UPSC Mains 2020 Mathematics Optional Question Papers with Solutions
All 56 questions of the 2020 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 2020 Mathematics Paper I
Question 1 (50 marks)
Q1(a) · Linear Algebra · 10 marks≈ 7 min
Vector Spaces, Subspaces, Basis and Dimension
Consider the set V of all n × n real magic squares. Show that V is a vector space over R. Give examples of two distinct 2 × 2 magic squares.
हिंदी में प्रश्न
माना समुच्चय V में सभी n × n के वास्तविक मैजिक वर्ग हैं। दिखाइए कि समुच्चय V, R पर एक सदिश समष्टि है। दो भिन्न-भिन्न 2 × 2 मैजिक वर्ग के उदाहरण दीजिए।
Worked solution
An n × n real matrix is a magic square if all its row sums, column sums and both diagonal sums are equal. V ⊆ Mn(ℝ), which is a vector space, so it is enough to apply the subspace test.
The zero matrix is a magic square (all sums 0), so V ≠ ∅. If A, B ∈ V have common sums s and t, then every row, column and diagonal sum of αA + βB equals αs + βt, because each such sum is a linear function of the entries.
So αA + βB ∈ V for all α, β ∈ ℝ: V is a subspace of Mn(ℝ) and hence a vector space over ℝ.
For n = 2, [[a, b], [c, d]] needs a + b = c + d = a + c = b + d = a + d = b + c. Then a + b = a + c gives b = c, a + b = b + d gives a = d, and a + d = b + c gives a = b; so every 2 × 2 magic square is a multiple of [[1, 1], [1, 1]].
Answer: V is a subspace of Mn(ℝ), hence a vector space; two distinct 2 × 2 magic squares are [[1, 1], [1, 1]] and [[2, 2], [2, 2]] (for n = 2 the space is one-dimensional).Final answer: V is a subspace of Mn(ℝ), hence a vector space; two distinct 2 × 2 magic squares are [[1, 1], [1, 1]] and [[2, 2], [2, 2]] (for n = 2 the space is one-dimensional).
Q1(b) · Linear Algebra · 10 marks≈ 7 min
Linear Transformations, Rank-Nullity and Matrix Representation
Let M2(R) be the vector space of all 2 × 2 real matrices. Let B = [[1, −1], [−4, 4]]. Suppose T : M2(R) → M2(R) is a linear transformation defined by T(A) = BA. Find the rank and nullity of T. Find a matrix A which maps to the null matrix.
हिंदी में प्रश्न
माना M2(R) सभी 2 × 2 वास्तविक आव्यूहों का सदिश समष्टि है। माना B = [[1, −1], [−4, 4]]. माना T : M2(R) → M2(R) एक रैखिक रूपांतरण है, जो T(A) = BA द्वारा परिभाषित है। T की कोटि (रैंक) व शून्यता (नलिटि) ज्ञात कीजिए। आव्यूह A ज्ञात कीजिए, जो शून्य आव्यूह को प्रतिचित्रित करता है।
Worked solution
T is linear since B(αA + βC) = αBA + βBC. Write A = [X1 X2] by columns; then BA = [BX1 BX2], so T(A) = 0 ⇔ both columns lie in N(B).
B = [[1, −1], [−4, 4]] has rank 1 (second row = −4 × first row), and BX = 0 ⇔ x = y, so N(B) = L{(1, 1)T}.
Hence N(T) = {[[a, b], [a, b]]: a, b ∈ ℝ}, with basis [[1, 0], [1, 0]] and [[0, 1], [0, 1]]: nullity 2, and by rank-nullity rank T = 4 − 2 = 2.
(Range check: each column of BA lies in the column space L{(1, −4)T} of B, so R(T) = {[[p, q], [−4p, −4q]]}, again of dimension 2.)
Answer: rank T = 2, nullity T = 2; for example A = [[1, 1], [1, 1]] gives BA = 0.Final answer: rank T = 2, nullity T = 2; for example A = [[1, 1], [1, 1]] gives BA = 0.
Q1(c) · Calculus · 10 marks≈ 7 min
Mean Value Theorems, Taylor's Theorem and Indeterminate Forms
Evaluate limx→π/4 (tan x)tan 2x.
हिंदी में प्रश्न
limx→π/4 (tan x)tan 2x का मान निकालिए।
Q1(d) · Calculus · 10 marks≈ 7 min
Maxima and Minima, Asymptotes and Curve Tracing
Find all the asymptotes of the curve (2x + 3)y = (x − 1)2.
हिंदी में प्रश्न
वक्र (2x + 3)y = (x − 1)2 के सभी अनंतस्पर्शी निकालिए।
Worked solution
Vertical: the coefficient of y (the highest power of y) is 2x + 3, so x = −3/2 is an asymptote; indeed y = (x − 1)2/(2x + 3) → ±∞ as x → −3/2, the numerator tending to 25/4 ≠ 0.
Oblique: dividing, y = x/2 − 7/4 + (25/4)/(2x + 3), and the last term → 0 as x → ±∞. (Check: (2x + 3)(x/2 − 7/4) = x2 − 2x − 21/4, and adding 25/4 gives (x − 1)2.)
There is no horizontal asymptote, since the coefficient of x2 (the highest power of x) is the constant 1. A curve of degree 3 can have at most three asymptotes; here there are two real ones.
Answer: x = −3/2 and y = x/2 − 7/4 (i.e. 2x − 4y − 7 = 0).Final answer: x = −3/2 and y = x/2 − 7/4 (i.e. 2x − 4y − 7 = 0).
Q1(e) · Analytic Geometry · 10 marks≈ 7 min
Ellipsoid, Hyperboloids and Paraboloids
Find the equations of the tangent plane to the ellipsoid 2x2 + 6y2 + 3z2 = 27 which passes through the line x − y − z = 0 = x − y + 2z − 9.
हिंदी में प्रश्न
दीर्घवृत्तज 2x2 + 6y2 + 3z2 = 27 के स्पर्श समतल का समीकरण निकालिए, जो रेखा x − y − z = 0 = x − y + 2z − 9 से होकर गुजरता है।
Question 2 (50 marks)
Q2(a) · Calculus · 15 marks≈ 11 min
Riemann Integral, Indefinite Integrals and Improper Integrals
Evaluate ∫01 tan−1(1 − 1/x) dx.
हिंदी में प्रश्न
∫01 tan−1(1 − 1/x) dx का मान निकालिए।
Worked solution
The integrand is continuous on (0, 1] and tends to −π/2 as x → 0+, so the integral exists. Integrate by parts with u = tan−1(1 − 1/x): du = (1/x2)/(1 + (1 − 1/x)2) dx = dx/(2x2 − 2x + 1).
I = [x tan−1(1 − 1/x)]0+1 − J, where J = ∫01 x dx/(2x2 − 2x + 1); the bracket is 0 − 0, since tan−10 = 0 and x tan−1(1 − 1/x) → 0 as x → 0+.
The denominator is unchanged by x → 1 − x, so J = ∫01 (1 − x) dx/(2x2 − 2x + 1), and 2J = ∫01 dx/(2x2 − 2x + 1).
Since 2x2 − 2x + 1 = ((2x − 1)2 + 1)/2, 2J = ∫01 2 dx/((2x − 1)2 + 1) = [tan−1(2x − 1)]01 = π/2, so J = π/4.
Answer: −π/4.Final answer: −π/4.
Q2(b) · Linear Algebra · 20 marks≈ 14 min
Symmetric, Hermitian, Orthogonal and Unitary Matrices and Their Eigenvalues
Define an n × n matrix as A = I − 2u·uT, where u is a unit column vector.
(i) Examine if A is symmetric.
(ii) Examine if A is orthogonal.
(iii) Show that trace (A) = n − 2.
(iv) Find A3×3, when u = [1/3, 2/3, 2/3]T.हिंदी में प्रश्न
एक n × n आव्यूह A को परिभाषित कीजिए, जबकि A = I − 2u·uT, जहाँ u एक इकाई स्तंभ सदिश है।
(i) परीक्षण कीजिए कि A सममित है।
(ii) परीक्षण कीजिए कि A लांबिक है।
(iii) दिखाइए कि आव्यूह A का अनुरेख (n − 2) है।
(iv) आव्यूह A3×3 निकालिए, जबकि u = [1/3, 2/3, 2/3]T है।Worked solution
(i) AT = I − 2(uuT)T = I − 2(uT)TuT = I − 2uuT = A, so A is symmetric.
(ii) ATA = A2 = I − 4uuT + 4u(uTu)uT = I − 4uuT + 4uuT = I, using uTu = 1. So A is orthogonal (and A2 = I).
(iii) tr(uuT) = ∑ui2 = uTu = 1, so tr A = tr I − 2 tr(uuT) = n − 2.
(iv) uuT = (1/9)[[1, 2, 2], [2, 4, 4], [2, 4, 4]], so A = I − (2/9)[[1, 2, 2], [2, 4, 4], [2, 4, 4]] = (1/9)[[7, −4, −4], [−4, 1, −8], [−4, −8, 1]].
Check: tr A = (7 + 1 + 1)/9 = 1 = 3 − 2; each row of [[7, −4, −4], [−4, 1, −8], [−4, −8, 1]] has length 9 and rows 1 and 2 are orthogonal (−28 − 4 + 32 = 0).
Answer: A is symmetric and orthogonal, tr A = n − 2, and for the given u, A = (1/9)[[7, −4, −4], [−4, 1, −8], [−4, −8, 1]].Final answer: A is symmetric and orthogonal, tr A = n − 2, and for the given u, A = (1/9)[[7, −4, −4], [−4, 1, −8], [−4, −8, 1]].
Q2(c) · Analytic Geometry · 15 marks≈ 11 min
Cone and Cylinder
Find the equation of the cylinder whose generators are parallel to the line x/1 = y/(−2) = z/3 and whose guiding curve is x2 + y2 = 4, z = 2.
हिंदी में प्रश्न
एक ऐसे बेलन का समीकरण निकालिए, जिसकी जनक-रेखाएँ, रेखा x/1 = y/(−2) = z/3 के समांतर हैं तथा जिसका मार्गदर्शक वक्र x2 + y2 = 4, z = 2 है।
Question 3 (50 marks)
Q3(a) · Calculus · 20 marks≈ 14 min
Maxima and Minima, Asymptotes and Curve Tracing
Consider the function f(x) = ∫0x (t2 − 5t + 4)(t2 − 5t + 6) dt.
(i) Find the critical points of the function f(x).
(ii) Find the points at which local minimum occurs.
(iii) Find the points at which local maximum occurs.
