UPSC Mains 2026 Mathematics Optional Question Papers with Solutions
All 56 questions of the 2026 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 2026 Mathematics Paper I
Question 1 (50 marks)
Q1(a) · Linear Algebra · 10 marks≈ 7 min
Systems of Linear Equations
Find all the values of a for which the system of homogeneous equations
ax + y + z = 0
x + y − z = 0
x + y − az = 0
has non-trivial solutions, and hence determine all the solutions for each value of a.हिंदी में प्रश्न
a के वे सभी मान ज्ञात कीजिए, जिनके लिए समघातीय समीकरणों के निकाय
ax + y + z = 0
x + y − z = 0
x + y − az = 0
के अतुच्छ हल हैं, और अतः a के प्रत्येक मान के लिए सभी हल निर्धारित कीजिए।Q1(b) · Linear Algebra · 10 marks≈ 7 min
Algebra of Matrices, Echelon Form, Rank, Inverse, Congruence and Similarity
If A and B are similar matrices, then show that A and B have the same rank, trace, characteristic polynomial and eigenvalues.
हिंदी में प्रश्न
यदि A और B समरूप आव्यूह हैं, तो दर्शाइए कि A और B की कोटि (रैंक), अनुरेख (ट्रेस), अभिलाक्षणिक बहुपद और अभिलाक्षणिक मान समान होते हैं।
Worked solution
Let B = P−1AP with P non-singular. Then B − λI = P−1(A − λI)P, so det(B − λI) = det P−1 · det(A − λI) · det P = det(A − λI): the characteristic polynomials, and hence the eigenvalues with their multiplicities, coincide.
Trace: tr B = tr(P−1(AP)) = tr((AP)P−1) = tr A, using tr(XY) = tr(YX).
Rank: ρ(XY) ≤ ρ(X) and ρ(XY) ≤ ρ(Y). So ρ(B) = ρ(P−1AP) ≤ ρ(A), and since A = PBP−1, also ρ(A) ≤ ρ(B). Hence ρ(A) = ρ(B).
Answer: Similar matrices have the same rank, trace, characteristic polynomial and eigenvalues (and determinant).Final answer: Similar matrices have the same rank, trace, characteristic polynomial and eigenvalues (and determinant).
Q1(c) · Calculus · 10 marks≈ 7 min
Functions of Several Variables: Partial Derivatives, Extrema, Lagrange Multipliers and Jacobians
If u(x, y, z) = (x2 + y2 + z2)−1/2, then find the value of ∂2u/∂x2 + ∂2u/∂y2 + ∂2u/∂z2.
हिंदी में प्रश्न
यदि u(x, y, z) = (x2 + y2 + z2)−1/2 है, तो ∂2u/∂x2 + ∂2u/∂y2 + ∂2u/∂z2 का मान ज्ञात कीजिए।
Worked solution
Let r = √(x2 + y2 + z2), so u = r−1 and ∂r/∂x = x/r. Then ux = −r−2(x/r) = −xr−3.
uxx = −r−3 + 3xr−4(x/r) = −r−3 + 3x2r−5; similarly uyy = −r−3 + 3y2r−5 and uzz = −r−3 + 3z2r−5.
Adding: −3r−3 + 3(x2 + y2 + z2)r−5 = −3r−3 + 3r−3 = 0.
Answer: 0 at every point other than the origin (u = 1/r satisfies Laplace's equation).Final answer: 0 at every point other than the origin (u = 1/r satisfies Laplace's equation).
Q1(d) · Calculus · 10 marks≈ 7 min
Mean Value Theorems, Taylor's Theorem and Indeterminate Forms
Using Maclaurin's theorem, obtain the first four terms of the expansion of ex cos x in ascending powers of x. Hence or otherwise find the limit of (ex − ex cos x)/(x − sin x) as x tends to zero.
