UPSC Civil Services (Main) Examination 2026 — Mathematics Optional Paper I: Questions with Model Answers | Plutus IAS

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 part (a) requires you to determine the parameter values that make the homogeneous system singular, i.e. det(A)=0. Examiners test (i) rank-based reasoning, (ii) null-space computation, and (iii) parametric solution writing. The common mistake is to stop after computing the determinant and forget to write the explicit solution set for each admissible a.

Model answer

Consider the homogeneous system

• Ax = 0 with

A =

⎡⎣⎢ a 1 1
1 1 −1
1 1 −a⎤⎦⎥.

For non-trivial solutions we need det(A)=0.

det(A) = a(1·(−a)−(−1)·1) − 1(1·(−a)−(−1)·1) + 1(1·1−1·1)
= a(−a+1) − 1(−a+1) + 0
= (a−1)(1−a) = −(a−1)²(a+2).

Hence a = 1 or a = −2.

Case 1: a = 1

The system collapses to

x + y + z = 0,
x + y − z = 0.

Subtracting gives 2z = 0 ⇒ z = 0; then x + y = 0. So the solution space is

x = −t, y = t, z = 0, t ∈ ℝ.

Case 2: a = −2

The system becomes

−2x + y + z = 0,
x + y − z = 0,
x + y + 2z = 0.

Adding the second and third equations gives x + y = 0 ⇒ y = −x; then from the second equation z = 0. Thus the solution space is

x = s, y = −s, z = 0, s ∈ ℝ.

Conclusion: non-trivial solutions exist exactly when a = 1 or a = −2; in each case the solution set is a one-dimensional subspace of ℝ³.

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” requires computing eigenvalues and eigenvectors of the linear operator T by first building its matrix with respect to the standard basis {1,x,x²}, then solving the characteristic equation det([T]–λI)=0. Examiners test (i) correct construction of the representing matrix, (ii) accurate computation of its characteristic polynomial and roots, and (iii) correct back-substitution to obtain eigenvectors. The common mistake is misreading the transformation rule or mixing up the order of coefficients when assembling the matrix.

Model answer

Let the standard ordered basis of P₂[x] be B = {1, x, x²}. For any polynomial p(x)=a₀+a₁x+a₂x², the given transformation reads

• constant term: –(a₀+2a₁+a₂),
• coefficient of x: 2a₀+3a₁,
• coefficient of x²: –2(a₀+a₁).

Hence the matrix of T relative to B is

[T] =
  [–1 –2 –1]
  [ 2 3 0]
  [–2 –2 0].

To find eigenvalues we solve det([T]–λI)=0:

det
  [–1–λ –2 –1 ]
  [ 2 3–λ 0 ]
  [–2 –2 –λ ] = 0.

Expanding along the third column gives (–λ)·(λ²–2λ–1) = 0, so the eigenvalues are

λ₁ = 0, λ₂ = 1+√2, λ₃ = 1–√2.

For each λ we solve ([T]–λI)v = 0.

• λ=0: v = t[3, –2, 2]ᵀ, t≠0; an eigenvector is [3, –2, 2]ᵀ.
• λ=1+√2: v = t[–1, 1, √2]ᵀ; an eigenvector is [–1, 1, √2]ᵀ.
• λ=1–√2: v = t[–1, 1, –√2]ᵀ; an eigenvector is [–1, 1, –√2]ᵀ.

These three eigenvectors span the eigenspaces of T, completing the solution.

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

The directive word “Prove” in (i) signals a formal verification of subspace axioms; the examiner tests understanding of integral constraints, basis computation, and direct-sum decomposition. A top answer must: (1) verify closure under addition and scalar multiplication for H, (2) compute a basis for H by translating the integral condition into a linear equation on coefficients, and (3) construct K explicitly so that P₃[x]=H⊕K. The common mistake is forgetting to check the zero-vector or misinterpreting the integral condition as a point-evaluation.

Model answer

(i) To prove H is a subspace of P₃[x], we check three axioms. First, the zero polynomial p(x)=0 satisfies ∫₋₁¹ 0 dx=0, so 0∈H. Second, for p,q∈H and λ∈ℝ, ∫₋₁¹ (p+q) dx=∫₋₁¹ p dx+∫₋₁¹ q dx=0+0=0 and ∫₋₁¹ (λp) dx=λ∫₋₁¹ p dx=0, hence p+q and λp lie in H. Thus H is a subspace.

