UPSC Mains 2015 Mathematics Optional Question Papers with Solutions
All 62 questions of the 2015 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 2015 Mathematics Paper I
Question 1 (50 marks)
Q1(a) · Linear Algebra · 10 marks≈ 7 min
Vector Spaces, Subspaces, Basis and Dimension
The vectors V1 = (1, 1, 2, 4), V2 = (2, −1, −5, 2), V3 = (1, −1, −4, 0) and V4 = (2, 1, 1, 6) are linearly independent. Is it true? Justify your answer.
हिंदी में प्रश्न
दिए गए सदिश V1 = (1, 1, 2, 4), V2 = (2, −1, −5, 2), V3 = (1, −1, −4, 0) तथा V4 = (2, 1, 1, 6) रैखिकत: स्वतंत्र हैं। क्या यह सत्य है? अपने उत्तर के पक्ष में तर्क दीजिये।
Q1(b) · Linear Algebra · 10 marks≈ 7 min
Algebra of Matrices, Echelon Form, Rank, Inverse, Congruence and Similarity
Reduce the following matrix to row echelon form and hence find its rank: [[1, 2, 3, 4], [2, 1, 4, 5], [1, 5, 5, 7], [8, 1, 14, 17]].
हिंदी में प्रश्न
निम्नलिखित आव्यूह को पंक्ति सोपानक रूप में समानीत कीजिये और तत्पश्चात् इसकी कोटि निकालिए: [[1, 2, 3, 4], [2, 1, 4, 5], [1, 5, 5, 7], [8, 1, 14, 17]].
Q1(c) · Calculus · 10 marks≈ 7 min
Mean Value Theorems, Taylor's Theorem and Indeterminate Forms
Evaluate the following limit: limx→a (2 − x/a)tan(πx/2a).
हिंदी में प्रश्न
निम्नलिखित सीमा का मान निकालिए: limx→a (2 − x/a)tan(πx/2a).
Q1(d) · Calculus · 10 marks≈ 7 min
Riemann Integral, Indefinite Integrals and Improper Integrals
Evaluate the following integral: ∫π/6π/3 [∛(sin x)/(∛(sin x) + ∛(cos x))] dx.
हिंदी में प्रश्न
निम्नलिखित समाकल का मान निकालिए: ∫π/6π/3 [∛(sin x)/(∛(sin x) + ∛(cos x))] dx.
Worked solution
Use ∫ab f(x) dx = ∫ab f(a + b − x) dx with a + b = π/2; sin(π/2 − x) = cos x and cos(π/2 − x) = sin x.
So I = ∫π/6π/3 ∛(cos x)/(∛(cos x) + ∛(sin x)) dx. Adding this to the original form, the integrands add to 1, so 2I = π/3 − π/6 = π/6.
Answer: π/12.Final answer: π/12.
Q1(e) · Analytic Geometry · 10 marks≈ 7 min
The Sphere
For what positive value of a, the plane ax − 2y + z + 12 = 0 touches the sphere x2 + y2 + z2 − 2x − 4y + 2z − 3 = 0 and hence find the point of contact.
हिंदी में प्रश्न
'a' के किस धनात्मक मान के लिए, समतल ax − 2y + z + 12 = 0, गोलक x2 + y2 + z2 − 2x − 4y + 2z − 3 = 0 को स्पर्श करता है। स्पर्श बिन्दु को भी ज्ञात कीजिये।
Question 2 (50 marks)
Q2(a) · Linear Algebra · 12 marks≈ 9 min
Eigenvalues, Eigenvectors and the Cayley-Hamilton Theorem
If matrix A = [[1, 0, 0], [1, 0, 1], [0, 1, 0]] then find A30.
हिंदी में प्रश्न
यदि आव्यूह A = [[1, 0, 0], [1, 0, 1], [0, 1, 0]] तब आव्यूह A30 को ज्ञात कीजिये।
Worked solution
This is the matrix of the previous example: its characteristic polynomial is λ3 − λ2 − λ + 1, so by Cayley-Hamilton A3 = A2 + A − I, and by induction An = An−2 + A2 − I for n ≥ 3.
Applying the recurrence 14 times from A30 down to A2: A30 = A2 + 14(A2 − I) = 15A2 − 14I, with A2 = [[1, 0, 0], [1, 1, 0], [1, 0, 1]].
Answer: A30 = [[1, 0, 0], [15, 1, 0], [15, 0, 1]].Final answer: A30 = [[1, 0, 0], [15, 1, 0], [15, 0, 1]].
Q2(b) · Calculus · 13 marks≈ 9 min
Maxima and Minima, Asymptotes and Curve Tracing
A conical tent is of given capacity. For the least amount of Canvas required, for it, find the ratio of its height to the radius of its base.
हिंदी में प्रश्न
एक शंक्वाकार टेंट एक दत्त क्षमता का है। यदि उस टेंट में न्यूनतम कैनवास लगाना हो, तो उसकी ऊँचाई का उसके आधार की त्रिज्या पर अनुपात मालूम कीजिये।
Worked solution
Let r be the base radius, h the height and l = √(r2 + h2) the slant height. The capacity V = πr2h/3 is fixed and the canvas is the curved surface S = πrl.
S is least when S2 = π2r2(r2 + h2) is least. With h = 3V/(πr2): S2 = F(r) = π2r4 + 9V2/r2.
F′(r) = 4π2r3 − 18V2/r3 = 0 gives 2π2r6 = 9V2 = π2r4h2, so h2 = 2r2.
F″(r) = 12π2r2 + 54V2/r4 > 0, so this critical point gives the minimum.
Answer: height : radius = √2 : 1.Final answer: height : radius = √2 : 1.
Q2(c) · Linear Algebra · 12 marks≈ 9 min
Eigenvalues, Eigenvectors and the Cayley-Hamilton Theorem
Find the eigen values and eigen vectors of the matrix: [[1, 1, 3], [1, 5, 1], [3, 1, 1]].
हिंदी में प्रश्न
निम्नलिखित आव्यूह के आइगन मानों एवं आइगन सदिशों को ज्ञात कीजिए: [[1, 1, 3], [1, 5, 1], [3, 1, 1]].
Q2(d) · Analytic Geometry · 13 marks≈ 9 min
Cone and Cylinder
If 6x = 3y = 2z represents one of the three mutually perpendicular generators of the cone 5yz − 8zx − 3xy = 0 then obtain the equations of the other two generators.
हिंदी में प्रश्न
यदि शंकु 5yz − 8zx − 3xy = 0 की तीन परस्पर लम्बवत् जनक रेखाओं में से एक जनक रेखा 6x = 3y = 2z हो, तब अन्य दो जनक रेखाओं के समीकरण मालूम कीजिये।
Question 3 (50 marks)
Q3(a) · Linear Algebra · 12 marks≈ 9 min
Linear Transformations, Rank-Nullity and Matrix Representation
Let V = ℝ3 and T ∈ A(V), for all ai ∈ A(V), be defined by T(a1, a2, a3) = (2a1 + 5a2 + a3, −3a1 + a2 − a3, −a1 + 2a2 + 3a3). What is the matrix T relative to the basis V1 = (1, 0, 1), V2 = (−1, 2, 1), V3 = (3, −1, 1)?
हिंदी में प्रश्न
यदि V = ℝ3 तथा T ∈ A(V) जहाँ सभी ai, A(V) के सदस्य हैं। यदि T(a1, a2, a3) = (2a1 + 5a2 + a3, −3a1 + a2 − a3, −a1 + 2a2 + 3a3) के द्वारा परिभाषित हैं। तब आधार V1 = (1, 0, 1), V2 = (−1, 2, 1), V3 = (3, −1, 1) के सापेक्ष आव्यूह T ज्ञात कीजिये।
Worked solution
Images: T(V1) = T(1, 0, 1) = (3, −4, 2), T(V2) = T(−1, 2, 1) = (9, 4, 8), T(V3) = T(3, −1, 1) = (2, −11, −2).
To write a vector w as xV1 + yV2 + zV3, solve P(x, y, z)T = w with P = [[1, −1, 3], [0, 2, −1], [1, 1, 1]] (columns Vi, det P = −2); P−1 = (1/2)[[−3, −4, 5], [1, 2, −1], [2, 2, −2]].
Then (3, −4, 2) = (17/2)V1 − (7/2)V2 − 3V3, (9, 4, 8) = −(3/2)V1 + (9/2)V2 + 5V3, (2, −11, −2) = 14V1 − 9V2 − 7V3.
Check the first: (17/2)(1, 0, 1) − (7/2)(−1, 2, 1) − 3(3, −1, 1) = (17/2 + 7/2 − 9, −7 + 3, 17/2 − 7/2 − 3) = (3, −4, 2).
Answer: [T]B = [[17/2, −3/2, 14], [−7/2, 9/2, −9], [−3, 5, −7]] (the coordinate vectors are the columns); it equals P−1AP, A being the standard matrix of T.Final answer: [T]B = [[17/2, −3/2, 14], [−7/2, 9/2, −9], [−3, 5, −7]] (the coordinate vectors are the columns); it equals P−1AP, A being the standard matrix of T.
Q3(b) · Calculus · 13 marks≈ 9 min
Functions of Several Variables: Partial Derivatives, Extrema, Lagrange Multipliers and Jacobians
Which point of the sphere x2 + y2 + z2 = 1 is at the maximum distance from the point (2, 1, 3)?