हिंदी में प्रश्न
मैकलॉरिन प्रमेय का प्रयोग करते हुए x की आरोही घातों में ex cos x के प्रसार के प्रथम चार पद प्राप्त कीजिए। अतः या अन्य विधि से (ex − ex cos x)/(x − sin x) की सीमा (लिमिट) ज्ञात कीजिए, जब x शून्य की ओर अग्रसर होता है।
Worked solution
u = x cos x = x − x3/2 + O(x5), so u2 = x2 + O(x4) and u3 = x3 + O(x5). In eu = 1 + u + u2/2 + u3/6 + O(u4):
ex cos x = 1 + (x − x3/2) + x2/2 + x3/6 + O(x4) = 1 + x + x2/2 − x3/3 + O(x4). By Maclaurin's theorem these are f(0), f′(0)x, f″(0)x2/2! and f‴(0)x3/3! for f(x) = ex cos x.
ex − ex cos x = (1 + x + x2/2 + x3/6) − (1 + x + x2/2 − x3/3) + O(x4) = x3/2 + O(x4), and x − sin x = x3/6 + O(x5).
Hence the ratio is (1/2 + O(x))/(1/6 + O(x2)) → 3.
Answer: ex cos x = 1 + x + x2/2 − x3/3 + ...; the limit is 3.Final answer: ex cos x = 1 + x + x2/2 − x3/3 + ...; the limit is 3.
Q1(e) · Analytic Geometry · 10 marks≈ 7 min
The Sphere
Obtain the equation of the sphere having its centre on the line 5y + 2z = 0 = 2x − 3y and passing through the points (0, −2, −4) and (2, −1, −1).
हिंदी में प्रश्न
उस गोले का समीकरण ज्ञात कीजिए जिसका केन्द्र, रेखा 5y + 2z = 0 = 2x − 3y पर है तथा जो बिंदुओं (0, −2, −4) और (2, −1, −1) से होकर जाता है।
Worked solution
The line 5y + 2z = 0 = 2x − 3y passes through the origin with direction (0, 5, 2) × (2, −3, 0) = (6, 4, −10), proportional to (3, 2, −5); so the centre is (3t, 2t, −5t).
Equal distances to the two points: 9t2 + (2t + 2)2 + (4 − 5t)2 = (3t − 2)2 + (2t + 1)2 + (1 − 5t)2, that is 38t2 − 32t + 20 = 38t2 − 18t + 6, so t = 1.
Centre (3, 2, −5); radius2 = 9 + 16 + 1 = 26.
Answer: x2 + y2 + z2 − 6x − 4y + 10z + 12 = 0.Final answer: x2 + y2 + z2 − 6x − 4y + 10z + 12 = 0.
Question 2 (50 marks)
Q2(a) · Linear Algebra · 20 marks≈ 14 min
Eigenvalues, Eigenvectors and the Cayley-Hamilton Theorem
Let T : P2[x] → P2[x] be a linear transformation defined by
T(a0 + a1x + a2x2) = −(a0 + 2a1 + a2) + (2a0 + 3a1)x − 2(a0 + a1)x2
where P2[x] denotes the set of all polynomials in x of degree ≤ 2 and a0, a1, a2 ∈ ℝ. Find the eigenvalues and corresponding eigenvectors of T using the matrix representation of T with respect to the standard basis of P2[x].हिंदी में प्रश्न
माना T : P2[x] → P2[x] एक रैखिक रूपांतरण है, जिसे
T(a0 + a1x + a2x2) = −(a0 + 2a1 + a2) + (2a0 + 3a1)x − 2(a0 + a1)x2
द्वारा परिभाषित किया गया है, जहाँ P2[x], x में सभी बहुपदों का समुच्चय है, जिनकी घात ≤ 2 है तथा a0, a1, a2 ∈ ℝ हैं। P2[x] के मानक आधार के सापेक्ष T के आव्यूह निरूपण का उपयोग करते हुए T के अभिलाक्षणिक मान तथा संगत अभिलाक्षणिक सदिश ज्ञात कीजिए।Worked solution
T(1) = −1 + 2x − 2x2, T(x) = −2 + 3x − 2x2, T(x2) = −1, so in the basis {1, x, x2}, [T] = [[−1, −2, −1], [2, 3, 0], [−2, −2, 0]] (columns are the coordinates of the images).
tr = 2; principal minors (−3 + 4) + (0 − 2) + 0 = −1; det = −1·det [[2, 3], [−2, −2]] = −2 (third column). So λ3 − 2λ2 − λ + 2 = (λ − 1)(λ + 1)(λ − 2) = 0, and λ = 1, −1, 2.
λ = 1: −2a0 − 2a1 − a2 = 0 and 2a0 + 2a1 = 0 give a1 = −a0, a2 = 0: eigenvector 1 − x.