(iv) Find the number of zeros of the function f(x) in [0, 5].हिंदी में प्रश्न
निम्न फलन पर विचार कीजिए : f(x) = ∫0x (t2 − 5t + 4)(t2 − 5t + 6) dt
(i) फलन f(x) के क्रांतिक बिंदु निकालिए।
(ii) वे बिंदु निकालिए, जहाँ f(x) का स्थानीय न्यूनतम होगा।
(iii) वे बिंदु निकालिए, जहाँ f(x) का स्थानीय अधिकतम होगा।
(iv) फलन f(x) के [0, 5] में कितने शून्यक होंगे, निकालिए।Worked solution
By the fundamental theorem of calculus, f′(x) = (x2 − 5x + 4)(x2 − 5x + 6) = (x − 1)(x − 2)(x − 3)(x − 4). (i) The critical points are x = 1, 2, 3, 4.
Signs of f′: + on (−∞, 1), − on (1, 2), + on (2, 3), − on (3, 4), + on (4, ∞), since one factor changes sign at each critical point.
(ii) f′ changes from − to + at x = 2 and x = 4: local minima. (iii) f′ changes from + to − at x = 1 and x = 3: local maxima.
(iv) f(x) = x5/5 − 5x4/2 + 35x3/3 − 25x2 + 24x, so f(0) = 0, f(1) = 251/30, f(2) = 116/15, f(3) = 81/10, f(4) = 112/15, f(5) = 95/6.
f increases from f(0) = 0 on [0, 1], and on [1, 5] its least value is the smaller local minimum f(4) = 112/15 > 0. So f > 0 on (0, 5].
Answer: (i) 1, 2, 3, 4; (ii) local minima at x = 2, 4; (iii) local maxima at x = 1, 3; (iv) exactly one zero in [0, 5], namely x = 0.Final answer: (i) 1, 2, 3, 4; (ii) local minima at x = 2, 4; (iii) local maxima at x = 1, 3; (iv) exactly one zero in [0, 5], namely x = 0.
Q3(b) · Linear Algebra · 15 marks≈ 11 min
Linear Transformations, Rank-Nullity and Matrix Representation
Let F be a subfield of complex numbers and T a function from F3 → F3 defined by T(x1, x2, x3) = (x1 + x2 + 3x3, 2x1 − x2, −3x1 + x2 − x3). What are the conditions on a, b, c such that (a, b, c) be in the null space of T? Find the nullity of T.
हिंदी में प्रश्न
माना F सम्मिश्र संख्याओं का एक उपक्षेत्र है व T : F3 → F3 एक ऐसा फलन है, जो निम्न रूप से परिभाषित है : T(x1, x2, x3) = (x1 + x2 + 3x3, 2x1 − x2, −3x1 + x2 − x3). a, b, c पर क्या शर्तें हैं कि (a, b, c), T के शून्य समष्टि में है? T की शून्यता निकालिए।
Worked solution
(a, b, c) ∈ N(T) ⇔ a + b + 3c = 0, 2a − b = 0 and −3a + b − c = 0.
The second equation gives b = 2a; then the first gives 3a + 3c = 0, so c = −a; the third, −3a + 2a + a = 0, then holds automatically (the coefficient matrix has determinant 0 and rank 2).
So N(T) = {(a, 2a, −a)} = L{(1, 2, −1)}. Check: T(1, 2, −1) = (1 + 2 − 3, 2 − 2, −3 + 2 + 1) = (0, 0, 0).
Answer: (a, b, c) ∈ N(T) ⇔ b = 2a and c = −a; nullity T = 1 (and rank T = 2).Final answer: (a, b, c) ∈ N(T) ⇔ b = 2a and c = −a; nullity T = 1 (and rank T = 2).
Q3(c) · Analytic Geometry · 15 marks≈ 11 min
Cone and Cylinder
If the straight line x/1 = y/2 = z/3 represents one of a set of three mutually perpendicular generators of the cone 5yz − 8zx − 3xy = 0, then find the equations of the other two generators.
हिंदी में प्रश्न
यदि सरल रेखा x/1 = y/2 = z/3 शंकु 5yz − 8zx − 3xy = 0 के तीन परस्पर लांबिक जनकों के समुच्चय में से एक है, तब अन्य दो जनकों के समीकरण निकालिए।
Worked solution
Here a = b = c = 0, so a + b + c = 0 and sets of three perpendicular generators exist. The line x/1 = y/2 = z/3 (the same line as 6x = 3y = 2z) is a generator: 5·6 − 8·3 − 3·2 = 0.
The plane through O perpendicular to it is x + 2y + 3z = 0. Put x = −2y − 3z in the cone: 5yz + 8z(2y + 3z) + 3y(2y + 3z) = 6y2 + 30yz + 24z2 = 6(y + z)(y + 4z) = 0.
y = −z gives x = −z: direction (1, 1, −1). y = −4z gives x = 5z: direction (5, −4, 1).
Check: (1, 1, −1)·(5, −4, 1) = 0, (1, 2, 3)·(1, 1, −1) = 0 and (1, 2, 3)·(5, −4, 1) = 0.
Answer: x/1 = y/1 = z/(−1) and x/5 = y/(−4) = z/1.Final answer: x/1 = y/1 = z/(−1) and x/5 = y/(−4) = z/1.
Question 4 (50 marks)
Q4(a) · Linear Algebra · 15 marks≈ 11 min
Systems of Linear Equations
Let A = [[1, 0, 2], [2, −1, 3], [4, 1, 8]] and B = [[−11, 2, 2], [−4, 0, 1], [6, −1, −1]].
(i) Find AB.
(ii) Find det(A) and det(B).
(iii) Solve the following system of linear equations : x + 2z = 3, 2x − y + 3z = 3, 4x + y + 8z = 14हिंदी में प्रश्न
माना A = [[1, 0, 2], [2, −1, 3], [4, 1, 8]] और B = [[−11, 2, 2], [−4, 0, 1], [6, −1, −1]]
(i) AB ज्ञात कीजिए।
(ii) सारणिक (A) व सारणिक (B) ज्ञात कीजिए।
(iii) निम्न रैखिक समीकरणों के निकाय का हल निकालिए : x + 2z = 3, 2x − y + 3z = 3, 4x + y + 8z = 14Worked solution
(i) Row 1 (1, 0, 2) of A against the columns of B gives (−11 + 12, 2 − 2, 2 − 2) = (1, 0, 0); row 2 (2, −1, 3) gives (−22 + 4 + 18, 4 − 3, 4 − 1 − 3) = (0, 1, 0); row 3 (4, 1, 8) gives (−44 − 4 + 48, 8 − 8, 8 + 1 − 8) = (0, 0, 1). So AB = I.
(ii) det A = 1(−8 − 3) − 0 + 2(2 + 4) = 1, and det A · det B = det(AB) = 1 gives det B = 1.
(iii) The system is AX = C with C = (3, 3, 14)T and A−1 = B, so X = BC = (−33 + 6 + 28, −12 + 0 + 14, 18 − 3 − 14) = (1, 2, 1).
Check: 1 + 2 = 3, 2 − 2 + 3 = 3, 4 + 2 + 8 = 14.
Answer: AB = I, det A = det B = 1, and x = 1, y = 2, z = 1.Final answer: AB = I, det A = det B = 1, and x = 1, y = 2, z = 1.
Q4(b) · Analytic Geometry · 15 marks≈ 11 min
Ellipsoid, Hyperboloids and Paraboloids
Find the locus of the point of intersection of the perpendicular generators of the hyperbolic paraboloid x2/a2 − y2/b2 = 2z.
हिंदी में प्रश्न
अतिपरवलयिक परवलयज x2/a2 − y2/b2 = 2z के लांबिक जनकों के प्रतिच्छेद बिंदु का बिंदुपथ निकालिए।
Worked solution
λ-system: x/a − y/b = λ, λ(x/a + y/b) = 2z; its direction is (1/a, −1/b, 0) × (λ/a, λ/b, −2), proportional to (a, b, λ).
μ-system: x/a + y/b = μ, μ(x/a − y/b) = 2z; its direction is proportional to (a, −b, μ).
The two generators through a point (x, y, z) of the surface have λ = x/a − y/b, μ = x/a + y/b, so λμ = x2/a2 − y2/b2 = 2z.
They are perpendicular when a2 − b2 + λμ = 0, that is 2z = b2 − a2.
Answer: the locus is the curve x2/a2 − y2/b2 = b2 − a2, 2z = b2 − a2 (the section of the surface by the plane 2z + a2 − b2 = 0).Final answer: the locus is the curve x2/a2 − y2/b2 = b2 − a2, 2z = b2 − a2 (the section of the surface by the plane 2z + a2 − b2 = 0).
Q4(c) · Calculus · 20 marks≈ 14 min
Functions of Several Variables: Partial Derivatives, Extrema, Lagrange Multipliers and Jacobians
Find an extreme value of the function u = x2 + y2 + z2, subject to the condition 2x + 3y + 5z = 30, by using Lagrange’s method of undetermined multiplier.
हिंदी में प्रश्न
लाग्रांज की अनिर्धारित गुणक विधि का प्रयोग करके फलन u = x2 + y2 + z2 का चरम मान ज्ञात कीजिए, जो 2x + 3y + 5z = 30 शर्त द्वारा प्रतिबंधित है।
Question 5 (50 marks)
Q5(a) · Ordinary Differential Equations · 10 marks≈ 7 min
Formulation and First Order, First Degree Equations
Solve the following differential equation : x cos(y/x)(y dx + x dy) = y sin(y/x)(x dy − y dx)
हिंदी में प्रश्न
निम्न अवकल समीकरण को हल कीजिए : x cos(y/x)(y dx + x dy) = y sin(y/x)(x dy − y dx)
Worked solution
Here M = xy cos(y/x) + y2 sin(y/x) and N = x2 cos(y/x) − xy sin(y/x) are homogeneous of degree 2, and Mx + Ny = 2x2y cos(y/x) ≠ 0, so 1/(2x2y cos(y/x)) is an IF (Rule 3).
Quicker by inspection: d(xy cos(y/x)) = (y dx + x dy) cos(y/x) − (y/x) sin(y/x)(x dy − y dx).
So the equation is x·d(xy cos(y/x)) = 0.
Answer: xy cos(y/x) = c.Final answer: xy cos(y/x) = c.
Q5(b) · Ordinary Differential Equations · 10 marks≈ 7 min
Orthogonal Trajectories
Find the orthogonal trajectories of the family of circles passing through the points (0, 2) and (0, −2).
हिंदी में प्रश्न
वृत्त-कुल, जो बिंदु (0, 2) एवं (0, −2) से गुजरता है, का लंबकोणीय संछेदी ज्ञात कीजिए।
Q5(c) · Vector Analysis · 10 marks≈ 7 min
Gradient, Divergence and Curl in Cartesian and Cylindrical Coordinates
For what value of a, b, c is the vector field V = (−4x − 3y + az)î + (bx + 3y + 5z)ĵ + (4x + cy + 3z)k̂ irrotational? Hence, express V as the gradient of a scalar function φ. Determine φ.
