UPSC Mains 2017 Mathematics Optional Question Papers with Solutions
All 59 questions of the 2017 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 2017 Mathematics Paper I
Question 1 (50 marks)
Q1(a) · Linear Algebra · 10 marks≈ 7 min
Eigenvalues, Eigenvectors and the Cayley-Hamilton Theorem
Let A = [[2, 2], [1, 3]]. Find a non-singular matrix P such that P−1AP is a diagonal matrix.
हिंदी में प्रश्न
मान लीजिए A = [[2, 2], [1, 3]]। एक व्युत्क्रमणीय आव्यूह P ज्ञात कीजिए ताकि P−1AP एक विकर्ण-आव्यूह हो।
Q1(b) · Linear Algebra · 10 marks≈ 7 min
Algebra of Matrices, Echelon Form, Rank, Inverse, Congruence and Similarity
Show that similar matrices have the same characteristic polynomial.
हिंदी में प्रश्न
दर्शाइए कि समरूप आव्यूहों के समान अभिलक्षणिक बहुपद होते हैं।
Q1(c) · Calculus · 10 marks≈ 7 min
Double and Triple Integrals; Areas, Surfaces and Volumes
Integrate the function f(x, y) = xy(x2 + y2) over the domain R : {−3 ≤ x2 − y2 ≤ 3, 1 ≤ xy ≤ 4}.
हिंदी में प्रश्न
प्रान्त R : {−3 ≤ x2 − y2 ≤ 3, 1 ≤ xy ≤ 4} पर फलन f(x, y) = xy(x2 + y2) का समाकलन कीजिए।
Worked solution
Put u = x2 − y2, v = xy. Then ∂(u, v)/∂(x, y) = det [[2x, −2y], [y, x]] = 2(x2 + y2), so dx dy = du dv/(2(x2 + y2)).
Hence xy(x2 + y2) dx dy = (v/2) du dv.
In the first quadrant the map is one-one onto the rectangle −3 ≤ u ≤ 3, 1 ≤ v ≤ 4 (it is z ↦ z2 with z = x + iy, since z2 = u + 2iv), so the integral is (1/2)∫−33 du ∫14 v dv = (1/2)(6)(15/2) = 45/2.
[The set 1 ≤ xy ≤ 4 also has a branch in the third quadrant, where the integrand takes the same values; the intended region is the first-quadrant one. If both branches are counted, the integral is 45.]
Answer: 45/2 (over the first-quadrant region).Final answer: 45/2 (over the first-quadrant region).
Q1(d) · Analytic Geometry · 10 marks≈ 7 min
Ellipsoid, Hyperboloids and Paraboloids
Find the equation of the tangent plane at point (1, 1, 1) to the conicoid 3x2 − y2 = 2z.
हिंदी में प्रश्न
बिन्दु (1, 1, 1) पर शांकवज 3x2 − y2 = 2z के स्पर्श-तल का समीकरण निकालिए।
Q1(e) · Analytic Geometry · 10 marks≈ 7 min
Straight Lines and Shortest Distance between Skew Lines
Find the shortest distance between the skew lines:
(x − 3)/3 = (8 − y)/1 = (z − 3)/1 and (x + 3)/(−3) = (y + 7)/2 = (z − 6)/4.हिंदी में प्रश्न
विषमतलीय रेखाओं (x − 3)/3 = (8 − y)/1 = (z − 3)/1 व (x + 3)/(−3) = (y + 7)/2 = (z − 6)/4 के बीच न्यूनतम-दूरी ज्ञात कीजिए।
Question 2 (50 marks)
Q2(a) · Calculus · 15 marks≈ 11 min
Double and Triple Integrals; Areas, Surfaces and Volumes
Find the volume of the solid above the xy-plane and directly below the portion of the elliptic paraboloid x2 + y2/4 = z which is cut off by the plane z = 9.
हिंदी में प्रश्न
xy-तल के ऊपर के और ठीक नीचे के दीर्घवृत्तीय पैराबोलौएड x2 + y2/4 = z, जो समतल z = 9 से कटा हुआ है, का आयतन मालूम कीजिए।
Worked solution
The paraboloid z = x2 + y2/4 meets z = 9 above the ellipse E: x2 + y2/4 ≤ 9 (semi-axes 3 and 6, area 18π). The solid above z = 0 and below this portion has V = ∬E (x2 + y2/4) dx dy.
Put x = r cos θ, y = 2r sin θ (Jacobian 2r): E becomes 0 ≤ r ≤ 3, 0 ≤ θ ≤ 2π, and the integrand is r2.
V = ∫02π∫03 r2·2r dr dθ = 2π · 2 · 81/4 = 81π. (The solid inside the paraboloid up to z = 9 has volume 9 · 18π − 81π = 81π as well, so both readings agree.)
Answer: 81π.Final answer: 81π.
Q2(b) · Analytic Geometry · 15 marks≈ 11 min
The Sphere
A plane passes through a fixed point (a, b, c) and cuts the axes at the points A, B, C respectively. Find the locus of the centre of the sphere which passes through the origin O and A, B, C.
हिंदी में प्रश्न
एक समतल, नियत बिन्दु (a, b, c) में से गुज़रता है तथा अक्षों को क्रमशः बिन्दुओं A, B व C पर काटता है। मूल बिन्दु O तथा A, B व C में से गुज़रने वाले गोले के केन्द्र का बिन्दु-पथ ज्ञात कीजिए।
Q2(c) · Analytic Geometry · 10 marks≈ 7 min
The Sphere
Show that the plane 2x − 2y + z + 12 = 0 touches the sphere x2 + y2 + z2 − 2x − 4y + 2z − 3 = 0. Find the point of contact.
हिंदी में प्रश्न
दर्शाइए कि समतल 2x − 2y + z + 12 = 0, गोले x2 + y2 + z2 − 2x − 4y + 2z − 3 = 0, को स्पर्श करता है। सम्पर्क बिन्दु ज्ञात कीजिए।
Q2(d) · Linear Algebra · 10 marks≈ 7 min
Vector Spaces, Subspaces, Basis and Dimension
Suppose U and W are distinct four dimensional subspaces of a vector space V, where dim V = 6. Find the possible dimensions of subspace U ∩ W.
हिंदी में प्रश्न
मान लीजिए कि U व W सदिश समष्टि V के चार सुस्पष्ट विमीय उप-आकाश जहाँ पर विमा V = 6। उप-आकाश (U ∩ W) की सम्भावित विमाएँ ज्ञात कीजिए।
Worked solution
U ⊆ U + W ⊆ V, so 4 ≤ dim(U + W) ≤ 6. If dim(U + W) = 4, then U + W = U (a subspace of U + W of the same finite dimension), and likewise U + W = W, so U = W, which is excluded. Hence dim(U + W) = 5 or 6.
dim(U ∩ W) = dim U + dim W − dim(U + W) = 8 − dim(U + W) = 3 or 2.
Both values occur: in ℝ6 take U = L{e1, e2, e3, e4} and W = L{e1, e2, e3, e5} (intersection of dimension 3) or W = L{e1, e2, e5, e6} (intersection of dimension 2).
Answer: dim(U ∩ W) = 2 or 3.Final answer: dim(U ∩ W) = 2 or 3.
Question 3 (50 marks)
Q3(a) · Linear Algebra · 15 marks≈ 11 min
Linear Transformations, Rank-Nullity and Matrix Representation
Consider the matrix mapping A : R4 → R3, where A = [[1, 2, 3, 1], [1, 3, 5, −2], [3, 8, 13, −3]]. Find a basis and dimension of the image of A and those of the kernel A.
हिंदी में प्रश्न
विचारिए आव्यूह-प्रतिरूपण A : R4 → R3 है, जहाँ पर A = [[1, 2, 3, 1], [1, 3, 5, −2], [3, 8, 13, −3]]। A की प्रतिछाया की विमा व एक आधार तथा कर्नल A की विमा व एक आधार भी ज्ञात कीजिए।
Q3(b) · Linear Algebra · 10 marks≈ 7 min
Eigenvalues, Eigenvectors and the Cayley-Hamilton Theorem
Prove that distinct non-zero eigenvectors of a matrix are linearly independent.
हिंदी में प्रश्न
सिद्ध कीजिए कि आव्यूह के विभिन्न अशून्य-अभिलक्षणिक सदिश रैखिक स्वतंत्र होते हैं।
Q3(c) · Calculus · 15 marks≈ 11 min
Functions of Several Variables: Partial Derivatives, Extrema, Lagrange Multipliers and Jacobians
If f(x, y) = xy(x2 − y2)/(x2 + y2), (x, y) ≠ (0, 0);
f(x, y) = 0, (x, y) = (0, 0),
calculate ∂2f/∂x∂y and ∂2f/∂y∂x at (0, 0).हिंदी में प्रश्न
यदि f(x, y) = xy(x2 − y2)/(x2 + y2), (x, y) ≠ (0, 0);
f(x, y) = 0, (x, y) = (0, 0),
तब ∂2f/∂x∂y व ∂2f/∂y∂x की (0, 0) का परिकलन कीजिए।Worked solution
Convention: ∂2f/∂x∂y = ∂/∂x(∂f/∂y) and ∂2f/∂y∂x = ∂/∂y(∂f/∂x).
For y ≠ 0: fx(0, y) = limh→0 f(h, y)/h = lim y(h2 − y2)/(h2 + y2) = −y; and fx(0, 0) = lim f(h, 0)/h = 0. So fx(0, y) = −y for all y.
For x ≠ 0: fy(x, 0) = limk→0 f(x, k)/k = lim x(x2 − k2)/(x2 + k2) = x; and fy(0, 0) = 0. So fy(x, 0) = x for all x.
∂2f/∂y∂x (0, 0) = limk→0 (fx(0, k) − fx(0, 0))/k = −1, and ∂2f/∂x∂y (0, 0) = limh→0 (fy(h, 0) − fy(0, 0))/h = 1.
The two are unequal; Schwarz's theorem does not apply because the mixed partial derivatives are not continuous at (0, 0).
