UPSC Mains 2025 Mathematics Optional Question Papers with Solutions
All 56 questions of the 2025 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 2025 Mathematics Paper I
Question 1 (50 marks)
Q1(a) · Linear Algebra · 10 marks≈ 7 min
Vector Spaces, Subspaces, Basis and Dimension
Can the set {(0, 0, 0, 3), (1, 1, 0, 0), (0, 1, −1, 0)} be extended to form a basis of the vector space ℝ4? Justify your answer.
हिंदी में प्रश्न
क्या समुच्चय {(0, 0, 0, 3), (1, 1, 0, 0), (0, 1, −1, 0)} को सदिश समष्टि ℝ4 का एक आधार बनाने के लिए विस्तारित किया जा सकता है? अपने उत्तर की पुष्टि कीजिए।
Q1(b) · Linear Algebra · 10 marks≈ 7 min
Linear Transformations, Rank-Nullity and Matrix Representation
Find the range, rank, kernel and nullity of the linear transformation T : ℝ4 → ℝ3 given by T(x, y, z, w) = (x − w, y + z, z − w).
हिंदी में प्रश्न
रैखिक रूपांतरण T : ℝ4 → ℝ3, जो T(x, y, z, w) = (x − w, y + z, z − w) द्वारा दिया गया है, का परिसर (रेन्ज), कोटि (रैंक), अष्टि (कर्नेल) और शून्यता ज्ञात कीजिए।
Q1(c) · Calculus · 10 marks≈ 7 min
Maxima and Minima, Asymptotes and Curve Tracing
A rectangular sheet of metal of length 6 meters and width 2 meters is given. Four equal squares are removed from the four corners. The sides of this sheet are now folded up to form an open rectangular box. Find approximately the height of the box, such that the volume of the box is maximum.
हिंदी में प्रश्न
लम्बाई 6 मीटर और चौड़ाई 2 मीटर की एक आयताकार धातु की चादर दी गई है। चारों कोनों से चार बराबर वर्गों को हटाया गया है। इस चादर के फलकों को मोड़कर एक खुला आयताकार सन्दूक बनाना है। सन्दूक की ऐसी सन्निकट ऊँचाई ज्ञात कीजिए कि सन्दूक का आयतन अधिकतम हो।
Worked solution
Let the squares have side x m; this is the height of the box, 0 < x < 1. The base is (6 − 2x) by (2 − 2x), so V = x(6 − 2x)(2 − 2x) = 4(x3 − 4x2 + 3x).
V′ = 4(3x2 − 8x + 3) = 0 gives x = (4 ± √7)/3; only (4 − √7)/3 ≈ 0.451 lies in (0, 1).
V″ = 4(6x − 8) < 0 for x < 4/3, so this is a maximum, with Vmax ≈ 2.52 m3.
Answer: Height = (4 − √7)/3 ≈ 0.45 m.Final answer: Height = (4 − √7)/3 ≈ 0.45 m.
Q1(d) · Calculus · 10 marks≈ 7 min
Real Numbers, Limits, Continuity and Differentiability
Given that f(x + y) = f(x) f(y) for all real x, y, f(x) ≠ 0 for any real x and f′(0) = 2. Show that for all real x, f′(x) = 2f(x). Hence find f(x).
हिंदी में प्रश्न
दिया गया है कि f(x + y) = f(x) f(y), सभी वास्तविक x, y के लिए, f(x) ≠ 0 किसी भी वास्तविक x के लिए और f′(0) = 2 है। सभी वास्तविक x के लिए दर्शाइए कि f′(x) = 2f(x) है। अतः f(x) ज्ञात कीजिए।
Worked solution
Put x = y = 0: f(0) = f(0)2, and f(0) ≠ 0, so f(0) = 1.
For any x: (f(x + h) − f(x))/h = f(x)(f(h) − 1)/h = f(x)(f(0 + h) − f(0))/h → f(x)f′(0) as h → 0. So f is differentiable everywhere and f′(x) = 2f(x).
Then (e−2xf(x))′ = e−2x(f′(x) − 2f(x)) = 0, so e−2xf(x) is constant, equal to its value f(0) = 1 at x = 0.
Answer: f′(x) = 2f(x) for all x, and f(x) = e2x.Final answer: f′(x) = 2f(x) for all x, and f(x) = e2x.
Q1(e) · Analytic Geometry · 10 marks≈ 7 min
Cone and Cylinder
Find the equation of the cone whose vertex is the point (1, 1, 0) and whose guiding curve is y = 0, x2 + z2 = 4.
हिंदी में प्रश्न
उस शंकु का समीकरण ज्ञात कीजिए जिसका शीर्ष बिंदु (1, 1, 0) है तथा जिसका निर्देशक वक्र y = 0, x2 + z2 = 4 है।
Question 2 (50 marks)
Q2(a) · Linear Algebra · 15 marks≈ 11 min
Linear Transformations, Rank-Nullity and Matrix Representation
Let T : ℝ3 → ℝ2 be a linear transformation such that T(1, 1, −1) = (1, 0), T(4, 1, 1) = (0, 1) and T(1, −1, 2) = (1, 1). Find T.
हिंदी में प्रश्न
माना T : ℝ3 → ℝ2 एक ऐसा रैखिक रूपांतरण है कि T(1, 1, −1) = (1, 0), T(4, 1, 1) = (0, 1) तथा T(1, −1, 2) = (1, 1) है। T ज्ञात कीजिए।
Q2(b) · Calculus · 15 marks≈ 11 min
Mean Value Theorems, Taylor's Theorem and Indeterminate Forms
Using Mean Value Theorem, prove that
π/6 + √3/15 < sin−1(3/5) < π/6 + 1/8.हिंदी में प्रश्न
माध्यमान प्रमेय का प्रयोग करते हुए सिद्ध कीजिए कि
π/6 + √3/15 < sin−1(3/5) < π/6 + 1/8.Worked solution
Apply Lagrange's MVT to f(x) = sin−1x on [1/2, 3/5]; f is continuous there and differentiable on the open interval with f′(x) = 1/√(1 − x2).
So sin−1(3/5) − sin−1(1/2) = (3/5 − 1/2)/√(1 − c2) = 1/(10√(1 − c2)) for some c ∈ (1/2, 3/5), and sin−1(1/2) = π/6.
1/√(1 − x2) increases on (0, 1), so 1/√(1 − 1/4) < 1/√(1 − c2) < 1/√(1 − 9/25), i.e. 2/√3 < 1/√(1 − c2) < 5/4.
Dividing by 10: √3/15 = 2/(10√3) < sin−1(3/5) − π/6 < 5/40 = 1/8.
Answer: π/6 + √3/15 < sin−1(3/5) < π/6 + 1/8.Final answer: π/6 + √3/15 < sin−1(3/5) < π/6 + 1/8.
Q2(c) · Analytic Geometry · 20 marks≈ 14 min
Cone and Cylinder
(i) Find the equation of the cylinder whose generators are parallel to the line x/1 = y/2 = z/3 and that passes through the curve x2 + y2 = 16, z = 0. (10 marks)
(ii) Find the shortest distance between the straight lines (x − 3)/3 = (y − 8)/(−1) = (z − 3)/1 and (x + 3)/(−3) = (y + 7)/2 = (z − 6)/4. (10 marks)हिंदी में प्रश्न
(i) उस बेलन का समीकरण ज्ञात कीजिए जिसके जनक, रेखा x/1 = y/2 = z/3 के समांतर हैं और जो वक्र x2 + y2 = 16, z = 0 से होकर गुज़रता है। (10 अंक)
(ii) सरल रेखाओं (x − 3)/3 = (y − 8)/(−1) = (z − 3)/1 और (x + 3)/(−3) = (y + 7)/2 = (z − 6)/4 के बीच की न्यूनतम दूरी ज्ञात कीजिए। (10 अंक)
Question 3 (50 marks)
Q3(a) · Linear Algebra · 15 marks≈ 11 min
Algebra of Matrices, Echelon Form, Rank, Inverse, Congruence and Similarity
Reduce the following matrix to echelon form:
A = [[2, −2, 2, 1], [−3, 6, 0, −1], [1, −7, 10, 2]]हिंदी में प्रश्न
निम्नलिखित आव्यूह को सोपानक (एशेलॉन) रूप में समानीत कीजिए:
A = [[2, −2, 2, 1], [−3, 6, 0, −1], [1, −7, 10, 2]]Q3(b) · Analytic Geometry · 15 marks≈ 11 min
The Sphere
Find the equations of the spheres which pass through the circle x2 + y2 + z2 − 2x + 2y + 4z − 3 = 0, 2x + y + z = 4 and touch the plane 3x + 4y = 14.
हिंदी में प्रश्न
उन गोलों के समीकरण ज्ञात कीजिए जो वृत्त x2 + y2 + z2 − 2x + 2y + 4z − 3 = 0, 2x + y + z = 4 से होकर गुज़रते हैं और समतल 3x + 4y = 14 को स्पर्श करते हैं।
Worked solution
S + λP = 0: x2 + y2 + z2 + (2λ − 2)x + (2 + λ)y + (4 + λ)z − 3 − 4λ = 0, with centre (1 − λ, −(2 + λ)/2, −(4 + λ)/2).
R2 = (1 − λ)2 + (2 + λ)2/4 + (4 + λ)2/4 + 3 + 4λ = (3/2)λ2 + 5λ + 9.
Distance of the centre from 3x + 4y − 14 = 0: |3(1 − λ) − 2(2 + λ) − 14|/5 = |λ + 3|. Tangency: (λ + 3)2 = (3/2)λ2 + 5λ + 9, that is λ2/2 − λ = 0, so λ = 0 or λ = 2.
