UPSC Mains 2023 Mathematics Optional Question Papers with Solutions

All 56 questions of the 2023 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 2023 Mathematics Paper I

Question 1 (50 marks)

Q1(a) · Linear Algebra · 10 marks≈ 7 min
Vector Spaces, Subspaces, Basis and Dimension
Let V₁ = (2, −1, 3, 2), V₂ = (−1, 1, 1, −3) and V₃ = (1, 1, 9, −5) be three vectors of the space ℝ⁴. Does (3, −1, 0, −1) ∈ span {V₁, V₂, V₃}? Justify your answer.
हिंदी में प्रश्न
मान लीजिए V₁ = (2, −1, 3, 2), V₂ = (−1, 1, 1, −3), V₃ = (1, 1, 9, −5) समष्टि ℝ⁴ के तीन सदिश हैं। क्या (3, −1, 0, −1) ∈ विस्तृति {V₁, V₂, V₃}? अपने उत्तर को तर्कसहित सिद्ध कीजिए।
Practise this question Linear Algebra notes
Worked solution
v = (3, −1, 0, −1) ∈ span ⇔ xV₁ + yV₂ + zV₃ = v is consistent. The augmented matrix with columns V₁, V₂, V₃ | v has rows (2, −1, 1 | 3), (−1, 1, 1 | −1), (3, 1, 9 | 0), (2, −3, −5 | −1).
R₁ ↔ R₂, then R₂ + 2R₁, R₃ + 3R₁, R₄ + 2R₁ give (−1, 1, 1 | −1), (0, 1, 3 | 1), (0, 4, 12 | −3), (0, −1, −3 | −3).
R₃ − 4R₂ and R₄ + R₂ give (0, 0, 0 | −7) and (0, 0, 0 | −2). So the coefficient matrix has rank 2 but the augmented matrix has rank 3, and the system is inconsistent.
(In fact V₃ = 2V₁ + 3V₂, so the span is the plane L{V₁, V₂}, which does not contain v.)
Answer: No: (3, −1, 0, −1) ∉ span{V₁, V₂, V₃}.
Final answer: No: (3, −1, 0, −1) ∉ span{V₁, V₂, V₃}.
Q1(b) · Linear Algebra · 10 marks≈ 7 min
Linear Transformations, Rank-Nullity and Matrix Representation
Find the rank and nullity of the linear transformation: T : ℝ³ → ℝ³ given by T(x, y, z) = (x + z, x + y + 2z, 2x + y + 3z)
हिंदी में प्रश्न
T(x, y, z) = (x + z, x + y + 2z, 2x + y + 3z) द्वारा दिए गए रैखिक रूपांतरण: T : ℝ³ → ℝ³ की कोटि तथा शून्यता ज्ञात कीजिए।
Practise this question Linear Algebra notes
Q1(c) · Calculus · 10 marks≈ 7 min
Mean Value Theorems, Taylor's Theorem and Indeterminate Forms
Find the values of p and q for which lim_x→0 [x(1 + p cos x) − q sin x]/x³ exists and equals 1.
हिंदी में प्रश्न
p तथा q के वो मान निकालिए जिसके लिए lim_x→0 [x(1 + p cos x) − q sin x]/x³ का अस्तित्व है एवं 1 के बराबर है।
Practise this question Calculus notes
Worked solution
With cos x = 1 − x²/2 + O(x⁴) and sin x = x − x³/6 + O(x⁵): x(1 + p cos x) − q sin x = (1 + p − q)x + (−p/2 + q/6)x³ + O(x⁵).
After dividing by x³, the limit is finite only if 1 + p − q = 0, and then it equals −p/2 + q/6.
So q = 1 + p and −p/2 + (1 + p)/6 = 1, i.e. −3p + 1 + p = 6, giving p = −5/2 and q = −3/2.
Check: x − (5/2)x cos x + (3/2) sin x = x − (5/2)x + (5/4)x³ + (3/2)x − (1/4)x³ + O(x⁵) = x³ + O(x⁵).
Answer: p = −5/2, q = −3/2.
Final answer: p = −5/2, q = −3/2.
Q1(d) · Calculus · 10 marks≈ 7 min
Riemann Integral, Indefinite Integrals and Improper Integrals
Examine the convergence of the integral ∫₀¹ (log x)/(1 + x) dx
हिंदी में प्रश्न
समाकल ∫₀¹ (log x)/(1 + x) dx की अभिसारिता का परीक्षण कीजिए।
Practise this question Calculus notes
Worked solution
The integrand is continuous on (0, 1] and log x → −∞ as x → 0+, so the integral is improper at 0. The integrand is ≤ 0, so test the non-negative function g(x) = −log x/(1 + x).
μ-test with μ = 1/2: x^1/2g(x) = −x^1/2 log x/(1 + x) → 0 as x → 0+, which is finite with μ < 1, so the integral converges. (Comparison: g(x) ≤ −log x and ∫₀¹ −log x dx = 1.)
Its value: log x/(1 + x) = ∑(−1)ⁿxⁿ log x and ∫₀¹ xⁿ log x dx = −1/(n + 1)², so the integral is −(1 − 1/4 + 1/9 − ...) = −π²/12.
Answer: The integral converges (its value is −π²/12).
Final answer: The integral converges (its value is −π²/12).
Q1(e) · Analytic Geometry · 10 marks≈ 7 min
Coordinates in Three Dimensions and the Plane
A variable plane which is at a constant distance 3p from the origin O cuts the axes in the points A, B, C respectively. Show that the locus of the centroid of the tetrahedron OABC is 9(1/x² + 1/y² + 1/z²) = 16/p².
हिंदी में प्रश्न
एक चर समतल, जो कि मूल-बिन्दु O से अचर दूरी 3p पर है, अक्षों को क्रमशः बिन्दुओं A, B, C पर काटता है। दर्शाइए कि चतुष्फलक OABC के केन्द्रक का बिन्दुपथ 9(1/x² + 1/y² + 1/z²) = 16/p² है।
Practise this question Analytic Geometry notes

