumferential velocity and solve it for two boundary-condition sets: (i) inner cylinder rotating with angular velocity Omega, outer at rest; (ii) outer rotating with Omega, inner at rest.

• Separation of variables V = R(r)Θ(θ), giving (r²R'' + rR')/R = -Θ''/Θ = n²
• Θ'' + n²Θ = 0, so Θ = e^(±inθ), and n an integer from 2π-periodicity
• Euler-Cauchy equation r²R'' + rR' - n²R = 0 with R = Ar^n + Br^-n; for n = 0, R = A + B ln r
• Finiteness at r = 0 gives B = 0 (also for the ln r term)
• V(r, θ) = Σ c_n (r/a)^n cos(nθ), with the constant term c_0
• RK4 formula y₁ = y₀ + (1/6)(k₁ + 2k₂ + 2k₃ + k₄)
• k₁ = 2, k₂ = 2.3333, k₃ = 2.4444, k₄ = 2.7222, so y(2) = 3.3796
• Euler: y(2) = 3 and y(3) = 5.5 (second step from (2, 3))
• Comparison with the exact solution y = x(1 + ln x): y(2) = 3.3863, y(3) = 6.2958
• Circular Couette flow: u_θ = v(r) only; governing equation d²v/dr² + (1/r)dv/dr - v/r² = 0
• General solution v = Ar + B/r
• Case (i) v(a) = Ωa, v(b) = 0: v = Ωa²(b² - r²)/[r(b² - a²)]
• Case (ii) v(a) = 0, v(b) = Ωb: v = Ωb²(r² - a²)/[r(b² - a²)]