Answer: x2 + y2 + z2 − 2x + 2y + 4z − 3 = 0 (centre (1, −1, −2), radius 3) and x2 + y2 + z2 + 2x + 4y + 6z − 11 = 0 (centre (−1, −2, −3), radius 5).Final answer: x2 + y2 + z2 − 2x + 2y + 4z − 3 = 0 (centre (1, −1, −2), radius 3) and x2 + y2 + z2 + 2x + 4y + 6z − 11 = 0 (centre (−1, −2, −3), radius 5).
Q3(c) · Calculus · 20 marks≈ 14 min
Double and Triple Integrals; Areas, Surfaces and Volumes
(i) Evaluate ∬R y dx dy, where R is the region bounded by y = x and y = 4x − x2. (10 marks)
(ii) If u(x, y) = x f(y/x) + g(y/x), where f and g are arbitrary functions, then show that
I. x ∂u/∂x + y ∂u/∂y = x f(y/x),
II. x2 ∂2u/∂x2 + 2xy ∂2u/∂x∂y + y2 ∂2u/∂y2 = 0. (10 marks)हिंदी में प्रश्न
(i) ∬R y dx dy का मान ज्ञात कीजिए, जहाँ R, y = x तथा y = 4x − x2 से परिबद्ध क्षेत्र है। (10 अंक)
(ii) यदि u(x, y) = x f(y/x) + g(y/x) है, जहाँ f और g स्वेच्छ फलन हैं, तो दर्शाइए कि
I. x ∂u/∂x + y ∂u/∂y = x f(y/x) है,
II. x2 ∂2u/∂x2 + 2xy ∂2u/∂x∂y + y2 ∂2u/∂y2 = 0 है। (10 अंक)Worked solution
(i) The curves meet where x = 4x − x2, i.e. x = 0 and x = 3, and 4x − x2 ≥ x on [0, 3]. So ∬R y dx dy = ∫03∫x4x−x2 y dy dx = (1/2)∫03 [(4x − x2)2 − x2] dx.
= (1/2)∫03 (15x2 − 8x3 + x4) dx = (1/2)(135 − 162 + 243/5) = (1/2)(108/5) = 54/5.
(ii) Assume f, g have continuous second derivatives. v = x f(y/x) is homogeneous of degree 1 and w = g(y/x) of degree 0 (replacing x, y by tx, ty multiplies them by t and by 1).
I. By Euler's theorem xvx + yvy = v and xwx + ywy = 0; adding, xux + yuy = v = x f(y/x).
II. The second-order form gives x2vxx + 2xyvxy + y2vyy = 1(1 − 1)v = 0, and the same expression in w equals 0(0 − 1)w = 0; adding gives the result.
Answer: (i) 54/5; (ii) both identities hold.Final answer: (i) 54/5; (ii) both identities hold.
Question 4 (50 marks)
Q4(a) · Analytic Geometry · 15 marks≈ 11 min
The Sphere
Show that there is no tangent plane to the sphere
x2 + y2 + z2 − 4x + 2y − 4z + 4 = 0
that can be passed through the straight line
(x + 6)/2 = y + 3 = z + 1.हिंदी में प्रश्न
दर्शाइए कि गोले
x2 + y2 + z2 − 4x + 2y − 4z + 4 = 0
का कोई ऐसा स्पर्श समतल नहीं है, जो कि सरल रेखा (x + 6)/2 = y + 3 = z + 1 से होकर गुज़र सके।Worked solution
The sphere has centre (2, −1, 2) and radius √(4 + 1 + 4 − 4) = √5. A general point of the line is (−6 + 2t, −3 + t, −1 + t).
Substituting in the sphere: 6t2 − 42t + 72 = 6(t − 3)(t − 4) = 0, so the line cuts the sphere in the two real points (0, 0, 2) and (2, 1, 3).
Every plane through the line contains these two points of the sphere, so it meets the sphere in a circle through two distinct points, which cannot be a single point of contact. Hence no plane through the line touches the sphere.
Answer: the line is a secant of the sphere (its distance from the centre, √(7/2), is less than √5), so no tangent plane passes through it.Final answer: the line is a secant of the sphere (its distance from the centre, √(7/2), is less than √5), so no tangent plane passes through it.
Q4(b) · Calculus · 15 marks≈ 11 min
Functions of Several Variables: Partial Derivatives, Extrema, Lagrange Multipliers and Jacobians
If f(x, y) = xy (x2 − y2)/(x2 + y2), when (x, y) ≠ (0, 0)
f(x, y) = 0, when (x, y) = (0, 0),
then find fxy(0, 0) and fyx(0, 0).हिंदी में प्रश्न
यदि f(x, y) = xy (x2 − y2)/(x2 + y2), जब (x, y) ≠ (0, 0)
f(x, y) = 0, जब (x, y) = (0, 0)
है, तो fxy(0, 0) और fyx(0, 0) ज्ञात कीजिए।Worked solution
This is the function of the previous example. From first principles, fx(0, y) = −y for all y and fy(x, 0) = x for all x.
With fxy = (fx)y: fxy(0, 0) = limk→0 (fx(0, k) − fx(0, 0))/k = lim (−k)/k = −1.
fyx = (fy)x: fyx(0, 0) = limh→0 (fy(h, 0) − fy(0, 0))/h = lim h/h = 1. (With the opposite convention the two values are interchanged; state the convention in the answer.)
Answer: fxy(0, 0) = −1, fyx(0, 0) = 1, so fxy(0, 0) ≠ fyx(0, 0).Final answer: fxy(0, 0) = −1, fyx(0, 0) = 1, so fxy(0, 0) ≠ fyx(0, 0).
Q4(c) · Linear Algebra · 20 marks≈ 14 min
Eigenvalues, Eigenvectors and the Cayley-Hamilton Theorem
(i) Find the eigenvalues and the corresponding eigenvectors of the matrix
A = [[1, 2, 0], [2, 1, −6], [2, −2, 3]]. (12 marks)
(ii) Let Pn denote the vector space of all polynomials of degree ≤ n over ℝ. Verify that
dim(P4/P2) = dim P4 − dim P2. (8 marks)हिंदी में प्रश्न
(i) आव्यूह A = [[1, 2, 0], [2, 1, −6], [2, −2, 3]] के अभिलक्षणिक मान और संगत अभिलक्षणिक सदिश ज्ञात कीजिए। (12 अंक)
(ii) माना Pn, ℝ पर घात ≤ n के सभी बहुपदों के सदिश समष्टि को दर्शाता है। सत्यापित कीजिए कि
dim(P4/P2) = dim P4 − dim P2 (8 अंक)Worked solution
(i) tr A = 5; principal minors (1 − 4) + (3 − 0) + (3 − 12) = −9; det A = 1(3 − 12) − 2(6 + 12) = −45. So λ3 − 5λ2 − 9λ + 45 = (λ − 5)(λ2 − 9) = 0, and λ = 5, 3, −3.
λ = 5: −4x + 2y = 0 and 2x − 4y − 6z = 0 give y = 2x, z = −x, X1 = (1, 2, −1). λ = 3: −2x + 2y = 0 and 2x − 2y − 6z = 0 give y = x, z = 0, X2 = (1, 1, 0).
λ = −3: 4x + 2y = 0 and 2x + 4y − 6z = 0 give y = −2x, z = −x, X3 = (1, −2, −1). Check: AX1 = (1 + 4, 2 + 2 + 6, 2 − 4 − 3) = (5, 10, −5) = 5X1.
(ii) dim P4 = 5 and dim P2 = 3. In P4/P2, p + P2 = q + P2 ⇔ p − q ∈ P2, so each coset is (a3x3 + a4x4) + P2 = a3(x3 + P2) + a4(x4 + P2).
These two cosets are independent: a3x3 + a4x4 ∈ P2 forces a3 = a4 = 0. So dim(P4/P2) = 2 = 5 − 3 = dim P4 − dim P2.
Answer: (i) λ = 5, 3, −3 with eigenvectors k(1, 2, −1), k(1, 1, 0), k(1, −2, −1); (ii) basis {x3 + P2, x4 + P2}, so both sides equal 2.Final answer: (i) λ = 5, 3, −3 with eigenvectors k(1, 2, −1), k(1, 1, 0), k(1, −2, −1); (ii) basis {x3 + P2, x4 + P2}, so both sides equal 2.
Question 5 (50 marks)
Q5(a) · Ordinary Differential Equations · 10 marks≈ 7 min
First Order Higher Degree Equations, Clairaut's Equation and Singular Solutions
Solve (1 − y2 + y4/x2)(dy/dx)2 − 2(y/x)(dy/dx) + y2/x2 = 0.
हिंदी में प्रश्न
(1 − y2 + y4/x2)(dy/dx)2 − 2(y/x)(dy/dx) + y2/x2 = 0 को हल कीजिए।
Worked solution
Multiply by x2 and regroup: (xp − y)2 = p2y2(x2 − y2). Put y = vx, so p = v + xv′ and xp − y = x2v′; the equation becomes v′ = ±v√(1 − v2)·p.
Taking v as the independent variable: dx/dv + x/v = ±1/(v2√(1 − v2)), a linear equation with IF v; so d(xv)/dv = ±1/(v√(1 − v2)).
Since xv = y and ∫dv/(v√(1 − v2)) = −sech−1v: y − c = ∓sech−1(y/x) for 0 < y/x ≤ 1, that is y/x = sech(y − c); since the equation is unchanged by x → −x, the curves with y/x < 0 are the mirror images, and together |y/x| = sech(y − c).
p-discriminant: (y/x)2 − (1 − y2 + y4/x2)(y2/x2) = (y4/x4)(x2 − y2) = 0. The lines y = ±x (p = ±1) satisfy the equation, and each curve x = ±y cosh(y − c) touches one of them at y = c (where dx/dy = ±1), so they are singular solutions; y = 0 also satisfies the equation trivially.
Answer: general solution x2 = y2 cosh2(y − c), that is x = ±y cosh(y − c); singular solutions y = ±x.Final answer: general solution x2 = y2 cosh2(y − c), that is x = ±y cosh(y − c); singular solutions y = ±x.