Question 2 (50 marks)

Q2(a) · Linear Algebra · 15 marks≈ 11 min
Linear Transformations, Rank-Nullity and Matrix Representation
If the matrix of a linear transformation T : ℝ³ → ℝ³ relative to the basis {(1, 0, 0), (0, 1, 0), (0, 0, 1)} is [[1, 1, 2], [−1, 2, 1], [0, 1, 3]], then find the matrix of T relative to the basis {(1, 1, 1), (0, 1, 1), (0, 0, 1)}.
हिंदी में प्रश्न
यदि आधार {(1, 0, 0), (0, 1, 0), (0, 0, 1)} के सापेक्ष रैखिक रूपांतरण T : ℝ³ → ℝ³ का आव्यूह [[1, 1, 2], [−1, 2, 1], [0, 1, 3]] है, तब आधार {(1, 1, 1), (0, 1, 1), (0, 0, 1)} के सापेक्ष T का आव्यूह ज्ञात कीजिए।
Practise this question Linear Algebra notes
Q2(b) · Calculus · 15 marks≈ 11 min
Double and Triple Integrals; Areas, Surfaces and Volumes
Evaluate the triple integral which gives the volume of the solid enclosed between the two paraboloids Z = 5(x² + y²) and Z = 6 − 7x² − y².
हिंदी में प्रश्न
दो परवलयजों Z = 5(x² + y²) और Z = 6 − 7x² − y² के बीच घिरे ठोस के आयतन को दर्शाने वाले त्रिशः समाकल का मान निकालिए।
Practise this question Calculus notes
Worked solution
The surfaces meet where 5(x² + y²) = 6 − 7x² − y², i.e. 2x² + y² = 1. The solid lies over the ellipse E: 2x² + y² ≤ 1, between z = 5(x² + y²) below and z = 6 − 7x² − y² above.
V = ∬_E ∫_5(x²+y²)^{6−7x²−y²} dz dx dy = ∬_E (6 − 12x² − 6y²) dx dy = 6∬_E (1 − 2x² − y²) dx dy.
Put x = r cos θ/√2, y = r sin θ (Jacobian r/√2): V = 6∫₀^2π∫₀¹ (1 − r²)(r/√2) dr dθ = (6/√2)(2π)(1/4).
Answer: V = 3π/√2 = (3√2/2)π ≈ 6.66 cubic units.
Final answer: V = 3π/√2 = (3√2/2)π ≈ 6.66 cubic units.
Q2(c) · Analytic Geometry · 20 marks≈ 14 min
Ellipsoid, Hyperboloids and Paraboloids
(i) Show that the equation 2x² + 3y² − 8x + 6y − 12z + 11 = 0 represents an elliptic paraboloid. Also find its principal axis and principal planes. (10)
(ii) The plane x/a + y/b + z/c = 1 meets the coordinate axes in A, B, C respectively. Prove that the equation of the cone generated by the lines drawn from the origin O to meet the circle ABC is yz(b/c + c/b) + zx(c/a + a/c) + xy(b/a + a/b) = 0. (10)
हिंदी में प्रश्न
(i) दर्शाइए कि समीकरण 2x² + 3y² − 8x + 6y − 12z + 11 = 0 एक दीर्घवृत्तीय परवलयज प्रदर्शित करता है। साथ ही मुख्य अक्ष और मुख्य समतलों को भी ज्ञात कीजिए। (10)
(ii) समतल x/a + y/b + z/c = 1, निर्देशांक अक्षों को क्रमशः A, B, C में मिलता है। सिद्ध कीजिये कि मूल बिन्दु O से वृत्त ABC को मिलाने वाली रेखाओं द्वारा जनित शंकु का समीकरण yz(b/c + c/b) + zx(c/a + a/c) + xy(b/a + a/b) = 0 है। (10)
Practise this question Analytic Geometry notes
Worked solution
(i)
Completing squares: 2(x − 2)² − 8 + 3(y + 1)² − 3 − 12z + 11 = 0, that is 2(x − 2)² + 3(y + 1)² = 12z.
With X = x − 2, Y = y + 1, Z = z: X²/6 + Y²/4 = Z, and both coefficients are positive, so it is an elliptic paraboloid with vertex (2, −1, 0).
Answer: principal axis x = 2, y = −1 (parallel to the z-axis); principal planes x = 2 and y = −1.
(ii)
The circle ABC is x² + y² + z² − ax − by − cz = 0, x/a + y/b + z/c = 1 (sphere OABC with the plane).
Make the sphere equation homogeneous with the plane: x² + y² + z² − (ax + by + cz)(x/a + y/b + z/c) = 0.
The product contains x² + y² + z², which cancels, plus xy(a/b + b/a) + yz(b/c + c/b) + zx(c/a + a/c).
Answer: the cone is yz(b/c + c/b) + zx(c/a + a/c) + xy(b/a + a/b) = 0; hence proved.
Final answer: (i) principal axis x = 2, y = −1 (parallel to the z-axis); principal planes x = 2 and y = −1; (ii) the cone is yz(b/c + c/b) + zx(c/a + a/c) + xy(b/a + a/b) = 0; hence proved

Question 3 (50 marks)