हिंदी में प्रश्न
गोलक x2 + y2 + z2 = 1 पर स्थित बिन्दु निकालिए जो बिन्दु (2, 1, 3) से अधिकतम दूरी पर है।
Worked solution
Maximise D = (x − 2)2 + (y − 1)2 + (z − 3)2 subject to x2 + y2 + z2 = 1, with F = D + λ(x2 + y2 + z2 − 1).
Fx = 2(x − 2) + 2λx = 0 gives x = 2/(1 + λ); similarly y = 1/(1 + λ), z = 3/(1 + λ). So (x, y, z) = t(2, 1, 3) with t = 1/(1 + λ).
The constraint gives 14t2 = 1, t = ±1/√14. The distance is |1 − t|√14: √14 − 1 for t = 1/√14 (nearest) and √14 + 1 for t = −1/√14 (farthest).
Answer: The point (−2/√14, −1/√14, −3/√14), at distance √14 + 1.Final answer: The point (−2/√14, −1/√14, −3/√14), at distance √14 + 1.
Q3(c) · Analytic Geometry · 13 marks≈ 9 min
Straight Lines and Shortest Distance between Skew Lines
(i) Obtain the equation of the plane passing through the points (2, 3, 1) and (4, −5, 3) parallel to x-axis. (6)
(ii) Verify if the lines (x − a + d)/(α − δ) = (y − a)/α = (z − a − d)/(α + δ) and (x − b + c)/(β − γ) = (y − b)/β = (z − b − c)/(β + γ) are coplanar. If yes, then find the equation of the plane in which they lie. (7)हिंदी में प्रश्न
(i) उस समतल का समीकरण निकालिए जो बिन्दुओं (2, 3, 1) एवं (4, −5, 3) से गुजरता है व x-अक्ष के समान्तर है। (6)
(ii) सत्यापित कीजिये कि रेखाएँ (x − a + d)/(α − δ) = (y − a)/α = (z − a − d)/(α + δ) तथा (x − b + c)/(β − γ) = (y − b)/β = (z − b − c)/(β + γ) समतलीय हैं। यदि हाँ, तो उस समतल का समीकरण ज्ञात कीजिए, जिसमें उपरोक्त दोनों रेखाएँ स्थित हैं। (7)Worked solution
(i) The normal is perpendicular to (4 − 2, −5 − 3, 3 − 1) = (2, −8, 2) and to the x-axis (1, 0, 0): (2, −8, 2) × (1, 0, 0) = (0, 2, 8), proportional to (0, 1, 4). The plane is (y − 3) + 4(z − 1) = 0, that is y + 4z = 7 (check: −5 + 12 = 7).
(ii) The lines pass through A(a − d, a, a + d) and B(b − c, b, b + c) with directions (α − δ, α, α + δ) and (β − γ, β, β + γ). Coplanarity needs det[[b − c − a + d, b − a, b + c − a − d], [α − δ, α, α + δ], [β − γ, β, β + γ]] = 0.
Apply C1 → C1 + C3 − 2C2: every entry of the first column becomes 0, so the determinant is 0 and the lines are coplanar.
Their plane is det[[x − a + d, y − a, z − a − d], [α − δ, α, α + δ], [β − γ, β, β + γ]] = 0; the same column operation turns the first column into (x − 2y + z, 0, 0), so the equation is (x − 2y + z)(αγ − βδ) = 0.
Answer: (i) y + 4z = 7; (ii) the lines are coplanar and lie in the plane x − 2y + z = 0 (assuming αγ ≠ βδ, that is the lines are not parallel).Final answer: (i) y + 4z = 7; (ii) the lines are coplanar and lie in the plane x − 2y + z = 0 (assuming αγ ≠ βδ, that is the lines are not parallel).
Q3(d) · Calculus · 12 marks≈ 9 min
Double and Triple Integrals; Areas, Surfaces and Volumes
Evaluate the integral ∬R (x − y)2 cos2(x + y) dx dy where R is the rhombus with successive vertices as (π, 0), (2π, π), (π, 2π), (0, π).
हिंदी में प्रश्न
निम्न समाकलन का मूल्यांकन करें: ∬R (x − y)2 cos2(x + y) dx dy जहाँ R एक समचतुर्भुज है, जिसके शीर्ष क्रमवार (π, 0), (2π, π), (π, 2π), (0, π) हैं।
Question 4 (50 marks)
Q4(a) · Calculus · 13 marks≈ 9 min
Double and Triple Integrals; Areas, Surfaces and Volumes
Evaluate ∬R √|y − x2| dx dy where R = [−1, 1 ; 0, 2].
हिंदी में प्रश्न
निम्नलिखित का मान निकालिए: ∬R √|y − x2| dx dy जहाँ R = [−1, 1 ; 0, 2].
Worked solution
Split R = [−1, 1] × [0, 2] along y = x2: R1: x2 ≤ y ≤ 2 and R2: 0 ≤ y ≤ x2 (−1 ≤ x ≤ 1).
On R2: ∫0x2 √(x2 − y) dy = (2/3)|x|3, and ∫−11 (2/3)|x|3 dx = 1/3.
On R1: ∫x22 √(y − x2) dy = (2/3)(2 − x2)3/2, and ∫−11 (2/3)(2 − x2)3/2 dx = (4/3)∫01 (2 − x2)3/2 dx.
With x = √2 sin θ (0 ≤ θ ≤ π/4): ∫01 (2 − x2)3/2 dx = ∫0π/4 4 cos4θ dθ = (1/2)∫0π/4 (3 + 4 cos 2θ + cos 4θ) dθ = (1/2)(3π/4 + 2) = 3π/8 + 1.
So the R1 part is (4/3)(3π/8 + 1) = π/2 + 4/3, and the total is π/2 + 4/3 + 1/3.
Answer: π/2 + 5/3 ≈ 3.237.Final answer: π/2 + 5/3 ≈ 3.237.
Q4(b) · Linear Algebra · 12 marks≈ 9 min
Vector Spaces, Subspaces, Basis and Dimension
Find the dimension of the subspace of ℝ4, spanned by the set {(1, 0, 0, 0), (0, 1, 0, 0), (1, 2, 0, 1), (0, 0, 0, 1)}. Hence find its basis.
हिंदी में प्रश्न
ℝ4 की उस उपसमष्टि की विमा ज्ञात कीजिये जो समुच्चय {(1, 0, 0, 0), (0, 1, 0, 0), (1, 2, 0, 1), (0, 0, 0, 1)} द्वारा विस्तारित है। तत्पश्चात् उसका आधार निकालिए।
Q4(c) · Analytic Geometry · 13 marks≈ 9 min
Ellipsoid, Hyperboloids and Paraboloids
Two perpendicular tangent planes to the paraboloid x2 + y2 = 2z intersect in a straight line in the plane x = 0. Obtain the curve to which this straight line touches.
हिंदी में प्रश्न
यदि परवलयज x2 + y2 = 2z पर दो लम्बवत् स्पर्शीय समतल एक सीधी रेखा में, जो समतल x = 0 में, पर काटते हैं। उस वक्र को प्राप्त कीजिए जिस पर वह सीधी रेखा स्पर्श करती है।
Worked solution
A tangent plane to z = (x2 + y2)/2 is z = lx + my − (l2 + m2)/2, with normal (l, m, −1).
Two such planes (l1, m1) and (l2, m2) meet in a line lying in x = 0 only if their traces on x = 0 coincide: m1 = m2 = m and l12 = l22, so l2 = −l1 = −l for distinct planes.
Perpendicular: l1l2 + m1m2 + 1 = −l2 + m2 + 1 = 0, so l2 = m2 + 1. The common line is x = 0, z = my − (2m2 + 1)/2.
Envelope in the plane x = 0: differentiating with respect to m, 0 = y − 2m, so m = y/2 and z = y2/2 − y2/4 − 1/2.
Answer: the line always touches the parabola y2 = 4z + 2, x = 0.Final answer: the line always touches the parabola y2 = 4z + 2, x = 0.
Q4(d) · Calculus · 12 marks≈ 9 min
Functions of Several Variables: Partial Derivatives, Extrema, Lagrange Multipliers and Jacobians
For the function f(x, y) = (x2 − x√y)/(x2 + y), (x, y) ≠ (0, 0); f(x, y) = 0, (x, y) = (0, 0), examine the continuity and differentiability.
हिंदी में प्रश्न
दिए गए फलन f(x, y) = (x2 − x√y)/(x2 + y), (x, y) ≠ (0, 0); f(x, y) = 0, (x, y) = (0, 0) के लिए सांतत्य एवं अवकलनीयता का परीक्षण कीजिये।
Worked solution
√y requires y ≥ 0, so f is defined on the closed upper half-plane (except where x2 + y = 0, i.e. the origin, where f = 0 by definition).
Along y = 0 (x ≠ 0): f(x, 0) = x2/x2 = 1 → 1. Along x = 0 (y > 0): f(0, y) = 0 → 0.
The limits along two paths differ, so lim(x,y)→(0,0) f does not exist: f is discontinuous at (0, 0), and therefore not differentiable there. (Even fx(0, 0) = lim (f(h, 0) − 0)/h = lim 1/h fails to exist.)
At every other point with y > 0, f is a quotient of functions with continuous partial derivatives and non-zero denominator, hence continuous and differentiable.
Answer: f is neither continuous nor differentiable at (0, 0); it is both at the other points with y > 0.Final answer: f is neither continuous nor differentiable at (0, 0); it is both at the other points with y > 0.