Q5(b) · Ordinary Differential Equations · 10 marks≈ 7 min
Formulation and First Order, First Degree Equations
Form the differential equation of all ellipses whose axes coincide with coordinate axes.
हिंदी में प्रश्न
सभी दीर्घवृत्तों, जिनके अक्ष निर्देशांक अक्षों के संपाती हैं, का अवकल समीकरण बनाइए।
Worked solution
The ellipses are x2/a2 + y2/b2 = 1 (two constants). Differentiating: x/a2 + yy′/b2 = 0, so yy′/x = −b2/a2, a constant.
Differentiating again: [x(y′2 + yy″) − yy′]/x2 = 0.
Answer: xyy″ + xy′2 − yy′ = 0.Final answer: xyy″ + xy′2 − yy′ = 0.
Q5(c) · Dynamics and Statics · 10 marks≈ 7 min
Central Orbits and Kepler's Laws
Prove that the time taken by the Earth to travel over half of its orbit, which is separated by the minor axis and is remote from the Sun, when the Sun is at the focus of the elliptic orbit, is two days more than half of the year. The eccentricity of the orbit is taken as 1/60.
हिंदी में प्रश्न
सिद्ध कीजिए कि पृथ्वी को अपनी कक्षा के आधे भाग, जो कि लघु अक्ष द्वारा अलग किया गया है और सूर्य से सुदूर है, जब सूर्य दीर्घवृत्तीय कक्षा की नाभि (फोकस) पर है, की यात्रा करने में लगने वाला समय आधे वर्ष से दो दिन अधिक है। कक्षा की उत्केन्द्रता 1/60 ली गई है।
Worked solution
Let the ellipse have semi-axes a, b and eccentricity e, with the sun at the focus S, at distance ae from the centre; BB′ is the minor axis.
The half of the orbit remote from the sun is the arc from B through the aphelion to B′; while the earth describes it, the radius vector from S sweeps the half-ellipse beyond BB′ together with the triangle SBB′.
Area swept = πab/2 + (1/2)(2b)(ae) = πab/2 + abe.
By Kepler's second law time is proportional to area, so the time is T(πab/2 + abe)/(πab) = T/2 + Te/π, T being the year.
With e = 1/60 and T = 365.25 days, Te/π = 365.25/(60π) ≈ 1.94 days.
Answer: the time exceeds half a year by Te/π ≈ 2 days.Final answer: the time exceeds half a year by Te/π ≈ 2 days.
Q5(d) · Dynamics and Statics · 10 marks≈ 7 min
Equilibrium, Friction and the Common Catenary
Given that A and B are two points in the same horizontal line distant 2a apart. AO and BO are two equal heavy strings tied together at O and carrying their weight at O. If l is length of each string and d is depth of O below AB, then show that the parameter c of this catenary, in which the strings hang, is given by
l2 − d2 = 2c2[cosh(a/c) − 1].हिंदी में प्रश्न
दिया गया है कि A और B एक ही क्षैतिज रेखा पर स्थित दो बिंदु हैं, जिनके बीच की दूरी 2a है। AO और BO दो समान भारी डोरी हैं जो O पर एक साथ बँधी हैं और जिनका भार O पर है। यदि प्रत्येक डोरी की लंबाई l है तथा d, AB से नीचे O की गहराई है, तो दर्शाइए कि इस कैटिनरी, जिसमें डोरी लटकी है, का प्राचल c
l2 − d2 = 2c2[cosh(a/c) − 1]
द्वारा दिया गया है।Worked solution
[Whether or not a weight hangs at O, each string is an arc of a catenary with the same parameter c, by symmetry, and the identity below uses only that.]
For OA let O and A have abscissae x1, x2 on y = c cosh(x/c), with x2 − x1 = a by symmetry.
Length l = c[sinh(x2/c) − sinh(x1/c)] and depth d = c[cosh(x2/c) − cosh(x1/c)].
l2 − d2 = c2[(sinh2 − cosh2)(x1/c) + (sinh2 − cosh2)(x2/c) + 2 cosh(x1/c)cosh(x2/c) − 2 sinh(x1/c)sinh(x2/c)].
This equals c2[−2 + 2 cosh((x2 − x1)/c)] = 2c2[cosh(a/c) − 1].
Answer: l2 − d2 = 2c2[cosh(a/c) − 1].Final answer: l2 − d2 = 2c2[cosh(a/c) − 1].
Q5(e) · Vector Analysis · 10 marks≈ 7 min
Gradient, Divergence and Curl in Cartesian and Cylindrical Coordinates
If u = x + y + z, v = x2 + y2 + z2 and w = xy + yz + zx, then show that grad u, grad v and grad w are coplanar.
हिंदी में प्रश्न
यदि u = x + y + z, v = x2 + y2 + z2 और w = xy + yz + zx है, तो दर्शाइए कि grad u, grad v और grad w समतलीय हैं।
Worked solution
∇u = (1, 1, 1), ∇v = 2(x, y, z) and ∇w = (y + z, z + x, x + y).
[∇u ∇v ∇w] = 2 det[[1, 1, 1], [x, y, z], [y + z, z + x, x + y]].
Adding the second row to the third gives (x + y + z)(1, 1, 1), a multiple of the first row, so the determinant is 0 at every point.
Answer: the scalar triple product vanishes identically, so grad u, grad v, grad w are coplanar; this reflects the functional relation v = u2 − 2w.Final answer: the scalar triple product vanishes identically, so grad u, grad v, grad w are coplanar; this reflects the functional relation v = u2 − 2w.
Question 6 (50 marks)
Q6(a) · Ordinary Differential Equations · 15 marks≈ 11 min
Laplace Transforms and Initial Value Problems
If F(s) and G(s) are Laplace transforms of f(t) and g(t) respectively, then prove that 𝓛(∫0t f(x) g(t − x) dx) = F(s) G(s). Using this result, solve the equation y(t) = t + ∫0t y(x) sin(t − x) dx.
हिंदी में प्रश्न
यदि f(t) और g(t) के लाप्लास रूपान्तर क्रमशः F(s) और G(s) हैं, तो सिद्ध कीजिए कि 𝓛(∫0t f(x) g(t − x) dx) = F(s) G(s) है। इस परिणाम का प्रयोग करते हुए, समीकरण y(t) = t + ∫0t y(x) sin(t − x) dx को हल कीजिए।
Worked solution
Proof: L{f * g} = ∫0∞ e−st∫0t f(x)g(t − x) dx dt, over the region 0 ≤ x ≤ t < ∞.
Change the order of integration (justified by absolute convergence for s large enough when f, g are of exponential order): = ∫0∞ f(x)[∫x∞ e−stg(t − x) dt] dx.
Put t − x = u in the inner integral: ∫0∞ e−s(x + u)g(u) du = e−sxG(s). Hence L{f * g} = G(s)∫0∞ e−sxf(x) dx = F(s)G(s).
Application: Y = 1/s2 + Y/(s2 + 1), so Y = (s2 + 1)/s4 = 1/s2 + 1/s4.
Answer: y(t) = t + t3/6.Final answer: y(t) = t + t3/6.
Q6(b) · Dynamics and Statics · 15 marks≈ 11 min
Rectilinear Motion and Simple Harmonic Motion
One end of an elastic string, having natural length a, is fixed at some point O and a heavy particle is attached to the other end of the string. The string is drawn vertically downward till it is four times its natural length at the point C and then released. If the modulus of elasticity of the string is equal to the weight of the particle, then show that the particle will return to the same point C in the time √(a/g) (2√3 + 4π/3).
हिंदी में प्रश्न
एक प्रत्यास्थ डोरी, जिसकी प्राकृतिक लंबाई a है, का एक छोर किसी बिंदु O पर स्थिर है और डोरी के दूसरे छोर पर एक भारी कण जुड़ा हुआ है। डोरी को ऊर्ध्वाधर नीचे की ओर बिंदु C तक तब तक खींचा जाता है जब तक वह अपनी प्राकृतिक लंबाई से चार गुना न हो जाए तथा फिर छोड़ दिया जाता है। यदि डोरी का प्रत्यास्थता गुणांक कण के भार के बराबर है, तो दर्शाइए कि कण √(a/g) (2√3 + 4π/3) समय में उसी बिंदु C पर वापस आ जाएगा।
Worked solution
λ = mg, so the equilibrium extension is e = mga/λ = a: the equilibrium position E is 2a below O, and while the string is taut the motion is SHM about E with n2 = g/e = g/a.
C is 4a below O, that is 2a below E, so the amplitude is 2a.
Measuring x upwards from E, x = −2a cos nt; the string becomes slack at its natural length, x = a, when cos nt = −1/2, that is after t1 = (2π/3)√(a/g).
There v2 = n2(4a2 − a2) = 3ga; the particle then moves freely under gravity and returns to the same point with the same speed after t2 = 2v/g = 2√(3a/g).
Below that point the string is taut again, and by symmetry the SHM brings the particle back to C in a further time t1.
Total time = 2t1 + t2 = √(a/g)(4π/3 + 2√3).
Answer: the particle returns to C after √(a/g)(2√3 + 4π/3).Final answer: the particle returns to C after √(a/g)(2√3 + 4π/3).
