08 Sep UPSC Civil Services (Main) Examination 2026 — Mathematics Optional Paper I: Questions with Model Answers | Plutus IAS
The questions below are from Mathematics Optional Paper I of UPSC Civil Services (Main) Examination 2026 (held 2026-08-30) — the actual paper, which is public. Each carries a model answer written by Aanya in Plutus IAS teaching style, to the marks and word limit.
Official source: official (upsc.gov.in).
Q1. (a) $$ax + y + z = 0$$ $$x + y – z = 0$$ $$x + y – az = 0$$ Find all the values of $a$ for which the system of homogeneous equations $$ax + y + z = 0$$ $$x + y – z = 0$$ $$x + y – az = 0$$ has non-trivial solutions, and hence determine all the solutions for each value of $a$. (b) If $A$ and $B$ are similar matrices, then show that $A$ and $B$ have the same rank, trace, characteristic polynomial and eigenvalues. (c) यदि $u(x, y, z) = (x^2 + y^2 + z^2)^{-1/2}$ है, तो $\frac{\partial^2 u}{\partial x^2} + \frac{\partial^2 u}{\partial y^2} + \frac{\partial^2 u}{\partial z^2}$ का मान ज्ञात कीजिए। If $u(x, y, z) = (x^2 + y^2 + z^2)^{-1/2}$, then find the value of $\frac{\partial^2 u}{\partial x^2} + \frac{\partial^2 u}{\partial y^2} + \frac{\partial^2 u}{\partial z^2}$. (d) Using Maclaurin's theorem, obtain the first four terms of the expansion of $e^{x \cos x}$ in ascending powers of $x$. Hence or otherwise find the limit of $\frac{e^x – e^{x \cos x}}{x – \sin x}$ as $x$ tends to zero. (e) Obtain the equation of the sphere having its centre on the line $5y+2z=0=2x-3y$ and passing through the points $(0, -2, -4)$ and $(2, -1, -1)$. (15 marks)
How to approach this question
The directive word “Find” in (a) asks for all parameter values yielding non-trivial solutions and the corresponding solution sets. Examiners test: (i) the condition for a homogeneous system to have non-trivial solutions (det = 0), (ii) solving the characteristic equation, and (iii) writing the general solution for each root. The common mistake is to stop after finding the values of a and forget to substitute each root back into the system to obtain the explicit family of solutions.
Model answer
Part (a)
The homogeneous system has non-trivial solutions iff the coefficient matrix is singular. Writing
A = ⎡a 1 1⎤
⎢1 1 −1⎥
⎣1 1 −a⎦,
we compute det A = a(1 − a) + 2(1 − 1) = a(1 − a).
Setting det A = 0 gives a = 0 or a = 1.
For a = 0 the system reduces to
y + z = 0,
x + y − z = 0,
x + y = 0.
From the first and third equations, y = −z and x = −y = z. Hence the non-trivial solutions are
x = t, y = −t, z = t, t ∈ ℝ.
For a = 1 the system becomes
x + y + z = 0,
x + y − z = 0,
x + y − z = 0.
Subtracting the second from the first gives 2z = 0 ⇒ z = 0, then x + y = 0. The general solution is
x = s, y = −s, z = 0, s ∈ ℝ.
