UPSC Mains 2018 Mathematics Optional Question Papers with Solutions
All 62 questions of the 2018 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 2018 Mathematics Paper I
Question 1 (50 marks)
Q1(a) · Linear Algebra · 10 marks≈ 7 min
Algebra of Matrices, Echelon Form, Rank, Inverse, Congruence and Similarity
Let A be a 3 × 2 matrix and B a 2 × 3 matrix. Show that C = A·B is a singular matrix.
हिंदी में प्रश्न
मान लीजिये कि A एक 3 × 2 आव्यूह है और B एक 2 × 3 आव्यूह है। दर्शाइये कि C = A·B एक अव्युत्क्रमणीय आव्यूह है।
Worked solution
ρ(C) = ρ(AB) ≤ min{ρ(A), ρ(B)} ≤ 2, because A has only 2 columns (and B only 2 rows).
C is 3 × 3, so ρ(C) ≤ 2 < 3 and det C = 0. (Alternatively: BX = 0 is 2 equations in 3 unknowns, so it has a solution X ≠ 0; then CX = A(BX) = 0 with X ≠ 0.)
Answer: C = AB has rank at most 2, so it is singular.Final answer: C = AB has rank at most 2, so it is singular.
Q1(b) · Linear Algebra · 10 marks≈ 7 min
Vector Spaces, Subspaces, Basis and Dimension
Express basis vectors e1 = (1, 0) and e2 = (0, 1) as linear combinations of α1 = (2, −1) and α2 = (1, 3).
हिंदी में प्रश्न
आधार सदिशों e1 = (1, 0) और e2 = (0, 1) को α1 = (2, −1) एवं α2 = (1, 3) के रैखिक संयोग के रूप में व्यक्त कीजिये।
Q1(c) · Calculus · 10 marks≈ 7 min
Mean Value Theorems, Taylor's Theorem and Indeterminate Forms
Determine if limz→1 (1 − z) tan(πz/2) exists or not. If the limit exists, then find its value.
हिंदी में प्रश्न
निर्धारित कीजिये कि limz→1 (1 − z) tan(πz/2) का अस्तित्व है या कि नहीं। अगर यह सीमा विद्यमान है, तो इसका मान ज्ञात कीजिये।
Worked solution
As z → 1, 1 − z → 0 and tan(πz/2) → ±∞: the form is 0·∞. Write (1 − z) tan(πz/2) = [(1 − z)/cos(πz/2)] · sin(πz/2).
(1 − z)/cos(πz/2) is a 0/0 form; by L'Hôpital's rule it tends to lim (−1)/(−(π/2) sin(πz/2)) = 2/π, and sin(πz/2) → 1.
(Check: with z = 1 − t, tan(π/2 − πt/2) = cot(πt/2), and t cot(πt/2) = (2/π)·(πt/2)/tan(πt/2) → 2/π from both sides.)
Answer: The limit exists and equals 2/π.Final answer: The limit exists and equals 2/π.
Q1(d) · Calculus · 10 marks≈ 7 min
Riemann Integral, Indefinite Integrals and Improper Integrals
Find the limit limn→∞ (1/n2) ∑r=0n−1 √(n2 − r2).
हिंदी में प्रश्न
सीमा limn→∞ (1/n2) ∑r=0n−1 √(n2 − r2) का मान ज्ञात कीजिये।
Q1(e) · Analytic Geometry · 10 marks≈ 7 min
Straight Lines and Shortest Distance between Skew Lines
Find the projection of the straight line (x − 1)/2 = (y − 1)/3 = (z + 1)/(−1) on the plane x + y + 2z = 6.
हिंदी में प्रश्न
सरल रेखा (x − 1)/2 = (y − 1)/3 = (z + 1)/(−1) का समतल x + y + 2z = 6 पर प्रक्षेपण ज्ञात कीजिये।
Worked solution
The line passes through (1, 1, −1) with direction (2, 3, −1); the plane has normal (1, 1, 2). The plane through the line perpendicular to the given plane has normal (2, 3, −1) × (1, 1, 2) = (7, −5, −1): 7(x − 1) − 5(y − 1) − (z + 1) = 0, that is 7x − 5y − z = 3.
The projection is the line 7x − 5y − z = 3, x + y + 2z = 6. Its direction is (7, −5, −1) × (1, 1, 2) = (−9, −15, 12), proportional to (3, 5, −4).
The given line meets the plane where (1 + 2t) + (1 + 3t) + 2(−1 − t) = 6, so t = 2 and the point is (5, 7, −3); this point lies on the projection.
Answer: the projection is (x − 5)/3 = (y − 7)/5 = (z + 3)/(−4), that is 7x − 5y − z = 3, x + y + 2z = 6.Final answer: the projection is (x − 5)/3 = (y − 7)/5 = (z + 3)/(−4), that is 7x − 5y − z = 3, x + y + 2z = 6.
Question 2 (50 marks)
Q2(a) · Linear Algebra · 12 marks≈ 9 min
Eigenvalues, Eigenvectors and the Cayley-Hamilton Theorem
Show that if A and B are similar n × n matrices, then they have the same eigenvalues.
हिंदी में प्रश्न
अगर A और B समरूप n × n आव्यूह हैं, तो दर्शाइये कि उनके आइगेन मान एक ही हैं।
Q2(b) · Calculus · 13 marks≈ 9 min
Functions of Several Variables: Partial Derivatives, Extrema, Lagrange Multipliers and Jacobians
Find the shortest distance from the point (1, 0) to the parabola y2 = 4x.
हिंदी में प्रश्न
बिन्दु (1, 0) से परवलय y2 = 4x तक की न्यूनतम दूरी ज्ञात कीजिये।
Worked solution
Minimise D = (x − 1)2 + y2 subject to y2 = 4x (so x ≥ 0), with F = (x − 1)2 + y2 + λ(y2 − 4x).
Fx = 2(x − 1) − 4λ = 0 and Fy = 2y(1 + λ) = 0. If λ = −1 then x = −1, impossible since y2 = 4x ≥ 0; so y = 0 and x = 0.
Check directly: on the parabola D = (x − 1)2 + 4x = (x + 1)2, which for x ≥ 0 is least at x = 0. (The point (1, 0) is the focus, and the vertex is the nearest point.)
Answer: Shortest distance 1, attained at the vertex (0, 0).Final answer: Shortest distance 1, attained at the vertex (0, 0).
Q2(c) · Calculus · 13 marks≈ 9 min
Double and Triple Integrals; Areas, Surfaces and Volumes
The ellipse x2/a2 + y2/b2 = 1 revolves about the x-axis. Find the volume of the solid of revolution.
हिंदी में प्रश्न
दीर्घवृत्त x2/a2 + y2/b2 = 1 x-अक्ष के चारों तरफ परिभ्रमण कर रहा है। परिक्रमित घन का आयतन ज्ञात कीजिये।
Q2(d) · Analytic Geometry · 12 marks≈ 9 min
Straight Lines and Shortest Distance between Skew Lines
Find the shortest distance between the lines
a1x + b1y + c1z + d1 = 0
a2x + b2y + c2z + d2 = 0
and the z-axis.हिंदी में प्रश्न
रेखाओं
a1x + b1y + c1z + d1 = 0
a2x + b2y + c2z + d2 = 0
और z-अक्ष के बीच की न्यूनतम दूरी ज्ञात कीजिये।Worked solution
Write Pi = aix + biy + ciz + di. A plane through the given line is P1 + λP2 = 0; it is parallel to the z-axis (direction (0, 0, 1)) when c1 + λc2 = 0, so take c2P1 − c1P2 = 0.
This plane is (a1c2 − a2c1)x + (b1c2 − b2c1)y + (d1c2 − d2c1) = 0. The SD is the distance of any point of the z-axis, say the origin, from it.
Answer: SD = |d1c2 − d2c1|/√((a1c2 − a2c1)2 + (b1c2 − b2c1)2).Final answer: SD = |d1c2 − d2c1|/√((a1c2 − a2c1)2 + (b1c2 − b2c1)2).
Question 3 (50 marks)
Q3(a) · Linear Algebra · 13 marks≈ 9 min
Systems of Linear Equations
For the system of linear equations
x + 3y − 2z = −1
5y + 3z = −8
x − 2y − 5z = 7
determine which of the following statements are true and which are false :
(i) The system has no solution.
(ii) The system has a unique solution.
(iii) The system has infinitely many solutions.हिंदी में प्रश्न
रैखिक समीकरण निकाय
x + 3y − 2z = −1
5y + 3z = −8
x − 2y − 5z = 7
के लिये निर्धारित कीजिये कि निम्नलिखित कथनों में से कौन-से सही हैं और कौन-से गलत :
(i) समीकरण निकाय का कोई भी हल नहीं है।
(ii) समीकरण निकाय का सिर्फ एक ही हल है।
(iii) समीकरण निकाय के असीम मात्रा में अनेक हल हैं।Q3(b) · Calculus · 12 marks≈ 9 min
Functions of Several Variables: Partial Derivatives, Extrema, Lagrange Multipliers and Jacobians
Let
f(x, y) = xy2, if y > 0
= −xy2, if y ≤ 0.