Q6(c) · Vector Analysis · 20 marks≈ 14 min
Vector Identities and Vector Equations
(i) Find the absolute value of the directional derivative of φ(x, y, z) = x2y2z2 at the point (1, 1, −1) in the direction of the tangent to the curve x = et, y = 2 sin t + 1, z = t − cos t, at t = 0. (10 marks)
(ii) If ∇ · E = 0, ∇ · H = 0, ∇ × E = −∂H/∂t and ∇ × H = ∂E/∂t, then show that ∇2H = ∂2H/∂t2 and ∇2E = ∂2E/∂t2. (10 marks)हिंदी में प्रश्न
(i) φ(x, y, z) = x2y2z2 का बिन्दु (1, 1, −1) पर, वक्र x = et, y = 2 sin t + 1, z = t − cos t, के बिन्दु t = 0 पर स्पर्श-रेखा की दिशा में दिक्-अवकलज का निरपेक्ष मान ज्ञात कीजिए। (10 अंक)
(ii) यदि ∇ · E = 0, ∇ · H = 0, ∇ × E = −∂H/∂t और ∇ × H = ∂E/∂t हैं, तो दर्शाइए कि ∇2H = ∂2H/∂t2 और ∇2E = ∂2E/∂t2 हैं। (10 अंक)Worked solution
(i) ∇φ = (2xy2z2, 2x2yz2, 2x2y2z) = (2, 2, −2) at (1, 1, −1).
At t = 0 the curve is at (1, 1, −1), and r′ = (et, 2 cos t, 1 + sin t) = (1, 2, 1), of length √6.
Directional derivative = (2 + 4 − 2)/√6 = 4/√6 = 2√6/3.
(ii) Take the curl of ∇ × H = ∂E/∂t: ∇ × (∇ × H) = ∂(∇ × E)/∂t = −∂2H/∂t2, since space and time derivatives commute for smooth fields.
The left side is ∇(∇·H) − ∇2H = −∇2H, because ∇·H = 0; hence ∇2H = ∂2H/∂t2.
Similarly ∇ × (∇ × E) = −∂(∇ × H)/∂t = −∂2E/∂t2, and the left side is −∇2E since ∇·E = 0, so ∇2E = ∂2E/∂t2.
Answer: (i) 2√6/3; (ii) both E and H satisfy the wave equation.Final answer: (i) 2√6/3; (ii) both E and H satisfy the wave equation.
Question 7 (50 marks)
Q7(a) · Dynamics and Statics · 15 marks≈ 11 min
Virtual Work, Stability and Forces in Three Dimensions
A solid sphere rests inside a fixed rough and hemispherical bowl of twice its radius. If a large amount of weight, whatsoever, is attached to the highest point of the sphere, then show that the equilibrium is stable.
हिंदी में प्रश्न
एक ठोस गोला अपनी त्रिज्या से दुगुनी त्रिज्या के स्थिर रूक्ष अर्धगोलीय कटोरे में रखा हुआ है। यदि एक बड़ा भार, कितना भी हो, गोले के सबसे ऊँचे बिंदु पर जुड़ा है, तो दर्शाइए कि संतुलन स्थिर है।
Worked solution
Let the sphere have radius a and weight W, the bowl radius 2a and centre O, and a weight W′ be fixed at the top point P of the sphere.
As the sphere rolls, its centre C moves on the circle of radius a about O; let OC make angle φ with the downward vertical, so C = O + a(sin φ, −cos φ) in the vertical plane.
Rolling without slipping inside the bowl turns the sphere through (2a − a)φ/a = φ in the opposite sense, so the radius CP, initially vertical, becomes a(sin φ, cos φ) and P = O + (2a sin φ, 0).
Hence P stays at the level of O for every φ (it moves along a horizontal diameter of the bowl), and the potential energy is V = −Wa cos φ + W′ × 0.
V has a minimum at φ = 0 whatever W′ is, so the equilibrium is stable however large W′ is.
Check by the rule for a body inside a concave surface: 1/h > 1/ρ1 − 1/ρ2 = 1/(2a) needs h < 2a, and h = (Wa + 2aW′)/(W + W′) is always less than 2a.
Answer: the equilibrium is stable for every attached weight.Final answer: the equilibrium is stable for every attached weight.
Q7(b) · Vector Analysis · 15 marks≈ 11 min
Gauss and Stokes' Theorems and Green's Identities
Verify Green's theorem in the plane for ∮C [(xy + y2) dx + x2 dy], where C is the boundary of the region bounded by the curves y = x and y = x2.
हिंदी में प्रश्न
∮C [(xy + y2) dx + x2 dy], जहाँ C, वक्रों y = x और y = x2 द्वारा परिबद्ध क्षेत्र की परिसीमा है, के लिए समतल में ग्रीन का प्रमेय सत्यापित कीजिए।
Worked solution
Double integral (Example 1): ∬(x − 2y) dA = −1/20.
Along y = x2 from (0, 0) to (1, 1): M dx + N dy = (x3 + x4) dx + x2(2x dx), with integral ∫01(3x3 + x4) dx = 19/20.
Along y = x from (1, 1) to (0, 0): M dx + N dy = 3x2 dx, with integral −1.
Answer: the line integral is 19/20 − 1 = −1/20, equal to the double integral.Final answer: the line integral is 19/20 − 1 = −1/20, equal to the double integral.
Q7(c) · Ordinary Differential Equations · 20 marks≈ 14 min
First Order Higher Degree Equations, Clairaut's Equation and Singular Solutions
(i) Find the general solution and singular solution of the differential equation (1 + dy/dx)3 = (27/8a)(x + y)(1 − dy/dx)3. (10 marks)
(ii) Find the complete solution of x3 d3y/dx3 + 3x2 d2y/dx2 + x dy/dx + y = x log x. (10 marks)हिंदी में प्रश्न
(i) अवकल समीकरण (1 + dy/dx)3 = (27/8a)(x + y)(1 − dy/dx)3 के व्यापक हल और विचित्र हल ज्ञात कीजिए। (10 अंक)
(ii) x3 d3y/dx3 + 3x2 d2y/dx2 + x dy/dx + y = x log x का पूर्ण हल ज्ञात कीजिए। (10 अंक)Worked solution
(i)
Put X = x + y, Y = x − y: P = dY/dX = (1 − p)/(1 + p). The equation becomes 1/P3 = (27/(8a))X, so P = (2/3)a1/3X−1/3.
Integrating: Y − c = a1/3X2/3, so (Y − c)3 = aX2.
c-discriminant: 3(Y − c)2 = 0 gives X2 = 0, that is x + y = 0. On x + y = 0, p = −1 and both sides of the equation are 0, so it is a solution; each curve (a semicubical parabola) meets it at its cusp with tangent along it, and it is not a member of the family.
Answer: general solution (x − y − c)3 = a(x + y)2; singular solution x + y = 0.
(ii)
With x = ez the operator is θ3 + 1 (as in Problem 9 of the Solved Problems), so CF = c1/x + √x[c2 cos((√3/2) log x) + c3 sin((√3/2) log x)].
PI = [1/(θ3 + 1)] zez = ez[1/((θ + 1)3 + 1)] z = ez[1/(2 + 3θ + 3θ2 + θ3)] z = (ez/2)(1 − 3θ/2)z = (ez/2)(z − 3/2).
Answer: y = c1/x + √x[c2 cos((√3/2) log x) + c3 sin((√3/2) log x)] + (x/2)(log x − 3/2).Final answer: (i) general solution (x − y − c)3 = a(x + y)2; singular solution x + y = 0; (ii) y = c1/x + √x[c2 cos((√3/2) log x) + c3 sin((√3/2) log x)] + (x/2)(log x − 3/2)
Question 8 (50 marks)
Q8(a) · Ordinary Differential Equations · 15 marks≈ 11 min
Second Order Equations with Variable Coefficients
Solve the differential equation (x + 2) d2y/dx2 − (2x + 5) dy/dx + 2y = (1 + x) ex by the method of variation of parameters.
हिंदी में प्रश्न
अवकल समीकरण (x + 2) d2y/dx2 − (2x + 5) dy/dx + 2y = (1 + x) ex को प्राचल विचरण विधि द्वारा हल कीजिए।
Worked solution
By inspection: y = e2x gives 4(x + 2) − 2(2x + 5) + 2 = 0, and y = 2x + 5 gives −2(2x + 5) + 2(2x + 5) = 0. So y1 = e2x, y2 = 2x + 5.
W = 2e2x − 2(2x + 5)e2x = −4(x + 2)e2x; standard form R = (1 + x)ex/(x + 2).
u1 = (1/4)∫e−x(2x + 5)(x + 1)/(x + 2)2 dx = (1/4)∫e−x[2 − 1/(x + 2) − 1/(x + 2)2] dx = (1/4)[−2e−x + e−x/(x + 2)].
u2 = −(1/4)∫ex(x + 1)/(x + 2)2 dx = −(1/4)∫ex[1/(x + 2) − 1/(x + 2)2] dx = −(1/4)ex/(x + 2).
PI = u1e2x + u2(2x + 5) = (ex/4)[−2 + (1 − 2x − 5)/(x + 2)] = −ex.
Answer: y = c1e2x + c2(2x + 5) − ex (check: ex[−(x + 2) + (2x + 5) − 2] = (1 + x)ex).Final answer: y = c1e2x + c2(2x + 5) − ex (check: ex[−(x + 2) + (2x + 5) − 2] = (1 + x)ex).
Q8(b) · Vector Analysis · 15 marks≈ 11 min
Gauss and Stokes' Theorems and Green's Identities
Verify Gauss's divergence theorem for F = [(x2 − yz) î + (y2 − zx) ĵ + (z2 − xy) k̂], taken over the rectangular parallelopiped 0 ≤ x ≤ a, 0 ≤ y ≤ b, 0 ≤ z ≤ c.
हिंदी में प्रश्न
समकोणिक समांतरषट्फलक 0 ≤ x ≤ a, 0 ≤ y ≤ b, 0 ≤ z ≤ c पर F = [(x2 − yz) î + (y2 − zx) ĵ + (z2 − xy) k̂] के लिए गाउस अपसरण प्रमेय सत्यापित कीजिए।
Worked solution
∇·F = 2(x + y + z), so ∭∇·F dV = 2[(a2/2)bc + (b2/2)ca + (c2/2)ab] = abc(a + b + c).
Face x = a (n = i): ∬(a2 − yz) dy dz = a2bc − b2c2/4; face x = 0 (n = −i): ∬yz dy dz = b2c2/4; together a2bc.