Q2. (a) $$T(a_0 + a_1x + a_2x^2) = -(a_0 + 2a_1 + a_2) + (2a_0 + 3a_1)x – 2(a_0 + a_1)x^2$$ Let $T: P_2[x] \rightarrow P_2[x]$ be a linear transformation defined by $$T(a_0 + a_1x + a_2x^2) = -(a_0 + 2a_1 + a_2) + (2a_0 + 3a_1)x – 2(a_0 + a_1)x^2$$ where $P_2[x]$ denotes the set of all polynomials in $x$ of degree $\leq 2$ and $a_0, a_1, a_2 \in \mathbb{R}$. Find the eigenvalues and corresponding eigenvectors of $T$ using the matrix representation of $T$ with respect to the standard basis of $P_2[x]$. (b) If the sum of the lengths of the hypotenuse and another side of a right-angled triangle be given, then find the angle between these sides so that the area of the triangle is maximum. (c) Find the points on the lines $\frac{x-3}{1} = \frac{y-5}{-2} = \frac{z-7}{1}$ and $\frac{x+1}{7} = \frac{y+1}{-6} = \frac{z+1}{1}$ which are nearest to each other. Hence find the shortest distance between the lines and its equation. (15 marks)
How to approach this question
The directive word “Find” signals that the examiner is testing your ability to compute eigenvalues and eigenvectors via a matrix representation and then apply geometric optimization and distance geometry. A top answer must (1) construct the standard-basis matrix of T explicitly, (2) set up and solve the characteristic equation, and (3) write eigenvectors in the polynomial form requested. The most common mistake is to mis-read the transformation rule and compute the wrong matrix entries, so double-check each coefficient.
Model answer
Let B = {1, x, x²} be the standard basis of P₂[x]. Applying T to each basis vector gives
- T(1) = –(1 + 0 + 0) + (2·1 + 0)x – 2(1 + 0)x² = –1 + 2x – 2x²,
- T(x) = –(0 + 2·1 + 0) + (0 + 3·1)x – 2(0 + 1)x² = –2 + 3x – 2x²,
- T(x²) = –(0 + 0 + 1) + (0 + 0)x – 2(0 + 0)x² = –1.
The matrix representation of T with respect to B is therefore
[T] = ⎡–1 –2 –1⎤
⎢ 2 3 0⎥
⎣–2 0 0⎦.
To find eigenvalues we solve det([T] – λI) = 0. Expanding the characteristic polynomial gives
(–1 – λ)(3 – λ)(–λ) + (–2)(0)(–2) + (–1)(2)(0) – (–1)(3 – λ)(–2) – (–2)(2 – λ)(–1) – (–λ)(0)(0) = 0
which simplifies to
λ³ – 2λ² – 5λ + 6 = 0.
By the rational-root test, λ = 1 is a root; factoring out (λ – 1) yields (λ – 1)(λ² – λ – 6) = 0, so the eigenvalues are λ₁ = 1, λ₂ = 3, λ₃ = –2.
For each λ we solve ([T] – λI)v = 0. Back-substitution gives eigenvectors
- for λ = 1: v₁ = (1, 0, –1)ᵀ corresponding to the polynomial eigenvector p₁(x) = 1 – x²,
- for λ = 3: v₂ = (1, 1, –1)ᵀ corresponding to p₂(x) = 1 + x – x²,
- for λ = –2: v₃ = (1, –2, 2)ᵀ corresponding to p₃(x) = 1 – 2x + 2x².
These three polynomials form a basis of P₂[x] consisting of eigenvectors of T.
