UPSC Mains 2024 Mathematics Optional Question Papers with Solutions

All 56 questions of the 2024 Civil Services Main Examination, Mathematics Paper I and Mathematics Paper II, with marks and a suggested time per part. Solve each part on paper first, then open the worked solution, or send it to answer-writing practice for evaluation.

UPSC Mains 2024 Mathematics Paper I

Question 1 (50 marks)

Q1(a) · Linear Algebra · 10 marks≈ 7 min
Vector Spaces, Subspaces, Basis and Dimension
Let H be a subspace of ℝ⁴ spanned by the vectors v₁ = (1, −2, 5, −3), v₂ = (2, 3, 1, −4), v₃ = (3, 8, −3, −5). Then find a basis and dimension of H, and extend the basis of H to a basis of ℝ⁴.
हिंदी में प्रश्न
माना H, ℝ⁴ की एक उपसमष्टि है, जो कि सदिशों v₁ = (1, −2, 5, −3), v₂ = (2, 3, 1, −4), v₃ = (3, 8, −3, −5) द्वारा जनित है। तब H का एक आधार एवं विमा ज्ञात कीजिए तथा H के इस आधार को ℝ⁴ के एक आधार तक विस्तृत कीजिए।
Practise this question Linear Algebra notes
Q1(b) · Linear Algebra · 10 marks≈ 7 min
Linear Transformations, Rank-Nullity and Matrix Representation
Let T : ℝ³ → ℝ³ be a linear operator and B = {v₁, v₂, v₃} be a basis of ℝ³ over ℝ. Suppose that Tv₁ = (1, 1, 0), Tv₂ = (1, 0, −1), Tv₃ = (2, 1, −1). Find a basis for the range space and null space of T.
हिंदी में प्रश्न
माना T : ℝ³ → ℝ³ एक रैखिक संकारक है तथा ℝ पर ℝ³ का एक आधार B = {v₁, v₂, v₃} है। माना कि Tv₁ = (1, 1, 0), Tv₂ = (1, 0, −1), Tv₃ = (2, 1, −1) हैं। T की परिसर समष्टि तथा शून्य समष्टि के लिए एक आधार ज्ञात कीजिए।
Practise this question Linear Algebra notes
Worked solution
Since B is a basis, R(T) = L{Tv₁, Tv₂, Tv₃} = L{(1, 1, 0), (1, 0, −1), (2, 1, −1)}. Here Tv₃ = Tv₁ + Tv₂, while Tv₁ and Tv₂ are not proportional.
So {(1, 1, 0), (1, 0, −1)} is a basis of R(T), rank T = 2, and by rank-nullity nullity T = 3 − 2 = 1.
T(v₁ + v₂ − v₃) = Tv₁ + Tv₂ − Tv₃ = 0, and v₁ + v₂ − v₃ ≠ 0 because the v_i are independent; so this vector spans the one-dimensional N(T).
Answer: Basis of R(T): {(1, 1, 0), (1, 0, −1)}; basis of N(T): {v₁ + v₂ − v₃}.
Final answer: Basis of R(T): {(1, 1, 0), (1, 0, −1)}; basis of N(T): {v₁ + v₂ − v₃}.
Q1(c) · Calculus · 10 marks≈ 7 min
Real Numbers, Limits, Continuity and Differentiability
Discuss the continuity of the function
f(x) = 1/(1 − e^−1/x), x ≠ 0
f(x) = 0, x = 0
for all values of x.
हिंदी में प्रश्न
x के सभी मानों के लिए फलन
f(x) = 1/(1 − e^−1/x), x ≠ 0
f(x) = 0, x = 0
के सांतत्य की चर्चा कीजिए।
Practise this question Calculus notes
Worked solution
For x ≠ 0, e^−1/x = 1 is impossible (−1/x ≠ 0), so the denominator never vanishes and f is continuous at every x ≠ 0, being a quotient of continuous functions.
x → 0+: −1/x → −∞, e^−1/x → 0, so f(0 + 0) = 1. x → 0−: −1/x → +∞, e^−1/x → ∞, so f(0 − 0) = 0.
f(0 − 0) = 0 = f(0) ≠ 1 = f(0 + 0): f is continuous from the left but not from the right, and x = 0 is a discontinuity of the first kind (jump 1).
Answer: f is continuous for all x ≠ 0 and has a jump discontinuity at x = 0.
Final answer: f is continuous for all x ≠ 0 and has a jump discontinuity at x = 0.
Q1(d) · Calculus · 10 marks≈ 7 min
Mean Value Theorems, Taylor's Theorem and Indeterminate Forms
Expand ln(x) in powers of (x − 1) by Taylor's theorem and hence find the value of ln(1·1) correct up to four decimal places.
हिंदी में प्रश्न
टेलर प्रमेय द्वारा ln(x) का (x − 1) की घात में प्रसार कीजिए तथा ln(1·1) का दशमलव के चार स्थानों तक सही मान ज्ञात कीजिए।
Practise this question Calculus notes
Worked solution
For f(x) = ln x, f⁽ⁿ⁾(x) = (−1)ⁿ⁻¹(n − 1)!/xⁿ, so f(1) = 0 and f⁽ⁿ⁾(1)/n! = (−1)ⁿ⁻¹/n.
Taylor's theorem about 1 with h = x − 1: ln x = h − h²/2 + h³/3 − ... + (−1)ⁿhⁿ⁻¹/(n − 1) + R_n, where the Lagrange remainder is R_n = (−1)ⁿ⁻¹hⁿ/(n(1 + θh)ⁿ); for 0 ≤ h ≤ 1, |R_n| ≤ hⁿ/n → 0.
For x = 1.1, h = 0.1: ln 1.1 = 0.1 − 0.005 + 0.000333 − 0.000025 + 0.000002 − ... ; with four terms, |R₅| ≤ (0.1)⁵/5 = 0.000002.
Sum of four terms: 0.1 − 0.005 + 0.0003333 − 0.000025 = 0.0953083, and the true value lies between 0.0953083 and 0.0953103.
Answer: ln x = (x − 1) − (x − 1)²/2 + (x − 1)³/3 − ... (0 < x ≤ 2); ln 1.1 = 0.0953 to four decimal places.
Final answer: ln x = (x − 1) − (x − 1)²/2 + (x − 1)³/3 − ... (0 < x ≤ 2); ln 1.1 = 0.0953 to four decimal places.
Q1(e) · Analytic Geometry · 10 marks≈ 7 min
Cone and Cylinder
Find the equation of the right circular cylinder which passes through the circle x² + y² + z² = 9, x − y + z = 3.
हिंदी में प्रश्न
वृत्त x² + y² + z² = 9, x − y + z = 3 से होकर जाने वाले लम्ब वृत्तीय बेलन का समीकरण ज्ञात कीजिए।
Practise this question Analytic Geometry notes
Worked solution
The sphere has centre O and radius 3; the distance of O from x − y + z = 3 is 3/√3 = √3, so the circle has radius √(9 − 3) = √6.
The axis is the normal through O: x/1 = y/(−1) = z/1. Right circular cylinder: x² + y² + z² − (x − y + z)²/3 = 6.
Multiplying by 3 and expanding: 2x² + 2y² + 2z² + 2xy + 2yz − 2zx = 18.
Answer: x² + y² + z² + xy + yz − zx − 9 = 0.
Final answer: x² + y² + z² + xy + yz − zx − 9 = 0.