Answer: ∂2f/∂x∂y (0, 0) = 1 and ∂2f/∂y∂x (0, 0) = −1.Final answer: ∂2f/∂x∂y (0, 0) = 1 and ∂2f/∂y∂x (0, 0) = −1.
Q3(d) · Analytic Geometry · 10 marks≈ 7 min
Ellipsoid, Hyperboloids and Paraboloids
Find the locus of the point of intersection of three mutually perpendicular tangent planes to ax2 + by2 + cz2 = 1.
हिंदी में प्रश्न
ax2 + by2 + cz2 = 1 के तीन परस्पर लम्बवत् स्पर्शतलों के प्रतिच्छेदन बिन्दु का बिन्दु-पथ ज्ञात कीजिए।
Question 4 (50 marks)
Q4(a) · Analytic Geometry · 15 marks≈ 11 min
General Second Degree Equation and Reduction to Canonical Form
Reduce the following equation to the standard form and hence determine the nature of the conicoid: x2 + y2 + z2 − yz − zx − xy − 3x − 6y − 9z + 21 = 0.
हिंदी में प्रश्न
समीकरण x2 + y2 + z2 − yz − zx − xy − 3x − 6y − 9z + 21 = 0 को प्रमाणिक-रूप में व्यक्त कीजिए अतः शांकवज की प्रकृति निर्धारित कीजिए।
Worked solution
The quadratic part has eigenvalues 0, 3/2, 3/2 (as in Example 5), with λ = 0 along n = (1, 1, 1)/√3. Take the orthonormal directions e1 = (1, −1, 0)/√2, e2 = (1, 1, −2)/√6, e3 = n.
The linear part −3x − 6y − 9z has components 3/√2, 9/√6 and −18/√3 = −6√3 along e1, e2, e3, so the equation becomes (3/2)(X2 + Y2) + (3/√2)X + (9/√6)Y − 6√3Z + 21 = 0.
Completing squares: (3/2)[(X + 1/√2)2 + (Y + 3/√6)2] − 3 − 6√3Z + 21 = 0, that is (3/2)(X′2 + Y′2) = 6√3(Z − √3).
The vertex X′ = Y′ = 0, Z = √3 is −(1/√2)e1 − (3/√6)e2 + √3n = (0, 1, 2), which satisfies the equation. Check by invariants: Δ = −243/4 and k2 = −Δ/(λ1λ2) = (243/4)/(9/4) = 27, so 2k = 6√3.
Answer: X2 + Y2 = 4√3 Z, a paraboloid of revolution with vertex (0, 1, 2) and axis x/1 = (y − 1)/1 = (z − 2)/1.Final answer: X2 + Y2 = 4√3 Z, a paraboloid of revolution with vertex (0, 1, 2) and axis x/1 = (y − 1)/1 = (z − 2)/1.
Q4(b) · Linear Algebra · 15 marks≈ 11 min
Systems of Linear Equations
Consider the following system of equations in x, y, z:
x + 2y + 2z = 1
x + ay + 3z = 3
x + 11y + az = b.
(i) For which values of a does the system have a unique solution?
(ii) For which pair of values (a, b) does the system have more than one solution?हिंदी में प्रश्न
x, y, z में समीकरणों के निम्नलिखित निकाय को विचारिए :
x + 2y + 2z = 1
x + ay + 3z = 3
x + 11y + az = b
(i) a के किन मानों के लिए निकाय का एकल-हल है?
(ii) युग्म जोड़ों (a, b) के किन मानों के लिए समुदाय के एक से अधिक हल हैं?Q4(c) · Calculus · 10 marks≈ 7 min
Riemann Integral, Indefinite Integrals and Improper Integrals
Examine if the improper integral ∫03 2x dx/(1 − x2)2/3 exists.
हिंदी में प्रश्न
परीक्षण कीजिए कि क्या अनंत समाकल ∫03 2x dx/(1 − x2)2/3 का अस्तित्व है।
Worked solution
In [0, 3] the integrand is unbounded only at x = 1, so the integral exists only if both ∫01 and ∫13 converge.
With u = 1 − x2, du = −2x dx, an antiderivative is −3(1 − x2)1/3 (real cube root), which is continuous on [0, 3].
∫01−ε = 3 − 3(1 − (1 − ε)2)1/3 → 3, and ∫1+ε3 = −3(−8)1/3 + 3(1 − (1 + ε)2)1/3 → 6.
Both limits are finite, as expected since near x = 1 the integrand behaves like a constant times |1 − x|−2/3, and 2/3 < 1.
Answer: The integral exists and equals 3 + 6 = 9.Final answer: The integral exists and equals 3 + 6 = 9.
Q4(d) · Calculus · 10 marks≈ 7 min
Double and Triple Integrals; Areas, Surfaces and Volumes
Prove that π/3 ≤ ∬D dx dy/√(x2 + (y − 2)2) ≤ π where D is the unit disc.
हिंदी में प्रश्न
सिद्ध कीजिए कि π/3 ≤ ∬D dx dy/√(x2 + (y − 2)2) ≤ π जहाँ पर D एकक डिस्क है।
Worked solution
f(x, y) = 1/√(x2 + (y − 2)2) = 1/d, where d is the distance from (x, y) to P(0, 2), and |OP| = 2.
For (x, y) in the unit disc D, the triangle inequality gives 2 − 1 ≤ d ≤ 2 + 1, i.e. 1 ≤ d ≤ 3 (attained at (0, 1) and (0, −1)); hence 1/3 ≤ f ≤ 1 on D.
With m = 1/3, M = 1 and area(D) = π: m·π ≤ ∬D f dx dy ≤ M·π.
Answer: π/3 ≤ ∬D dx dy/√(x2 + (y − 2)2) ≤ π.Final answer: π/3 ≤ ∬D dx dy/√(x2 + (y − 2)2) ≤ π.
Question 5 (50 marks)
Q5(a) · Ordinary Differential Equations · 10 marks≈ 7 min
Formulation and First Order, First Degree Equations
Find the differential equation representing all the circles in the x-y plane.
हिंदी में प्रश्न
x-y समतल में सभी वृत्तों को निरूपित करने वाला अवकल समीकरण ज्ञात कीजिए।
Worked solution
A circle is (x − a)2 + (y − b)2 = r2, with three arbitrary constants. Its radius of curvature is the constant r: (1 + y′2)3/2/y″ = ±r.
Differentiating with respect to x: [3y′y″2(1 + y′2)1/2 − (1 + y′2)3/2y‴]/y″2 = 0.
(Directly: differentiating three times, 1 + y′2 + (y − b)y″ = 0 and 3y′y″ + (y − b)y‴ = 0; eliminating y − b gives the same equation.)
Answer: (1 + y′2)y‴ − 3y′y″2 = 0, a third order equation of degree 1.Final answer: (1 + y′2)y‴ − 3y′y″2 = 0, a third order equation of degree 1.
Q5(b) · Ordinary Differential Equations · 10 marks≈ 7 min
Orthogonal Trajectories
Suppose that the streamlines of the fluid flow are given by a family of curves xy = c. Find the equipotential lines, that is, the orthogonal trajectories of the family of curves representing the streamlines.
हिंदी में प्रश्न
मान लीजिए किसी द्रव्य-प्रवाह की धारा-रेखाएं, वक्र समुदाय xy = c के द्वारा प्रदर्शित हैं। सम-विभव रेखाओं, अर्थात् धारा-रेखाओं को प्रदर्शित करने वाले वक्र-समुदाय के लंबकोणीय संवेदियों को ज्ञात कीजिए।
Q5(c) · Dynamics and Statics · 10 marks≈ 7 min
Constrained Motion, Work and Energy
A fixed wire is in the shape of the cardioid r = a(1 + cos θ), the initial line being the downward vertical. A small ring of mass m can slide on the wire and is attached to the point r = 0 of the cardioid by an elastic string of natural length a and modulus of elasticity 4mg. The string is released from rest when the string is horizontal. Show by using the laws of conservation of energy that
aθ̇2(1 + cos θ) − g cos θ (1 − cos θ) = 0, g being the acceleration due to gravity.हिंदी में प्रश्न
एक तार, कार्डिऑइड r = a(1 + cos θ), जिसकी प्रारम्भिक रेखा अधोगत ऊर्ध्वाधर है, के आकार में स्थिर है। m द्रव्यमान का एक छोटा छल्ला तार पर फिसल सकता है तथा कार्डिऑइड के बिन्दु r = 0 से स्वाभाविक लम्बाई a की प्रत्यास्थ (elastic) डोरी से, जिसका प्रत्यास्थता गुणांक 4mg है, बंधा है। छल्ले को विरामावस्था से, जबकि डोरी क्षैतिज है, छोड़ा जाता है। ऊर्जा-संरक्षण के नियमों का प्रयोग कर दर्शाइये कि aθ̇2(1 + cos θ) − g cos θ (1 − cos θ) = 0, g गुरुत्वाकर्षण के कारण त्वरण है।
Worked solution
Take the pole O as origin with θ measured from the downward vertical; at the start θ = π/2 and r = a, so the string has its natural length and the ring is at rest.
Speed: ṙ = −a sin θ θ̇, so v2 = ṙ2 + r2θ̇2 = a2θ̇2[sin2θ + (1 + cos θ)2] = 2a2θ̇2(1 + cos θ).
The ring is at depth r cos θ = a(1 + cos θ) cos θ below O, so its gravitational potential energy is −mga(1 + cos θ) cos θ.
For θ < π/2 the extension is r − a = a cos θ, so the elastic energy is λx2/(2a) = 4mga2cos2θ/(2a) = 2mga cos2θ.
All three energies vanish at the start, so ma2θ̇2(1 + cos θ) − mga(1 + cos θ) cos θ + 2mga cos2θ = 0.
Dividing by ma: aθ̇2(1 + cos θ) − g cos θ − g cos2θ + 2g cos2θ = 0, that is aθ̇2(1 + cos θ) − g cos θ (1 − cos θ) = 0.
Answer: conservation of energy gives aθ̇2(1 + cos θ) − g cos θ (1 − cos θ) = 0, as required.Final answer: conservation of energy gives aθ̇2(1 + cos θ) − g cos θ (1 − cos θ) = 0, as required.