In the same way the faces y = b, y = 0 give b2ca and the faces z = c, z = 0 give c2ab.
Answer: both sides equal abc(a + b + c).Final answer: both sides equal abc(a + b + c).
Q8(c) · Dynamics and Statics · 20 marks≈ 14 min
Constrained Motion, Work and Energy
A particle is projected inside a fixed smooth cylinder with circular cross-section in a vertical plane from the lowest point with initial horizontal velocity u. Show that for
(i) (u2 ≤ 2ag); the particle oscillates about the mean position in the lower half,
(ii) (u2 ≥ 5ag); the particle executes complete circular motion, and
(iii) (2ag < u2 < 5ag); the particle will leave the curve in a tangential direction, making an angle α with the horizontal such that cos α = (u2 − 2ag)/3ag.हिंदी में प्रश्न
एक कण को ऊर्ध्वाधर तल में वृत्ताकार अनुप्रस्थ-परिच्छेद वाले स्थिर चिकने बेलन के अन्दर प्रारम्भिक क्षैतिज वेग u के साथ सबसे निचले बिंदु से प्रक्षेपित किया जाता है। दर्शाइए कि
(i) (u2 ≤ 2ag) के लिए; कण निचले आधे भाग में माध्य स्थिति के आसपास (about) दोलन करता है,
(ii) (u2 ≥ 5ag) के लिए; कण पूर्णतः वृत्तीय गति करता है, और
(iii) (2ag < u2 < 5ag) के लिए; कण, वक्र को एक स्पर्शी की दिशा में, जो क्षैतिज के साथ कोण α बनाती है, छोड़ देगा, जबकि cos α = (u2 − 2ag)/3ag है।Worked solution
With θ measured from the lowest point: v2 = u2 − 2ga(1 − cos θ) and R = (m/a)(u2 − 2ga + 3ga cos θ).
(i) If u2 ≤ 2ga, v = 0 where cos θ = 1 − u2/(2ga) ≥ 0, at or below the level of the centre, and there R = mg cos θ ≥ 0; so the particle never leaves the surface and oscillates in the lower half.
(ii) R is least at the top (θ = π), where R = (m/a)(u2 − 5ga) ≥ 0 if u2 ≥ 5ga; then v2 ≥ ga > 0 at the top and the full circle is described.
(iii) If 2ga < u2 < 5ga, R vanishes at cos θ = −(u2 − 2ga)/(3ga), above the centre, where v2 = −ga cos θ > 0; beyond this R would be negative, so the particle leaves the surface there.
It leaves along the tangent, which is inclined to the horizontal at α = π − θ, so cos α = −cos θ = (u2 − 2ga)/(3ga).
Answer: the three cases follow from the signs of v2 and R; in case (iii) the particle leaves tangentially at α to the horizontal with cos α = (u2 − 2ag)/(3ag).Final answer: the three cases follow from the signs of v2 and R; in case (iii) the particle leaves tangentially at α to the horizontal with cos α = (u2 − 2ag)/(3ag).
UPSC Mains 2025 Mathematics Paper II
Question 1 (50 marks)
Q1(a) · Abstract Algebra · 10 marks≈ 7 min
Groups, Subgroups, Cyclic Groups, Cosets and Lagrange's Theorem
Let H and K be two subgroups of a group G such that o(H) > √o(G) and o(K) > √o(G). Show that H ∩ K ≠ {e}, where e is the identity element. Here o(H), o(K) and o(G) denote the order of H, K and G respectively.
हिंदी में प्रश्न
माना एक समूह G के दो उपसमूह H और K इस प्रकार हैं कि o(H) > √o(G) और o(K) > √o(G) हैं। दर्शाइए कि H ∩ K ≠ {e} है, जहाँ e तत्समक अवयव है। यहाँ o(H), o(K) और o(G) क्रमशः H, K और G की कोटि को दर्शाते हैं।
Worked solution
HK = {hk : h ∈ H, k ∈ K} is a subset of G (not necessarily a subgroup), and |HK| = o(H)o(K)/o(H ∩ K) for finite subgroups H, K.
Suppose H ∩ K = {e}. Then |HK| = o(H)o(K) > √o(G)·√o(G) = o(G).
But HK ⊆ G, so |HK| ≤ o(G), a contradiction.
Answer: H ∩ K ≠ {e}; H and K have a common non-identity element.Final answer: H ∩ K ≠ {e}; H and K have a common non-identity element.
Q1(b) · Abstract Algebra · 10 marks≈ 7 min
Groups, Subgroups, Cyclic Groups, Cosets and Lagrange's Theorem
Let G = {e, x, x2, y, yx, yx2} be a non-Abelian group with o(x) = 3 and o(y) = 2. Show that xy = yx2 (where e is the identity element of G and o(x), o(y) denote the order of the elements x, y respectively).
हिंदी में प्रश्न
माना G = {e, x, x2, y, yx, yx2} एक अन्-आबेली समूह है तथा o(x) = 3 और o(y) = 2 हैं। दर्शाइए कि xy = yx2 है (जहाँ e, समूह G का तत्समक अवयव है और o(x), o(y) क्रमशः अवयवों x, y की कोटि को दर्शाते हैं)।
Worked solution
xy ∈ G. If xy ∈ {e, x, x2} = ⟨x⟩, then y = x−1(xy) ∈ ⟨x⟩, which is false; so xy ∈ {y, yx, yx2}.
xy = y gives x = e, impossible since o(x) = 3.
xy = yx would make x and y commute; since every element of G is a product of powers of x and y, G would be abelian, contrary to hypothesis.
Hence xy = yx2 = yx−1, the defining relation of D3 ≅ S3 (one model: x = (1 2 3), y = (1 2), where both sides equal (1 3)).
Answer: xy = yx2.Final answer: xy = yx2.
Q1(c) · Real Analysis · 10 marks≈ 7 min
Series: Convergence Tests, Absolute and Conditional Convergence, Rearrangement
Examine whether the series ∑n=1∞ (−1)n−1/n is absolutely or conditionally convergent.
हिंदी में प्रश्न
श्रेणी ∑n=1∞ (−1)n−1/n के निरपेक्षतः या सापेक्ष अभिसारी होने की जाँच कीजिए।
Q1(d) · Complex Analysis · 10 marks≈ 7 min
Power Series, Taylor's Series and Laurent's Series
Expand f(z) = 1/[(z + 1)(z + 3)] in a Laurent series valid for 1 < |z| < 3.
हिंदी में प्रश्न
1 < |z| < 3 के लिए f(z) = 1/[(z + 1)(z + 3)] का एक लौराँ श्रेणी में प्रसार कीजिए।
Q1(e) · Linear Programming · 10 marks≈ 7 min
LPP Formulation, Basic Solutions and the Graphical Method
How many basic solutions are there for the following system of equations?
2x1 − x2 + 3x3 + x4 = 6
4x1 − 2x2 − x3 + 2x4 = 10
Find all of them. Furthermore, find the number of basic solutions, which are feasible/non-feasible/non-degenerate.हिंदी में प्रश्न
समीकरण निकाय
2x1 − x2 + 3x3 + x4 = 6
4x1 − 2x2 − x3 + 2x4 = 10
के कितने आधारी हल हैं? उन सभी को ज्ञात कीजिए। उन आधारी हलों की संख्या भी ज्ञात कीजिए जो सुसंगत/असुसंगत/अनपभ्रष्ट हैं।Worked solution
This is the system of Example 9: 3 of the 6 column pairs are singular, so there are 3 basic solutions.
(18/7, 0, 2/7, 0) and (0, 0, 2/7, 36/7) have both basic variables positive, so they are basic feasible and non-degenerate.
(0, −36/7, 2/7, 0) has x2 < 0, so it is basic but not feasible; it is also non-degenerate, since no basic variable is 0.
Answer: 3 basic solutions: 2 feasible, 1 non-feasible, and all 3 non-degenerate (none is degenerate).Final answer: 3 basic solutions: 2 feasible, 1 non-feasible, and all 3 non-degenerate (none is degenerate).
Question 2 (50 marks)
Q2(a) · Real Analysis · 15 marks≈ 11 min
Real Number System, Sequences, Cauchy Sequences and Completeness
Define Cauchy sequence and prove that every convergent sequence of real numbers is a Cauchy sequence. What is the importance of Cauchy condition?
हिंदी में प्रश्न
कौशी अनुक्रम की परिभाषा दीजिए और सिद्ध कीजिए कि वास्तविक संख्याओं का प्रत्येक अभिसारी अनुक्रम एक कौशी अनुक्रम है। कौशी की शर्त का क्या महत्त्व है?
Q2(b) · Abstract Algebra · 15 marks≈ 11 min
Integral Domains, PID, Euclidean Domains, UFD, Fields and Quotient Fields
Show that 3 is an irreducible element in the integral domain Z[i].
हिंदी में प्रश्न
दर्शाइए कि पूर्णांकीय प्रांत Z[i] में 3 एक अखंडनीय अवयव है।
Worked solution
Use the norm N(a + bi) = a2 + b2. N(αβ) = N(α)N(β), and α is a unit iff N(α) = 1, i.e. α ∈ {±1, ±i}.
3 ≠ 0 and N(3) = 9 ≠ 1, so 3 is a non-zero non-unit.
If 3 = αβ, then N(α)N(β) = 9, so N(α) ∈ {1, 3, 9}.
N(α) = 3 would need a2 + b2 = 3 in integers, impossible since squares are 0, 1, 4, 9, .... So N(α) = 1 or N(β) = 1: α or β is a unit.
Answer: 3 is irreducible in ℤ[i]; since ℤ[i] is a PID, it is also prime, and ℤ[i]/⟨3⟩ is a field with 9 elements.Final answer: 3 is irreducible in ℤ[i]; since ℤ[i] is a PID, it is also prime, and ℤ[i]/⟨3⟩ is a field with 9 elements.