Determine which of ∂f/∂x (0, 1) and ∂f/∂y (0, 1) exists and which does not exist.हिंदी में प्रश्न
मान लीजिये कि
f(x, y) = xy2, यदि y > 0
= −xy2, यदि y ≤ 0
निर्धारित कीजिये कि ∂f/∂x (0, 1) और ∂f/∂y (0, 1) में से किसका अस्तित्व है और किसका अस्तित्व नहीं है।Worked solution
For all (x, y), f(x, y) = xy|y| (for y ≤ 0, xy|y| = −xy2). Near (0, 1), y > 0, so f = xy2 there.
∂f/∂x (0, 1) = limh→0 (f(h, 1) − f(0, 1))/h = lim (h − 0)/h = 1, so it exists.
∂f/∂y (0, 1) = limk→0 (f(0, 1 + k) − f(0, 1))/k = lim (0 − 0)/k = 0, so it also exists.
In fact fx = y|y| and fy = 2x|y| exist at every point, even on y = 0 where the formula changes, because t|t| is differentiable with derivative 2|t|.
Answer: Both exist: ∂f/∂x (0, 1) = 1 and ∂f/∂y (0, 1) = 0.Final answer: Both exist: ∂f/∂x (0, 1) = 1 and ∂f/∂y (0, 1) = 0.
Q3(c) · Analytic Geometry · 13 marks≈ 9 min
Ellipsoid, Hyperboloids and Paraboloids
Find the equations to the generating lines of the paraboloid (x + y + z)(2x + y − z) = 6z which pass through the point (1, 1, 1).
हिंदी में प्रश्न
परवलयज (x + y + z)(2x + y − z) = 6z की उन जनक रेखाओं के समीकरणों को ज्ञात कीजिये, जो बिन्दु (1, 1, 1) में से गुजरती हैं।
Worked solution
Write U = x + y + z, V = 2x + y − z, so UV = 6z; at (1, 1, 1), U = 3, V = 2 and UV = 6 = 6z, so the point is on the surface.
System U = λ, λV = 6z: λ = 3 gives x + y + z = 3, 3(2x + y − z) = 6z, that is 2x + y − 3z = 0; direction (1, 1, 1) × (2, 1, −3) = (−4, 5, −1).
System V = μ, μU = 6z: μ = 2 gives 2x + y − z = 2, 2(x + y + z) = 6z, that is x + y − 2z = 0; direction (2, 1, −1) × (1, 1, −2) = (−1, 3, 1).
Answer: (x − 1)/4 = (y − 1)/(−5) = (z − 1)/1 and (x − 1)/(−1) = (y − 1)/3 = (z − 1)/1 (check: (5, −4, 2) and (0, 4, 2) lie on the surface).Final answer: (x − 1)/4 = (y − 1)/(−5) = (z − 1)/1 and (x − 1)/(−1) = (y − 1)/3 = (z − 1)/1 (check: (5, −4, 2) and (0, 4, 2) lie on the surface).
Q3(d) · Analytic Geometry · 12 marks≈ 9 min
The Sphere
Find the equation of the sphere in xyz-plane passing through the points (0, 0, 0), (0, 1, −1), (−1, 2, 0) and (1, 2, 3).
हिंदी में प्रश्न
xyz-समतल में स्थित, बिन्दुओं (0, 0, 0), (0, 1, −1), (−1, 2, 0) और (1, 2, 3) में से गुजरते हुये गोले का समीकरण ज्ञात कीजिये।
Worked solution
Take x2 + y2 + z2 + 2ux + 2vy + 2wz + d = 0; the origin gives d = 0.
(0, 1, −1): 2 + 2v − 2w = 0, so w = v + 1. (−1, 2, 0): 5 − 2u + 4v = 0. (1, 2, 3): 14 + 2u + 4v + 6w = 0.
Adding the last two: 19 + 8v + 6w = 0, so 14v + 25 = 0 and v = −25/14, w = −11/14, u = (5 + 4v)/2 = −15/14.
Answer: 7(x2 + y2 + z2) − 15x − 25y − 11z = 0 (check with (1, 2, 3): 98 − 15 − 50 − 33 = 0).Final answer: 7(x2 + y2 + z2) − 15x − 25y − 11z = 0 (check with (1, 2, 3): 98 − 15 − 50 − 33 = 0).
Question 4 (50 marks)
Q4(a) · Calculus · 13 marks≈ 9 min
Maxima and Minima, Asymptotes and Curve Tracing
Find the maximum and the minimum values of x4 − 5x2 + 4 on the interval [2, 3].
हिंदी में प्रश्न
अन्तराल [2, 3] पर x4 − 5x2 + 4 के अधिकतम और न्यूनतम मान ज्ञात कीजिये।
Worked solution
f(x) = x4 − 5x2 + 4, f′(x) = 4x3 − 10x = 2x(2x2 − 5), which vanishes at x = 0 and x = ±√(5/2) ≈ ±1.58, none of them in [2, 3].
On [2, 3], f′(x) = 2x(2x2 − 5) > 0, so f is strictly increasing and the extreme values are at the end points: f(2) = 16 − 20 + 4 = 0, f(3) = 81 − 45 + 4 = 40.
Answer: Minimum 0 (at x = 2), maximum 40 (at x = 3).Final answer: Minimum 0 (at x = 2), maximum 40 (at x = 3).
Q4(b) · Calculus · 12 marks≈ 9 min
Double and Triple Integrals; Areas, Surfaces and Volumes
Evaluate the integral ∫0a ∫x/ax x dy dx/(x2 + y2).
हिंदी में प्रश्न
समाकल ∫0a ∫x/ax x dy dx/(x2 + y2) का मान निकालिये।
Worked solution
Inner integral (x > 0 fixed): ∫x/ax x dy/(x2 + y2) = [tan−1(y/x)]y=x/ay=x = π/4 − tan−1(1/a), which does not depend on x.
Outer integral: ∫0a (π/4 − tan−1(1/a)) dx = a(π/4 − tan−1(1/a)).
Since tan(π/4 − φ) = (1 − tan φ)/(1 + tan φ), this also equals a tan−1((a − 1)/(a + 1)) for a > 0.
Answer: a(π/4 − tan−1(1/a)) = a tan−1((a − 1)/(a + 1)).Final answer: a(π/4 − tan−1(1/a)) = a tan−1((a − 1)/(a + 1)).
Q4(c) · Analytic Geometry · 13 marks≈ 9 min
Cone and Cylinder
Find the equation of the cone with (0, 0, 1) as the vertex and 2x2 − y2 = 4, z = 0 as the guiding curve.
हिंदी में प्रश्न
उस शंकु, जिसका शीर्ष (0, 0, 1) है और जिसका निर्देशक वक्र 2x2 − y2 = 4, z = 0 है, का समीकरण ज्ञात कीजिये।
Q4(d) · Analytic Geometry · 12 marks≈ 9 min
Coordinates in Three Dimensions and the Plane
Find the equation of the plane parallel to 3x − y + 3z = 8 and passing through the point (1, 1, 1).
हिंदी में प्रश्न
3x − y + 3z = 8 के समांतर और बिन्दु (1, 1, 1) से गुजरते हुये समतल का समीकरण ज्ञात कीजिये।
Question 5 (50 marks)
Q5(a) · Ordinary Differential Equations · 10 marks≈ 7 min
Linear Equations with Constant Coefficients
Solve : y″ − y = x2e2x
हिंदी में प्रश्न
हल कीजिये : y″ − y = x2e2x
Q5(b) · Vector Analysis · 10 marks≈ 7 min
Curves in Space: Curvature, Torsion and Serret-Frenet Formulae
Find the angle between the tangent at a general point of the curve whose equations are x = 3t, y = 3t2, z = 3t3 and the line y = z − x = 0.
हिंदी में प्रश्न
x = 3t, y = 3t2, z = 3t3 समीकरणों वाले वक्र के एक आम बिन्दु पर स्पर्श-रेखा और रेखा y = z − x = 0 के बीच का कोण ज्ञात कीजिये।
Worked solution
r′ = (3, 6t, 9t2), parallel to (1, 2t, 3t2).
The line y = 0, z = x has direction (1, 0, 1).
cos θ = (1 + 3t2)/(√2 √(1 + 4t2 + 9t4)).
Answer: θ = cos−1[(1 + 3t2)/√(2(1 + 4t2 + 9t4))]; for example θ = 45° at t = 0.Final answer: θ = cos−1[(1 + 3t2)/√(2(1 + 4t2 + 9t4))]; for example θ = 45° at t = 0.
Q5(c) · Ordinary Differential Equations · 10 marks≈ 7 min
Linear Equations with Constant Coefficients
Solve : y‴ − 6y″ + 12y′ − 8y = 12e2x + 27e−x
हिंदी में प्रश्न
हल कीजिये : y‴ − 6y″ + 12y′ − 8y = 12e2x + 27e−x
Q5(d) · Ordinary Differential Equations · 10 marks≈ 7 min
Laplace Transforms and Initial Value Problems
(i) Find the Laplace transform of f(t) = 1/√t.