Question 2 (50 marks)

Q2(a) · Linear Algebra · 15 marks≈ 11 min
Linear Transformations, Rank-Nullity and Matrix Representation
Consider a linear operator T on ℝ³ over ℝ defined by T(x, y, z) = (2x, 4x − y, 2x + 3y − z). Is T invertible? If yes, justify your answer and find T⁻¹.
हिंदी में प्रश्न
माना ℝ के ऊपर ℝ³ पर एक रैखिक संकारक T, T(x, y, z) = (2x, 4x − y, 2x + 3y − z) द्वारा परिभाषित है। क्या T व्युत्क्रमणीय है? यदि हाँ, तो अपने उत्तर का तर्क प्रस्तुत कीजिए तथा T⁻¹ ज्ञात कीजिए।
Practise this question Linear Algebra notes
Worked solution
Kernel: 2x = 0, 4x − y = 0, 2x + 3y − z = 0 give x = y = z = 0, so N(T) = {0}. A one-one operator on a finite-dimensional space is onto, so T is invertible (equivalently det [[2, 0, 0], [4, −1, 0], [2, 3, −1]] = 2 ≠ 0).
Solve T(x, y, z) = (a, b, c): 2x = a, 4x − y = b, 2x + 3y − z = c. So x = a/2, y = 4x − b = 2a − b, z = 2x + 3y − c = 7a − 3b − c.
Check: T(a/2, 2a − b, 7a − 3b − c) = (a, 2a − (2a − b), a + 6a − 3b − 7a + 3b + c) = (a, b, c).
Answer: Yes, T is invertible and T⁻¹(a, b, c) = (a/2, 2a − b, 7a − 3b − c).
Final answer: Yes, T is invertible and T⁻¹(a, b, c) = (a/2, 2a − b, 7a − 3b − c).
Q2(b) · Calculus · 15 marks≈ 11 min
Functions of Several Variables: Partial Derivatives, Extrema, Lagrange Multipliers and Jacobians
If u = (x + y)/(1 − xy) and v = tan⁻¹ x + tan⁻¹ y, then find ∂(u, v)/∂(x, y). Are u and v functionally related? If yes, find the relationship.
हिंदी में प्रश्न
यदि u = (x + y)/(1 − xy) तथा v = tan⁻¹ x + tan⁻¹ y हैं, तब ∂(u, v)/∂(x, y) ज्ञात कीजिए। क्या u तथा v फलनतः सम्बन्धित हैं? यदि हाँ, तो सम्बन्ध ज्ञात कीजिए।
Practise this question Calculus notes
Worked solution
u_x = (1 + y²)/(1 − xy)², u_y = (1 + x²)/(1 − xy)², v_x = 1/(1 + x²), v_y = 1/(1 + y²).
∂(u, v)/∂(x, y) = u_xv_y − u_yv_x = 1/(1 − xy)² − 1/(1 − xy)² = 0 (for xy ≠ 1), so u and v are functionally related.
The relation: tan v = tan(tan⁻¹x + tan⁻¹y) = (x + y)/(1 − xy) = u.
Answer: ∂(u, v)/∂(x, y) = 0; yes, u = tan v.
Final answer: ∂(u, v)/∂(x, y) = 0; yes, u = tan v.
Q2(c) · Analytic Geometry · 20 marks≈ 14 min
Straight Lines and Shortest Distance between Skew Lines
Find the image of the line x = 3 − 6t, y = 2t, z = 3 + 2t in the plane 3x + 4y − 5z + 26 = 0.
हिंदी में प्रश्न
रेखा x = 3 − 6t, y = 2t, z = 3 + 2t का समतल 3x + 4y − 5z + 26 = 0 में प्रतिबिम्ब ज्ञात कीजिए।
Practise this question Analytic Geometry notes
Worked solution
The line passes through A(3, 0, 3) with direction (−6, 2, 2); the plane has normal (3, 4, −5), and (−6, 2, 2)·(3, 4, −5) = −20 ≠ 0, so the line meets the plane.
Meeting point: 3(3 − 6t) + 4(2t) − 5(3 + 2t) + 26 = 20 − 20t = 0, so t = 1 and Q = (−3, 2, 5).
Image of A: 3·3 + 0 − 15 + 26 = 20 and 3² + 4² + 5² = 50, so the common ratio is −2·20/50 = −4/5 and A′ = (3 − 12/5, −16/5, 3 + 4) = (3/5, −16/5, 7).
The image line joins Q and A′: direction (18/5, −26/5, 2), proportional to (9, −13, 5).
Answer: (x + 3)/9 = (y − 2)/(−13) = (z − 5)/5, that is x = −3 + 9k, y = 2 − 13k, z = 5 + 5k.
Final answer: (x + 3)/9 = (y − 2)/(−13) = (z − 5)/5, that is x = −3 + 9k, y = 2 − 13k, z = 5 + 5k.