Q2(c) · Complex Analysis · 20 marks≈ 14 min
Contour Integration: Evaluation of Real Integrals
Use the method of contour integration to prove that ∫−∞∞ (x2 − x + 2)/(x4 + 10x2 + 9) dx = 5π/12.
हिंदी में प्रश्न
कन्टूर समाकलन की विधि से सिद्ध कीजिए कि ∫−∞∞ (x2 − x + 2)/(x4 + 10x2 + 9) dx = 5π/12 है।
Question 3 (50 marks)
Q3(a) · Complex Analysis · 15 marks≈ 11 min
Singularities and Cauchy's Residue Theorem
Evaluate the integral ∮C ez/[z2(z + 1)3] dz, C : |z| = 2.
हिंदी में प्रश्न
समाकल ∮C ez/[z2(z + 1)3] dz, C : |z| = 2 का मान ज्ञात कीजिए।
Q3(b) · Real Analysis · 20 marks≈ 14 min
Functions of Several Variables: Partial Derivatives, Maxima and Minima
Show that the volume of the greatest rectangular parallelopiped that can be inscribed in the ellipsoid x2/a2 + y2/b2 + z2/c2 = 1 is 8abc/(3√3).
हिंदी में प्रश्न
सिद्ध कीजिए कि दीर्घवृत्तज x2/a2 + y2/b2 + z2/c2 = 1 के अंतर्गत सबसे बड़े समकोणिक समांतरषट्फलक का आयतन 8abc/(3√3) है।
Worked solution
By symmetry the vertices are (±x, ±y, ±z) with x, y, z > 0 on the ellipsoid, and the volume is V = 8xyz.
F = xyz + λ(x2/a2 + y2/b2 + z2/c2 − 1). Fx = yz + 2λx/a2 = 0, Fy = xz + 2λy/b2 = 0, Fz = xy + 2λz/c2 = 0.
Multiplying by x, y, z: xyz = −2λx2/a2 = −2λy2/b2 = −2λz2/c2. As xyz ≠ 0, x2/a2 = y2/b2 = z2/c2, and their sum is 1, so each equals 1/3.
So x = a/√3, y = b/√3, z = c/√3 and V = 8abc/(3√3).
This is the maximum: V is continuous on the compact part of the ellipsoid with x, y, z ≥ 0, vanishes on its edges (where a coordinate is 0), and has only this stationary point inside.
Answer: the greatest volume is 8abc/(3√3).Final answer: the greatest volume is 8abc/(3√3).
Q3(c) · Linear Programming · 15 marks≈ 11 min
Duality in Linear Programming
Apply the principle of duality to solve the following linear programming problem:
Maximize Z = 3x1 + 4x2
subject to the constraints
x1 − x2 ≤ 1
x1 + x2 ≥ 4
x1 − 3x2 ≤ 3
x1, x2 ≥ 0हिंदी में प्रश्न
द्वैतता (ड्यूअलिटी) के सिद्धान्त का उपयोग कर निम्न रैखिक प्रोग्रामन समस्या को हल कीजिए:
अधिकतमीकरण कीजिए Z = 3x1 + 4x2
बशर्ते कि
x1 − x2 ≤ 1
x1 + x2 ≥ 4
x1 − 3x2 ≤ 3
x1, x2 ≥ 0Worked solution
Primal in Max canonical form: x1 − x2 ≤ 1, −x1 − x2 ≤ −4, x1 − 3x2 ≤ 3, x ≥ 0.
Dual: Min W = y1 − 4y2 + 3y3 subject to y1 − y2 + y3 ≥ 3, −y1 − y2 − 3y3 ≥ 4, y1, y2, y3 ≥ 0.
The second dual constraint is impossible: for y ≥ 0 its left side is ≤ 0, but it must be ≥ 4. So the dual has no feasible solution.
By the existence theorem the primal is then infeasible or unbounded. It is feasible, since (2, 2) gives 0 ≤ 1, 4 ≥ 4, −4 ≤ 3; hence it is unbounded. Directly, x1 = 0, x2 = t ≥ 4 is feasible for every t, with Z = 4t → ∞.
Answer: the dual is infeasible and the primal is feasible, so the problem has an unbounded solution (no finite maximum).Final answer: the dual is infeasible and the primal is feasible, so the problem has an unbounded solution (no finite maximum).
Question 4 (50 marks)
Q4(a) · Abstract Algebra · 15 marks≈ 11 min
Rings, Subrings, Ideals and Ring Homomorphisms
Examine whether the mapping φ : Z[x] → Z defined by φ(f(x)) = f(0), for f(x) ∈ Z[x], is a homomorphism. Deduce that the ideal ⟨x⟩ is a prime ideal in Z[x], but not a maximal ideal in Z[x].
हिंदी में प्रश्न
जाँचिए कि क्या f(x) ∈ Z[x] के लिए φ(f(x)) = f(0) द्वारा परिभाषित प्रतिचित्रण φ : Z[x] → Z एक समाकारिता है। निगमन कीजिए कि गुणजावली ⟨x⟩, Z[x] में एक अभाज्य गुणजावली है, किन्तु Z[x] में एक उच्चिष्ठ गुणजावली नहीं है।
Q4(b) · Real Analysis · 15 marks≈ 11 min
Riemann Integral, Improper Integrals and Fundamental Theorems of Integral Calculus
Prove that every continuous function is Riemann integrable.
हिंदी में प्रश्न
सिद्ध कीजिए कि प्रत्येक सतत फलन रीमान समाकलनीय है।
Q4(c) · Linear Programming · 20 marks≈ 14 min
Transportation Problem
The following table shows all the necessary information on the available supply to each warehouse, the requirement of each market and the unit transportation cost from each warehouse to each market:
Warehouse / Market: I, II, III, IV, Supply
A: 5, 2, 4, 3, 22
B: 4, 8, 1, 6, 15
C: 4, 6, 7, 5, 8
Requirement: 7, 12, 17, 9
The shipping clerk has worked out the following schedule from experience:
12 units from A to II, 1 unit from A to III, 9 units from A to IV, 15 units from B to III, 7 units from C to I and 1 unit from C to III
Find the optimal schedule and minimum total shipping cost.हिंदी में प्रश्न
निम्न सारणी में प्रत्येक गोदाम में उपलब्ध सप्लाई, प्रत्येक बाजार की आवश्यकता और प्रत्येक गोदाम से प्रत्येक बाजार की इकाई परिवहन लागत की सभी आवश्यक जानकारी दी गई है:
गोदाम / बाजार: I, II, III, IV, सप्लाई
A: 5, 2, 4, 3, 22
B: 4, 8, 1, 6, 15
C: 4, 6, 7, 5, 8
आवश्यकता: 7, 12, 17, 9
अनुभव के आधार पर शिपिंग क्लर्क ने निम्न अनुसूची (शेड्यूल) तैयार की है:
A से II पर 12 इकाई, A से III पर 1 इकाई, A से IV पर 9 इकाई, B से III पर 15 इकाई, C से I पर 7 इकाई और C से III पर 1 इकाई
इष्टतम अनुसूची और निम्नतम कुल परिवहन लागत ज्ञात कीजिए।
Question 5 (50 marks)
Q5(a) · Partial Differential Equations · 10 marks≈ 7 min
Linear Second-Order PDEs with Constant Coefficients
Find the solution of the equation (D2 + DD′ − 2D2) z = y sin x, where D ≡ ∂/∂x and D′ ≡ ∂/∂y.
हिंदी में प्रश्न
समीकरण (D2 + DD′ − 2D2) z = y sin x, जहाँ D ≡ ∂/∂x और D′ ≡ ∂/∂y हैं, का हल ज्ञात कीजिए।
Worked solution
[The paper prints (D2 + DD′ − 2D2); the intended operator is D2 + DD′ − 2D′2.]
m2 + m − 2 = 0 gives m = 1, −2: CF = φ1(y + x) + φ2(y − 2x), F = (D − D′)(D + 2D′).
(D + 2D′)−1y sin x = ∫(c + 2x)sin x dx (c = y − 2x) = −(c + 2x)cos x + 2 sin x = −y cos x + 2 sin x.
(D − D′)−1(−y cos x + 2 sin x) = ∫[−(c − x)cos x + 2 sin x] dx (c = y + x) = −c sin x + x sin x − cos x = −y sin x − cos x.
Answer: z = φ1(y + x) + φ2(y − 2x) − y sin x − cos x (check: zxx + zxy − 2zyy = y sin x + cos x − cos x = y sin x).Final answer: z = φ1(y + x) + φ2(y − 2x) − y sin x − cos x (check: zxx + zxy − 2zyy = y sin x + cos x − cos x = y sin x).
Q5(b) · Numerical Analysis and Computer Programming · 10 marks≈ 7 min
Systems of Linear Equations: Direct and Iterative Methods
Solve the following system of linear equations by Gauss-Seidel method:
10x + 2y + z = 9
2x + 20y − 2z = −44
−2x + 3y + 10z = 22हिंदी में प्रश्न
निम्न रैखिक समीकरण निकाय को गाउस-सीडल विधि से हल कीजिए:
10x + 2y + z = 9
2x + 20y − 2z = −44
−2x + 3y + 10z = 22Q5(c) · Numerical Analysis and Computer Programming · 10 marks≈ 7 min
Number Systems and Representation of Numbers in a Computer
(i) Convert the number (3479)10 into binary system and the number (7AE·9F)16 into decimal system.