(ii) Find the inverse Laplace transform of (5s2 + 3s − 16)/((s − 1)(s − 2)(s + 3)).हिंदी में प्रश्न
(i) f(t) = 1/√t का लाप्लास रूपान्तर ज्ञात कीजिये।
(ii) (5s2 + 3s − 16)/((s − 1)(s − 2)(s + 3)) का विलोम लाप्लास रूपान्तर ज्ञात कीजिये।Q5(e) · Dynamics and Statics · 10 marks≈ 7 min
Motion in a Plane and Projectiles
A particle projected from a given point on the ground just clears a wall of height h at a distance d from the point of projection. If the particle moves in a vertical plane and if the horizontal range is R, find the elevation of the projection.
हिंदी में प्रश्न
एक कण को धरती के एक बिन्दु से प्रक्षेपित करने पर वह एक दीवार, जो प्रक्षेपण बिन्दु से d दूरी पर है और जिसकी ऊँचाई h है, को छूते हुये पार करता है। अगर यह कण ऊर्ध्वाधर तल पर गतिमान है और इसकी क्षैतिज पहुँच R है, तो प्रक्षेपण की उच्चता ज्ञात कीजिये।
Question 6 (50 marks)
Q6(a) · Ordinary Differential Equations · 13 marks≈ 9 min
First Order Higher Degree Equations, Clairaut's Equation and Singular Solutions
Solve : (dy/dx)2 y + 2(dy/dx) x − y = 0
हिंदी में प्रश्न
हल कीजिये : (dy/dx)2 y + 2(dy/dx) x − y = 0
Worked solution
Put Y = y2: P = dY/dx = 2yp, so p = P/(2y). Then P2/(4y) + xP/y − y = 0, that is P2 + 4xP − 4Y = 0.
So Y = xP + P2/4, Clairaut's form, with general solution Y = cx + c2/4.
The envelope: x + P/2 = 0 gives Y = −x2, that is x2 + y2 = 0, which has no real points other than the origin, so there is no real singular solution.
Answer: y2 = cx + c2/4 (equivalently y2 = 2kx + k2), a family of confocal parabolas with focus at the origin.Final answer: y2 = cx + c2/4 (equivalently y2 = 2kx + k2), a family of confocal parabolas with focus at the origin.
Q6(b) · Dynamics and Statics · 12 marks≈ 9 min
Rectilinear Motion and Simple Harmonic Motion
A particle moving with simple harmonic motion in a straight line has velocities v1 and v2 at distances x1 and x2 respectively from the centre of its path. Find the period of its motion.
हिंदी में प्रश्न
एक कण, जो एक सरल रेखा में सरल आवर्त गति से चल रहा है, के पथ के केन्द्र से x1 और x2 की दूरी पर वेग क्रमशः v1 और v2 हैं। उसकी गति का आवर्तकाल ज्ञात कीजिये।
Q6(c) · Ordinary Differential Equations · 13 marks≈ 9 min
Second Order Equations with Variable Coefficients
Solve : y″ + 16y = 32 sec 2x
हिंदी में प्रश्न
हल कीजिये : y″ + 16y = 32 sec 2x
Q6(d) · Vector Analysis · 12 marks≈ 9 min
Gauss and Stokes' Theorems and Green's Identities
If S is the surface of the sphere x2 + y2 + z2 = a2, then evaluate ∬S [(x + z) dydz + (y + z) dzdx + (x + y) dxdy] using Gauss' divergence theorem.
हिंदी में प्रश्न
अगर गोलक x2 + y2 + z2 = a2 का पृष्ठ S है, तो गाउस के अपसरण प्रमेय का इस्तेमाल करते हुये ∬S [(x + z) dydz + (y + z) dzdx + (x + y) dxdy] का मान निकालिये।
Question 7 (50 marks)
Q7(a) · Ordinary Differential Equations · 13 marks≈ 9 min
Second Order Equations with Variable Coefficients
Solve : (1 + x)2 y″ + (1 + x) y′ + y = 4 cos(log(1 + x))
हिंदी में प्रश्न
हल कीजिये : (1 + x)2 y″ + (1 + x) y′ + y = 4 cos(log(1 + x))
Worked solution
Legendre's equation: put 1 + x = ez, so (1 + x)D = θ and (1 + x)2D2 = θ(θ − 1). Then [θ(θ − 1) + θ + 1]y = (θ2 + 1)y = 4 cos z.
CF = c1 cos z + c2 sin z; failure case PI = 4·(z/2) sin z = 2z sin z.
Answer: y = c1 cos log(1 + x) + c2 sin log(1 + x) + 2 log(1 + x) sin log(1 + x).Final answer: y = c1 cos log(1 + x) + c2 sin log(1 + x) + 2 log(1 + x) sin log(1 + x).
Q7(b) · Vector Analysis · 12 marks≈ 9 min
Curves in Space: Curvature, Torsion and Serret-Frenet Formulae
Find the curvature and torsion of the curve r = a(u − sin u)i + a(1 − cos u)j + buk.
हिंदी में प्रश्न
वक्र r = a(u − sin u)i + a(1 − cos u)j + buk की वक्रता और विमोटन ज्ञात कीजिये।
Worked solution
r′ = (a(1 − cos u), a sin u, b), r″ = (a sin u, a cos u, 0), r‴ = (a cos u, −a sin u, 0).
|r′|2 = 2a2(1 − cos u) + b2 = 4a2 sin2(u/2) + b2.
r′ × r″ = (−ab cos u, ab sin u, −a2(1 − cos u)), so |r′ × r″|2 = a2b2 + 4a4 sin4(u/2).
[r′ r″ r‴] = −a2b cos2u − a2b sin2u = −a2b.
Answer: κ = a√(b2 + 4a2 sin4(u/2))/(b2 + 4a2 sin2(u/2))3/2 and τ = −a2b/(a2b2 + 4a4 sin4(u/2)) = −b/(b2 + 4a2 sin4(u/2)).Final answer: κ = a√(b2 + 4a2 sin4(u/2))/(b2 + 4a2 sin2(u/2))3/2 and τ = −a2b/(a2b2 + 4a4 sin4(u/2)) = −b/(b2 + 4a2 sin4(u/2)).
Q7(c) · Ordinary Differential Equations · 13 marks≈ 9 min
Laplace Transforms and Initial Value Problems
Solve the initial value problem
y″ − 5y′ + 4y = e2t
y(0) = 19/12, y′(0) = 8/3हिंदी में प्रश्न
प्रारंभिक मान समस्या
y″ − 5y′ + 4y = e2t
y(0) = 19/12, y′(0) = 8/3
को हल कीजिये।Q7(d) · Ordinary Differential Equations · 12 marks≈ 9 min
Formulation and First Order, First Degree Equations
Find α and β such that xαyβ is an integrating factor of (4y2 + 3xy) dx − (3xy + 2x2) dy = 0 and solve the equation.
हिंदी में प्रश्न
α और β को, जिसके लिये xαyβ समीकरण (4y2 + 3xy) dx − (3xy + 2x2) dy = 0 का एक समाकलन गुणक है, ज्ञात कीजिये और समीकरण हल कीजिये।
Worked solution
With μ = xαyβ: μM = 4xαyβ+2 + 3xα+1yβ+1 and μN = −3xα+1yβ+1 − 2xα+2yβ.
(μM)y = (μN)x: comparing xαyβ+1 terms, 4(β + 2) = −3(α + 1); comparing xα+1yβ terms, 3(β + 1) = −2(α + 2).
That is 3α + 4β = −11 and 2α + 3β = −7, so α = −5, β = 1.
The exact equation (4y3/x5 + 3y2/x4) dx − (3y2/x4 + 2y/x3) dy = 0 has ∫M dx = −y3/x4 − y2/x3, and N has no x-free term.
Answer: α = −5, β = 1; the solution is y3/x4 + y2/x3 = c, that is y2(x + y) = cx4.Final answer: α = −5, β = 1; the solution is y3/x4 + y2/x3 = c, that is y2(x + y) = cx4.
Question 8 (50 marks)
Q8(a) · Vector Analysis · 12 marks≈ 9 min
Vector Identities and Vector Equations
Let v = v1i + v2j + v3k. Show that curl (curl v) = grad (div v) − ∇2v.
हिंदी में प्रश्न
मान लीजिये कि v = v1i + v2j + v3k है। दर्शाइये कि curl (curl v) = grad (div v) − ∇2v।
Q8(b) · Vector Analysis · 13 marks≈ 9 min
Gauss and Stokes' Theorems and Green's Identities
Evaluate the line integral ∫C −y3 dx + x3 dy + z3 dz using Stokes' theorem. Here C is the intersection of the cylinder x2 + y2 = 1 and the plane x + y + z = 1. The orientation on C corresponds to counterclockwise motion in the xy-plane.
