UPSC Mains 2021 Mathematics Optional Question Papers with Solutions
All 56 questions of the 2021 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 2021 Mathematics Paper I
Question 1 (50 marks)
Q1(a) · Linear Algebra · 10 marks≈ 7 min
Eigenvalues, Eigenvectors and the Cayley-Hamilton Theorem
If A = [[1, −1, 1], [2, −1, 0], [1, 0, 0]], then show that A2 = A−1 (without finding A−1).
हिंदी में प्रश्न
यदि A = [[1, −1, 1], [2, −1, 0], [1, 0, 0]] है, तो A−1 को ज्ञात किए बिना दर्शाइए कि A2 = A−1
Q1(b) · Linear Algebra · 10 marks≈ 7 min
Linear Transformations, Rank-Nullity and Matrix Representation
Find the matrix associated with the linear operator on V3(R) defined by T(a, b, c) = (a + b, a − b, 2c) with respect to the ordered basis B = {(0, 1, 1), (1, 0, 1), (1, 1, 0)}.
हिंदी में प्रश्न
क्रमित आधारक B = {(0, 1, 1), (1, 0, 1), (1, 1, 0)} के सापेक्ष V3(R) पर परिभाषित रैखिक संकारक : T(a, b, c) = (a + b, a − b, 2c) से संबन्धित आव्यूह ज्ञात कीजिए।
Worked solution
Let b1 = (0, 1, 1), b2 = (1, 0, 1), b3 = (1, 1, 0). Then T(b1) = (1, −1, 2), T(b2) = (1, 1, 2), T(b3) = (2, 0, 0).
For (p, q, r) = xb1 + yb2 + zb3 we need y + z = p, x + z = q, x + y = r; with s = (p + q + r)/2 this gives x = s − p, y = s − q, z = s − r.
(1, −1, 2): s = 1, coordinates (0, 2, −1). (1, 1, 2): s = 2, coordinates (1, 1, 0). (2, 0, 0): s = 1, coordinates (−1, 1, 1).
Check: 2b2 − b3 = (2, 0, 2) − (1, 1, 0) = (1, −1, 2) = T(b1).
Answer: [T]B = [[0, 1, −1], [2, 1, 1], [−1, 0, 1]].Final answer: [T]B = [[0, 1, −1], [2, 1, 1], [−1, 0, 1]].
Q1(c) · Calculus · 10 marks≈ 7 min
Real Numbers, Limits, Continuity and Differentiability
Given : Δ(x) = |[f(x + α), f(x + 2α), f(x + 3α)], [f(α), f(2α), f(3α)], [f′(α), f′(2α), f′(3α)]| where f is a real valued differentiable function and α is a constant. Find limx→0 Δ(x)/x.
हिंदी में प्रश्न
दिया गया है : Δ(x) = |[f(x + α), f(x + 2α), f(x + 3α)], [f(α), f(2α), f(3α)], [f′(α), f′(2α), f′(3α)]| जहाँ f एक वास्तविक-मान अवकलनीय फलन है तथा α एक अचर है। limx→0 Δ(x)/x को ज्ञात कीजिए।
Worked solution
Δ(0) has its first two rows equal, so Δ(0) = 0 and limx→0 Δ(x)/x = lim (Δ(x) − Δ(0))/(x − 0) = Δ′(0).
Only the first row depends on x, and f is differentiable, so Δ′(x) is the determinant with first row (f′(x + α), f′(x + 2α), f′(x + 3α)) and the other two rows unchanged.
At x = 0 this first row is (f′(α), f′(2α), f′(3α)), identical to the third row, so Δ′(0) = 0.
Answer: limx→0 Δ(x)/x = 0.Final answer: limx→0 Δ(x)/x = 0.
Q1(d) · Calculus · 10 marks≈ 7 min
Mean Value Theorems, Taylor's Theorem and Indeterminate Forms
Show that between any two roots of ex cos x = 1, there exists at least one root of ex sin x − 1 = 0.
हिंदी में प्रश्न
दर्शाइए कि ex cos x = 1 के किन्हीं दो मूलों के बीच में ex sin x − 1 = 0 का कम से कम एक मूल विद्यमान है।
Worked solution
Let ψ(x) = e−x − cos x. Then ex cos x = 1 ⇔ cos x = e−x ⇔ ψ(x) = 0.
If α < β are roots of ex cos x = 1, then ψ(α) = ψ(β) = 0 and ψ is differentiable on ℝ, so Rolle's theorem gives c ∈ (α, β) with ψ′(c) = 0.
ψ′(x) = −e−x + sin x = e−x(ex sin x − 1), so ψ′(c) = 0 gives ec sin c − 1 = 0.
Answer: Between any two roots of ex cos x = 1 there is a root c of ex sin x − 1 = 0.Final answer: Between any two roots of ex cos x = 1 there is a root c of ex sin x − 1 = 0.
Q1(e) · Analytic Geometry · 10 marks≈ 7 min
Cone and Cylinder
Find the equation of the cylinder whose generators are parallel to the line x = −y/2 = z/3 and whose guiding curve is x2 + 2y2 = 1, z = 0.
हिंदी में प्रश्न
उस बेलन का समीकरण ज्ञात कीजिए जिसके जनक, रेखा : x = −y/2 = z/3 के समानान्तर हैं तथा जिसका निर्देशक-वक्र x2 + 2y2 = 1, z = 0 है।
Question 2 (50 marks)
Q2(a) · Analytic Geometry · 20 marks≈ 14 min
Cone and Cylinder
Show that the planes, which cut the cone ax2 + by2 + cz2 = 0 in perpendicular generators, touch the cone x2/(b + c) + y2/(c + a) + z2/(a + b) = 0.
हिंदी में प्रश्न
दर्शाइए कि वे समतल, जो कि शंकु ax2 + by2 + cz2 = 0 को लंब जनकों में काटते हैं, शंकु x2/(b + c) + y2/(c + a) + z2/(a + b) = 0 को स्पर्श करते हैं।
Worked solution
By the previous example, ux + vy + wz = 0 cuts the cone in perpendicular generators if and only if (b + c)u2 + (c + a)v2 + (a + b)w2 = 0.
The plane ux + vy + wz = 0 touches the cone αx2 + βy2 + γz2 = 0 if and only if u2/α + v2/β + w2/γ = 0 (the tangency condition with A = βγ, B = γα, C = αβ, divided by αβγ).
For α = 1/(b + c), β = 1/(c + a), γ = 1/(a + b) this condition reads (b + c)u2 + (c + a)v2 + (a + b)w2 = 0, the same as the perpendicularity condition.
Answer: every such plane touches x2/(b + c) + y2/(c + a) + z2/(a + b) = 0; hence proved.Final answer: every such plane touches x2/(b + c) + y2/(c + a) + z2/(a + b) = 0; hence proved.
Q2(b) · Calculus · 15 marks≈ 11 min
Functions of Several Variables: Partial Derivatives, Extrema, Lagrange Multipliers and Jacobians
Given that f(x, y) = |x2 − y2|. Find fxy(0, 0) and fyx(0, 0). Hence show that fxy(0, 0) = fyx(0, 0).
हिंदी में प्रश्न
दिया गया है : f(x, y) = |x2 − y2|, तब fxy(0, 0) तथा fyx(0, 0) ज्ञात कीजिए। अतः दर्शाइए कि fxy(0, 0) = fyx(0, 0)।
Worked solution
For y ≠ 0 and |h| < |y|, |h2 − y2| = y2 − h2, so fx(0, y) = limh→0 (y2 − h2 − y2)/h = lim (−h) = 0. At y = 0, fx(0, 0) = lim h2/h = 0. So fx(0, y) = 0 for all y.
Hence fxy(0, 0) = limk→0 (fx(0, k) − fx(0, 0))/k = 0.
f(x, y) = f(y, x), so in the same way fy(x, 0) = 0 for all x and fyx(0, 0) = limh→0 (fy(h, 0) − fy(0, 0))/h = 0.
Answer: fxy(0, 0) = fyx(0, 0) = 0.Final answer: fxy(0, 0) = fyx(0, 0) = 0.
Q2(c) · Linear Algebra · 15 marks≈ 11 min
Vector Spaces, Subspaces, Basis and Dimension
Show that S = {(x, 2y, 3x) : x, y are real numbers} is a subspace of R3(R). Find two bases of S. Also find the dimension of S.
हिंदी में प्रश्न
दर्शाइए कि S = {(x, 2y, 3x) : x, y वास्तविक संख्याएँ हैं} R3(R) का एक उपसमष्टि है। S के दो आधार ज्ञात कीजिए। S की विमा भी ज्ञात कीजिए।
Question 3 (50 marks)
Q3(a) · Calculus · 20 marks≈ 14 min
Riemann Integral, Indefinite Integrals and Improper Integrals
(i) If u = x2 + y2, v = x2 − y2, where x = r cos θ, y = r sin θ, then find ∂(u, v)/∂(r, θ). (7 marks)
(ii) If ∫0x f(t) dt = x + ∫x1 t f(t) dt, then find the value of f(1). (5 marks)
(iii) Express ∫ab (x − a)m(b − x)n dx in terms of Beta function. (8 marks)हिंदी में प्रश्न
(i) यदि u = x2 + y2, v = x2 − y2, जहाँ पर x = r cos θ, y = r sin θ हैं, तब ∂(u, v)/∂(r, θ) ज्ञात कीजिए। (7 अंक)
(ii) यदि ∫0x f(t) dt = x + ∫x1 t f(t) dt है, तो f(1) का मान ज्ञात कीजिए। (5 अंक)
(iii) ∫ab (x − a)m(b − x)n dx को बीटा-फलन के रूप में व्यक्त कीजिए। (8 अंक)Worked solution
(i) In polar form u = r2 and v = r2(cos2θ − sin2θ) = r2 cos 2θ, so ∂(u, v)/∂(r, θ) = det [[2r, 0], [2r cos 2θ, −2r2 sin 2θ]] = −4r3 sin 2θ.