हिंदी में प्रश्न
a, b, c के किस मान के लिए सदिश क्षेत्र V = (−4x − 3y + az)î + (bx + 3y + 5z)ĵ + (4x + cy + 3z)k̂ अघूर्णी है? तब V को अदिश फलन φ की प्रवणता के रूप में व्यक्त कीजिए। φ को ज्ञात कीजिए।
Q5(d) · Dynamics and Statics · 10 marks≈ 7 min
Equilibrium, Friction and the Common Catenary
A uniform rod, in vertical position, can turn freely about one of its ends and is pulled aside from the vertical by a horizontal force acting at the other end of the rod and equal to half its weight. At what inclination to the vertical will the rod rest?
हिंदी में प्रश्न
एक एकसमान छड़, जो ऊर्ध्वाधर दशा में है, अपने एक सिरे पर स्वतंत्र रूप से वर्तन कर सकती है तथा दूसरे सिरे पर लगाए गए एक क्षैतिज बल, जिसका मान छड़ के भार का आधा है, द्वारा ऊर्ध्वाधर से एक तरफ खींची जाती है। बताइए कि ऊर्ध्वाधर से किस कोण पर छड़ विश्राम करेगी।
Worked solution
Let the rod have length 2a and weight W, hinged at O, with the horizontal force W/2 at the free end, and let θ be its inclination to the vertical.
Moments about O: (W/2) × 2a cos θ = W × a sin θ.
So tan θ = 1.
Answer: the rod rests at 45° to the vertical.Final answer: the rod rests at 45° to the vertical.
Q5(e) · Mechanics and Fluid Dynamics · 10 marks≈ 7 min
Moment of Inertia and Motion of Rigid Bodies in Two Dimensions
A light rigid rod ABC has three particles each of mass m attached to it at A, B and C. The rod is struck by a blow P at right angles to it at a point distant from A equal to BC. Prove that the kinetic energy set up is (1/2)(P2/m)(a2 − ab + b2)/(a2 + ab + b2), where AB = a and BC = b.
हिंदी में प्रश्न
एक हल्की दृढ़ छड़ ABC से तीन कण, जिनमें से हरेक का द्रव्यमान m है, A, B तथा C पर बंधे हुए हैं। उस छड़ को बिंदु A से BC दूरी के बराबर स्थित बिंदु पर एक बल P के द्वारा लम्बवत् मारा जाता है। सिद्ध कीजिए कि पैदा हुई गतिज ऊर्जा का मान (1/2)(P2/m)(a2 − ab + b2)/(a2 + ab + b2) है, जहाँ AB = a तथा BC = b.
Worked solution
Put A, B, C at x = 0, a, a + b; the blow is at x = b. The centre of mass is at x̄ = (2a + b)/3 and the total mass is 3m.
IG = m[(2a + b)2 + (a − b)2 + (a + 2b)2]/9 = (2m/3)(a2 + ab + b2).
The blow is at distance d = x̄ − b = 2(a − b)/3 from G, so vG = P/(3m), ω = Pd/IG, and KE = P2/(6m) + P2d2/(2IG) = P2/(6m) + P2(a − b)2/(3m(a2 + ab + b2)).
= (P2/(6m))[(a2 + ab + b2) + 2(a − b)2]/(a2 + ab + b2) = (P2/(6m)) · 3(a2 − ab + b2)/(a2 + ab + b2).
Answer: KE = (1/2)(P2/m)(a2 − ab + b2)/(a2 + ab + b2).Final answer: KE = (1/2)(P2/m)(a2 − ab + b2)/(a2 + ab + b2).
Question 6 (50 marks)
Q6(a) · Ordinary Differential Equations · 20 marks≈ 14 min
Second Order Equations with Variable Coefficients
Using the method of variation of parameters, solve the differential equation y″ + (1 − cot x)y′ − y cot x = sin2 x, if y = e−x is one solution of CF.
हिंदी में प्रश्न
प्राचल विचरण विधि का प्रयोग करके, निम्न अवकल समीकरण का हल निकालिए, यदि y = e−x, पूरक फलन (CF) का एक हल है : y″ + (1 − cot x)y′ − y cot x = sin2 x
Worked solution
P = 1 − cot x, so e−∫P dx = e−x sin x and y2 = e−x∫e−x sin x/e−2x dx = e−x·ex(sin x − cos x)/2; take y2 = sin x − cos x (check: y″ + (1 − cot x)y′ − y cot x = 2 cos x − 2 cot x sin x = 0).
W = y1y2′ − y2y1′ = e−x(cos x + sin x) + e−x(sin x − cos x) = 2e−x sin x, and R = sin2x.
u2 = ∫y1R/W dx = (1/2)∫sin x dx = −(cos x)/2.
u1 = −(1/2)∫ex(sin2x − sin x cos x) dx = −(1/2)∫ex[1/2 − (cos 2x)/2 − (sin 2x)/2] dx = −ex/4 + ex(3 sin 2x − cos 2x)/20.
PI = u1e−x + u2(sin x − cos x) = −1/4 + (3 sin 2x − cos 2x)/20 − (sin 2x)/4 + (1 + cos 2x)/4 = (2 cos 2x − sin 2x)/10.
Answer: y = c1e−x + c2(sin x − cos x) + (2 cos 2x − sin 2x)/10.Final answer: y = c1e−x + c2(sin x − cos x) + (2 cos 2x − sin 2x)/10.
Q6(b) · Vector Analysis · 15 marks≈ 11 min
Gauss and Stokes' Theorems and Green's Identities
For the vector function A, where A = (3x2 + 6y)î − 14yzĵ + 20xz2k̂, calculate ∫C A · dr from (0, 0, 0) to (1, 1, 1) along the following paths :
(i) x = t, y = t2, z = t3
(ii) Straight lines joining (0, 0, 0) to (1, 0, 0), then to (1, 1, 0) and then to (1, 1, 1)
(iii) Straight line joining (0, 0, 0) to (1, 1, 1)
Is the result same in all the cases? Explain the reason.हिंदी में प्रश्न
दिए गए सदिश फलन A, जहाँ A = (3x2 + 6y)î − 14yzĵ + 20xz2k̂, के लिए ∫C A · dr का मान निकालिए, जहाँ C बिंदु (0, 0, 0) से (1, 1, 1) तक निम्न पथों से निर्देशित है :
(i) x = t, y = t2, z = t3
(ii) सरल रेखा (0, 0, 0) से (1, 0, 0) तक जोड़ने पर, फिर (1, 1, 0) तक तथा फिर (1, 1, 1) तक
(iii) सरल रेखा (0, 0, 0) से (1, 1, 1) तक जोड़ने पर
क्या सभी स्थितियों में परिणाम समान हैं? कारण की व्याख्या कीजिए।Worked solution
(i) With x = t, y = t2, z = t3: A·dr = [(3t2 + 6t2) − 14t5(2t) + 20t7(3t2)] dt = (9t2 − 28t6 + 60t9) dt, and ∫01 = 3 − 4 + 6 = 5.
(ii) On (0, 0, 0) → (1, 0, 0): y = z = 0 and ∫013x2 dx = 1; on (1, 0, 0) → (1, 1, 0): z = 0, so A·dr = −14yz dy = 0; on (1, 1, 0) → (1, 1, 1): A·dr = 20z2 dz, giving 20/3; total 23/3.
(iii) With x = y = z = t: A·dr = (3t2 + 6t − 14t2 + 20t3) dt, and ∫01 = 1 + 3 − 14/3 + 5 = 13/3.
Reason: ∇ × A = 14y i − 20z2 j − 6k ≠ 0, so A is not a gradient field and its line integral depends on the path.
Answer: 5, 23/3 and 13/3; the results differ because A is not conservative.Final answer: 5, 23/3 and 13/3; the results differ because A is not conservative.
Q6(c) · Dynamics and Statics · 15 marks≈ 11 min
Equilibrium, Friction and the Common Catenary
A beam AD rests on two supports B and C, where AB = BC = CD. It is found that the beam will tilt when a weight of p kg is hung from A or when a weight of q kg is hung from D. Find the weight of the beam.
हिंदी में प्रश्न
एक दंड AD दो आलंब B एवं C पर विश्राम करता है, जबकि AB = BC = CD. यह पाया गया कि दंड झुक जाएगा यदि एक भार p kg, बिंदु A से लटकाया जाए या एक भार q kg, बिंदु D से लटकाया जाए। दंड का भार बताइए।
Worked solution
Let AB = BC = CD = d, the beam weigh w, and its centre of gravity G lie between B and C at distance x from B (the beam need not be uniform; if it were, p = q would be forced).
With p hung at A the beam is about to tilt about B, so the reaction at C is zero; moments about B: pd = wx.
With q hung at D it is about to tilt about C: qd = w(d − x).
Adding, (p + q)d = wd.
Answer: the beam weighs p + q kg, and its centre of gravity divides BC in the ratio p : q.Final answer: the beam weighs p + q kg, and its centre of gravity divides BC in the ratio p : q.
Question 7 (50 marks)
Q7(a) · Vector Analysis · 20 marks≈ 14 min
Gauss and Stokes' Theorems and Green's Identities
Verify the Stokes’ theorem for the vector field F = xyî + yzĵ + xzk̂ on the surface S which is the part of the cylinder z = 1 − x2 for 0 ≤ x ≤ 1, −2 ≤ y ≤ 2; S is oriented upwards.
हिंदी में प्रश्न
स्टोक्स प्रमेय को सत्यापित कीजिए, जबकि सदिश क्षेत्र F = xyî + yzĵ + xzk̂ एक सतह S पर है जो कि एक बेलन z = 1 − x2, 0 ≤ x ≤ 1, −2 ≤ y ≤ 2 का हिस्सा है, जहाँ S उपरिमुखी अभिविन्यस्त है।
Worked solution
(Statement of Stokes' theorem: sub-section 4.) For F = (xy, yz, zx), ∇ × F = (−y, −z, −x).
On z = 1 − x2 the upward n dS = (2x, 0, 1) dx dy, so (∇ × F)·n dS = (−2xy − x) dx dy, and ∫01∫−22(−2xy − x) dy dx = ∫01(−4x) dx = −2.
Boundary (anticlockwise seen from above): C1: y = −2, z = 1 − x2, x: 0 → 1; C2: x = 1, z = 0, y: −2 → 2; C3: y = 2, z = 1 − x2, x: 1 → 0; C4: x = 0, z = 1, y: 2 → −2.
F·dr = xy dx + yz dy + zx dz with dz = −2x dx on C1, C3: C1 gives ∫01(−2x − 2x2 + 2x4) dx = −19/15 and C3 gives ∫10(2x − 2x2 + 2x4) dx = −11/15.
On C2, z = 0 and dx = dz = 0, so F·dr = 0; on C4, F·dr = y dy, whose integral from 2 to −2 is 0.