Q3(a) · Linear Algebra · 20 marks≈ 14 min
Eigenvalues, Eigenvectors and the Cayley-Hamilton Theorem
Let A = [[1, 0, 0], [1, 0, 1], [0, 1, 0]]
(i) Verify the Cayley-Hamilton theorem for the matrix A.
(ii) Show that Aⁿ = Aⁿ⁻² + A² − I for n ≥ 3, where I is the identity matrix of order 3. Hence, find A⁴⁰. (10+10)
हिंदी में प्रश्न
दिया गया है A = [[1, 0, 0], [1, 0, 1], [0, 1, 0]]
(i) आव्यूह A के लिये कैले-हैमिल्टन प्रमेय को सत्यापित कीजिए।
(ii) दर्शाइए कि n ≥ 3 के लिये Aⁿ = Aⁿ⁻² + A² − I; जहाँ I कोटि 3 का तत्समक आव्यूह है। अतएव A⁴⁰ ज्ञात कीजिए। (10+10)
Practise this question Linear Algebra notes
Worked solution
(i) λI − A = [[λ − 1, 0, 0], [−1, λ, −1], [0, −1, λ]]; expanding along the first row, det(λI − A) = (λ − 1)(λ² − 1) = λ³ − λ² − λ + 1.
A² = [[1, 0, 0], [1, 1, 0], [1, 0, 1]] and A³ = A·A² = [[1, 0, 0], [2, 0, 1], [1, 1, 0]]. Entry by entry, A³ − A² − A + I = 0, which verifies the theorem.
(ii) The theorem gives A³ = A² + A − I, which is the claim for n = 3. If Aⁿ = Aⁿ⁻² + A² − I, then Aⁿ⁺¹ = Aⁿ⁻¹ + A³ − A = Aⁿ⁻¹ + (A² + A − I) − A = Aⁿ⁻¹ + A² − I; by induction the relation holds for all n ≥ 3.
For even powers, A^2k = A^2k−2 + (A² − I) = ... = A² + (k − 1)(A² − I) = kA² − (k − 1)I = [[1, 0, 0], [k, 1, 0], [k, 0, 1]].
Answer: A⁴⁰ = 20A² − 19I = [[1, 0, 0], [20, 1, 0], [20, 0, 1]].
Final answer: A⁴⁰ = 20A² − 19I = [[1, 0, 0], [20, 1, 0], [20, 0, 1]].
Q3(b) · Calculus · 15 marks≈ 11 min
Functions of Several Variables: Partial Derivatives, Extrema, Lagrange Multipliers and Jacobians
Justify whether (0, 0) is an extreme point for the function f(x, y) = 2x⁴ − 3x²y + y².
हिंदी में प्रश्न
तर्क सहित दर्शाइये कि (0, 0), फलन f(x, y) = 2x⁴ − 3x²y + y² का चरम-बिन्दु है अथवा नहीं।
Practise this question Calculus notes
Worked solution
f_x = 8x³ − 6xy and f_y = −3x² + 2y vanish at (0, 0). There r = 24x² − 6y = 0, s = −6x = 0, t = 2, so rt − s² = 0 and the test is inconclusive.
Factorise: f(x, y) = (2x² − y)(x² − y), and f(0, 0) = 0.
On the y-axis, f(0, y) = y² > 0 (y ≠ 0). On the parabola y = (3/2)x², which lies between y = x² and y = 2x², f = (x²/2)(−x²/2) = −x⁴/4 < 0 (x ≠ 0).
So every neighbourhood of (0, 0) has points with f > 0 and points with f < 0. (Along every straight line through the origin f has a minimum at O, which shows that testing lines alone is not enough.)
Answer: (0, 0) is a stationary point but not an extreme point of f.
Final answer: (0, 0) is a stationary point but not an extreme point of f.
Q3(c) · Analytic Geometry · 15 marks≈ 11 min
The Sphere
Find the equation of the sphere through the circle x² + y² + z² − 4x − 6y + 2z − 16 = 0; 3x + y + 3z − 4 = 0 in the following two cases.
(i) the point (1, 0, −3) lies on the sphere.
(ii) the given circle is a great circle of the sphere.
हिंदी में प्रश्न
वृत्त x² + y² + z² − 4x − 6y + 2z − 16 = 0; 3x + y + 3z − 4 = 0 से होकर गुजरने वाले गोले का समीकरण निम्न दो स्थितियों में ज्ञात कीजिए।
(i) बिन्दु (1, 0, −3) गोले पर हो।
(ii) दिया गया वृत्त गोले का एक वृहत् वृत्त हो।
Practise this question Analytic Geometry notes
Worked solution
Let S = x² + y² + z² − 4x − 6y + 2z − 16 and P = 3x + y + 3z − 4; every sphere through the circle is S + λP = 0.
(i) At (1, 0, −3): S = 1 + 9 − 4 − 6 − 16 = −16 and P = 3 − 9 − 4 = −10, so λ = −8/5 and 5S − 8P = 0: 5(x² + y² + z²) − 44x − 38y − 14z − 48 = 0.
(ii) The centre of S + λP = 0 is (2 − 3λ/2, 3 − λ/2, −1 − 3λ/2). For a great circle it lies on P = 0: 3(2 − 3λ/2) + (3 − λ/2) + 3(−1 − 3λ/2) − 4 = 2 − 19λ/2 = 0, so λ = 4/19.
Answer: (i) 5(x² + y² + z²) − 44x − 38y − 14z − 48 = 0; (ii) 19(x² + y² + z²) − 64x − 110y + 50z − 320 = 0.
Final answer: (i) 5(x² + y² + z²) − 44x − 38y − 14z − 48 = 0; (ii) 19(x² + y² + z²) − 64x − 110y + 50z − 320 = 0.