(Chain-rule check: ∂(u, v)/∂(x, y) = det [[2x, 2y], [2x, −2y]] = −8xy and ∂(x, y)/∂(r, θ) = r, so the product is −8r3 sin θ cos θ = −4r3 sin 2θ.)
(ii) Differentiate both sides (f continuous): f(x) = 1 − x f(x), because d/dx ∫x1 t f(t) dt = −x f(x). So f(x) = 1/(1 + x) and f(1) = 1/2.
(iii) Put x = a + (b − a)t, dx = (b − a) dt, so x − a = (b − a)t and b − x = (b − a)(1 − t); t runs from 0 to 1.
∫ab (x − a)m(b − x)n dx = (b − a)m+n+1∫01 tm(1 − t)n dt = (b − a)m+n+1 β(m + 1, n + 1), for m, n > −1.
Answer: (i) −4r3 sin 2θ; (ii) f(1) = 1/2; (iii) (b − a)m+n+1 β(m + 1, n + 1).Final answer: (i) −4r3 sin 2θ; (ii) f(1) = 1/2; (iii) (b − a)m+n+1 β(m + 1, n + 1).
Q3(b) · Analytic Geometry · 15 marks≈ 11 min
The Sphere
A sphere of constant radius r passes through the origin O and cuts the axes at the points A, B and C. Find, the locus of the foot of the perpendicular drawn from O to the plane ABC.
हिंदी में प्रश्न
अचर त्रिज्या r का एक गोला मूल-बिंदु O से गुजरता है तथा अक्षों को A, B, C बिन्दुओं पर काटता है। O से समतल ABC पर खींचे गए लंब-पाद का बिन्दुपथ ज्ञात कीजिए।
Worked solution
Let A(α, 0, 0), B(0, β, 0), C(0, 0, γ). The sphere OABC is x2 + y2 + z2 − αx − βy − γz = 0, and its radius is r, so α2 + β2 + γ2 = 4r2.
The foot F(x, y, z) of the perpendicular from O to x/α + y/β + z/γ = 1 lies on the normal through O, so (x, y, z) = t(1/α, 1/β, 1/γ). Then x2 + y2 + z2 = t2(1/α2 + 1/β2 + 1/γ2), while F on the plane gives t(1/α2 + 1/β2 + 1/γ2) = 1.
Hence t = x2 + y2 + z2 and α = t/x = (x2 + y2 + z2)/x, β = (x2 + y2 + z2)/y, γ = (x2 + y2 + z2)/z.
Answer: substituting in α2 + β2 + γ2 = 4r2, the locus is (x2 + y2 + z2)2(x−2 + y−2 + z−2) = 4r2.Final answer: substituting in α2 + β2 + γ2 = 4r2, the locus is (x2 + y2 + z2)2(x−2 + y−2 + z−2) = 4r2.
Q3(c) · Linear Algebra · 15 marks≈ 11 min
Symmetric, Hermitian, Orthogonal and Unitary Matrices and Their Eigenvalues
(i) Prove that the eigen vectors, corresponding to two distinct eigen values of a real symmetric matrix, are orthogonal. (8 marks)
(ii) For two square matrices A and B of order 2, show that trace (AB) = trace (BA). Hence show that AB − BA ≠ I2, where I2 is an identity matrix of order 2. (7 marks)हिंदी में प्रश्न
(i) सिद्ध कीजिए कि एक वास्तविक सममित आव्यूह के दो भिन्न अभिलक्षणिक मानों के संगत अभिलक्षणिक सदिश, लांबिक हैं। (8 अंक)
(ii) दो वर्ग आव्यूह A तथा B जिनकी कोटि, 2 है के लिए दर्शाइए कि अनुरेख (AB) = अनुरेख (BA)। अतैव दर्शाइए कि AB − BA ≠ I2 जहाँ I2 एक 2-कोटि का तत्समक आव्यूह है। (7 अंक)Worked solution
(i) Let A be real symmetric with AX = λX, AY = μY, X, Y ≠ 0, λ ≠ μ (both real by Theorem 1). Then λXTY = (AX)TY = XTATY = XTAY = μXTY.
So (λ − μ)XTY = 0, and λ ≠ μ gives XTY = 0: X and Y are orthogonal.
(ii) For A = [aij], B = [bij] of order 2, tr(AB) = a11b11 + a12b21 + a21b12 + a22b22, and tr(BA) = b11a11 + b12a21 + b21a12 + b22a22 is the same sum.
Hence tr(AB − BA) = tr(AB) − tr(BA) = 0, while tr I2 = 2; so AB − BA = I2 is impossible.
Answer: Eigenvectors for distinct eigenvalues of a real symmetric matrix are orthogonal; AB − BA ≠ I2 because their traces are 0 and 2.Final answer: Eigenvectors for distinct eigenvalues of a real symmetric matrix are orthogonal; AB − BA ≠ I2 because their traces are 0 and 2.
Question 4 (50 marks)
Q4(a) · Linear Algebra · 20 marks≈ 14 min
Algebra of Matrices, Echelon Form, Rank, Inverse, Congruence and Similarity
(i) Reduce the following matrix to a row-reduced echelon form and hence also, find its rank : A = [[1, 3, 2, 4, 1], [0, 0, 2, 2, 0], [2, 6, 2, 6, 2], [3, 9, 1, 10, 6]] (10 marks)
(ii) Find the eigen values and the corresponding eigen vectors of the matrix A = [[0, −i], [i, 0]], over the complex-number field. (10 marks)हिंदी में प्रश्न
(i) निम्नलिखित आव्यूह का पंक्ति-समानीत सोपानक रूप में समानयन कीजिए एवं अतैव इसकी कोटि भी ज्ञात कीजिए : A = [[1, 3, 2, 4, 1], [0, 0, 2, 2, 0], [2, 6, 2, 6, 2], [3, 9, 1, 10, 6]] (10 अंक)
(ii) सम्मिश्र संख्या क्षेत्र पर आव्यूह A = [[0, −i], [i, 0]] के अभिलक्षणिक मान तथा संगत अभिलक्षणिक सदिशों को ज्ञात कीजिए। (10 अंक)Q4(b) · Calculus · 15 marks≈ 11 min
Double and Triple Integrals; Areas, Surfaces and Volumes
Show that the entire area of the Astroid : x2/3 + y2/3 = a2/3 is (3/8)πa2.
हिंदी में प्रश्न
दर्शाइए कि ऐस्ट्रॉइड : x2/3 + y2/3 = a2/3 का पूरा क्षेत्रफल (3/8)πa2 है।
Worked solution
Parametrise the astroid by x = a cos3t, y = a sin3t. It is symmetric about both axes, so A = 4∫0a y dx (first quadrant).
As x goes from 0 to a, t goes from π/2 to 0, and dx = −3a cos2t sin t dt; so A = 4∫π/20 a sin3t(−3a cos2t sin t) dt = 12a2∫0π/2 sin4t cos2t dt.
By Wallis' formula ∫0π/2 sin4t cos2t dt = (3·1·1)/(6·4·2)·(π/2) = π/32.
Answer: A = 12a2(π/32) = (3/8)πa2.Final answer: A = 12a2(π/32) = (3/8)πa2.
Q4(c) · Analytic Geometry · 15 marks≈ 11 min
Straight Lines and Shortest Distance between Skew Lines
Find equation of the plane containing the lines (x + 1)/3 = (y + 3)/5 = (z + 5)/7, (x − 2)/1 = (y − 4)/3 = (z − 6)/5. Also find the point of intersection of the given lines.
हिंदी में प्रश्न
रेखाओं (x + 1)/3 = (y + 3)/5 = (z + 5)/7, (x − 2)/1 = (y − 4)/3 = (z − 6)/5 को अंतर्विष्ट करने वाले समतल का समीकरण ज्ञात कीजिए। दी गई रेखाओं के प्रतिच्छेद बिंदु को भी ज्ञात कीजिए।
Question 5 (50 marks)
Q5(a) · Ordinary Differential Equations · 10 marks≈ 7 min
Linear Equations with Constant Coefficients
Solve the differential equation : d2y/dx2 + 2y = x2e3x + ex cos 2x
हिंदी में प्रश्न
अवकल समीकरण : d2y/dx2 + 2y = x2e3x + ex cos 2x को हल कीजिए।
Q5(b) · Ordinary Differential Equations · 10 marks≈ 7 min
Laplace Transforms and Initial Value Problems
Solve the initial value problem : d2y/dx2 + 4y = e−2x sin 2x; y(0) = y′(0) = 0 using Laplace transform method.
हिंदी में प्रश्न
लाप्लास रूपान्तर विधि का उपयोग करते हुए प्रारम्भिक मान समस्या : d2y/dx2 + 4y = e−2x sin 2x; y(0) = y′(0) = 0 को हल कीजिए।
Q5(c) · Dynamics and Statics · 10 marks≈ 7 min
Equilibrium, Friction and the Common Catenary
Two rods LM and MN are joined rigidly at the point M such that (LM)2 + (MN)2 = (LN)2 and they are hanged freely in equilibrium from a fixed point L. Let ω be the weight per unit length of both the rods which are uniform. Determine the angle, which the rod LM makes with the vertical direction, in terms of lengths of the rods.
हिंदी में प्रश्न
दो छड़े LM व MN बिन्दु M पर दृढ़ता से इस प्रकार जुड़ी हैं कि (LM)2 + (MN)2 = (LN)2 तथा वे स्वतन्त्र रूप से साम्यावस्था में स्थिर बिन्दु L पर टँगी हैं। माना कि दोनों एकसमान छड़ों का प्रति एकांक लम्बाई, भार ω है। छड़ LM का ऊर्ध्वाधर दिशा के साथ बने कोण को छड़ों की लम्बाई के रूप में ज्ञात कीजिए।
Worked solution
Let LM = a and MN = b, with ∠LMN = 90° (since LM2 + MN2 = LN2); the weights ωa and ωb act at the mid-points of LM and MN.