Q5(d) · Vector Analysis · 10 marks≈ 7 min
Gradient, Divergence and Curl in Cartesian and Cylindrical Coordinates
For what values of the constants a, b and c the vector V̄ = (x + y + az)î + (bx + 2y − z)ĵ + (−x + cy + 2z)k̂ is irrotational. Find the divergence in cylindrical coordinates of this vector with these values.
हिंदी में प्रश्न
स्थिरांकों a, b व c के किन मानों के लिए सदिश V̄ = (x + y + az)î + (bx + 2y − z)ĵ + (−x + cy + 2z)k̂ अघूर्णी है। इन मानों के साथ इस सदिश के बेलनी-निर्देशांकों में अपसारिता ज्ञात कीजिए।
Worked solution
curl V = (c + 1)i + (a + 1)j + (b − 1)k = 0 gives a = −1, b = 1, c = −1.
Then V = (x + y − z)i + (x + 2y − z)j + (−x − y + 2z)k, and in Cartesian form ∇·V = 1 + 2 + 2 = 5.
With x = ρ cos φ, y = ρ sin φ and C = cos φ, S = sin φ: Vρ = VxC + VyS = ρ(C2 + 2CS + 2S2) − z(C + S), Vφ = −VxS + VyC = ρ(C2 − S2 + CS) + z(S − C), Vz = −ρ(C + S) + 2z.
(1/ρ)∂(ρVρ)/∂ρ = 2(C2 + 2CS + 2S2) − z(C + S)/ρ and (1/ρ)∂Vφ/∂φ = C2 − S2 − 4CS + z(C + S)/ρ, while ∂Vz/∂z = 2.
Adding: 3C2 + 3S2 + 2 = 5.
Answer: a = −1, b = 1, c = −1, and the divergence in cylindrical coordinates is 5, as in Cartesian form.Final answer: a = −1, b = 1, c = −1, and the divergence in cylindrical coordinates is 5, as in Cartesian form.
Q5(e) · Vector Analysis · 10 marks≈ 7 min
Scalar and Vector Fields; Differentiation of Vector Functions
The position vector of a moving point at time t is r̄ = sin t î + cos 2t ĵ + (t2 + 2t)k̂. Find the components of acceleration ā in the directions parallel to the velocity vector v̄ and perpendicular to the plane of r̄ and v̄ at time t = 0.
हिंदी में प्रश्न
समय t पर एक गतिमान बिन्दु का स्थिति सदिश r̄ = sin t î + cos 2t ĵ + (t2 + 2t)k̂ है। इसके त्वरण ā के अवयव वेग-सदिश v̄ के समान्तर दिशा में तथा r̄ व v̄ के तल के लम्बवत् दिशा में समय t = 0 पर ज्ञात कीजिए।
Worked solution
v = dr/dt = cos t i − 2 sin 2t j + (2t + 2)k and a = d2r/dt2 = −sin t i − 4 cos 2t j + 2k.
At t = 0: r = j, v = i + 2k and a = −4j + 2k.
Component along v: a·v/|v| = (0 + 0 + 4)/√5 = 4/√5.
The normal to the plane of r and v is r × v = j × (i + 2k) = 2i − k, with |r × v| = √5.
Component perpendicular to that plane: a·(r × v)/|r × v| = (0 + 0 − 2)/√5 = −2/√5.
Answer: 4/√5 along v, and −2/√5 along the normal 2i − k (that is, magnitude 2/√5 in the direction −(2i − k)).Final answer: 4/√5 along v, and −2/√5 along the normal 2i − k (that is, magnitude 2/√5 in the direction −(2i − k)).
Question 6 (50 marks)
Q6(a) · Ordinary Differential Equations · 16 marks≈ 12 min
Linear Equations with Constant Coefficients
(i) Solve the following simultaneous linear differential equations:
(D + 1)y = z + ex and (D + 1)z = y + ex where y and z are functions of independent variable x and D ≡ d/dx. (8 marks)
(ii) If the growth rate of the population of bacteria at any time t is proportional to the amount present at that time and population doubles in one week, then how much bacterias can be expected after 4 weeks? (8 marks)हिंदी में प्रश्न
(i) निम्नलिखित युगपत रेखीय अवकल समीकरणों को हल कीजिए :
(D + 1)y = z + ex व (D + 1)z = y + ex, जहाँ y व z स्वतंत्र चर x के फलन हैं तथा D ≡ d/dx। (8 अंक)
(ii) यदि किसी भी समय t पर बैक्टीरिया संख्या की वृद्धि दर उस समय विद्यमान संख्या के समानुपाती है तथा संख्या एक सप्ताह में दो गुणी हो जाती है तब 4 सप्ताह के बाद बैक्टीरिया की कितनी संख्या अपेक्षित है? (8 अंक)Worked solution
(i)
Subtracting the equations: (D + 1)(y − z) = z − y, that is (D + 2)(y − z) = 0, so y − z = Ae−2x.
Adding: (D + 1)(y + z) = (y + z) + 2ex, that is D(y + z) = 2ex, so y + z = 2ex + B.
Answer: y = ex + B/2 + (A/2)e−2x, z = ex + B/2 − (A/2)e−2x (two arbitrary constants, as the determinant (D + 1)2 − 1 has degree 2).
(ii)
Let N(t) be the population at time t (weeks): dN/dt = kN, so N = N0ekt.
Doubling in one week: 2N0 = N0ek, so k = log 2 and N = N02t.
Answer: after 4 weeks N = 24N0 = 16N0, sixteen times the initial number.Final answer: (i) y = ex + B/2 + (A/2)e−2x, z = ex + B/2 − (A/2)e−2x (two arbitrary constants, as the determinant (D + 1)2 − 1 has degree 2); (ii) after 4 weeks N = 24N0 = 16N0, sixteen times the initial number
Q6(b) · Ordinary Differential Equations · 17 marks≈ 12 min
First Order Higher Degree Equations, Clairaut's Equation and Singular Solutions
(i) Consider the differential equation xyp2 − (x2 + y2 − 1)p + xy = 0 where p = dy/dx. Substituting u = x2 and v = y2 reduce the equation to Clairaut's form in terms of u, v and p′ = dv/du. Hence, or otherwise solve the equation. (10 marks)
(ii) Solve the following initial value differential equations:
20y″ + 4y′ + y = 0, y(0) = 3·2 and y′(0) = 0. (7 marks)हिंदी में प्रश्न
(i) अवकल समीकरण : xyp2 − (x2 + y2 − 1)p + xy = 0, जहाँ पर p = dy/dx है, पर विचार कीजिए। u = x2 तथा v = y2 द्वारा प्रतिस्थापन कर क्लेराउट्स रूप (Clairaut's form) में u, v तथा p′ = dv/du में व्यक्त कीजिए। अतएव या अन्यथा समीकरण को हल कीजिए। (10 अंक)
(ii) निम्नलिखित प्रारम्भिक-मान अवकल समीकरणों को हल कीजिए :
20y″ + 4y′ + y = 0, y(0) = 3·2 व y′(0) = 0। (7 अंक)Worked solution
(i)
With u = x2, v = y2: p′ = dv/du = yp/x, so p = xp′/y. Substituting and multiplying by y/x: x2p′2 − (x2 + y2 − 1)p′ + y2 = 0, that is up′2 − (u + v − 1)p′ + v = 0.
Rearranging: v(1 − p′) = up′(1 − p′) − p′, that is v = up′ − p′/(1 − p′), Clairaut's form.
General solution: v = cu − c/(1 − c), that is y2 = cx2 − c/(1 − c).
Singular solution: u − 1/(1 − p′)2 = 0 gives 1 − p′ = ±1/√u and then v = (√u ∓ 1)2, that is y2 = (x ∓ 1)2.
Answer: y2 = cx2 − c/(1 − c); singular solutions the four lines x ± y = ±1 (check y = x − 1, p = 1: 2xy − x2 − y2 + 1 = 1 − (x − y)2 = 0).
(ii)
Auxiliary equation 20m2 + 4m + 1 = 0: m = (−4 ± √(16 − 80))/40 = −1/10 ± i/5, so y = e−x/10(A cos(x/5) + B sin(x/5)).
y(0) = A = 3.2 (written 3·2 in the paper) and y′(0) = −A/10 + B/5 = 0, so B = 1.6.
Answer: y = e−0.1x(3.2 cos 0.2x + 1.6 sin 0.2x).Final answer: (i) y2 = cx2 − c/(1 − c); singular solutions the four lines x ± y = ±1 (check y = x − 1, p = 1: 2xy − x2 − y2 + 1 = 1 − (x − y)2 = 0); (ii) y = e−0.1x(3.2 cos 0.2x + 1.6 sin 0.2x)
Q6(c) · Dynamics and Statics · 17 marks≈ 12 min
Virtual Work, Stability and Forces in Three Dimensions
A uniform solid hemisphere rests on a rough plane inclined to the horizon at an angle φ with its curved surface touching the plane. Find the greatest admissible value of the inclination φ for equilibrium. If φ be less than this value, is the equilibrium stable?
हिंदी में प्रश्न
एक एकसमान अर्धगोला, क्षैतिज से φ कोण पर झुके हुए रुक्ष समतल पर अपने वक्रीय पृष्ठ को समतल से स्पर्श करते हुए रखा है। सन्तुलन के लिए φ का अधिकतम ग्राह्य मान निकालिए। यदि φ का मान इससे कम है तब क्या यह साम्यावस्था स्थिर है?
Worked solution
Let O be the centre and a the radius; G is on the axis with OG = 3a/8, and the contact point P is the foot of the normal from O to the plane, so OP makes angle φ with the vertical.
Only the weight and the reaction at P act, so G must be vertically above P: the horizontal offset a sin φ of O from P must be cancelled by OG, so (3a/8) sin β = a sin φ, β being the angle of OG with the vertical.
This needs sin φ ≤ 3/8, so the greatest inclination is sin−1(3/8) ≈ 22°, provided friction can hold the body (μ ≥ tan φ, which at the limit is μ ≥ 3/√55).