(ii) Determine the truth table for the Boolean function
F(x, y, z) = (x + y + z′)(x′ + y′)
Also derive the full disjunctive normal form of F(x, y, z) from the truth table.हिंदी में प्रश्न
(i) संख्या (3479)10 को द्वि-आधारी पद्धति और संख्या (7AE·9F)16 को दशमलव पद्धति में बदलिए।
(ii) बूलीय फलन F(x, y, z) = (x + y + z′)(x′ + y′) के लिए सत्यमान सारणी ज्ञात कीजिए। सत्यमान सारणी से F(x, y, z) का पूर्ण वियोजनीय प्रसामान्य रूप भी प्राप्त कीजिए।Q5(d) · Mechanics and Fluid Dynamics · 10 marks≈ 7 min
Generalized Coordinates, D'Alembert's Principle and Lagrange's Equations
A bead of mass m slides on a frictionless wire in the shape of a cycloid given by x = a(θ − sin θ), y = a(1 + cos θ), (0 ≤ θ ≤ 2π). Find the Lagrangian function. Hence show that the equation of motion can be written as
d2u/dt2 + (g/4a) u = 0
where u = cos(θ/2). (4 + 6 = 10 marks)हिंदी में प्रश्न
x = a(θ − sin θ), y = a(1 + cos θ), (0 ≤ θ ≤ 2π) द्वारा दिए गए एक चक्रज के रूप में एक घर्षणहीन तार पर m द्रव्यमान का एक मनका फिसलता है। लग्रांजी फलन ज्ञात कीजिए। अतः दर्शाइए कि गति का समीकरण
d2u/dt2 + (g/4a) u = 0
के रूप में लिखा जा सकता है, जहाँ u = cos(θ/2) है। (4 + 6 = 10 अंक)Worked solution
ẋ = a(1 − cos θ)θ̇, ẏ = −a sin θ θ̇, so v2 = 2a2(1 − cos θ)θ̇2 = 4a2sin2(θ/2)θ̇2.
With y vertically upward, V = mgy = 2mga cos2(θ/2), so L = 2ma2sin2(θ/2)θ̇2 − 2mga cos2(θ/2).
Put u = cos(θ/2), a valid coordinate since it is monotonic on 0 ≤ θ ≤ 2π: u̇ = −(1/2)sin(θ/2)θ̇, so sin2(θ/2)θ̇2 = 4u̇2 and L = 8ma2u̇2 − 2mgau2.
Lagrange's equation in u: 16ma2ü + 4mgau = 0.
Answer: L = 2ma2sin2(θ/2)θ̇2 − 2mga cos2(θ/2); d2u/dt2 + (g/(4a))u = 0, simple harmonic with period 4π√(a/g) whatever the amplitude (the cycloid is tautochronous).Final answer: L = 2ma2sin2(θ/2)θ̇2 − 2mga cos2(θ/2); d2u/dt2 + (g/(4a))u = 0, simple harmonic with period 4π√(a/g) whatever the amplitude (the cycloid is tautochronous).
Q5(e) · Mechanics and Fluid Dynamics · 10 marks≈ 7 min
Two-Dimensional and Axisymmetric Motion: Sources, Sinks and Vortices
A source and a sink of equal strength are placed at points (±a/2, 0) within a fixed circular boundary x2 + y2 = a2. Show that the streamlines are given by
(r2 − a2/4)(r2 − 4a2) − 4a2y2 = ky(r2 − a2)
where k is a constant and r2 = x2 + y2.हिंदी में प्रश्न
बराबर सामर्थ्य के एक स्रोत और एक अभिगम एक निश्चित वृत्तीय सीमा x2 + y2 = a2 के अंतर्गत बिंदुओं (±a/2, 0) पर रखे हैं। दर्शाइए कि धारारेखाएँ
(r2 − a2/4)(r2 − 4a2) − 4a2y2 = ky(r2 − a2)
द्वारा दी जाती हैं, जहाँ k एक अचर है और r2 = x2 + y2 है।Worked solution
f(z) = −m log(z − a/2) + m log(z + a/2); by the circle theorem f̄(a2/z) = −m log(z − 2a) + m log(z + 2a) + constant (the log z terms cancel): images a source at 2a and a sink at −2a.
w = −m log[(P − Q)/(P + Q)] with P = z2 + a2, Q = (5a/2)z, since (z − a/2)(z − 2a) = P − Q and (z + a/2)(z + 2a) = P + Q.
arg[(P − Q)/(P + Q)] = arg[(P − Q)(P̄ + Q̄)] and (P − Q)(P̄ + Q̄) = |P|2 − |Q|2 + 2i Im(PQ̄), with Im(PQ̄) = (5a/2)y(r2 − a2).
|P|2 − |Q|2 = r4 + 2a2(x2 − y2) + a4 − (25/4)a2r2 = (r2 − a2/4)(r2 − 4a2) − 4a2y2.
Answer: ψ = constant gives (r2 − a2/4)(r2 − 4a2) − 4a2y2 = ky(r2 − a2); the boundary r = a and the x-axis form the streamline y(r2 − a2) = 0.Final answer: ψ = constant gives (r2 − a2/4)(r2 − 4a2) − 4a2y2 = ky(r2 − a2); the boundary r = a and the x-axis form the streamline y(r2 − a2) = 0.
Question 6 (50 marks)
Q6(a) · Partial Differential Equations · 20 marks≈ 14 min
Vibrating String, Heat Equation and Laplace Equation
Solve ∂2u/∂x2 + ∂2u/∂y2 = 0 for a rectangular plate subject to the boundary conditions
u(0, y) = 0, u(a, y) = 0
u(x, 0) = 0, u(x, b) = f(x)हिंदी में प्रश्न
एक आयताकार प्लेट के लिए ∂2u/∂x2 + ∂2u/∂y2 = 0 को परिसीमा प्रतिबन्धों
u(0, y) = 0, u(a, y) = 0
u(x, 0) = 0, u(x, b) = f(x)
के अधीन हल कीजिए।Q6(b) · Numerical Analysis and Computer Programming · 15 marks≈ 11 min
Computer Systems, Logic Gates, Boolean Algebra and Algorithms
Simplify the Boolean function
F(x, y, z) = xyz + x′yz + xy′z + xyz′
and draw the corresponding GATE network.हिंदी में प्रश्न
बूलीय फलन
F(x, y, z) = xyz + x′yz + xy′z + xyz′
का सरलीकरण कीजिए और संगत GATE परिपथ को रेखांकित कीजिए।Q6(c) · Mechanics and Fluid Dynamics · 15 marks≈ 11 min
Moment of Inertia and Motion of Rigid Bodies in Two Dimensions
Calculate the moment of inertia of a uniform solid cylinder of mass M, radius R and length L with respect to a set of axes passing through the centre of the cylinder, where z-axis is the axis of the cylinder and ρ is the constant density at any point of the cylinder. Also find L/R for which the moment of inertia about x- or y-axis will be minimum for a given mass of the cylinder.
हिंदी में प्रश्न
द्रव्यमान M, त्रिज्या R और लम्बाई L के एक एकसमान ठोस बेलन का, बेलन के केन्द्र से होकर जाने वाले अक्षों के एक समुच्चय के सापेक्ष जड़त्व आघूर्ण की गणना कीजिए, जहाँ z-अक्ष, बेलन का अक्ष है और ρ, बेलन के किसी भी बिन्दु पर अचर घनत्व है। बेलन के एक दिए गए द्रव्यमान के लिए L/R, जिसके लिए x- या y-अक्ष के सापेक्ष जड़त्व आघूर्ण न्यूनतम होगा, भी ज्ञात कीजिए।
Worked solution
Cylindrical coordinates (s, θ, z), −L/2 ≤ z ≤ L/2, M = ρπR2L: Iz = ρ∭s2 · s ds dθ dz = ρ · 2π · L · R4/4 = MR2/2.
∫x2 dm = ∫y2 dm = Iz/2 = MR2/4 and ∫z2 dm = ρπR2∫z2 dz = ML2/12, so Ix = Iy = M(R2/4 + L2/12); all products of inertia vanish (principal axes).
For fixed M, R2 = M/(ρπL), so Ix = M[M/(4ρπL) + L2/12] and dIx/dL = M[−M/(4ρπL2) + L/6] = 0 gives L3 = 3M/(2ρπ) = (3/2)R2L.
The second derivative M[M/(2ρπL3) + 1/6] is positive, so this is a minimum.
Answer: Iz = MR2/2, Ix = Iy = M(3R2 + L2)/12; minimum when L/R = √(3/2) ≈ 1.225.Final answer: Iz = MR2/2, Ix = Iy = M(3R2 + L2)/12; minimum when L/R = √(3/2) ≈ 1.225.
Question 7 (50 marks)
Q7(a) · Partial Differential Equations · 15 marks≈ 11 min
Cauchy's Method of Characteristics and Nonlinear First-Order PDEs
Find the complete integral of z(p2 − q2) = x − y; p ≡ ∂z/∂x, q ≡ ∂z/∂y.
हिंदी में प्रश्न
z(p2 − q2) = x − y; p ≡ ∂z/∂x, q ≡ ∂z/∂y का पूर्ण समाकल ज्ञात कीजिए।
Worked solution
Write the equation as (√z p)2 − (√z q)2 = x − y and put Z = (2/3)z3/2, so that P = Zx = √z p and Q = Zy = √z q.
Then P2 − x = Q2 − y = a (separable form), P = √(x + a), Q = √(y + a).
dZ = √(x + a) dx + √(y + a) dy, so Z = (2/3)(x + a)3/2 + (2/3)(y + a)3/2 + (2/3)b.
Answer: z3/2 = (x + a)3/2 + (y + a)3/2 + b.Final answer: z3/2 = (x + a)3/2 + (y + a)3/2 + b.
Q7(b) · Numerical Analysis and Computer Programming · 15 marks≈ 11 min
Interpolation: Newton's Forward, Backward and Lagrange's Formulae
Find the unique polynomial of degree 2 or less which fits the following data:
x : 0, 1, 3
f(x) : 1, 3, 55
Also obtain the bound on the truncation error.हिंदी में प्रश्न
घात 2 या 2 से कम का वह अद्वितीय बहुपद, जो आँकड़ों
x : 0, 1, 3
f(x) : 1, 3, 55
पर ठीक बैठता है, ज्ञात कीजिए। रुंडन त्रुटि पर परिबंध भी प्राप्त कीजिए।Worked solution
Lagrange: P(x) = 1 · (x − 1)(x − 3)/3 + 3 · x(x − 3)/(−2) + 55 · x(x − 1)/6.
