UPSC Mains 2014 Mathematics Optional Question Papers with Solutions
All 56 questions of the 2014 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 2014 Mathematics Paper I
Question 1 (50 marks)
Q1(a) · Linear Algebra · 10 marks≈ 7 min
Vector Spaces, Subspaces, Basis and Dimension
Find one vector in R3 which generates the intersection of V and W, where V is the xy plane and W is the space generated by the vectors (1, 2, 3) and (1, −1, 1).
हिंदी में प्रश्न
एक सदिश R3 में ज्ञात कीजिए, जो कि V तथा W के प्रतिच्छेद का जनक है, जहाँ कि V एक xy समतल है तथा W सदिश (1, 2, 3) तथा सदिश (1, −1, 1) के द्वारा जनित किया गया आकाश (स्पेस) है।
Worked solution
V = {(x, y, 0)}. A vector of W is a(1, 2, 3) + b(1, −1, 1) = (a + b, 2a − b, 3a + b); it lies in V if and only if 3a + b = 0, that is b = −3a.
Then the vector is (a − 3a, 2a + 3a, 0) = a(−2, 5, 0). So V ∩ W = L{(−2, 5, 0)}, a line; check: dim(V + W) = 2 + 2 − 1 = 3, and indeed V + W contains the xy-plane and (1, 2, 3).
Answer: (−2, 5, 0), or any non-zero multiple of it, generates V ∩ W.Final answer: (−2, 5, 0), or any non-zero multiple of it, generates V ∩ W.
Q1(b) · Linear Algebra · 10 marks≈ 7 min
Algebra of Matrices, Echelon Form, Rank, Inverse, Congruence and Similarity
Using elementary row or column operations, find the rank of the matrix [[0, 1, −3, −1], [0, 0, 1, 1], [3, 1, 0, 2], [1, 1, −2, 0]].
हिंदी में प्रश्न
प्रारंभिक पंक्ति या स्तंभ संक्रियाओं का प्रयोग करके, आव्यूह (मैट्रिक्स) [[0, 1, −3, −1], [0, 0, 1, 1], [3, 1, 0, 2], [1, 1, −2, 0]] की कोटि ज्ञात कीजिए।
Q1(c) · Calculus · 10 marks≈ 7 min
Mean Value Theorems, Taylor's Theorem and Indeterminate Forms
Prove that between two real roots of ex cos x + 1 = 0, a real root of ex sin x + 1 = 0 lies.
हिंदी में प्रश्न
सिद्ध कीजिए कि ex cos x + 1 = 0 के दो वास्तविक मूलों के बीच ex sin x + 1 = 0 का एक वास्तविक मूल स्थित है।
Worked solution
Let φ(x) = e−x + cos x. Since ex > 0, ex cos x + 1 = exφ(x) = 0 ⇔ φ(x) = 0.
Let α < β be real roots of ex cos x + 1 = 0. Then φ(α) = φ(β) = 0, and φ is continuous on [α, β] and differentiable on (α, β); by Rolle's theorem φ′(c) = 0 for some c ∈ (α, β).
φ′(x) = −e−x − sin x = −e−x(1 + ex sin x), so φ′(c) = 0 gives ec sin c + 1 = 0.
Answer: c ∈ (α, β) is a root of ex sin x + 1 = 0, as required.Final answer: c ∈ (α, β) is a root of ex sin x + 1 = 0, as required.
Q1(d) · Calculus · 10 marks≈ 7 min
Riemann Integral, Indefinite Integrals and Improper Integrals
Evaluate: ∫01 [loge(1 + x)/(1 + x2)] dx
हिंदी में प्रश्न
मूल्यांकन कीजिए: ∫01 [loge(1 + x)/(1 + x2)] dx
Worked solution
Put x = tan θ, dx = sec2θ dθ, and 1 + x2 = sec2θ: I = ∫0π/4 log(1 + tan θ) dθ.
Replace θ by π/4 − θ: 1 + tan(π/4 − θ) = 1 + (1 − tan θ)/(1 + tan θ) = 2/(1 + tan θ).
So I = ∫0π/4 [log 2 − log(1 + tan θ)] dθ = (π/4) log 2 − I, and 2I = (π/4) log 2.
Answer: (π/8) loge2.Final answer: (π/8) loge2.
Q1(e) · Analytic Geometry · 10 marks≈ 7 min
Cone and Cylinder
Examine whether the plane x + y + z = 0 cuts the cone yz + zx + xy = 0 in perpendicular lines.
हिंदी में प्रश्न
परीक्षण कीजिए कि क्या समतल x + y + z = 0 शंकु yz + zx + xy = 0 को समकोणीय (लंब) रेखाओं में काटता है।
Question 2 (50 marks)
Q2(a) · Linear Algebra · 15 marks≈ 11 min
Vector Spaces, Subspaces, Basis and Dimension
Let V and W be the following subspaces of R4:
V = {(a, b, c, d) : b − 2c + d = 0} and
W = {(a, b, c, d) : a = d, b = 2c}.
Find a basis and the dimension of (i) V, (ii) W, (iii) V ∩ W.हिंदी में प्रश्न
मान लीजिए कि V और W निम्न उपसमष्टियाँ हैं R4 की:
V = {(a, b, c, d) : b − 2c + d = 0} और
W = {(a, b, c, d) : a = d, b = 2c}.
(i) V, (ii) W, (iii) V ∩ W का एक आधार और विस्तार ज्ञात कीजिए।Worked solution
(i) V: b = 2c − d with a, c, d free, so (a, b, c, d) = a(1, 0, 0, 0) + c(0, 2, 1, 0) + d(0, −1, 0, 1). These three vectors are independent (compare the 1st, 3rd and 4th coordinates), so dim V = 3.
(ii) W: a = d and b = 2c with c, d free, so (a, b, c, d) = c(0, 2, 1, 0) + d(1, 0, 0, 1); these are independent and dim W = 2.
(iii) V ∩ W: put a = d, b = 2c into b − 2c + d = 0 to get d = 0, hence a = 0. So V ∩ W = {(0, 2c, c, 0)} = L{(0, 2, 1, 0)}, of dimension 1.
Check: dim(V + W) = 3 + 2 − 1 = 4, so V + W = ℝ4; this agrees with (1, 0, 0, 1) ∈ W not lying in V.
Answer: dim V = 3, basis {(1, 0, 0, 0), (0, 2, 1, 0), (0, −1, 0, 1)}; dim W = 2, basis {(0, 2, 1, 0), (1, 0, 0, 1)}; dim(V ∩ W) = 1, basis {(0, 2, 1, 0)}.Final answer: dim V = 3, basis {(1, 0, 0, 0), (0, 2, 1, 0), (0, −1, 0, 1)}; dim W = 2, basis {(0, 2, 1, 0), (1, 0, 0, 1)}; dim(V ∩ W) = 1, basis {(0, 2, 1, 0)}.
Q2(b) · Linear Algebra · 20 marks≈ 14 min
Systems of Linear Equations
(i) Investigate the values of λ and μ so that the equations x + y + z = 6, x + 2y + 3z = 10, x + 2y + λz = μ have (1) no solution, (2) a unique solution, (3) an infinite number of solutions. (10)
(ii) Verify Cayley-Hamilton theorem for the matrix A = [[1, 4], [2, 3]] and hence find its inverse. Also, find the matrix represented by A5 − 4A4 − 7A3 + 11A2 − A − 10I. (10)हिंदी में प्रश्न
(i) λ तथा μ के मान जाँच कीजिए ताकि समीकरण x + y + z = 6, x + 2y + 3z = 10, x + 2y + λz = μ का (1) कोई हल नहीं है, (2) एक अद्वितीय हल है, (3) अपरिमित हल हैं। (10)
(ii) आव्यूह A = [[1, 4], [2, 3]] के लिए कैली-हैमिल्टन प्रमेय सत्यापित कीजिए और अतएव इसका व्युत्क्रम ज्ञात कीजिए। साथ ही, A5 − 4A4 − 7A3 + 11A2 − A − 10I के द्वारा निरूपित आव्यूह भी ज्ञात कीजिए। (10)Q2(c) · Calculus · 15 marks≈ 11 min
Double and Triple Integrals; Areas, Surfaces and Volumes
By using the transformation x + y = u, y = uv, evaluate the integral ∬ {xy(1 − x − y)}1/2 dx dy taken over the area enclosed by the straight lines x = 0, y = 0 and x + y = 1.