Stability: if the body rolls down the plane through a small angle χ, O moves parallel to the plane through aχ, and the height of G is z(χ) = constant − aχ sin φ − (3a/8) cos(β + χ).
dz/dχ = 0 at χ = 0 gives back sin β = (8/3) sin φ, and d2z/dχ2 = (3a/8) cos β > 0 because G is below O (β acute).
Answer: the greatest inclination is sin−1(3/8); for smaller φ the equilibrium is stable.Final answer: the greatest inclination is sin−1(3/8); for smaller φ the equilibrium is stable.
Question 7 (50 marks)
Q7(a) · Vector Analysis · 16 marks≈ 12 min
Curves in Space: Curvature, Torsion and Serret-Frenet Formulae
Find the curvature vector and its magnitude at any point r̄ = (θ) of the curve r̄ = (a cos θ, a sin θ, aθ). Show that the locus of the feet of the perpendicular from the origin to the tangent is a curve that completely lies on the hyperboloid x2 + y2 − z2 = a2.
हिंदी में प्रश्न
वक्र r̄ = (a cos θ, a sin θ, aθ) के किसी भी बिन्दु r̄ = (θ) पर वक्रता-सदिश व इसका परिमाण निकालिए। दर्शाइये कि मूल बिन्दु से स्पर्श रेखा पर डाले गये लम्बपाद का बिन्दु-पथ एक वक्र है जो पूर्णरूप से अतिपरवलयज x2 + y2 − z2 = a2 पर स्थित है।
Worked solution
This is the helix of Example 1 with b = a: |r′| = √2a, so ds = √2a dθ and T = (−sin θ, cos θ, 1)/√2.
dT/ds = (1/(√2a)) dT/dθ = −(cos θ, sin θ, 0)/(2a): this is the curvature vector, of magnitude κ = 1/(2a), directed horizontally towards the axis.
The foot F of the perpendicular from O to the tangent at P is F = r − (r·T)T, and r·T = (−a cos θ sin θ + a sin θ cos θ + aθ)/√2 = aθ/√2.
So F = r − (aθ/2)(−sin θ, cos θ, 1) = (a cos θ + (aθ/2) sin θ, a sin θ − (aθ/2) cos θ, aθ/2).
Then x2 + y2 = a2 + a2θ2/4 (the cross terms cancel) and z2 = a2θ2/4.
Answer: curvature vector −(cos θ i + sin θ j)/(2a), magnitude 1/(2a); x2 + y2 − z2 = a2 for every θ, so the locus of F lies on the hyperboloid.Final answer: curvature vector −(cos θ i + sin θ j)/(2a), magnitude 1/(2a); x2 + y2 − z2 = a2 for every θ, so the locus of F lies on the hyperboloid.
Q7(b) · Ordinary Differential Equations · 17 marks≈ 12 min
Second Order Equations with Variable Coefficients
(i) Solve the differential equation:
x d2y/dx2 − dy/dx − 4x3y = 8x3 sin(x2). (9 marks)
(ii) Solve the following differential equation using method of variation of parameters:
d2y/dx2 − dy/dx − 2y = 44 − 76x − 48x2. (8 marks)हिंदी में प्रश्न
(i) निम्नलिखित अवकल समीकरण को हल कीजिए :
x d2y/dx2 − dy/dx − 4x3y = 8x3 sin(x2)। (9 अंक)
(ii) निम्नलिखित अवकल समीकरण को प्राचल-विचरण विधि के द्वारा हल कीजिए :
d2y/dx2 − dy/dx − 2y = 44 − 76x − 48x2। (8 अंक)Q7(c) · Dynamics and Statics · 17 marks≈ 12 min
Constrained Motion, Work and Energy
A particle is free to move on a smooth vertical circular wire of radius a. At time t = 0 it is projected along the circle from its lowest point A with velocity just sufficient to carry it to the highest point B. Find the time T at which the reaction between the particle and the wire is zero.
हिंदी में प्रश्न
एक कण त्रिज्या a के चिकनी ऊर्ध्वाधर वृत्ताकार तार पर चलने में स्वतंत्र है। प्रारम्भिक समय t = 0 पर इसे वृत्त के निम्नतम बिन्दु A से ऐसे वेग से, वृत्त के अनुदिश फेंका जाता है जो इसे मात्र उच्चतम बिन्दु B तक ले जाने में ही सक्षम है। समय T को ज्ञात कीजिए जिस पर कण व तार के बीच प्रतिक्रिया शून्य हो।
Worked solution
The particle is on a wire, so the reaction can act either way; 'just sufficient to reach B' means v = 0 at B, so u2 = 4ga.
With θ measured from A: v2 = 4ga − 2ga(1 − cos θ) = 2ga(1 + cos θ) = 4ga cos2(θ/2).
Normal equation (R towards the centre): R − mg cos θ = mv2/a, so R = mg(2 + 3 cos θ), which vanishes when cos θ = −2/3.
Time: a dθ/dt = 2√(ga) cos(θ/2), so dt = (1/2)√(a/g) sec(θ/2) dθ and t = √(a/g) log[sec(θ/2) + tan(θ/2)].
At cos θ = −2/3, cos2(θ/2) = (1 + cos θ)/2 = 1/6, so sec(θ/2) = √6 and tan(θ/2) = √5.
Answer: T = √(a/g) log(√6 + √5) ≈ 1.54√(a/g) (the particle itself only approaches B as t → ∞).Final answer: T = √(a/g) log(√6 + √5) ≈ 1.54√(a/g) (the particle itself only approaches B as t → ∞).
Question 8 (50 marks)
Q8(a) · Dynamics and Statics · 16 marks≈ 12 min
Constrained Motion, Work and Energy
A spherical shot of W gm weight and radius r cm, lies at the bottom of cylindrical bucket of radius R cm. The bucket is filled with water up to a depth of h cm (h > 2r). Show that the minimum amount of work done in lifting the shot just clear of the water must be [W(h − 4r3/(3R2)) + W′(r − h + 2r3/(3R2))] cm gm. W′ gm is the weight of water displaced by the shot.
हिंदी में प्रश्न
W ग्राम भार व r त्रिज्या का एक गोला R से.मी. त्रिज्या की बेलनाकार बाल्टी की तली पर स्थित है। बाल्टी में h से.मी. (h > 2r) की गहराई तक पानी भरा हुआ है। दर्शाइये कि गोले को पानी की सतह के ठीक ऊपर लाने में किया गया न्यूनतम कार्य [W(h − 4r3/(3R2)) + W′(r − h + 2r3/(3R2))] से.मी. ग्राम होना चाहिए। W′ ग्राम गोले द्वारा विस्थापित पानी का भार है।
Worked solution
Let ρ be the weight of water per unit volume, so W′ = (4/3)πr3ρ; heights are measured from the bottom of the bucket.
When the shot leaves the water the level falls by (4/3)πr3/(πR2) = 4r3/(3R2), to h′ = h − 4r3/(3R2).
Shot: its centre rises from height r to h′ + r (just clear of the new surface), so the work against its weight is Wh′ = W(h − 4r3/(3R2)).
Water at the start fills the cylinder of depth h except the sphere, so its potential energy is ρπR2h2/2 − W′r; at the end it is ρπR2h′2/2.
Change = (ρπR2/2)(h′ − h)(h′ + h) + W′r = −(W′/2)(2h − 4r3/(3R2)) + W′r, using ρπR2 × 4r3/(3R2) = W′.
So the water's potential energy changes by W′(r − h + 2r3/(3R2)), a decrease since h > 2r; the least work, done slowly so that no kinetic energy is given, is the total increase in potential energy.
Answer: least work = W(h − 4r3/(3R2)) + W′(r − h + 2r3/(3R2)) cm gm.Final answer: least work = W(h − 4r3/(3R2)) + W′(r − h + 2r3/(3R2)) cm gm.
Q8(b) · Ordinary Differential Equations · 17 marks≈ 12 min
Laplace Transforms and Initial Value Problems
Solve the following initial value problem using Laplace transform:
d2y/dx2 + 9y = r(x), y(0) = 0, y′(0) = 4
where r(x) = 8 sin x if 0 < x < π; r(x) = 0 if x ≥ π.हिंदी में प्रश्न
निम्नलिखित प्रारम्भिक-मान समस्या को लैपलास रूपान्तरण के द्वारा हल कीजिए :
d2y/dx2 + 9y = r(x), y(0) = 0, y′(0) = 4
जहाँ पर r(x) = 8 sin x यदि 0 < x < π; r(x) = 0 यदि x ≥ π।Worked solution
r(x) = 8 sin x − 8 sin x·u(x − π) = 8 sin x + 8 sin(x − π)u(x − π), so L{r} = 8(1 + e−πs)/(s2 + 1).
(s2 + 9)Y − 4 = L{r}, so Y = 4/(s2 + 9) + 8(1 + e−πs)/((s2 + 1)(s2 + 9)), with 8/((s2 + 1)(s2 + 9)) = 1/(s2 + 1) − 1/(s2 + 9).
Inverting: y = (4/3) sin 3x + sin x − (1/3) sin 3x + [sin(x − π) − (1/3) sin 3(x − π)]u(x − π) = sin x + sin 3x + [−sin x + (1/3) sin 3x]u(x − π).
Answer: y = sin x + sin 3x for 0 ≤ x ≤ π and y = (4/3) sin 3x for x ≥ π (y and y′ are continuous at π: both pieces give 0 and −4).Final answer: y = sin x + sin 3x for 0 ≤ x ≤ π and y = (4/3) sin 3x for x ≥ π (y and y′ are continuous at π: both pieces give 0 and −4).