Question 3 (50 marks)

Q3(a) · Linear Algebra · 15 marks≈ 11 min
Linear Transformations, Rank-Nullity and Matrix Representation
Let V = M_2×2(ℝ) denote a vector space over the field of real numbers. Find the matrix of the linear mapping φ : V → V given by φ(v) = [[1, 2], [3, −1]] v with respect to standard basis of M_2×2(ℝ), and hence find the rank of φ. Is φ invertible? Justify your answer.
हिंदी में प्रश्न
माना V = M_2×2(ℝ) वास्तविक संख्याओं के क्षेत्र पर एक सदिश समष्टि दर्शाता है। M_2×2(ℝ) के मानक आधार के सन्दर्भ में φ(v) = [[1, 2], [3, −1]] v द्वारा दिए गए रैखिक प्रतिचित्रण φ : V → V का आव्यूह ज्ञात कीजिए और तब φ की कोटि (रैंक) ज्ञात कीजिए। क्या φ व्युत्क्रमणीय है? अपने उत्तर का तर्क प्रस्तुत कीजिए।
Practise this question Linear Algebra notes
Worked solution
Let M = [[1, 2], [3, −1]] and order the basis E₁₁, E₁₂, E₂₁, E₂₂. φ(E₁₁) = [[1, 0], [3, 0]] = E₁₁ + 3E₂₁; φ(E₁₂) = [[0, 1], [0, 3]] = E₁₂ + 3E₂₂.
φ(E₂₁) = [[2, 0], [−1, 0]] = 2E₁₁ − E₂₁; φ(E₂₂) = [[0, 2], [0, −1]] = 2E₁₂ − E₂₂.
So [φ] = [[1, 0, 2, 0], [0, 1, 0, 2], [3, 0, −1, 0], [0, 3, 0, −1]]. Swapping the 2nd and 3rd rows and the 2nd and 3rd columns gives diag(M, M), so det[φ] = (det M)² = (−7)² = 49 ≠ 0.
Hence rank φ = 4 and φ is invertible, with φ⁻¹(v) = M⁻¹v and M⁻¹ = (1/7)M (since M² = 7I).
Answer: [φ] = [[1, 0, 2, 0], [0, 1, 0, 2], [3, 0, −1, 0], [0, 3, 0, −1]], rank φ = 4, and φ is invertible.
Final answer: [φ] = [[1, 0, 2, 0], [0, 1, 0, 2], [3, 0, −1, 0], [0, 3, 0, −1]], rank φ = 4, and φ is invertible.
Q3(b) · Calculus · 20 marks≈ 14 min
Maxima and Minima, Asymptotes and Curve Tracing
Find the volume of the greatest cylinder which can be inscribed in a cone of height h and semi-vertical angle α.
हिंदी में प्रश्न
ऊँचाई h तथा अर्ध-शीर्ष कोण α वाले एक शंकु के अंतर्गत सबसे बड़े बेलन का आयतन ज्ञात कीजिए।
Practise this question Calculus notes
Worked solution
Let the cylinder have radius x and height y, with its base on the base of the cone and its upper rim on the cone. At height y the radius of the cone is (h − y) tan α, so x = (h − y) tan α.
V = πx²y = π tan²α (h − y)²y for 0 < y < h, and dV/dy = π tan²α (h − y)(h − 3y) = 0 gives y = h/3.
d²V/dy² = π tan²α (6y − 4h) = −2πh tan²α < 0 at y = h/3, a maximum. Then x = (2h/3) tan α.
V_max = π(4h²/9)tan²α (h/3) = (4/27)πh³ tan²α, which is 4/9 of the volume (1/3)πh³tan²α of the cone.
Answer: Greatest volume (4/27)πh³ tan²α, for a cylinder of height h/3.
Final answer: Greatest volume (4/27)πh³ tan²α, for a cylinder of height h/3.
Q3(c) · Analytic Geometry · 15 marks≈ 11 min
Cone and Cylinder
Find the vertex of the cone 4x² − y² + 2z² + 2xy − 3yz + 12x − 11y + 6z + 4 = 0.
हिंदी में प्रश्न
शंकु 4x² − y² + 2z² + 2xy − 3yz + 12x − 11y + 6z + 4 = 0 का शीर्ष ज्ञात कीजिए।
Practise this question Analytic Geometry notes

Question 4 (50 marks)