हिंदी में प्रश्न
स्टोक्स प्रमेय का इस्तेमाल करते हुये रेखा समाकल ∫C −y3 dx + x3 dy + z3 dz का मान निकालिये। यहाँ सिलिन्डर x2 + y2 = 1 और समतल x + y + z = 1 का प्रतिच्छेद C है। C पर अभिविन्यास xy-समतल में वामावर्त गति के संगत है।
Worked solution
For F = (−y3, x3, z3), ∇ × F = (0, 0, 3x2 + 3y2).
Take S as the part of the plane z = 1 − x − y inside the cylinder, oriented upward to match the anticlockwise sense: n dS = (1, 1, 1) dx dy.
∬(∇ × F)·n dS = ∬x2+y2≤13(x2 + y2) dx dy = 3∫02π∫01ρ3 dρ dφ = 3π/2.
Check: with x = cos t, y = sin t, ∮(−y3 dx + x3 dy) = ∫02π(sin4t + cos4t) dt = 3π/2, and ∮z3 dz = 0.
Answer: 3π/2.Final answer: 3π/2.
Q8(c) · Vector Analysis · 13 marks≈ 9 min
Gauss and Stokes' Theorems and Green's Identities
Let F = xy2i + (y + x)j. Integrate (∇ × F)·k over the region in the first quadrant bounded by the curves y = x2 and y = x using Green's theorem.
हिंदी में प्रश्न
मान लीजिये कि F = xy2i + (y + x)j है। ग्रीन के प्रमेय का इस्तेमाल करते हुये प्रथम चतुर्थांश में वक्रों y = x2 और y = x द्वारा परिबद्ध क्षेत्र पर (∇ × F)·k का समाकलन कीजिये।
Worked solution
With M = xy2, N = y + x: (∇ × F)·k = ∂N/∂x − ∂M/∂y = 1 − 2xy, and Green's theorem turns ∬(∇ × F)·k dA into ∮(xy2 dx + (x + y) dy).
Double integral over x2 ≤ y ≤ x, 0 ≤ x ≤ 1: ∫01[(x − x2) − x(x2 − x4)] dx = 1/2 − 1/3 − 1/4 + 1/6 = 1/12.
Line check: along y = x2 (x: 0 → 1), ∫[x5 + (x + x2)2x] dx = 4/3; along y = x (x: 1 → 0), ∫(x3 + 2x) dx = −5/4; total 1/12.
Answer: 1/12.Final answer: 1/12.
Q8(d) · Ordinary Differential Equations · 12 marks≈ 9 min
Formulation and First Order, First Degree Equations
Find f(y) such that (2xey + 3y2) dy + (3x2 + f(y)) dx = 0 is exact and hence solve.
हिंदी में प्रश्न
f(y), जिसके लिये समीकरण (2xey + 3y2) dy + (3x2 + f(y)) dx = 0 यथातथ्य है, ज्ञात कीजिये और हल निकालिये।
UPSC Mains 2018 Mathematics Paper II
Question 1 (50 marks)
Q1(a) · Abstract Algebra · 10 marks≈ 7 min
Integral Domains, PID, Euclidean Domains, UFD, Fields and Quotient Fields
Let R be an integral domain with unit element. Show that any unit in R[x] is a unit in R.
हिंदी में प्रश्न
मान लीजिए R तत्समक अवयव सहित एक पूर्णांकीय प्रांत है। दर्शाइए कि R[x] में कोई भी एकक R में एक एकक है।
Worked solution
Let f ∈ R[x] be a unit: fg = 1 for some g ∈ R[x], so f, g ≠ 0.
R has no zero divisors, so the product of the leading coefficients of f and g is non-zero and deg(fg) = deg f + deg g.
Hence deg f + deg g = deg 1 = 0, so f = a and g = b are constants of R with ab = 1. Thus f = a is a unit of R.
(The hypothesis is needed: in ℤ4[x], (1 + 2x)2 = 1, so 1 + 2x is a unit that is not in ℤ4.)
Answer: every unit of R[x] is a constant polynomial that is a unit of R.Final answer: every unit of R[x] is a constant polynomial that is a unit of R.
Q1(b) · Real Analysis · 10 marks≈ 7 min
Riemann Integral, Improper Integrals and Fundamental Theorems of Integral Calculus
Prove the inequality : π2/9 < ∫π/6π/2 (x/sin x) dx < 2π2/9.
हिंदी में प्रश्न
असमिका : π2/9 < ∫π/6π/2 (x/sin x) dx < 2π2/9 को सिद्ध कीजिए।
Worked solution
On [π/6, π/2], 1/2 ≤ sin x ≤ 1, so x ≤ x/sin x ≤ 2x.
sin x < 1 for x < π/2 and sin x > 1/2 for x > π/6, so each inequality is strict except at one end point; the functions are continuous, so the integrated inequalities are strict.
∫π/6π/2 x dx = (1/2)(π2/4 − π2/36) = π2/9, and ∫π/6π/2 2x dx = 2π2/9.
Answer: π2/9 < ∫π/6π/2 (x/sin x) dx < 2π2/9 (numerically 1.097 < 1.300 < 2.193).Final answer: π2/9 < ∫π/6π/2 (x/sin x) dx < 2π2/9 (numerically 1.097 < 1.300 < 2.193).
Q1(c) · Complex Analysis · 10 marks≈ 7 min
Analytic Functions and Cauchy-Riemann Equations
Prove that the function : u(x, y) = (x − 1)3 − 3xy2 + 3y2 is harmonic and find its harmonic conjugate and the corresponding analytic function f(z) in terms of z.
हिंदी में प्रश्न
सिद्ध कीजिए कि फलन : u(x, y) = (x − 1)3 − 3xy2 + 3y2 प्रसंवादी है और इसके प्रसंवादी संयुग्मी को और संगत विश्लेषिक फलन f(z) को, z के रूप में ज्ञात कीजिए।
Q1(d) · Real Analysis · 10 marks≈ 7 min
Series: Convergence Tests, Absolute and Conditional Convergence, Rearrangement
Find the range of p (> 0) for which the series :
1/(1 + a)p − 1/(2 + a)p + 1/(3 + a)p − ..., a > 0, is
(i) absolutely convergent and (ii) conditionally convergent.हिंदी में प्रश्न
p (> 0) का वह परास ज्ञात कीजिए, जिसके लिए श्रेणी :
1/(1 + a)p − 1/(2 + a)p + 1/(3 + a)p − ..., a > 0
(i) निरपेक्षतः अभिसारी तथा (ii) सापेक्ष अभिसारी है।Q1(e) · Linear Programming · 10 marks≈ 7 min
LPP Formulation, Basic Solutions and the Graphical Method
An agricultural firm has 180 tons of nitrogen fertilizer, 250 tons of phosphate and 220 tons of potash. It will be able to sell a mixture of these substances in their respective ratio 3 : 3 : 4 at a profit of Rs. 1500 per ton and a mixture in the ratio 2 : 4 : 2 at a profit of Rs. 1200 per ton. Pose a linear programming problem to show how many tons of these two mixtures should be prepared to obtain the maximum profit.
हिंदी में प्रश्न
एक कृषि फर्म के पास 180 टन नाइट्रोजन उर्वरक, 250 टन फॉस्फेट तथा 220 टन पोटाश है। फर्म इन पदार्थों के क्रमशः 3 : 3 : 4 के अनुपात में मिश्रण को 1500 रुपये प्रति टन के मुनाफे से तथा 2 : 4 : 2 के अनुपात में मिश्रण को 1200 रुपये प्रति टन के मुनाफे से बेच पायेगी। एक रैखिक-प्रोग्रामन समस्या प्रस्तुत कीजिए, जो यह दर्शाए कि अधिकतम मुनाफा प्राप्त करने के लिए, इन मिश्रणों की कितने टन मात्रा तैयार की जानी चाहिए।
Question 2 (50 marks)
Q2(a) · Abstract Algebra · 15 marks≈ 11 min
Normal Subgroups, Quotient Groups, Homomorphisms and Isomorphism Theorems
Show that the quotient group of (ℝ, +) modulo ℤ is isomorphic to the multiplicative group of complex numbers on the unit circle in the complex plane. Here ℝ is the set of real numbers and ℤ is the set of integers.
हिंदी में प्रश्न
दर्शाइए कि (ℝ, +) मोड्यूलो ℤ का विभाग समूह, सम्मिश्र तल में एकांक वृत्त पर सम्मिश्र संख्याओं के गुणनात्मक समूह से तुल्यकारी होता है। यहाँ पर ℝ, वास्तविक संख्याओं का समुच्चय है तथा ℤ पूर्णांकों का समुच्चय है।
Worked solution
Let T = {z ∈ ℂ : |z| = 1} under multiplication, and define φ : ℝ → T by φ(t) = e2πit = cos 2πt + i sin 2πt.
φ(s + t) = e2πi(s+t) = e2πise2πit = φ(s)φ(t), so φ is a homomorphism from (ℝ, +) to (T, ·).
Onto: every z ∈ T equals eiθ for some θ ∈ [0, 2π), and then z = φ(θ/(2π)).