Q8(c) · Vector Analysis · 17 marks≈ 12 min
Gauss and Stokes' Theorems and Green's Identities
(i) Evaluate the integral: ∬S F̄ · n̂ ds where F̄ = 3xy2 î + (yx2 − y3)ĵ + 3zx2 k̂ and S is a surface of the cylinder y2 + z2 ≤ 4, −3 ≤ x ≤ 3, using divergence theorem. (9 marks)
(ii) Using Green's theorem, evaluate the ∫C F(r̄) · dr̄ counterclockwise where F(r̄) = (x2 + y2)î + (x2 − y2)ĵ and dr̄ = dx î + dy ĵ and the curve C is the boundary of the region R = {(x, y) | 1 ≤ y ≤ 2 − x2}. (8 marks)हिंदी में प्रश्न
(i) समाकलन ∬S F̄ · n̂ ds का अभिसरण प्रमेय से हल कीजिए :
जहाँ पर F̄ = 3xy2 î + (yx2 − y3)ĵ + 3zx2 k̂ तथा S बेलन y2 + z2 ≤ 4, −3 ≤ x ≤ 3 का पृष्ठ है। (9 अंक)
(ii) ग्रीन्स प्रमेय का प्रयोग कर ∫C F(r̄) · dr̄ का वामावर्ती दिशा में मूल्यांकन कीजिए :
जहाँ पर F(r̄) = (x2 + y2)î + (x2 − y2)ĵ तथा dr̄ = dx î + dy ĵ और वक्र C, क्षेत्र R = {(x, y) | 1 ≤ y ≤ 2 − x2} की परिधि है। (8 अंक)Worked solution
(i) ∇·F = 3y2 + (x2 − 3y2) + 3x2 = 4x2; the solid is a cylinder along the x-axis with cross-section y2 + z2 ≤ 4 of area 4π.
∬F·n dS = ∭4x2 dV = 4π∫−334x2 dx = 4π × 72 = 288π.
(ii) M = x2 + y2, N = x2 − y2; R lies between y = 1 and y = 2 − x2, which meet at x = ±1, and ∂N/∂x − ∂M/∂y = 2x − 2y.
The 2x term is odd in x and integrates to 0; the rest is −2∫−11∫12−x2y dy dx = −∫−11(3 − 4x2 + x4) dx = −(6 − 8/3 + 2/5) = −56/15.
Answer: (i) 288π; (ii) −56/15.Final answer: (i) 288π; (ii) −56/15.
UPSC Mains 2017 Mathematics Paper II
Question 1 (50 marks)
Q1(a) · Real Analysis · 10 marks≈ 7 min
Real Number System, Sequences, Cauchy Sequences and Completeness
Let x1 = 2 and xn+1 = √(xn + 20), n = 1, 2, 3, .... Show that the sequence x1, x2, x3, ... is convergent.
हिंदी में प्रश्न
मान लीजिए कि x1 = 2 और xn+1 = √(xn + 20), n = 1, 2, 3, ... है। दर्शाइए कि अनुक्रम x1, x2, x3, ... अभिसारी है।
Q1(b) · Abstract Algebra · 10 marks≈ 7 min
Permutation Groups and Cayley's Theorem
Let G be a group of order n. Show that G is isomorphic to a subgroup of the permutation group Sn.
हिंदी में प्रश्न
मान लीजिए कि G कोटि n का एक समूह है। दर्शाइए कि G क्रमचय समूह Sn के एक उपसमूह के समरूपी है।
Q1(c) · Real Analysis · 10 marks≈ 7 min
Real Number System, Sequences, Cauchy Sequences and Completeness
Find the supremum and the infimum of x/sin x on the interval (0, π/2].
हिंदी में प्रश्न
x/sin x का अन्तराल (0, π/2] पर उच्चक और निम्नक मान ज्ञात कीजिए।
Worked solution
Let f(x) = x/sin x on (0, π/2]. Then f′(x) = (sin x − x cos x)/sin2x, which equals cos x(tan x − x)/sin2x for x < π/2, and f′(π/2) = 1.
For 0 < x < π/2, tan x > x (g(x) = tan x − x has g(0) = 0 and g′(x) = sec2x − 1 > 0), so f′ > 0 and f is strictly increasing on (0, π/2].
Hence sup f = max f = f(π/2) = π/2.
inf f = limx→0+ x/sin x = 1, and it is not attained, because f is strictly increasing, so f(x) > 1 for every x in (0, π/2].
Answer: sup = π/2 (attained at x = π/2) and inf = 1 (not attained).Final answer: sup = π/2 (attained at x = π/2) and inf = 1 (not attained).
Q1(d) · Complex Analysis · 10 marks≈ 7 min
Singularities and Cauchy's Residue Theorem
Determine all entire functions f(z) such that 0 is a removable singularity of f(1/z).
हिंदी में प्रश्न
वे सभी सर्वत्र वैश्लेषिक फलनें f(z) ज्ञात कीजिए जिनके लिए 0 फलन f(1/z) की अपनेय विचित्रता है।
Worked solution
0 is a removable singularity of f(1/z), so limz→0 f(1/z) exists and f(1/z) is bounded on some 0 < |z| < δ; that is, |f(w)| ≤ M1 for |w| > 1/δ.
On the compact disc |w| ≤ 1/δ the continuous function f is bounded, say by M2.
So f is a bounded entire function and, by Liouville's theorem, constant. Conversely, if f is constant then f(1/z) is constant and 0 is removable.
Answer: exactly the constant functions f(z) = c.Final answer: exactly the constant functions f(z) = c.
Q1(e) · Linear Programming · 10 marks≈ 7 min
LPP Formulation, Basic Solutions and the Graphical Method
Using graphical method, find the maximum value of
2x + y
subject to
4x + 3y ≤ 12
4x + y ≤ 8
4x − y ≤ 8
x, y ≥ 0.हिंदी में प्रश्न
ग्राफी विधि के इस्तेमाल के द्वारा
2x + y
का उच्चतम मान, बशर्ते
4x + 3y ≤ 12
4x + y ≤ 8
4x − y ≤ 8
x, y ≥ 0
ज्ञात कीजिए।
Question 2 (50 marks)
Q2(a) · Real Analysis · 15 marks≈ 11 min
Riemann Integral, Improper Integrals and Fundamental Theorems of Integral Calculus
Let f(t) = ∫0t [x] dx, where [x] denotes the largest integer less than or equal to x.
(i) Determine all the real numbers t at which f is differentiable.
(ii) Determine all the real numbers t at which f is continuous but not differentiable.हिंदी में प्रश्न
मान लीजिए कि f(t) = ∫0t [x] dx, जहाँ [x] सबसे बड़ी पूर्ण संख्या है, जो x से छोटी या x के बराबर है, को दर्शाता है।
(i) निर्धारित कीजिए सभी वास्तविक संख्याएँ t, जहाँ f अवकलनीय है।
(ii) निर्धारित कीजिए सभी वास्तविक संख्याएँ t, जहाँ f संतत है लेकिन अवकलनीय नहीं है।Worked solution
[x] is bounded on bounded intervals and continuous except at the integers, so f(t) = ∫0t [x] dx is defined for every real t (for t < 0 it means −∫t0 [x] dx).
By the first fundamental theorem f is continuous on ℝ, and f′(t) = [t] at every t where [x] is continuous, i.e. at every t ∉ ℤ.
At an integer n: for n ≤ t < n + 1, f(t) = f(n) + n(t − n), so the right derivative is n; for n − 1 ≤ t < n, f(t) = f(n) − (n − 1)(n − t), so the left derivative is n − 1.
The one-sided derivatives differ, so f is not differentiable at any integer.
Answer: (i) f is differentiable exactly at t ∈ ℝ∖ℤ, with f′(t) = [t]; (ii) f is continuous everywhere, so it is continuous but not differentiable exactly at the integers.Final answer: (i) f is differentiable exactly at t ∈ ℝ∖ℤ, with f′(t) = [t]; (ii) f is continuous everywhere, so it is continuous but not differentiable exactly at the integers.
Q2(b) · Complex Analysis · 15 marks≈ 11 min
Contour Integration: Evaluation of Real Integrals
Using contour integral method, prove that
∫0∞ x sin mx/(a2 + x2) dx = (π/2) e−ma.हिंदी में प्रश्न
परिरेखा (कन्टूर) समाकल विधि के इस्तेमाल के द्वारा सिद्ध कीजिए कि
∫0∞ x sin mx/(a2 + x2) dx = (π/2) e−ma।Q2(c) · Abstract Algebra · 20 marks≈ 14 min
Integral Domains, PID, Euclidean Domains, UFD, Fields and Quotient Fields
Let F be a field and F[X] denote the ring of polynomials over F in a single variable X. For f(X), g(X) ∈ F[X] with g(X) ≠ 0, show that there exist q(X), r(X) ∈ F[X] such that degree (r(X)) < degree (g(X)) and
f(X) = q(X) . g(X) + r(X).हिंदी में प्रश्न
मान लीजिए कि F एक क्षेत्र है और F[X] सभी बहुपद, जो F में एकल चर X में हैं, का वलय द्योतित करता है। f(X), g(X) ∈ F[X] बशर्ते g(X) ≠ 0 के लिए, दर्शाइए कि ऐसे q(X), r(X) ∈ F[X] हैं जिनके लिए (r(X)) का घात (g(X)) के घात से छोटा है और
f(X) = q(X) . g(X) + r(X)।
Question 3 (50 marks)
Q3(a) · Abstract Algebra · 15 marks≈ 11 min
Normal Subgroups, Quotient Groups, Homomorphisms and Isomorphism Theorems
Show that the groups ℤ5 × ℤ7 and ℤ35 are isomorphic.
हिंदी में प्रश्न
दर्शाइए कि समूहें ℤ5 × ℤ7 एवं ℤ35 समरूपी हैं।
Worked solution
ℤ5 × ℤ7 has order 35, and o((1, 1)) = lcm(5, 7) = 35.
So ℤ5 × ℤ7 = ⟨(1, 1)⟩ is cyclic of order 35, and every cyclic group of order 35 is isomorphic to ℤ35.
Explicitly, φ(k) = (k mod 5, k mod 7) is a well-defined homomorphism ℤ35 → ℤ5 × ℤ7 (5 and 7 divide 35). φ(k) = (0, 0) ⇔ 5 | k and 7 | k ⇔ 35 | k, so φ is one-one, hence onto, as both groups have 35 elements.
Answer: ℤ5 × ℤ7 ≅ ℤ35, because gcd(5, 7) = 1.Final answer: ℤ5 × ℤ7 ≅ ℤ35, because gcd(5, 7) = 1.