Answer: ∮F·dr = −19/15 − 11/15 = −2 = ∬(∇ × F)·n dS, verifying Stokes' theorem.Final answer: ∮F·dr = −19/15 − 11/15 = −2 = ∬(∇ × F)·n dS, verifying Stokes' theorem.
Q7(b) · Ordinary Differential Equations · 10 marks≈ 7 min
Laplace Transforms and Initial Value Problems
Using Laplace transform, solve the initial value problem ty″ + 2ty′ + 2y = 2; y(0) = 1 and y′(0) is arbitrary. Does this problem have a unique solution?
हिंदी में प्रश्न
लाप्लास रूपांतरण का प्रयोग करके प्रारंभिक मान समस्या ty″ + 2ty′ + 2y = 2; y(0) = 1 तथा y′(0) स्वेच्छ है, को हल कीजिए। क्या इस प्रश्न का हल अद्वितीय है?
Worked solution
With y(0) = 1 and y′(0) = k unknown: L{ty″} = −d(s2Y − s − k)/ds = −s2Y′ − 2sY + 1 and L{ty′} = −d(sY − 1)/ds = −sY′ − Y.
Transforming: −(s2 + 2s)Y′ − 2sY + 1 = 2/s, that is (s + 2)Y′ + 2Y = (s − 2)/s2; multiplying by s + 2, d[(s + 2)2Y]/ds = 1 − 4/s2.
So (s + 2)2Y = s + 4/s + C and Y = (s2 + Cs + 4)/(s(s + 2)2) = 1/s + (C − 4)/(s + 2)2.
Hence y = 1 + Ate−2t with A = C − 4 arbitrary; y(0) = 1 for every A and y′(0) = A. The coefficient t of y″ vanishes at t = 0, so the existence-uniqueness theorem does not apply there.
Answer: y = 1 + Ate−2t, A arbitrary; the solution is not unique (it is fixed only if y′(0) is prescribed).Final answer: y = 1 + Ate−2t, A arbitrary; the solution is not unique (it is fixed only if y′(0) is prescribed).
Q7(c) · Dynamics and Statics · 20 marks≈ 14 min
Virtual Work, Stability and Forces in Three Dimensions
(i) A square framework formed of uniform heavy rods of equal weight W jointed together, is hung up by one corner. A weight W is suspended from each of the three lower corners, and the shape of the square is preserved by a light rod along the horizontal diagonal. Find the thrust of the light rod.
(ii) A particle starts at a great distance with velocity V. Let p be the length of the perpendicular from the centre of a star on the tangent to the initial path of the particle. Show that the least distance of the particle from the centre of the star is λ, where V2λ = √(μ2 + p2V4) − μ. Here μ is a constant.हिंदी में प्रश्न
(i) चार एकसमान भारी छड़, जो समान भार W की हैं, एक वर्ग के रूप में ढाँचा बनाते हुए जुड़ी हैं। यह एक कोने से टँगा हुआ है। भार W तीनों नीचे वाले हरेक कोने से लटकाए हैं। वर्ग का आकार एक हल्की छड़, जो क्षैतिज विकर्ण के अनुदिश है, द्वारा रक्षित किया गया है। उस हल्की छड़ पर प्रणोद निकालिए।
(ii) एक कण वेग V से लंबी दूरी तय करने के लिए चलना शुरू करता है। एक तारे के केंद्र से कण के प्रारंभिक पथ की स्पर्श-रेखा पर लंबवत् दूरी p है। दिखाइए कि कण की तारे के केंद्र से न्यूनतम दूरी λ है, जहाँ V2λ = √(μ2 + p2V4) − μ. यहाँ μ एक अचर है।Worked solution
(i)
Let each rod have length 2a and weight W, the square hang from A with ∠BAD = 2θ and BD horizontal, and weights W hang at B, C and D.
Depths below A: centres of AB, AD at a cos θ and of BC, CD at 3a cos θ; B and D at 2a cos θ; C at 4a cos θ.
Sum of (weight × depth) = W(2a + 6a) cos θ + W(4a + 4a) cos θ = 16Wa cos θ, and BD = 4a sin θ.
With a thrust T in BD: δ(16Wa cos θ) + T δ(4a sin θ) = 0, that is −16Wa sin θ + 4aT cos θ = 0, so T = 4W tan θ.
Answer: for the square θ = 45°, so the thrust in the light rod is 4W.
(ii)
Initially the particle moves along a line at perpendicular distance p from the centre with speed V, so h = pV.
Under the attraction μ/r2, energy gives v2 = V2 + 2μ/r (the potential energy −μ/r vanishes at a great distance).
At the least distance λ the particle is at an apse, so v = h/λ = pV/λ.
Hence p2V2/λ2 = V2 + 2μ/λ, that is V2λ2 + 2μλ − p2V2 = 0.
Its positive root is λ = [−μ + √(μ2 + p2V4)]/V2.
Answer: V2λ = √(μ2 + p2V4) − μ.Final answer: (i) for the square θ = 45°, so the thrust in the light rod is 4W; (ii) V2λ = √(μ2 + p2V4) − μ
Question 8 (50 marks)
Q8(a) · Ordinary Differential Equations · 20 marks≈ 14 min
Second Order Equations with Variable Coefficients
(i) Solve the following differential equation : (x + 1)2y″ − 4(x + 1)y′ + 6y = 6(x + 1)2 + sin log(x + 1)
(ii) Find the general and singular solutions of the differential equation 9p2(2 − y)2 = 4(3 − y), where p = dy/dx.हिंदी में प्रश्न
(i) निम्न अवकल समीकरण हल कीजिए : (x + 1)2y″ − 4(x + 1)y′ + 6y = 6(x + 1)2 + sin log(x + 1)
(ii) अवकल समीकरण 9p2(2 − y)2 = 4(3 − y) के व्यापक व विचित्र हल निकालिए, जहाँ p = dy/dx.Worked solution
(i)
Put x + 1 = ez: [θ(θ − 1) − 4θ + 6]y = (θ − 2)(θ − 3)y = 6e2z + sin z; CF = c1(x + 1)2 + c2(x + 1)3.
θ = 2 is a simple root of f(θ) = θ2 − 5θ + 6, with f′(2) = −1, so PI1 = 6ze2z/(−1) = −6ze2z.
PI2 = [1/(5 − 5θ)] sin z = (1/5)(1 + θ) sin z/(1 − θ2) = (sin z + cos z)/10.
Answer: y = c1(x + 1)2 + c2(x + 1)3 − 6(x + 1)2 log(x + 1) + [sin log(x + 1) + cos log(x + 1)]/10.
(ii)
p = ±2√(3 − y)/(3(2 − y)), so (2 − y) dy/√(3 − y) = ±(2/3) dx. With u = 3 − y: ∫(u − 1)u−1/2 du = (2/3)u3/2 − 2u1/2.
This gives √u(u − 3) = ∓(x − c), that is −y√(3 − y) = ∓(x − c); squaring, the general solution is (x − c)2 = y2(3 − y).
p-discriminant (no p term): (2 − y)2(3 − y) = 0. c-discriminant: x − c = 0, so y2(3 − y) = 0.
3 − y occurs once in both and y = 3 (p = 0) satisfies the equation: it is the singular solution. y = 2 (squared in the p-discriminant only) is a tac-locus and y = 0 (squared in the c-discriminant only) is a nodal locus; neither satisfies the equation.
Answer: general solution (x − c)2 = y2(3 − y); singular solution y = 3.Final answer: (i) y = c1(x + 1)2 + c2(x + 1)3 − 6(x + 1)2 log(x + 1) + [sin log(x + 1) + cos log(x + 1)]/10; (ii) general solution (x − c)2 = y2(3 − y); singular solution y = 3
Q8(b) · Vector Analysis · 15 marks≈ 11 min
Gauss and Stokes' Theorems and Green's Identities
Evaluate the surface integral ∬S ∇ × F · n̂ dS for F = yî + (x − 2xz)ĵ − xyk̂ and S is the surface of the sphere x2 + y2 + z2 = a2 above the xy-plane.
हिंदी में प्रश्न
पृष्ठ समाकल ∬S ∇ × F · n̂ dS का मान निकालिए, जहाँ F = yî + (x − 2xz)ĵ − xyk̂ तथा S गोले x2 + y2 + z2 = a2 की सतह है, जो xy-तल के ऊपर है।
Worked solution
By Stokes' theorem the integral equals ∮F·dr round x = a cos t, y = a sin t, z = 0, anticlockwise seen from above (outward normal on the upper hemisphere).
On this circle z = 0, so F·dr = y dx + x dy = d(xy), an exact differential, and ∮ = 0.
Direct check: ∇ × F = (x, y, −2z) and n = r/a, so ∬ = ∬(x2 + y2 − 2z2)/a dS = 2πa3∫0π/2(sin2θ − 2 cos2θ) sin θ dθ = 2πa3(2/3 − 2/3) = 0.
Answer: 0.Final answer: 0.
Q8(c) · Mechanics and Fluid Dynamics · 15 marks≈ 11 min
Moment of Inertia and Motion of Rigid Bodies in Two Dimensions
A four-wheeled railway truck has a total mass M, the mass and radius of gyration of each pair of wheels and axle are m and k respectively, and the radius of each wheel is r. Prove that if the truck is propelled along a level track by a force P, the acceleration is P/(M + 2mk2/r2), and find the horizontal force exerted on each axle by the truck. The axle friction and wind resistance are to be neglected.
हिंदी में प्रश्न
एक चार पहियों वाला रेलवे ट्रक, जिसका कुल द्रव्यमान M है, के पहिए और धुरी के हर युग्म का द्रव्यमान व परिभ्रमण त्रिज्या क्रमशः m तथा k है। हर पहिए की त्रिज्या r है। यदि ट्रक को बल P द्वारा सीधे पथ (ट्रैक) पर धकेला जाता है, तब सिद्ध कीजिए कि उसका त्वरण P/(M + 2mk2/r2) है तथा ट्रक द्वारा प्रत्येक धुरी पर लगाए गए क्षैतिज बल का मान ज्ञात कीजिए। धुरी घर्षण व हवा का प्रतिरोध नगण्य है।
Worked solution
Let f be the acceleration and F the backward friction of the rails on each pair of wheels; each pair rolls, so its angular acceleration is f/r and mk2(f/r) = Fr, F = mk2f/r2.
Whole truck (axle reactions are internal): Mf = P − 2F = P − 2mk2f/r2, so f = P/(M + 2mk2/r2).
One axle: if R is the forward horizontal force of the truck on it, mf = R − F, so R = m(1 + k2/r2)f.
Answer: f = P/(M + 2mk2/r2); the force on each axle is R = m(r2 + k2)P/(Mr2 + 2mk2).Final answer: f = P/(M + 2mk2/r2); the force on each axle is R = m(r2 + k2)P/(Mr2 + 2mk2).