Take axes at L along LM (unit vector e1) and perpendicular to it towards N (e2); the mid-points are (a/2)e1 and ae1 + (b/2)e2.
The centre of gravity is G = [(a2/2 + ab)e1 + (b2/2)e2]/(a + b).
Hanging freely from L, LG is vertical, so the angle θ between LM and the vertical satisfies tan θ = (b2/2)/(a2/2 + ab).
Answer: tan θ = MN2/(LM2 + 2 LM·MN), with LM on the side of the vertical away from N.Final answer: tan θ = MN2/(LM2 + 2 LM·MN), with LM on the side of the vertical away from N.
Q5(d) · Dynamics and Statics · 10 marks≈ 7 min
Rectilinear Motion and Simple Harmonic Motion
If a planet, which revolves around the Sun in a circular orbit, is suddenly stopped in its orbit, then find the time in which it would fall into the Sun. Also, find the ratio of its falling time to the period of revolution of the planet.
हिंदी में प्रश्न
यदि एक ग्रह, जो सूर्य के परितः वृत्तीय कक्षा में परिभ्रमण करता है, अचानक अपनी कक्षा में रोक दिया जाता है, तो वह समय, जिसमें वह सूर्य में गिर जाएगा, ज्ञात कीजिए। इसके गिरने के समय का ग्रह के परिभ्रमण आवर्तकाल से अनुपात भी ज्ञात कीजिए।
Worked solution
Let the orbit have radius a and the attraction be μ/r2; in the circle v2 = μ/a, so the period is T = 2πa/v = 2πa3/2/√μ.
When stopped, the planet falls from rest at distance a, with v2 = 2μ(1/x − 1/a).
As in §1, with x = a sin2φ, t = √(a/(2μ)) ∫0a √(x/(a − x)) dx = √(a/(2μ)) × πa/2 = (π/2)√(a3/(2μ)).
Ratio: t/T = (π/2)(a3/2/√(2μ)) × √μ/(2πa3/2) = 1/(4√2) = √2/8 ≈ 0.177.
Answer: the fall takes (π/2)√(a3/(2μ)), which is √2/8 of the period of revolution (about 65 days for the earth).Final answer: the fall takes (π/2)√(a3/(2μ)), which is √2/8 of the period of revolution (about 65 days for the earth).
Q5(e) · Vector Analysis · 10 marks≈ 7 min
Vector Identities and Vector Equations
Show that ∇2[∇ · (r/r)] = 2/r4, where r = xî + yĵ + zk̂.
हिंदी में प्रश्न
दर्शाइए कि ∇2[∇ · (r/r)] = 2/r4, जहाँ r = xî + yĵ + zk̂ है।
Worked solution
[As printed the result is false: ∇·(r/r) = 2/r and ∇2(2/r) = 0 for r ≠ 0; the intended field is r/r2, solved below.]
By ∇·(rnr) = (n + 3)rn with n = −2: ∇·(r/r2) = 1/r2.
By ∇2rn = n(n + 1)rn−2 with n = −2: ∇2(r−2) = (−2)(−1)r−4 = 2/r4.
Answer: ∇2[∇·(r/r2)] = 2/r4, while ∇2[∇·(r/r)] = 0 (r ≠ 0).Final answer: ∇2[∇·(r/r2)] = 2/r4, while ∇2[∇·(r/r)] = 0 (r ≠ 0).
Question 6 (50 marks)
Q6(a) · Dynamics and Statics · 20 marks≈ 14 min
Equilibrium, Friction and the Common Catenary
A heavy string, which is not of uniform density, is hung up from two points. Let T1, T2, T3 be the tensions at the intermediate points A, B, C of the catenary respectively where its inclinations to the horizontal are in arithmetic progression with common difference β. Let ω1 and ω2 be the weights of the parts AB and BC of the string respectively. Prove that
(i) Harmonic mean of T1, T2 and T3 = 3T2/(1 + 2 cos β)
(ii) T1/T3 = ω1/ω2हिंदी में प्रश्न
एक भारी डोरी, जिसका घनत्व एक समान नहीं है, दो बिन्दुओं से टँगी हुई है। माना कि T1, T2, T3 क्रमशः कैटिनरी के बीच के बिन्दुओं A, B, C पर तनाव हैं, जिन पर इसके क्षैतिज के साथ आनति कोण, सार्व अंतर β के साथ समांतर श्रेढ़ी में हैं। माना कि डोरी के AB तथा BC भागों के भार क्रमशः ω1 तथा ω2 हैं। सिद्ध कीजिए
(i) T1, T2 तथा T3 का हरात्मक माध्य = 3T2/(1 + 2 cos β)
(ii) T1/T3 = ω1/ω2Worked solution
For any heavy string the horizontal tension T0 = T cos ψ is constant; let the inclinations at A, B, C be ψ − β, ψ, ψ + β.
(i) 1/T1 + 1/T2 + 1/T3 = [cos(ψ − β) + cos ψ + cos(ψ + β)]/T0 = cos ψ(1 + 2 cos β)/T0 = (1 + 2 cos β)/T2.
So the harmonic mean 3/(1/T1 + 1/T2 + 1/T3) = 3T2/(1 + 2 cos β).
(ii) Resolving vertically for the portion AB, ω1 = T0[tan ψ − tan(ψ − β)] = T0 sin β/(cos(ψ − β) cos ψ).
Similarly ω2 = T0 sin β/(cos ψ cos(ψ + β)).
So ω1/ω2 = cos(ψ + β)/cos(ψ − β) = (T0/T3)/(T0/T1) = T1/T3.
Answer: both results follow from the constancy of the horizontal tension; uniform density is never used.Final answer: both results follow from the constancy of the horizontal tension; uniform density is never used.
Q6(b) · Ordinary Differential Equations · 15 marks≈ 11 min
Second Order Equations with Variable Coefficients
Solve the equation : d2y/dx2 + (tan x − 3 cos x) dy/dx + 2y cos2x = cos4x completely by demonstrating all the steps involved.
हिंदी में प्रश्न
सभी अन्तर्ग्रस्त (शामिल) चरणों को दर्शाते हुए समीकरण : d2y/dx2 + (tan x − 3 cos x) dy/dx + 2y cos2x = cos4x को पूर्ण रूप से हल कीजिए।
Q6(c) · Vector Analysis · 15 marks≈ 11 min
Gauss and Stokes' Theorems and Green's Identities
Evaluate ∫C F · dr, where C is an arbitrary closed curve in the xy-plane and F = (−yî + xĵ)/(x2 + y2).
हिंदी में प्रश्न
∫C F · dr का मान निकालिए, जहाँ C, xy-समतल में एक स्वैच्छिक संवृत वक्र है तथा F = (−yî + xĵ)/(x2 + y2) है।
Worked solution
M = −y/(x2 + y2), N = x/(x2 + y2); for (x, y) ≠ (0, 0), ∂N/∂x = (y2 − x2)/(x2 + y2)2 = ∂M/∂y.
Origin outside C: F is smooth on C and its interior, so Green's theorem gives ∮F·dr = ∬0 dA = 0.
Origin inside C (C anticlockwise): take a small circle Cε: x = ε cos t, y = ε sin t inside C; Green's theorem on the ring between them gives ∮C = ∮Cε, both anticlockwise.
On Cε, F·dr = (ε2 sin2t + ε2 cos2t)/ε2 dt = dt, so the integral is 2π.
If C passes through the origin, F is undefined there and the integral does not exist as an ordinary integral.
Answer: 0 if the origin is outside C, 2π if it is inside (−2π for clockwise C); in general 2π times the number of times C winds round the origin.Final answer: 0 if the origin is outside C, 2π if it is inside (−2π for clockwise C); in general 2π times the number of times C winds round the origin.
Question 7 (50 marks)
Q7(a) · Vector Analysis · 20 marks≈ 14 min
Gauss and Stokes' Theorems and Green's Identities
Verify Gauss divergence theorem for F = 2x2yî − y2ĵ + 4xz2k̂ taken over the region in the first octant bounded by y2 + z2 = 9 and x = 2.
हिंदी में प्रश्न
प्रथम अष्टांशक में y2 + z2 = 9 तथा x = 2 द्वारा परिबद्ध क्षेत्र पर F = 2x2yî − y2ĵ + 4xz2k̂ के लिए गाउस अपसरण प्रमेय को सत्यापित कीजिए।
Worked solution
∇·F = 4xy − 2y + 8xz; the region is 0 ≤ x ≤ 2 over the quarter disc y, z ≥ 0, y2 + z2 ≤ 9, for which ∬y dA = ∬z dA = 9.
Volume integral: 4(2)(9) − 2(2)(9) + 8(2)(9) = 72 − 36 + 144 = 180, using ∫02x dx = 2.
Faces x = 0, y = 0, z = 0: F·n is −2x2y, y2 and −4xz2 respectively, all zero there.
Face x = 2: n = i, F·n = 8y, flux ∬8y dy dz = 72.
Curved face y = 3 cos θ, z = 3 sin θ (0 ≤ θ ≤ π/2): n = (0, cos θ, sin θ), dS = 3 dθ dx, F·n = −9 cos3θ + 36x sin3θ, flux ∫02∫0π/2(−27 cos3θ + 108x sin3θ) dθ dx = −36 + 144 = 108.
Answer: total flux 72 + 108 = 180 = ∭∇·F dV, verifying the theorem.Final answer: total flux 72 + 108 = 180 = ∭∇·F dV, verifying the theorem.
Q7(b) · Ordinary Differential Equations · 15 marks≈ 11 min
First Order Higher Degree Equations, Clairaut's Equation and Singular Solutions
Find all possible solutions of the differential equation : y2 log y = xy dy/dx + (dy/dx)2.
हिंदी में प्रश्न
अवकल समीकरण : y2 log y = xy dy/dx + (dy/dx)2 के सभी सम्भव हल ज्ञात कीजिए।
Worked solution
Divide by y2: log y = x(p/y) + (p/y)2. Put v = log y, so P = dv/dx = p/y: v = xP + P2, Clairaut's form.