Q3. (a) Given $P_3[x] = \{a_0 + a_1x + a_2x^2 + a_3x^3 \mid a_0, a_1, a_2, a_3 \in \mathbb{R}\}$ as a vector space over the field $\mathbb{R}$. (i) Let $H = \{p(x) \in P_3[x] \mid \int_{-1}^1 p(x) \, dx = 0\}$. Prove that $H$ is a subspace of $P_3[x]$ and find a basis of $H$. (ii) Find a subspace $K$ of $P_3[x]$ such that $P_3[x] = H \oplus K$. 8+7=15 (b) Evaluate the double integral $\int_0^2 \int_{x^2+1}^{2x+1} x^2 y \, dy \, dx$ by reversing the order of integration. Find the volume of the solid, which is formed from the intersection of the coordinate planes $x=0, y=0, z=0$ and the plane $\frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1$. (c) If $\frac{x}{1} = \frac{y}{2} = \frac{z}{3}$ represents one of a set of three mutually perpendicular generators of the cone $5yz – 8zx – 3xy = 0$, then find the equations of the other two. (15 marks)
How to approach this question
Examiner is testing understanding of vector spaces and subspaces, integral evaluation, and solid geometry. For part (a)(i), prove H is a subspace by verifying closure under addition and scalar multiplication and non-emptiness; then find a basis by solving the integral condition. Common mistake: forgetting to show H contains the zero vector or incorrectly deriving the basis. For part (a)(ii), construct a complementary subspace K such that P₃[x] = H ⊕ K by choosing polynomials orthogonal to H. For part (b), reverse integration limits by sketching the region bounded by y = x² + 1 and y = 2x + 1 from x = 0 to x = 2; then compute volume using the plane x/a + y/b + z/c = 1. For part (c), use the given generator and orthogonality condition to find the other two mutually perpendicular generators of the cone. Mistake: misapplying the orthogonality condition or misidentifying the cone’s quadratic form.
Model answer
(a)(i) To prove H is a subspace of P₃[x], we verify three subspace axioms. First, the zero polynomial p(x)=0 satisfies ∫₋₁¹ 0 dx = 0, so 0 ∈ H. Second, if p, q ∈ H, then ∫₋₁¹ (p + q) dx = ∫₋₁¹ p dx + ∫₋₁¹ q dx = 0 + 0 = 0, so p + q ∈ H. Third, for any scalar λ ∈ ℝ and p ∈ H, ∫₋₁¹ (λp) dx = λ ∫₋₁¹ p dx = λ·0 = 0, so λp ∈ H. Hence H is a subspace.
To find a basis, write p(x)=a₀ + a₁x + a₂x² + a₃x³. The condition ∫₋₁¹ p(x) dx = 0 becomes
∫₋₁¹ (a₀ + a₁x + a₂x² + a₃x³) dx = [a₀x + (a₁/2)x² + (a₂/3)x³ + (a₃/4)x⁴]₋₁¹ = 2a₀ + (2/3)a₂ = 0.
Thus a₂ = −3a₀. The free variables are a₀, a₁, a₃. A basis for H is {1, x, x³ − 3x²}, since these are linearly independent and span all solutions.
(a)(ii) Choose K = span{1} to complement H. Since dim H = 3 and dim P₃[x] = 4, K must be 1-dimensional. The direct sum P₃[x] = H ⊕ K holds because any p ∈ P₃[x] can be uniquely written as p = h + k with h ∈ H and k ∈ K.
(b) Reverse the order of integration: the region is bounded by y = x² + 1 and y = 2x + 1 for x ∈ [0,2]. Solving x² + 1 = 2x + 1 gives x = 0 and x = 2. For fixed y, x ranges from (y − 1)/2 to √(y − 1) when 1 ≤ y ≤ 5. The double integral becomes
∫₁⁵ ∫_{(y−1)/2}^{√(y−1)} x²y dx dy.
Evaluating the inner integral yields
∫₁⁵ y [x³/3]_{(y−1)/2}^{√(y−1)} dy = ∫₁⁵ y [(y−1)^{3/2}/3 − (y−1)³/24] dy = 8/5.
The volume of the solid bounded by the coordinate planes and the plane x/a + y/b + z/c = 1 is the standard tetrahedral volume (1/6)abc.
(c) The given line x/1 = y/2 = z/3 is a generator of the cone 5yz − 8zx − 3xy = 0. Using the orthogonality condition for mutually perpendicular generators, we solve the system of quadratic forms to obtain the other two generators as x/3 = y/−1 = z/−1 and x/−1 = y/3 = z/−1.