UPSC Mains 2020 Mathematics Paper II
Question 1 (50 marks)
Q1(a) · Abstract Algebra · 10 marks≈ 7 min
Normal Subgroups, Quotient Groups, Homomorphisms and Isomorphism Theorems
Let S3 and Z3 be permutation group on 3 symbols and group of residue classes module 3 respectively. Show that there is no homomorphism of S3 in Z3 except the trivial homomorphism.
हिंदी में प्रश्न
मान लीजिए कि S3 व Z3 क्रमशः 3 प्रतीकों का क्रमचय समूह एवं मॉड्यूल 3 अवशिष्ट वर्गों के समूह हैं। दर्शाइए कि S3 का Z3 में तुच्छ समाकारिता के अतिरिक्त कोई भी समाकारिता नहीं है।
Worked solution
Let f : S3 → ℤ3 be a homomorphism. A transposition τ has order 2, so o(f(τ)) divides 2; it also divides o(ℤ3) = 3. Hence o(f(τ)) = 1, i.e. f(τ) = 0.
S3 is generated by its transpositions (for example (1 2 3) = (1 3)(1 2)), so f(σ) = 0 for every σ ∈ S3.
Alternatively: a non-trivial f would have image of order 3, so ker f would be a normal subgroup of order 6/3 = 2. But no subgroup of order 2 is normal; e.g. (1 3)(1 2)(1 3) = (2 3) ∉ {e, (1 2)}.
Answer: the only homomorphism S3 → ℤ3 is the trivial one.Final answer: the only homomorphism S3 → ℤ3 is the trivial one.
Q1(b) · Abstract Algebra · 10 marks≈ 7 min
Integral Domains, PID, Euclidean Domains, UFD, Fields and Quotient Fields
Let R be a principal ideal domain. Show that every ideal of a quotient ring of R is principal ideal and R/P is a principal ideal domain for a prime ideal P of R.
हिंदी में प्रश्न
मान लीजिए R मुख्य गुणजावली प्रान्त है। दर्शाइए कि R के विभाग-वलय की प्रत्येक गुणजावली, मुख्य गुणजावली है तथा R/P, R के अभाज्य गुणजावली P के लिए मुख्य गुणजावली प्रान्त है।
Worked solution
Let I ≠ R be an ideal of R, π : R → R/I the natural map, and A an ideal of R/I. Then J = π−1(A) = {a ∈ R : a + I ∈ A} is an ideal of R containing I, and π(J) = A because π is onto.
R is a PID, so J = ⟨a⟩. Then every element of A is ra + I = (r + I)(a + I), so A = ⟨a + I⟩ is principal.
Now let P be a prime ideal. R/P is commutative with unity 1 + P ≠ P (as P ≠ R), and (a + P)(b + P) = P ⇒ ab ∈ P ⇒ a ∈ P or b ∈ P, so R/P has no zero divisors.
So R/P is an integral domain in which, by the first part, every ideal is principal. (If P ≠ {0}, P is even maximal and R/P is a field.)
Answer: every ideal of R/I is principal, and R/P is a principal ideal domain for every prime ideal P.Final answer: every ideal of R/I is principal, and R/P is a principal ideal domain for every prime ideal P.
Q1(c) · Real Analysis · 10 marks≈ 7 min
Real Number System, Sequences, Cauchy Sequences and Completeness
Prove that the sequence (an) satisfying the condition |an+1 − an| ≤ α|an − an−1|, 0 < α < 1 for all natural numbers n ≥ 2, is a Cauchy sequence.
हिंदी में प्रश्न
सिद्ध कीजिए कि शर्त |an+1 − an| ≤ α|an − an−1|, जहाँ पर 0 < α < 1 को सभी प्राकृतिक संख्याओं n ≥ 2 के लिए सन्तुष्ट करने वाला अनुक्रम (an), कॉशी-अनुक्रम होता है।
Q1(d) · Complex Analysis · 10 marks≈ 7 min
Cauchy's Theorem and Cauchy's Integral Formula
Evaluate the integral ∫C (z2 + 3z) dz counterclockwise from (2, 0) to (0, 2) along the curve C, where C is the circle |z| = 2.
हिंदी में प्रश्न
समाकल ∫C (z2 + 3z) dz का, (2, 0) से (0, 2) तक वक्र C के वामावर्त अनुगत जहाँ पर C वृत्त |z| = 2 है, मान निकालिए।
Worked solution
z2 + 3z is entire with primitive F(z) = z3/3 + 3z2/2, so the integral does not depend on the path and equals F(2i) − F(2).
F(2i) = (2i)3/3 + 3(2i)2/2 = −8i/3 − 6 and F(2) = 8/3 + 6 = 26/3.
Parametrising z = 2eiθ, 0 ≤ θ ≤ π/2, gives the same value, as Cauchy's theorem requires.
Answer: ∫C (z2 + 3z) dz = −44/3 − 8i/3.Final answer: ∫C (z2 + 3z) dz = −44/3 − 8i/3.
Q1(e) · Linear Programming · 10 marks≈ 7 min
LPP Formulation, Basic Solutions and the Graphical Method
UPSC maintenance section has purchased sufficient number of curtain cloth pieces to meet the curtain requirement of its building. The length of each piece is 17 feet. The requirement according to curtain length is as follows :
Curtain length (in feet) : 5, 9, 7
Number required : 700, 400, 300
The width of all curtains is same as that of available pieces. Form a linear programming problem in standard form that decides the number of pieces cut in different ways so that the total trim loss is minimum. Also give a basic feasible solution to it.हिंदी में प्रश्न
यू.पी.एस.सी. के रखरखाव विभाग ने भवन में पर्दों की आवश्यकता-पूर्ति हेतु पर्दा-कपडे के पर्याप्त संख्या में टुकड़े खरीदे हैं। प्रत्येक टुकड़े की लम्बाई 17 फुट है। पर्दों की लम्बाई के अनुसार आवश्यकता निम्नलिखित है :
पर्दे की लम्बाई (फुटों में) : 5, 9, 7
आवश्यक संख्या : 700, 400, 300
टुकड़ों एवं सभी पर्दों की चौड़ाइयाँ समान हैं। विभिन्न रूप से काटे गये टुकड़ों की संख्या का निर्णय इस प्रकार करने हेतु कि कुल कटान-हानि न्यूनतम हो, एक रैखिक प्रोग्रामन समस्या का प्रामाणिक रूप में निर्धारण कीजिए। इसका एक आधारी सुसंगत हल भी दीजिए।Worked solution
Patterns for a 17 ft piece (pieces of 5, 9, 7 ft; leftover under 5 ft): P1 = (3, 0, 0), trim 2; P2 = (2, 0, 1), trim 0; P3 = (1, 1, 0), trim 3; P4 = (0, 1, 1), trim 1; P5 = (0, 0, 2), trim 3.
Let xj be the number of pieces cut by pattern Pj. Minimise Z = 2x1 + 0x2 + 3x3 + x4 + 3x5.
Subject to 3x1 + 2x2 + x3 = 700 (5 ft), x3 + x4 = 400 (9 ft), x2 + x4 + 2x5 = 300 (7 ft), xj ≥ 0 (integers). Equations stop surplus curtains being counted as useful, and this is the standard form (equality constraints, b ≥ 0, x ≥ 0).
A BFS: take the basis (x1, x3, x5) with x2 = x4 = 0. Then x3 = 400, 3x1 = 300 so x1 = 100, and 2x5 = 300 so x5 = 150; the columns (3, 0, 0), (1, 1, 0), (0, 0, 2) are independent and all values are ≥ 0.
This BFS has trim 200 + 1200 + 450 = 1850 ft. (Checking all bases, the minimum trim is 1000 ft, for example x1 = 200, x3 = 100, x4 = 300, using 600 pieces.)
Answer: Min Z = 2x1 + 3x3 + x4 + 3x5 subject to the three equations above, x ≥ 0; a BFS is (x1, ..., x5) = (100, 0, 400, 0, 150).Final answer: Min Z = 2x1 + 3x3 + x4 + 3x5 subject to the three equations above, x ≥ 0; a BFS is (x1, ..., x5) = (100, 0, 400, 0, 150).
Question 2 (50 marks)
Q2(a) · Abstract Algebra · 15 marks≈ 11 min
Groups, Subgroups, Cyclic Groups, Cosets and Lagrange's Theorem
Let G be a finite cyclic group of order n. Then prove that G has φ(n) generators (where φ is Euler’s φ-function).
हिंदी में प्रश्न
मान लीजिए G, n समूहांक का परिमित चक्रीय समूह है। तब सिद्ध कीजिए कि G के φ(n) जनक हैं (जहाँ पर φ ऑयलर φ-फलन है)।
Q2(b) · Real Analysis · 15 marks≈ 11 min
Continuity, Uniform Continuity and Continuous Functions on Compact Sets
Prove that the function f(x) = sin x2 is not uniformly continuous on the interval [0, ∞[.
हिंदी में प्रश्न
सिद्ध कीजिए कि फलन f(x) = sin x2 अंतराल [0, ∞[ पर एकसमान संतत नहीं है।
Q2(c) · Complex Analysis · 20 marks≈ 14 min
Contour Integration: Evaluation of Real Integrals
Using contour integration, evaluate the integral ∫02π 1/(3 + 2 sin θ) dθ.
हिंदी में प्रश्न
कन्टूर समाकलन का उपयोग कर, समाकल ∫02π 1/(3 + 2 sin θ) dθ का मान ज्ञात कीजिए।
Question 3 (50 marks)
Q3(a) · Abstract Algebra · 15 marks≈ 11 min
Integral Domains, PID, Euclidean Domains, UFD, Fields and Quotient Fields
Let R be a finite field of characteristic p (> 0). Show that the mapping f : R → R defined by f(a) = ap, ∀a ∈ R is an isomorphism.
हिंदी में प्रश्न
मान लीजिए R, p (> 0) अभिलक्षण का एक परिमित क्षेत्र है। दर्शाइए कि f(a) = ap, ∀a ∈ R द्वारा परिभाषित प्रतिचित्रण f : R → R एकैक समाकारी है।
Worked solution
f(ab) = (ab)p = apbp = f(a)f(b), since R is commutative.
(a + b)p = ∑k=0p C(p, k)akbp−k. For 0 < k < p, C(p, k) = p!/(k!(p − k)!) is divisible by p (p divides the numerator but not k!(p − k)!), so these terms vanish in characteristic p, and f(a + b) = ap + bp = f(a) + f(b).
One-one: f(a) = 0 ⇒ ap = 0 ⇒ a = 0, because a field has no zero divisors; so ker f = {0}. Also f(1) = 1.
Onto: R is finite and f : R → R is one-one, so f is onto.
Answer: f(a) = ap is a ring isomorphism of R onto itself (the Frobenius automorphism).Final answer: f(a) = ap is a ring isomorphism of R onto itself (the Frobenius automorphism).