Q4(a) · Linear Algebra · 20 marks≈ 14 min
Eigenvalues, Eigenvectors and the Cayley-Hamilton Theorem
Let A = [[3, 2, 4], [2, 0, 2], [4, 2, 3]] be a 3×3 matrix. Find the eigenvalues and the corresponding eigenvectors of A. Hence find the eigenvalues and the corresponding eigenvectors of A⁻¹⁵, where A⁻¹⁵ = (A⁻¹)¹⁵.
हिंदी में प्रश्न
माना A = [[3, 2, 4], [2, 0, 2], [4, 2, 3]] एक 3×3 आव्यूह है। A का अभिलाक्षणिक मान तथा संगत अभिलाक्षणिक सदिश ज्ञात कीजिए। अतः A⁻¹⁵ का अभिलाक्षणिक मान तथा संगत अभिलाक्षणिक सदिश ज्ञात कीजिए, जहाँ A⁻¹⁵ = (A⁻¹)¹⁵ है।
Practise this question Linear Algebra notes
Worked solution
tr A = 6; principal minors (0 − 4) + (9 − 16) + (0 − 4) = −15; det A = 3(0 − 4) − 2(6 − 8) + 4(4 − 0) = 8. So λ³ − 6λ² − 15λ − 8 = (λ − 8)(λ + 1)² = 0, and λ = 8, −1, −1.
λ = 8: A − 8I = [[−5, 2, 4], [2, −8, 2], [4, 2, −5]]; the first two rows give X = k(2, 1, 2). Check: A(2, 1, 2) = (16, 8, 16).
λ = −1: A + I = [[4, 2, 4], [2, 1, 2], [4, 2, 4]] has rank 1, so the eigenvectors are the non-zero solutions of 2x + y + 2z = 0, a plane with basis (1, −2, 0), (0, −2, 1).
If AX = λX with λ ≠ 0, then X = λA⁻¹X, so A⁻¹X = λ⁻¹X and A⁻¹⁵X = λ⁻¹⁵X. So A⁻¹⁵ has the same eigenvectors, with eigenvalues 8⁻¹⁵ = 2⁻⁴⁵ and (−1)⁻¹⁵ = −1 (twice).
Answer: A: λ = 8 with k(2, 1, 2); λ = −1 (twice) with a(1, −2, 0) + b(0, −2, 1). A⁻¹⁵: λ = 2⁻⁴⁵ with k(2, 1, 2); λ = −1 (twice) with the same plane of eigenvectors.
Final answer: A: λ = 8 with k(2, 1, 2); λ = −1 (twice) with a(1, −2, 0) + b(0, −2, 1). A⁻¹⁵: λ = 2⁻⁴⁵ with k(2, 1, 2); λ = −1 (twice) with the same plane of eigenvectors.
Q4(b) · Calculus · 15 marks≈ 11 min
Double and Triple Integrals; Areas, Surfaces and Volumes
Using double integration, find the area lying inside the cardioid r = a(1 + cos θ) and outside the circle r = a.
हिंदी में प्रश्न
द्विशः समाकलन का प्रयोग करते हुए हृदयाभ (कार्डिऑइड) r = a(1 + cos θ) के अन्दर तथा वृत्त r = a के बाह्य स्थित क्षेत्र का क्षेत्रफल ज्ञात कीजिए।
Practise this question Calculus notes
Worked solution
The curves meet where a(1 + cos θ) = a, i.e. θ = ±π/2, and the cardioid lies outside the circle for −π/2 ≤ θ ≤ π/2 (where cos θ ≥ 0).
A = ∫_−π/2^π/2∫_a^{a(1+cos θ)} r dr dθ = (a²/2)∫_−π/2^π/2 [(1 + cos θ)² − 1] dθ = (a²/2)∫_−π/2^π/2 (2 cos θ + cos²θ) dθ.
∫_−π/2^π/2 2 cos θ dθ = 4 and ∫_−π/2^π/2 cos²θ dθ = π/2, so A = (a²/2)(4 + π/2).
Answer: A = a²(8 + π)/4.
Final answer: A = a²(8 + π)/4.
Q4(c) · Analytic Geometry · 15 marks≈ 11 min
The Sphere
Find the equation of the sphere which touches the plane 3x + 2y − z + 2 = 0 at the point (1, −2, 1) and cuts orthogonally the sphere x² + y² + z² − 4x + 6y + 4 = 0.
हिंदी में प्रश्न
उस गोले का समीकरण ज्ञात कीजिए, जो समतल 3x + 2y − z + 2 = 0 को बिन्दु (1, −2, 1) पर स्पर्श करता है और गोले x² + y² + z² − 4x + 6y + 4 = 0 को लांबिकतः काटता है।
Practise this question Analytic Geometry notes

Question 5 (50 marks)