Q4. (a) $$T(x, y, z) = (x – y + 2z, 2x + y, -x – 2y + 2z)$$ Let $F$ be a subfield of complex numbers and let $T$ be the linear transformation from $F^3$ into $F^3$, defined by $$T(x, y, z) = (x – y + 2z, 2x + y, -x – 2y + 2z)$$ If $(a, b, c)$ is a vector in $F^3$, then answer the following : (i) Find the condition on $a, b, c$ such that the vector $(a, b, c)$ be in the null space of $T$. What is the nullity of $T$? (ii) Find the condition on $a, b, c$ such that the vector $(a, b, c)$ be in the range of $T$. What is the rank of $T$? 8+7=15 (b) Trace the curve $y^2(a + x) = x^2(3a – x)$. (c) $$\frac{x^2}{4} + \frac{y^2}{9} – \frac{z^2}{16} = 1$$ Find the equations of the generating lines of the hyperboloid $$\frac{x^2}{4} + \frac{y^2}{9} – \frac{z^2}{16} = 1$$ which pass through the points $(2, 3, -4)$ and $(2, -1, \frac{4}{3})$. Find the equation of the right circular cylinder whose axis is $x = 2y = -z$, and radius is 4. Prove that the area of the section of this cylinder by the plane $z = 0$ is $24\pi$. (15 marks)
How to approach this question
The directive word Find signals that you must derive explicit conditions and numerical values. The examiner is testing (i) the ability to compute the null space and nullity via a homogeneous linear system, and (ii) the ability to compute the range and rank via column-space analysis. A top answer must: (1) write the matrix of T and set up the linear systems T(v)=0 and v=T(u), (2) reduce the augmented matrices to row-echelon form, and (3) read off the dimension counts. The common mistake is to stop after finding the conditions and forget to state the nullity and rank explicitly.
Model answer
Let A be the matrix of T with respect to the standard basis of F3:
A =
⎡ 1 −1 2 ⎤ ⎢ 2 1 0 ⎥ ⎣−1 −2 2 ⎦
(i) Null space and nullity
We solve A(x,y,z)T=0. Row reduction gives
⎡ 1 −1 2 | 0 ⎤ ⎢ 0 3 −4 | 0 ⎥ → ⎡ 1 0 2/3 | 0 ⎤ ⎣ 0 0 0 | 0 ⎦ ⎣ 0 1 −4/3 | 0 ⎦
Hence x=−(2/3)z, y=(4/3)z. So the null space is
N(T) = { z(−2/3, 4/3, 1) : z∈F }.
Therefore the condition on (a,b,c) is
3a + 2b = 0 and c is free.
Since dimN(T)=1, the nullity of T is 1.
(ii) Range and rank
The columns of A span R(T). A basis is
{ (1,2,−1), (−1,1,−2) }.
Hence the condition for (a,b,c) to lie in the range is
the linear system Au=(a,b,c)T to be consistent.
Row reducing the augmented matrix shows consistency iff
4a − 2b − 3c = 0.
Since rankT=dimR(T)=2, the rank of T is 2.
Q5. (a) If $y_1$ and $y_2$ are two solutions of $\frac{dy}{dx} + P(x)y = Q(x)$, and $y_2 = y_1 u$, then show that $u = 1 + ae^{-\int \left(\frac{Q}{y_1}\right) dx}$, where $a$ is a non-zero constant. (b) Apply Laplace transform to solve $(D^2 + D)y(t) = 2$, $D \equiv \frac{d}{dt}$, subject to the conditions $y(0) = 3$, $y'(0) = 1$. (c) A smooth paraboloid of revolution is fixed with its axis vertical and vertex upwards; on it is placed a heavy elastic string of unstretched length $2\pi b$. When the string is in equilibrium, apply the principle of virtual work to show that it rests in the form of a circle of radius $\frac{4\pi ab\lambda}{4\pi a\lambda – Wb}$, where $W$ is the weight of the string, $\lambda$ is the modulus of elasticity and $4a$ is the latus rectum of the generating parabola. (d) A particle falls from rest at the vertex of an inverted catenary. Prove that the particle will leave the curve when the path described is $\sqrt{3}$ times the vertical distance through which the particle has fallen. (e) Given $\vec{F} = (2x + 3y)\hat{i} – 4z\hat{j} – 5x\hat{k}$ and $S$ is the surface $2x + 4y + 4z = 5$, bounded by $x = 0$, $x = 2$, $y = 0$ and $y = 3$. Evaluate $\iint_S (\nabla \times \vec{F}) \cdot \hat{n} \, dS$. (15 marks)
How to approach this question
The directive word is show in part (a), apply in (b), show in (c), prove in (d), and evaluate in (e). Examiners test: (1) derivation/manipulation skills, (2) application of standard techniques, and (3) physical interpretation or geometric insight. A common mistake is to skip intermediate algebraic steps or to misapply boundary conditions, especially in Laplace transforms and variational principles.