Q3(b) · Complex Analysis · 15 marks≈ 11 min
Analytic Functions and Cauchy-Riemann Equations
Let f = u + iv be an analytic function on the unit disc D = {z ∈ ℂ : |z| < 1}. Show that
∂2u/∂x2 + ∂2u/∂y2 = 0 = ∂2v/∂x2 + ∂2v/∂y2
at all points of D.हिंदी में प्रश्न
मान लीजिए कि f = u + iv एकक डिस्क D = {z ∈ ℂ : |z| < 1} पर एक विश्लेषिक फलन है। दर्शाइए कि D के सभी बिन्दुओं पर
∂2u/∂x2 + ∂2u/∂y2 = 0 = ∂2v/∂x2 + ∂2v/∂y2
है।Q3(c) · Linear Programming · 20 marks≈ 14 min
The Simplex Method
Solve the following linear programming problem by simplex method:
Maximize
z = 3x1 + 5x2 + 4x3
subject to
2x1 + 3x2 ≤ 8
2x2 + 5x3 ≤ 10
3x1 + 2x2 + 4x3 ≤ 15
x1, x2, x3 ≥ 0.हिंदी में प्रश्न
एकधा विधि के द्वारा निम्नलिखित रैखिक प्रोग्रामन समस्या को हल कीजिए :
अधिकतमीकरण कीजिए
z = 3x1 + 5x2 + 4x3
बशर्ते कि
2x1 + 3x2 ≤ 8
2x2 + 5x3 ≤ 10
3x1 + 2x2 + 4x3 ≤ 15
x1, x2, x3 ≥ 0.
Question 4 (50 marks)
Q4(a) · Complex Analysis · 15 marks≈ 11 min
Power Series, Taylor's Series and Laurent's Series
For a function f : ℂ → ℂ and n ≥ 1, let f(n) denote the nth derivative of f and f(0) = f. Let f be an entire function such that for some n ≥ 1, f(n)(1/k) = 0 for all k = 1, 2, 3, .... Show that f is a polynomial.
हिंदी में प्रश्न
फलन f : ℂ → ℂ एवं n ≥ 1 के लिए, मान लीजिए कि f(n) फलन f का nवाँ अवकलज दर्शाता है और f(0) = f है। मान लीजिए कि f एक ऐसा सर्वत्र वैश्लेषिक फलन है जिसके लिए किसी n ≥ 1 के लिए, f(n)(1/k) = 0, सभी k = 1, 2, 3, ..., के लिए। दर्शाइए कि f एक बहुपद है।
Worked solution
g = f(n) is entire and g(1/k) = 0 for k = 1, 2, 3, ...; the zeros 1/k are distinct and converge to 0, a point of the domain ℂ.
Zeros of a non-zero analytic function are isolated, so by the identity theorem g ≡ 0 on ℂ.
The Taylor series f(z) = ∑ akzk, ak = f(k)(0)/k!, converges on all of ℂ; for k ≥ n, f(k) = g(k−n) ≡ 0, so ak = 0.
Answer: f(z) = a0 + a1z + ... + an−1zn−1, a polynomial of degree at most n − 1.Final answer: f(z) = a0 + a1z + ... + an−1zn−1, a polynomial of degree at most n − 1.
Q4(b) · Linear Programming · 15 marks≈ 11 min
Transportation Problem
Find the initial basic feasible solution of the following transportation problem using Vogel's approximation method and find the cost.
[Table: costs from origins O1, O2, O3 to destinations D1 to D5]
O1: 4, 7, 0, 3, 6; supply 14
O2: 1, 2, −3, 3, 8; supply 9
O3: 3, −1, 4, 0, 5; supply 17
Demand: 8, 3, 8, 13, 8हिंदी में प्रश्न
निम्नलिखित परिवहन समस्या के लिए, वोगेल की सन्निकटन विधि के द्वारा, आरंभिक आधारिक सुसंगत हल ज्ञात कीजिए तथा लागत ज्ञात कीजिए।
[सारणी: उद्गम O1, O2, O3 से गन्तव्य D1 से D5 तक लागत]
O1: 4, 7, 0, 3, 6; पूर्ति 14
O2: 1, 2, −3, 3, 8; पूर्ति 9
O3: 3, −1, 4, 0, 5; पूर्ति 17
माँग: 8, 3, 8, 13, 8Q4(c) · Real Analysis · 20 marks≈ 14 min
Series: Convergence Tests, Absolute and Conditional Convergence, Rearrangement
Let ∑n=1∞ xn be a conditionally convergent series of real numbers. Show that there is a rearrangement ∑n=1∞ xπ(n) of the series ∑n=1∞ xn that converges to 100.
हिंदी में प्रश्न
मान लीजिए कि ∑n=1∞ xn एक वास्तविक संख्याओं की सप्रतिबन्ध अभिसारी श्रेणी है। दर्शाइए कि श्रेणी ∑n=1∞ xn का एक पुनर्विन्यास ∑n=1∞ xπ(n) होगा जो 100 को अभिसरित होता है।
Question 5 (50 marks)
Q5(a) · Partial Differential Equations · 10 marks≈ 7 min
Linear Second-Order PDEs with Constant Coefficients
Solve (D2 − 2DD′ + D′2)z = ex+2y + x3 + sin 2x,
where D ≡ ∂/∂x, D′ ≡ ∂/∂y, D2 ≡ ∂2/∂x2, D′2 ≡ ∂2/∂y2.हिंदी में प्रश्न
(D2 − 2DD′ + D′2)z = ex+2y + x3 + sin 2x को हल कीजिए,
जहाँ D ≡ ∂/∂x, D′ ≡ ∂/∂y, D2 ≡ ∂2/∂x2, D′2 ≡ ∂2/∂y2 हैं।Worked solution
Auxiliary (m − 1)2 = 0, so CF = φ1(y + x) + xφ2(y + x).
ex + 2y: F(1, 2) = (1 − 2)2 = 1, PI1 = ex + 2y; x3: PI2 = (1/D2)(1 − D′/D)−2x3 = (1/D2)x3 = x5/20 since D′x3 = 0.
sin 2x: a = 2, b = 0, so D2 → −4, DD′ → 0, D′2 → 0 and PI3 = −(1/4)sin 2x.
Answer: z = φ1(y + x) + xφ2(y + x) + ex + 2y + x5/20 − (1/4)sin 2x.Final answer: z = φ1(y + x) + xφ2(y + x) + ex + 2y + x5/20 − (1/4)sin 2x.
Q5(b) · Numerical Analysis and Computer Programming · 10 marks≈ 7 min
Systems of Linear Equations: Direct and Iterative Methods
Explain the main steps of the Gauss-Jordan method and apply this method to find the inverse of the matrix
[[2, 6, 6], [2, 8, 6], [2, 6, 8]].हिंदी में प्रश्न
गाउस-जॉर्डन विधि के मुख्य सोपानों की व्याख्या कीजिए तथा इस विधि का इस्तेमाल करते हुए आव्यूह
[[2, 6, 6], [2, 8, 6], [2, 6, 8]]
का प्रतिलोम ज्ञात कीजिए।Q5(c) · Numerical Analysis and Computer Programming · 10 marks≈ 7 min
Computer Systems, Logic Gates, Boolean Algebra and Algorithms
Write the Boolean expression
z(y + z)(x + y + z)
in its simplest form using Boolean postulate rules. Mention the rules used during simplification. Verify your result by constructing the truth table for the given expression and for its simplest form.हिंदी में प्रश्न
बूलीय अभिगृहीत नियमों का इस्तेमाल करते हुए बूलीय व्यंजक
z(y + z)(x + y + z)
को इसके सरलतम रूप में लिखिए। सरलीकरण के दौरान इस्तेमाल होने वाले नियमों का उल्लेख कीजिए। दिए गए व्यंजक तथा उसके सरलतम रूप के लिए सत्यमान सारणी बनाकर अपने परिणाम को सत्यापित कीजिए।Worked solution
z(y + z) = z(z + y) (commutative law) = z (absorption law a(a + b) = a).
Then z(x + y + z) = z(z + (x + y)) (commutative and associative laws) = z (absorption law again).
Truth table over xyz = 000, ..., 111: y + z = 0, 1, 1, 1, 0, 1, 1, 1 and x + y + z = 0, 1, 1, 1, 1, 1, 1, 1, so z(y + z)(x + y + z) = 0, 1, 0, 1, 0, 1, 0, 1, which is exactly the column of z.
Answer: z(y + z)(x + y + z) = z.Final answer: z(y + z)(x + y + z) = z.
Q5(d) · Partial Differential Equations · 10 marks≈ 7 min
Vibrating String, Heat Equation and Laplace Equation
Let Γ be a closed curve in xy-plane and let S denote the region bounded by the curve Γ. Let
∂2w/∂x2 + ∂2w/∂y2 = f(x, y) ∀ (x, y) ∈ S.
If f is prescribed at each point (x, y) of S and w is prescribed on the boundary Γ of S, then prove that any solution w = w(x, y), satisfying these conditions, is unique.हिंदी में प्रश्न
माना xy-समतल में Γ एक संवृत वक्र है तथा S उसके वक्र Γ द्वारा सीमाबद्ध क्षेत्र को निर्दिष्ट करता है। माना कि
∂2w/∂x2 + ∂2w/∂y2 = f(x, y) ∀ (x, y) ∈ S है।
यदि S के प्रत्येक बिन्दु (x, y) पर f निर्धारित है तथा S की सीमा Γ पर w निर्धारित है, तो सिद्ध कीजिए कि इन शर्तों को पूरा करने वाला कोई भी हल w = w(x, y) एकमात्र हल है।Worked solution
Let w1, w2 be two solutions and w = w1 − w2; then ∇2w = f − f = 0 in S and w = 0 on Γ.
Green's first identity (the divergence theorem for w∇w): ∬S (w∇2w + wx2 + wy2) dx dy = ∮Γ w ∂w/∂n ds.
Both ∇2w = 0 in S and w = 0 on Γ, so ∬S (wx2 + wy2) dx dy = 0.
The integrand is continuous and non-negative, so wx = wy = 0 in S; w is constant, and continuity up to Γ, where w = 0, gives w ≡ 0 (alternatively, apply the maximum principle to the harmonic w).