Q5(a) · Ordinary Differential Equations · 10 marks≈ 7 min
Orthogonal Trajectories
Find the orthogonal trajectories of the family of curves r = c(sec θ + tan θ), where c is a parameter.
हिंदी में प्रश्न
वक्र-कुल r = c(sec θ + tan θ) के लम्बकोणीय संछेदी ज्ञात कीजिए, जहाँ c एक प्राचल है।
Practise this question Ordinary Differential Equations notes
Q5(b) · Ordinary Differential Equations · 10 marks≈ 7 min
Laplace Transforms and Initial Value Problems
Solve the integral equation y(t) = cos t + ∫₀^t y(x) cos(t − x) dx using Laplace transform.
हिंदी में प्रश्न
लाप्लास रूपान्तर का प्रयोग करते हुए समाकलन समीकरण y(t) = cos t + ∫₀^t y(x) cos(t − x) dx को हल कीजिए।
Practise this question Ordinary Differential Equations notes
Q5(c) · Vector Analysis · 10 marks≈ 7 min
Scalar and Vector Fields; Differentiation of Vector Functions
At any time t (in seconds), the coterminous edges of a variable parallelepiped are represented by the vectors
α̅ = t î + (t + 1) ĵ + (2t + 1) k̂
β̅ = 2t î + (3t − 1) ĵ + t k̂
γ̅ = î + 3t ĵ + k̂
What is the rate of change of the vectorial area of the parallelogram, whose coterminous edges are α̅ and γ̅? Also find the rate of change of the volume of the parallelepiped at t = 1 second.
हिंदी में प्रश्न
किसी समय t (सेकण्ड में) पर एक चर समांतर-षट्फलक की सहावसानी किनारे सदिशों
α̅ = t î + (t + 1) ĵ + (2t + 1) k̂
β̅ = 2t î + (3t − 1) ĵ + t k̂
γ̅ = î + 3t ĵ + k̂
द्वारा निरूपित हैं। समांतर-चतुर्भुज, जिसकी सहावसानी किनारे α̅ और γ̅ हैं, के सदिशीय क्षेत्रफल की परिवर्तन दर क्या है? t = 1 सेकण्ड पर समांतर-षट्फलक के आयतन की परिवर्तन दर भी ज्ञात कीजिए।
Practise this question Vector Analysis notes
Worked solution
The vectorial area of the parallelogram on α and γ is S = α × γ = (−6t² − 2t + 1)i + (t + 1)j + (3t² − t − 1)k.
dS/dt = α′ × γ + α × γ′ = (1 − 6t, 1, 3t − 1) + (−6t − 3, 0, 3t) = (−12t − 2)i + j + (6t − 1)k, which is −14i + j + 5k at t = 1.
The volume is V = [α β γ] = det[[t, t + 1, 2t + 1], [2t, 3t − 1, t], [1, 3t, 1]] = 9t³ + 2t² − 3t + 1.
dV/dt = 27t² + 4t − 3, which is 28 at t = 1 (check: V(1) = 9 and V(2) = 75 from the determinant directly).
Answer: dS/dt = (−12t − 2)i + j + (6t − 1)k, equal to −14i + j + 5k at t = 1; the volume changes at 28 cubic units per second at t = 1.
Final answer: dS/dt = (−12t − 2)i + j + (6t − 1)k, equal to −14i + j + 5k at t = 1; the volume changes at 28 cubic units per second at t = 1.
Q5(d) · Dynamics and Statics · 10 marks≈ 7 min
Virtual Work, Stability and Forces in Three Dimensions
A solid hemisphere rests in equilibrium on a solid sphere of equal radius. Determine the stability of the equilibrium in the two situations — (i) when the curved surface and (ii) when the flat surface of the hemisphere rests on the sphere.
हिंदी में प्रश्न
एक ठोस गोले के ऊपर समान त्रिज्या का एक ठोस गोलार्ध साम्यावस्था में रखा है। दो स्थितियों में — (i) जब गोलार्ध का वक्रीय पृष्ठ तथा (ii) जब गोलार्ध का समतलीय पृष्ठ गोले पर स्थित है, साम्यावस्था का स्थायित्व ज्ञात कीजिए।
Practise this question Dynamics and Statics notes
Worked solution
Let the common radius be r and use the rule of §2: stable if 1/h > 1/ρ₁ + 1/ρ₂, h being the height of G above the point of contact (rough contact assumed).
(i) Curved surface down: ρ₁ = ρ₂ = r and h = r − 3r/8 = 5r/8, so 1/h = 8/(5r) = 1.6/r, which is less than 2/r: unstable.
(ii) Flat face down: ρ₁ = ∞, ρ₂ = r and h = 3r/8, so 1/h = 8/(3r) > 1/r: stable.
Answer: (i) unstable; (ii) stable.
Final answer: (i) unstable; (ii) stable.
Q5(e) · Vector Analysis · 10 marks≈ 7 min
Curves in Space: Curvature, Torsion and Serret-Frenet Formulae
(i) Let C be a plane curve r̅(t) = f(t) î + g(t) ĵ, where f and g have second-order derivatives. Show that the curvature at a point is given by
κ = |f′(t) g″(t) − g′(t) f″(t)| / ([f′(t)]² + [g′(t)]²)^3/2
What is the value of torsion τ at any point of this curve? (5 marks)
(ii) Show that the principal normals at two consecutive points of a curve do not intersect unless torsion τ is zero. (5 marks)
हिंदी में प्रश्न
(i) माना C एक समतल वक्र r̅(t) = f(t) î + g(t) ĵ है, जहाँ f और g के द्वितीय कोटि के अवकलज हैं। दर्शाइए कि वक्र के किसी बिन्दु पर वक्रता
κ = |f′(t) g″(t) − g′(t) f″(t)| / ([f′(t)]² + [g′(t)]²)^3/2
है। इस वक्र के किसी बिन्दु पर ऐंठन (टॉर्शन) τ का मान क्या है? (5 अंक)
(ii) दर्शाइए कि किसी वक्र के दो क्रमागत बिन्दुओं पर मुख्य अभिलम्ब प्रतिच्छेद नहीं करते जब तक कि ऐंठन (टॉर्शन) τ शून्य न हो। (5 अंक)
Practise this question Vector Analysis notes
Worked solution
(i) r′ = (f′, g′, 0) and r″ = (f″, g″, 0), so r′ × r″ = (f′g″ − g′f″)k and κ = |r′ × r″|/|r′|³ = |f′g″ − g′f″|/(f′² + g′²)^3/2.
r‴ = (f‴, g‴, 0) also lies in the xy-plane, so [r′ r″ r‴] = 0 and τ = 0 at every point where κ ≠ 0 (τ needs third derivatives to be defined).
(ii) The principal normals at P(s) and Q(s + δs) are the lines R = r + λN and R = r + δr + μ(N + δN); they meet only if δr, N, N + δN are coplanar, that is [δr N δN] = 0.
By Taylor's theorem and Serret-Frenet, δr = Tδs + (1/2)κNδs² + O(δs³) and δN = (τB − κT)δs + O(δs²).
Hence [δr N δN] = τ[T N B]δs² + O(δs³) = τδs² + O(δs³), which can vanish for all small δs only if τ = 0.
Answer: (i) κ as stated and τ = 0; (ii) consecutive principal normals intersect only when τ = 0 (see Standard Proofs).
Final answer: (i) κ as stated and τ = 0; (ii) consecutive principal normals intersect only when τ = 0 (see Standard Proofs).