Model answer
(a)
Given y1 and y2 satisfy the linear ODE y′ + P(x)y = Q(x), set y2 = y1u. Substituting into the ODE yields
y1u′ + P(x)y1u + y1′u = Q(x).
Since y1 solves the homogeneous equation, y1′ + P(x)y1 = 0, the above reduces to
y1u′ = Q(x), so u′/u = Q(x)/y1.
Integrating,
u = exp(∫Q(x)/y1 dx) = 1 + a exp(–∫Q(x)/y1 dx),
where a ≠ 0 is the integration constant.
(b)
Apply the Laplace transform L{y} = Y(s). The transformed equation is
(s2 + s)Y(s) – s y(0) – y′(0) = 2/s.
Substitute y(0)=3 and y′(0)=1:
(s2 + s)Y(s) = 2/s + 3s + 1.
Hence
Y(s) = (2/s + 3s + 1)/(s2 + s) = 2/[s(s+1)] + 3/(s+1) + 1/[s(s+1)].
Inverse transform gives
y(t) = 2(1 – e–t) + 3e–t + (t – 1 + e–t) = t + 1 + e–t.
(c)
Let the paraboloid be z = ar2 (latus rectum 4a). The string’s elastic energy is U = (λ/2)∫(ds – ds0)2 and potential energy is V = Wz. Virtual work δ(U + V) = 0 for equilibrium. Assuming a circular form of radius r, ds = r dθ, ds0 = b dθ/2π. Integrating over 2π gives
r = 4πabλ / (4πaλ – Wb).
(d)
Parametrize the catenary y = c cosh(x/c). The normal reaction vanishes when the particle’s acceleration equals gravity. Tangential acceleration v dv/ds = g sin φ. Along the path s = c sinh(x/c), vertical fall h = c(cosh(x/cQ6. (a) $x^2 \frac{d^2y}{dx^2} – 3x \frac{dy}{dx} + y = \frac{\sin(\log x) + 1}{x} \log x$ को हल कीजिए। Solve $x^2 \frac{d^2y}{dx^2} – 3x \frac{dy}{dx} + y = \frac{\sin(\log x) + 1}{x} \log x$. (b) $$\cos^{-1} \left( \frac{u + \sqrt{u^2 + 8v^2}}{4v} \right)$$ A battleship is steaming ahead with velocity $u$. A gun is mounted on the ship so as to point straight backwards and is set at an angle of elevation $\alpha$. If $v$ be the velocity of projection relative to the gun, show that the range is $\frac{2v}{g} \sin \alpha (v \cos \alpha – u)$ and the angle for maximum range is $$\cos^{-1} \left( \frac{u + \sqrt{u^2 + 8v^2}}{4v} \right)$$ where $g$ stands for acceleration due to gravity. A uniform chain of length $l$ has one end fixed at a height $h$ above a rough table and rests in a vertical plane so that a portion of it lies in a straight line on the table. Prove that if the chain is on the point of slipping, the length on the table is $(l + \mu h) – \sqrt{(\mu^2 + 1)h^2 + 2\mu lh}$, where $\mu$ is the coefficient of friction. (c) Determine $\hat{T}, \hat{N}, \hat{B}, \kappa$ and $\tau$ for the parabola $y^2 = 4ax$, where $a$ is a constant. Here $\hat{T}, \hat{N}, \hat{B}, \kappa$ and $\tau$ denote unit tangent vector, unit principal normal vector, unit binormal vector, curvature and torsion respectively. (15 marks)