हिंदी में प्रश्न
रूपांतर x + y = u, y = uv का प्रयोग करते हुए, समाकल ∬ {xy(1 − x − y)}1/2 dx dy का सीधी रेखाएँ x = 0, y = 0 तथा x + y = 1 के द्वारा परिबद्ध क्षेत्र पर मूल्यांकन कीजिए।
Worked solution
x + y = u, y = uv give x = u(1 − v), y = uv; the triangle x ≥ 0, y ≥ 0, x + y ≤ 1 becomes the square 0 ≤ u ≤ 1, 0 ≤ v ≤ 1.
∂(x, y)/∂(u, v) = det [[1 − v, −u], [v, u]] = u(1 − v) + uv = u, so dx dy = u du dv.
xy(1 − x − y) = u2v(1 − v)(1 − u), so the integral is ∫01∫01 u2(1 − u)1/2v1/2(1 − v)1/2 du dv = β(3, 3/2)β(3/2, 3/2).
β(3, 3/2) = Γ(3)Γ(3/2)/Γ(9/2) = 2(√π/2)/((105/16)√π) = 16/105 and β(3/2, 3/2) = (√π/2)2/Γ(3) = π/8.
Answer: (16/105)(π/8) = 2π/105.Final answer: (16/105)(π/8) = 2π/105.
Question 3 (50 marks)
Q3(a) · Calculus · 15 marks≈ 11 min
Maxima and Minima, Asymptotes and Curve Tracing
Find the height of the cylinder of maximum volume that can be inscribed in a sphere of radius a.
हिंदी में प्रश्न
एक ऐसे महत्तम आयतन के बेलन की ऊँचाई ज्ञात कीजिए जो कि a त्रिज्या के गोले के भीतर आ सके।
Q3(b) · Calculus · 20 marks≈ 14 min
Functions of Several Variables: Partial Derivatives, Extrema, Lagrange Multipliers and Jacobians
Find the maximum or minimum values of x2 + y2 + z2 subject to the conditions ax2 + by2 + cz2 = 1 and lx + my + nz = 0. Interpret the result geometrically.
हिंदी में प्रश्न
ax2 + by2 + cz2 = 1 तथा lx + my + nz = 0 के द्वारा प्रतिबंधित x2 + y2 + z2 के अधिकतम या निम्नतम मान ज्ञात कीजिए। परिणाम की ज्यामितीय व्याख्या कीजिए।
Worked solution
F = x2 + y2 + z2 + λ1(ax2 + by2 + cz2 − 1) + λ2(lx + my + nz). Fx = 0 gives 2x + 2λ1ax + λ2l = 0, and similarly 2y + 2λ1by + λ2m = 0, 2z + 2λ1cz + λ2n = 0.
Multiply by x, y, z and add, using both constraints: 2u + 2λ1 = 0, where u = x2 + y2 + z2 is the stationary value. So λ1 = −u.
Then 2x(1 − au) = −λ2l, so x = −λ2l/(2(1 − au)), y = −λ2m/(2(1 − bu)), z = −λ2n/(2(1 − cu)) (in the generic case 1 − au, 1 − bu, 1 − cu ≠ 0, and λ2 ≠ 0 since (x, y, z) ≠ 0).
Substituting into lx + my + nz = 0: l2/(1 − au) + m2/(1 − bu) + n2/(1 − cu) = 0, a quadratic in u (after clearing denominators) whose two roots are the maximum and minimum values.
Interpretation: ax2 + by2 + cz2 = 1 is a central conicoid and lx + my + nz = 0 a plane through its centre. Their section is a central conic, and x2 + y2 + z2 is the squared distance from the centre, so the two roots are the squares of the semi-axes of this section.
Answer: The extreme values u are the roots of l2/(1 − au) + m2/(1 − bu) + n2/(1 − cu) = 0; they are the squares of the semi-axes of the central section.Final answer: The extreme values u are the roots of l2/(1 − au) + m2/(1 − bu) + n2/(1 − cu) = 0; they are the squares of the semi-axes of the central section.
Q3(c) · Linear Algebra · 15 marks≈ 11 min
Eigenvalues, Eigenvectors and the Cayley-Hamilton Theorem
(i) Let A = [[−2, 2, −3], [2, 1, −6], [−1, −2, 0]]. Find the eigen values of A and the corresponding eigen vectors. (8)
(ii) Prove that the eigen values of a unitary matrix have absolute value 1. (7)हिंदी में प्रश्न
(i) मान लीजिए कि A = [[−2, 2, −3], [2, 1, −6], [−1, −2, 0]]. A के आइगेन मानों और संगत आइगेन सदिशों को ज्ञात कीजिए। (8)
(ii) सिद्ध कीजिए कि ऐकिक आव्यूह के आइगेन मानों का निरपेक्ष मान 1 होता है। (7)
Question 4 (50 marks)
Q4(a) · Analytic Geometry · 20 marks≈ 14 min
The Sphere
(i) Find the co-ordinates of the points on the sphere x2 + y2 + z2 − 4x + 2y = 4, the tangent planes at which are parallel to the plane 2x − y + 2z = 1. (10)
(ii) Prove that the equation ax2 + by2 + cz2 + 2ux + 2vy + 2wz + d = 0 represents a cone if u2/a + v2/b + w2/c = d. (10)हिंदी में प्रश्न
(i) गोलक x2 + y2 + z2 − 4x + 2y = 4 के बिन्दुओं के निर्देशांक ज्ञात कीजिए जिसके स्पर्शी समतल, समतल 2x − y + 2z = 1 के समांतर हैं। (10)
(ii) सिद्ध कीजिए कि समीकरण ax2 + by2 + cz2 + 2ux + 2vy + 2wz + d = 0 एक शंकु निरूपित करता है, यदि u2/a + v2/b + w2/c = d हो तो। (10)Q4(b) · Analytic Geometry · 15 marks≈ 11 min
Ellipsoid, Hyperboloids and Paraboloids
Show that the lines drawn from the origin parallel to the normals to the central conicoid ax2 + by2 + cz2 = 1, at its points of intersection with the plane lx + my + nz = p generate the cone p2(x2/a + y2/b + z2/c) = (lx/a + my/b + nz/c)2.
हिंदी में प्रश्न
दर्शाइए कि उद्गम (मूल-बिन्दु) से खींची हुई रेखाएँ, जो केन्द्रीय शांकवज ax2 + by2 + cz2 = 1 के समतल lx + my + nz = p के साथ प्रतिच्छेदन बिन्दुओं पर लम्बों के समान्तर हैं, शंकु p2(x2/a + y2/b + z2/c) = (lx/a + my/b + nz/c)2 का जनन करती हैं।
Worked solution
Let (α, β, γ) be a point of the conicoid on the plane: aα2 + bβ2 + cγ2 = 1 and lα + mβ + nγ = p. The normal there has direction (aα, bβ, cγ).
The parallel line through O is x/(aα) = y/(bβ) = z/(cγ) = 1/k (say), so α = kx/a, β = ky/b, γ = kz/c for a point (x, y, z) of it.
The plane condition gives k(lx/a + my/b + nz/c) = p; the conicoid gives k2(x2/a + y2/b + z2/c) = 1.
Eliminating k (square the first and divide by the second): p2(x2/a + y2/b + z2/c) = (lx/a + my/b + nz/c)2.
Answer: this is homogeneous of degree 2, so the lines generate the cone stated; hence proved.Final answer: this is homogeneous of degree 2, so the lines generate the cone stated; hence proved.
Q4(c) · Analytic Geometry · 15 marks≈ 11 min
Ellipsoid, Hyperboloids and Paraboloids
Find the equations of the two generating lines through any point (a cos θ, b sin θ, 0), of the principal elliptic section x2/a2 + y2/b2 = 1, z = 0, of the hyperboloid by the plane z = 0.