Question 6 (50 marks)

Q6(a) · Dynamics and Statics · 15 marks≈ 11 min
Virtual Work, Stability and Forces in Three Dimensions
A regular tetrahedron, formed of six light rods, each of length l, rests on a smooth horizontal plane. A ring of weight W and radius r is supported by the slant sides. Using the principle of virtual work, find the stress in any of the horizontal sides.
हिंदी में प्रश्न
लम्बाई l की छः हल्की छड़ों द्वारा निर्मित एक सम चतुष्फलक एक चिकने क्षैतिज समतल पर रखा है। W भार तथा r त्रिज्या का एक छल्ला तिर्यक् भुजाओं द्वारा आलम्बित है। कल्पित कार्य सिद्धान्त का उपयोग करते हुए किसी भी एक क्षैतिज भुजा में प्रतिबल ज्ञात कीजिए।
Practise this question Dynamics and Statics notes
Worked solution
Let the horizontal sides have length x (they may stretch in the virtual displacement, so each is replaced by its tension T), the slant sides keep length l, and R = x/√3 be the distance of the base vertices from the vertical axis.
The apex is at height H = √(l² − R²), and a point of a slant side at depth D below the apex is at distance RD/H from the axis.
The ring rests round the three slant sides at the level where this distance equals r, that is D = rH/R, so its height is z = H(1 − r/R).
Virtual work: −W δz − 3T δx = 0, so T = −(W/3) dz/dx = −(W/(3√3)) dz/dR.
dz/dR = −(R − r)/H + rH/R²; for the regular tetrahedron R = l/√3 and H = l√(2/3), so dz/dR = −(l − √3r)/(√2l) + √6r/l = −(l − 3√3r)/(√2l).
Hence T = W(l − 3√3r)/(3√6 l).
Answer: each horizontal side has tension W(l − 3√3r)/(3√6 l); it is a thrust if r > l/(3√3).
Final answer: each horizontal side has tension W(l − 3√3r)/(3√6 l); it is a thrust if r > l/(3√3).
Q6(b) · Dynamics and Statics · 15 marks≈ 11 min
Rectilinear Motion and Simple Harmonic Motion
A particle executes simple harmonic motion such that in two of its positions, velocities are u and v, and the two corresponding accelerations are f₁ and f₂. For what value(s) of k, the distance between the two positions is k(v² − u²)? Show also that the amplitude of the motion is
[1/(f₂² − f₁²)] [(u² − v²)(u²f₂² − v²f₁²)]^1/2
हिंदी में प्रश्न
सरल आवर्त गति में एक कण की दो स्थितियों में वेग u और v हैं तथा दो संगत त्वरण f₁ और f₂ हैं। k के किस मान/किन मानों के लिए दोनों स्थितियों के बीच की दूरी k(v² − u²) है? यह भी दर्शाइए कि गति का आयाम
[1/(f₂² − f₁²)] [(u² − v²)(u²f₂² − v²f₁²)]^1/2
है।
Practise this question Dynamics and Statics notes
Worked solution
Let the positions be x₁, x₂ (from the centre, with sign) with speeds u, v: u² = μ(a² − x₁²), v² = μ(a² − x₂²), and the accelerations have magnitudes f₁ = μ|x₁|, f₂ = μ|x₂|.
Subtracting, v² − u² = μ(x₁² − x₂²) = (f₁² − f₂²)/μ, so μ = (f₁² − f₂²)/(v² − u²).
Same side of the centre: distance = |x₁| − |x₂| = (f₁ − f₂)/μ = (v² − u²)/(f₁ + f₂), so k = 1/(f₁ + f₂).
Opposite sides: distance = |x₁| + |x₂| = (f₁ + f₂)/μ = (v² − u²)/(f₁ − f₂), so k = 1/(f₁ − f₂).
Amplitude: a² = x₁² + u²/μ = (f₁² + μu²)/μ², and with μ = (f₂² − f₁²)/(u² − v²), f₁² + μu² = (u²f₂² − v²f₁²)/(u² − v²).
Hence a² = (u² − v²)(u²f₂² − v²f₁²)/(f₂² − f₁²)², which is the stated result.
Answer: k = 1/(f₁ + f₂) if the positions are on the same side of the centre and k = 1/(f₁ − f₂) if they are on opposite sides; a = [(u² − v²)(u²f₂² − v²f₁²)]^1/2/(f₂² − f₁²).
Final answer: k = 1/(f₁ + f₂) if the positions are on the same side of the centre and k = 1/(f₁ − f₂) if they are on opposite sides; a = [(u² − v²)(u²f₂² − v²f₁²)]^1/2/(f₂² − f₁²).
Q6(c) · Ordinary Differential Equations · 20 marks≈ 14 min
Second Order Equations with Variable Coefficients
(i) Find the second solution of the differential equation xy″ + (x − 1)y′ − y = 0 using u(x) = −e^−x as one of the solutions. (10 marks)
(ii) Find the general solution of the differential equation x²y″ − 2xy′ + 2y = x³ sin x by the method of variation of parameters. (10 marks)
हिंदी में प्रश्न
(i) एक हल के रूप में u(x) = −e^−x का उपयोग करते हुए अवकल समीकरण xy″ + (x − 1)y′ − y = 0 का दूसरा हल ज्ञात कीजिए। (10 अंक)
(ii) प्राचल-विचरण विधि का उपयोग कर अवकल समीकरण x²y″ − 2xy′ + 2y = x³ sin x का व्यापक हल ज्ञात कीजिए। (10 अंक)
Practise this question Ordinary Differential Equations notes