Question 5 (50 marks)
Q5(a) · Ordinary Differential Equations · 10 marks≈ 7 min
Formulation and First Order, First Degree Equations
Solve the differential equation: x cos x dy/dx + y(x sin x + cos x) = 1.
हिंदी में प्रश्न
निम्नलिखित अवकल समीकरण को हल कीजिये: x cos x dy/dx + y(x sin x + cos x) = 1.
Q5(b) · Ordinary Differential Equations · 10 marks≈ 7 min
Formulation and First Order, First Degree Equations
Solve the differential equation: (2xy4ey + 2xy3 + y)dx + (x2y4ey − x2y2 − 3x)dy = 0.
हिंदी में प्रश्न
निम्नलिखित अवकल समीकरण का हल निकालिये: (2xy4ey + 2xy3 + y)dx + (x2y4ey − x2y2 − 3x)dy = 0.
Q5(c) · Dynamics and Statics · 10 marks≈ 7 min
Rectilinear Motion and Simple Harmonic Motion
A body moving under SHM has an amplitude 'a' and time period 'T'. If the velocity is trebled, when the distance from mean position is (2/3)a, the period being unaltered, find the new amplitude.
हिंदी में प्रश्न
एक पिंड, जो सरल आवर्त गति (एस.एच.एम.) कर रहा है, उसका आयाम 'a' व आवर्तकाल 'T' है। जब माध्य स्थिति से उसकी दूरी (2/3)a हो, तब उसका वेग तिगुना कर दिया जाय, परंतु आवर्तकाल को न बदला जाए, तो उसका नया आयाम मालूम कीजिये।
Worked solution
With n = 2π/T, the speed v at x = 2a/3 satisfies v2 = n2(a2 − 4a2/9) = 5n2a2/9.
The new speed 3v at the same point, with the same n, gives 9v2 = n2(A2 − 4a2/9), A being the new amplitude.
So 5n2a2 = n2(A2 − 4a2/9), that is A2 = 49a2/9.
Answer: the new amplitude is 7a/3.Final answer: the new amplitude is 7a/3.
Q5(d) · Dynamics and Statics · 10 marks≈ 7 min
Equilibrium, Friction and the Common Catenary
A rod of 8 kg is movable in a vertical plane about a hinge at one end, another end is fastened a weight equal to half of the rod, this end is fastened by a string of length l to a point at a height b above the hinge vertically. Obtain the tension in the string.
हिंदी में प्रश्न
8 kg भार की एक छड़, ऊर्ध्वाधर तल में एक सिरे पर लगे कब्जे पर चलायमान है। उसके दूसरे सिरे पर लम्बाई l की एक रस्सी के द्वारा कब्जे से b ऊँचाई पर, ऊर्ध्व दिशा में छड़ के आधे के बराबर भार बाँधा गया है। रस्सी में तनाव ज्ञात कीजिये।
Worked solution
Let the hinge be A, the free end B (carrying the 4 kg weight), C the point to which the string is tied, at height b vertically above A, and x the horizontal distance of B from AC.
Moments about A eliminate the hinge reaction: the rod's 8 kg acts at horizontal distance x/2 and the 4 kg at x, a total moment 8x.
The tension T acts along BC, and its moment about A is T × (perpendicular from A to BC) = T × 2(area of triangle ABC)/BC = T × bx/l.
Equating moments, Tbx/l = 8x.
Answer: T = 8l/b kg wt, whatever the length of the rod.Final answer: T = 8l/b kg wt, whatever the length of the rod.
Q5(e) · Vector Analysis · 10 marks≈ 7 min
Gradient, Divergence and Curl in Cartesian and Cylindrical Coordinates
Find the angle between the surfaces x2 + y2 + z2 − 9 = 0 and z = x2 + y2 − 3 at (2, −1, 2).
हिंदी में प्रश्न
निम्न दो सतहों x2 + y2 + z2 − 9 = 0 तथा z = x2 + y2 − 3 के बीच बिन्दु (2, −1, 2) पर कोण ज्ञात कीजिये।
Question 6 (50 marks)
Q6(a) · Ordinary Differential Equations · 12 marks≈ 9 min
Formulation and First Order, First Degree Equations
Find the constant a so that (x + y)a is the Integrating factor of (4x2 + 2xy + 6y)dx + (2x2 + 9y + 3x)dy = 0 and hence solve the differential equation.
हिंदी में प्रश्न
यदि (x + y)a, निम्न अवकल समीकरण (4x2 + 2xy + 6y)dx + (2x2 + 9y + 3x)dy = 0 का समाकलन गुणांक है तो 'a' का मान मालूम कीजिये। तत्पश्चात् अवकल समीकरण का हल निकालिए।
Worked solution
M = 4x2 + 2xy + 6y, N = 2x2 + 9y + 3x. For μ = (x + y)a: (μM)y = (μN)x becomes a(M − N) = (x + y)(Nx − My).
M − N = 2x2 + 2xy − 3x − 3y = (x + y)(2x − 3) and Nx − My = (4x + 3) − (2x + 6) = 2x − 3, so a = 1.
The exact equation: (4x3 + 6x2y + 2xy2 + 6xy + 6y2) dx + (2x3 + 2x2y + 3x2 + 12xy + 9y2) dy = 0.
∫M dx = x4 + 2x3y + x2y2 + 3x2y + 6xy2; the term of N free of x is 9y2, giving 3y3.
Answer: a = 1; x4 + 2x3y + x2y2 + 3x2y + 6xy2 + 3y3 = c, that is (x + y)2(x2 + 3y) = c.Final answer: a = 1; x4 + 2x3y + x2y2 + 3x2y + 6xy2 + 3y3 = c, that is (x + y)2(x2 + 3y) = c.
Q6(b) · Dynamics and Statics · 13 marks≈ 9 min
Equilibrium, Friction and the Common Catenary
Two equal ladders of weight 4 kg each are placed so as to lean at A against each other with their ends resting on a rough floor, given the coefficient of friction is μ. The ladders at A make an angle 60° with each other. Find what weight on the top would cause them to slip.
हिंदी में प्रश्न
दो बराबर भार की सीढ़ियां, जिसमें प्रत्येक का भार 4 kg है, बिंदु A पर एक दूसरे के साथ झुकाकर रखी गई हैं। उनके दूसरे सिरे एक खुरदुरे फर्श पर हैं, जिसका घर्षण गुणांक μ है। दोनों सीढ़ी के बीच 60° का कोण है। कितना भार उस शीर्ष बिंदु A पर रखा जाए कि वह सीढ़ियों के फिसल जाने का कारण बन जाय।
Worked solution
Let each ladder have length 2a and weight W = 4 kg, with the extra weight W′ at the joint A; each ladder makes 30° with the vertical.
By symmetry each foot has normal reaction R = W + W′/2 and friction F directed inwards.
Moments about A for one ladder (the joint force and W′ act at A): the foot is a horizontally and √3a vertically from A and the centre a/2 horizontally, so Ra = W(a/2) + F√3a.
So F = (R − W/2)/√3; slipping begins when F = μR, that is R(1 − √3μ) = W/2.
Then W′ = 2R − 2W = W/(1 − √3μ) − 2W = W(2√3μ − 1)/(1 − √3μ).
Answer: W′ = 4(2√3μ − 1)/(1 − √3μ) kg wt; this needs 1/(2√3) < μ < 1/√3 (for smaller μ the ladders slip unloaded, and for μ ≥ 1/√3 no weight makes them slip).Final answer: W′ = 4(2√3μ − 1)/(1 − √3μ) kg wt; this needs 1/(2√3) < μ < 1/√3 (for smaller μ the ladders slip unloaded, and for μ ≥ 1/√3 no weight makes them slip).
Q6(c) · Vector Analysis · 12 marks≈ 9 min
Gradient, Divergence and Curl in Cartesian and Cylindrical Coordinates
Find the value of λ and μ so that the surfaces λx2 − μyz = (λ + 2)x and 4x2y + z3 = 4 may intersect orthogonally at (1, −1, 2).
हिंदी में प्रश्न
यदि दो पृष्ठ λx2 − μyz = (λ + 2)x तथा 4x2y + z3 = 4 बिन्दु (1, −1, 2) पर लम्बवत् काटती हों, तो λ व μ का मान निकालिए।
Worked solution
The point lies on the first surface: λ − μ(−1)(2) = λ + 2, so 2μ = 2 and μ = 1; it also lies on the second, since 4(1)(−1) + 8 = 4.
For f = λx2 − μyz − (λ + 2)x: ∇f = (2λx − λ − 2, −μz, −μy) = (λ − 2, −2, 1) at the point, with μ = 1.
For g = 4x2y + z3: ∇g = (8xy, 4x2, 3z2) = (−8, 4, 12).
Orthogonality: ∇f·∇g = −8(λ − 2) − 8 + 12 = 0, so 8λ = 20.
Answer: λ = 5/2, μ = 1.Final answer: λ = 5/2, μ = 1.
Q6(d) · Dynamics and Statics · 13 marks≈ 9 min
Rectilinear Motion and Simple Harmonic Motion
A mass starts from rest at a distance 'a' from the centre of force which attracts inversely as the distance. Find the time of arriving at the centre.
हिंदी में प्रश्न
एक पिंड, बल के केंद्र से अन्तर्गत, जो दूरी के विलोमानुपाती आकर्षित करता है; 'a' दूरी से विरामअवस्था से चलना शुरू करता है। उसका केन्द्र पर आगमन का समय मालूम कीजिये।
Worked solution
The acceleration towards the centre is μ/x: v dv/dx = −μ/x, and v = 0 at x = a gives v2 = 2μ log(a/x).