हिंदी में प्रश्न
अतिपरवलयज के समतल z = 0 के द्वारा मुख्य दीर्घवृत्तीय खण्ड x2/a2 + y2/b2 = 1, z = 0 के कोई बिन्दु (a cos θ, b sin θ, 0) में से दो जनक रेखाओं के समीकरण ज्ञात कीजिए।
Worked solution
The point P(a cos θ, b sin θ, 0) lies on x2/a2 + y2/b2 − z2/c2 = 1. Put (a cos θ + lr, b sin θ + mr, nr) in the equation.
Coefficient of r: (l cos θ)/a + (m sin θ)/b = 0, so l/a : m/b = sin θ : −cos θ, that is l = ka sin θ, m = −kb cos θ.
Coefficient of r2: l2/a2 + m2/b2 − n2/c2 = 0 gives k2 = n2/c2, so n = ±kc.
Answer: (x − a cos θ)/(a sin θ) = (y − b sin θ)/(−b cos θ) = z/(±c).Final answer: (x − a cos θ)/(a sin θ) = (y − b sin θ)/(−b cos θ) = z/(±c).
Question 5 (50 marks)
Q5(a) · Ordinary Differential Equations · 10 marks≈ 7 min
Formulation and First Order, First Degree Equations
Justify that a differential equation of the form [y + x f(x2 + y2)] dx + [y f(x2 + y2) − x] dy = 0, where f(x2 + y2) is an arbitrary function of (x2 + y2), is not an exact differential equation and 1/(x2 + y2) is an integrating factor for it. Hence solve this differential equation for f(x2 + y2) = (x2 + y2)2.
हिंदी में प्रश्न
उचित ठहराइए कि [y + x f(x2 + y2)] dx + [y f(x2 + y2) − x] dy = 0 की भांति अवकल समीकरण जहाँ कि f(x2 + y2), (x2 + y2) का एक स्वेच्छ फलन है, एक यथातथ अवकल समीकरण नहीं है तथा इसका 1/(x2 + y2) एक समाकलन गुणक है। अतएव इस अवकल समीकरण को f(x2 + y2) = (x2 + y2)2 के लिए हल कीजिए।
Worked solution
Write r2 = x2 + y2, f = f(r2). M = y + xf, N = yf − x: My = 1 + 2xyf′, Nx = 2xyf′ − 1, so My − Nx = 2 ≠ 0 and the equation is not exact.
Regroup: (y dx − x dy) + f(r2)(x dx + y dy) = 0. Dividing by r2: −d(tan−1(y/x)) + [f(r2)/r2]·(1/2)d(r2) = 0, each term an exact differential, so 1/(x2 + y2) is an IF.
For f = (x2 + y2)2: −d(tan−1(y/x)) + (1/2)r2 d(r2) = 0, so −tan−1(y/x) + r4/4 = c.
Answer: (x2 + y2)2 = 4 tan−1(y/x) + C.Final answer: (x2 + y2)2 = 4 tan−1(y/x) + C.
Q5(b) · Ordinary Differential Equations · 10 marks≈ 7 min
Formulation and First Order, First Degree Equations
Find the curve for which the part of the tangent cut-off by the axes is bisected at the point of tangency.
हिंदी में प्रश्न
वह वक्र ज्ञात कीजिए जिसके लिए कि स्पर्शी का भाग जो अक्षों द्वारा काटा हुआ है, स्पर्शी बिन्दु पर द्विभाजित हो।
Worked solution
The tangent at P(x, y) meets the axes at A(x − y/p, 0) and B(0, y − px). P is the midpoint of AB, so (x − y/p)/2 = x and (y − px)/2 = y.
Either condition gives p = −y/x, that is dy/y + dx/x = 0, so log xy = log c.
Answer: xy = c, a family of rectangular hyperbolas with the axes as asymptotes.Final answer: xy = c, a family of rectangular hyperbolas with the axes as asymptotes.
Q5(c) · Dynamics and Statics · 10 marks≈ 7 min
Rectilinear Motion and Simple Harmonic Motion
A particle is performing a simple harmonic motion (S.H.M.) of period T about a centre O with amplitude a and it passes through a point P, where OP = b in the direction OP. Prove that the time which elapses before it returns to P is (T/π) cos−1(b/a).
हिंदी में प्रश्न
एक कण एक सरल आवर्त गति (S.H.M.), केन्द्र O पर कालांक (आवर्तकाल) T, आयाम a के साथ प्रस्तुत कर रहा है तथा यह एक बिन्दु P से गुज़रता है, जहाँ कि OP = b, OP की दिशा में। सिद्ध कीजिए कि गुज़रा हुआ समय जब यह P पर वापस लौटता है, (T/π) cos−1(b/a) है।
Worked solution
Take n = 2π/T and measure t from the extreme position A on the side of P, so that x = a cos nt.
The particle passes P moving away from O, so it goes on to A, stops, and comes back to P; by symmetry the two halves of this journey take equal times.
Time from A to P: a cos nt1 = b, so t1 = (1/n) cos−1(b/a).
Required time = 2t1 = (2/n) cos−1(b/a) = (T/π) cos−1(b/a).
Answer: the particle returns to P after time (T/π) cos−1(b/a).Final answer: the particle returns to P after time (T/π) cos−1(b/a).
Q5(d) · Dynamics and Statics · 10 marks≈ 7 min
Virtual Work, Stability and Forces in Three Dimensions
Two equal uniform rods AB and AC, each of length l, are freely jointed at A and rest on a smooth fixed vertical circle of radius r. If 2θ is the angle between the rods, then find the relation between l, r and θ, by using the principle of virtual work.
हिंदी में प्रश्न
दो बराबर एकसमान दंड AB तथा AC, प्रत्येक l लंबे, A पर मुक्त रूप से जुड़े हैं तथा त्रिज्या r के चिकने नियत ऊर्ध्वाधर वृत्त पर टिके हैं। यदि दंडों के बीच कोण 2θ है, तो कल्पित कार्य सिद्धांत के प्रयोग से l, r तथा θ में संबंध ज्ञात कीजिए।
Worked solution
Let O be the centre of the circle; by symmetry A is vertically above O and each rod touches the circle, making angle θ with the vertical.
The distance from O to each rod is r, so OA = r cosec θ, and the centre of each rod is (l/2) cos θ below A.
Height of the centres of gravity above O: z = r cosec θ − (l/2) cos θ.
The reactions of the smooth circle and at the joint do no work, so the virtual work −2W δz = 0 gives dz/dθ = −r cos θ/sin2θ + (l/2) sin θ = 0.
Answer: l sin3θ = 2r cos θ.Final answer: l sin3θ = 2r cos θ.
Q5(e) · Vector Analysis · 10 marks≈ 7 min
Curves in Space: Curvature, Torsion and Serret-Frenet Formulae
Find the curvature vector at any point of the curve r̄(t) = t cos t î + t sin t ĵ, 0 ≤ t ≤ 2π. Give its magnitude also.
हिंदी में प्रश्न
वक्र r̄(t) = t cos t î + t sin t ĵ, 0 ≤ t ≤ 2π के किसी भी बिन्दु पर वक्रता सदिश ज्ञात कीजिए। उसका परिमाण भी बताइए।
Worked solution
r′ = (cos t − t sin t, sin t + t cos t, 0) with |r′|2 = 1 + t2, and r″ = (−2 sin t − t cos t, 2 cos t − t sin t, 0).
r′·r″ = t and r′ × r″ = (2 + t2)k.
Curvature vector κN = [(1 + t2)r″ − t r′]/(1 + t2)2.
The bracket has x-component (1 + t2)(−2 sin t − t cos t) − t(cos t − t sin t) = −(2 + t2)(sin t + t cos t) and y-component (1 + t2)(2 cos t − t sin t) − t(sin t + t cos t) = (2 + t2)(cos t − t sin t).
So κN = [(2 + t2)/(1 + t2)2][−(sin t + t cos t)i + (cos t − t sin t)j].
The last vector has length √(1 + t2), so κ = (2 + t2)/(1 + t2)3/2, agreeing with |r′ × r″|/|r′|3.
Answer: κN = [(2 + t2)/(1 + t2)2][−(sin t + t cos t)i + (cos t − t sin t)j], with magnitude (2 + t2)/(1 + t2)3/2.Final answer: κN = [(2 + t2)/(1 + t2)2][−(sin t + t cos t)i + (cos t − t sin t)j], with magnitude (2 + t2)/(1 + t2)3/2.