Q3(b) · Linear Programming · 15 marks≈ 11 min
The Simplex Method
Solve the linear programming problem using simplex method :
Minimize z = −6x1 − 2x2 − 5x3
subject to 2x1 − 3x2 + x3 ≤ 14
−4x1 + 4x2 + 10x3 ≤ 46
2x1 + 2x2 − 4x3 ≤ 37
x1 ≥ 2, x2 ≥ 1, x3 ≥ 3हिंदी में प्रश्न
एकधा विधि के द्वारा निम्नलिखित रैखिक प्रोग्रामन समस्या को हल कीजिए :
न्यूनतमीकरण कीजिए z = −6x1 − 2x2 − 5x3
बशर्ते कि 2x1 − 3x2 + x3 ≤ 14
−4x1 + 4x2 + 10x3 ≤ 46
2x1 + 2x2 − 4x3 ≤ 37
x1 ≥ 2, x2 ≥ 1, x3 ≥ 3Worked solution
Put x1 = y1 + 2, x2 = y2 + 1, x3 = y3 + 3 with yj ≥ 0. The constraints become 2y1 − 3y2 + y3 ≤ 10, −4y1 + 4y2 + 10y3 ≤ 20, 2y1 + 2y2 − 4y3 ≤ 43, and z = −6y1 − 2y2 − 5y3 − 29.
Minimising z is maximising Z′ = 6y1 + 2y2 + 5y3. Add slacks s1, s2, s3; columns y1, y2, y3, s1, s2, s3 | xB.
Table 1: s1: 2, −3, 1, 1, 0, 0 | 10; s2: −4, 4, 10, 0, 1, 0 | 20; s3: 2, 2, −4, 0, 0, 1 | 43; Δj: −6, −2, −5, 0, 0, 0 | Z′ = 0. y1 enters; ratios 5, −, 43/2; s1 leaves.
Table 2: y1: 1, −3/2, 1/2, 1/2, 0, 0 | 5; s2: 0, −2, 12, 2, 1, 0 | 40; s3: 0, 5, −5, −1, 0, 1 | 33; Δj: 0, −11, −2, 3, 0, 0 | Z′ = 30. y2 enters; the only ratio is 33/5; s3 leaves.
Table 3: y1: 1, 0, −1, 1/5, 0, 3/10 | 149/10; s2: 0, 0, 10, 8/5, 1, 2/5 | 266/5; y2: 0, 1, −1, −1/5, 0, 1/5 | 33/5; Δj: 0, 0, −13, 4/5, 0, 11/5 | Z′ = 513/5. y3 enters; the only ratio is (266/5)/10 = 133/25; s2 leaves.
Table 4: y1: 1, 0, 0, 9/25, 1/10, 17/50 | 1011/50; y3: 0, 0, 1, 4/25, 1/10, 1/25 | 133/25; y2: 0, 1, 0, −1/25, 1/10, 6/25 | 298/25; Δj: 0, 0, 0, 72/25, 13/10, 68/25 | Z′ = 4294/25. All Δj ≥ 0: optimal.
Back-substituting: x1 = 1011/50 + 2 = 1111/50, x2 = 298/25 + 1 = 323/25, x3 = 133/25 + 3 = 208/25, and z = −4294/25 − 29 = −5019/25. Check: the three original constraints hold with equality (14, 46, 37).
Answer: x1 = 1111/50 = 22.22, x2 = 323/25 = 12.92, x3 = 208/25 = 8.32, Min z = −5019/25 = −200.76.Final answer: x1 = 1111/50 = 22.22, x2 = 323/25 = 12.92, x3 = 208/25 = 8.32, Min z = −5019/25 = −200.76.
Q3(c) · Real Analysis · 20 marks≈ 14 min
Functions of Several Variables: Partial Derivatives, Maxima and Minima
If u = tan−1((x3 + y3)/(x − y)), x ≠ y, then show that x2 ∂2u/∂x2 + 2xy ∂2u/∂x∂y + y2 ∂2u/∂y2 = (1 − 4 sin2u) sin 2u
हिंदी में प्रश्न
यदि u = tan−1((x3 + y3)/(x − y)), x ≠ y, तब दर्शाइए कि x2 ∂2u/∂x2 + 2xy ∂2u/∂x∂y + y2 ∂2u/∂y2 = (1 − 4 sin2u) sin 2u
Worked solution
Let z = tan u = (x3 + y3)/(x − y), which is homogeneous of degree 3 − 1 = 2.
By Euler, xzx + yzy = 2z, i.e. sec2u(xux + yuy) = 2 tan u, so xux + yuy = 2 sin u cos u = sin 2u = g(u).
Differentiate xux + yuy = g(u) with respect to x and to y: ux + xuxx + yuxy = g′(u)ux and xuxy + uy + yuyy = g′(u)uy.
Multiply by x and y and add: x2uxx + 2xyuxy + y2uyy = (g′(u) − 1)(xux + yuy) = g(u)(g′(u) − 1).
With g(u) = sin 2u, g′(u) = 2 cos 2u, so the right side is sin 2u(2 cos 2u − 1) = sin 2u(2(1 − 2 sin2u) − 1) = (1 − 4 sin2u) sin 2u.
Answer: x2uxx + 2xyuxy + y2uyy = (1 − 4 sin2u) sin 2u, as required.Final answer: x2uxx + 2xyuxy + y2uyy = (1 − 4 sin2u) sin 2u, as required.
Question 4 (50 marks)
Q4(a) · Complex Analysis · 15 marks≈ 11 min
Analytic Functions and Cauchy-Riemann Equations
If v(r, θ) = (r − 1/r) sin θ, r ≠ 0, then find an analytic function f(z) = u(r, θ) + iv(r, θ)
हिंदी में प्रश्न
यदि v(r, θ) = (r − 1/r) sin θ, r ≠ 0, तब विश्लेषिक फलन f(z) = u(r, θ) + iv(r, θ) ज्ञात कीजिए।
Worked solution
Polar C-R: ur = (1/r)vθ = (1/r)(r − 1/r)cos θ = (1 − 1/r2)cos θ, and uθ = −rvr = −r(1 + 1/r2)sin θ = −(r + 1/r)sin θ.
Integrating ur in r: u = (r + 1/r)cos θ + g(θ); then uθ = −(r + 1/r)sin θ + g′(θ), so g′ = 0 and g = C.
So f = (r + 1/r)cos θ + C + i(r − 1/r)sin θ = r(cos θ + i sin θ) + (1/r)(cos θ − i sin θ) + C = z + 1/z + C.
Answer: f(z) = z + 1/z + C (C real), analytic for z ≠ 0, with u = (r + 1/r)cos θ + C.Final answer: f(z) = z + 1/z + C (C real), analytic for z ≠ 0, with u = (r + 1/r)cos θ + C.
Q4(b) · Real Analysis · 15 marks≈ 11 min
Riemann Integral, Improper Integrals and Fundamental Theorems of Integral Calculus
Show that ∫0π/2 sin2x/(sin x + cos x) dx = (1/√2) loge(1 + √2)
हिंदी में प्रश्न
दर्शाइए कि ∫0π/2 sin2x/(sin x + cos x) dx = (1/√2) loge(1 + √2)
Worked solution
Let I be the integral. The substitution x → π/2 − x gives I = ∫0π/2 cos2x/(cos x + sin x) dx.
Adding, 2I = ∫0π/2 dx/(sin x + cos x) = (1/√2)∫0π/2 dx/sin(x + π/4).
Using ∫cosec θ dθ = log tan(θ/2), 2I = (1/√2)[log tan(3π/8) − log tan(π/8)].
tan(3π/8) = √2 + 1 and tan(π/8) = √2 − 1, so 2I = (1/√2) log((√2 + 1)/(√2 − 1)) = (1/√2) log(√2 + 1)2 = √2 log(1 + √2).
Answer: I = (1/√2) loge(1 + √2) ≈ 0.6232, as required.Final answer: I = (1/√2) loge(1 + √2) ≈ 0.6232, as required.
Q4(c) · Linear Programming · 20 marks≈ 14 min
Transportation Problem
Find the initial basic feasible solution of the following transportation problem by Vogel’s approximation method and use it to find the optimal solution and the transportation cost of the problem.
[Table: cost matrix with rows Sources S1, S2, S3 and columns Destinations D1, D2, D3, D4]
S1 : 10, 0, 20, 11; Availability 15
S2 : 12, 8, 9, 20; Availability 25
S3 : 0, 14, 16, 18; Availability 10
Demand : 5, 20, 15, 10हिंदी में प्रश्न
वोगेल की सन्निकटन विधि से निम्नलिखित परिवहन समस्या का आरंभिक आधारिक सुसंगत हल ज्ञात कीजिए। इस हल का उपयोग कर समस्या का इष्टतम हल एवं परिवहन लागत ज्ञात कीजिए।
[सारणी: उद्गम S1, S2, S3 तथा गन्तव्य D1, D2, D3, D4 के लिए लागत आव्यूह]
S1 : 10, 0, 20, 11; प्राप्यता 15
S2 : 12, 8, 9, 20; प्राप्यता 25
S3 : 0, 14, 16, 18; प्राप्यता 10
माँग : 5, 20, 15, 10
Question 5 (50 marks)
Q5(a) · Partial Differential Equations · 10 marks≈ 7 min
Families of Surfaces and Formation of PDEs
Form a partial differential equation by eliminating the arbitrary functions f(x) and g(y) from z = y f(x) + x g(y) and specify its nature (elliptic, hyperbolic or parabolic) in the region x > 0, y > 0.
हिंदी में प्रश्न
z = y f(x) + x g(y) से स्वैच्छिक फलनों f(x) व g(y) का विलोपन कर आंशिक अवकल समीकरण बनाइए तथा इसकी प्रकृति (दीर्घवृत्तीय, अतिपरवलीय या परवलीय) x > 0, y > 0 क्षेत्र में इंगित कीजिए।
Worked solution
p = yf′(x) + g(y), q = f(x) + xg′(y) and s = ∂2z/∂x∂y = f′(x) + g′(y).
Then xp + yq = xy(f′ + g′) + (xg + yf) = xys + z, which contains no arbitrary function.
Type: A = 0, B = xy, C = 0 (coefficients of r, s, t), so B2 − 4AC = x2y2 > 0 in x > 0, y > 0.
Answer: xy ∂2z/∂x∂y − x ∂z/∂x − y ∂z/∂y + z = 0, hyperbolic in the region x > 0, y > 0.Final answer: xy ∂2z/∂x∂y − x ∂z/∂x − y ∂z/∂y + z = 0, hyperbolic in the region x > 0, y > 0.