Question 4 (50 marks)

Q4(a) · Linear Algebra · 15 marks≈ 11 min
Algebra of Matrices, Echelon Form, Rank, Inverse, Congruence and Similarity
Find the rank of the matrix A = [[1, 2, −1, 0], [−1, 3, 0, −4], [2, 1, 3, −2], [1, 1, 1, −1]] by reducing it to row-reduced echelon form.
हिंदी में प्रश्न
आव्यूह A = [[1, 2, −1, 0], [−1, 3, 0, −4], [2, 1, 3, −2], [1, 1, 1, −1]] का पंक्ति समानीत सोपानक रूप में समानयन करके उसकी कोटि ज्ञात कीजिए।
Practise this question Linear Algebra notes
Q4(b) · Calculus · 20 marks≈ 14 min
Maxima and Minima, Asymptotes and Curve Tracing
Trace the curve y²(x² − 1) = 2x − 1.
हिंदी में प्रश्न
वक्र y²(x² − 1) = 2x − 1 को अनुरेखित कीजिए।
Practise this question Calculus notes
Worked solution
y² = g(x) = (2x − 1)/(x² − 1). Symmetry: only y² occurs, so the curve is symmetric about the x-axis; it misses the origin, and g(0) = 1 gives the points (0, ±1).
Region: x < −1: g < 0, no curve; −1 < x ≤ 1/2: g ≥ 0, curve exists; 1/2 < x < 1: g < 0, no curve; x > 1: g > 0, curve exists.
Asymptotes: the coefficient of y² is x² − 1, giving x = 1 and x = −1; the coefficient of x² is y², giving y = 0 (y² ≈ 2/x → 0 as x → +∞; the curve does not exist for x < −1).
Monotonicity: g′(x) = −2(x² − x + 1)/(x² − 1)² < 0, so y² decreases on each interval where the curve exists.
Shape: on (−1, 1/2], y² falls from +∞ (x → −1+) through 1 (x = 0) to 0 at x = 1/2, where the curve crosses the x-axis vertically: one branch, shaped like ')' , whose two ends go up and down along x = −1. On (1, ∞), y² falls from +∞ to 0: two arms come down from the asymptote x = 1 and approach the x-axis from above and below.
Answer: Symmetric about the x-axis; exists only for −1 < x ≤ 1/2 and x > 1; asymptotes x = ±1 and y = 0; passes through (0, ±1) and (1/2, 0).
Final answer: Symmetric about the x-axis; exists only for −1 < x ≤ 1/2 and x > 1; asymptotes x = ±1 and y = 0; passes through (0, ±1) and (1/2, 0).
Q4(c) · Analytic Geometry · 15 marks≈ 11 min
Straight Lines and Shortest Distance between Skew Lines
Prove that the locus of a line which meets the lines y = mx, z = c; y = −mx, z = −c and the circle x² + y² = a², z = 0 is c²m²(cy − mzx)² + c²(yz − cmx)² = a²m²(z² − c²)².
हिंदी में प्रश्न
सिद्ध कीजिए कि रेखाओं y = mx, z = c; y = −mx, z = −c और वृत्त x² + y² = a², z = 0 से मिलने वाली रेखा का बिन्दु-पथ c²m²(cy − mzx)² + c²(yz − cmx)² = a²m²(z² − c²)² है।
Practise this question Analytic Geometry notes
Worked solution
A line meeting the first two lines is (y − mx) + λ(z − c) = 0, (y + mx) + μ(z + c) = 0. It meets z = 0 where y − mx = λc and y + mx = −μc, that is at x = −c(λ + μ)/(2m), y = c(λ − μ)/2.
This point lies on the circle: c²(λ + μ)²/(4m²) + c²(λ − μ)²/4 = a².
With λ = −(y − mx)/(z − c) and μ = −(y + mx)/(z + c): λ + μ = −2(yz − cmx)/(z² − c²) and λ − μ = −2(cy − mzx)/(z² − c²).
Substituting and multiplying by m²(z² − c²)²: c²(yz − cmx)² + c²m²(cy − mzx)² = a²m²(z² − c²)².
Answer: the locus is c²m²(cy − mzx)² + c²(yz − cmx)² = a²m²(z² − c²)², as required.
Final answer: the locus is c²m²(cy − mzx)² + c²(yz − cmx)² = a²m²(z² − c²)², as required.

Question 5 (50 marks)