Question 6 (50 marks)
Q6(a) · Ordinary Differential Equations · 10 marks≈ 7 min
Second Order Equations with Variable Coefficients
Solve by the method of variation of parameters: dy/dx − 5y = sin x
हिंदी में प्रश्न
प्राचलों के विचरण की विधि के द्वारा हल कीजिए: dy/dx − 5y = sin x
Worked solution
The homogeneous equation y′ − 5y = 0 has yh = e5x. Put y = u(x)e5x: u′e5x = sin x, so u = ∫e−5x sin x dx.
∫e−5x sin x dx = e−5x(−5 sin x − cos x)/26 + c.
Answer: y = ce5x − (5 sin x + cos x)/26.Final answer: y = ce5x − (5 sin x + cos x)/26.
Q6(b) · Ordinary Differential Equations · 20 marks≈ 14 min
Second Order Equations with Variable Coefficients
Solve the differential equation: x3 d3y/dx3 + 3x2 d2y/dx2 + x dy/dx + 8y = 65 cos(loge x)
हिंदी में प्रश्न
अवकल समीकरण हल कीजिए: x3 d3y/dx3 + 3x2 d2y/dx2 + x dy/dx + 8y = 65 cos(loge x)
Q6(c) · Vector Analysis · 20 marks≈ 14 min
Gauss and Stokes' Theorems and Green's Identities
Evaluate by Stokes' theorem ∫Γ (y dx + z dy + x dz) where Γ is the curve given by x2 + y2 + z2 − 2ax − 2ay = 0, x + y = 2a, starting from (2a, 0, 0) and then going below the z-plane.
हिंदी में प्रश्न
स्टोक्स के प्रमेय के द्वारा मूल्यांकन कीजिए: ∫Γ (y dx + z dy + x dz) जहाँ Γ वक्र है x2 + y2 + z2 − 2ax − 2ay = 0, x + y = 2a, जो कि (2a, 0, 0) से शुरू होता है और फिर z-समतल के नीचे से होकर जाता है।
Worked solution
The sphere is (x − a)2 + (y − a)2 + z2 = 2a2; its centre (a, a, 0) lies on the plane x + y = 2a, so Γ is a great circle of radius √2a about (a, a, 0).
For F = (y, z, x), ∇ × F = −(i + j + k).
At the start (2a, 0, 0) the radius from the centre is (a, −a, 0) and the motion is along −k (below the xy-plane), so n ∥ (1, −1, 0) × (0, 0, −1) = (1, 1, 0) and n = (i + j)/√2.
Taking S as the flat disc bounded by Γ: ∬(∇ × F)·n dS = (−2/√2)(π · 2a2) = −2√2πa2.
Check with r = (a + a cos t, a − a cos t, −√2a sin t), 0 ≤ t ≤ 2π, which starts at (2a, 0, 0) going below the plane: the line integral is also −2√2πa2.
Answer: −2√2πa2 (the opposite sense would give +2√2πa2).Final answer: −2√2πa2 (the opposite sense would give +2√2πa2).
Question 7 (50 marks)
Q7(a) · Ordinary Differential Equations · 15 marks≈ 11 min
Second Order Equations with Variable Coefficients
Solve the following differential equation: x d2y/dx2 − 2(x + 1) dy/dx + (x + 2)y = (x − 2)e2x, when ex is a solution to its corresponding homogeneous differential equation.
हिंदी में प्रश्न
निम्नलिखित अवकल समीकरण हल कीजिए: x d2y/dx2 − 2(x + 1) dy/dx + (x + 2)y = (x − 2)e2x जबकि ex इसके संगत समघात अवकल समीकरण का एक हल है।
Q7(b) · Dynamics and Statics · 15 marks≈ 11 min
Constrained Motion, Work and Energy
A particle of mass m, hanging vertically from a fixed point by a light inextensible cord of length l, is struck by a horizontal blow which imparts to it a velocity 2√(gl). Find the velocity and height of the particle from the level of its initial position when the cord becomes slack.
हिंदी में प्रश्न
द्रव्यमान m का एक कण जो ऊर्ध्वाधर में एक स्थिर बिन्दु से एक हल्की अवितान्य (न खिंचने वाली) l लंबाई की डोरी से लटका हुआ है, को एक क्षैतिज आघात दिया जाता है जो कि उसको वेग 2√(gl) प्रदान करता है। कण का वेग तथा जब डोरी शैथिल्य हो जाए तब प्रारंभिक स्थान के स्तर से उसकी ऊँचाई ज्ञात कीजिए।
Q7(c) · Dynamics and Statics · 20 marks≈ 14 min
Virtual Work, Stability and Forces in Three Dimensions
A regular pentagon ABCDE, formed of equal heavy uniform bars jointed together, is suspended from the joint A, and is maintained in form by a light rod joining the middle points of BC and DE. Find the stress in this rod.
हिंदी में प्रश्न
एक सम पंचभुज (रेग्यूलर पैंटागन) ABCDE बराबर भारी एकसमान छड़ों को जोड़कर बनाया हुआ है, जोड़ A से लटक रहा है तथा BC एवं DE के मध्य बिन्दुओं को जोड़ते हुए एक हल्की छड़ के द्वारा वह अपने रूप में अनुरक्षित है। इस छड़ में प्रतिबल ज्ञात कीजिए।
Worked solution
Let each bar have length 2a and weight W; by symmetry AB, AE make angle θ with the vertical, BC, ED make angle φ, and CD is horizontal; for the regular pentagon θ = 54° and φ = 18°.
Closure: the half-width at C is a, so 2a sin θ − 2a sin φ = a, which gives cos θ δθ = cos φ δφ.
Depths below A: AB, AE at a cos θ; BC, ED at 2a cos θ + a cos φ; CD at 2a cos θ + 2a cos φ; in total z = 8a cos θ + 4a cos φ.
The light rod joins the mid-points of BC and DE, so its length is L = 2(2a sin θ − a sin φ), and δL = 4a cos θ δθ − 2a cos φ δφ = 2a cos θ δθ.
With a thrust T in the rod, W δz + T δL = 0: −W(8a sin θ + 4a sin φ cos θ/cos φ)δθ + 2aT cos θ δθ = 0.
So T = W(4 tan θ + 2 tan φ) = 2W(2 tan 54° + tan 18°) = 2W cot 18°, using cot x − tan x = 2 cot 2x.
Answer: the rod is in compression, with thrust 2W cot 18° ≈ 6.16W, W being the weight of one bar.Final answer: the rod is in compression, with thrust 2W cot 18° ≈ 6.16W, W being the weight of one bar.
Question 8 (50 marks)
Q8(a) · Ordinary Differential Equations · 15 marks≈ 11 min
Formulation and First Order, First Degree Equations
Find the sufficient condition for the differential equation M(x, y) dx + N(x, y) dy = 0 to have an integrating factor as a function of (x + y). What will be the integrating factor in that case? Hence find the integrating factor for the differential equation (x2 + xy) dx + (y2 + xy) dy = 0, and solve it.
हिंदी में प्रश्न
अवकल समीकरण M(x, y) dx + N(x, y) dy = 0 के लिए पर्याप्त शर्त ज्ञात कीजिए ताकि उसका समाकलन गुणक, (x + y) का फलन हो। उस दशा में समाकलन गुणक क्या होगा? अतएव अवकल समीकरण (x2 + xy) dx + (y2 + xy) dy = 0 का समाकलन गुणक ज्ञात कीजिए तथा हल कीजिए।
Worked solution
Let μ = μ(z), z = x + y, so μx = μy = μ′. Exactness of μM dx + μN dy: μ′M + μMy = μ′N + μNx, that is μ′/μ = (Nx − My)/(M − N).
Sufficient condition: (Nx − My)/(M − N) = φ(x + y), a function of x + y alone; then IF = e∫φ(z) dz.
Here M = x2 + xy, N = y2 + xy: Nx − My = y − x and M − N = (x − y)(x + y), so φ = −1/(x + y) and IF = e−log z = 1/(x + y).
Multiplying: x dx + y dy = 0, which is exact.
Answer: IF = 1/(x + y); the solution is x2 + y2 = c.Final answer: IF = 1/(x + y); the solution is x2 + y2 = c.