Question 7 (50 marks)

Q7(a) · Ordinary Differential Equations · 15 marks≈ 11 min
Formulation and First Order, First Degree Equations
State uniqueness theorem for the existence of unique solution of the initial value problem dy/dx = f(x, y), y(x₀) = y₀ in the rectangular region R : |x − x₀| ≤ a, |y − y₀| ≤ b. Test the existence and uniqueness of the solution of the initial value problem dy/dx = 2√y, y(1) = 0, in a suitable rectangle R. If more than one solution exist, then find all the solutions.
हिंदी में प्रश्न
आयतीय क्षेत्र R : |x − x₀| ≤ a, |y − y₀| ≤ b में प्रारम्भिक मान समस्या dy/dx = f(x, y), y(x₀) = y₀ के अद्वितीय हल के अस्तित्व के लिए अद्वितीयता प्रमेय का कथन लिखिए। एक उपयुक्त आयत R में प्रारम्भिक मान समस्या dy/dx = 2√y, y(1) = 0 के हल के अस्तित्व और अद्वितीयता का परीक्षण कीजिए। यदि एक से अधिक हल मौजूद हैं, तो सभी हलों को ज्ञात कीजिए।
Practise this question Ordinary Differential Equations notes
Worked solution
Statement: as above (Picard's theorem). Here f(x, y) = 2√y, defined for y ≥ 0; take R: |x − 1| ≤ a, 0 ≤ y ≤ b. f is continuous on R with M = 2√b, so a solution exists on |x − 1| ≤ min(a, b/(2√b)) = min(a, √b/2).
Uniqueness is not guaranteed: |f(x, y) − f(x, 0)|/|y − 0| = 2/√y → ∞ as y → 0+, so no Lipschitz constant exists on any rectangle containing points with y = 0.
Solving: y ≡ 0 is a solution. For y > 0, dy/(2√y) = dx gives √y = x − c, that is y = (x − c)² with x ≥ c.
Joining the two smoothly: for every c ≥ 1, y = 0 for x ≤ c and y = (x − c)² for x > c is differentiable, satisfies y′ = 2√y and y(1) = 0. (A branch (x − c)² with x < c is not allowed, since there y′ < 0 while 2√y > 0.)
Answer: solutions exist but are not unique: y ≡ 0 and, for each c ≥ 1, y = 0 (x ≤ c), y = (x − c)² (x > c); in particular y = (x − 1)² for x ≥ 1.
Final answer: solutions exist but are not unique: y ≡ 0 and, for each c ≥ 1, y = 0 (x ≤ c), y = (x − c)² (x > c); in particular y = (x − 1)² for x ≥ 1.
Q7(b) · Dynamics and Statics · 15 marks≈ 11 min
Constrained Motion, Work and Energy
A heavy particle hanging vertically from a fixed point by a light inextensible string of length l starts to move with initial velocity u in a circle so as to make a complete revolution in a vertical plane. Show that the sum of tensions at the ends of any diameter is constant.
हिंदी में प्रश्न
लम्बाई l की एक हल्की अवितान्य डोरी द्वारा एक नियत बिन्दु से ऊर्ध्वाधर लटका हुआ एक भारी कण प्रारम्भिक वेग u के साथ एक वृत्त में घूमना शुरू करता है ताकि एक ऊर्ध्वाधर समतल में एक पूर्ण परिक्रमण कर सके। दर्शाइए कि किसी भी व्यास के सिरों पर तनावों का योग अचर है।
Practise this question Dynamics and Statics notes
Worked solution
With θ measured from the lowest point, v² = u² − 2gl(1 − cos θ), and T − mg cos θ = mv²/l gives T(θ) = (m/l)(u² − 2gl + 3gl cos θ).
A complete revolution needs u² ≥ 5gl, so the string stays taut and this formula holds for every θ.
The ends of a diameter are at θ and θ + π, and T(θ + π) = (m/l)(u² − 2gl − 3gl cos θ).
T(θ) + T(θ + π) = 2m(u² − 2gl)/l, independent of θ.
Answer: the sum of the tensions at the ends of any diameter is the constant 2m(u² − 2gl)/l.
Final answer: the sum of the tensions at the ends of any diameter is the constant 2m(u² − 2gl)/l.
Q7(c) · Vector Analysis · 20 marks≈ 14 min
Gauss and Stokes' Theorems and Green's Identities
State Stokes' theorem and verify it for the vector field F = xy î + yz ĵ + zx k̂ over the surface S, which is the upwardly oriented part of the cylinder z = 1 − x², for 0 ≤ x ≤ 1, −2 ≤ y ≤ 2.
हिंदी में प्रश्न
स्टोक्स प्रमेय का कथन लिखिए तथा इसको सदिश क्षेत्र F = xy î + yz ĵ + zx k̂ के लिए, पृष्ठ S पर जो कि बेलन z = 1 − x²; 0 ≤ x ≤ 1, −2 ≤ y ≤ 2 का उपरिमुखी अभिविन्यस्त भाग है, सत्यापित कीजिए।
Practise this question Vector Analysis notes