Next, the condition ∫₋₁¹ p(x) dx=0 becomes a linear constraint on the coefficients. Compute ∫₋₁¹ (a₀+a₁x+a₂x²+a₃x³) dx = 2a₀ + (2/3)a₂ = 0, i.e. 3a₀ + a₂ = 0. Solving gives a₂ = −3a₀. Hence any p∈H has the form p(x)=a₀ + a₁x − 3a₀x² + a₃x³ = a₀(1−3x²) + a₁x + a₃x³. The set {1−3x², x, x³} is linearly independent and spans H, so it is a basis of H.

(ii) To construct K with P₃[x]=H⊕K, choose a complementary subspace spanned by basis vectors not in H. From the constraint 3a₀+a₂=0, the monomials x and x³ already satisfy the condition, so we supplement with a polynomial violating it, for example x². Define K=span{x²}. Then any q∈K has q(x)=b x² with b∈ℝ, and the decomposition p(x)=h(x)+q(x) with h∈H and q∈K is unique because the coefficients of 1, x, x³ go into h while the coefficient of x² goes into q. Hence P₃[x]=H⊕K.

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 is “find” and the examiner is testing your ability to (i) compute the null space and range conditions explicitly, (ii) extract their dimensions via rank–nullity, and (iii) justify every step with matrix algebra. A top answer must: (1) write the homogeneous system T(x,y,z)=0 and solve for the constraints on a,b,c; (2) compute the rank by counting the number of free variables; (3) state the nullity from rank–nullity. The common mistake is to stop after finding the conditions and forget to state the explicit numerical values of rank and nullity.

Model answer

Let T: F³→F³ be given by T(x,y,z)=(x−y+2z, 2x+y, −x−2y+2z). We represent T by the matrix

A =

(i) Null space and nullity

To find the null space we solve A·[x; y; z]=0:

• Row-reduce A to R =
• The system reduces to x+(2/3)z=0, y−(4/3)z=0, so x=−(2/3)z, y=(4/3)z. Hence every null vector is z(−2/3, 4/3, 1), z∈F.

Therefore (a,b,c) lies in the null space iff 3a+2c=0 and 3b−4c=0. The null space is one-dimensional, so nullity(T)=1.

(ii) Range and rank

The reduced row-echelon form R has two non-zero rows, so the column space has dimension 2. Equivalently, the range consists of all vectors (u,v,w) such that 3u−2v−4w=0. Thus (a,b,c) lies in the range iff 3a−2b−4c=0. By the rank–nullity theorem,

rank(T) = dim(F³) − nullity(T) = 3 − 1 = 2.

Conclusion: the conditions on (a,b,c) are linear constraints derived directly from the row-reduced matrix, and the dimensions follow from rank–nullity, giving a complete algebraic description of both kernel and image.

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 examiner tests your ability to (i) manipulate linear ODEs and Laplace transforms, (ii) apply variational principles to elastic systems, (iii) use energy/work arguments in dynamics, and (iv) compute surface curls via Stokes’ theorem. A top answer must:

• Part (a): Derive the integrating-factor–based relation between two solutions, showing every step.
• Part (b): Apply Laplace transforms to a second-order ODE with constant coefficients and initial data.
• Part (c): Set up the virtual-work equation for an elastic string on a paraboloid and solve for the equilibrium radius.

The common mistake is to skip the derivation in (a) or to misapply the Laplace transform in (b) by forgetting initial conditions.

Model answer

(a) Let y₁ and y₂ satisfy the same linear ODE

y′ + P(x)y = Q(x), y₂ = y₁u.

Substitute y₂ into the ODE:

y₁′u + y₁u′ + Py₁u = Q.

Use y₁′ + Py₁ = Q to cancel terms, leaving

y₁u′ = Q(1 − u).

Separate variables:

du⁄(1 − u) = (Q⁄y₁)dx.

Integrate:

−ln|1 − u| = ∫(Q⁄y₁)dx + C.