Q8(b) · Dynamics and Statics · 15 marks≈ 11 min
Motion in a Plane and Projectiles
A particle is acted on by a force parallel to the axis of y whose acceleration (always towards the axis of x) is μy−2 and when y = a, it is projected parallel to the axis of x with velocity √(2μ/a). Find the parametric equation of the path of the particle. Here μ is a constant.
हिंदी में प्रश्न
एक कण पर y-अक्ष के समांतर एक बल लगा हुआ है, जिसका त्वरण (सदैव x-अक्ष की ओर) μy−2 है तथा जब y = a है, तब x-अक्ष के समांतर वेग √(2μ/a) से प्रक्षेपित है। कण के पथ का प्राचलिक समीकरण ज्ञात कीजिए। यहाँ μ एक अचर है।
Worked solution
The force has no x-component, so dx/dt = √(2μ/a) throughout.
Along y: d2y/dt2 = −μ/y2 with dy/dt = 0 at y = a, so (dy/dt)2 = 2μ(1/y − 1/a) = 2μ(a − y)/(ay).
Put y = a cos2φ (φ = 0 at the start): (dy/dt)2 = 2μ tan2φ/a and dy/dt = −2a sin φ cos φ dφ/dt, so dφ/dt = √(2μ/a)/(2a cos2φ).
Hence dx/dφ = (dx/dt)/(dφ/dt) = 2a cos2φ = a(1 + cos 2φ), and x = a(φ + (1/2) sin 2φ), taking x = 0 at the start.
Writing ψ = 2φ: x = (a/2)(ψ + sin ψ), y = (a/2)(1 + cos ψ).
Answer: x = (a/2)(ψ + sin ψ), y = (a/2)(1 + cos ψ): a cycloid with its vertex at the point of projection and its cusps on the x-axis.Final answer: x = (a/2)(ψ + sin ψ), y = (a/2)(1 + cos ψ): a cycloid with its vertex at the point of projection and its cusps on the x-axis.
Q8(c) · Ordinary Differential Equations · 20 marks≈ 14 min
Laplace Transforms and Initial Value Problems
Solve the initial value problem d2y/dt2 + y = 8e−2t sin t, y(0) = 0, y′(0) = 0 by using Laplace-transform.
हिंदी में प्रश्न
प्रारंभिक मान समस्या d2y/dt2 + y = 8e−2t sin t, y(0) = 0, y′(0) = 0 को लाप्लास-रूपांतर के प्रयोग से हल कीजिए।
UPSC Mains 2014 Mathematics Paper II
Question 1 (50 marks)
Q1(a) · Abstract Algebra · 10 marks≈ 7 min
Normal Subgroups, Quotient Groups, Homomorphisms and Isomorphism Theorems
Let G be the set of all real 2 × 2 matrices [[x, y], [0, z]], where xz ≠ 0. Show that G is a group under matrix multiplication. Let N denote the subset {[[1, a], [0, 1]] : a ∈ ℝ}. Is N a normal subgroup of G? Justify your answer.
हिंदी में प्रश्न
मान लीजिए G सभी 2 × 2 के वास्तविक आव्यूहों [[x, y], [0, z]] का समुच्चय है, जहाँ कि xz ≠ 0. दर्शाइए कि G आव्यूह गुणन के अन्तर्गत एक समूह है। मान लीजिए N उपसमुच्चय {[[1, a], [0, 1]] : a ∈ ℝ} को द्योतित करता है। क्या N समूह G का सामान्य (नॉर्मल) उपसमूह है? अपने उत्तर का तर्क प्रस्तुत कीजिए।
Worked solution
Closure: [[x, y], [0, z]]·[[x′, y′], [0, z′]] = [[xx′, xy′ + yz′], [0, zz′]], and (xx′)(zz′) = (xz)(x′z′) ≠ 0.
Matrix multiplication is associative; I = [[1, 0], [0, 1]] ∈ G; and [[1/x, −y/(xz)], [0, 1/z]] ∈ G is the inverse of [[x, y], [0, z]]. So G is a (non-abelian) group.
Define φ : G → ℝ* × ℝ* by φ([[x, y], [0, z]]) = (x, z). By the product formula φ(AB) = (xx′, zz′) = φ(A)φ(B), so φ is a homomorphism.
ker φ = {A ∈ G : x = z = 1} = {[[1, a], [0, 1]] : a ∈ ℝ} = N. The kernel of a homomorphism is a normal subgroup, so N ⊴ G.
Direct check: for A = [[x, y], [0, z]], A[[1, a], [0, 1]]A−1 = [[1, xa/z], [0, 1]] ∈ N.
Answer: G is a group and N is a normal subgroup of G (the kernel of A ↦ (x, z)); moreover G/N ≅ ℝ* × ℝ*, since φ is onto.Final answer: G is a group and N is a normal subgroup of G (the kernel of A ↦ (x, z)); moreover G/N ≅ ℝ* × ℝ*, since φ is onto.
Q1(b) · Real Analysis · 10 marks≈ 7 min
Riemann Integral, Improper Integrals and Fundamental Theorems of Integral Calculus
Test the convergence of the improper integral ∫1∞ dx/[x2(1 + e−x)].
हिंदी में प्रश्न
अनंत समाकल ∫1∞ dx/[x2(1 + e−x)] के अभिसरण का परीक्षण कीजिए।
Q1(c) · Complex Analysis · 10 marks≈ 7 min
Analytic Functions and Cauchy-Riemann Equations
Prove that the function f(z) = u + iv, where f(z) = [x3(1 + i) − y3(1 − i)]/(x2 + y2), z ≠ 0; f(0) = 0 satisfies Cauchy-Riemann equations at the origin, but the derivative of f at z = 0 does not exist.
हिंदी में प्रश्न
सिद्ध कीजिए कि फलन f(z) = u + iv, जहाँ f(z) = [x3(1 + i) − y3(1 − i)]/(x2 + y2), z ≠ 0; f(0) = 0 मूलबिन्दु पर कौशी-रीमान समीकरणों को सन्तुष्ट करता है, परन्तु z = 0 पर f के अवकलज का अस्तित्व नहीं है।
Q1(d) · Complex Analysis · 10 marks≈ 7 min
Power Series, Taylor's Series and Laurent's Series
Expand in Laurent series the function f(z) = 1/[z2(z − 1)] about z = 0 and z = 1.
हिंदी में प्रश्न
फलन f(z) = 1/[z2(z − 1)] का z = 0 तथा z = 1 के इर्द-गिर्द लौराँ श्रेणी में प्रसार कीजिए।
Worked solution
About z = 0, for 0 < |z| < 1: 1/(z − 1) = −(1 + z + z2 + ...), so f = −1/z2 − 1/z − 1 − z − z2 − ...; the residue at 0 is −1.
About z = 0, for |z| > 1: 1/(z − 1) = (1/z)∑n=0∞ z−n, so f = ∑n=3∞ 1/zn = 1/z3 + 1/z4 + ....
About z = 1, put w = z − 1, so f = 1/(w(1 + w)2). For 0 < |w| < 1: f = (1/w)∑n=0∞ (−1)n(n + 1)wn = 1/(z − 1) − 2 + 3(z − 1) − 4(z − 1)2 + ...; the residue at 1 is 1.
About z = 1, for |w| > 1: 1/(1 + w)2 = (1/w2)(1 + 1/w)−2, so f = ∑n=0∞ (−1)n(n + 1)/(z − 1)n+3 = 1/(z − 1)3 − 2/(z − 1)4 + 3/(z − 1)5 − ....
Answer: 0 < |z| < 1: −∑n=−2∞ zn; |z| > 1: ∑n=3∞ z−n; 0 < |z − 1| < 1: ∑n=0∞ (−1)n(n + 1)(z − 1)n−1; |z − 1| > 1: ∑n=0∞ (−1)n(n + 1)(z − 1)−n−3.Final answer: 0 < |z| < 1: −∑n=−2∞ zn; |z| > 1: ∑n=3∞ z−n; 0 < |z − 1| < 1: ∑n=0∞ (−1)n(n + 1)(z − 1)n−1; |z − 1| > 1: ∑n=0∞ (−1)n(n + 1)(z − 1)−n−3.