λ = −1: −2a1 − a2 = 0 and 2a0 + 4a1 = 0 give a2 = −2a1, a0 = −2a1; with a1 = −1, eigenvector 2 − x + 2x2.
λ = 2: −3a0 − 2a1 − a2 = 0 and 2a0 + a1 = 0 give a1 = −2a0, a2 = a0: eigenvector 1 − 2x + x2. Check: T(1 − x) = −(1 − 2) + (2 − 3)x − 2(0)x2 = 1 − x.
Answer: λ = 1: k(1 − x); λ = −1: k(2 − x + 2x2); λ = 2: k(1 − x)2 (k ≠ 0).Final answer: λ = 1: k(1 − x); λ = −1: k(2 − x + 2x2); λ = 2: k(1 − x)2 (k ≠ 0).
Q2(b) · Calculus · 15 marks≈ 11 min
Maxima and Minima, Asymptotes and Curve Tracing
If the sum of the lengths of the hypotenuse and another side of a right-angled triangle be given, then find the angle between these sides so that the area of the triangle is maximum.
हिंदी में प्रश्न
यदि एक समकोण त्रिभुज के कर्ण तथा एक अन्य भुजा की लंबाइयों का योग दिया गया हो, तो इन भुजाओं के बीच का कोण ज्ञात कीजिए, जिसके लिए त्रिभुज का क्षेत्रफल अधिकतम है।
Worked solution
Let the given side be a, the hypotenuse c, with a + c = k (constant), and θ the angle between them. The third side is √(c2 − a2), so the area is A = (1/2)a√((k − a)2 − a2) = (1/2)a√(k2 − 2ka).
Maximise F = A2 = (1/4)(k2a2 − 2ka3): F′(a) = (ka/2)(k − 3a) = 0 gives a = k/3.
F″(a) = (1/4)(2k2 − 12ka) = −k2/2 < 0 at a = k/3, a maximum. Then c = 2k/3 and cos θ = a/c = 1/2.
Answer: θ = π/3 (60°).Final answer: θ = π/3 (60°).
Q2(c) · Analytic Geometry · 15 marks≈ 11 min
Straight Lines and Shortest Distance between Skew Lines
Find the points on the lines (x − 3)/1 = (y − 5)/(−2) = (z − 7)/1 and (x + 1)/7 = (y + 1)/(−6) = (z + 1)/1 which are nearest to each other. Hence find the shortest distance between the lines and its equation.
हिंदी में प्रश्न
रेखाओं (x − 3)/1 = (y − 5)/(−2) = (z − 7)/1 और (x + 1)/7 = (y + 1)/(−6) = (z + 1)/1 पर वे बिंदु ज्ञात कीजिए, जो एक-दूसरे के निकटतम हैं। अतः इन रेखाओं के बीच की न्यूनतम दूरी तथा इसका समीकरण ज्ञात कीजिए।
Question 3 (50 marks)
Q3(a) · Linear Algebra · 15 marks≈ 11 min
Vector Spaces, Subspaces, Basis and Dimension
Given P3[x] = {a0 + a1x + a2x2 + a3x3 | a0, a1, a2, a3 ∈ ℝ} as a vector space over the field ℝ.
(i) Let H = {p(x) ∈ P3[x] | ∫−11 p(x) dx = 0}. Prove that H is a subspace of P3[x] and find a basis of H.
(ii) Find a subspace K of P3[x] such that P3[x] = H ⊕ K. (8 + 7 = 15)हिंदी में प्रश्न
क्षेत्र ℝ पर एक सदिश समष्टि के रूप में P3[x] = {a0 + a1x + a2x2 + a3x3 | a0, a1, a2, a3 ∈ ℝ} दिया गया है।
(i) माना H = {p(x) ∈ P3[x] | ∫−11 p(x) dx = 0} है। सिद्ध कीजिए कि H, P3[x] की एक उपसमष्टि है तथा H का एक आधार ज्ञात कीजिए।
(ii) P3[x] की एक उपसमष्टि K ज्ञात कीजिए, जिसके लिए P3[x] = H ⊕ K है। (8 + 7 = 15)Worked solution
(i) For p = a0 + a1x + a2x2 + a3x3, φ(p) = ∫−11 p dx = 2a0 + (2/3)a2, since odd powers integrate to 0. φ is linear, so H = ker φ is a subspace (directly: 0 ∈ H, and ∫(αp + βq) dx = α∫p dx + β∫q dx = 0 for p, q ∈ H).
p ∈ H ⇔ a2 = −3a0, so p = a0(1 − 3x2) + a1x + a3x3. The polynomials 1 − 3x2, x, x3 lie in H, span it and are independent, so they form a basis and dim H = 3.