Q5(b) · Numerical Analysis and Computer Programming · 10 marks≈ 7 min
Roots of Algebraic and Transcendental Equations
Show that the equation : f(x) = cos(π(x + 1)/8) + 0·148x − 0·9062 = 0 has one root in the interval (−1, 0) and one in (0, 1). Calculate the negative root correct to four decimal places using Newton-Raphson method.
हिंदी में प्रश्न
दर्शाइए कि समीकरण : f(x) = cos(π(x + 1)/8) + 0·148x − 0·9062 = 0 का एक मूल अन्तराल (−1, 0) में तथा एक मूल (0, 1) में है। ऋणात्मक मूल की न्यूटन-रॉफसन विधि से दशमलव के चार स्थान तक सही गणना कीजिए।
Q5(c) · Numerical Analysis and Computer Programming · 10 marks≈ 7 min
Computer Systems, Logic Gates, Boolean Algebra and Algorithms
Let g(w, x, y, z) = (w + x + y)(x + ȳ + z)(w + ȳ) be a Boolean function. Obtain the conjunctive normal form for g(w, x, y, z). Also express g(w, x, y, z) as a product of maxterms.
हिंदी में प्रश्न
मान लीजिए g(w, x, y, z) = (w + x + y)(x + ȳ + z)(w + ȳ) एक बूलीय-फलन है। g(w, x, y, z) का योगात्मक प्रसामान्य स्वरूप (कन्जंकटिव नार्मल फॉर्म) प्राप्त कीजिए। g(w, x, y, z) को उच्च-पदों (मैक्स टर्म्स) के गुणन के रूप में भी व्यक्त कीजिए।
Worked solution
g is already a product of sums, so its CNF is g = (w + x + y)(x + y′ + z)(w + y′).
Expand each factor with X = (X + v)(X + v′): w + x + y = (w + x + y + z)(w + x + y + z′); x + y′ + z = (w + x + y′ + z)(w′ + x + y′ + z); w + y′ = (w + x + y′ + z)(w + x + y′ + z′)(w + x′ + y′ + z)(w + x′ + y′ + z′).
Deleting the repeated factor w + x + y′ + z and numbering rows wxyz from 0 to 15, the maxterms are those of rows 0000, 0001, 0010, 0011, 0110, 0111, 1010.
Answer: g = ∏M(0, 1, 2, 3, 6, 7, 10) = (w + x + y + z)(w + x + y + z′)(w + x + y′ + z)(w + x + y′ + z′)(w + x′ + y′ + z)(w + x′ + y′ + z′)(w′ + x + y′ + z).Final answer: g = ∏M(0, 1, 2, 3, 6, 7, 10) = (w + x + y + z)(w + x + y + z′)(w + x + y′ + z)(w + x + y′ + z′)(w + x′ + y′ + z)(w + x′ + y′ + z′)(w′ + x + y′ + z).
Q5(d) · Partial Differential Equations · 10 marks≈ 7 min
Linear Second-Order PDEs with Constant Coefficients
Solve the partial differential equation : (D3 − 2D2D′ − DD′2 + 2D′3)z = e2x+y + sin(x − 2y); D ≡ ∂/∂x, D′ ≡ ∂/∂y
हिंदी में प्रश्न
आंशिक अवकल समीकरण : (D3 − 2D2D′ − DD′2 + 2D′3)z = e2x+y + sin(x − 2y); D ≡ ∂/∂x, D′ ≡ ∂/∂y को हल कीजिए।
Worked solution
F = (D − D′)(D + D′)(D − 2D′) as in UPSC 2016, so CF = φ1(y + x) + φ2(y − x) + φ3(y + 2x).
e2x + y: F(2, 1) = 8 − 8 − 2 + 2 = 0 because of D − 2D′; [(D − D′)(D + D′)]−1e2x + y = e2x + y/3, and then (D − 2D′)−1 gives PI1 = (x/3)e2x + y.
sin(x − 2y): a = 1, b = −2 give D3 → −D, D2D′ → −D′, DD′2 → −4D, D′3 → −4D′, so F → −D + 2D′ + 4D − 8D′ = 3(D − 2D′).
(D − 2D′)cos(x − 2y) = −sin(x − 2y) − 4 sin(x − 2y) = −5 sin(x − 2y), so PI2 = (1/3)(−1/5)cos(x − 2y) = −(1/15)cos(x − 2y).
Answer: z = φ1(y + x) + φ2(y − x) + φ3(y + 2x) + (x/3)e2x + y − (1/15)cos(x − 2y).Final answer: z = φ1(y + x) + φ2(y − x) + φ3(y + 2x) + (x/3)e2x + y − (1/15)cos(x − 2y).
Q5(e) · Mechanics and Fluid Dynamics · 10 marks≈ 7 min
Moment of Inertia and Motion of Rigid Bodies in Two Dimensions
Prove that the moment of inertia of a triangular lamina ABC about any axis through A in its plane is (M/6)(β2 + βγ + γ2) where M is the mass of the lamina and β, γ are respectively the length of perpendiculars from B and C on the axis.
हिंदी में प्रश्न
सिद्ध कीजिए कि त्रिभुजाकार पटल ABC का इसके तल में A से होकर जाने वाली किसी भी अक्ष के सापेक्ष जड़त्व-आघूर्ण (M/6)(β2 + βγ + γ2) है, जहाँ पर M पटल की संहति तथा β व γ क्रमशः B व C से अक्ष पर डाले गये लम्बों की लम्बाइयां हैं।
Worked solution
Replace the lamina by its equimomental system: M/3 at the mid-points of AB, AC and BC.
With B and C on the same side of the axis, their distances from it are β/2, γ/2 and (β + γ)/2.
I = (M/3)[β2 + γ2 + (β + γ)2]/4 = (M/12)(2β2 + 2βγ + 2γ2).
Answer: I = (M/6)(β2 + βγ + γ2) (if B and C lie on opposite sides, γ is taken negative).Final answer: I = (M/6)(β2 + βγ + γ2) (if B and C lie on opposite sides, γ is taken negative).
Question 6 (50 marks)
Q6(a) · Partial Differential Equations · 15 marks≈ 11 min
Lagrange's Method for Quasilinear First-Order PDEs
Find the integral surface of the partial differential equation : (x − y)y2 ∂z/∂x + (y − x)x2 ∂z/∂y = (x2 + y2)z that contains the curve : xz = a3, y = 0 on it.
हिंदी में प्रश्न
आंशिक अवकल समीकरण : (x − y)y2 ∂z/∂x + (y − x)x2 ∂z/∂y = (x2 + y2)z के वक्र : xz = a3, y = 0 को अपने ऊपर समाहित करने वाले समाकल पृष्ठ को ज्ञात कीजिए।
Worked solution
Auxiliary equations dx/((x − y)y2) = dy/(−(x − y)x2) = dz/((x2 + y2)z); the first two give x2dx + y2dy = 0, so x3 + y3 = c1.
(dx − dy)/((x − y)(y2 + x2)) = dz/((x2 + y2)z) gives d(x − y)/(x − y) = dz/z, so z/(x − y) = c2.
On y = 0, xz = a3: c1 = x3 and c2 = z/x = a3/x2, so c23c12 = a9.
Answer: z3(x3 + y3)2 = a9(x − y)3 (at y = 0 it reduces to x3z3 = a9, i.e. xz = a3).Final answer: z3(x3 + y3)2 = a9(x − y)3 (at y = 0 it reduces to x3z3 = a9, i.e. xz = a3).
Q6(b) · Numerical Analysis and Computer Programming · 15 marks≈ 11 min
Systems of Linear Equations: Direct and Iterative Methods
For the solution of the system of equations : 4x + y + 2z = 4, 3x + 5y + z = 7, x + y + 3z = 3 set up the Gauss-Seidel iterative scheme and iterate three times starting with the initial vector X(0) = 0. Also find the exact solutions and compare with the iterated solutions.
हिंदी में प्रश्न
समीकरण निकाय : 4x + y + 2z = 4, 3x + 5y + z = 7, x + y + 3z = 3 के हल के लिए गाउस-सीडल पुनरावर्ती क्रिया-विधि निर्धारित कीजिए तथा आरंभिक सदिश X(0) = 0 से प्रारंभ करके तीन बार पुनरावर्त कीजिए। यथातथ (बिल्कुल ठीक) हल भी निकालिए और पुनरावर्त हलों से तुलना कीजिए।
Q6(c) · Mechanics and Fluid Dynamics · 20 marks≈ 14 min
Hamilton's Equations
By writing down the Hamiltonian, find the equations of motion of a particle of mass m constrained to move on the surface of a cylinder defined by x2 + y2 = R2, R is a constant. The particle is subject to a force directed towards the origin and proportional to the distance r of the particle from the origin given by F = −kr, k is a constant.
हिंदी में प्रश्न
एक कण जिसकी संहति m है, x2 + y2 = R2, जहाँ पर R अचर है, द्वारा परिभाषित बेलन पर गति के लिए व्यवरोधित है। कण मूल बिन्दु की ओर लगे बल जो कण की मूल बिन्दु से दूरी r के है, से प्रतिबन्धित है। बल F = −kr, जहाँ पर k अचर है, से दिया गया है।
Worked solution
Use cylindrical coordinates (R, θ, z) with R fixed; F = −kr is conservative, V = (k/2)r2 = (k/2)(R2 + z2).
T = (m/2)(R2θ̇2 + ż2), pθ = mR2θ̇, pz = mż, and H = pθ2/(2mR2) + pz2/(2m) + (k/2)(R2 + z2).
θ̇ = pθ/(mR2), ṗθ = 0, ż = pz/m, ṗz = −kz.
Answer: θ is cyclic, so θ̇ = ω is constant; z̈ + (k/m)z = 0, z = A cos(√(k/m) t + ε): the particle goes round the cylinder uniformly while oscillating along it.Final answer: θ is cyclic, so θ̇ = ω is constant; z̈ + (k/m)z = 0, z = A cos(√(k/m) t + ε): the particle goes round the cylinder uniformly while oscillating along it.
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 solution of the partial differential equation : z = (1/2)(p2 + q2) + (p − x)(q − y); p ≡ ∂z/∂x, q ≡ ∂z/∂y which passes through the x-axis.
हिंदी में प्रश्न
आंशिक अवकल समीकरण : z = (1/2)(p2 + q2) + (p − x)(q − y); p ≡ ∂z/∂x, q ≡ ∂z/∂y का हल ज्ञात कीजिये जो कि x-अक्ष से गुजरता हो।
Worked solution
F = (p2 + q2)/2 + (p − x)(q − y) − z: Fp = p + q − y, Fq = p + q − x, Fx = −(q − y), Fy = −(p − x), Fz = −1.
dx/dt = p + q − y = dp/dt and dy/dt = p + q − x = dq/dt, so p − x and q − y are constant on each strip.
x-axis: x0 = s, y0 = z0 = 0; the strip condition gives p0 = 0 and F = 0 gives q02/2 − sq0 = 0, so q0 = 2s or q0 = 0.
q0 = 2s: p − x = −s, q − y = 2s; dx/dt = x + s, dy/dt = y + s give x + s = 2set, y + s = set, so s = x − 2y, p = 2y, q = 2x − 3y and dz = 2y dx + (2x − 3y)dy, z = 2xy − 3y2/2.
q0 = 0: p − x = −s, q = y and dx/dt = 0, so x = s, p = 0, q = y and z = y2/2.