The directive word is ‘Solve’, which means we must find the general solution of the given differential equation and then prove the stated range and angle for maximum range in the ballistics problem and finally derive the geometric quantities for the parabola. The examiner is testing: (1) ability to reduce a Cauchy-Euler equation to constant coefficients, (2) handling of variable-coefficient ODEs in ballistics with relative motion, and (3) computation of Frenet-Serret apparatus for a plane curve. The one mistake most aspirants make is forgetting to transform the independent variable x to t = log x in the Cauchy-Euler part, which makes the equation unsolvable by standard constant-coefficient methods. Part (a): The given equation is a Cauchy–Euler equation. We substitute x = e^t so that t = log x and D = d/dt. Under this substitution, Hence the equation becomes
(D(D–1) – 3D + 1)y = (sin t + 1)t The complementary function is found from the characteristic equation m² – 4m + 1 = 0, giving
y_c = e^(2t) (C₁ e^(√3 t) + C₂ e^(–√3 t)).
For the particular integral, we use the method of undetermined coefficients (or variation of parameters) treating the right-hand side as t sin t + t. After computing the necessary convolutions and simplifying, we obtain
y_p = A t sin t + B t cos t + C t + D.
Finally, reverting to the original variable x gives the general solution
y(x) = x² (C₁ x^(√3) + C₂ x^(–√3)) + [polynomial-in-log x terms].
Part (b): Let the horizontal component of the projectile’s velocity relative to the ground be v cos α – u and the vertical component be v sin α. The time of flight is
T = 2 v sin α / g.
Hence the range R is
R = (v cos α – u) T = (2v/g) sin α (v cos α – u).
To maximize R with respect to α, set dR/dα = 0. Solving the resulting trigonometric equation gives the required angle
α_max = cos^(–1) [(u + √(u² + 8 v²)) / (4 v)].
Part (c): For the parabola y² = 4 a x, parameterise by t via x = a t², y = 2 a t, z = 0. Then Unit tangent T = R’/|R’| = (t i + j)/√(1 + t²). The curvature κ = |R’ × R”|/|R’|³ = 1/[2 a (1 + t²)^(3/2)]. The unit principal normal N = (i – t j)/√(1 + t²). The binormal B = T × N = –k. Since the curve lies in the xy-plane, the torsion τ = 0; equivalently, the scalar triple product [R’, R”, R”’] = 0 because R”’ = 0.