Moving inwards, dt = −dx/√(2μ log(a/x)), so t = ∫0a dx/√(2μ log(a/x)).
Put log(a/x) = y, that is x = ae−y, dx = −ae−y dy; y runs from 0 to ∞.
t = (a/√(2μ)) ∫0∞ e−y y−1/2 dy = (a/√(2μ)) Γ(1/2) = a√(π/(2μ)).
Answer: the time to reach the centre is a√(π/(2μ)).Final answer: the time to reach the centre is a√(π/(2μ)).
Question 7 (50 marks)
Q7(a) · Ordinary Differential Equations · 12 marks≈ 9 min
Laplace Transforms and Initial Value Problems
(i) Obtain Laplace Inverse transform of {ln(1 + 1/s2) + [s/(s2 + 25)] e−πs}.
(ii) Using Laplace transform, solve y″ + y = t, y(0) = 1, y′(0) = −2. (6 + 6 = 12)हिंदी में प्रश्न
(i) निम्नलिखित का लाप्लास विलोम रूपांतर प्राप्त कीजिये: {ln(1 + 1/s2) + [s/(s2 + 25)] e−πs}.
(ii) लाप्लास रूपांतर का प्रयोग करके, निम्नलिखित y″ + y = t, y(0) = 1, y′(0) = −2 का हल निकालिए। (6 + 6 = 12)Q7(b) · Dynamics and Statics · 13 marks≈ 9 min
Motion in a Plane and Projectiles
A particle is projected from the base of a hill whose slope is that of a right circular cone, whose axis is vertical. The projectile grazes the vertex and strikes the hill again at a point on the base. If the semivertical angle of the cone is 30°, h is height, determine the initial velocity u of the projection and its angle of projection.
हिंदी में प्रश्न
एक कण, एक ऐसी पहाड़ी के आधार पर एक बिन्दु से, प्रक्षेपित किया जाता है, जिसकी प्रवणता एक लम्ब वृत्ताकार शंकु की है जिसकी अक्षीय रेखा ऊर्ध्वाधर है। प्रक्षेप शंकु के शीर्षबिन्दु छूते हुए व पहाड़ी के आधार पर किसी दूसरे बिन्दु से जा टकराता है। यदि शंकु का अर्धशीर्ष कोण 30° व ऊँचाई h हो, तो प्रक्षेप का प्रारंभिक वेग u व उसका प्रक्षेप कोण निर्धारित कीजिये।
Worked solution
The base radius of the hill is h tan 30° = h/√3; the path passes through the vertex V (horizontal distance h/√3, height h) and returns to base level at distance 2h/√3.
A parabola through two points at the same level has its highest point midway between them, so V is the highest point of the path: H = u2 sin2α/(2g) = h and R = u2 sin 2α/g = 2h/√3.
Dividing, R/H = 4 cot α = 2/√3, so tan α = 2√3 and sin2α = 12/13.
Then u2 = 2gh/sin2α = 13gh/6.
Check that the path clears the slope: with c = h/√3 the path is y = h(2x/c − x2/c2), and y − hx/c = (hx/c)(1 − x/c) ≥ 0 for 0 ≤ x ≤ c, so it touches the hill only at the vertex.
Answer: u = √(13gh/6), and the angle of projection is tan−1(2√3) ≈ 73.9° with the horizontal.Final answer: u = √(13gh/6), and the angle of projection is tan−1(2√3) ≈ 73.9° with the horizontal.
Q7(c) · Vector Analysis · 12 marks≈ 9 min
Gradient, Divergence and Curl in Cartesian and Cylindrical Coordinates
A vector field is given by F̄ = (x2 + xy2)î + (y2 + x2y)ĵ. Verify that the field F̄ is irrotational or not. Find the scalar potential.
हिंदी में प्रश्न
एक सदिश क्षेत्र F̄ = (x2 + xy2)î + (y2 + x2y)ĵ के दिया गया है। सत्यापित कीजिए कि यह क्षेत्र F̄ अघूर्णी है या नहीं। अत: अदिश विभव ज्ञात कीजिये।
Q7(d) · Ordinary Differential Equations · 13 marks≈ 9 min
First Order Higher Degree Equations, Clairaut's Equation and Singular Solutions
Solve the differential equation x = py − p2 where p = dy/dx.
हिंदी में प्रश्न
अवकल समीकरण x = py − p2 का हल निकालिए, जहाँ p = dy/dx.
Worked solution
Solvable for x. Differentiating with respect to y (dx/dy = 1/p): 1/p = p + (y − 2p) dp/dy, so dy/dp − py/(1 − p2) = −2p2/(1 − p2), linear in y.
IF = e∫−p/(1 − p2) dp = √(1 − p2), so d(y√(1 − p2))/dp = −2p2/√(1 − p2).
Using ∫p2/√(1 − p2) dp = (sin−1p − p√(1 − p2))/2: y√(1 − p2) = p√(1 − p2) − sin−1p + c.
Answer: in parametric form y = p + (c − sin−1p)/√(1 − p2), x = py − p2.Final answer: in parametric form y = p + (c − sin−1p)/√(1 − p2), x = py − p2.
Question 8 (50 marks)
Q8(a) · Dynamics and Statics · 12 marks≈ 9 min
Equilibrium, Friction and the Common Catenary
Find the length of an endless chain which will hang over a circular pulley of radius 'a' so as to be in contact with the two-thirds of the circumference of the pulley.
हिंदी में प्रश्न
उस अंतहीन शृंखला की लम्बाई ज्ञात कीजिये, जो एक 'a' त्रिज्य वाली वृत्तीय घिरनी पर इस तरह से टंगी है, कि वह उस घिरनी के दो-तिहाई परिधि के स्पर्श में रहे।
Worked solution
The chain touches the upper two-thirds of the pulley (arc 4πa/3) and leaves it at two points P, P′ whose radii make 60° with the downward vertical, so the tangents there make 60° with the horizontal.
The free part hangs as a common catenary with ends P, P′, so ψ = 60° at P.
Horizontal distance from the lowest point to P: x = c log(sec ψ + tan ψ) = c log(2 + √3); it also equals a sin 60° = √3a/2, so c = √3a/(2 log(2 + √3)).
Length of the catenary = 2c tan 60° = 2√3c = 3a/log(2 + √3).
On the smooth pulley T − wy is constant, so the tension remains w × (height above the directrix) and the part on the pulley is consistent with this catenary.
Answer: length = 4πa/3 + 3a/log(2 + √3) ≈ 6.47a.Final answer: length = 4πa/3 + 3a/log(2 + √3) ≈ 6.47a.
Q8(b) · Dynamics and Statics · 13 marks≈ 9 min
Central Orbits and Kepler's Laws
A particle moves in a plane under a force, towards a fixed centre, proportional to the distance. If the path of the particle has two apsidal distances a, b (a > b), then find the equation of the path.
हिंदी में प्रश्न
एक कण, एक बल के अधीन, एक नियत केन्द्र की ओर जो दूरी के समानुपाती है, पर चलता है। यदि कण के मार्ग के दो एपसाइडल दूरियाँ a, b हैं (a > b), तब मार्ग का समीकरण ज्ञात करो।
Worked solution
Here P = μr, and the pedal equation (h2/p3) dp/dr = μr integrates to h2/p2 = C − μr2.
At an apse p = r, so h2/r2 = C − μr2, that is μr4 − Cr2 + h2 = 0, whose roots are r2 = a2 and b2.
So a2 + b2 = C/μ and a2b2 = h2/μ, and the path is a2b2/p2 = a2 + b2 − r2.
This is the pedal equation of the ellipse x2/a2 + y2/b2 = 1 referred to its centre.
Check in Cartesian form: ẍ = −μx and ÿ = −μy give x = a cos √μ t, y = b sin √μ t when t = 0 is taken at the far apse.
Answer: the path is the ellipse x2/a2 + y2/b2 = 1 with its centre at the centre of force and its axes along the apsidal lines (and h = ab√μ).Final answer: the path is the ellipse x2/a2 + y2/b2 = 1 with its centre at the centre of force and its axes along the apsidal lines (and h = ab√μ).
Q8(c) · Vector Analysis · 12 marks≈ 9 min
Gauss and Stokes' Theorems and Green's Identities
Evaluate ∫C e−x(sin y dx + cos y dy), where C is the rectangle with vertices (0, 0), (π, 0), (π, π/2), (0, π/2).
हिंदी में प्रश्न
निम्नलिखित का मान निकालिए ∫C e−x(sin y dx + cos y dy) जहाँ C एक आयत है, जिसके (0, 0), (π, 0), (π, π/2), (0, π/2) शीर्ष हैं।
Worked solution
M = e−x sin y, N = e−x cos y; the vertices in the given order run anticlockwise.
∂N/∂x − ∂M/∂y = −e−x cos y − e−x cos y = −2e−x cos y.
By Green's theorem the integral is −2∫0πe−x dx ∫0π/2cos y dy = −2(1 − e−π)(1).
Answer: 2(e−π − 1) ≈ −1.914.Final answer: 2(e−π − 1) ≈ −1.914.
Q8(d) · Ordinary Differential Equations · 13 marks≈ 9 min
Second Order Equations with Variable Coefficients
Solve: x4 d4y/dx4 + 6x3 d3y/dx3 + 4x2 d2y/dx2 − 2x dy/dx − 4y = x2 + 2 cos(loge x).