(ii) Take K = L{1}, the constants. φ(c) = 2c = 0 only for c = 0, so H ∩ K = {0}; and dim H + dim K = 3 + 1 = 4 = dim P3[x], so H + K = P3[x].
Explicitly p = (p − φ(p)/2) + φ(p)/2, where the first part lies in H (its integral is φ(p) − φ(p) = 0) and the second is a constant.
Answer: H is a subspace with basis {1 − 3x2, x, x3} (dim 3); K = L{1} gives P3[x] = H ⊕ K.Final answer: H is a subspace with basis {1 − 3x2, x, x3} (dim 3); K = L{1} gives P3[x] = H ⊕ K.
Q3(b) · Calculus · 20 marks≈ 14 min
Double and Triple Integrals; Areas, Surfaces and Volumes
(i) Evaluate the double integral ∫02 ∫x2+12x+1 x2y dy dx by reversing the order of integration. (10)
(ii) Find the volume of the solid, which is formed from the intersection of the coordinate planes x = 0, y = 0, z = 0 and the plane x/a + y/b + z/c = 1. (10)हिंदी में प्रश्न
(i) समाकलन के क्रम को उलटकर द्विशः समाकल ∫02 ∫x2+12x+1 x2y dy dx का मान ज्ञात कीजिए। (10)
(ii) उस ठोस का आयतन ज्ञात कीजिए, जो निर्देशांक समतलों x = 0, y = 0, z = 0 तथा समतल x/a + y/b + z/c = 1 के प्रतिच्छेदन से बनता है। (10)Worked solution
(i) Region: 0 ≤ x ≤ 2, x2 + 1 ≤ y ≤ 2x + 1; the parabola and the line meet at (0, 1) and (2, 5). For 1 ≤ y ≤ 5, x runs from the line, x = (y − 1)/2, to the parabola, x = √(y − 1).
I = ∫15∫(y−1)/2√(y−1) x2y dx dy = ∫15 (y/3)[(y − 1)3/2 − (y − 1)3/8] dy.
With t = y − 1: I = (1/3)∫04 (t + 1)(t3/2 − t3/8) dt = (1/3)[2t7/2/7 + 2t5/2/5 − t5/40 − t4/32]04 = (1/3)(256/7 + 64/5 − 128/5 − 8) = (1/3)(552/35) = 184/35.
Check in the original order: (1/2)∫02 x2[(2x + 1)2 − (x2 + 1)2] dx = (1/2)∫02 (4x3 + 2x4 − x6) dx = (1/2)(16 + 64/5 − 128/7) = 184/35.
(ii) The tetrahedron is 0 ≤ x ≤ a, 0 ≤ y ≤ b(1 − x/a), 0 ≤ z ≤ c(1 − x/a − y/b) (a, b, c > 0). V = ∫0a∫0b(1−x/a) c(1 − x/a − y/b) dy dx = ∫0a (bc/2)(1 − x/a)2 dx = abc/6.
Answer: (i) 184/35; (ii) abc/6.Final answer: (i) 184/35; (ii) abc/6.
Q3(c) · Analytic Geometry · 15 marks≈ 11 min
Cone and Cylinder
If x/1 = y/2 = z/3 represents one of a set of three mutually perpendicular generators of the cone 5yz − 8zx − 3xy = 0, then find the equations of the other two.
हिंदी में प्रश्न
यदि x/1 = y/2 = z/3, शंकु 5yz − 8zx − 3xy = 0 के तीन परस्पर लंबवत् जनकों के समुच्चय में से किसी एक को निरूपित करता है, तब अन्य दो के समीकरण ज्ञात कीजिए।
Question 4 (50 marks)
Q4(a) · Linear Algebra · 15 marks≈ 11 min
Linear Transformations, Rank-Nullity and Matrix Representation
Let F be a subfield of complex numbers and let T be the linear transformation from F3 into F3, defined by
T(x, y, z) = (x − y + 2z, 2x + y, −x − 2y + 2z)
If (a, b, c) is a vector in F3, then answer the following:
(i) Find the condition on a, b, c such that the vector (a, b, c) be in the null space of T. What is the nullity of T?