Q1(e) · Linear Programming · 10 marks≈ 7 min
LPP Formulation, Basic Solutions and the Graphical Method
Solve graphically:
Maximize Z = 6x1 + 5x2
subject to
2x1 + x2 ≤ 16
x1 + x2 ≤ 11
x1 + 2x2 ≥ 6
5x1 + 6x2 ≤ 90
x1, x2 ≥ 0हिंदी में प्रश्न
आलेखीय (ग्राफीय) विधि के द्वारा हल कीजिए:
अधिकतमीकृत कीजिए Z = 6x1 + 5x2
बशर्ते कि
2x1 + x2 ≤ 16
x1 + x2 ≤ 11
x1 + 2x2 ≥ 6
5x1 + 6x2 ≤ 90
x1, x2 ≥ 0
Question 2 (50 marks)
Q2(a) · Abstract Algebra · 15 marks≈ 11 min
Integral Domains, PID, Euclidean Domains, UFD, Fields and Quotient Fields
Show that ℤ7 is a field. Then find ([5] + [6])−1 and (−[4])−1 in ℤ7.
हिंदी में प्रश्न
दर्शाइए कि ℤ7 एक क्षेत्र है। तब ℤ7 में ([5] + [6])−1 तथा (−[4])−1 ज्ञात कीजिए।
Q2(b) · Real Analysis · 15 marks≈ 11 min
Riemann Integral, Improper Integrals and Fundamental Theorems of Integral Calculus
Integrate ∫01 f(x) dx, where
f(x) = 2x sin(1/x) − cos(1/x), x ∈ ]0, 1]
f(x) = 0, x = 0हिंदी में प्रश्न
∫01 f(x) dx का समाकलन कीजिए, जहाँ
f(x) = 2x sin(1/x) − cos(1/x), x ∈ ]0, 1]
f(x) = 0, x = 0Worked solution
f is bounded on [0, 1] (|f(x)| ≤ 2x + 1 ≤ 3) and continuous except at x = 0, so f ∈ R[0, 1].
Let F(x) = x2 sin(1/x) for x ≠ 0 and F(0) = 0. For 0 < x ≤ 1, F′(x) = 2x sin(1/x) − cos(1/x) = f(x).
At 0, F′(0) = limh→0+ h sin(1/h) = 0 = f(0). So F is differentiable on [0, 1] with F′ = f.
By the second fundamental theorem (F′ = f ∈ R[0, 1]), ∫01 f(x) dx = F(1) − F(0) = sin 1.
Answer: ∫01 f(x) dx = sin 1 ≈ 0.8415.Final answer: ∫01 f(x) dx = sin 1 ≈ 0.8415.
Q2(c) · Linear Programming · 20 marks≈ 14 min
Transportation Problem
Find the initial basic feasible solution to the following transportation problem by Vogel's approximation method. Also, find its optimal solution and the minimum transportation cost:
Costs (rows Origins O1, O2, O3; columns Destinations D1, D2, D3, D4):
O1: 6, 4, 1, 5; Supply 14
O2: 8, 9, 2, 7; Supply 16
O3: 4, 3, 6, 2; Supply 5
Demand: 6, 10, 15, 4हिंदी में प्रश्न
निम्नलिखित परिवहन समस्या के लिए वोगेल की सन्निकटन विधि के द्वारा आरंभिक आधारिक साध्य हल ज्ञात कीजिए। इसका इष्टतम हल तथा न्यूनतम परिवहन लागत भी ज्ञात कीजिए:
लागत (पंक्तियाँ उद्गम O1, O2, O3; स्तंभ गन्तव्य D1, D2, D3, D4):
O1: 6, 4, 1, 5; उपलब्धता 14
O2: 8, 9, 2, 7; उपलब्धता 16
O3: 4, 3, 6, 2; उपलब्धता 5
माँग: 6, 10, 15, 4
Question 3 (50 marks)
Q3(a) · Abstract Algebra · 15 marks≈ 11 min
Integral Domains, PID, Euclidean Domains, UFD, Fields and Quotient Fields
Show that the set {a + bω : ω3 = 1}, where a and b are real numbers, is a field with respect to usual addition and multiplication.
हिंदी में प्रश्न
दर्शाइए कि समुच्चय {a + bω : ω3 = 1}, जहाँ a तथा b वास्तविक संख्याएँ हैं, सामान्य योग तथा गुणन के अन्तर्गत एक क्षेत्र है।
Worked solution
(Here ω is a non-real cube root of unity, ω = (−1 + i√3)/2; with ω = 1 the set would just be ℝ.)
Let F = {a + bω : a, b ∈ ℝ} ⊆ ℂ. Since ω2 + ω + 1 = 0, ω2 = −1 − ω. F is closed under subtraction: (a + bω) − (c + dω) = (a − c) + (b − d)ω.
(a + bω)(c + dω) = ac + (ad + bc)ω + bdω2 = (ac − bd) + (ad + bc − bd)ω ∈ F. The operations are those of ℂ, so they are associative, commutative and distributive, and 0, 1 ∈ F.
Inverses: (a + bω)((a − b) − bω) = a2 − ab − b2(ω + ω2) = a2 − ab + b2, using ω + ω2 = −1.
a2 − ab + b2 = (a − b/2)2 + 3b2/4 > 0 unless a = b = 0. So for a + bω ≠ 0, (a + bω)−1 = ((a − b) − bω)/(a2 − ab + b2) ∈ F.
Answer: F is a field; in fact F = ℂ, since a + bω = (a − b/2) + i(√3b/2) takes every complex value.Final answer: F is a field; in fact F = ℂ, since a + bω = (a − b/2) + i(√3b/2) takes every complex value.
Q3(b) · Real Analysis · 15 marks≈ 11 min
Functions of Several Variables: Partial Derivatives, Maxima and Minima
Obtain ∂2f(0, 0)/∂x∂y and ∂2f(0, 0)/∂y∂x for the function
f(x, y) = xy(3x2 − 2y2)/(x2 + y2), (x, y) ≠ (0, 0)
f(x, y) = 0, (x, y) = (0, 0).
Also, discuss the continuity of ∂2f/∂x∂y and ∂2f/∂y∂x at (0, 0).हिंदी में प्रश्न
फलन
f(x, y) = xy(3x2 − 2y2)/(x2 + y2), (x, y) ≠ (0, 0)
f(x, y) = 0, (x, y) = (0, 0)
के लिए ∂2f(0, 0)/∂x∂y तथा ∂2f(0, 0)/∂y∂x ज्ञात कीजिए। इसके साथ, बिन्दु (0, 0) पर ∂2f/∂x∂y तथा ∂2f/∂y∂x के सांतत्य की भी विवेचना कीजिए।Worked solution
f = 0 on both axes, so fx(0, 0) = fy(0, 0) = 0. For y ≠ 0, fx(0, y) = limh→0 f(h, y)/h = lim y(3h2 − 2y2)/(h2 + y2) = −2y, which also holds at y = 0.
For x ≠ 0, fy(x, 0) = limk→0 f(x, k)/k = lim x(3x2 − 2k2)/(x2 + k2) = 3x, also true at x = 0.
Hence ∂2f/∂y∂x (0, 0) = ∂/∂y(fx) at (0, 0) = limk→0 (fx(0, k) − fx(0, 0))/k = −2, and ∂2f/∂x∂y (0, 0) = ∂/∂x(fy) at (0, 0) = limh→0 (fy(h, 0) − fy(0, 0))/h = 3.
Away from the origin f is a rational function with non-zero denominator, so both mixed partials exist near (0, 0). If either were continuous at (0, 0), Schwarz's theorem would make the two equal there; since −2 ≠ 3, neither is continuous at (0, 0).
(Convention: ∂2f/∂x∂y means ∂/∂x(∂f/∂y); with the opposite convention the two values are interchanged.)
Answer: ∂2f/∂x∂y (0, 0) = 3 and ∂2f/∂y∂x (0, 0) = −2; both mixed partials are discontinuous at (0, 0).Final answer: ∂2f/∂x∂y (0, 0) = 3 and ∂2f/∂y∂x (0, 0) = −2; both mixed partials are discontinuous at (0, 0).