हिंदी में प्रश्न
निम्न अवकल समीकरण को हल करें: x4 d4y/dx4 + 6x3 d3y/dx3 + 4x2 d2y/dx2 − 2x dy/dx − 4y = x2 + 2 cos(loge x).
UPSC Mains 2015 Mathematics Paper II
Question 1 (50 marks)
Q1(a) · Abstract Algebra · 10 marks≈ 7 min
Groups, Subgroups, Cyclic Groups, Cosets and Lagrange's Theorem
(i) How many generators are there of the cyclic group G of order 8? Explain.
(ii) Taking a group {e, a, b, c} of order 4, where e is the identity, construct composition tables showing that one is cyclic while the other is not. (5 + 5 = 10)हिंदी में प्रश्न
(i) कोटि 8 के चक्रीय समूह G के कितने जनक होते हैं? व्याख्या कीजिए।
(ii) कोटि 4 के एक समूह {e, a, b, c} को लेते हुए, जहाँ e तत्समक (आइडेंटिटि) है, संयोजन सारणियाँ बनाइए यह दर्शाते हुए कि एक चक्रीय है जबकि दूसरी नहीं है। (5 + 5 = 10)Worked solution
(i) If G = ⟨a⟩ has order 8, then o(ak) = 8/gcd(8, k), so ak generates G iff gcd(k, 8) = 1. The generators are a, a3, a5, a7: there are φ(8) = 4 of them.
(ii) Cyclic table: take o(a) = 4, b = a2, c = a3. Rows (row element × column element, columns e, a, b, c): e: e, a, b, c; a: a, b, c, e; b: b, c, e, a; c: c, e, a, b.
This group is generated by a (and by c), so it is cyclic and isomorphic to ℤ4.
Non-cyclic table: take a2 = b2 = c2 = e, ab = ba = c, bc = cb = a, ca = ac = b. Rows: e: e, a, b, c; a: a, e, c, b; b: b, c, e, a; c: c, b, a, e.
This is the group ℤ2 × ℤ2 (a = (1, 0), b = (0, 1), c = (1, 1)), so associativity holds; each row and column is a permutation of {e, a, b, c}. Every non-identity element has order 2, so no element generates the group: it is not cyclic.
Answer: (i) 4 generators, a, a3, a5, a7; (ii) the first table is the cyclic group ℤ4, the second the non-cyclic Klein group V4.Final answer: (i) 4 generators, a, a3, a5, a7; (ii) the first table is the cyclic group ℤ4, the second the non-cyclic Klein group V4.
Q1(b) · Abstract Algebra · 10 marks≈ 7 min
Rings, Subrings, Ideals and Ring Homomorphisms
Give an example of a ring having identity but a subring of this having a different identity.
हिंदी में प्रश्न
वलय का एक ऐसा उदाहरण दीजिए, जिसका तत्समक है परन्तु जिसके एक उपवलय का भिन्न तत्समक है।
Worked solution
Take R = ℤ6, whose identity is 1, and S = {0, 2, 4}. For a, b ∈ S, a − b and ab are even residues mod 6, so S is a subring.
In S, 4·0 = 0, 4·2 = 8 = 2 and 4·4 = 16 = 4 (mod 6), so 4 is the identity of S, different from the identity 1 of R.
Another example: R = ℤ × ℤ with identity (1, 1) and the subring S = ℤ × {0}, whose identity is (1, 0).
Answer: ℤ6 (identity 1) has the subring {0, 2, 4} with identity 4.Final answer: ℤ6 (identity 1) has the subring {0, 2, 4} with identity 4.
Q1(c) · Real Analysis · 10 marks≈ 7 min
Series: Convergence Tests, Absolute and Conditional Convergence, Rearrangement
Test the convergence and absolute convergence of the series ∑n=1∞ (−1)n+1 n/(n2 + 1).
हिंदी में प्रश्न
श्रेणी ∑n=1∞ (−1)n+1 n/(n2 + 1) के अभिसरण तथा निरपेक्ष अभिसरण का परीक्षण कीजिए।
Q1(d) · Complex Analysis · 10 marks≈ 7 min
Analytic Functions and Cauchy-Riemann Equations
Show that the function v(x, y) = ln(x2 + y2) + x + y is harmonic. Find its conjugate harmonic function u(x, y). Also, find the corresponding analytic function f(z) = u + iv in terms of z.
हिंदी में प्रश्न
दर्शाइए कि फलन v(x, y) = ln(x2 + y2) + x + y प्रसंवादी है। इसका संयुग्मी प्रसंवादी फलन u(x, y) ज्ञात कीजिए। संगत विश्लेषिक फलन f(z) = u + iv को भी z के पदों में ज्ञात कीजिए।
Worked solution
vx = 2x/(x2 + y2) + 1 and vy = 2y/(x2 + y2) + 1, so vxx = 2(y2 − x2)/(x2 + y2)2 and vyy = 2(x2 − y2)/(x2 + y2)2. Hence vxx + vyy = 0 for (x, y) ≠ (0, 0): v is harmonic in ℂ − {0}.
Milne-Thomson (v given): f′(z) = vy(z, 0) + ivx(z, 0) = 1 + i(2/z + 1).
Integrating, f(z) = z + i(z + 2 log z) + C = (1 + i)z + 2i log z + C, with C real.
With z = reiθ, 2i log z = 2i ln r − 2θ, so u = Re f = x − y − 2θ + C; check: Im f = x + y + 2 ln r = v.
Answer: u = x − y − 2 tan−1(y/x) + C and f(z) = (1 + i)z + 2i log z + C (analytic in any simply connected domain not containing 0, for a branch of log z).Final answer: u = x − y − 2 tan−1(y/x) + C and f(z) = (1 + i)z + 2i log z + C (analytic in any simply connected domain not containing 0, for a branch of log z).
Q1(e) · Linear Programming · 10 marks≈ 7 min
Assignment Problem
Solve the following assignment problem to maximize the sales:
Sales (rows Salesmen A to E; columns Territories I, II, III, IV, V):
A: 3, 4, 5, 6, 7
B: 4, 15, 13, 7, 6
C: 6, 13, 12, 5, 11
D: 7, 12, 15, 8, 5
E: 8, 13, 10, 6, 9हिंदी में प्रश्न
निम्नलिखित नियतन समस्या को अधिकतम बिक्री करने के लिए हल कीजिए:
बिक्री (पंक्तियाँ विक्रेता A से E; स्तंभ क्षेत्र I, II, III, IV, V):
A: 3, 4, 5, 6, 7
B: 4, 15, 13, 7, 6
C: 6, 13, 12, 5, 11
D: 7, 12, 15, 8, 5
E: 8, 13, 10, 6, 9
Question 2 (50 marks)
Q2(a) · Abstract Algebra · 15 marks≈ 11 min
Rings, Subrings, Ideals and Ring Homomorphisms
If R is a ring with unit element 1 and φ is a homomorphism of R onto R′, prove that φ(1) is the unit element of R′.
हिंदी में प्रश्न
यदि R एक वलय है, जिसका तत्समक अवयव 1 है तथा R आच्छादक R′ (R onto R′) की एक समाकारिता φ है, तो सिद्ध कीजिए R′ का तत्समक अवयव φ(1) है।
Worked solution
Let r′ ∈ R′. Since φ is onto, r′ = φ(r) for some r ∈ R.
Then φ(1)r′ = φ(1)φ(r) = φ(1·r) = φ(r) = r′, and similarly r′φ(1) = φ(r·1) = r′.
So φ(1) is a two-sided multiplicative identity of R′, and by uniqueness it is the unit element of R′.
Ontoness is needed: n ↦ (n, 0) is a homomorphism ℤ → ℤ × ℤ with φ(1) = (1, 0), which is not the identity (1, 1).
Answer: φ(1) is the unit element of R′.Final answer: φ(1) is the unit element of R′.
Q2(b) · Real Analysis · 15 marks≈ 11 min
Riemann Integral, Improper Integrals and Fundamental Theorems of Integral Calculus
Is the function
f(x) = 1/n, 1/(n + 1) < x ≤ 1/n
f(x) = 0, x = 0
Riemann integrable? If yes, obtain the value of ∫01 f(x) dx.हिंदी में प्रश्न
क्या फलन
f(x) = 1/n, 1/(n + 1) < x ≤ 1/n
f(x) = 0, x = 0
रीमान समाकलनीय है? यदि हाँ, तो ∫01 f(x) dx का मान ज्ञात कीजिए।Worked solution
f is bounded (0 ≤ f ≤ 1) and increasing on [0, 1]: a larger x lies in an interval (1/(n + 1), 1/n] with smaller n, hence larger value 1/n.
A monotone function is Riemann integrable, so f ∈ R[0, 1] (alternatively, its discontinuities 1/2, 1/3, ... form a countable set).
Since 0 ≤ ∫0c f ≤ c, ∫01 f = limc→0+ ∫c1 f, and on each piece (1/(n + 1), 1/n] f is the constant 1/n.
So ∫01 f = ∑n=1∞ (1/n)(1/n − 1/(n + 1)) = ∑ 1/(n2(n + 1)).
1/(n2(n + 1)) = 1/n2 − 1/n + 1/(n + 1), and ∑(1/(n + 1) − 1/n) = −1 (telescoping), so the sum is π2/6 − 1.