Answer: w1 = w2 throughout S, so the solution is unique. Hence proved.Final answer: w1 = w2 throughout S, so the solution is unique. Hence proved.
Q5(e) · Mechanics and Fluid Dynamics · 10 marks≈ 7 min
Moment of Inertia and Motion of Rigid Bodies in Two Dimensions
Show that the moment of inertia of an elliptic area of mass M and semi-axis a and b about a semi-diameter of length r is (1/4)M a2b2/r2. Further, prove that the moment of inertia about a tangent is (5M/4)p2, where p is the perpendicular distance from the centre of the ellipse to the tangent.
हिंदी में प्रश्न
दर्शाइए कि एक दीर्घवृत्ताकार क्षेत्रफल, जिसका द्रव्यमान M है तथा अर्धाक्ष a व b हैं, का अर्ध-व्यास, जिसकी लम्बाई r है, के इर्दगिर्द जड़त्व आघूर्ण (1/4)M a2b2/r2 होगा। आगे सिद्ध कीजिए कि स्पर्शरेखा के इर्दगिर्द जड़त्व आघूर्ण (5M/4)p2 होगा, जहाँ p दीर्घवृत्त के केन्द्र से स्पर्शरेखा की लम्बवत् दूरी है।
Worked solution
About the axes A = Mb2/4, B = Ma2/4, F = 0, so about a diameter at angle θ to the major axis I = A cos2θ + B sin2θ = (M/4)(b2cos2θ + a2sin2θ).
The semi-diameter ends at (r cos θ, r sin θ) on the ellipse, so 1/r2 = cos2θ/a2 + sin2θ/b2, i.e. b2cos2θ + a2sin2θ = a2b2/r2; hence I = Ma2b2/(4r2).
The tangent parallel to this diameter touches at an end of the conjugate diameter, and its distance p from the centre satisfies pr = ab (area of the parallelogram on conjugate semi-diameters).
Parallel axes: Itangent = Ma2b2/(4r2) + Mp2 = Mp2/4 + Mp2.
Answer: Ma2b2/(4r2) about the semi-diameter and 5Mp2/4 about the tangent.Final answer: Ma2b2/(4r2) about the semi-diameter and 5Mp2/4 about the tangent.
Question 6 (50 marks)
Q6(a) · Partial Differential Equations · 15 marks≈ 11 min
Cauchy's Method of Characteristics and Nonlinear First-Order PDEs
Find a complete integral of the partial differential equation
2(pq + yp + qx) + x2 + y2 = 0.हिंदी में प्रश्न
आंशिक अवकल समीकरण
2(pq + yp + qx) + x2 + y2 = 0
का पूर्ण समाकल ज्ञात कीजिए।Worked solution
F = 2pq + 2yp + 2qx + x2 + y2: Fx = 2(q + x), Fy = 2(p + y), Fz = 0, Fp = 2(q + y), Fq = 2(p + x).
Charpit: dx/(2(q + y)) = dy/(2(p + x)) = dp/(−2(q + x)) = dq/(−2(p + y)); then d(p + x) has denominator 2(y − x) and d(q + y) has 2(x − y), so d(p + x) = −d(q + y) and (p + x) + (q + y) = a.
The PDE is 2(p + x)(q + y) = 2xy − x2 − y2, so P = p + x and Q = q + y satisfy P + Q = a, PQ = −(x − y)2/2, giving P, Q = [a ± √(a2 + 2(x − y)2)]/2.
dz = (P − x)dx + (Q − y)dy = (a/2)d(x + y) + (1/2)√(a2 + 2v2) dv − x dx − y dy with v = x − y.
∫√(a2 + 2v2) dv = (v/2)√(a2 + 2v2) + (a2/(2√2)) log(√2v + √(a2 + 2v2)).
Answer: z = a(x + y)/2 − (x2 + y2)/2 + ((x − y)/4)√(a2 + 2(x − y)2) + (a2/(4√2)) log[√2(x − y) + √(a2 + 2(x − y)2)] + b (verified numerically).Final answer: z = a(x + y)/2 − (x2 + y2)/2 + ((x − y)/4)√(a2 + 2(x − y)2) + (a2/(4√2)) log[√2(x − y) + √(a2 + 2(x − y)2)] + b (verified numerically).
Q6(b) · Numerical Analysis and Computer Programming · 15 marks≈ 11 min
Interpolation: Newton's Forward, Backward and Lagrange's Formulae
For given equidistant values u−1, u0, u1 and u2, a value is interpolated by Lagrange's formula. Show that it may be written in the form
ux = yu0 + xu1 + (y(y2 − 1)/3!) Δ2u−1 + (x(x2 − 1)/3!) Δ2u0,
where x + y = 1.हिंदी में प्रश्न
दिए गए समदूरस्थ मान u−1, u0, u1 और u2 के लिए, एक मान लग्रांज फॉर्मूला से अंतर्वेशित किया जाता है। दर्शाइए कि इसको
ux = yu0 + xu1 + (y(y2 − 1)/3!) Δ2u−1 + (x(x2 − 1)/3!) Δ2u0
के रूप में लिखा जा सकता है, जहाँ x + y = 1 है।Worked solution
Take h = 1, nodes −1, 0, 1, 2 and y = 1 − x; the right side R(x) is then a polynomial of degree ≤ 3 in x. By uniqueness, R is Lagrange's cubic through the four points if R(k) = uk for k = −1, 0, 1, 2.
x = 0 (y = 1): both cubic terms vanish since y(y2 − 1) = 0 = x(x2 − 1), so R = u0; x = 1 (y = 0): R = u1.
x = −1 (y = 2): R = 2u0 − u1 + (2 · 3/6)Δ2u−1 = 2u0 − u1 + (u1 − 2u0 + u−1) = u−1.
x = 2 (y = −1): R = −u0 + 2u1 + (2 · 3/6)Δ2u0 = −u0 + 2u1 + (u2 − 2u1 + u0) = u2.
Answer: R agrees with u at all four nodes, so it is the Lagrange interpolating cubic; this is Everett's formula up to second differences.Final answer: R agrees with u at all four nodes, so it is the Lagrange interpolating cubic; this is Everett's formula up to second differences.
Q6(c) · Mechanics and Fluid Dynamics · 20 marks≈ 14 min
Generalized Coordinates, D'Alembert's Principle and Lagrange's Equations
Two uniform rods AB, AC, each of mass m and length 2a, are smoothly hinged together at A and move on a horizontal plane. At time t, the mass centre of the rods is at the point (ξ, η) referred to fixed perpendicular axes Ox, Oy in the plane, and the rods make angles θ ± φ with Ox. Prove that the kinetic energy of the system is
m[ξ̇2 + η̇2 + (1/3 + sin2φ)a2θ̇2 + (1/3 + cos2φ)a2φ̇2].
Also derive Lagrange's equations of motion for the system if an external force with components [X, Y] along the axes acts at A.हिंदी में प्रश्न
दो एकसमान छड़ों AB, AC को, जिसमें प्रत्येक का द्रव्यमान m तथा लम्बाई 2a है, मसृण तरीक़े से A पर कब्ज़े के साथ जोड़ा गया है तथा वे क्षैतिज समतल पर गतिमान हैं। समय t पर, समतल में स्थिर लंब अक्ष Ox, Oy के सन्दर्भ में छड़ों का द्रव्यमान केन्द्र बिन्दु (ξ, η) पर है तथा छड़ें Ox के साथ θ ± φ का कोण बनाती हैं। सिद्ध कीजिए कि तंत्र की गतिज ऊर्जा
m[ξ̇2 + η̇2 + (1/3 + sin2φ)a2θ̇2 + (1/3 + cos2φ)a2φ̇2] है।
साथ ही तंत्र के लिए गति के लग्रांज समीकरणों को भी व्युत्पन्न कीजिए, यदि घटकों [X, Y] सहित एक बाहरी बल अक्षों के साथ-साथ, A बिन्दु पर क्रिया करता है।Worked solution
Let A = (xA, yA). The centres of AB and AC are A + a(cos(θ ± φ), sin(θ ± φ)), so their mean is G = (ξ, η) = A + a cos φ(cos θ, sin θ).
Relative to G the rod-centres are at ±a sin φ(−sin θ, cos θ), with squared speed a2(φ̇2cos2φ + θ̇2sin2φ); the rods turn with angular velocities θ̇ ± φ̇.
Summing (1/2)mv2 + (1/2)(ma2/3)ω2 (the cross terms cancel): T = m(ξ̇2 + η̇2) + ma2(φ̇2cos2φ + θ̇2sin2φ) + (ma2/3)(θ̇2 + φ̇2), which is the stated form.
Generalized forces: xA = ξ − a cos φ cos θ, yA = η − a cos φ sin θ, so Qξ = X, Qη = Y, Qθ = a cos φ(X sin θ − Y cos θ), Qφ = a sin φ(X cos θ + Y sin θ).
Lagrange's equations: 2mξ̈ = X, 2mη̈ = Y, d/dt[2ma2(1/3 + sin2φ)θ̇] = a cos φ(X sin θ − Y cos θ).
and 2ma2(1/3 + cos2φ)φ̈ − ma2sin 2φ(φ̇2 + θ̇2) = a sin φ(X cos θ + Y sin θ).
Answer: T as stated, and the four equations above; the first two say that G moves like a particle of mass 2m under the force (X, Y).Final answer: T as stated, and the four equations above; the first two say that G moves like a particle of mass 2m under the force (X, Y).
Question 7 (50 marks)
Q7(a) · Partial Differential Equations · 15 marks≈ 11 min
Classification and Canonical Forms of Second-Order PDEs
Reduce the equation
y2 ∂2z/∂x2 − 2xy ∂2z/∂x∂y + x2 ∂2z/∂y2 = (y2/x) ∂z/∂x + (x2/y) ∂z/∂y
to canonical form and hence solve it.हिंदी में प्रश्न
समीकरण
y2 ∂2z/∂x2 − 2xy ∂2z/∂x∂y + x2 ∂2z/∂y2 = (y2/x) ∂z/∂x + (x2/y) ∂z/∂y
को विहित रूप में समानीत कीजिए और अतएव इसका हल ज्ञात कीजिए।Worked solution
A = y2, B = −2xy, C = x2: B2 − 4AC = 0, parabolic; (y dy/dx + x)2 = 0 gives ξ = x2 + y2; take η = x.
zx = 2xzξ + zη, zy = 2yzξ, zxx = 2zξ + 4x2zξξ + 4xzξη + zηη, zxy = 4xyzξξ + 2yzξη, zyy = 2zξ + 4y2zξξ.