General solution: log y = cx + c2.
Singular solution: x + 2P = 0 gives P = −x/2 and v = −x2/2 + x2/4 = −x2/4.
Answer: y = ecx + c2 (general) and y = e−x2/4 (singular, the envelope).Final answer: y = ecx + c2 (general) and y = e−x2/4 (singular, the envelope).
Q7(c) · Dynamics and Statics · 15 marks≈ 11 min
Constrained Motion, Work and Energy
A heavy particle hangs by an inextensible string of length a from a fixed point and is then projected horizontally with a velocity √(2gh). If 5a/2 > h > a, then prove that the circular motion ceases when the particle has reached the height (1/3)(a + 2h) from the point of projection. Also, prove that the greatest height ever reached by the particle above the point of projection is (4a − h)(a + 2h)2/(27a2).
हिंदी में प्रश्न
एक भारी कण a लम्बाई की अवितान्य डोरी से एक स्थिर बिन्दु से टँगा है तथा √(2gh) वेग से क्षैतिज दिशा में प्रक्षेपित किया जाता है। यदि 5a/2 > h > a है, तो सिद्ध कीजिए कि प्रक्षेपण बिन्दु से (1/3)(a + 2h) ऊँचाई पहुँचने पर कण की वृत्तीय गति समाप्त हो जाती है। यह भी सिद्ध कीजिए कि उस कण द्वारा प्रक्षेपण बिंदु से ऊपर प्राप्य अधिकतम ऊँचाई (4a − h)(a + 2h)2/(27a2) है।
Worked solution
With θ measured from the lowest point, v2 = 2gh − 2ga(1 − cos θ), and the normal equation T − mg cos θ = mv2/a shows that T = 0 when v2 = −ga cos θ.
So 2gh − 2ga + 2ga cos θ = −ga cos θ, cos θ = −2(h − a)/(3a); since a < h < 5a/2, −1 < cos θ < 0 and the string slackens above the level of the centre.
Height above the point of projection: a(1 − cos θ) = a + 2(h − a)/3 = (a + 2h)/3.
There v2 = −ga cos θ = 2g(h − a)/3, and the velocity is along the tangent, inclined to the horizontal at α = π − θ with cos α = 2(h − a)/(3a).
As a projectile it rises a further v2sin2α/(2g) = ((h − a)/3)(1 − 4(h − a)2/(9a2)) = (h − a)(5a − 2h)(a + 2h)/(27a2).
Greatest height = (a + 2h)/3 + (h − a)(5a − 2h)(a + 2h)/(27a2) = (a + 2h)(4a2 + 7ah − 2h2)/(27a2), and 4a2 + 7ah − 2h2 = (4a − h)(a + 2h).
Answer: the circular motion stops at height (a + 2h)/3, and the greatest height reached is (4a − h)(a + 2h)2/(27a2).Final answer: the circular motion stops at height (a + 2h)/3, and the greatest height reached is (4a − h)(a + 2h)2/(27a2).
Question 8 (50 marks)
Q8(a) · Ordinary Differential Equations · 20 marks≈ 14 min
Orthogonal Trajectories
(i) Find the orthogonal trajectories of the family of confocal conics x2/(a2 + λ) + y2/(b2 + λ) = 1; a > b > 0 are constants and λ is a parameter. Show that the given family of curves is self orthogonal. (10 marks)
(ii) Find the general solution of the differential equation : x2 d2y/dx2 − 2x(1 + x) dy/dx + 2(1 + x)y = 0. Hence, solve the differential equation : x2 d2y/dx2 − 2x(1 + x) dy/dx + 2(1 + x)y = x3 by the method of variation of parameters. (10 marks)हिंदी में प्रश्न
(i) संनाभि शांकव कुल x2/(a2 + λ) + y2/(b2 + λ) = 1; a > b > 0 अचर हैं तथा λ एक प्राचल है, के लंबकोणीय संछेदी ज्ञात कीजिए। दर्शाइए कि दिया गया वक्र-कुल स्वलांबिक है। (10 अंक)
(ii) अवकल समीकरण : x2 d2y/dx2 − 2x(1 + x) dy/dx + 2(1 + x)y = 0 का व्यापक हल ज्ञात कीजिए। अतः अवकल समीकरण : x2 d2y/dx2 − 2x(1 + x) dy/dx + 2(1 + x)y = x3 को प्राचल विचरण विधि द्वारा हल कीजिए। (10 अंक)Worked solution
(i)
Let A = a2 + λ, B = b2 + λ, so A − B = a2 − b2. Differentiating: x/A + yp/B = 0, so B = −ypA/x.
Then A(1 + yp/x) = a2 − b2, giving A = (a2 − b2)x/(x + yp) and B = −(a2 − b2)yp/(x + yp).
Substituting in x2/A + y2/B = 1: (x + yp)(x − y/p) = a2 − b2, the ODE of the family.
Replace p by −1/p: (x − y/p)(x + yp) = a2 − b2, the same equation. So the orthogonal trajectories are the family itself.
Answer: the orthogonal trajectories are x2/(a2 + μ) + y2/(b2 + μ) = 1; the family is self-orthogonal.
(ii)
Standard form: P = −2(1 + x)/x, Q = 2(1 + x)/x2; P + Qx = 0, so y1 = x. Then e−∫P dx = x2e2x and y2 = x∫e2x dx, so take y2 = xe2x; the general solution of the homogeneous equation is c1x + c2xe2x.
For the right side x3: R = x and W = x(e2x + 2xe2x) − xe2x = 2x2e2x.
u1 = −∫xe2x·x/(2x2e2x) dx = −x/2; u2 = ∫x·x/(2x2e2x) dx = −e−2x/4.
PI = −x2/2 − x/4; the term −x/4 belongs to the CF.
Answer: y = c1x + c2xe2x for the homogeneous equation, and y = c1x + c2xe2x − x2/2 for the given one.Final answer: (i) the orthogonal trajectories are x2/(a2 + μ) + y2/(b2 + μ) = 1; the family is self-orthogonal; (ii) y = c1x + c2xe2x for the homogeneous equation, and y = c1x + c2xe2x − x2/2 for the given one
Q8(b) · Dynamics and Statics · 15 marks≈ 11 min
Motion in a Plane and Projectiles
Describe the motion and path of a particle of mass m which is projected in a vertical plane through a point of projection with velocity u in a direction making an angle θ with the horizontal direction. Further, if particles are projected from that point in the same vertical plane with velocity 4√g, then determine the locus of vertices of their paths.
हिंदी में प्रश्न
द्रव्यमान m का एक कण, जो की प्रक्षेपण बिन्दु से वेग u के साथ क्षैतिज दिशा के साथ θ कोण बनाने वाली दिशा में प्रक्षेपण बिन्दु से गुजरने वाले ऊर्ध्वाधर समतल में प्रक्षेपित किया जाता है, उसकी गति तथा पथ का वर्णन कीजिए। यदि कणों को उसी बिन्दु से उसी ऊर्ध्वाधर समतल में वेग 4√g के साथ प्रक्षेपित किया जाता है, तो उनके पथों के शीर्षों के बिन्दुपथ को भी निर्धारित कीजिए।
Worked solution
Motion: the horizontal velocity u cos θ is constant and the vertical motion has constant retardation g: x = ut cos θ, y = ut sin θ − gt2/2.
Path: y = x tan θ − gx2/(2u2cos2θ), a parabola with vertical axis, latus rectum 2u2cos2θ/g and directrix at height u2/(2g); time of flight 2u sin θ/g, range u2 sin 2θ/g (the mass m does not enter).
Vertex: X = u2 sin θ cos θ/g = (u2/(2g)) sin 2θ and Y = u2 sin2θ/(2g) = (u2/(4g))(1 − cos 2θ).
With u = 4√g, u2/g = 16: X = 8 sin 2θ, Y = 4 − 4 cos 2θ.
Eliminating θ: X2/64 + (Y − 4)2/16 = 1, that is X2 + 4Y2 = 32Y.
Answer: the vertices lie on the ellipse x2 + 4y2 = 32y, with centre (0, 4) and semi-axes 8 (horizontal) and 4 (vertical).Final answer: the vertices lie on the ellipse x2 + 4y2 = 32y, with centre (0, 4) and semi-axes 8 (horizontal) and 4 (vertical).
Q8(c) · Vector Analysis · 15 marks≈ 11 min
Gauss and Stokes' Theorems and Green's Identities
Using Stokes’ theorem, evaluate ∬S (∇ × F) · n̂ dS, where F = (x2 + y − 4)î + 3xyĵ + (2xy + z2)k̂ and S is the surface of the paraboloid z = 4 − (x2 + y2) above the xy-plane. Here, n̂ is the unit outward normal vector on S.
हिंदी में प्रश्न
स्टोक्स प्रमेय का उपयोग करते हुए ∬S (∇ × F) · n̂ dS का मान निकालिए, जहाँ पर F = (x2 + y − 4)î + 3xyĵ + (2xy + z2)k̂ तथा S, परवलयज z = 4 − (x2 + y2) का xy-समतल से ऊपर का पृष्ठ है। यहाँ n̂, S पर एकक बहिर्मुखी अभिलंब सदिश है।
UPSC Mains 2021 Mathematics Paper II
Question 1 (50 marks)
Q1(a) · Abstract Algebra · 10 marks≈ 7 min
Integral Domains, PID, Euclidean Domains, UFD, Fields and Quotient Fields
Let m1, m2, ⋯, mk be positive integers and d > 0 the greatest common divisor of m1, m2, ⋯, mk. Show that there exist integers x1, x2, ⋯, xk such that d = x1m1 + x2m2 + ⋯ + xkmk
हिंदी में प्रश्न
मान लीजिए कि m1, m2, ⋯, mk धनात्मक पूर्णांक हैं तथा d > 0, m1, m2, ⋯, mk का महत्तम समापवर्तक है। दर्शाइए कि ऐसे पूर्णांक x1, x2, ⋯, xk अस्तित्व में हैं ताकि d = x1m1 + x2m2 + ⋯ + xkmk
Worked solution
Let I = {a1m1 + ... + akmk : ai ∈ ℤ}. I is closed under subtraction and under multiplication by integers, so it is an ideal of ℤ, and it contains the positive integer m1.