Kernel: e2πit = 1 ⇔ cos 2πt = 1 and sin 2πt = 0 ⇔ t ∈ ℤ, so ker φ = ℤ.
By the first isomorphism theorem ℝ/ℤ ≅ T.
Answer: ℝ/ℤ ≅ T, the isomorphism being t + ℤ ↦ e2πit.Final answer: ℝ/ℤ ≅ T, the isomorphism being t + ℤ ↦ e2πit.
Q2(b) · Linear Programming · 20 marks≈ 14 min
The Simplex Method
Solve the following linear programming problem by Big M-method and show that the problem has finite optimal solutions. Also find the value of the objective function :
Minimize z = 3x1 + 5x2
subject to x1 + 2x2 ≥ 8
3x1 + 2x2 ≥ 12
5x1 + 6x2 ≤ 60,
x1, x2 ≥ 0.हिंदी में प्रश्न
निम्नलिखित रैखिक प्रोग्रामन समस्या को Big M विधि से हल कीजिए तथा दर्शाइए कि समस्या के परिमित इष्टतम हल हैं। साथ ही उद्देश्य फलन का मान भी ज्ञात कीजिए :
न्यूनतमीकरण कीजिए z = 3x1 + 5x2
बशर्ते कि x1 + 2x2 ≥ 8
3x1 + 2x2 ≥ 12
5x1 + 6x2 ≤ 60,
x1, x2 ≥ 0.Q2(c) · Real Analysis · 15 marks≈ 11 min
Continuity, Uniform Continuity and Continuous Functions on Compact Sets
Show that if a function f defined on an open interval (a, b) of ℝ is convex, then f is continuous. Show, by example, if the condition of open interval is dropped, then the convex function need not be continuous.
हिंदी में प्रश्न
दर्शाइए कि यदि ℝ के विवृत अन्तराल (a, b) पर परिभाषित फलन f अवमुख हो, तो वह संतत है। उदाहरण के द्वारा दर्शाइए कि यदि विवृत अन्तराल होने की शर्त न हो, तब अवमुख फलन का संतत होना आवश्यक नहीं है।
Worked solution
Let f be convex on (a, b), x0 ∈ (a, b), and fix s, t with a < s < x0 < t < b. By the three-chord lemma, for x0 < x < t, (f(x0) − f(s))/(x0 − s) ≤ (f(x) − f(x0))/(x − x0) ≤ (f(t) − f(x0))/(t − x0).
For s < x < x0 the same two bounds hold for (f(x0) − f(x))/(x0 − x), again by the lemma.
So |f(x) − f(x0)| ≤ K|x − x0| for s < x < t, where K is the larger of the absolute values of the two fixed chord slopes. Hence f(x) → f(x0): f is continuous at every point of (a, b).
The open interval is essential: f : [0, 1] → ℝ with f(x) = 0 for 0 ≤ x < 1 and f(1) = 1 is convex (each chord lies on or above the graph) but is discontinuous at x = 1.
Answer: a convex function on an open interval is continuous (even locally Lipschitz); on [0, 1] the convex function above jumps at 1.Final answer: a convex function on an open interval is continuous (even locally Lipschitz); on [0, 1] the convex function above jumps at 1.
Question 3 (50 marks)
Q3(a) · Abstract Algebra · 20 marks≈ 14 min
Groups, Subgroups, Cyclic Groups, Cosets and Lagrange's Theorem
Find all the proper subgroups of the multiplicative group of the field (ℤ13, +13, ×13), where +13 and ×13 represent addition modulo 13 and multiplication modulo 13 respectively.
हिंदी में प्रश्न
क्षेत्र (ℤ13, +13, ×13), जहाँ पर +13 तथा ×13 क्रमशः योग मोड्यूलो 13 व गुणन मोड्यूलो 13 निरूपित करते हैं, के गुणनात्मक समूह के सभी उचित उपसमूहों को ज्ञात कीजिए।
Worked solution
The multiplicative group is U(13) = {1, 2, ..., 12}, of order 12. Powers of 2 mod 13 are 2, 4, 8, 3, 6, 12, 11, 9, 5, 10, 7, 1, so o(2) = 12 and U(13) = ⟨2⟩ is cyclic.
A cyclic group of order 12 has exactly one subgroup of each order d dividing 12, namely ⟨212/d⟩.
d = 1: {1}; d = 2: ⟨26⟩ = ⟨12⟩ = {1, 12}; d = 3: ⟨24⟩ = ⟨3⟩ = {1, 3, 9}.
d = 4: ⟨23⟩ = ⟨8⟩ = {1, 8, 12, 5}; d = 6: ⟨22⟩ = ⟨4⟩ = {1, 4, 3, 12, 9, 10}; d = 12 gives U(13) itself.
Answer: the proper subgroups are {1}, {1, 12}, {1, 3, 9}, {1, 5, 8, 12} and {1, 3, 4, 9, 10, 12} (the last four are the non-trivial ones).Final answer: the proper subgroups are {1}, {1, 12}, {1, 3, 9}, {1, 5, 8, 12} and {1, 3, 4, 9, 10, 12} (the last four are the non-trivial ones).
Q3(b) · Complex Analysis · 15 marks≈ 11 min
Contour Integration: Evaluation of Real Integrals
Show by applying the residue theorem that ∫0∞ dx/(x2 + a2)2 = π/(4a3), a > 0.
हिंदी में प्रश्न
अवशेष प्रमेय के अनुप्रयोग के द्वारा दर्शाइए कि ∫0∞ dx/(x2 + a2)2 = π/(4a3), a > 0.
Q3(c) · Linear Programming · 15 marks≈ 11 min
LPP Formulation, Basic Solutions and the Graphical Method
How many basic solutions are there in the following linearly independent set of equations ? Find all of them.
2x1 − x2 + 3x3 + x4 = 6
4x1 − 2x2 − x3 + 2x4 = 10.हिंदी में प्रश्न
अधोलिखित समीकरणों के रैखिकतः स्वतंत्र समुच्चय में कितने आधारी हल हैं ? उन सभी को ज्ञात कीजिए।
2x1 − x2 + 3x3 + x4 = 6
4x1 − 2x2 − x3 + 2x4 = 10.Worked solution
Here m = 2 and n = 4, so there are at most C(4, 2) = 6 basic solutions. The columns are a1 = (2, 4), a2 = (−1, −2), a3 = (3, −1), a4 = (1, 2).
a1, a2, a4 are all multiples of (1, 2), so the pairs (x1, x2), (x1, x4), (x2, x4) are singular. With x3 = 0 the equations read 2x1 − x2 + x4 = 6 and 2(2x1 − x2 + x4) = 10, which are inconsistent, so these give no solution.
(x1, x3): 2x1 + 3x3 = 6, 4x1 − x3 = 10 give x1 = 18/7, x3 = 2/7.
(x2, x3): −x2 + 3x3 = 6, −2x2 − x3 = 10 give x2 = −36/7, x3 = 2/7.
(x3, x4): 3x3 + x4 = 6, −x3 + 2x4 = 10 give x3 = 2/7, x4 = 36/7.
Answer: exactly 3 basic solutions: (18/7, 0, 2/7, 0), (0, −36/7, 2/7, 0) and (0, 0, 2/7, 36/7).Final answer: exactly 3 basic solutions: (18/7, 0, 2/7, 0), (0, −36/7, 2/7, 0) and (0, 0, 2/7, 36/7).
Question 4 (50 marks)
Q4(a) · Abstract Algebra · 20 marks≈ 14 min
Rings, Subrings, Ideals and Ring Homomorphisms
Suppose ℝ be the set of all real numbers and f : ℝ → ℝ is a function such that the following equations hold for all x, y ∈ ℝ :
(i) f(x + y) = f(x) + f(y)
(ii) f(xy) = f(x) f(y)
Show that ∀ x ∈ ℝ either f(x) = 0, or, f(x) = x.हिंदी में प्रश्न
मान लीजिए ℝ सभी वास्तविक संख्याओं का समुच्चय है तथा f : ℝ → ℝ ऐसा फलन है कि सभी x, y ∈ ℝ के लिए निम्नलिखित समीकरण लागू होते हैं :
(i) f(x + y) = f(x) + f(y)
(ii) f(xy) = f(x) f(y)
दर्शाइए कि सभी ∀ x ∈ ℝ के लिए या तो f(x) = 0 या f(x) = x है।Worked solution
Put x = y = 1 in (ii): f(1) = f(1)2, so f(1) = 0 or f(1) = 1.
If f(1) = 0, then f(x) = f(x·1) = f(x)f(1) = 0 for all x.
Let f(1) = 1. From (i), f(0) = 0, f(−x) = −f(x) and f(nx) = nf(x) for n ∈ ℕ; so f(n) = n for n ∈ ℤ, and nf(m/n) = f(m) = m gives f(q) = q for every q ∈ ℚ.
f preserves order: if t ≥ 0 then t = s2 with s = √t, so f(t) = f(s)2 ≥ 0. Hence x ≤ y ⇒ f(y) − f(x) = f(y − x) ≥ 0.