LHS = 2(x2 + y2)zξ + y2zηη (the zξξ, zξη terms cancel); RHS = (y2/x)(2xzξ + zη) + (x2/y)(2yzξ) = 2(x2 + y2)zξ + (y2/x)zη.
So zηη = zη/η (with η = x2 instead, this becomes zηη = 0); then ∂(zη/η)/∂η = 0, zη = ηf(ξ) and z = (η2/2)f(ξ) + g(ξ).
Answer: canonical form zηη = (1/η)zη (ξ = x2 + y2, η = x); solution z = x2f(x2 + y2) + g(x2 + y2).Final answer: canonical form zηη = (1/η)zη (ξ = x2 + y2, η = x); solution z = x2f(x2 + y2) + g(x2 + y2).
Q7(b) · Numerical Analysis and Computer Programming · 20 marks≈ 14 min
Numerical Integration and Numerical Solution of ODEs
Derive the formula
∫ab y dx = (3h/8)[(y0 + yn) + 3(y1 + y2 + y4 + y5 + ... + yn−1) + 2(y3 + y6 + ... + yn−3)].
Is there any restriction on n? State that condition. What is the error bound in the case of Simpson's 3/8 rule?हिंदी में प्रश्न
सूत्र
∫ab y dx = (3h/8)[(y0 + yn) + 3(y1 + y2 + y4 + y5 + ... + yn−1) + 2(y3 + y6 + ... + yn−3)]
को व्युत्पन्न कीजिए। क्या n पर कोई प्रतिबन्ध है? उस शर्त को लिखिए। सिम्प्सन के 3/8 नियम के मामले में त्रुटि परिबंध क्या है?Worked solution
With x = x0 + ph and Newton's forward formula up to Δ3: ∫x0x3 y dx = h∫03[y0 + pΔy0 + p(p − 1)/2 Δ2y0 + p(p − 1)(p − 2)/6 Δ3y0] dp = h[3y0 + (9/2)Δy0 + (9/4)Δ2y0 + (3/8)Δ3y0].
Putting Δy0 = y1 − y0, Δ2y0 = y2 − 2y1 + y0, Δ3y0 = y3 − 3y2 + 3y1 − y0, the coefficients are y0: 3 − 9/2 + 9/4 − 3/8 = 3/8; y1: 9/2 − 9/2 + 9/8 = 9/8; y2: 9/4 − 9/8 = 9/8; y3: 3/8.
So ∫x0x3 y dx = (3h/8)(y0 + 3y1 + 3y2 + y3); adding the same rule over [x3, x6], ..., [xn−3, xn], the junction ordinates y3, y6, ... appear twice, which gives the stated composite formula.
Restriction: n must be a multiple of 3, since the strips are used three at a time.
Answer: one-panel error −(3/80)h5y⁗(ξ); over n/3 panels the error is −((b − a)/80)h4y⁗(ξ), so |E| ≤ ((b − a)h4/80) max|y⁗| on [a, b] (the rule is exact for cubics).Final answer: one-panel error −(3/80)h5y⁗(ξ); over n/3 panels the error is −((b − a)/80)h4y⁗(ξ), so |E| ≤ ((b − a)h4/80) max|y⁗| on [a, b] (the rule is exact for cubics).
Q7(c) · Mechanics and Fluid Dynamics · 15 marks≈ 11 min
Euler's Equation, Bernoulli's Theorem and the Navier–Stokes Equation
A stream is rushing from a boiler through a conical pipe, the diameters of the ends of which are D and d. If V and v be the corresponding velocities of the stream and if the motion is assumed to be steady and diverging from the vertex of the cone, then prove that
v/V = (D2/d2) e(v2 − V2)/2K,
where K is the pressure divided by the density and is constant.हिंदी में प्रश्न
एक बॉयलर से एक शंक्वाकार नलिका में से जिसके सिरों के व्यास D तथा d हैं, धारा तेज़ी से निकल रही है। यदि V तथा v धारा के संगत-वेग हों तथा यदि माना जाए कि गति अपरिवर्ती है तथा शंकु-शीर्ष से अपसारित हो रही है, तो सिद्ध कीजिए कि
v/V = (D2/d2) e(v2 − V2)/2K होगा,
जहाँ K घनत्व द्वारा विभाजित दाब है तथा अपरिवर्ती है।Worked solution
The flow is radial from the vertex, so the cross-section at distance r is proportional to r2, i.e. to (diameter)2; continuity gives ρ1VD2 = ρ2vd2.
Bernoulli with p = Kρ (steady, no body force): K log ρ + q2/2 is constant, so K log(ρ1/ρ2) = (v2 − V2)/2 and ρ1/ρ2 = e(v2 − V2)/(2K).
From continuity v/V = (ρ1/ρ2)(D2/d2).
Answer: v/V = (D2/d2)e(v2 − V2)/(2K).Final answer: v/V = (D2/d2)e(v2 − V2)/(2K).
Question 8 (50 marks)
Q8(a) · Partial Differential Equations · 20 marks≈ 14 min
Vibrating String, Heat Equation and Laplace Equation
Given the one-dimensional wave equation
∂2y/∂t2 = c2 ∂2y/∂x2; t > 0,
where c2 = T/m, T is the constant tension in the string and m is the mass per unit length of the string.
(i) Find the appropriate solution of the above wave equation.
(ii) Find also the solution under the conditions
y(0, t) = 0, y(l, t) = 0 for all t
and [∂y/∂t]t=0 = 0, y(x, 0) = a sin(πx/l), 0 < x < l, a > 0.हिंदी में प्रश्न
दिया हुआ है एकविमीय तरंग समीकरण,
∂2y/∂t2 = c2 ∂2y/∂x2; t > 0,
जहाँ c2 = T/m, T तार में अपरिवर्ती तनाव है तथा m तार का प्रति इकाई लम्बाई द्रव्यमान है।
(i) उपर्युक्त तरंग समीकरण का यथोचित हल ज्ञात कीजिए।
(ii) प्रतिबन्धों
y(0, t) = 0, y(l, t) = 0 सभी t के लिए
तथा [∂y/∂t]t=0 = 0, y(x, 0) = a sin(πx/l), 0 < x < l, a > 0
के साथ भी हल निकालिए।Q8(b) · Numerical Analysis and Computer Programming · 15 marks≈ 11 min
Computer Systems, Logic Gates, Boolean Algebra and Algorithms
Write an algorithm in the form of a flow chart for Newton-Raphson method. Describe the cases of failure of this method.
हिंदी में प्रश्न
न्यूटन-रैफ्सन विधि के लिए प्रवाह चार्ट के रूप में एक ऐल्गोरिथ्म लिखिए। इस विधि की विफलता की स्थितियों का वर्णन कीजिए।
Worked solution
Flow chart: START → READ x0, ε, N → i = 1 → compute f(x0), f′(x0) → diamond '|f′(x0)| very small?' (YES: PRINT 'slope zero, choose another x0', STOP).
→ x1 = x0 − f(x0)/f′(x0) → diamond '|x1 − x0| < ε?' (YES: PRINT x1, STOP) → x0 = x1, i = i + 1 → diamond 'i > N?' (YES: PRINT 'no convergence', STOP; NO: back to computing f, f′).
Failure cases: (i) f′(xn) = 0 or nearly 0, so the tangent is horizontal and the next iterate is undefined or very far away; (ii) cycling, e.g. f(x) = x3 − 2x + 2 with x0 = 0 gives 0, 1, 0, 1, ...
(iii) Divergence, e.g. f(x) = x1/3 gives xn+1 = −2xn; (iv) a start near a turning point or inflection sends the iterates to a far root or away; (v) at a multiple root convergence is only linear (use xn+1 = xn − mf/f′), and complex roots cannot be reached from a real x0.
Answer: the flow chart above, with the five failure cases listed.Final answer: the flow chart above, with the five failure cases listed.
Q8(c) · Mechanics and Fluid Dynamics · 15 marks≈ 11 min
Kinematics of Fluids: Continuity, Streamlines and Potential Flow
If the velocity of an incompressible fluid at the point (x, y, z) is given by
(3xz/r5, 3yz/r5, (3z2 − r2)/r5), r2 = x2 + y2 + z2,
then prove that the liquid motion is possible and that the velocity potential is z/r3. Further, determine the streamlines.हिंदी में प्रश्न
यदि (x, y, z) बिन्दु पर असंपीड्य तरल का वेग
(3xz/r5, 3yz/r5, (3z2 − r2)/r5), r2 = x2 + y2 + z2
के द्वारा दत्त है, तो सिद्ध कीजिए कि द्रव गति सम्भव है तथा वेग विभव z/r3 है। और आगे, धारा-रेखाओं का भी निर्धारण कीजिए।Worked solution
With φ = z/r3: −∂φ/∂x = 3xz/r5, −∂φ/∂y = 3yz/r5, −∂φ/∂z = −1/r3 + 3z2/r5 = (3z2 − r2)/r5, which are u, v, w; so q = −∇φ.
∇·q = −∇2φ = 0, since z/r3 = −∂(1/r)/∂z and 1/r is harmonic for r ≠ 0; so the liquid motion is possible (and irrotational).
Streamlines: dx/(3xz) = dy/(3yz) gives y = c1x (planes through Oz); in spherical polars φ = cos θ/r2, qr = 2cos θ/r3, qθ = sin θ/r3, so dr/(2cos θ) = r dθ/sin θ and r = c2sin2θ.
Answer: φ = z/r3 (a doublet along Oz); streamlines y = c1x, (x2 + y2 + z2)3 = c(x2 + y2)2.Final answer: φ = z/r3 (a doublet along Oz); streamlines y = c1x, (x2 + y2 + z2)3 = c(x2 + y2)2.