Let d′ = x1m1 + ... + xkmk be the least positive element of I. For a ∈ I write a = qd′ + r with 0 ≤ r < d′; then r = a − qd′ ∈ I, so r = 0 by minimality, and I = d′ℤ.
Each mi ∈ I, so d′ | mi for every i: d′ is a common divisor, so d′ ≤ d.
Conversely d divides each mi, hence divides d′ = ∑ximi, so d ≤ d′. Therefore d = d′.
Answer: d = x1m1 + x2m2 + ... + xkmk for suitable integers xi; equivalently m1ℤ + ... + mkℤ = dℤ.Final answer: d = x1m1 + x2m2 + ... + xkmk for suitable integers xi; equivalently m1ℤ + ... + mkℤ = dℤ.
Q1(b) · Real Analysis · 10 marks≈ 7 min
Uniform Convergence of Sequences and Series of Functions
Test the uniform convergence of the series x4 + x4/(1 + x4) + x4/(1 + x4)2 + x4/(1 + x4)3 + ⋯ on [0, 1].
हिंदी में प्रश्न
श्रेणी x4 + x4/(1 + x4) + x4/(1 + x4)2 + x4/(1 + x4)3 + ⋯ के [0, 1] पर एकसमान अभिसरण की जाँच कीजिए।
Worked solution
At x = 0 every term is 0, so S(0) = 0. For 0 < x ≤ 1 the series is geometric with first term x4 and ratio r = 1/(1 + x4) < 1, so S(x) = x4/(1 − r) = 1 + x4.
So S(x) → 1 as x → 0+, while S(0) = 0: the sum is discontinuous at 0.
Every term is continuous on [0, 1], and a uniformly convergent series of continuous functions has a continuous sum; hence the convergence is not uniform on [0, 1].
Directly: for x > 0, S(x) − Sn(x) = (1 + x4)1−n → 1 as x → 0+, so sup|S − Sn| ≥ 1 for every n. On [a, 1] with a > 0 the sup is (1 + a4)1−n → 0, so the convergence is uniform there.
Answer: the series converges pointwise on [0, 1] but not uniformly (uniformly on [a, 1] for every a > 0).Final answer: the series converges pointwise on [0, 1] but not uniformly (uniformly on [a, 1] for every a > 0).
Q1(c) · Real Analysis · 10 marks≈ 7 min
Riemann Integral, Improper Integrals and Fundamental Theorems of Integral Calculus
If a function f is monotonic in the interval [a, b], then prove that f is Riemann integrable in [a, b].
हिंदी में प्रश्न
यदि एक फलन f, अन्तराल [a, b] में एकदिष्ट है, तब सिद्ध कीजिए कि f, [a, b] में रीमान समाकलनीय है।
Q1(d) · Complex Analysis · 10 marks≈ 7 min
Singularities and Cauchy's Residue Theorem
Let c : [0, 1] → ℂ be the curve, where c(t) = e4πit, 0 ≤ t ≤ 1. Evaluate the contour integral ∫c dz/(2z2 − 5z + 2).
हिंदी में प्रश्न
मान लीजिए कि c : [0, 1] → ℂ, c(t) = e4πit, 0 ≤ t ≤ 1 के द्वारा परिभाषित एक वक्र है। कन्टूर समाकल ∫c dz/(2z2 − 5z + 2) का मान निकालिए।
Worked solution
c(t) = e4πit, 0 ≤ t ≤ 1, runs round the unit circle twice anticlockwise (its argument 4πt goes from 0 to 4π), so its winding number about every point inside |z| = 1 is 2.
2z2 − 5z + 2 = (2z − 1)(z − 2): simple poles at z = 1/2 (inside) and z = 2 (outside).
Res at 1/2 = 1/(4z − 5) at z = 1/2 = −1/3.
∫c = 2πi·n(c, 1/2)·Res = 2πi·2·(−1/3) = −4πi/3. A common error is −2πi/3, which ignores the second turn.
Answer: ∫c dz/(2z2 − 5z + 2) = −4πi/3.Final answer: ∫c dz/(2z2 − 5z + 2) = −4πi/3.
Q1(e) · Linear Programming · 10 marks≈ 7 min
Assignment Problem
A department of a company has five employees with five jobs to be performed. The time (in hours) that each man takes to perform each job is given in the effectiveness matrix. Assign all the jobs to these five employees to minimize the total processing time :
[Table: rows Jobs A–E, columns Employees I–V]
A : 10, 5, 13, 15, 16
B : 3, 9, 18, 13, 6
C : 10, 7, 2, 2, 2
D : 7, 11, 9, 7, 12
E : 7, 9, 10, 4, 12हिंदी में प्रश्न
एक कम्पनी के एक विभाग के पाँच कर्मचारियों को पाँच कार्य सम्पन्न करने हैं। जितना समय (घंटों में) एक व्यक्ति एक कार्य को सम्पन्न करने के लिए लेता है, वह प्रभाविता आव्यूह में दिया गया है। इन पाँच कर्मचारियों को इन सभी कार्यों को इस तरह निर्धारित कीजिए जिससे कि समस्त कार्य सम्पन्न करने का समय न्यूनतम हो :
[सारणी: पंक्तियाँ कार्य A–E, स्तंभ कर्मचारी I–V]
A : 10, 5, 13, 15, 16
B : 3, 9, 18, 13, 6
C : 10, 7, 2, 2, 2
D : 7, 11, 9, 7, 12
E : 7, 9, 10, 4, 12
Question 2 (50 marks)
Q2(a) · Calculus · 15 marks≈ 11 min
Maxima and Minima, Asymptotes and Curve Tracing
Find the maximum and minimum values of f(x) = x3 − 9x2 + 26x − 24 for 0 ≤ x ≤ 1.
हिंदी में प्रश्न
f(x) = x3 − 9x2 + 26x − 24 का, 0 ≤ x ≤ 1 के लिए, अधिकतम तथा न्यूनतम मान निकालिए।
Worked solution
f′(x) = 3x2 − 18x + 26 = 3(x − 3)2 − 1, which is zero only at x = 3 ± 1/√3 (≈ 2.42 and 3.58), both outside [0, 1].
On [0, 1], (x − 3)2 ≥ 4, so f′(x) ≥ 11 > 0 and f is increasing: f(0) = −24, f(1) = 1 − 9 + 26 − 24 = −6.
Answer: Minimum −24 (at x = 0), maximum −6 (at x = 1).Final answer: Minimum −24 (at x = 0), maximum −6 (at x = 1).
Q2(b) · Abstract Algebra · 15 marks≈ 11 min
Integral Domains, PID, Euclidean Domains, UFD, Fields and Quotient Fields
Let F be a field and f(x) ∈ F[x] a polynomial of degree > 0 over F. Show that there is a field F′ and an imbedding q : F → F′ s.t. the polynomial fq ∈ F′[x] has a root in F′, where fq is obtained by replacing each coefficient a of f by q(a).
हिंदी में प्रश्न
मान लीजिए कि F एक क्षेत्र है तथा f(x) ∈ F[x], क्षेत्र F के ऊपर घात > 0 का एक बहुपद है। दर्शाइए कि एक क्षेत्र F′ तथा एक अंतःस्थापन q : F → F′ इस प्रकार से अस्तित्व में हैं कि बहुपद fq ∈ F′[x] का एक मूल F′ में है, जहाँ fq, f के प्रत्येक गुणांक a को q(a) द्वारा प्रतिस्थापित करने से प्राप्त होता है।
Worked solution
F[x] is a UFD and deg f > 0, so f = p·g with p ∈ F[x] irreducible.
F[x] is a PID and p is irreducible, so ⟨p⟩ is a maximal ideal and F′ = F[x]/⟨p⟩ is a field.
Define q : F → F′ by q(a) = a + ⟨p⟩. It is a ring homomorphism with q(1) = 1 + ⟨p⟩ ≠ 0 (deg p ≥ 1), so its kernel is an ideal of the field F other than F, i.e. {0}: q is an imbedding.
Let α = x + ⟨p⟩ ∈ F′. If p = ∑cixi, then pq(α) = ∑(ci + ⟨p⟩)(x + ⟨p⟩)i = p(x) + ⟨p⟩ = 0 in F′.
h ↦ hq is a ring homomorphism F[x] → F′[x], so fq = pqgq and fq(α) = pq(α)gq(α) = 0.
Answer: F′ = F[x]/⟨p(x)⟩ with the imbedding a ↦ a + ⟨p⟩ works, and α = x + ⟨p(x)⟩ is a root of fq (Kronecker's theorem).Final answer: F′ = F[x]/⟨p(x)⟩ with the imbedding a ↦ a + ⟨p⟩ works, and α = x + ⟨p(x)⟩ is a root of fq (Kronecker's theorem).
Q2(c) · Complex Analysis · 20 marks≈ 14 min
Power Series, Taylor's Series and Laurent's Series
Find the Laurent series expansion of f(z) = (z2 − z + 1)/(z(z2 − 3z + 2)) in the powers of (z + 1) in the region |z + 1| > 3.
हिंदी में प्रश्न
क्षेत्र |z + 1| > 3 में f(z) = (z2 − z + 1)/(z(z2 − 3z + 2)) का लौराँ श्रेणी प्रसार, (z + 1) की घातों में ज्ञात कीजिए।
Question 3 (50 marks)
Q3(a) · Complex Analysis · 15 marks≈ 11 min
Singularities and Cauchy's Residue Theorem
Let f be an entire function whose Taylor series expansion with centre z = 0 has infinitely many terms. Show that z = 0 is an essential singularity of f(1/z).