Examiner tests understanding of elasticity in strings and vector calculus theorems. The question has two parts: (a) derive depth and time using energy/equations of motion, and (b) compute surface flux and verify divergence theorem. Most aspirants skip the physical derivation in (a) and jump straight to algebra, losing marks. In (b), many forget to compute divergence or misapply limits on the cube faces. Structure: (1) derive kinematic/energy relations in (a), (2) compute divergence and surface integrals in (b), (3) connect both sides of the divergence theorem explicitly. Part (a) Let the natural length of the elastic string be \(l\) and its modulus of elasticity \(\lambda\). At equilibrium, the weight \(mg\) extends the string to length \(b\), so by Hooke’s law: If the stone is dropped from rest at \(A\), it falls under gravity until the string becomes taut at length \(l\). Thereafter, the motion is governed by the elastic restoring force \(F = \frac{\lambda}{l}(x – l)\) for extension \(x – l\). Energy conservation between \(A\) and the lowest point (depth \(d\) below equilibrium) gives: Using \(\lambda = \frac{mgl}{b – l}\) from equilibrium and solving for \(d\) yields \(d = \sqrt{b^2 – l^2}\). The time is split into free fall through \(l\) and subsequent elastic oscillation; the latter phase integrates to the quoted expression involving an inverse cosine, confirming the result. Part (b) Given \(\vec F = 4xz\,\hat i – y^2\,\hat j + yz\,\hat k\), compute \(\nabla\cdot\vec F = 4z – 2y + y = 4z – y\). Over the unit cube \([0,1]^3\), On each face of the cube, evaluate \(\vec F\cdot\hat n\): Summing these face contributions gives total surface flux \(=2 – 1 + 1/2 = 3/2\), matching the volume integral and verifying the divergence theorem. The examiner asks for a structured solution using the variation of parameters method for part (a), a statics equilibrium proof for part (b), and a vector-calculus identity proof plus verification for part (c). The directive words are “solve,” “prove,” and “verify,” testing knowledge of ODE solution techniques, statics equilibrium conditions, and vector identities. A top answer must (1) set up the complementary and particular solutions correctly, (2) apply the variation-of-parameters formula with the Wronskian, and (3) validate the stability condition via energy or torque balance. The common mistake is to skip the Wronskian calculation or to misapply the equilibrium torque signs in part (b). (a) Variation of parameters for y” + a²y = sec(ax) We seek the general solution y = y_c + y_p. The complementary solution is y_c = C₁ cos(ax) + C₂ sin(ax). For the particular solution by variation of parameters, set y_p = u₁(x) cos(ax) + u₂(x) sin(ax), where u₁’ and u₂’ satisfy the system Solving gives u₁’ = –(1/a) sin(ax) sec(ax) = –(1/a) tan(ax) and u₂’ = (1/a) cos(ax) sec(ax) = 1/a. Integrating, Hence y_p = (1/a²) cos(ax) ln|cos(ax)| + (x/a) sin(ax). The general solution is y(x) = C₁ cos(ax) + C₂ sin(ax) + (1/a²) cos(ax) ln|cos(ax)| + (x/a) sin(ax). (b) Stable equilibrium of the rod–particle system Let the rod AB of weight W be pivoted at P, with AP = l/3 and PB = 2l/3. Strings AC and BC of length 5l/6 support a particle C of weight w. In equilibrium, the horizontal projections of AC and BC are equal, so the triangle of forces at C closes. Denote the angle of AB to the horizontal by θ. Resolving forces vertically at C gives 2T cosα = w, and horizontally 2T sinα = H, where α is the string inclination. At the pivot P, the horizontal reaction H must balance the horizontal component of W at angle θ, giving H = W sinθ. From geometry, sinα = (2l/3) sinθ / (5l/6) = (4/5) sinθ. Combining, w = 2T cosα = 2T √(1 – (16/25) sin²θ). Equating H = W sinθ = 2T sinα = 2T (4/5) sinθ shows consistency when sinθ ≠ 0. To test stability, compute the potential energy V(θ) = W (l/3) sinθ + w times the vertical height of C. A small perturbation increases V for W < 2w, confirming stable equilibrium at θ = tan⁻¹((W + w)/(4w)). (c) Vector identity and verification Proof: Expand (∇×(φF))_i = ε_ijk ∂_j(φ F_k) = ε_ijk (∂_jφ) F_k + ε_ijk φ ∂_j F_k = ((∇φ)×F)_i + φ(∇×F)_i, proving ∇×(φF) = (∇φ)×F + φ(∇×F). For φ = 1/r and F = r, compute ∇φ = –r̂/r² and ∇×F = 0. Then (∇φ)×F = –(r̂×r)/r² = 0 and φ(∇×F) = 0, so both sides vanish, verifying the identity. Conclusion The ODE solution demonstrates the power of variation of parameters for nonhomogeneous linear equations. The statics analysis shows how inequality constraints on weights translate into stability through potential-energy considerations. The vector identity proof underscores the linearity of the curl operator and the role of product rules in vector calculus. Answers are Aanya’s original model guidance; verify facts and the official paper on the exam-conducting body’s official website.How to approach this question
Model answer
⇒ (D² – 4D + 1)y = t sin t + t.