Q3(c) · Complex Analysis · 20 marks≈ 14 min
Contour Integration: Evaluation of Real Integrals
Evaluate the integral ∫0π dθ/(1 + (1/2) cos θ)2 using residues.
हिंदी में प्रश्न
अवशेषों का इस्तेमाल करते हुए समाकल ∫0π dθ/(1 + (1/2) cos θ)2 का मान निकालिए।
Worked solution
(1 + (1/2)cos θ)2 = (2 + cos θ)2/4 and the integrand depends only on cos θ, so I = 4∫0π dθ/(2 + cos θ)2 = 2∫02π dθ/(2 + cos θ)2.
With z = eiθ: 2 + cos θ = (z2 + 4z + 1)/(2z) and dθ = dz/(iz), so ∫02π dθ/(2 + cos θ)2 = ∮|z|=1 4z2/(z2 + 4z + 1)2·dz/(iz) = (4/i)∮ z dz/(z2 + 4z + 1)2.
z2 + 4z + 1 = (z − α)(z − β) with α = −2 + √3 (|α| ≈ 0.27, inside) and β = −2 − √3 (outside); α is a double pole.
Res at α = d/dz[z/(z − β)2] at z = α = −(α + β)/(α − β)3 = 4/(2√3)3 = 1/(6√3).
So ∫02π dθ/(2 + cos θ)2 = (4/i)·2πi/(6√3) = 4π/(3√3), agreeing with 2πa/(a2 − b2)3/2 for a = 2, b = 1; hence I = 8π/(3√3).
Answer: ∫0π dθ/(1 + (1/2)cos θ)2 = 8π/(3√3) = 8√3π/9 ≈ 4.837.Final answer: ∫0π dθ/(1 + (1/2)cos θ)2 = 8π/(3√3) = 8√3π/9 ≈ 4.837.
Question 4 (50 marks)
Q4(a) · Abstract Algebra · 15 marks≈ 11 min
Rings, Subrings, Ideals and Ring Homomorphisms
Prove that the set ℚ(√5) = {a + b√5 : a, b ∈ ℚ} is a commutative ring with identity.
हिंदी में प्रश्न
सिद्ध कीजिए कि समुच्चय ℚ(√5) = {a + b√5 : a, b ∈ ℚ} तत्समक सहित एक क्रमविनिमेय वलय है।
Q4(b) · Real Analysis · 15 marks≈ 11 min
Functions of Several Variables: Partial Derivatives, Maxima and Minima
Find the minimum value of x2 + y2 + z2 subject to the condition xyz = a3 by the method of Lagrange multipliers.
हिंदी में प्रश्न
लग्रांज गुणकों की विधि के द्वारा x2 + y2 + z2 का न्यूनतम मान ज्ञात कीजिए बशर्ते कि xyz = a3.
Q4(c) · Linear Programming · 20 marks≈ 14 min
The Simplex Method
Find all optimal solutions of the following linear programming problem by the simplex method:
Maximize Z = 30x1 + 24x2
subject to
5x1 + 4x2 ≤ 200
x1 ≤ 32
x2 ≤ 40
x1, x2 ≥ 0हिंदी में प्रश्न
एकधा विधि के द्वारा निम्नलिखित रैखिक प्रोग्रामन समस्या के सभी इष्टतम हल ज्ञात कीजिए:
अधिकतमीकृत कीजिए Z = 30x1 + 24x2
बशर्ते कि
5x1 + 4x2 ≤ 200
x1 ≤ 32
x2 ≤ 40
x1, x2 ≥ 0
Question 5 (50 marks)
Q5(a) · Partial Differential Equations · 10 marks≈ 7 min
Linear Second-Order PDEs with Constant Coefficients
Solve the partial differential equation (2D2 − 5DD′ + 2D′2)z = 24(y − x).
हिंदी में प्रश्न
आंशिक अवकल समीकरण (2D2 − 5DD′ + 2D′2)z = 24(y − x) को हल कीजिए।
Worked solution
Auxiliary 2m2 − 5m + 2 = 0 gives m = 2, 1/2, so CF = φ1(y + 2x) + φ2(2y + x).
24(y − x) = 24v with v = −x + y, so a = −1, b = 1 and F(a, b) = 2 + 5 + 2 = 9 ≠ 0.
PI = (1/9)∫∫24v dv dv = (1/9)(4v3) = (4/9)(y − x)3; check: zxx = zyy = (8/3)v, zxy = −(8/3)v give (2 + 5 + 2)(8/3)v = 24v.
Answer: z = φ1(y + 2x) + φ2(2y + x) + (4/9)(y − x)3.Final answer: z = φ1(y + 2x) + φ2(2y + x) + (4/9)(y − x)3.
Q5(b) · Numerical Analysis and Computer Programming · 10 marks≈ 7 min
Roots of Algebraic and Transcendental Equations
Apply Newton-Raphson method to determine a root of the equation cos x − xex = 0 correct up to four decimal places.
हिंदी में प्रश्न
समीकरण cos x − xex = 0 का, चार दशमलव स्थानों तक सही, मूल ज्ञात करने के लिए न्यूटन-रैफसन विधि का इस्तेमाल कीजिए।
Q5(c) · Numerical Analysis and Computer Programming · 10 marks≈ 7 min
Numerical Integration and Numerical Solution of ODEs
Use five subintervals to integrate ∫01 dx/(1 + x2) using trapezoidal rule.
हिंदी में प्रश्न
समलंबी नियम के द्वारा ∫01 dx/(1 + x2) का समाकलन करने के लिए पाँच उपांतरालों का इस्तेमाल कीजिए।
Q5(d) · Numerical Analysis and Computer Programming · 10 marks≈ 7 min
Computer Systems, Logic Gates, Boolean Algebra and Algorithms
Use only AND and OR logic gates to construct a logic circuit for the Boolean expression z = xy + uv.
हिंदी में प्रश्न
केवल AND तथा OR तर्कसंगत द्वार इस्तेमाल करते हुए बूलीय व्यंजक z = xy + uv के लिए एक तर्क परिपथ की रचना कीजिए।
Q5(e) · Mechanics and Fluid Dynamics · 10 marks≈ 7 min
Hamilton's Equations
Find the equation of motion of a compound pendulum using Hamilton's equations.
हिंदी में प्रश्न
हैमिल्टन के समीकरणों का इस्तेमाल करते हुए असरल लोलक की गति का समीकरण ज्ञात कीजिए।
Worked solution
Let the body (mass M, radius of gyration k about G) turn about a fixed horizontal axis through O with OG = h, and let θ be the angle of OG with the downward vertical.
T = (1/2)M(k2 + h2)θ̇2, V = −Mgh cos θ, p = M(k2 + h2)θ̇, and H = T + V = p2/(2M(k2 + h2)) − Mgh cos θ.
θ̇ = ∂H/∂p = p/(M(k2 + h2)), ṗ = −∂H/∂θ = −Mgh sin θ.
Answer: θ̈ + (gh/(k2 + h2))sin θ = 0; small oscillations have period 2π√((k2 + h2)/(gh)).Final answer: θ̈ + (gh/(k2 + h2))sin θ = 0; small oscillations have period 2π√((k2 + h2)/(gh)).
Question 6 (50 marks)
Q6(a) · Partial Differential Equations · 15 marks≈ 11 min
Classification and Canonical Forms of Second-Order PDEs
Reduce the equation ∂2z/∂x2 = x2 ∂2z/∂y2 to canonical form.
हिंदी में प्रश्न
समीकरण ∂2z/∂x2 = x2 ∂2z/∂y2 को विहित रूप में समानीत कीजिए।
Q6(b) · Numerical Analysis and Computer Programming · 15 marks≈ 11 min
Systems of Linear Equations: Direct and Iterative Methods
Solve the system of equations
2x1 − x2 = 7
−x1 + 2x2 − x3 = 1
−x2 + 2x3 = 1
using Gauss-Seidel iteration method (Perform three iterations).हिंदी में प्रश्न
गाउस-साइडल पुनरावृत्ति विधि के द्वारा समीकरण निकाय
2x1 − x2 = 7
−x1 + 2x2 − x3 = 1
−x2 + 2x3 = 1
का हल कीजिए (तीन पुनरावृत्तियाँ कीजिए)।Q6(c) · Numerical Analysis and Computer Programming · 20 marks≈ 14 min
Numerical Integration and Numerical Solution of ODEs
Use Runge-Kutta formula of fourth order to find the value of y at x = 0.8, where dy/dx = √(x + y), y(0.4) = 0.41. Take the step length h = 0.2.