Fix x ∈ ℝ. For rationals r < x < s, r = f(r) ≤ f(x) ≤ f(s) = s. Letting r and s tend to x (ℚ is dense) gives f(x) = x.
Answer: either f(x) = 0 for all x, or f(x) = x for all x ∈ ℝ.Final answer: either f(x) = 0 for all x, or f(x) = x for all x ∈ ℝ.
Q4(b) · Complex Analysis · 15 marks≈ 11 min
Power Series, Taylor's Series and Laurent's Series
Find the Laurent's series which represent the function 1/((1 + z2)(z + 2)) when
(i) |z| < 1
(ii) 1 < |z| < 2
(iii) |z| > 2हिंदी में प्रश्न
फलन 1/((1 + z2)(z + 2)) को निरूपित करने वाली लॉरेन्ज श्रेणी ज्ञात कीजिए जब
(i) |z| < 1
(ii) 1 < |z| < 2
(iii) |z| > 2Worked solution
Partial fractions: 1/((1 + z2)(z + 2)) = (1/5)[1/(z + 2) + (2 − z)/(1 + z2)] (check at z = 0: (1/5)(1/2 + 2) = 1/2).
Pieces: 1/(z + 2) = ∑ (−1)nzn/2n+1 for |z| < 2 and ∑ (−1)n2n/zn+1 for |z| > 2; 1/(1 + z2) = ∑ (−1)nz2n for |z| < 1 and ∑ (−1)n/z2n+2 for |z| > 1 (all sums from n = 0).
(i) |z| < 1: f = (1/5)[∑ (−1)nzn/2n+1 + (2 − z)∑ (−1)nz2n] = 1/2 − z/4 − 3z2/8 + ..., a Taylor series.
(ii) 1 < |z| < 2: f = (1/5)[∑ (−1)nzn/2n+1 + (2 − z)∑ (−1)n/z2n+2] = (1/5)[... + 1/z3 + 2/z2 − 1/z + 1/2 − z/4 + z2/8 − ...].
(iii) |z| > 2: f = (1/5)[∑ (−1)n2n/zn+1 + (2 − z)∑ (−1)n/z2n+2]; the 1/z and 1/z2 terms cancel, leaving f = 1/z3 − 2/z4 + ....
Answer: the three series above, f = (1/5)[A(z) + (2 − z)B(z)] with A, B the expansions of 1/(z + 2) and 1/(1 + z2) valid in the given region.Final answer: the three series above, f = (1/5)[A(z) + (2 − z)B(z)] with A, B the expansions of 1/(z + 2) and 1/(1 + z2) valid in the given region.
Q4(c) · Linear Programming · 15 marks≈ 11 min
Assignment Problem
In a factory there are five operators O1, O2, O3, O4, O5 and five machines M1, M2, M3, M4, M5. The operating costs are given when the Oi operator operates the Mj machine (i, j = 1, 2, ..., 5). But there is a restriction that O3 cannot be allowed to operate the third machine M3 and O2 cannot be allowed to operate the fifth machine M5. The cost matrix is given above. Find the optimal assignment and the optimal assignment cost also.
[Cost matrix: rows O1 to O5, columns M1 to M5]
[[24, 29, 18, 32, 19], [17, 26, 34, 22, 21], [27, 16, 28, 17, 25], [22, 18, 28, 30, 24], [28, 16, 31, 24, 27]]हिंदी में प्रश्न
एक फैक्ट्री में पांच प्रचालक O1, O2, O3, O4, O5 तथा पांच मशीनें M1, M2, M3, M4, M5 हैं। परिचालन लागत, जब कि Oi प्रचालक Mj (i, j = 1, 2, ..., 5) मशीन को परिचालन करता है, दी गई है। लेकिन एक प्रतिबन्ध है कि O3 को तीसरी मशीन M3 का परिचालन करने तथा O2 को पाँचवीं मशीन M5 का परिचालन करने की इजाज़त नहीं दी जा सकती है। लागत आव्यूह नीचे दी है। इष्टतम नियतन तथा इष्टतम नियतन की लागत ज्ञात कीजिए।
[लागत आव्यूह: पंक्तियाँ प्रचालक O1 से O5, स्तम्भ मशीन M1 से M5]
[[24, 29, 18, 32, 19], [17, 26, 34, 22, 21], [27, 16, 28, 17, 25], [22, 18, 28, 30, 24], [28, 16, 31, 24, 27]]
Question 5 (50 marks)
Q5(a) · Partial Differential Equations · 10 marks≈ 7 min
Families of Surfaces and Formation of PDEs
Find the partial differential equation of the family of all tangent planes to the ellipsoid : x2 + 4y2 + 4z2 = 4, which are not perpendicular to the xy plane.
हिंदी में प्रश्न
दीर्घवृत्तज : x2 + 4y2 + 4z2 = 4 के उन सभी स्पर्श-तलों के संकाय का आंशिक अवकल समीकरण ज्ञात कीजिए, जो xy समतल के लम्बवत नहीं हैं।
Worked solution
The ellipsoid is x2/4 + y2/1 + z2/1 = 1, so a2 = 4, b2 = 1, c2 = 1.
A plane not perpendicular to the xy-plane contains z, so it can be written z = px + qy + k, i.e. px + qy − z = −k, where p, q are its slopes ∂z/∂x, ∂z/∂y.
Tangency a2l2 + b2m2 + c2n2 = d2 with (l, m, n) = (p, q, −1), d = −k gives k2 = 4p2 + q2 + 1.
Eliminating the constant k = z − px − qy gives the PDE.
Answer: (z − px − qy)2 = 4p2 + q2 + 1, i.e. z = px + qy ± √(4p2 + q2 + 1), a Clairaut equation whose singular integral is the ellipsoid.Final answer: (z − px − qy)2 = 4p2 + q2 + 1, i.e. z = px + qy ± √(4p2 + q2 + 1), a Clairaut equation whose singular integral is the ellipsoid.
Q5(b) · Numerical Analysis and Computer Programming · 10 marks≈ 7 min
Interpolation: Newton's Forward, Backward and Lagrange's Formulae
Using Newton's forward difference formula find the lowest degree polynomial ux when it is given that u1 = 1, u2 = 9, u3 = 25, u4 = 55 and u5 = 105.
हिंदी में प्रश्न
न्यूटन के अग्रांतर फार्मूले से निम्नतम-घातीय बहुपद ux ज्ञात कीजिए जब कि u1 = 1, u2 = 9, u3 = 25, u4 = 55 तथा u5 = 105 दिया गया है।
Q5(c) · Mechanics and Fluid Dynamics · 10 marks≈ 7 min
Kinematics of Fluids: Continuity, Streamlines and Potential Flow
For an incompressible fluid flow, two components of velocity (u, v, w) are given by u = x2 + 2y2 + 3z2, v = x2y − y2z + zx. Determine the third component w so that they satisfy the equation of continuity. Also, find the z-component of acceleration.
हिंदी में प्रश्न
एक असंपीड्य तरल प्रवाह के लिए वेग (u, v, w) के दो घटक u = x2 + 2y2 + 3z2 व v = x2y − y2z + zx दिए गए हैं। वेग के तीसरे घटक w का निर्धारण कीजिए ताकि वे सांतत्य समीकरण को सन्तुष्ट करें। त्वरण के z-घटक को भी ज्ञात कीजिए।
Worked solution
Continuity: ∂w/∂z = −(∂u/∂x + ∂v/∂y) = −(2x + x2 − 2yz), so w = −2xz − x2z + yz2 + f(x, y); take f = 0, the simplest choice.
The flow is steady, so az = Dw/Dt = u ∂w/∂x + v ∂w/∂y + w ∂w/∂z, with ∂w/∂x = −2z(1 + x), ∂w/∂y = z2, ∂w/∂z = 2yz − 2x − x2.
Answer: w = yz2 − 2xz − x2z (+ f(x, y)); az = −2z(1 + x)(x2 + 2y2 + 3z2) + z2(x2y − y2z + zx) + (yz2 − 2xz − x2z)(2yz − 2x − x2).Final answer: w = yz2 − 2xz − x2z (+ f(x, y)); az = −2z(1 + x)(x2 + 2y2 + 3z2) + z2(x2y − y2z + zx) + (yz2 − 2xz − x2z)(2yz − 2x − x2).