Q5(a) · Ordinary Differential Equations · 10 marks≈ 7 min
Formulation and First Order, First Degree Equations
Obtain the solution of the initial-value problem dy/dx − 2xy = 2, y(0) = 1 in the form y = e^x²[1 + √π erf(x)].
हिंदी में प्रश्न
प्रारम्भिक-मान समस्या: dy/dx − 2xy = 2, y(0) = 1 का हल y = e^x²[1 + √π erf(x)] के रूप में प्राप्त कीजिए।
Practise this question Ordinary Differential Equations notes
Worked solution
IF = e^{∫−2x dx} = e^−x², so d(ye^−x²)/dx = 2e^−x².
Integrating from 0 to x and using y(0) = 1: ye^−x² − 1 = 2∫₀^x e^−t² dt = √π erf(x), since erf(x) = (2/√π)∫₀^x e^−t² dt.
Answer: y = e^x²[1 + √π erf(x)].
Final answer: y = e^x²[1 + √π erf(x)].
Q5(b) · Ordinary Differential Equations · 10 marks≈ 7 min
Laplace Transforms and Initial Value Problems
Given that L{f(t); p} = F(p). Show that ∫₀^∞ (f(t)/t) dt = ∫₀^∞ F(x) dx. Hence evaluate the integral ∫₀^∞ (e^−t − e^−3t)/t dt.
हिंदी में प्रश्न
दिया गया है L{f(t); p} = F(p). दर्शाइए कि ∫₀^∞ (f(t)/t) dt = ∫₀^∞ F(x) dx. अतः समाकल ∫₀^∞ (e^−t − e^−3t)/t dt का मान ज्ञात कीजिए।
Practise this question Ordinary Differential Equations notes
Worked solution
By the previous result L{f(t)/t; p} = ∫_p^∞ F(x) dx. Let p → 0+: ∫₀^∞ (f(t)/t) dt = ∫₀^∞ F(x) dx, whenever both integrals converge.
For f = e^−t − e^−3t: F(x) = 1/(x + 1) − 1/(x + 3), so the integral is [log((x + 1)/(x + 3))]₀^∞ = 0 − log(1/3).
Answer: ∫₀^∞ (e^−t − e^−3t)/t dt = log 3.
Final answer: ∫₀^∞ (e^−t − e^−3t)/t dt = log 3.
Q5(c) · Dynamics and Statics · 10 marks≈ 7 min
Equilibrium, Friction and the Common Catenary
A cylinder of radius 'a' touches a vertical wall along a generating line. Axis of the cylinder is fixed horizontally. A uniform flat beam of length 'l' and weight 'W' rests with its extremities in contact with the wall and the cylinder, making an angle of 45° with the vertical. If frictional forces are neglected, then show that a/l = (√5 + 5)/(4√2). Also, find the reactions of the cylinder and wall.
हिंदी में प्रश्न
अर्द्धव्यास 'a' का एक बेलन (सिलिंडर) एक जनक रेखा के अनुदिश एक ऊर्ध्वाधर दीवार को स्पर्श किया हुआ है। बेलन का अक्ष क्षैतिजतः स्थिर है। लम्बाई 'l' तथा भार 'W' का एक एक समान समतल दंड ऊर्ध्वाधर से 45° का कोण बनाते हुए अपने सिरों को दीवार के सहारे तथा बेलन पर टिकाए हैं। अगर घर्षण बल नगण्य हैं, तब दर्शाइए कि a/l = (√5 + 5)/(4√2). दीवार और बेलन की प्रतिक्रियायें भी ज्ञात कीजिए।
Practise this question Dynamics and Statics notes
Worked solution
Take the wall as x = 0 and the cylinder's axis at C(a, 0); the smooth wall's reaction S at the upper end A is horizontal, and the cylinder's reaction R at the lower end B is along the normal CB.
Write L = l/√2 for the horizontal and vertical projections of the beam; B = (a − a cos φ, a sin φ) lies at horizontal distance L from the wall, so a(1 − cos φ) = L.
The three forces are concurrent: S (the line y = y_B + L) and W (the line x = L/2) meet at K(L/2, y_B + L), which must lie on CB, the line y = (a − x) tan φ.
So (a − L/2) tan φ = a sin φ + L; with k = L/a and cos φ = 1 − k this gives (k/2)/(1 − k) = k/sin φ, so sin φ = 2(1 − k) = 2 cos φ.
Then cos²φ = 1/5, k = 1 − 1/√5 and a/L = √5/(√5 − 1) = (5 + √5)/4, so a/l = (5 + √5)/(4√2).
Resolving: R sin φ = W and S = R cos φ, with sin φ = 2/√5 and cos φ = 1/√5.
Answer: a/l = (√5 + 5)/(4√2); the cylinder's reaction is (√5/2)W along the normal and the wall's reaction is W/2.
Final answer: a/l = (√5 + 5)/(4√2); the cylinder's reaction is (√5/2)W along the normal and the wall's reaction is W/2.
Q5(d) · Dynamics and Statics · 10 marks≈ 7 min
Rectilinear Motion and Simple Harmonic Motion
A particle is moving under Simple Harmonic Motion of period T about a centre O. It passes through the point P with velocity v along the direction OP and OP = p. Find the time that elapses before the particle returns to the point P. What will be the value of p when the elapsed time is T/2?
हिंदी में प्रश्न
कोई कण केन्द्र 'O' के सापेक्ष आवर्त काल T के साथ सरल आवर्त गति में गतिशील है। कण बिन्दु P से OP के अनुदिश दिशा में v वेग से गुजरता है तथा OP = p है। कण का बिन्दु P पर पुनः लौटने में लगा समय ज्ञात कीजिए। यदि लगा समय T/2 हो, तो p का मान क्या होगा?
Practise this question Dynamics and Statics notes
Worked solution
With n = 2π/T, v² = n²(a² − p²) gives the amplitude a = √(p² + v²T²/(4π²)).
As in Example 7, the particle goes out to the extreme point and back, taking time t = (2/n) cos⁻¹(p/a) = (T/π) cos⁻¹(p/a).
The angle cos⁻¹(p/a) has cosine p/a and sine vT/(2πa), so t = (T/π) tan⁻¹(vT/(2πp)).
t = T/2 requires cos⁻¹(p/a) = π/2, that is p = 0.
Answer: the time is (T/π) tan⁻¹(vT/(2πp)); it equals T/2 only when p = 0, that is when P is the centre O itself.
Final answer: the time is (T/π) tan⁻¹(vT/(2πp)); it equals T/2 only when p = 0, that is when P is the centre O itself.
Q5(e) · Vector Analysis · 10 marks≈ 7 min
Scalar and Vector Fields; Differentiation of Vector Functions
If a = sin θ î + cos θ ĵ + θ k̂, b = cos θ î − sin θ ĵ − 3k̂, c = 2î + 3ĵ − 3k̂, then find the values of the derivative of the vector function a × (b × c) w.r.t. θ at θ = π/2 and θ = π.
हिंदी में प्रश्न
यदि a = sin θ î + cos θ ĵ + θ k̂, b = cos θ î − sin θ ĵ − 3k̂, c = 2î + 3ĵ − 3k̂, तो सदिश फलन a × (b × c) के θ के सापेक्ष अवकलज के मान, θ = π/2 और θ = π पर ज्ञात कीजिए।
Practise this question Vector Analysis notes
Worked solution
Expand first: a × (b × c) = (a·c)b − (a·b)c, with a·c = 2 sin θ + 3 cos θ − 3θ and a·b = sin θ cos θ − sin θ cos θ − 3θ = −3θ.
So F(θ) = (2 sin θ + 3 cos θ − 3θ)b + 3θc and dF/dθ = (2 cos θ − 3 sin θ − 3)b + (2 sin θ + 3 cos θ − 3θ)b′ + 3c, where b′ = −sin θ i − cos θ j.
At θ = π/2: b = −j − 3k, b′ = −i and the coefficients are −6 and 2 − 3π/2, so dF/dθ = (6j + 18k) − (2 − 3π/2)i + (6i + 9j − 9k) = (4 + 3π/2)i + 15j + 9k.
At θ = π: b = −i − 3k, b′ = j and the coefficients are −5 and −3 − 3π, so dF/dθ = (5i + 15k) − (3 + 3π)j + (6i + 9j − 9k) = 11i + (6 − 3π)j + 6k.
Answer: (4 + 3π/2)i + 15j + 9k at θ = π/2 and 11i + (6 − 3π)j + 6k at θ = π.
Final answer: (4 + 3π/2)i + 15j + 9k at θ = π/2 and 11i + (6 − 3π)j + 6k at θ = π.