हिंदी में प्रश्न
x = 0.8 पर y का मान ज्ञात करने के लिए, जहाँ dy/dx = √(x + y), y(0.4) = 0.41 चतुष्कोटि के रुन्गे-कुट्टा फॉर्म्यूले का इस्तेमाल कीजिए। पग लंबाई h = 0.2 लीजिए।
Question 7 (50 marks)
Q7(a) · Partial Differential Equations · 15 marks≈ 11 min
Vibrating String, Heat Equation and Laplace Equation
Find the deflection of a vibrating string (length = π, ends fixed, ∂2u/∂t2 = ∂2u/∂x2) corresponding to zero initial velocity and initial deflection f(x) = k(sin x − sin 2x).
हिंदी में प्रश्न
एक कंपमान डोरी (लम्बाई = π, स्थिर सिरे, ∂2u/∂t2 = ∂2u/∂x2) का विक्षेप ज्ञात कीजिए, यदि प्रारंभिक वेग शून्य हो और प्रारंभिक विक्षेप f(x) = k(sin x − sin 2x) हो।
Q7(b) · Numerical Analysis and Computer Programming · 15 marks≈ 11 min
Computer Systems, Logic Gates, Boolean Algebra and Algorithms
Draw a flowchart for Simpson's one-third rule.
हिंदी में प्रश्न
सिम्प्सन के एक-तिहाई नियम के लिए एक प्रवाह-चार्ट बनाइए।
Worked solution
(1) Oval START; (2) parallelogram READ a, b, n; (3) diamond 'n even?', and if NO print 'n must be even' and go to STOP.
(4) Rectangle h = (b − a)/n, S = f(a) + f(b), i = 1; (5) diamond 'i ≤ n − 1?', and if NO go to (8).
(6) Diamond 'i odd?': YES gives S = S + 4f(a + ih), NO gives S = S + 2f(a + ih); (7) rectangle i = i + 1 with an arrow back to (5).
(8) Rectangle I = (h/3)S; (9) parallelogram PRINT I; (10) oval STOP.
Answer: the flow chart above computes I ≈ (h/3)[y0 + 4(y1 + y3 + ...) + 2(y2 + y4 + ...) + yn].Final answer: the flow chart above computes I ≈ (h/3)[y0 + 4(y1 + y3 + ...) + 2(y2 + y4 + ...) + yn].
Q7(c) · Mechanics and Fluid Dynamics · 20 marks≈ 14 min
Two-Dimensional and Axisymmetric Motion: Sources, Sinks and Vortices
Given the velocity potential φ = (1/2) log[((x + a)2 + y2)/((x − a)2 + y2)], determine the streamlines.
हिंदी में प्रश्न
दत्त वेग विभव φ = (1/2) log[((x + a)2 + y2)/((x − a)2 + y2)] के लिए प्रवाह-रेखाएँ ज्ञात कीजिए।
Worked solution
φ = log r1 − log r2, with r1, r2 the distances from (−a, 0) and (a, 0), is the real part of w = log(z + a) − log(z − a): a unit source at (a, 0) and a unit sink at (−a, 0) (q = −∇φ).
ψ = θ1 − θ2, the angles at (−a, 0) and (a, 0); tan(θ1 − θ2) = [y/(x + a) − y/(x − a)]/[1 + y2/(x2 − a2)] = −2ay/(x2 + y2 − a2).
Answer: the streamlines are x2 + y2 − a2 = λy (λ a parameter), the coaxal circles through (±a, 0).Final answer: the streamlines are x2 + y2 − a2 = λy (λ a parameter), the coaxal circles through (±a, 0).
Question 8 (50 marks)
Q8(a) · Partial Differential Equations · 15 marks≈ 11 min
Vibrating String, Heat Equation and Laplace Equation
Solve ∂2u/∂t2 = ∂2u/∂x2, 0 < x < 1, t > 0, given that
(i) u(x, 0) = 0, 0 ≤ x ≤ 1
(ii) ∂u/∂t (x, 0) = x2, 0 ≤ x ≤ 1
(iii) u(0, t) = u(1, t) = 0, for all tहिंदी में प्रश्न
हल कीजिए ∂2u/∂t2 = ∂2u/∂x2, 0 < x < 1, t > 0, दिया है कि
(i) u(x, 0) = 0, 0 ≤ x ≤ 1
(ii) ∂u/∂t (x, 0) = x2, 0 ≤ x ≤ 1
(iii) u(0, t) = u(1, t) = 0, सभी t के लिएWorked solution
Here c = 1, L = 1, u(x, 0) = 0, so An = 0 and u = ∑Bnsin nπt sin nπx with nπBn = 2∫01 x2sin nπx dx.
∫01 x2sin nπx dx = (−1)n+1/(nπ) + 2((−1)n − 1)/(nπ)3 (integrate by parts twice).
So Bn = (2/(n2π2))[(−1)n+1 + 2((−1)n − 1)/(n2π2)]: 2/(n2π2) − 8/(n4π4) for odd n and −2/(n2π2) for even n.
Answer: u(x, t) = ∑n=1∞ (2/(n2π2))[(−1)n+1 + 2((−1)n − 1)/(n2π2)] sin nπx sin nπt.Final answer: u(x, t) = ∑n=1∞ (2/(n2π2))[(−1)n+1 + 2((−1)n − 1)/(n2π2)] sin nπx sin nπt.
Q8(b) · Numerical Analysis and Computer Programming · 15 marks≈ 11 min
Computer Systems, Logic Gates, Boolean Algebra and Algorithms
For any Boolean variables x and y, show that x + xy = x.
हिंदी में प्रश्न
किसी भी बूलीय चर x और y के लिए दर्शाइए कि x + xy = x.
Worked solution
x + xy = x · 1 + x · y (identity law) = x(1 + y) (distributive law) = x · 1 = x (identity law), provided 1 + y = 1.
Proof of 1 + y = 1 from the postulates: y + 1 = (y + 1) · 1 = (y + 1)(y + y′) = y + 1 · y′ (distributive law of + over ·) = y + y′ = 1.
Truth-table check: if x = 0, x + xy = 0 + 0 = 0; if x = 1, x + xy = 1 + y = 1; in both cases the value equals x.
Answer: x + xy = x (the absorption law).Final answer: x + xy = x (the absorption law).
Q8(c) · Mechanics and Fluid Dynamics · 20 marks≈ 14 min
Euler's Equation, Bernoulli's Theorem and the Navier–Stokes Equation
Find Navier-Stokes equation for a steady laminar flow of a viscous incompressible fluid between two infinite parallel plates.
हिंदी में प्रश्न
दो अनंत समांतर प्लेटों के बीच, एक श्यान असंपीड्य तरल के अपरिवर्ती स्तरीय (लैमिनर) प्रवाह के लिए, नैवियर-स्टोक्स समीकरण ज्ञात कीजिए।
Worked solution
Take fixed plates y = 0 and y = h with flow along Ox: q = (u(y), 0, 0), steady, no body force; continuity ∂u/∂x = 0 holds and (q·∇)q = u ∂q/∂x = 0.
The Navier–Stokes equations become 0 = −∂p/∂x + μ d2u/dy2, 0 = −∂p/∂y, 0 = −∂p/∂z, so p = p(x) and dp/dx = μu″(y) = constant = −G.
Integrating, u = −(G/(2μ))y2 + C1y + C2; no slip u(0) = u(h) = 0 gives u = (G/(2μ))y(h − y).
Discharge per unit width Q = Gh3/(12μ); maximum speed Gh2/(8μ) at y = h/2; wall shear stress Gh/2.
Answer: the equations reduce to μ d2u/dy2 = dp/dx with ∂p/∂y = ∂p/∂z = 0, giving the parabolic profile u = (G/(2μ))y(h − y) (plane Poiseuille flow).Final answer: the equations reduce to μ d2u/dy2 = dp/dx with ∂p/∂y = ∂p/∂z = 0, giving the parabolic profile u = (G/(2μ))y(h − y) (plane Poiseuille flow).