(ii) Find the condition on a, b, c such that the vector (a, b, c) be in the range of T. What is the rank of T? (8 + 7 = 15)हिंदी में प्रश्न
माना F सम्मिश्र संख्याओं का एक उपक्षेत्र है तथा T, F3 से F3 तक रैखिक रूपांतरण है, जिसे
T(x, y, z) = (x − y + 2z, 2x + y, −x − 2y + 2z)
द्वारा परिभाषित किया गया है। यदि F3 में (a, b, c) एक सदिश है, तो निम्नलिखित के उत्तर दीजिए:
(i) a, b, c पर वह शर्त ज्ञात कीजिए, जिसके लिए सदिश (a, b, c), T की शून्य समष्टि में हो। T की शून्यता क्या है?
(ii) a, b, c पर वह शर्त ज्ञात कीजिए, जिसके लिए सदिश (a, b, c), T के परिसर (रेंज) में हो। T की कोटि (रैंक) क्या है? (8 + 7 = 15)Worked solution
T(X) = AX with A = [[1, −1, 2], [2, 1, 0], [−1, −2, 2]]. (i) (a, b, c) ∈ N(T) ⇔ a − b + 2c = 0, 2a + b = 0, −a − 2b + 2c = 0.
From 2a + b = 0, b = −2a; the first gives 3a + 2c = 0, c = −3a/2; the third, −a + 4a − 3a = 0, holds. So N(T) = L{(2, −4, −3)} and nullity T = 1; check T(2, −4, −3) = (2 + 4 − 6, 4 − 4, −2 + 8 − 6) = 0.
(ii) (a, b, c) ∈ R(T) ⇔ x − y + 2z = a, 2x + y = b, −x − 2y + 2z = c is consistent. The coefficient rows satisfy (row 3) = (row 1) − (row 2), so R3 − R1 + R2 turns the augmented row 3 into (0, 0, 0 | c − a + b).
The first two rows are independent, so the system is consistent ⇔ a − b − c = 0. Thus R(T) is the plane a = b + c and rank T = 2; check 2 + 1 = 3 = dim F3.
Answer: (i) b = −2a and c = −3a/2, i.e. (a, b, c) is a multiple of (2, −4, −3); nullity 1. (ii) a − b − c = 0; rank 2.Final answer: (i) b = −2a and c = −3a/2, i.e. (a, b, c) is a multiple of (2, −4, −3); nullity 1. (ii) a − b − c = 0; rank 2.
Q4(b) · Calculus · 15 marks≈ 11 min
Maxima and Minima, Asymptotes and Curve Tracing
Trace the curve y2(a + x) = x2(3a − x).
हिंदी में प्रश्न
वक्र y2(a + x) = x2(3a − x) का अनुरेखण कीजिए।
Worked solution
Take a > 0; y2 = x2(3a − x)/(a + x). Symmetry: only y2 occurs, so the curve is symmetric about the x-axis.
Origin: the curve passes through O, and the lowest-degree terms ay2 − 3ax2 = 0 give two real tangents y = ±√3 x: O is a node.
Region: y2 ≥ 0 requires −a < x ≤ 3a; the curve meets the x-axis at x = 0 and x = 3a, with a vertical tangent at (3a, 0).
Asymptote: the coefficient of y2 is a + x, so x = −a is an asymptote, with y2 → +∞ as x → −a+. The cubic terms x(x2 + y2) give φ3(m) = 1 + m2 ≠ 0, so there are no oblique asymptotes.
Shape: for 0 ≤ x ≤ 3a the curve forms a loop from the node O to (3a, 0); for −a < x < 0 two branches leave O (touching y = ±√3 x) and run off to infinity along x = −a, one upwards and one downwards.
Answer: Symmetric about the x-axis, node at O with tangents y = ±√3x, a loop on [0, 3a], asymptote x = −a, curve confined to −a < x ≤ 3a.Final answer: Symmetric about the x-axis, node at O with tangents y = ±√3x, a loop on [0, 3a], asymptote x = −a, curve confined to −a < x ≤ 3a.