Exponentiate and rearrange:

u = 1 + a e^{−∫(Q⁄y₁)dx}, a≠0.

(b) Apply the Laplace transform L to (D² + D)y(t) = 2:

[s²Y(s) − sy(0) − y′(0)] + [sY(s) − y(0)] = 2⁄s.

Insert y(0)=3, y′(0)=1:

s²Y + sY − 3s − 1 − 3 = 2⁄s.

Y(s) = (3s + 4 + 2⁄s)⁄(s²+s) = 4⁄s − 1⁄(s+1) + 2⁄s².

Invert term-wise:

y(t) = 4 − e^−t + 2t.

(c) For a paraboloid z = x²⁄(4a), latus rectum 4a. Let the string lie on the horizontal circle z=h. Its stretched length is

L = ∫₀^2π√(r² + (dr⁄dθ)²)dθ = 2π√(r² + h²⁄a²).

Virtual work: δW = Wδh − TδL = 0. With T = λ(L−2πb)⁄2πb, solve for r to obtain

r = 4πabλ⁄(4πaλ − Wb).

Q6. (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)

How to approach this question

The directive word is solve, so the examiner tests your ability to reduce a non-constant-coefficient Cauchy-Euler ODE to a constant-coefficient ODE, integrate twice, and impose boundary conditions. A top answer must (i) identify the Cauchy-Euler structure, (ii) transform to a constant-coefficient equation in t, (iii) solve the homogeneous part plus find a particular integral, and (iv) impose the physical or implicit boundary conditions. The mistake most aspirants make is to forget the x-dependent right-hand side and try a power-series guess without first checking that the forcing term is a product of log x and trigonometric functions.

Model answer

We solve the Cauchy–Euler equation

$x^2\frac{d^2y}{dx^2}-3x\frac{dy}{dx}+y=\frac{\sin(\log x)+1}{x}\log x\,.$

Step 1. Substitution to constant coefficients.
Set x = e^t so that t = log x. Then

$\displaystyle
x\frac{dy}{dx}=\frac{dy}{dt},\quad
x^2\frac{d^2y}{dx^2}=\frac{d^2y}{dt^2}-\frac{dy}{dt}.
$

The ODE becomes

$\displaystyle
\frac{d^2y}{dt^2}-4\frac{dy}{dt}+y
=\bigl(\sin t+1\bigr)\,t\,.$

Step 2. Homogeneous solution.
The characteristic equation r²–4r+1=0 gives r=2±√3. Hence

$\displaystyle
y_h(t)=Ae^{(2+\sqrt3)t}+Be^{(2-\sqrt3)t}.
$

Step 3. Particular integral.
The right-hand side is t sin t + t. We use the method of undetermined coefficients: try

$\displaystyle
y_p(t)=t\bigl(C_1\sin t+C_2\cos t\bigr)+C_3t+C_4.
$

Substituting and matching coefficients gives

$\displaystyle
C_1=\tfrac12,\quad C_2=-\tfrac12,\quad C_3=1,\quad C_4=4.
$

Thus

$\displaystyle
y_p(t)=\tfrac12t(\sin t-\cos t)+t+4.
$

Step 4. General solution and back-substitution.
The general solution is y=y_h+y_p. Reverting to x gives

$\displaystyle
y(x)=A\,x^{2+\sqrt3}+B\,x^{2-\sqrt3}
+\tfrac12\log x\bigl(\sin(\log x)-\cos(\log x)\bigr)
+\log x+4.
$

Conclusion. The complete solution is the sum of the complementary function and the particular integral above, expressed in terms of x and the arbitrary constants A,B.

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)

How to approach this question

The directive word “Show” signals a proof-oriented task where the examiner tests your ability to derive physical consequences from a given elastic-string model and to verify an integral identity. A top answer must (i) set up the energy or equation-of-motion framework, (ii) solve for the turning point and time of fall, and (iii) connect the maximum depth to the modulus of elasticity via Hooke’s law. The one mistake aspirants make is to skip the dimensional check or the explicit time integration, thereby losing marks on rigour.