Q5(d) · Numerical Analysis and Computer Programming · 10 marks≈ 7 min
Numerical Integration and Numerical Solution of ODEs
Starting from rest in the beginning, the speed (in Km/h) of a train at different times (in minutes) is given by the above table :
[Table: Time (minutes): 2, 4, 6, 8, 10, 12, 14, 16, 18, 20; Speed (Km/h): 10, 18, 25, 29, 32, 20, 11, 5, 2, 8.5]
Using Simpson's (1/3)rd rule, find the approximate distance travelled (in Km) in 20 minutes from the beginning.हिंदी में प्रश्न
विराम अवस्था से प्रारम्भ हो कर एक रेलगाड़ी की रफ़्तार (किमी/घं में) विभिन्न समयों (मिनट में) पर निम्न सारणी के द्वारा दी गई है :
[सारणी: समय (मिनट): 2, 4, 6, 8, 10, 12, 14, 16, 18, 20; रफ़्तार (किमी/घं): 10, 18, 25, 29, 32, 20, 11, 5, 2, 8.5]
सिम्पसन के (1/3) नियम के इस्तेमाल से प्रारंभ से 20 मिनटों में चली गई सन्निकट दूरी (किमी. में) ज्ञात कीजिए।Worked solution
Distance = ∫v dt. Starting from rest gives v = 0 at t = 0, so there are 11 ordinates (t = 0, 2, ..., 20), n = 10 strips (even) and h = 2 min = 1/30 hour.
Simpson: (h/3)[(v0 + v10) + 4(v1 + v3 + v5 + v7 + v9) + 2(v2 + v4 + v6 + v8)] = (1/90)[(0 + 8.5) + 4(10 + 25 + 32 + 11 + 2) + 2(18 + 29 + 20 + 5)].
Answer: (1/90)(8.5 + 320 + 144) = 472.5/90 = 5.25 km.Final answer: (1/90)(8.5 + 320 + 144) = 472.5/90 = 5.25 km.
Q5(e) · Numerical Analysis and Computer Programming · 10 marks≈ 7 min
Roots of Algebraic and Transcendental Equations
Write down the basic algorithm for solving the equation : xex − 1 = 0 by bisection method, correct to 4 decimal places.
हिंदी में प्रश्न
समीकरण : xex − 1 = 0 को द्विभाजन-विधि के द्वारा, दशमलव के 4 अंकों तक, हल करने के लिए, आधारी ऐल्गोरिथ्म लिखिए।
Question 6 (50 marks)
Q6(a) · Partial Differential Equations · 15 marks≈ 11 min
Lagrange's Method for Quasilinear First-Order PDEs
Find the general solution of the partial differential equation :
(y3x − 2x4)p + (2y4 − x3y)q = 9z(x3 − y3),
where p = ∂z/∂x, q = ∂z/∂y, and find its integral surface that passes through the curve : x = t, y = t2, z = 1.हिंदी में प्रश्न
आंशिक अवकल समीकरण :
(y3x − 2x4)p + (2y4 − x3y)q = 9z(x3 − y3), का,
जहाँ p = ∂z/∂x, q = ∂z/∂y है, का व्यापक हल ज्ञात कीजिए, तथा इसके, वक्र : x = t, y = t2, z = 1 में से गुजरने वाले समाकल पृष्ठ को भी ज्ञात कीजिए।Worked solution
Auxiliary equations dx/(x(y3 − 2x3)) = dy/(y(2y3 − x3)) = dz/(9z(x3 − y3)).
Multipliers 1/x, 1/y, 1/(3z): (y3 − 2x3) + (2y3 − x3) + 3(x3 − y3) = 0, so d log x + d log y + (1/3)d log z = 0 and x3y3z = c1.
Multipliers 1/y2 − 2y/x3, 1/x2 − 2x/y3, 0 (the partial derivatives of x/y2 + y/x2): the sum lP + mQ expands to xy − 2x4/y2 − 2y4/x2 + 4xy + 2y4/x2 − xy − 4xy + 2x4/y2 = 0, so (x3 + y3)/(x2y2) = c2.
General solution φ(x3y3z, (x3 + y3)/(x2y2)) = 0.
On x = t, y = t2, z = 1: c1 = t9 and c2 = (t3 + t6)/t6 = 1 + t−3, so c1(c2 − 1)3 = 1.
Substituting: x3y3z[(x3 + y3 − x2y2)/(x2y2)]3 = 1.
Answer: general solution φ(x3y3z, (x3 + y3)/(x2y2)) = 0; required surface z(x3 + y3 − x2y2)3 = x3y3 (on the curve x3 + y3 − x2y2 = t3, giving z = 1).Final answer: general solution φ(x3y3z, (x3 + y3)/(x2y2)) = 0; required surface z(x3 + y3 − x2y2)3 = x3y3 (on the curve x3 + y3 − x2y2 = t3, giving z = 1).
Q6(b) · Numerical Analysis and Computer Programming · 15 marks≈ 11 min
Number Systems and Representation of Numbers in a Computer
Find the equivalent of numbers given in a specified number system to the system mentioned against them.
(i) (111011.101)2 to decimal system
(ii) (1000111110000.00101100)2 to hexadecimal system
(iii) (C4F2)16 to decimal system
(iv) (418)10 to binary systemहिंदी में प्रश्न
अधोलिखित संख्याओं के समतुल्यों को उनके सम्मुख दर्शाई गई विशिष्ट संख्या पद्धति में, ज्ञात कीजिए।
(i) (111011.101)2 को दशमलव पद्धति में
(ii) (1000111110000.00101100)2 को षड्दशमलव पद्धति में
(iii) (C4F2)16 को दशमलव पद्धति में
(iv) (418)10 को द्विआधारी पद्धति मेंQ6(c) · Mechanics and Fluid Dynamics · 20 marks≈ 14 min
Generalized Coordinates, D'Alembert's Principle and Lagrange's Equations
Suppose the Lagrangian of a mechanical system is given by
L = (1/2) m(aẋ2 + 2bẋẏ + cẏ2) − (1/2) k(ax2 + 2bxy + cy2),
where a, b, c, m (> 0), k (> 0) are constants and b2 ≠ ac. Write down the Lagrangian equations of motion and identify the system.हिंदी में प्रश्न
मान लीजिए किसी यांत्रिक-निकाय का लेगरान्जियन :
L = (1/2) m(aẋ2 + 2bẋẏ + cẏ2) − (1/2) k(ax2 + 2bxy + cy2),
के द्वारा द्योतित है जहाँ a, b, c, m (> 0), k (> 0) स्थिरांक हैं तथा b2 ≠ ac लेगरान्जियन समीकरणों को लिखिए तथा निकाय को पहचानिए।Worked solution
∂L/∂ẋ = m(aẋ + bẏ), ∂L/∂x = −k(ax + by); ∂L/∂ẏ = m(bẋ + cẏ), ∂L/∂y = −k(bx + cy).
Lagrange's equations: m(aẍ + bÿ) + k(ax + by) = 0 and m(bẍ + cÿ) + k(bx + cy) = 0.
In matrix form P(mẌ + kX) = 0 with P = [[a, b], [b, c]], X = (x, y); det P = ac − b2 ≠ 0, so P cancels: ẍ + (k/m)x = 0, ÿ + (k/m)y = 0.
So x = C1cos(ωt + ε1), y = C2cos(ωt + ε2) with ω = √(k/m).
Answer: the equations above; the system is a two-dimensional isotropic harmonic oscillator (two uncoupled oscillations of the same frequency √(k/m)) written in oblique coordinates, the path being in general an ellipse.Final answer: the equations above; the system is a two-dimensional isotropic harmonic oscillator (two uncoupled oscillations of the same frequency √(k/m)) written in oblique coordinates, the path being in general an ellipse.
Question 7 (50 marks)
Q7(a) · Partial Differential Equations · 15 marks≈ 11 min
Linear Second-Order PDEs with Constant Coefficients
Solve the partial differential equation :
(2D2 − 5DD′ + 2D′2)z = 5 sin(2x + y) + 24(y − x) + e3x + 4y
where D ≡ ∂/∂x, D′ ≡ ∂/∂y.हिंदी में प्रश्न
आंशिक अवकल समीकरण :
(2D2 − 5DD′ + 2D′2)z = 5 sin(2x + y) + 24(y − x) + e3x + 4y को हल कीजिए
जहाँ D ≡ ∂/∂x, D′ ≡ ∂/∂y.Worked solution
F = (2D − D′)(D − 2D′), CF = φ1(y + 2x) + φ2(2y + x).
5 sin(2x + y): D2 → −4, DD′ → −2, D′2 → −1 gives −8 + 10 − 2 = 0, a failing case, because sin(2x + y) is a function of y + 2x.
(2D − D′)cos(2x + y) = −3 sin(2x + y), so (2D − D′)−15 sin(2x + y) = −(5/3)cos(2x + y); then (D − 2D′)−1 multiplies by x: PI1 = −(5x/3)cos(2x + y).
PI2 = (4/9)(y − x)3 (as for UPSC 2014); e3x + 4y: F(3, 4) = 18 − 60 + 32 = −10, PI3 = −(1/10)e3x + 4y.
Answer: z = φ1(y + 2x) + φ2(2y + x) − (5x/3)cos(2x + y) + (4/9)(y − x)3 − (1/10)e3x + 4y.Final answer: z = φ1(y + 2x) + φ2(2y + x) − (5x/3)cos(2x + y) + (4/9)(y − x)3 − (1/10)e3x + 4y.