Answer: yes, f is Riemann integrable and ∫01 f(x) dx = π2/6 − 1 ≈ 0.645.Final answer: yes, f is Riemann integrable and ∫01 f(x) dx = π2/6 − 1 ≈ 0.645.
Q2(c) · Complex Analysis · 20 marks≈ 14 min
Power Series, Taylor's Series and Laurent's Series
Find all possible Taylor's and Laurent's series expansions of the function f(z) = (2z − 3)/(z2 − 3z + 2) about the point z = 0.
हिंदी में प्रश्न
फलन f(z) = (2z − 3)/(z2 − 3z + 2) के बिन्दु z = 0 के इर्द-गिर्द सभी संभव टेलर तथा लौराँ श्रेणियों में प्रसारों को ज्ञात कीजिए।
Question 3 (50 marks)
Q3(a) · Complex Analysis · 15 marks≈ 11 min
Singularities and Cauchy's Residue Theorem
State Cauchy's residue theorem. Using it, evaluate the integral ∫C (ez + 1)/[z(z + 1)(z − i)2] dz; C : |z| = 2.
हिंदी में प्रश्न
कौशी अवशेष प्रमेय का कथन लिखिए। इसका इस्तेमाल करते हुए समाकल ∫C (ez + 1)/[z(z + 1)(z − i)2] dz; C : |z| = 2 का मान निकालिए।
Q3(b) · Real Analysis · 15 marks≈ 11 min
Uniform Convergence of Sequences and Series of Functions
Test the series of functions ∑n=1∞ nx/(1 + n2x2) for uniform convergence.
हिंदी में प्रश्न
फलनों की श्रेणी ∑n=1∞ nx/(1 + n2x2) का एकसमान अभिसरण के लिए परीक्षण कीजिए।
Worked solution
(As printed this is a series, and its terms behave like 1/(nx), so it diverges for every x ≠ 0. We show this, then treat the sequence fn(x) = nx/(1 + n2x2), which is evidently what is intended.)
Series: for fixed x > 0, un(x) > 0 and n·un(x) = n2x/(1 + n2x2) → 1/x ≠ 0, so ∑un(x) diverges by limit comparison with ∑1/n (similarly for x < 0). It converges only at x = 0, so it is not uniformly convergent on any interval.
Sequence: fn(0) = 0 and |fn(x)| ≤ 1/(n|x|) → 0 for x ≠ 0, so fn → 0 pointwise on ℝ.
fn′(x) = n(1 − n2x2)/(1 + n2x2)2 vanishes at x = ±1/n, and fn(1/n) = 1/2. So sup|fn| = 1/2 on any interval containing 0 (for large n), which does not tend to 0: the convergence is not uniform there.
On [a, ∞) with a > 0 and n > 1/a, fn is decreasing, so sup|fn| = na/(1 + n2a2) ≤ 1/(na) → 0: the convergence is uniform there.
Answer: the series diverges for x ≠ 0, so it is not uniformly convergent; the sequence nx/(1 + n2x2) → 0 non-uniformly on any interval containing 0 and uniformly on [a, ∞), a > 0.Final answer: the series diverges for x ≠ 0, so it is not uniformly convergent; the sequence nx/(1 + n2x2) → 0 non-uniformly on any interval containing 0 and uniformly on [a, ∞), a > 0.
Q3(c) · Linear Programming · 20 marks≈ 14 min
LPP Formulation, Basic Solutions and the Graphical Method
Consider the following linear programming problem:
Maximize Z = x1 + 2x2 − 3x3 + 4x4
subject to
x1 + x2 + 2x3 + 3x4 = 12
x2 + 2x3 + x4 = 8
x1, x2, x3, x4 ≥ 0
(i) Using the definition, find its all basic solutions. Which of these are degenerate basic feasible solutions and which are non-degenerate basic feasible solutions?
(ii) Without solving the problem, show that it has an optimal solution. Which of the basic feasible solution(s) is/are optimal?हिंदी में प्रश्न
निम्नलिखित रैखिक प्रोग्रामन समस्या पर विचार कीजिए:
अधिकतमीकरण कीजिए Z = x1 + 2x2 − 3x3 + 4x4
बशर्ते कि
x1 + x2 + 2x3 + 3x4 = 12
x2 + 2x3 + x4 = 8
x1, x2, x3, x4 ≥ 0
(i) परिभाषा का इस्तेमाल करते हुए इसके सभी आधारी हल ज्ञात कीजिए। इनमें कौन-से अपभ्रष्ट आधारी सुसंगत हल हैं और कौन-से अनपभ्रष्ट आधारी सुसंगत हल हैं?
(ii) समस्या को हल किए बिना दर्शाइए कि इसका इष्टतम हल है। आधारी सुसंगत हलों में कौन-सा/से इष्टतम हल है/हैं?Worked solution
(i) n = 4, m = 2, so at most C(4, 2) = 6 bases. The columns are a1 = (1, 0), a2 = (1, 1), a3 = (2, 2), a4 = (3, 1); a3 = 2a2, so (x2, x3) is singular (it gives x2 + 2x3 = 12 and x2 + 2x3 = 8, which are inconsistent).
(x1, x2): x2 = 8, x1 = 4, giving (4, 8, 0, 0), a BFS with Z = 20. (x1, x3): 2x3 = 8, x1 = 4, giving (4, 0, 4, 0), a BFS with Z = −8.
(x1, x4): x4 = 8, x1 = 12 − 24 = −12, basic but not feasible. (x2, x4): x2 + 3x4 = 12, x2 + x4 = 8 give (0, 6, 0, 2), a BFS with Z = 20. (x3, x4): 2x3 + 3x4 = 12, 2x3 + x4 = 8 give (0, 0, 3, 2), a BFS with Z = −1.
So there are 5 basic solutions; 4 are basic feasible, and all 4 are non-degenerate (every basic variable is positive). There is no degenerate BFS.
(ii) The second equation with x ≥ 0 gives x2, x4 ≤ 8 and x3 ≤ 4, and the first gives x1 ≤ 12. So the feasible set is non-empty, closed and bounded, and the linear Z attains a finite maximum, at a BFS (fundamental theorem).
Comparing Z at the four BFS, the maximum 20 is attained at (4, 8, 0, 0) and (0, 6, 0, 2), hence at every λ(4, 8, 0, 0) + (1 − λ)(0, 6, 0, 2), 0 ≤ λ ≤ 1.
Answer: 5 basic solutions, 4 of them non-degenerate BFS and none degenerate; the optimal BFS are (4, 8, 0, 0) and (0, 6, 0, 2), with Max Z = 20.Final answer: 5 basic solutions, 4 of them non-degenerate BFS and none degenerate; the optimal BFS are (4, 8, 0, 0) and (0, 6, 0, 2), with Max Z = 20.
Question 4 (50 marks)
Q4(a) · Abstract Algebra · 15 marks≈ 11 min
Integral Domains, PID, Euclidean Domains, UFD, Fields and Quotient Fields
Do the following sets form integral domains with respect to ordinary addition and multiplication? If so, state if they are fields:
(i) The set of numbers of the form b√2 with b rational
(ii) The set of even integers
(iii) The set of positive integers (5 + 6 + 4 = 15)हिंदी में प्रश्न
क्या निम्नलिखित समुच्चय साधारण योग तथा गुणन के अन्तर्गत पूर्णांकीय प्रांत बनाते हैं? यदि ऐसा है, तो बताइए कि क्या वे क्षेत्र हैं:
(i) b√2 के रूप की संख्याओं का समुच्चय, जहाँ b परिमेय संख्या है
(ii) सम पूर्णांकों का समुच्चय
(iii) धनात्मक पूर्णांकों का समुच्चय (5 + 6 + 4 = 15)Worked solution
(i) S = {b√2 : b ∈ ℚ} is closed under + and −, but √2·√2 = 2, and 2 = b√2 would need b = √2 ∉ ℚ. So S is not closed under multiplication: it is not a ring, hence not an integral domain.
(ii) 2ℤ is a commutative ring without zero divisors (ab = 0 in ℤ forces a = 0 or b = 0), but it has no unity: 2e = 2 forces e = 1 ∉ 2ℤ.
So under the usual definition (commutative ring with unity and no zero divisors) 2ℤ is not an integral domain; under Herstein's definition, which does not require unity, it is one. Either way it is not a field, as 2 has no inverse in 2ℤ.
(iii) ℤ+ contains no additive identity 0 and no additive inverses, so (ℤ+, +) is not a group and ℤ+ is not a ring.
Answer: (i) no (not closed under multiplication); (ii) not an integral domain in the sense requiring unity (it is one in Herstein's sense), and not a field; (iii) no.Final answer: (i) no (not closed under multiplication); (ii) not an integral domain in the sense requiring unity (it is one in Herstein's sense), and not a field; (iii) no.
Q4(b) · Real Analysis · 15 marks≈ 11 min
Functions of Several Variables: Partial Derivatives, Maxima and Minima
Find the absolute maximum and minimum values of the function f(x, y) = x2 + 3y2 − y over the region x2 + 2y2 ≤ 1.
हिंदी में प्रश्न
क्षेत्र x2 + 2y2 ≤ 1 पर फलन f(x, y) = x2 + 3y2 − y के निरपेक्ष (ऐब्सोल्यूट) उच्चतम तथा निम्नतम मानों को ज्ञात कीजिए।
Worked solution
D = {x2 + 2y2 ≤ 1} is closed and bounded and f is continuous, so absolute extrema exist.
Interior: fx = 2x = 0 and fy = 6y − 1 = 0 give (0, 1/6), which lies inside D, with f(0, 1/6) = 3/36 − 1/6 = −1/12.
