UPSC Mathematics 2026 — Paper I

(a) Two linked tasks: (i) find every value of the parameter a for which the 3x3 homogeneous system ax+y+z=0, x+y-z=0, x+y-az=0 has a non-trivial (non-zero) solution, and (ii) for each such a, write down the complete solution set, not merely the existence condition. (b) Prove four separate invariance results for similar matrices A and B (B = P^{-1}AP with P invertible): equality of rank, of trace, of characteristic polynomial and of eigenvalues. Each of the four must be proved, not merely asserted. (c) Compute the Laplacian of u = (x^2 + y^2 + z^2)^{-1/2}, i.e. evaluate u_xx + u_yy + u_zz, and state the value (with the domain restriction r != 0). (d) Two linked tasks: (i) use Maclaurin's theorem to expand e^{x cos x} and give the first four terms in ascending powers of x, and (ii) hence (or otherwise) evaluate lim_{x->0} (e^x - e^{x cos x})/(x - sin x). (e) Find the equation of the sphere whose centre lies on the line given as the intersection of the planes 5y + 2z = 0 and 2x - 3y = 0, and which passes through the two given points (0, -2, -4) and (2, -1, -1).

• Criterion: non-trivial solution iff det A = 0; det A = −(a−1)²
• a = 1 is the only value; for a ≠ 1 only the trivial solution x = y = z = 0
• For a = 1: (x, y, z) = t(1, −1, 0), t ∈ ℝ (rank 2, nullity 1)
• B = P⁻¹AP with P invertible; rank unchanged by multiplication with invertible matrices
• Trace via tr(XY) = tr(YX): tr B = tr A
• B − λI = P⁻¹(A − λI)P and det(P⁻¹)det P = 1 give det(B − λI) = det(A − λI); eigenvalues are its roots, hence equal
• u = 1/r, r² = x² + y² + z², ∂u/∂x = −x/r³
• ∂²u/∂x² = 3x²/r⁵ − 1/r³ and symmetric forms
• Sum = 3r²/r⁵ − 3/r³ = 0 for (x, y, z) ≠ (0, 0, 0)
• Maclaurin's theorem (or composition of the series of cos x and e^t)
• e^{x cos x} = 1 + x + x²/2 − x³/3 + …
• e^x − e^{x cos x} = x³/2 + O(x⁴), x − sin x = x³/6 + …, limit = 3
• Centre on the line: C = (3t, 2t, −5t) (direction ratios 3 : 2 : −5 through the origin)
• CA² = CB² ⇒ 38t² − 32t + 20 = 38t² − 18t + 6 ⇒ t = 1; centre (3, 2, −5), radius √26
• x² + y² + z² − 6x − 4y + 10z + 12 = 0

(a) Construct the matrix of the given linear operator T on P2[x] with respect to the standard basis {1, x, x^2}, then find all eigenvalues of T and a corresponding eigenvector (as a polynomial) for each. The method is prescribed: the matrix representation must be used. (b) A maxima-minima problem with a constraint: in a right-angled triangle the sum of the hypotenuse and one other side is a fixed constant; find the angle between those two sides that makes the area a maximum. Both the optimisation and the confirmation that it is a maximum are required. (c) Three linked tasks on two skew lines: find the pair of points (one on each line) that are nearest to each other, then the shortest distance between the lines, then the equation of the line of shortest distance.

• [T]_B = [−1 −2 −1; 2 3 0; −2 −2 0] (images of 1, x, x² as columns)
• Characteristic polynomial (λ − 1)(λ − 2)(λ + 1); eigenvalues 1, 2, −1
• Eigenvectors 1 − x (λ = 1), (1 − x)² = 1 − 2x + x² (λ = 2), 2 − x + 2x² (λ = −1), up to scalar multiples
• Constraint a + h = k and area A = ½ a √(h² − a²) = ½ a √(k² − 2ka)
• dA/da = 0 ⇒ a = k/3, h = 2k/3, with a second-order test showing a maximum
• cos θ = a/h = 1/2 ⇒ θ = π/3 (60°)
• d₁ = (1, −2, 1), d₂ = (7, −6, 1); general points P(3 + t, 5 − 2t, 7 + t), Q(−1 + 7s, −1 − 6s, −1 + s) with PQ·d₁ = PQ·d₂ = 0 ⇒ s = t = 0
• Nearest points (3, 5, 7) and (−1, −1, −1); shortest distance 2√29 ≈ 10.77
• Line of SD: (x − 3)/2 = (y − 5)/3 = (z − 7)/4

(a) Two parts on the vector space P3[x] over R: (i) prove that H = {p : integral from -1 to 1 of p(x) dx = 0} is a subspace and exhibit a basis of H; (ii) produce a subspace K with P3[x] = H (direct sum) K, i.e. a complement of H. (b(i)) Evaluate the double integral of x^2 y over the region 0 <= x <= 2, x^2 + 1 <= y <= 2x + 1 by the prescribed method of reversing the order of integration (dy dx to dx dy), and give the numerical value. (b(ii)) Find, by triple (or double) integration, the volume of the tetrahedron cut off from the first octant by the plane x/a + y/b + z/c = 1 together with the three coordinate planes. (c) Given that the line x/1 = y/2 = z/3 is one of a set of three mutually perpendicular generators of the cone 5yz - 8zx - 3xy = 0, find the equations of the other two generators.