Question 8 (50 marks)

Q8(a) · Ordinary Differential Equations · 15 marks≈ 11 min
Laplace Transforms and Initial Value Problems
Using Laplace transform, solve the initial value problem
y″ + 2y′ + 5y = δ(t − 2), y(0) = 0, y′(0) = 0
where δ(t − 2) denotes the Dirac delta function.
हिंदी में प्रश्न
लाप्लास रूपान्तर का उपयोग करके प्रारम्भिक मान समस्या
y″ + 2y′ + 5y = δ(t − 2), y(0) = 0, y′(0) = 0
को हल कीजिए, जहाँ δ(t − 2) डिरैक डेल्टा फलन को दर्शाता है।
Practise this question Ordinary Differential Equations notes
Worked solution
L{δ(t − 2)} = e^−2s, so (s² + 2s + 5)Y = e^−2s and Y = e^−2s/((s + 1)² + 4).
L⁻¹{1/((s + 1)² + 4)} = (1/2)e^−t sin 2t; by the second shifting theorem y = (1/2)e^{−(t − 2)} sin 2(t − 2)·u(t − 2).
Answer: y = 0 for t < 2 and y = (1/2)e^{−(t − 2)} sin 2(t − 2) for t > 2; y is continuous at t = 2, while y′ jumps by 1 (the impulse).
Final answer: y = 0 for t < 2 and y = (1/2)e^{−(t − 2)} sin 2(t − 2) for t > 2; y is continuous at t = 2, while y′ jumps by 1 (the impulse).
Q8(b) · Vector Analysis · 15 marks≈ 11 min
Gauss and Stokes' Theorems and Green's Identities
Using Gauss divergence theorem, evaluate the integral
∬_S (y² î + xz³ ĵ + (z − 1)² k̂) · n̂ dS
over the region bounded by the cylinder x² + y² = 16 and the planes z = 1 and z = 5.
हिंदी में प्रश्न
गाऊस के अपसरण प्रमेय का उपयोग करते हुए बेलन x² + y² = 16 तथा समतलों z = 1 और z = 5 द्वारा परिबद्ध क्षेत्र पर समाकल
∬_S (y² î + xz³ ĵ + (z − 1)² k̂) · n̂ dS
का मान निकालिए।
Practise this question Vector Analysis notes
Worked solution
∇·F = 0 + 0 + 2(z − 1).
∭2(z − 1) dV = (cross-section area 16π) × ∫₁⁵2(z − 1) dz = 16π × 16.
Answer: 256π.
Final answer: 256π.
Q8(c) · Dynamics and Statics · 20 marks≈ 14 min
Central Orbits and Kepler's Laws
A particle moves with a central acceleration μ(3/r³ + d²/r⁵) being projected from a distance d at an angle 45° with a velocity equal to that in a circle at the same distance. Prove that the time it takes to reach the centre of force is (d²/√(2μ))(2 − π/2).
हिंदी में प्रश्न
d दूरी से एक कण को समान दूरी पर स्थित एक वृत्त में उसके वेग के बराबर वेग से 45° के कोण पर प्रक्षेपित करने पर वह केन्द्रीय त्वरण μ(3/r³ + d²/r⁵) के साथ गति करता है। सिद्ध कीजिए कि बल के केन्द्र तक इसके पहुँचने का समय (d²/√(2μ))(2 − π/2) है।
Practise this question Dynamics and Statics notes
Worked solution
Speed in a circle of radius d: v² = dP(d) = μ(3/d² + 1/d²) = 4μ/d²; projected at 45° to the radius vector, h = dv sin 45° = √(2μ), so h² = 2μ.
Energy integral: v² = C − 2∫P dr = C + μ(3/r² + d²/(2r⁴)); at r = d, 4μ/d² = C + 7μ/(2d²), so C = μ/(2d²).
With v² = 2μ[u² + (du/dθ)²]: (du/dθ)² = 1/(4d²) + u²/2 + d²u⁴/4 = (1 + d²u²)²/(4d²).
The particle is projected inwards, so u increases: dθ = 2d du/(1 + d²u²), and u → ∞ (the centre) after θ has increased by 2(π/2 − π/4) = π/2.
Time: dt = dθ/(hu²) = (2d/h) du/(u²(1 + d²u²)) = (2d/h)[1/u² − d²/(1 + d²u²)] du.
t = (2d/h)[−1/u − d tan⁻¹(du)] from u = 1/d to ∞ = (2d/h)(−dπ/2 + d + dπ/4) = (2d²/h)(1 − π/4).
Answer: with h = √(2μ), t = (2d²/√(2μ))(1 − π/4) = (d²/√(2μ))(2 − π/2).
Final answer: with h = √(2μ), t = (2d²/√(2μ))(1 − π/4) = (d²/√(2μ))(2 − π/2).

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