Coefficient of x2: 1/3 − 3/2 + 55/6 = 8; of x: −4/3 + 9/2 − 55/6 = −6; constant 1. So P(x) = 8x2 − 6x + 1 (check: P(0) = 1, P(1) = 3, P(3) = 55).
Uniqueness: two such polynomials would differ by a polynomial of degree ≤ 2 with the three roots 0, 1, 3, hence by zero.
Truncation error E(x) = f‴(ξ)x(x − 1)(x − 3)/3!. For w(x) = x3 − 4x2 + 3x, w′ = 3x2 − 8x + 3 = 0 at x = (4 ± √7)/3 = 0.4514, 2.2153, where |w| = 0.6311 and 2.1126.
Answer: P(x) = 8x2 − 6x + 1; for 0 ≤ x ≤ 3, |E(x)| ≤ 2.1126 M3/6 ≈ 0.352 M3, where M3 = max|f‴| on [0, 3].Final answer: P(x) = 8x2 − 6x + 1; for 0 ≤ x ≤ 3, |E(x)| ≤ 2.1126 M3/6 ≈ 0.352 M3, where M3 = max|f‴| on [0, 3].
Q7(c) · Mechanics and Fluid Dynamics · 20 marks≈ 14 min
Euler's Equation, Bernoulli's Theorem and the Navier–Stokes Equation
Show that for an incompressible steady flow with constant viscosity, the velocity components
u(y) = (U/h) y − (hy/2μ)(dp/dx)(1 − y/h)
v = 0 = w, with p = p(x), satisfy the equation of motion in the absence of body force. Given that U, h and dp/dx are constants.हिंदी में प्रश्न
दर्शाइए कि अचर विस्कासिता के एक असंपीड्य अपरिवर्ती प्रवाह के लिए वेग घटक
u(y) = (U/h) y − (hy/2μ)(dp/dx)(1 − y/h)
v = 0 = w, p = p(x) के साथ, पिंड बल की अनुपस्थिति में गति के समीकरण को सन्तुष्ट करते हैं। यह दिया गया है कि U, h और dp/dx अचर हैं।Worked solution
Here u = (U/h)y − (1/(2μ))(dp/dx)(hy − y2), v = w = 0, so ∂u/∂x = 0 and continuity holds; the inertia terms u∂u/∂x + v∂u/∂y + w∂u/∂z all vanish.
d2u/dy2 = (1/(2μ))(dp/dx) · 2 = (1/μ)dp/dx, and ∂2u/∂x2 = ∂2u/∂z2 = 0, so the x-equation 0 = −dp/dx + μ∇2u holds.
The y- and z-equations reduce to ∂p/∂y = ∂p/∂z = 0, true since p = p(x).
Answer: the field satisfies all the equations of motion; it is Couette–Poiseuille flow with u(0) = 0 and u(h) = U (lower plate fixed, upper plate moving with speed U).Final answer: the field satisfies all the equations of motion; it is Couette–Poiseuille flow with u(0) = 0 and u(h) = U (lower plate fixed, upper plate moving with speed U).
Question 8 (50 marks)
Q8(a) · Partial Differential Equations · 15 marks≈ 11 min
Cauchy's Method of Characteristics and Nonlinear First-Order PDEs
Find the characteristics of the partial differential equation
p2 + q2 = 2; p ≡ ∂z/∂x, q ≡ ∂z/∂y
and determine the integral surface which passes through x = 0, z = y.हिंदी में प्रश्न
आंशिक अवकल समीकरण
p2 + q2 = 2; p ≡ ∂z/∂x, q ≡ ∂z/∂y
के अभिलक्षण (कैरेक्टरिस्टिक्स) ज्ञात कीजिए और x = 0, z = y से होकर जाने वाला समाकल पृष्ठ (सरफेस) प्राप्त कीजिए।Worked solution
F = p2 + q2 − 2: dx/dt = 2p, dy/dt = 2q, dz/dt = 2(p2 + q2) = 4, dp/dt = dq/dt = 0; p, q are constant on each strip, so the characteristics are straight lines.
Initial curve x0 = 0, y0 = s, z0 = s: the strip condition gives q0 = 1 and F = 0 gives p0 = ±1.
x = 2p0t, y = s + 2t, z = s + 4t, so z − y = 2t = x/p0.
Answer: characteristics x = ±2t, y = s + 2t, z = s + 4t (with p = ±1, q = 1); integral surfaces z = y + x and z = y − x.Final answer: characteristics x = ±2t, y = s + 2t, z = s + 4t (with p = ±1, q = 1); integral surfaces z = y + x and z = y − x.
Q8(b) · Numerical Analysis and Computer Programming · 15 marks≈ 11 min
Numerical Integration and Numerical Solution of ODEs
Find the constant p and error term for the quadrature formula
∫x0x1 f(x) dx = (h/2)(f0 + f1) + ph2(f′0 − f′1)
where x0 + h = x1, f0 = f(x0), f1 = f(x1) and prime (′) represents derivative with respect to x. Hence deduce the composite rule for integrating
∫ab f(x) dx, a = x0 < x1 < ⋯ < xN = bहिंदी में प्रश्न
क्षेत्रकलन-सूत्र
∫x0x1 f(x) dx = (h/2)(f0 + f1) + ph2(f′0 − f′1)
जहाँ x0 + h = x1, f0 = f(x0), f1 = f(x1) हैं और प्राइम (′), x के सापेक्ष अवकलज को निरूपित करता है, के लिए अचर p और त्रुटि-पद ज्ञात कीजिए। अतः समाकलन
∫ab f(x) dx, a = x0 < x1 < ⋯ < xN = b
का मान ज्ञात करने के लिए संयुक्त नियम का निगमन कीजिए।Worked solution
Take x0 = 0 (the rule is unchanged by translation). For f = 1 and f = x the rule is exact for every p (the trapezoidal part is exact and f′0 − f′1 = 0).
f = x2: h3/3 = (h/2)h2 + ph2(0 − 2h) = h3/2 − 2ph3, so p = 1/12.
f = x3: rule h4/2 − (h2/12)(3h2) = h4/4, exact; f = x4: rule h5/2 − h5/3 = h5/6 against h5/5, so the degree is 3, C = h5/30 and E = Cf⁗(ξ)/4! = (h5/720)f⁗(ξ).
Composite: with h = (b − a)/N, xi = a + ih, adding the rule over each [xi−1, xi], the derivative terms telescope to (h2/12)[f′(a) − f′(b)].
Answer: p = 1/12, error (h5/720)f⁗(ξ); ∫ab f dx = h[f0/2 + f1 + ... + fN−1 + fN/2] + (h2/12)[f′(a) − f′(b)] + ((b − a)h4/720)f⁗(η), the corrected trapezoidal rule.Final answer: p = 1/12, error (h5/720)f⁗(ξ); ∫ab f dx = h[f0/2 + f1 + ... + fN−1 + fN/2] + (h2/12)[f′(a) − f′(b)] + ((b − a)h4/720)f⁗(η), the corrected trapezoidal rule.
Q8(c) · Mechanics and Fluid Dynamics · 20 marks≈ 14 min
Hamilton's Equations
(i) A particle of mass m moves in a force field of potential
V(r) = −k cos θ/r2, k is constant
Find the Hamiltonian and the Hamilton's equations in spherical polar coordinates (r, θ, φ). (10 marks)
(ii) Consider the Lagrangian
L = m ẋ ẏ − mω02 xy
where m and ω0 are constants. Find the Hamiltonian and Hamilton's equations of motion. Identify the system. (10 marks)हिंदी में प्रश्न
(i) विभव
V(r) = −k cos θ/r2, k अचर है
के एक बल क्षेत्र में द्रव्यमान m का एक कण गतिमान है। गोलीय ध्रुवीय निर्देशांकों (r, θ, φ) में हैमिल्टोनियन और हैमिल्टन का समीकरण ज्ञात कीजिए। (10 अंक)
(ii) लग्रांजी L = m ẋ ẏ − mω02 xy, जहाँ m और ω0 अचर हैं, का विचार कीजिए। हैमिल्टोनियन और गति का हैमिल्टन समीकरण ज्ञात कीजिए। तंत्र (सिस्टम) की पहचान बताइए। (10 अंक)Worked solution
(i) T = (m/2)(ṙ2 + r2θ̇2 + r2sin2θ φ̇2); pr = mṙ, pθ = mr2θ̇, pφ = mr2sin2θ φ̇; H = (1/(2m))(pr2 + pθ2/r2 + pφ2/(r2sin2θ)) − k cos θ/r2.
ṙ = pr/m, θ̇ = pθ/(mr2), φ̇ = pφ/(mr2sin2θ), ṗr = (pθ2 + pφ2/sin2θ)/(mr3) − 2k cos θ/r3.
ṗθ = pφ2cos θ/(mr2sin3θ) − k sin θ/r2, ṗφ = 0 (φ is cyclic).
(ii) px = ∂L/∂ẋ = mẏ, py = mẋ; H = pxẋ + pyẏ − L = mẋẏ + mω02xy = pxpy/m + mω02xy.
ẋ = py/m, ẏ = px/m, ṗx = −mω02y, ṗy = −mω02x, so ẍ + ω02x = 0 and ÿ + ω02y = 0.
Answer: (i) H and the six equations above; (ii) H = pxpy/m + mω02xy, and the system is a two-dimensional isotropic harmonic oscillator of frequency ω0 (H is conserved, though not of the form T + V).Final answer: (i) H and the six equations above; (ii) H = pxpy/m + mω02xy, and the system is a two-dimensional isotropic harmonic oscillator of frequency ω0 (H is conserved, though not of the form T + V).