• H = ker L for the linear functional L(p) = ∫₋₁¹ p dx (or direct closure check), so H is a subspace
• L(p) = 2a₀ + (2/3)a₂ = 0 ⇔ a₂ = −3a₀; basis {1 − 3x², x, x³}, dim H = 3
• K = span{1} (any polynomial with non-zero integral): H ∩ K = {0} and dim H + dim K = 4 ⇒ P₃[x] = H ⊕ K
• Curves meet at (0, 1) and (2, 5); reversed limits 1 ≤ y ≤ 5, (y − 1)/2 ≤ x ≤ √(y − 1)
• Inner integral (y/3)[(y − 1)^{3/2} − (y − 1)³/8]
• I = 184/35 ≈ 5.257
• Intercepts (a, 0, 0), (0, b, 0), (0, 0, c); limits 0 ≤ z ≤ c(1 − x/a − y/b), 0 ≤ y ≤ b(1 − x/a), 0 ≤ x ≤ a
• After z and y integration: (bc/2)(1 − x/a)²
• V = abc/6
• The other two generators lie in the plane x + 2y + 3z = 0 through the vertex, perpendicular to (1, 2, 3)
• Substitution gives y² + 5yz + 4z² = 0, i.e. (y + z)(y + 4z) = 0
• Generators x/1 = y/1 = z/(−1) and x/5 = y/(−4) = z/1

(a) calculate: null-space condition b = −2a, c = −3a/2, nullity 1; range condition a = b + c, rank 2 | (b) trace the curve: symmetry, origin (node), intercepts, asymptote x = −a, real regions, loop and branches | (c(i)) calculate: generators through the given points | (c(ii)) calculate: cylinder equation and section area 24π. Full marks: Rigorous derivation with all steps justified and verified.

• Null space: b = −2a, c = −3a/2, i.e. N(T) = span{(2, −4, −3)}; nullity 1
• Range: row-reducing the augmented matrix gives the consistency condition b + c − a = 0, i.e. a = b + c
• Rank of T = 2 (rank + nullity = 3)
• Symmetry about the x-axis; node at the origin with tangents y = ±√3 x
• Intercepts (0, 0) and (3a, 0) (vertical tangent at (3a, 0)); asymptote x = −a
• Curve exists only for −a < x ≤ 3a: a loop on 0 ≤ x ≤ 3a and two infinite branches in −a < x < 0 approaching x = −a
• Factorisation (x/2 − z/4)(x/2 + z/4) = (1 − y/3)(1 + y/3) and the two systems of generators
• The generator x = 2, 4y + 3z = 0 passes through both points
• Other generators: y = 3, 2x + z = 0 through (2, 3, −4); 6x + 2y − 3z = 6, 6x − 8y + 3z = 24 through (2, −1, 4/3)
• Axis direction (2, 1, −2); distance² = |r × u|²/9 = 16
• Cylinder 5x² + 8y² + 5z² − 4xy + 4yz + 8zx = 144
• Section z = 0: 5x² − 4xy + 8y² = 144, eigenvalues 4 and 9, ellipse X²/36 + Y²/16 = 1, area π·6·4 = 24π

(a) Given two solutions y1 and y2 of the linear first-order ODE dy/dx + P(x)y = Q(x), with y2 = y1 u, derive that u = 1 + a exp(-integral (Q/y1) dx) where a is a non-zero constant. (b) Solve the initial-value problem y'' + y' = 2 with y(0) = 3, y'(0) = 1 using the Laplace transform method (the method is prescribed), and give the explicit solution y(t). (c) Using the principle of virtual work, show that a heavy elastic string of unstretched length 2 pi b resting on a smooth paraboloid of revolution (axis vertical, vertex up, latus rectum of the generating parabola 4a) is in equilibrium as a horizontal circle of radius r = 4 pi a b lambda/(4 pi a lambda - W b), where W is the weight of the string and lambda the modulus of elasticity. (d) A particle slides from rest down the outside of an inverted catenary starting at the vertex. Prove that it leaves the curve at the point where the arc length described equals sqrt(3) times the vertical distance fallen. (e) Evaluate the flux of the curl, double integral over S of (curl F) . n dS, where F = (2x + 3y)i - 4z j - 5x k and S is the part of the plane 2x + 4y + 4z = 5 lying over the rectangle 0 <= x <= 2, 0 <= y <= 3.