हिंदी में प्रश्न
मान लीजिए कि f एक सर्वत्र वैश्लेषिक फलन है जिसके केन्द्र z = 0 पर टेलर श्रेणी प्रसार में अपरिमित रूप से अनेक पद हैं। दर्शाइए कि f(1/z) की z = 0 एक अनिवार्य विचित्रता है।
Worked solution
f is entire, so f(z) = ∑n=0∞ anzn for every z ∈ ℂ, with infinitely many an ≠ 0.
For every z ≠ 0, f(1/z) = a0 + a1/z + a2/z2 + ..., a series in powers of z converging in the annulus 0 < |z| < ∞, so z = 0 is an isolated singularity of f(1/z).
By uniqueness of Laurent expansions this is the Laurent series of f(1/z) about 0, and its principal part ∑n≥1 anz−n has infinitely many non-zero terms.
Answer: a principal part with infinitely many terms means z = 0 is an essential singularity of f(1/z) (for example f = ez gives e1/z).Final answer: a principal part with infinitely many terms means z = 0 is an essential singularity of f(1/z) (for example f = ez gives e1/z).
Q3(b) · Real Analysis · 20 marks≈ 14 min
Functions of Several Variables: Partial Derivatives, Maxima and Minima
Find the stationary 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
शर्तों ax2 + by2 + cz2 = 1 तथा lx + my + nz = 0 से प्रतिबन्धित x2 + y2 + z2 के स्तब्ध (अचर) मान निकालिए। परिणाम की ज्यामितीय व्याख्या कीजिए।
Worked solution
F = x2 + y2 + z2 + λ(ax2 + by2 + cz2 − 1) + μ(lx + my + nz). Fx = 0 gives 2x + 2λax + μl = 0, and similarly 2y + 2λby + μm = 0, 2z + 2λcz + μn = 0.
Multiply by x, y, z and add. With u = x2 + y2 + z2 and the two constraints, 2u + 2λ = 0, so λ = −u.
Then 2x(1 − au) = −μl, so x = −μl/(2(1 − au)); similarly y = −μm/(2(1 − bu)) and z = −μn/(2(1 − cu)) (in the general case 1 − au, 1 − bu, 1 − cu ≠ 0; then μ ≠ 0, otherwise x = y = z = 0).
Substituting in lx + my + nz = 0 gives l2/(1 − au) + m2/(1 − bu) + n2/(1 − cu) = 0.
Clearing denominators, l2(1 − bu)(1 − cu) + m2(1 − cu)(1 − au) + n2(1 − au)(1 − bu) = 0, a quadratic in u whose two roots are the stationary (maximum and minimum) values. (Check: a = 1, b = 2, c = 3, l = m = n = 1 gives u ≈ 0.388 and 0.703, matching a direct numerical search.)
Geometrically, the plane through the centre cuts the ellipsoid in an ellipse, and u is the squared distance of a point of this ellipse from the centre; so the roots are the squares of the semi-axes of this central section.
Answer: the stationary values u of x2 + y2 + z2 are the roots of l2/(1 − au) + m2/(1 − bu) + n2/(1 − cu) = 0; they are the squared semi-axes of the section of the ellipsoid by the plane.Final answer: the stationary values u of x2 + y2 + z2 are the roots of l2/(1 − au) + m2/(1 − bu) + n2/(1 − cu) = 0; they are the squared semi-axes of the section of the ellipsoid by the plane.
Q3(c) · Linear Programming · 15 marks≈ 11 min
Duality in Linear Programming
Convert the following LPP into dual LPP :
Minimize Z = x1 − 3x2 − 2x3
subject to 3x1 − x2 + 2x3 ≤ 7
2x1 − 4x2 ≥ 12
−4x1 + 3x2 + 8x3 = 10
where x1, x2 ≥ 0 and x3 is unrestricted in sign.हिंदी में प्रश्न
निम्न रैखिक प्रोग्रामन समस्या को द्वैती रैखिक प्रोग्रामन समस्या में परिवर्तित कीजिए :
न्यूनतमीकरण कीजिए Z = x1 − 3x2 − 2x3
बशर्ते कि 3x1 − x2 + 2x3 ≤ 7
2x1 − 4x2 ≥ 12
−4x1 + 3x2 + 8x3 = 10
जहाँ x1, x2 ≥ 0 तथा x3 का चिह्न अप्रतिबंधित है।Worked solution
Write the Min problem in canonical form (all ≥): row 1 times −1 gives −3x1 + x2 − 2x3 ≥ −7; row 2 is 2x1 − 4x2 ≥ 12; row 3 stays an equation.
Dual variables: y1 ≥ 0 and y2 ≥ 0 for the two ≥ rows, y3 unrestricted for the equation. Dual objective: Max W = −7y1 + 12y2 + 10y3.
Dual constraints: from x1 ≥ 0, −3y1 + 2y2 − 4y3 ≤ 1; from x2 ≥ 0, y1 − 4y2 + 3y3 ≤ −3; from x3 unrestricted, −2y1 + 8y3 = −2.
Equivalently, with w1 = −y1 ≤ 0 attached to the original ≤ row: Max W = 7w1 + 12y2 + 10y3, 3w1 + 2y2 − 4y3 ≤ 1, −w1 − 4y2 + 3y3 ≤ −3, 2w1 + 8y3 = −2.
Answer: Max W = −7y1 + 12y2 + 10y3 subject to −3y1 + 2y2 − 4y3 ≤ 1, y1 − 4y2 + 3y3 ≤ −3, −2y1 + 8y3 = −2, y1, y2 ≥ 0, y3 unrestricted in sign.Final answer: Max W = −7y1 + 12y2 + 10y3 subject to −3y1 + 2y2 − 4y3 ≤ 1, y1 − 4y2 + 3y3 ≤ −3, −2y1 + 8y3 = −2, y1, y2 ≥ 0, y3 unrestricted in sign.
Question 4 (50 marks)
Q4(a) · Abstract Algebra · 15 marks≈ 11 min
Groups, Subgroups, Cyclic Groups, Cosets and Lagrange's Theorem
Show that there are infinitely many subgroups of the additive group ℚ of rational numbers.
हिंदी में प्रश्न
दर्शाइए कि परिमेय संख्याओं के योज्य समूह ℚ के अपरिमित रूप से अनेक उपसमूह हैं।
Worked solution
For each n ∈ ℕ, Hn = nℤ = {nk : k ∈ ℤ} is a subgroup of (ℚ, +): it is non-empty and nk − nl = n(k − l) ∈ Hn.
If m < n, then m ∈ Hm but m ∉ Hn, because every positive element of Hn is at least n. So the subgroups H1, H2, H3, ... are pairwise distinct.
Similarly ⟨1/n⟩ = {k/n : k ∈ ℤ}, n ≥ 1, are infinitely many distinct subgroups not contained in ℤ. Note that ℚ itself is not cyclic: if ℚ = ⟨r⟩ with r ≠ 0, then r/2 would be an integer multiple of r, which is false.
Answer: the subgroups nℤ (n = 1, 2, 3, ...) are infinitely many distinct subgroups of (ℚ, +).Final answer: the subgroups nℤ (n = 1, 2, 3, ...) are infinitely many distinct subgroups of (ℚ, +).
Q4(b) · Complex Analysis · 20 marks≈ 14 min
Contour Integration: Evaluation of Real Integrals
Using contour integration, evaluate the integral ∫−∞∞ sin x dx/(x(x2 + a2)), a > 0.
हिंदी में प्रश्न
कन्टूर समाकलन का उपयोग कर समाकल ∫−∞∞ sin x dx/(x(x2 + a2)), a > 0 का मान ज्ञात कीजिए।
Worked solution
Take f(z) = eiz/(z(z2 + a2)), whose imaginary part on the real axis is the integrand. Poles: 0 (on the real axis, simple), ia (upper half-plane), −ia.
Contour: [−R, −ε], the small upper semicircle γε (clockwise), [ε, R] and the large semicircle ΓR. On ΓR, |1/(z(z2 + a2))| ≤ 1/(R(R2 − a2)) → 0, so by Jordan's lemma ∫ΓR f dz → 0, while ∫γε f dz → −πi·Res(f, 0).
Res(f, 0) = 1/a2 and Res(f, ia) = e−a/(ia·2ia) = −e−a/(2a2).
Letting ε → 0, R → ∞: PV∫−∞∞ eix dx/(x(x2 + a2)) = 2πi(−e−a/(2a2)) + πi/a2 = πi(1 − e−a)/a2.
Equate imaginary parts; sin x/(x(x2 + a2)) is continuous at 0, so no principal value is needed for it.
Answer: ∫−∞∞ sin x dx/(x(x2 + a2)) = π(1 − e−a)/a2.Final answer: ∫−∞∞ sin x dx/(x(x2 + a2)) = π(1 − e−a)/a2.
Q4(c) · Linear Programming · 15 marks≈ 11 min
The Simplex Method
Solve the following linear programming problem using Big M method :
Maximize Z = 4x1 + 5x2 + 2x3
subject to 2x1 + x2 + x3 ≥ 10
x1 + 3x2 + x3 ≤ 12
x1 + x2 + x3 = 6
x1, x2, x3 ≥ 0हिंदी में प्रश्न
बड़ा M (बिग M) विधि का उपयोग करके निम्नलिखित रैखिक प्रोग्रामन समस्या को हल कीजिए :
अधिकतमीकरण कीजिए Z = 4x1 + 5x2 + 2x3
बशर्ते कि 2x1 + x2 + x3 ≥ 10
x1 + 3x2 + x3 ≤ 12
x1 + x2 + x3 = 6
x1, x2, x3 ≥ 0
Question 5 (50 marks)
Q5(a) · Partial Differential Equations · 10 marks≈ 7 min
Families of Surfaces and Formation of PDEs
Obtain the partial differential equation by eliminating arbitrary function f from the equation f(x + y + z, x2 + y2 + z2) = 0.