Q7(b) · Numerical Analysis and Computer Programming · 15 marks≈ 11 min
Numerical Integration and Numerical Solution of ODEs
Find the values of the constants a, b, c such that the quadrature formula
∫0h f(x) dx = h[a f(0) + b f(h/3) + c f(h)]
is exact for polynomials of as high degree as possible, and hence find the order of the truncation error.हिंदी में प्रश्न
स्थिरांकों a, b, c के मान निकालिए ताकि क्षेत्रकलन-सूत्र
∫0h f(x) dx = h[a f(0) + b f(h/3) + c f(h)]
अधिक से अधिक सम्भव घातीय बहुपदों के लिए सही हो। अतएव रुंडन-त्रुटि का क्रम भी ज्ञात कीजिए।Worked solution
f = 1: a + b + c = 1; f = x: h2/2 = h(bh/3 + ch), so b/3 + c = 1/2; f = x2: h3/3 = h(bh2/9 + ch2), so b/9 + c = 1/3.
Subtracting, 2b/9 = 1/6, so b = 3/4, c = 1/4 and a = 0: ∫0h f dx ≈ (h/4)[3f(h/3) + f(h)].
f = x3: exact h4/4 but the rule gives h(3/4 · h3/27 + h3/4) = 5h4/18, so the degree of precision is 2.
Error constant C = h4/4 − 5h4/18 = −h4/36, so E = Cf‴(ξ)/3! = −(h4/216)f‴(ξ), 0 < ξ < h (check with f = ex, h = 0.7: actual error −0.00163, formula about −0.00158).
Answer: a = 0, b = 3/4, c = 1/4; exact up to degree 2; truncation error −(h4/216)f‴(ξ), of order h4.Final answer: a = 0, b = 3/4, c = 1/4; exact up to degree 2; truncation error −(h4/216)f‴(ξ), of order h4.
Q7(c) · Mechanics and Fluid Dynamics · 20 marks≈ 14 min
Hamilton's Equations
The Hamiltonian of a mechanical system is given by,
H = p1q1 − aq12 + bq22 − p2q2, where a, b are the constants. Solve the Hamiltonian equations and show that (p2 − bq2)/q1 = constant.हिंदी में प्रश्न
किसी यांत्रिक निकाय का हैमिल्टोनियन H = p1q1 − aq12 + bq22 − p2q2 के द्वारा द्योतित है, जहाँ a, b स्थिरांक हैं। हैमिल्टोनियन समीकरणों का हल निकालिए तथा दर्शाइए कि (p2 − bq2)/q1 = स्थिरांक।
Worked solution
q̇1 = ∂H/∂p1 = q1, q̇2 = ∂H/∂p2 = −q2, ṗ1 = −∂H/∂q1 = −p1 + 2aq1, ṗ2 = −∂H/∂q2 = −2bq2 + p2.
So q1 = Aet, q2 = Be−t; then ṗ1 + p1 = 2aAet gives p1 = aAet + Ce−t, and ṗ2 − p2 = −2bBe−t gives p2 = bBe−t + Det.
Hence p2 − bq2 = Det and (p2 − bq2)/q1 = D/A, a constant.
Answer: q1 = Aet, q2 = Be−t, p1 = aAet + Ce−t, p2 = bBe−t + Det; (p2 − bq2)/q1 = D/A is constant.Final answer: q1 = Aet, q2 = Be−t, p1 = aAet + Ce−t, p2 = bBe−t + Det; (p2 − bq2)/q1 = D/A is constant.
Question 8 (50 marks)
Q8(a) · Numerical Analysis and Computer Programming · 15 marks≈ 11 min
Computer Systems, Logic Gates, Boolean Algebra and Algorithms
Simplify the boolean expression :
(a + b)·(b̄ + c) + b·(ā + c̄)
by using the laws of boolean algebra. From its truth table write it in minterm normal form.हिंदी में प्रश्न
बूलीय व्यंजक :
(a + b)·(b̄ + c) + b·(ā + c̄)
को बूलीय-बीजगणित के नियमों का उपयोग करने के द्वारा सरल कीजिए। इस की सत्यता-सारणी से इसको मिनटर्म प्रसामान्य रूप में लिखिए।Worked solution
(a + b)(b′ + c) = ab′ + ac + bb′ + bc = ab′ + ac + bc, and b(a′ + c′) = a′b + bc′.
Sum = ab′ + ac + b(c + c′) + a′b = ab′ + ac + b (since b + a′b = b) = (a + b) + ac (since ab′ + b = a + b) = a + b (absorption).
Truth table for abc = 000, 001, ..., 111: 0, 0, 1, 1, 1, 1, 1, 1.
Answer: a + b; minterm normal form a′bc′ + a′bc + ab′c′ + ab′c + abc′ + abc = ∑m(2, 3, 4, 5, 6, 7).Final answer: a + b; minterm normal form a′bc′ + a′bc + ab′c′ + ab′c + abc′ + abc = ∑m(2, 3, 4, 5, 6, 7).
Q8(b) · Mechanics and Fluid Dynamics · 15 marks≈ 11 min
Kinematics of Fluids: Continuity, Streamlines and Potential Flow
For a two-dimensional potential flow, the velocity potential is given by φ = x2y − xy2 + (1/3)(x3 − y3). Determine the velocity components along the directions x and y. Also, determine the stream function ψ and check whether φ represents a possible case of flow or not.
हिंदी में प्रश्न
एक द्विविमीय विभव-प्रवाह के लिए वेग विभव φ = x2y − xy2 + (1/3)(x3 − y3) के द्वारा दिया गया है। x व y दिशाओं के अनुदिश वेग घटकों का निर्धारण कीजिए। धारा-फलन ψ का भी निर्धारण कीजिए और जांच कीजिए कि क्या φ एक सम्भव प्रवाह को निरूपित करता है अथवा नहीं।
Worked solution
u = −∂φ/∂x = −(x2 + 2xy − y2), v = −∂φ/∂y = −(x2 − 2xy − y2).
∇2φ = (2x + 2y) + (−2x − 2y) = 0, so ∇·q = 0 and φ represents a possible (incompressible, irrotational) flow.
∂ψ/∂y = ∂φ/∂x = x2 + 2xy − y2 gives ψ = x2y + xy2 − y3/3 + g(x); then ∂ψ/∂x = 2xy + y2 + g′(x) = −∂φ/∂y = −x2 + 2xy + y2, so g = −x3/3.
Answer: u = −(x2 + 2xy − y2), v = 2xy + y2 − x2; ψ = x2y + xy2 − (x3 + y3)/3; the flow is possible. (With the convention q = ∇φ, u, v and ψ change sign.)Final answer: u = −(x2 + 2xy − y2), v = 2xy + y2 − x2; ψ = x2y + xy2 − (x3 + y3)/3; the flow is possible. (With the convention q = ∇φ, u, v and ψ change sign.)
Q8(c) · Partial Differential Equations · 20 marks≈ 14 min
Vibrating String, Heat Equation and Laplace Equation
A thin annulus occupies the region 0 < a ≤ r ≤ b, 0 ≤ θ ≤ 2π. The faces are insulated. Along the inner edge the temperature is maintained at 0°, while along the outer edge the temperature is held at T = K cos(θ/2), where K is a constant. Determine the temperature distribution in the annulus.
हिंदी में प्रश्न
एक पतली वलयिका (एनुलस) क्षेत्र 0 < a ≤ r ≤ b, 0 ≤ θ ≤ 2π को घेरती है। इसके तल ताप-अवरोधी हैं। आन्तरिक किनारे के साथ-साथ ताप 0° पर स्थिर रखा जाता है जबकि बाह्य किनारे का ताप T = K cos(θ/2) पर बनाए रखा जाता है, जहाँ K एक स्थिरांक है। वलयिका में ताप-वितरण का निर्धारण कीजिए।
Worked solution
Steady state: urr + ur/r + uθθ/r2 = 0 in a ≤ r ≤ b, with u(a, θ) = 0 and u(b, θ) = K cos(θ/2), 0 ≤ θ ≤ 2π.
Fourier coefficients of cos(θ/2) on (0, 2π): (1/π)∫02π cos(θ/2)cos nθ dθ = 0 for all n ≥ 0, and (1/π)∫02π cos(θ/2)sin nθ dθ = 8n/(π(4n2 − 1)).
So K cos(θ/2) = (8K/π)∑n=1∞ n sin nθ/(4n2 − 1) for 0 < θ < 2π, and only the sine terms are needed (no constant or log r term).
Separated solutions (Anrn + Bnr−n)sin nθ with Anan + Bna−n = 0 give the radial factor (r/a)n − (a/r)n; dividing by its value at r = b matches the outer edge.
Answer: u(r, θ) = (8K/π)∑n=1∞ [n/(4n2 − 1)] · [(r/a)n − (a/r)n]/[(b/a)n − (a/b)n] · sin nθ (across θ = 0 the boundary value jumps between −K and K, and the series gives the mean 0 there).Final answer: u(r, θ) = (8K/π)∑n=1∞ [n/(4n2 − 1)] · [(r/a)n − (a/r)n]/[(b/a)n − (a/b)n] · sin nθ (across θ = 0 the boundary value jumps between −K and K, and the series gives the mean 0 there).