Question 6 (50 marks)

Q6(a) · Ordinary Differential Equations · 15 marks≈ 11 min
Linear Equations with Constant Coefficients
Solve the differential equation: d³y/dx³ − 3 d²y/dx² + 4 dy/dx − 2y = e^x + cos x.
हिंदी में प्रश्न
अवकल समीकरण: d³y/dx³ − 3 d²y/dx² + 4 dy/dx − 2y = e^x + cos x का हल कीजिए।
Practise this question Ordinary Differential Equations notes
Q6(b) · Dynamics and Statics · 15 marks≈ 11 min
Motion in a Plane and Projectiles
When a particle is projected from a point O₁ on the sea level with a velocity v and angle of projection θ with the horizon in a vertical plane, its horizontal range is R₁. If it is further projected from a point O₂, which is vertically above O₁ at a height h in the same vertical plane, with the same velocity v and same angle θ with the horizon, its horizontal range is R₂. Prove that R₂ > R₁ and (R₂ − R₁) : R₁ is equal to (1/2){√(1 + 2gh/(v² sin² θ)) − 1} : 1
हिंदी में प्रश्न
एक कण को समुद्र तल पर बिन्दु O₁ से वेग v तथा क्षैतिज से प्रक्षेप कोण θ पर ऊर्ध्वाधर तल में प्रक्षेपित किया जाता है तो क्षैतिज परास R₁ है। यदि इसको पुनः बिन्दु O₂, जो उसी ऊर्ध्वाधर तल में O₁ के ऊर्ध्वाधरतः h ऊँचाई पर है, से उसी वेग v तथा क्षैतिज से समान कोण θ पर प्रक्षेपित किया जाता है तो क्षैतिज परास R₂ है। सिद्ध कीजिए R₂ > R₁ तथा (R₂ − R₁) : R₁ = (1/2){√(1 + 2gh/(v² sin² θ)) − 1} : 1
Practise this question Dynamics and Statics notes
Worked solution
From O₁: R₁ = v² sin 2θ/g = 2v² sin θ cos θ/g.
From O₂ the particle lands when its height is −h: −h = vt sin θ − gt²/2, whose positive root is t = [v sin θ + √(v²sin²θ + 2gh)]/g.
R₂ = vt cos θ = (v² sin θ cos θ/g)[1 + √(1 + 2gh/(v²sin²θ))].
So R₂/R₁ = (1/2)[1 + √(1 + 2gh/(v²sin²θ))] > 1 since h > 0, that is R₂ > R₁.
Subtracting 1: (R₂ − R₁)/R₁ = (1/2)[√(1 + 2gh/(v²sin²θ)) − 1].
Answer: R₂ > R₁ and (R₂ − R₁) : R₁ = (1/2){√(1 + 2gh/(v²sin²θ)) − 1} : 1.
Final answer: R₂ > R₁ and (R₂ − R₁) : R₁ = (1/2){√(1 + 2gh/(v²sin²θ)) − 1} : 1.
Q6(c) · Vector Analysis · 20 marks≈ 14 min
Gauss and Stokes' Theorems and Green's Identities
Evaluate the integral ∬_S (3y²z² î + 4z²x² ĵ + z²y² k̂) · n̂ dS, where S is the upper part of the surface 4x² + 4y² + 4z² = 1 above the plane z = 0 and bounded by the xy-plane. Hence, verify Gauss-Divergence theorem.
हिंदी में प्रश्न
समाकल ∬_S (3y²z² î + 4z²x² ĵ + z²y² k̂) · n̂ dS का मान ज्ञात कीजिए; जहाँ S समतल z = 0 के ऊपर पृष्ठ 4x² + 4y² + 4z² = 1 का ऊपरी भाग है और xy-समतल द्वारा परिबद्ध है। अतैव गॉस-अपसरण प्रमेय को सत्यापित कीजिए।
Practise this question Vector Analysis notes
Worked solution
S is the upper hemisphere of radius R = 1/2: n = r/R and dS = R² sin θ dθ dφ.
F·n = (3xy²z² + 4x²yz² + y²z³)/R; the first two terms are odd in x and in y and integrate to 0.
∬y²z³/R dS = R⁶∫₀^2πsin²φ dφ ∫₀^π/2cos³θ sin³θ dθ = R⁶π/12 = π/768.
Close S by the disc z = 0, where F = 0 (every component contains z²), so the disc adds no flux.
∇·F = 2y²z, and ∭2y²z dV = 2∫₀^Rr⁵ dr ∫₀^π/2cos θ sin³θ dθ ∫₀^2πsin²φ dφ = 2(R⁶/6)(1/4)π = π/768.
Answer: the surface integral is π/768, equal to the volume integral, which verifies the theorem.
Final answer: the surface integral is π/768, equal to the volume integral, which verifies the theorem.