हिंदी में प्रश्न
समीकरण f(x + y + z, x2 + y2 + z2) = 0 से स्वैच्छिक फलन f का विलोपन कर आंशिक अवकल समीकरण को प्राप्त कीजिए।
Q5(b) · Numerical Analysis and Computer Programming · 10 marks≈ 7 min
Roots of Algebraic and Transcendental Equations
Find a positive root of the equation 3x = 1 + cos x by a numerical technique using initial values 0, π/2; and further improve the result using Newton-Raphson method correct to 8 significant figures.
हिंदी में प्रश्न
प्रारंभिक मानों 0, π/2 का उपयोग करके एक संख्यात्मक तकनीक के द्वारा समीकरण 3x = 1 + cos x का एक धनात्मक मूल ज्ञात कीजिए, तथा न्यूटन-राफ्सन विधि के द्वारा परिणाम को 8 सार्थक अंकों तक और शुद्ध मान के निकट लाइए।
Worked solution
f(x) = 3x − 1 − cos x is continuous, f(0) = −2 < 0 and f(π/2) = 3π/2 − 1 = 3.712389 > 0, so a root lies in (0, π/2); since f′(x) = 3 + sin x > 0, it is the only real root.
Regula-falsi with a = 0, b = π/2: c1 = (π/2)(2)/5.712389 = 0.549961 (f = −0.202661, so a = c1), c2 = 0.602804 (f = −0.015336), c3 = 0.606787 (f = −0.001125), c4 = 0.607079 (f = −0.000082).
Newton-Raphson xn+1 = xn − (3xn − 1 − cos xn)/(3 + sin xn) from x0 = 0.607079: f(x0) = −0.0000809, f′(x0) = 3.570471, so x1 = 0.60710165.
x2 = 0.60710165 again: the quadratic convergence roughly doubles the number of correct digits at each step.
Answer: x = 0.60710165 (8 significant figures).Final answer: x = 0.60710165 (8 significant figures).
Q5(c) · Numerical Analysis and Computer Programming · 10 marks≈ 7 min
Number Systems and Representation of Numbers in a Computer
(i) Convert (3798·3875)10 into octal and hexadecimal equivalents.
(ii) Obtain the principal conjunctive normal form of (¬P → R) ∧ (Q ⇄ P).हिंदी में प्रश्न
(i) (3798·3875)10 को अष्टाधारी तथा षोडशाधारी तुल्यमानों में बदलिए।
(ii) (¬P → R) ∧ (Q ⇄ P) का मुख्य संयोजक सामान्य रूप (प्रिंसिपल कंजंक्टिव नॉर्मल फॉर्म) प्राप्त कीजिए।Worked solution
(i) 3798 = 8 × 474 + 6, 474 = 8 × 59 + 2, 59 = 8 × 7 + 3, so 3798 = (7326)8; 3798 = 16 × 237 + 6, 237 = 16 × 14 + 13, so 3798 = (ED6)16.
0.3875 × 8 gives 3.1, 0.8, 6.4, 3.2, 1.6, 4.8, 6.4, ... (digits 3, 0, then 6, 3, 1, 4 recurring); 0.3875 × 16 gives 6.2, 3.2, 3.2, ... So (3798.3875)10 = (7326.30631463146...)8 = (ED6.6333...)16.
(ii) ¬P → R ≡ P ∨ R, and Q ⇄ P is true exactly when P, Q have the same value; so the formula is true only for (P, Q, R) = (T, T, T), (T, T, F), (F, F, T).
The PCNF is the product of the maxterms of the five false rows FFF, FTF, FTT, TFF, TFT (each maxterm contains a variable when it is F in that row and its negation when it is T).
Answer: (i) (7326.3063146...)8, (ED6.6333...)16; (ii) (P ∨ Q ∨ R) ∧ (P ∨ ¬Q ∨ R) ∧ (P ∨ ¬Q ∨ ¬R) ∧ (¬P ∨ Q ∨ R) ∧ (¬P ∨ Q ∨ ¬R).Final answer: (i) (7326.3063146...)8, (ED6.6333...)16; (ii) (P ∨ Q ∨ R) ∧ (P ∨ ¬Q ∨ R) ∧ (P ∨ ¬Q ∨ ¬R) ∧ (¬P ∨ Q ∨ R) ∧ (¬P ∨ Q ∨ ¬R).
Q5(d) · Mechanics and Fluid Dynamics · 10 marks≈ 7 min
Generalized Coordinates, D'Alembert's Principle and Lagrange's Equations
A particle is constrained to move along a circle lying in the vertical xy-plane. With the help of the D’Alembert’s principle, show that its equation of motion is ẍy − ÿx − gx = 0, where g is the acceleration due to gravity.
हिंदी में प्रश्न
ऊर्ध्वाधर xy-तल में स्थित एक वृत्त के अनुदिश एक कण गति के लिए व्यवरुद्ध है। डी’एलम्बर्ट के नियम की सहायता से दर्शाइए कि इसकी गति का समीकरण ẍy − ÿx − gx = 0 है, जहाँ g गुरुत्वीय त्वरण है।
Worked solution
Take y vertically upward, so F = (0, −mg); the constraint x2 + y2 = a2 gives x δx + y δy = 0, so we may take δx = yε, δy = −xε.
D'Alembert: (Fx − mẍ)δx + (Fy − mÿ)δy = 0, i.e. (−mẍ)(yε) + (−mg − mÿ)(−xε) = 0.
Dividing by −mε gives ẍy − ÿx − gx = 0; with x = a sin θ, y = −a cos θ this becomes θ̈ + (g/a)sin θ = 0, the simple pendulum.
Answer: ẍy − ÿx − gx = 0.Final answer: ẍy − ÿx − gx = 0.
Q5(e) · Mechanics and Fluid Dynamics · 10 marks≈ 7 min
Two-Dimensional and Axisymmetric Motion: Sources, Sinks and Vortices
What arrangements of sources and sinks can have the velocity potential w = loge(z − a2/z)? Draw the corresponding sketch of the streamlines and prove that two of them subdivide into the circle r = a and the axis of y.
हिंदी में प्रश्न
उद्गमों (स्रोतों) व अभिगमों (सिंकों) के किस विन्यास से वेग विभव w = loge(z − a2/z) हो सकता है? संगत धारा-रेखाओं का ख़ाका खींचिए और सिद्ध कीजिए कि उनमें से दो, वृत्त r = a तथा y-अक्ष में प्रविभाजित होती हैं।
Worked solution
w = log(z − a2/z) = log(z − a) + log(z + a) − log z; with q = −∇φ (source m: w = −m log(z − z0)) this is a unit sink at each of z = ±a and a unit source at z = 0 (with q = ∇φ: sources at ±a and a sink at O).
ψ = arg(z − a2/z) and z − a2/z = x(1 − a2/r2) + iy(1 + a2/r2), so tan ψ = y(r2 + a2)/(x(r2 − a2)).
ψ = 0, π gives y = 0 (the x-axis); ψ = ±π/2 needs x(1 − a2/r2) = 0, i.e. x = 0 or r = a. Sketch: curves symmetric in both axes, leaving the source O (or coming from infinity) and ending at the sinks ±a.
Answer: sinks at ±a and a source at O (strength 1 in this convention); the streamline ψ = ±π/2 splits into the circle r = a and the y-axis.Final answer: sinks at ±a and a source at O (strength 1 in this convention); the streamline ψ = ±π/2 splits into the circle r = a and the y-axis.
Question 6 (50 marks)
Q6(a) · Partial Differential Equations · 20 marks≈ 14 min
Vibrating String, Heat Equation and Laplace Equation
Solve the wave equation a2 ∂2u/∂x2 = ∂2u/∂t2, 0 < x < L, t > 0 subject to the conditions u(0, t) = 0, u(L, t) = 0, u(x, 0) = (1/4)x(L − x), ∂u/∂t|t=0 = 0
हिंदी में प्रश्न
तरंग समीकरण a2 ∂2u/∂x2 = ∂2u/∂t2, 0 < x < L, t > 0 का शर्तों u(0, t) = 0, u(L, t) = 0, u(x, 0) = (1/4)x(L − x), ∂u/∂t|t=0 = 0 से प्रतिबन्धित हल ज्ञात कीजिए।
Q6(b) · Numerical Analysis and Computer Programming · 15 marks≈ 11 min
Computer Systems, Logic Gates, Boolean Algebra and Algorithms
Obtain the Boolean function F(x, y, z) based on the table given below. Then simplify F(x, y, z) and draw the corresponding GATE network :
[Truth table: x, y, z, F(x, y, z)]
1 1 1 : 1
1 1 0 : 1
1 0 1 : 1
1 0 0 : 0
0 1 1 : 1
0 1 0 : 0
0 0 1 : 0
0 0 0 : 0हिंदी में प्रश्न
नीचे दी गई सारणी पर आधारित बूलीय फलन F(x, y, z) को निकालिए और तब F(x, y, z) को सरल कीजिए तथा उसके अनुरूप GATE परिपथ खींचिए :
[सत्य सारणी: x, y, z, F(x, y, z)]
1 1 1 : 1
1 1 0 : 1
1 0 1 : 1
1 0 0 : 0
0 1 1 : 1
0 1 0 : 0
0 0 1 : 0
0 0 0 : 0Q6(c) · Mechanics and Fluid Dynamics · 15 marks≈ 11 min
Generalized Coordinates, D'Alembert's Principle and Lagrange's Equations
Obtain the Lagrangian equation for the motion of a system of two particles of unequal masses connected by an inextensible string passing over a small smooth pulley.
हिंदी में प्रश्न
एक छोटी चिकनी घिरनी के ऊपर से गुजरने वाली एक अवितान्य डोरी के सिरों से बंधे असमान संहति वाले दो कणों के निकाय की गति के लिए लग्रांजी समीकरण ज्ञात कीजिए।
Question 7 (50 marks)
Q7(a) · Partial Differential Equations · 15 marks≈ 11 min
Linear Second-Order PDEs with Constant Coefficients
Find the general solution of the partial differential equation (D2 − D′2 − 3D + 3D′)z = xy + ex+2y where D ≡ ∂/∂x and D′ ≡ ∂/∂y.