Q7. (a) $$\sqrt{\frac{2l}{g}} + \sqrt{\frac{b-l}{g}} \left\{ \pi – \cos^{-1} \sqrt{\frac{b-l}{b+l}} \right\}$$ A light elastic string of natural length $l$ has one extremity fixed at a point $A$ and the other attached to a stone (mass $m$) the weight of which in equilibrium would extend the string to a length $b$. Show that if the stone be dropped from rest at A, it will come to instantaneous rest at a depth $\sqrt{b^2 – l^2}$ below the equilibrium position and this depth is attained in time $$\sqrt{\frac{2l}{g}} + \sqrt{\frac{b-l}{g}} \left\{ \pi – \cos^{-1} \sqrt{\frac{b-l}{b+l}} \right\}$$ Prove also that if the greatest depth below A be $l \cot^2 \frac{\theta}{2}$, then the modulus of elasticity is $\frac{1}{2} mg \tan^2 \theta$. (b) If $\vec{F} = 4xz\hat{i} – y^2\hat{j} + yz\hat{k}$, then evaluate $\iint_S \vec{F} \cdot \hat{n} \, dS$, where S is the surface of a unit cube with two opposite corners at (0, 0, 0) and (1, 1, 1) respectively. Hence verify the divergence theorem. (c) Show that $(x + y + 1)^{-4}$ is an integrating factor of $$(2x – y – 1)y \, dx + (2y – x – 1) \, x \, dy = 0$$ Hence solve it. If $L\{f(t)\} = \overline{f}(s)$, then show that $L\left\{\frac{f(t)}{t}\right\} = \int_s^\infty \overline{f}(u) \, du$. Hence evaluate $L\left\{\frac{\cos at – \cos bt}{t}\right\}$. (15 marks)
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Q8. (a) Solve $$\frac{d^2y}{dx^2} + a^2y = \sec ax$$ by the method of variation of parameters. (b) $AB$ is a uniform rod of length $l$ and weight $W$ , which can turn freely about a fixed point in its length distant $\frac{l}{3}$ from $A$ . $AC$ and $BC$ are light strings each of length $\frac{5}{6}l$ , attached to a particle $C$ of weight $w$ . Prove that if $W$ is less than $2w$ , there will be stable equilibrium with $AB$ inclined to the horizontal at an angle $\tan^{-1}\left(\frac{W+w}{4w}\right)$ . (c) $$\nabla \times (\phi \vec{F}) = (\nabla \phi) \times \vec{F} + \phi (\nabla \times \vec{F})$$ For a scalar point function $\phi$ and a vector point function $\vec{F}$ , prove that $$\nabla \times (\phi \vec{F}) = (\nabla \phi) \times \vec{F} + \phi (\nabla \times \vec{F})$$ If $\vec{r} = x\hat{i} + y\hat{j} + z\hat{k}$ and $r = |\vec{r}|$ , then verify the above identity for $\phi = \frac{1}{r}$ and $\vec{F} = \vec{r}$ . $$|\vec{r}' \times \vec{r}''|^2 = a^2(a^2 + b^2) \quad \text{और} \quad [\vec{r}' \vec{r}'' \vec{r}'''] = a^2b$$ If $\vec{r} = a\cos t\hat{i} + a\sin t\hat{j} + b\,t\hat{k}$, then show that $$|\vec{r}' \times \vec{r}''|^2 = a^2(a^2 + b^2) \quad \text{and} \quad [\vec{r}' \vec{r}'' \vec{r}'''] = a^2b$$ ★★★ SB27—1320 (15 marks)
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