Question 7 (50 marks)

Q7(a) · Ordinary Differential Equations · 20 marks≈ 14 min
First Order Higher Degree Equations, Clairaut's Equation and Singular Solutions
(i) Find the solution of the differential equation: dy/dx = −(2xy³ + 2)/(3x²y² + 8e^4y) (10)
(ii) Reduce the equation x²p² + y(2x + y)p + y² = 0 to Clairaut's form by the substitution y = u and xy = v. Hence solve the equation and show that y + 4x = 0 is a singular solution of the differential equation. (10)
हिंदी में प्रश्न
(i) अवकल समीकरण: dy/dx = −(2xy³ + 2)/(3x²y² + 8e^4y) का हल ज्ञात कीजिए। (10)
(ii) समीकरण x²p² + y(2x + y)p + y² = 0 का प्रतिस्थापन y = u और xy = v द्वारा क्लेरो रूप में समानयन कीजिए। अतः समीकरण का हल निकालिए और दर्शाइए कि y + 4x = 0 अवकल समीकरण का एक विचित्र हल है। (10)
Practise this question Ordinary Differential Equations notes
Q7(b) · Dynamics and Statics · 15 marks≈ 11 min
Equilibrium, Friction and the Common Catenary
A solid hemisphere is supported by a string fixed to a point on its rim and to a point on a smooth vertical wall with which the curved surface is in contact. If θ is the angle of inclination of the string with vertical and φ is the angle of inclination of the plane base of the hemisphere to the vertical, then find the value of (tan φ − tan θ).
हिंदी में प्रश्न
एक ठोस अर्द्ध-गोलक एक डोरी द्वारा, जिसका एक सिरा एक चिकनी ऊर्ध्वाधर दीवार पर एक बिन्दु से और दूसरा सिरा अर्द्धगोलक के किनारे (रिम) पर स्थित एक बिन्दु से बंधा है, ऊर्ध्वाधर दीवार के सहारे टिका है। ठोस अर्द्धगोलक का वक्रित पृष्ठ दीवार को स्पर्श करता है। अगर ऊर्ध्वाधर के साथ डोरी का आनति कोण θ है और अर्द्धगोलक के समतल आधार (बेस) का आनति कोण φ है तो (tan φ − tan θ) का मान ज्ञात कीजिए।
Practise this question Dynamics and Statics notes
Worked solution
Let O be the centre of the plane face, a the radius and G the centre of gravity, on the axis with OG = 3a/8.
The wall is smooth and its reaction is along the normal to the sphere at the contact, so it is the horizontal line through O.
The three forces are concurrent, so the line of the string passes through K, the point at the level of O vertically above G; OG makes angle φ with the horizontal (the base is at φ to the vertical), so K is (3a/8) cos φ from O towards the wall.
The string is tied at the upper rim point R, which is a sin φ nearer the wall than O and a cos φ above it.
Along the string from K to R the rise is a cos φ and the horizontal shift is a sin φ − (3a/8) cos φ, so tan θ = tan φ − 3/8.
Answer: tan φ − tan θ = 3/8.
Final answer: tan φ − tan θ = 3/8.
Q7(c) · Vector Analysis · 15 marks≈ 11 min
Curves in Space: Curvature, Torsion and Serret-Frenet Formulae
If the tangent to a curve makes a constant angle θ with a fixed line, then prove that the ratio of radius of torsion to radius of curvature is proportional to tan θ. Further prove that if this ratio is constant, then the tangent makes a constant angle with a fixed direction.
हिंदी में प्रश्न
अगर एक वक्र की स्पर्श रेखा एक नियत रेखा के साथ एक स्थिर कोण θ बनाती है तो सिद्ध कीजिए कि वक्रता की त्रिज्या के साथ व्यावर्तन त्रिज्या का अनुपात tan θ के समानुपाती है। और आगे सिद्ध कीजिए कि अगर यह अनुपात एक स्थिरांक है, तो स्पर्श रेखा एक नियत दिशा के साथ एक स्थिर कोण बनाती है।
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
Solve the following initial value problem by using Laplace transform technique: d²y/dt² − 4 dy/dt + 3y(t) = f(t), y(0) = 1, y′(0) = 0 and f(t) is a given function of t.
हिंदी में प्रश्न
लाप्लास रूपान्तर प्रविधि का उपयोग कर निम्नलिखित प्रारम्भिक मान समस्या को हल कीजिए। d²y/dt² − 4 dy/dt + 3y(t) = f(t), y(0) = 1, y′(0) = 0 और f(t), t का एक दिया गया फलन है।