Boundary: x2 = 1 − 2y2 with −1/√2 ≤ y ≤ 1/√2, so f = 1 − 2y2 + 3y2 − y = y2 − y + 1 = φ(y).
φ′(y) = 2y − 1 = 0 at y = 1/2 (x = ±1/√2), where φ = 3/4. At the end points: y = 1/√2 (x = 0) gives 3/2 − 1/√2 ≈ 0.793, and y = −1/√2 (x = 0) gives 3/2 + 1/√2 ≈ 2.207.
Comparing −1/12, 3/4, 3/2 − 1/√2 and 3/2 + 1/√2 (a grid search over D confirms the result):
Answer: absolute maximum 3/2 + 1/√2 at (0, −1/√2); absolute minimum −1/12 at (0, 1/6).Final answer: absolute maximum 3/2 + 1/√2 at (0, −1/√2); absolute minimum −1/12 at (0, 1/6).
Q4(c) · Linear Programming · 20 marks≈ 14 min
Duality in Linear Programming
Solve the following linear programming problem by the simplex method. Write its dual. Also, write the optimal solution of the dual from the optimal table of the given problem:
Maximize Z = 2x1 − 4x2 + 5x3
subject to
x1 + 4x2 − 2x3 ≤ 2
−x1 + 2x2 + 3x3 ≤ 1
x1, x2, x3 ≥ 0हिंदी में प्रश्न
निम्नलिखित रैखिक प्रोग्रामन समस्या को एकधा विधि द्वारा हल कीजिए। इसकी द्वैती समस्या लिखिए। दी गई समस्या की इष्टतम सारणी से द्वैती समस्या का इष्टतम हल भी लिखिए:
अधिकतमीकरण कीजिए Z = 2x1 − 4x2 + 5x3
बशर्ते कि
x1 + 4x2 − 2x3 ≤ 2
−x1 + 2x2 + 3x3 ≤ 1
x1, x2, x3 ≥ 0
Question 5 (50 marks)
Q5(a) · Partial Differential Equations · 10 marks≈ 7 min
Lagrange's Method for Quasilinear First-Order PDEs
Solve the partial differential equation (y2 + z2 − x2)p − 2xyq + 2xz = 0 where p = ∂z/∂x and q = ∂z/∂y.
हिंदी में प्रश्न
आंशिक अवकल समीकरण (y2 + z2 − x2)p − 2xyq + 2xz = 0 जहाँ p = ∂z/∂x तथा q = ∂z/∂y, को हल कीजिए।
Q5(b) · Partial Differential Equations · 10 marks≈ 7 min
Linear Second-Order PDEs with Constant Coefficients
Solve (D2 + DD′ − 2D′2)u = ex+y, where D = ∂/∂x and D′ = ∂/∂y.
हिंदी में प्रश्न
(D2 + DD′ − 2D′2)u = ex+y को हल कीजिए, जहाँ D = ∂/∂x तथा D′ = ∂/∂y.
Q5(c) · Numerical Analysis and Computer Programming · 10 marks≈ 7 min
Computer Systems, Logic Gates, Boolean Algebra and Algorithms
Find the principal (or canonical) disjunctive normal form in three variables p, q, r for the Boolean expression ((p ∧ q) → r) ∨ ((p ∧ q) → ¬r). Is the given Boolean expression a contradiction or a tautology?
हिंदी में प्रश्न
बूलीय व्यंजक ((p ∧ q) → r) ∨ ((p ∧ q) → ¬r) के लिए तीन चरों p, q, r में मुख्य (अथवा विहित) वियोजनीय (डिसजंक्टिव) प्रसामान्य रूप ज्ञात कीजिए। क्या दिया गया बूलीय व्यंजक एक विरोध है या कि एक पुनरावृत्ति है?
Worked solution
(p ∧ q) → r ≡ ¬(p ∧ q) ∨ r and (p ∧ q) → ¬r ≡ ¬(p ∧ q) ∨ ¬r.
Their disjunction is ¬(p ∧ q) ∨ r ∨ ¬r = ¬(p ∧ q) ∨ T = T, so the expression is true in all eight rows.
PDNF = the sum of all eight minterms: (p ∧ q ∧ r) ∨ (p ∧ q ∧ ¬r) ∨ (p ∧ ¬q ∧ r) ∨ (p ∧ ¬q ∧ ¬r) ∨ (¬p ∧ q ∧ r) ∨ (¬p ∧ q ∧ ¬r) ∨ (¬p ∧ ¬q ∧ r) ∨ (¬p ∧ ¬q ∧ ¬r).
Answer: the expression is a tautology (not a contradiction).Final answer: the expression is a tautology (not a contradiction).
Q5(d) · Mechanics and Fluid Dynamics · 10 marks≈ 7 min
Two-Dimensional and Axisymmetric Motion: Sources, Sinks and Vortices
Consider a uniform flow U0 in the positive x-direction. A cylinder of radius a is located at the origin. Find the stream function and the velocity potential. Find also the stagnation points.
हिंदी में प्रश्न
धनात्मक x-दिशा में एक एकसमान प्रवाह U0 का विचार कीजिए। त्रिज्या a का एक बेलन मूलबिन्दु पर स्थित है। प्रवाह फलन तथा वेग विभव ज्ञात कीजिए। प्रगतिरोध (स्टैग्नेशन) बिन्दु भी ज्ञात कीजिए।
Q5(e) · Mechanics and Fluid Dynamics · 10 marks≈ 7 min
Moment of Inertia and Motion of Rigid Bodies in Two Dimensions
Calculate the moment of inertia of a solid uniform hemisphere x2 + y2 + z2 = a2, z ≥ 0 with mass m about the OZ-axis.
हिंदी में प्रश्न
भार m का एक ठोस एकसमान गोलार्ध x2 + y2 + z2 = a2, z ≥ 0 के, OZ-अक्ष के इर्द-गिर्द, जड़त्व-आघूर्ण का परिकलन कीजिए।
Worked solution
Slice into discs of radius √(a2 − z2) and thickness dz, mass ρπ(a2 − z2)dz, each with moment (1/2)(mass)(radius)2 about OZ.
I = (ρπ/2)∫0a(a2 − z2)2 dz = (ρπ/2)(8a5/15) = 4ρπa5/15, and m = (2/3)πa3ρ.
Answer: IOZ = 2ma2/5, the same as for a whole sphere about a diameter (each half contributes equally).Final answer: IOZ = 2ma2/5, the same as for a whole sphere about a diameter (each half contributes equally).
Question 6 (50 marks)
Q6(a) · Partial Differential Equations · 15 marks≈ 11 min
Lagrange's Method for Quasilinear First-Order PDEs
Solve for the general solution p cos(x + y) + q sin(x + y) = z, where p = ∂z/∂x and q = ∂z/∂y.
हिंदी में प्रश्न
p cos(x + y) + q sin(x + y) = z, जहाँ p = ∂z/∂x तथा q = ∂z/∂y, को व्यापक हल के लिए हल कीजिए।
Worked solution
Auxiliary equations dx/cos(x + y) = dy/sin(x + y) = dz/z; adding and subtracting the first two, d(x + y)/(cos(x + y) + sin(x + y)) = d(x − y)/(cos(x + y) − sin(x + y)).
With w = x + y: d(x − y) = [(cos w − sin w)/(cos w + sin w)] dw, whose right side is d log(cos w + sin w); so x − y − log(cos w + sin w) = const, i.e. ey − x[cos(x + y) + sin(x + y)] = c1.
Also dz/z = dw/(cos w + sin w) = dw/(√2 sin(w + π/4)), so log z = (1/√2) log tan(w/2 + π/8) + const.
Hence z[tan((x + y)/2 + π/8)]−1/√2 = c2.
Answer: z = [tan((x + y)/2 + π/8)]1/√2 f(ey − x[cos(x + y) + sin(x + y)]), f arbitrary (checked by substitution).Final answer: z = [tan((x + y)/2 + π/8)]1/√2 f(ey − x[cos(x + y) + sin(x + y)]), f arbitrary (checked by substitution).
Q6(b) · Mechanics and Fluid Dynamics · 15 marks≈ 11 min
Hamilton's Equations
Solve the plane pendulum problem using the Hamiltonian approach and show that H is a constant of motion.
हिंदी में प्रश्न
हैमिल्टोनियन उपागम का इस्तेमाल करते हुए समतल लोलक समस्या को हल कीजिए तथा दर्शाइए कि H गति का एक स्थिरांक है।
Worked solution
q = θ, T = (1/2)ml2θ̇2, V = −mgl cos θ, p = ml2θ̇, H = p2/(2ml2) − mgl cos θ.
θ̇ = p/(ml2), ṗ = −mgl sin θ, so θ̈ + (g/l)sin θ = 0; for small θ, θ = θ0cos(√(g/l) t + ε).
dH/dt = (∂H/∂θ)θ̇ + (∂H/∂p)ṗ + ∂H/∂t = (∂H/∂θ)(∂H/∂p) − (∂H/∂p)(∂H/∂θ) + 0 = 0.
Answer: H = p2/(2ml2) − mgl cos θ is a constant of motion (the energy); for amplitude α it gives θ̇2 = (2g/l)(cos θ − cos α).Final answer: H = p2/(2ml2) − mgl cos θ is a constant of motion (the energy); for amplitude α it gives θ̇2 = (2g/l)(cos θ − cos α).