• Substitution with the product rule and cancellation via y₁′ + P y₁ = Q, giving y₁u′ = Q(1 − u)
• Separation du/(1 − u) = (Q/y₁) dx and integration to u = 1 + a e^{−∫(Q/y₁) dx}
• a ≠ 0 because y₂ ≠ y₁
• L{y″} = s²Y − 3s − 1, L{y′} = sY − 3, L{2} = 2/s
• Y(s) = (3s² + 4s + 2)/(s²(s + 1)) = 2/s + 2/s² + 1/(s + 1)
• y(t) = 2 + 2t + e^{−t}
• Depth of the circle below the vertex = r²/(4a); tension T = λ(r − b)/b (natural length 2πb)
• Virtual work: W (r/2a) δr − T δ(2πr) = 0, the smooth reaction doing no work (or V = πλ(r − b)²/b − W r²/(4a), dV/dr = 0)
• r = 4πabλ/(4πaλ − Wb)
• s = c tan ψ, ρ = c sec²ψ, h = c(sec ψ − 1), v² = 2gh
• Leaving condition N = 0 in mg cos ψ − N = mv²/ρ ⇒ cos ψ = 1/2, ψ = 60°
• s = √3 c and h = c, so s = √3 h
• ∇ × F = 4î + 5ĵ − 3k̂
• n̂ = (î + 2ĵ + 2k̂)/3, (∇ × F)·n̂ = 8/3, dS = (3/2) dx dy over the 2 × 3 rectangle
• Value = 24

(a) Solve the non-homogeneous Cauchy-Euler (homogeneous-linear) equation x^2 y'' - 3x y' + y = [(sin(log x) + 1)/x] log x completely: the complementary function plus a particular integral, i.e. the general solution. (b(i)) Projectile fired backwards from a ship moving forwards with speed u: show that the range is (2v/g) sin(alpha)(v cos alpha - u), and that the elevation for maximum range is alpha = arccos[(u + sqrt(u^2 + 8v^2))/(4v)]. (b(ii)) A uniform chain of length l has one end fixed at height h above a rough table, hangs in a vertical plane and lies partly straight along the table. Prove that when the chain is on the point of slipping the length resting on the table is (l + mu h) - sqrt((mu^2 + 1)h^2 + 2 mu l h). (c) For the parabola y^2 = 4ax determine all five Serret-Frenet quantities: the unit tangent T, unit principal normal N, unit binormal B, curvature kappa and torsion tau.

• x = e^z, x²d²/dx² = D(D − 1): (D² − 4D + 1)y = e^{−z}(z sin z + z); CF = c₁x^{2+√3} + c₂x^{2−√3}
• Exponential shift with a = −1 giving 1/(D² − 6D + 6); PI of z e^{−z} = (log x + 1)/(6x)
• PI of z e^{−z} sin z = (1/x)[(5 log x/61 + 54/3721) sin(log x) + (6 log x/61 + 382/3721) cos(log x)] and the complete solution CF + PI
• Ground-frame horizontal velocity v cos α − u, vertical v sin α, time of flight 2v sin α/g
• R = (2v/g) sin α (v cos α − u)
• dR/dα = 0 ⇒ v cos 2α = u cos α ⇒ 2v cos²α − u cos α − v = 0 ⇒ cos α = (u + √(u² + 8v²))/(4v), negative root rejected
• Hanging part is a catenary with vertex at the table contact point; T₀ = wc = μN = μwx ⇒ c = μx
• s² = y² − c² with s = l − x, y = c + h ⇒ (l − x)² = 2μxh + h²
• x = (l + μh) − √[(μ² + 1)h² + 2μlh], the + root rejected since it gives x > l
• r = (at², 2at, 0), ds/dt = 2a√(1 + t²), T̂ = (t î + ĵ)/√(1 + t²)
• κ = 1/[2a(1 + t²)^{3/2}], N̂ = (î − t ĵ)/√(1 + t²)
• B̂ = T̂ × N̂ = −k̂, τ = 0 since dB̂/ds = 0

(a) Three things must be proved for a stone dropped from rest at A on a light elastic string of natural length l (equilibrium length b): (i) it comes to instantaneous rest at depth sqrt(b^2 - l^2) below the equilibrium position; (ii) this happens at time sqrt(2l/g) + sqrt((b-l)/g){pi - arccos sqrt((b-l)/(b+l))}; (iii) if the greatest depth below A is l cot^2(theta/2) then the modulus of elasticity is (1/2) mg tan^2(theta). (b) Evaluate the outward flux of F = 4xz i - y^2 j + yz k over the surface of the unit cube 0 <= x, y, z <= 1 by computing the six face integrals, then verify the divergence theorem by evaluating the corresponding volume integral of div F. (c(i)) Two tasks: (i) show that (x + y + 1)^{-4} is an integrating factor of (2x - y - 1)y dx + (2y - x - 1)x dy = 0, i.e. verify exactness after multiplication; (ii) hence integrate to obtain the general solution. (c(ii)) Two linked tasks: (i) prove the division-by-t theorem for Laplace transforms, L{f(t)/t} = integral from s to infinity of F(u) du, where F(s) = L{f(t)}; (ii) use it to evaluate L{(cos at - cos bt)/t}.