Practise this question Ordinary Differential Equations notes
Worked solution
Transforming with y(0) = 1, y′(0) = 0: (s² − 4s + 3)Y − s + 4 = F(s), so Y = (s − 4)/((s − 1)(s − 3)) + F(s)/((s − 1)(s − 3)).
(s − 4)/((s − 1)(s − 3)) = (3/2)/(s − 1) − (1/2)/(s − 3), and 1/((s − 1)(s − 3)) = L{(e^3t − e^t)/2}.
By the convolution theorem the second term inverts to (1/2)∫₀^t f(u)[e^{3(t − u)} − e^{t − u}] du.
Answer: y(t) = (3/2)e^t − (1/2)e^3t + (1/2)∫₀^t f(u)[e^{3(t − u)} − e^{t − u}] du.
Final answer: y(t) = (3/2)e^t − (1/2)e^3t + (1/2)∫₀^t f(u)[e^{3(t − u)} − e^{t − u}] du.
Q8(b) · Dynamics and Statics · 20 marks≈ 14 min
Central Orbits and Kepler's Laws
A particle is projected from an apse at a distance √c from the centre of force with a velocity √((2λ/3)c³) and is moving with central acceleration λ(r⁵ − c²r). Find the path of motion of this particle. Will that be the curve x⁴ + y⁴ = c²?
हिंदी में प्रश्न
एक कण, बल-केन्द्र से √c दूरी पर स्थित एक स्तब्धिका से √((2λ/3)c³) वेग से प्रक्षेपित किया जाता है और यह केन्द्रीय त्वरण λ(r⁵ − c²r) से गतिशील है। इस कण की गति का पथ ज्ञात कीजिए। क्या यह वक्र x⁴ + y⁴ = c² होगा?
Practise this question Dynamics and Statics notes
Worked solution
At the apse r = √c, h = √c × √(2λc³/3) = c²√(2λ/3), so h² = 2λc⁴/3.
Energy integral: v² = C − 2∫λ(r⁵ − c²r) dr = C − λr⁶/3 + λc²r²; at r² = c, v² = 2λc³/3 gives C = 0.
With v = h/p: 2λc⁴/(3p²) = λr²(3c² − r⁴)/3, so the pedal equation of the path is 1/p² = r²(3c² − r⁴)/(2c⁴).
For the curve x⁴ + y⁴ = c²: in polar form c²u⁴ = f(θ) = cos⁴θ + sin⁴θ = (3 + cos 4θ)/4, so u = f^1/4/√c and du/dθ = −f^−3/4 sin 4θ/(4√c).
Then 1/p² = u² + (du/dθ)² = f^−3/2[f² + sin²4θ/16]/c = f^−3/2(3f − 1)/(2c), since f² + sin²4θ/16 = (5 + 3 cos 4θ)/8 = (3f − 1)/2.
Putting f = c²/r⁴ gives 1/p² = r²(3c² − r⁴)/(2c⁴), the same pedal equation, and the curve has an apse at θ = 0 with r = √c, like the path.
A curve satisfying the energy integral satisfies the orbit equation wherever du/dθ ≠ 0 (differentiate the integral), hence everywhere by continuity; by uniqueness of the solution with the same initial apse, the path is this curve.
Answer: yes; with the initial line through the point of projection the path is x⁴ + y⁴ = c², whose pedal equation is 2c⁴ = p²r²(3c² − r⁴).
Final answer: yes; with the initial line through the point of projection the path is x⁴ + y⁴ = c², whose pedal equation is 2c⁴ = p²r²(3c² − r⁴).
Q8(c) · Vector Analysis · 15 marks≈ 11 min
Vector Identities and Vector Equations
For a scalar point function φ and vector point function f, prove the identity ∇ · (φf) = ∇φ · f + φ(∇ · f). Also find the value of ∇ · ((f(r)/r) r) and then verify stated identity.
हिंदी में प्रश्न
एक अदिश बिन्दु फलन φ और सदिश बिन्दु फलन f के लिये निम्नलिखित सर्वसमिका सिद्ध कीजिए ∇ · (φf) = ∇φ · f + φ(∇ · f). ∇ · ((f(r)/r) r) का मान भी ज्ञात कीजिए और तब उल्लेखित सर्वसमिका का सत्यापन कीजिए।
Practise this question Vector Analysis notes
Worked solution
Proof: with f = (f₁, f₂, f₃), ∇·(φf) = Σ ∂(φf₁)/∂x = Σ(φ ∂f₁/∂x + f₁ ∂φ/∂x) = φ(∇·f) + f·∇φ, the sums running over the three coordinates.
Take φ = f(r)/r and the field r: ∇φ = φ′(r) r/r with φ′ = f′/r − f/r², and ∇·r = 3.
By the identity: ∇·((f/r) r) = 3f/r + (f′/r − f/r²)(r·r)/r = 3f/r + f′ − f/r = f′ + 2f/r.
Directly: using ∂r/∂x = x/r, ∂(fx/r)/∂x = f/r + x²(f′/r² − f/r³); adding the three such terms gives 3f/r + r²(f′/r² − f/r³) = f′ + 2f/r.
Answer: ∇·((f(r)/r) r) = f′(r) + 2f(r)/r by both methods, which verifies the identity.
Final answer: ∇·((f(r)/r) r) = f′(r) + 2f(r)/r by both methods, which verifies the identity.

UPSC Mains 2023 Mathematics Paper II