Model answer

Consider the light elastic string of natural length l fixed at A. In equilibrium the stone of mass m stretches the string to length b, so by Hooke’s law mg = k(b – l), giving the modulus k = mg/(b – l).

When the stone is released from rest at A, let x(t) be its downward displacement from A. Energy conservation yields the equation of motion

½mẋ² + ½k(x – l)² – mgx = 0.

At the lowest point the kinetic energy vanishes, so

½k(x – l)² – mgx = 0.

Substituting k = mg/(b – l) and solving the quadratic gives x = b. Hence the stone comes to instantaneous rest at a depth b – l below equilibrium, i.e. at depth √(b² – l²) below A.

To compute the time of fall, split the motion into two phases: free fall through the slack length l (time √(2l/g)), followed by simple harmonic motion of amplitude b – l about the equilibrium. The SHM phase contributes an interval

√((b – l)/g) [π – cos⁻¹√((b – l)/(b + l))].

Adding both parts gives the required total time.

Finally, if the greatest depth below A is l cot²(θ/2), then the extension at that point is l cot²(θ/2) – l. Equating the elastic restoring force to the weight yields k(l cot²(θ/2) – l) = mg. Solving for k gives the modulus as ½ mg tan²θ.

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)

How to approach this question

The examiner tests your ability to apply the variation of parameters method to a forced second-order linear ODE (part a), your competence in statics and stability analysis for a constrained rigid body (part b), and your vector calculus skills in proving and verifying a curl identity (part c). Each part must be treated as a self-contained proof or derivation. The common mistake is to skip the verification step in part (c) or to misapply the variation-of-parameters formula in part (a) by forgetting the Wronskian denominator.

Model answer

(a) The directive word “Solve” requires a general solution. The examiner is testing the variation-of-parameters technique for an inhomogeneous linear second-order ODE with constant coefficients.

Step 1 – Complementary solution:
The homogeneous equation
dy²/dx² + a²y = 0
has characteristic roots ±ia, so
y_c = C₁ cos(ax) + C₂ sin(ax).

Step 2 – Variation of parameters:
Set y_p = u₁(x) cos(ax) + u₂(x) sin(ax).
The Wronskian W = |cos(ax) sin(ax); –a sin(ax) a cos(ax)| = a.
Hence
u₁′ = –(1/a) sec(ax) sin(ax) = –(1/a) tan(ax),
u₂′ = (1/a) sec(ax) cos(ax) = 1/a.

Integrating,
u₁ = (1/a²) ln|cos(ax)|,
u₂ = x/a.

Therefore the particular integral is
y_p = (1/a²) cos(ax) ln|cos(ax)| + (x/a) sin(ax),
and the general solution is
y = C₁ cos(ax) + C₂ sin(ax) + (1/a²) cos(ax) ln|cos(ax)| + (x/a) sin(ax).

(b) The examiner asks for a proof of stable equilibrium under given geometric and weight constraints.

Let the rod AB make angle θ with the horizontal. Taking moments about the pivot (l/3 from A) and resolving tensions in AC and BC gives
T₁ = (W l/6) cos θ / (5l/6 sin φ),
T₂ = (W l/6) cos θ / (5l/6 sin ψ),
where φ and ψ are the inclinations of AC and BC to the vertical.

For equilibrium of the particle C,
w = T₁ sin φ + T₂ sin ψ.
Substituting T₁, T₂ and simplifying yields
tan θ = (W + w)/(4w).
A second derivative test on the potential energy
V(θ) = –(W l/6) sin θ – w (5l/6) cos(θ – α)
shows d²V/dθ² > 0 whenever W < 2w, proving stability.

(c) The examiner requires proof of the vector identity and its verification for φ = 1/r and F = r.

Proof (standard):
∇ × (φ F) = ∇φ × F + φ ∇ × F
follows from the product rule for curl and linearity of the differential operator.

Verification:
For φ = 1/r and F = r = x i + y j + z k,
∇φ = –r̂/r²,
∇ × F = 0,
so
∇ × (φ F) = (–r̂/r²) × r = 0.
Direct computation of ∇ × (r/r²) also gives 0, confirming the identity.

Answers are Aanya’s original model guidance; verify facts and the official paper on the exam-conducting body’s official website.

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