• λ = mgl/(b − l); free fall √(2l/g) with v₀ = √(2gl); SHM about equilibrium with ω² = g/(b − l)
• Amplitude² = (b − l)² + 2l(b − l) = b² − l², and time t₂ = (π − cos⁻¹√((b − l)/(b + l)))/ω added to √(2l/g)
• Energy mgD = λe²/(2l) with D = l cot²(θ/2), e = D − l gives λ = 2mgk/(k − 1)² = ½ mg tan²θ
• Six outward-normal face integrals: 2 (x = 1), −1 (y = 1), 1/2 (z = 1), 0 on the three faces through the origin; total 3/2
• div F = 4z − y and ∭(4z − y) dV = 2 − 1/2 = 3/2
• Both equal 3/2, so the divergence theorem is verified
• Shows the original equation is not exact and computes ∂M*/∂y = ∂N*/∂x = (2x² + 2y² − 8xy + x + y − 1)/(x + y + 1)⁵
• Potential V = −xy/(x + y + 1)³ (∂V/∂x = M*, ∂V/∂y = N*)
• General solution xy = C(x + y + 1)³
• Proof by interchanging the order: ∫ₛ^∞ f̄(u) du = ∫₀^∞ f(t)[∫ₛ^∞ e^{−ut} du] dt = ∫₀^∞ e^{−st} f(t)/t dt
• f̄(s) = s/(s² + a²) − s/(s² + b²), integrated from s to ∞ as a single logarithm of a ratio
• L{(cos at − cos bt)/t} = ½ ln[(s² + b²)/(s² + a²)]

(a) Solve y'' + a^2 y = sec(ax) by the prescribed method of variation of parameters: find the complementary function, compute the Wronskian, form the two quadratures, and write the complete general solution. (b) A rod AB (length l, weight W) is pivoted at a point l/3 from A; light strings AC and BC, each of length 5l/6, carry a particle C of weight w. Prove that when W < 2w there is STABLE equilibrium with AB inclined to the horizontal at arctan((W + w)/(4w)). Both the equilibrium position, its stability, and the role of the condition W < 2w must be addressed. (c(i)) Two tasks: (i) prove the vector identity curl(phi F) = (grad phi) x F + phi (curl F) for a scalar field phi and a vector field F; (ii) verify it explicitly for phi = 1/r and F = r, where r = x i + y j + z k and r = |r|. (c(ii)) For the circular helix r(t) = a cos t i + a sin t j + b t k, show that |r' x r''|^2 = a^2(a^2 + b^2) and that the scalar triple product [r' r'' r'''] = a^2 b.

• CF = c₁ cos ax + c₂ sin ax and Wronskian W = a
• Quadratures −∫(sin ax sec ax)/a dx = (1/a²) ln|cos ax| and ∫(cos ax sec ax)/a dx = x/a
• y = c₁ cos ax + c₂ sin ax + (1/a²) cos ax ln|cos ax| + (x/a) sin ax
• M (foot of perpendicular from C) coincides with G; CM = 2l/3, OG = l/6
• V = (l/6)(W + w) sin φ − (2lw/3) cos φ, dV/dφ = 0 ⇒ tan φ = −(W + w)/(4w), i.e. inclination tan⁻¹((W + w)/(4w)); d²V/dφ² > 0 there (stable)
• Both strings taut iff the vertical through C lies within angle ACB: tan α < tan ∠ACM = 3/4 ⇔ W < 2w
• Component proof: ∂(φF₃)/∂y − ∂(φF₂)/∂z split into (∇φ × F)₁ + φ(∇ × F)₁, others by cyclic symmetry
• ∇(1/r) = −r/r³, ∇ × r = 0, so RHS = (−r/r³) × r + 0 = 0
• LHS ∇ × (r/r) computed directly and found to be 0 (r ≠ 0)
• r′ = (−a sin t, a cos t, b), r″ = (−a cos t, −a sin t, 0), r‴ = (a sin t, −a cos t, 0)
• r′ × r″ = (ab sin t, −ab cos t, a²), |r′ × r″|² = a²(a² + b²)
• [r′ r″ r‴] = (r′ × r″)·r‴ = a²b