हिंदी में प्रश्न
आंशिक अवकल समीकरण (D2 − D′2 − 3D + 3D′)z = xy + ex+2y का व्यापक हल ज्ञात कीजिए, जहाँ D ≡ ∂/∂x तथा D′ ≡ ∂/∂y हैं।
Worked solution
F = (D − D′)(D + D′) − 3(D − D′) = (D − D′)(D + D′ − 3), so CF = φ1(y + x) + e3xφ2(y − x).
xy: (D + D′ − 3)−1xy = −(1/3)[1 + (D + D′)/3 + (D + D′)2/9]xy = −(1/3)[xy + (x + y)/3 + 2/9].
(D − D′)−1 = (1/D)(1 + D′/D + ...) applied to xy + (x + y)/3 + 2/9 gives x2y/2 + x3/6 + x2/3 + xy/3 + 2x/9, so PI1 = −(1/54)(9x2y + 3x3 + 6x2 + 6xy + 4x).
ex + 2y: F(1, 2) = 0 since D + D′ − 3 → 0; (D + D′ − 3)−1ex + 2y = ex + 2y(D + D′)−11 = xex + 2y, then (D − D′)−1(xex + 2y) = ex + 2y(D − D′ − 1)−1x = −(x + 1)ex + 2y.
The part −ex + 2y = −e3xe2(y − x) lies in the CF, so PI2 = −xex + 2y.
Answer: z = φ1(y + x) + e3xφ2(y − x) − (1/54)(9x2y + 3x3 + 6x2 + 6xy + 4x) − xex + 2y.Final answer: z = φ1(y + x) + e3xφ2(y − x) − (1/54)(9x2y + 3x3 + 6x2 + 6xy + 4x) − xex + 2y.
Q7(b) · Numerical Analysis and Computer Programming · 15 marks≈ 11 min
Systems of Linear Equations: Direct and Iterative Methods
Solve the system of equations 3x1 + 9x2 − 2x3 = 11, 4x1 + 2x2 + 13x3 = 24, 4x1 − 2x2 + x3 = −8 correct up to 4 significant figures by using Gauss-Seidel method after verifying whether the method is applicable in your transformed form of the system.
हिंदी में प्रश्न
समीकरणों के निकाय 3x1 + 9x2 − 2x3 = 11, 4x1 + 2x2 + 13x3 = 24, 4x1 − 2x2 + x3 = −8 का गाउस-सीडल विधि द्वारा 4 सार्थक अंकों तक सही हल ज्ञात कीजिए, यह सत्यापन करने के बाद कि क्या यह विधि आपके द्वारा निकाय के रूपांतरित रूप में अनुप्रयोज्य है।
Worked solution
As written no row is diagonally dominant. Rearrange: the third equation for x1 (|4| > |−2| + |1|), the first for x2 (|9| > |3| + |−2|) and the second for x3 (|13| > |4| + |2|).
The system 4x1 − 2x2 + x3 = −8, 3x1 + 9x2 − 2x3 = 11, 4x1 + 2x2 + 13x3 = 24 is strictly diagonally dominant, so Gauss-Seidel is applicable and converges from any start.
Scheme: x1 = (−8 + 2x2 − x3)/4, x2 = (11 − 3x1 + 2x3)/9, x3 = (24 − 4x1 − 2x2)/13, starting from (0, 0, 0).
Iterates: (−2, 1.88889, 2.17094), (−1.59829, 2.23742, 1.99372), (−1.37972, 2.12518, 1.94373), (−1.42334, 2.12861, 1.95663), (−1.42485, 2.13198, 1.95657), (−1.42315, 2.13140, 1.95614), (−1.42333, 2.13136, 1.95620), (−1.42337, 2.13139, 1.95621).
Answer: x1 = −1.423, x2 = 2.131, x3 = 1.956 (4 significant figures); the exact solution is (−195/137, 292/137, 268/137).Final answer: x1 = −1.423, x2 = 2.131, x3 = 1.956 (4 significant figures); the exact solution is (−195/137, 292/137, 268/137).
Q7(c) · Mechanics and Fluid Dynamics · 20 marks≈ 14 min
Kinematics of Fluids: Continuity, Streamlines and Potential Flow
Show that q = λ(−yî + xĵ)/(x2 + y2), (λ = constant) is a possible incompressible fluid motion. Determine the streamlines. Is the kind of the motion potential? If yes, then find the velocity potential.
हिंदी में प्रश्न
दर्शाइए कि q = λ(−yî + xĵ)/(x2 + y2), (λ = स्थिरांक) एक संभाव्य असंपीड्य तरल गति है। धारा-रेखाएँ निकालिए। क्या गति का प्रकार विभव है? यदि हाँ, तो वेग विभव निकालिए।
Question 8 (50 marks)
Q8(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 p = (z + qy)2 by using Charpit’s method.
हिंदी में प्रश्न
चार्पिट विधि का उपयोग करके आंशिक अवकल समीकरण p = (z + qy)2 का पूर्ण समाकल ज्ञात कीजिए।
Worked solution
F = (z + qy)2 − p: Fy = 2q(z + qy), Fz = 2(z + qy), Fp = −1, Fq = 2y(z + qy).
Charpit: dy/Fq = −dq/(Fy + qFz) gives dy/(2y(z + qy)) = −dq/(4q(z + qy)), so dy/y = −dq/(2q) and qy2 is constant; write q = −a/y2.
Then z + qy = z − a/y and p = (z − a/y)2; with w = z − a/y, dw = dz + (a/y2)dy = p dx = w2 dx.
So −1/w = x + b, i.e. z − a/y = −1/(x + b).
Answer: complete integral z = a/y − 1/(x + b) (check: z + qy = −1/(x + b), whose square is p).Final answer: complete integral z = a/y − 1/(x + b) (check: z + qy = −1/(x + b), whose square is p).
Q8(b) · Numerical Analysis and Computer Programming · 15 marks≈ 11 min
Interpolation: Newton's Forward, Backward and Lagrange's Formulae
Derive Newton’s backward difference interpolation formula and also do error analysis.
हिंदी में प्रश्न
न्यूटन के पश्चांतर अंतर्वेशन सूत्र की व्युत्पत्ति कीजिए तथा त्रुटि-विश्लेषण भी कीजिए।
Worked solution
Let xi = x0 + ih (i = 0, ..., n) and x = xn + ph. Since E−1 = 1 − ∇, y(x) = Epyn = (1 − ∇)−pyn = [1 + p∇ + p(p + 1)/2! ∇2 + ...]yn.
Stopping at ∇n: P(x) = yn + p∇yn + p(p + 1)/2! ∇2yn + ... + p(p + 1)...(p + n − 1)/n! ∇nyn, a polynomial of degree n in x.
Check at x = xn−k (p = −k, 0 ≤ k ≤ n): the j-th coefficient becomes (−1)jC(k, j), zero for j > k, so P = (1 − ∇)kyn = E−kyn = yn−k; P interpolates all n + 1 points.
Error: let w(t) = (t − x0)...(t − xn) and g(t) = f(t) − P(t) − [f(x) − P(x)]w(t)/w(x) for fixed x not a node; g vanishes at the n + 2 points x0, ..., xn, x, so by Rolle's theorem applied repeatedly g(n+1)(ξ) = 0 for some ξ in the interval containing them.
Since P(n+1) = 0 and w(n+1) = (n + 1)!, f(x) − P(x) = w(x)f(n+1)(ξ)/(n + 1)!; with x − xn−k = (p + k)h this is E(x) = p(p + 1)...(p + n)hn+1f(n+1)(ξ)/(n + 1)! ≈ [p(p + 1)...(p + n)/(n + 1)!]∇n+1yn.
Answer: Newton's backward formula as above, with error p(p + 1)...(p + n)hn+1f(n+1)(ξ)/(n + 1)!, which is small for small |p|, i.e. near the end of the table.Final answer: Newton's backward formula as above, with error p(p + 1)...(p + n)hn+1f(n+1)(ξ)/(n + 1)!, which is small for small |p|, i.e. near the end of the table.
Q8(c) · Mechanics and Fluid Dynamics · 20 marks≈ 14 min
Two-Dimensional and Axisymmetric Motion: Sources, Sinks and Vortices
Show that for the complex potential tan−1z, the streamlines and equipotential curves are circles. Find the velocity at any point and check the singularities at z = ±i.
हिंदी में प्रश्न
दर्शाइए कि सम्मिश्र विभव tan−1z के लिए धारा-रेखाएँ तथा समविभव वक्र, वृत्त हैं। किसी भी बिन्दु पर वेग निकालिए तथा z = ±i पर विचित्रता जाँचिए।
Worked solution
tan−1z = (1/(2i))log((1 + iz)/(1 − iz)) = −(i/2)log((z − i)/(z + i)) + constant; with z − i = r1eiθ1, z + i = r2eiθ2, φ = (θ1 − θ2)/2 and ψ = −(1/2)log(r1/r2).
Streamlines r1/r2 = λ: x2 + (y − 1)2 = λ2[x2 + (y + 1)2], circles with limiting points ±i; equipotentials θ1 − θ2 = constant: x2 + y2 − 1 = μx, circles through ±i.
dw/dz = 1/(1 + z2), so the speed is 1/|1 + z2| = 1/(r1r2), infinite at z = ±i.
Near z = i, w ≈ −(i/2)log(z − i): φ is many-valued and ψ single-valued, so this is a point vortex, with circulation −π (clockwise) round i and +π round −i.
Answer: streamlines and equipotentials are two orthogonal families of coaxal circles; speed 1/(r1r2); the singularities at ±i are equal and opposite point vortices of circulation π (they would be a source and a sink for the potential i tan−1z).Final answer: streamlines and equipotentials are two orthogonal families of coaxal circles; speed 1/(r1r2); the singularities at ±i are equal and opposite point vortices of circulation π (they would be a source and a sink for the potential i tan−1z).