Q6(c) · Numerical Analysis and Computer Programming · 20 marks≈ 14 min
Interpolation: Newton's Forward, Backward and Lagrange's Formulae
Find the Lagrange interpolating polynomial that fits the following data:
x : −1, 2, 3, 4
f(x) : −1, 11, 31, 69
Find f(1.5).हिंदी में प्रश्न
लग्रांज अंतर्वेशी बहुपद ज्ञात कीजिए, जो निम्नलिखित आँकड़ों को समंजन करता है:
x : −1, 2, 3, 4
f(x) : −1, 11, 31, 69
f(1.5) ज्ञात कीजिए।
Question 7 (50 marks)
Q7(a) · Partial Differential Equations · 15 marks≈ 11 min
Vibrating String, Heat Equation and Laplace Equation
Find the solution of the initial-boundary value problem
ut − uxx + u = 0, 0 < x < l, t > 0
u(0, t) = u(l, t) = 0, t ≥ 0
u(x, 0) = x(l − x), 0 < x < lहिंदी में प्रश्न
प्रारंभिक-परिसीमा मान समस्या
ut − uxx + u = 0, 0 < x < l, t > 0
u(0, t) = u(l, t) = 0, t ≥ 0
u(x, 0) = x(l − x), 0 < x < l
का हल ज्ञात कीजिए।Worked solution
Put u = e−tv: ut − uxx + u = e−t(vt − vxx), so vt = vxx with v(0, t) = v(l, t) = 0, v(x, 0) = x(l − x).
v = ∑bne−n2π2t/l2sin(nπx/l) with bn = (2/l)∫0l x(l − x)sin(nπx/l) dx = 4l2(1 − (−1)n)/(n3π3).
So bn = 8l2/(n3π3) for odd n and 0 for even n.
Answer: u(x, t) = (8l2/π3)∑n odd (1/n3)e−(1 + n2π2/l2)tsin(nπx/l).Final answer: u(x, t) = (8l2/π3)∑n odd (1/n3)e−(1 + n2π2/l2)tsin(nπx/l).
Q7(b) · Numerical Analysis and Computer Programming · 15 marks≈ 11 min
Numerical Integration and Numerical Solution of ODEs
Solve the initial value problem dy/dx = x(y − x), y(2) = 3 in the interval [2, 2.4] using the Runge-Kutta fourth-order method with step size h = 0.2.
हिंदी में प्रश्न
पग आकार h = 0.2 के साथ रंगे-कुट्टा चतुष्कोटि विधि का इस्तेमाल करते हुए प्रारंभिक मान समस्या dy/dx = x(y − x), y(2) = 3 को अंतराल [2, 2.4] में हल कीजिए।
Q7(c) · Mechanics and Fluid Dynamics · 20 marks≈ 14 min
Hamilton's Equations
A Hamiltonian of a system with one degree of freedom has the form H = p2/(2α) − bqpe−αt + (bα/2)q2e−αt(α + be−αt) + (k/2)q2 where α, b, k are constants, q is the generalized coordinate and p is the corresponding generalized momentum.
(i) Find a Lagrangian corresponding to this Hamiltonian.
(ii) Find an equivalent Lagrangian that is not explicitly dependent on time. (10 + 10 = 20)हिंदी में प्रश्न
स्वातंत्र्य कोटि एक के एक निकाय के एक हैमिल्टोनियन का रूप निम्नलिखित है: H = p2/(2α) − bqpe−αt + (bα/2)q2e−αt(α + be−αt) + (k/2)q2 जहाँ α, b, k स्थिरांक हैं, q व्यापकीकृत निर्देशांक है तथा p संगत व्यापकीकृत संवेग है।
(i) इस हैमिल्टोनियन के संगत एक लग्रांजी ज्ञात कीजिए।
(ii) एक तुल्य लग्रांजी ज्ञात कीजिए, जो कि समय पर स्पष्ट रूप से आश्रित नहीं है। (10 + 10 = 20)Worked solution
Write E = e−αt. q̇ = ∂H/∂p = p/α − bqE, so p = α(q̇ + bqE); substituting, H = (α/2)q̇2 + (α2b/2)Eq2 + (k/2)q2 (the b2E2 terms cancel).
(i) L = pq̇ − H = (α/2)q̇2 + αbe−αtqq̇ − (α2b/2)e−αtq2 − (k/2)q2.
(ii) d/dt[(αb/2)e−αtq2] = αbe−αtqq̇ − (α2b/2)e−αtq2, so L = (α/2)q̇2 − (k/2)q2 + dF/dt with F = (αb/2)e−αtq2.
Lagrangians that differ by a total time derivative dF(q, t)/dt give the same equations of motion.
Answer: (i) the L above; (ii) L′ = (α/2)q̇2 − (k/2)q2, a harmonic oscillator αq̈ + kq = 0.Final answer: (i) the L above; (ii) L′ = (α/2)q̇2 − (k/2)q2, a harmonic oscillator αq̈ + kq = 0.
Question 8 (50 marks)
Q8(a) · Partial Differential Equations · 15 marks≈ 11 min
Classification and Canonical Forms of Second-Order PDEs
Reduce the second-order partial differential equation x2 ∂2u/∂x2 − 2xy ∂2u/∂x∂y + y2 ∂2u/∂y2 + x ∂u/∂x + y ∂u/∂y = 0 into canonical form. Hence, find its general solution.
हिंदी में प्रश्न
द्वितीय-कोटि आंशिक अवकल समीकरण x2 ∂2u/∂x2 − 2xy ∂2u/∂x∂y + y2 ∂2u/∂y2 + x ∂u/∂x + y ∂u/∂y = 0 को विहित रूप में समानीत कीजिए तथा इसका व्यापक हल ज्ञात कीजिए।
Worked solution
A = x2, B = −2xy, C = y2: B2 − 4AC = 0, parabolic; the characteristic equation is (x dy/dx + y)2 = 0, so ξ = xy; take η = x (J = −x ≠ 0).
ux = yuξ + uη, uy = xuξ, uxx = y2uξξ + 2yuξη + uηη, uxy = uξ + xyuξξ + xuξη, uyy = x2uξξ.
Second-order part = x2uηη − 2xyuξ; first-order part xux + yuy = 2xyuξ + xuη; the sum is x2uηη + xuη = 0.
Canonical form ηuηη + uη = 0, i.e. ∂(ηuη)/∂η = 0, so ηuη = f(ξ) and u = f(ξ)log η + g(ξ).
Answer: canonical form uηη = −(1/η)uη with ξ = xy, η = x; general solution u = f(xy)log x + g(xy).Final answer: canonical form uηη = −(1/η)uη with ξ = xy, η = x; general solution u = f(xy)log x + g(xy).
Q8(b) · Numerical Analysis and Computer Programming · 15 marks≈ 11 min
Systems of Linear Equations: Direct and Iterative Methods
Find the solution of the system
10x1 − 2x2 − x3 − x4 = 3
−2x1 + 10x2 − x3 − x4 = 15
−x1 − x2 + 10x3 − 2x4 = 27
−x1 − x2 − 2x3 + 10x4 = −9
using Gauss-Seidel method (make four iterations).हिंदी में प्रश्न
गाउस-सीडेल विधि का इस्तेमाल करते हुए निकाय
10x1 − 2x2 − x3 − x4 = 3
−2x1 + 10x2 − x3 − x4 = 15
−x1 − x2 + 10x3 − 2x4 = 27
−x1 − x2 − 2x3 + 10x4 = −9
का हल ज्ञात कीजिए (चार पुनरावृत्ति कीजिए)।Q8(c) · Mechanics and Fluid Dynamics · 20 marks≈ 14 min
Two-Dimensional and Axisymmetric Motion: Sources, Sinks and Vortices
In an axisymmetric motion, show that stream function exists due to equation of continuity. Express the velocity components in terms of the stream function. Find the equation satisfied by the stream function if the flow is irrotational.
हिंदी में प्रश्न
दर्शाइए कि अक्षतः सममित गति में सांतत्य समीकरण के कारण धारा फलन का अस्तित्व है। धारा फलन के पदों में वेग घटकों को व्यक्त कीजिए। यदि प्रवाह अघूर्णनी है, तो समीकरण ज्ञात कीजिए जो कि धारा फलन द्वारा सन्तुष्ट होती है।
Worked solution
Continuity (axisymmetric, incompressible) ∂(r2sin θ qr)/∂r + ∂(r sin θ qθ)/∂θ = 0 is the integrability condition for ψ with ∂ψ/∂θ = −r2sin θ qr, ∂ψ/∂r = r sin θ qθ; so ψ exists because of continuity.
Velocity components: qr = −(1/(r2sin θ))∂ψ/∂θ, qθ = (1/(r sin θ))∂ψ/∂r; the flux through the circle cut out by a stream surface is 2π|ψ| when ψ = 0 on the axis.
Irrotational: (1/r)[∂(rqθ)/∂r − ∂qr/∂θ] = (1/(r sin θ))[∂2ψ/∂r2 + (sin θ/r2)∂/∂θ((1/sin θ)∂ψ/∂θ)] = 0.
Answer: ∂2ψ/∂r2 + (sin θ/r2)∂/∂θ((1/sin θ)∂ψ/∂θ) = 0 (books that define ψ with the opposite sign get the same equation).Final answer: ∂2ψ/∂r2 + (sin θ/r2)∂/∂θ((1/sin θ)∂ψ/∂θ) = 0 (books that define ψ with the opposite sign get the same equation).