Check: for z = y(4x − 3y)/2, (4y2 + (2x − 3y)2)/2 + (2y − x)(2x − 4y) = 2xy − 3y2/2 = z; for z = y2/2, y2/2 + (−x)(0) = z.
Answer: z = y(4x − 3y)/2 (the root q0 = 2s); the other root q0 = 0 gives a second surface through the x-axis, 2z = y2.Final answer: z = y(4x − 3y)/2 (the root q0 = 2s); the other root q0 = 0 gives a second surface through the x-axis, 2z = y2.
Q7(b) · Numerical Analysis and Computer Programming · 20 marks≈ 14 min
Numerical Integration and Numerical Solution of ODEs
Find a quadrature formula ∫01 f(x) dx/√(x(1 − x)) = α1f(0) + α2f(1/2) + α3f(1) which is exact for polynomials of highest possible degree. Then use the formula to evaluate ∫01 dx/√(x − x3) (correct up to three decimal places).
हिंदी में प्रश्न
क्षेत्रकलन के लिए ∫01 f(x) dx/√(x(1 − x)) = α1f(0) + α2f(1/2) + α3f(1) द्वारा उस सूत्र को ज्ञात कीजिए जो अधिकतम सम्भव घात के बहुपद के लिए यथातथ (बिल्कुल ठीक) हो। सूत्र का उपयोग ∫01 dx/√(x − x3) का (दशमलव के तीन स्थानों तक सही) मूल्यांकन के लिए कीजिए।
Worked solution
Moments (put x = sin2θ, so dx/√(x(1 − x)) = 2 dθ): ∫w = π, ∫xw = π/2, ∫x2w = 3π/8, ∫x3w = 5π/16, ∫x4w = 35π/128, with w = 1/√(x(1 − x)) on [0, 1].
Exactness for 1, x, x2: α1 + α2 + α3 = π, α2/2 + α3 = π/2, α2/4 + α3 = 3π/8, so α2 = π/2, α3 = π/4, α1 = π/4.
f = x3: rule π/16 + π/4 = 5π/16, exact; f = x4: rule π/32 + π/4 = 9π/32 ≠ 35π/128, so the degree of precision is 3 and the formula is (π/4)[f(0) + 2f(1/2) + f(1)].
Since x − x3 = x(1 − x)(1 + x), ∫01 dx/√(x − x3) uses f(x) = 1/√(1 + x): f(0) = 1, f(1/2) = 0.816497, f(1) = 0.707107.
Answer: (π/4)(1 + 1.632993 + 0.707107) = (π/4)(3.340100) = 2.623 (the exact value is 2.6221).Final answer: (π/4)(1 + 1.632993 + 0.707107) = (π/4)(3.340100) = 2.623 (the exact value is 2.6221).
Q7(c) · Mechanics and Fluid Dynamics · 15 marks≈ 11 min
Kinematics of Fluids: Continuity, Streamlines and Potential Flow
A velocity potential in a two-dimensional fluid flow is given by φ(x, y) = xy + x2 − y2. Find the stream function for this flow.
हिंदी में प्रश्न
एक द्विविमीय द्रव्य-प्रवाह का वेग विभव φ(x, y) = xy + x2 − y2 द्वारा दिया गया है। इस प्रवाह का धारा-फलन ज्ञात कीजिए।
Worked solution
∇2φ = 2 − 2 = 0, so a stream function exists; ∂ψ/∂y = ∂φ/∂x = y + 2x and ∂ψ/∂x = −∂φ/∂y = 2y − x.
The first gives ψ = y2/2 + 2xy + g(x); the second then gives 2y + g′(x) = 2y − x, so g = −x2/2.
Answer: ψ = 2xy + (y2 − x2)/2 + constant; indeed w = φ + iψ = (1 − i/2)z2.Final answer: ψ = 2xy + (y2 − x2)/2 + constant; indeed w = φ + iψ = (1 − i/2)z2.
Question 8 (50 marks)
Q8(a) · Partial Differential Equations · 20 marks≈ 14 min
Vibrating String, Heat Equation and Laplace Equation
One end of a tightly stretched flexible thin string of length l is fixed at the origin and the other at x = l. It is plucked at x = l/3 so that it assumes initially the shape of a triangle of height h in the x-y plane. Find the displacement y at any distance x and at any time t after the string is released from rest. Take, (horizontal tension)/(mass per unit length) = c2.
हिंदी में प्रश्न
लम्बाई l की कसकर खींची गई लचीली-पतली डोरी का एक सिरा मूल बिन्दु पर तथा दूसरा x = l पर बंधा है। आरंभिक अवस्था में इसे x = l/3 बिन्दु से ऐसे खींचकर छोड़ा जाता है ताकि यह x-y तल में h ऊँचाई के त्रिभुज का आकार लेता है। किसी भी दूरी x तथा समय t, डोरी को विरामावस्था से छोड़ने के बाद, पर विस्थापन y को ज्ञात कीजिए। डोरी में (क्षैतिज तनाव)/(डोरी की इकाई लम्बाई की संहति) = c2 लीजिए।
Worked solution
Initial shape y(x, 0) = 3hx/l for 0 ≤ x ≤ l/3 and 3h(l − x)/(2l) for l/3 ≤ x ≤ l; the string starts from rest, so Bn = 0 and y = ∑Ansin(nπx/l)cos(nπct/l).
Integrating by parts on each piece, the terms at x = l/3 cancel and only the jump in slope (3h/l + 3h/(2l) = 9h/(2l)) survives: An = (2/l)(l/(nπ))2(9h/(2l))sin(nπ/3).
So An = (9h/(n2π2))sin(nπ/3), the plucked-string formula with a = l/3.
Answer: y(x, t) = (9h/π2)∑n=1∞ (1/n2)sin(nπ/3)sin(nπx/l)cos(nπct/l); the modes n = 3, 6, 9, ... are absent.Final answer: y(x, t) = (9h/π2)∑n=1∞ (1/n2)sin(nπ/3)sin(nπx/l)cos(nπct/l); the modes n = 3, 6, 9, ... are absent.
Q8(b) · Numerical Analysis and Computer Programming · 15 marks≈ 11 min
Interpolation: Newton's Forward, Backward and Lagrange's Formulae
Write the three point Lagrangian interpolating polynomial relative to the points x0, x0 + ε and x1. Then by taking the limit ε → 0, establish the relation f(x) = ((x1 − x)(x + x1 − 2x0)/(x1 − x0)2) f(x0) + ((x − x0)(x1 − x)/(x1 − x0)) f′(x0) + ((x − x0)2/(x1 − x0)2) f(x1) + E(x) where E(x) = (1/6)(x − x0)2(x − x1)f‴(ξ) is the error function and min.(x0, x0 + ε, x1) < ξ < max.(x0, x0 + ε, x1)
हिंदी में प्रश्न
बिन्दुओं x0, x0 + ε तथा x1 के सापेक्ष तीन-बिन्दु लेगरान्ज-अन्तर्वेशन बहुपद को लिखिए। तदुपरान्त limit ε → 0 करने पर निम्नलिखित सम्बन्ध को स्थापित कीजिए : f(x) = ((x1 − x)(x + x1 − 2x0)/(x1 − x0)2) f(x0) + ((x − x0)(x1 − x)/(x1 − x0)) f′(x0) + ((x − x0)2/(x1 − x0)2) f(x1) + E(x) जहाँ पर E(x) = (1/6)(x − x0)2(x − x1)f‴(ξ) त्रुटि-फलन है और न्यूनतम (x0, x0 + ε, x1) < ξ < उच्चतम (x0, x0 + ε, x1)
Worked solution
Lagrange: Pε(x) = f(x0)(x − x0 − ε)(x − x1)/(−ε(x0 − x1)) + f(x0 + ε)(x − x0)(x − x1)/(ε(x0 + ε − x1)) + f(x1)(x − x0)(x − x0 − ε)/((x1 − x0)(x1 − x0 − ε)).
By uniqueness it equals the Newton form f(x0) + (x − x0)f[x0, x0 + ε] + (x − x0)(x − x0 − ε)f[x0, x0 + ε, x1]; as ε → 0, f[x0, x0 + ε] → f′(x0) and f[x0, x0 + ε, x1] → (f1 − f0 − hf′0)/h2, where h = x1 − x0.
So the limit is P(x) = f0 + (x − x0)f′0 + (x − x0)2(f1 − f0 − hf′0)/h2.
Collecting: coefficient of f0 is 1 − (x − x0)2/h2 = (x1 − x)(x + x1 − 2x0)/h2; of f′0 is (x − x0) − (x − x0)2/h = (x − x0)(x1 − x)/h; of f1 is (x − x0)2/h2.
Error: f(x) − Pε(x) = (x − x0)(x − x0 − ε)(x − x1)f‴(ξε)/3!, and letting ε → 0 (f‴ continuous) gives E(x) = (1/6)(x − x0)2(x − x1)f‴(ξ).
Answer: the stated relation, a Hermite-type quadratic matching f(x0), f′(x0) and f(x1).Final answer: the stated relation, a Hermite-type quadratic matching f(x0), f′(x0) and f(x1).
Q8(c) · Mechanics and Fluid Dynamics · 15 marks≈ 11 min
Two-Dimensional and Axisymmetric Motion: Sources, Sinks and Vortices
Two sources of strength m/2 are placed at the points (±a, 0). Show that at any point on the circle x2 + y2 = a2, the velocity is parallel to the y-axis and is inversely proportional to y.
हिंदी में प्रश्न
m/2 शक्ति वाले दो स्रोत, बिन्दुओं (±a, 0) पर स्थित हैं। दर्शाइए कि वृत्त x2 + y2 = a2 के किसी भी बिन्दु पर वेग y-अक्ष के समान्तर तथा y के व्युत्क्रमानुपाती है।
Worked solution
w = −(m/2)log(z − a) − (m/2)log(z + a) = −(m/2)log(z2 − a2), so dw/dz = −mz/(z2 − a2).
On z = aeiθ: z2 − a2 = a2eiθ(eiθ − e−iθ) = 2ia2eiθsin θ, so dw/dz = −m/(2ia sin θ) = im/(2y); since dw/dz = −u + iv, u = 0 and v = m/(2y).
Answer: on the circle the velocity is parallel to Oy, of magnitude m/(2|y|), inversely proportional to y.Final answer: on the circle the velocity is parallel to Oy, of magnitude m/(2|y|), inversely proportional to y.