Q4(c) · Analytic Geometry · 20 marks≈ 14 min
Ellipsoid, Hyperboloids and Paraboloids
(i) Find the equations of the generating lines of the hyperboloid x2/4 + y2/9 − z2/16 = 1 which pass through the points (2, 3, −4) and (2, −1, 4/3). (10)
(ii) Find the equation of the right circular cylinder whose axis is x = 2y = −z, and radius is 4. Prove that the area of the section of this cylinder by the plane z = 0 is 24π. (10)हिंदी में प्रश्न
(i) अतिपरवलयज x2/4 + y2/9 − z2/16 = 1 की उन जनक रेखाओं के समीकरण ज्ञात कीजिए, जो बिंदुओं (2, 3, −4) और (2, −1, 4/3) से होकर जाती हैं। (10)
(ii) उस लंबवृत्तीय बेलन का समीकरण ज्ञात कीजिए, जिसका अक्ष x = 2y = −z है तथा त्रिज्या 4 है। सिद्ध कीजिए कि समतल z = 0 द्वारा इस बेलन के परिच्छेद (सेक्शन) का क्षेत्रफल 24π है। (10)Worked solution
(i)
Through (2, 3, −4) (Example 2 of this topic): x = 2, (y − 3)/(−3) = (z + 4)/4 and (x − 2)/1 = (y − 3)/0 = (z + 4)/(−2).
At (2, −1, 4/3): x/2 = 1, y/3 = −1/3, z/4 = 1/3. λ-system x/2 − z/4 = λ(1 − y/3): 2/3 = (4/3)λ, so λ = 1/2; the line is 6x + 2y − 3z = 6, 6x − 8y + 3z = 24, with direction (3, 6, 10).
μ-system x/2 − z/4 = μ(1 + y/3): 2/3 = (2/3)μ, so μ = 1; the line is x/2 − z/4 = 1 + y/3, x/2 + z/4 = 1 − y/3, that is x = 2, 4y + 3z = 0.
The second line is the same as the generator x = 2, (y − 3)/(−3) = (z + 4)/4 through (2, 3, −4): the two given points lie on one common generator.
Answer: through (2, 3, −4): x = 2, (y − 3)/(−3) = (z + 4)/4 and (x − 2)/1 = (y − 3)/0 = (z + 4)/(−2); through (2, −1, 4/3): (x − 2)/3 = (y + 1)/6 = (z − 4/3)/10 and x = 2, (y + 1)/3 = (z − 4/3)/(−4).
(ii)
Writing x = 2y = −z = t, the axis passes through O with direction (1, 1/2, −1), that is (2, 1, −2), of length 3.
Cylinder: x2 + y2 + z2 − (2x + y − 2z)2/9 = 16, that is 5x2 + 8y2 + 5z2 − 4xy + 4yz + 8zx − 144 = 0.
Section z = 0: 5x2 − 4xy + 8y2 = 144, an ellipse with A = 5, H = −2, B = 8, K = 144; area = πK/√(AB − H2) = 144π/√36 = 24π.
Check by geometry: a plane whose normal makes an angle φ with the axis cuts a cylinder of radius r in an ellipse of area πr2/cos φ; here cos φ = |(2, 1, −2)·(0, 0, 1)|/3 = 2/3 and 16π/(2/3) = 24π.
Answer: 5x2 + 8y2 + 5z2 − 4xy + 4yz + 8zx − 144 = 0; the section by z = 0 has area 24π.Final answer: (i) through (2, 3, −4): x = 2, (y − 3)/(−3) = (z + 4)/4 and (x − 2)/1 = (y − 3)/0 = (z + 4)/(−2); through (2, −1, 4/3): (x − 2)/3 = (y + 1)/6 = (z − 4/3)/10 and x = 2, (y + 1)/3 = (z − 4/3)/(−4); (ii) 5x2 + 8y2 + 5z2 − 4xy + 4yz + 8zx − 144 = 0; the section by z = 0 has area 24π
Question 5 (50 marks)
Q5(a) · Ordinary Differential Equations · 10 marks≈ 7 min
Formulation and First Order, First Degree Equations
If y1 and y2 are two solutions of dy/dx + P(x)y = Q(x), and y2 = y1u, then show that u = 1 + a e−∫(Q/y1) dx, where a is a non-zero constant.
हिंदी में प्रश्न
यदि y1 और y2, dy/dx + P(x)y = Q(x) के दो हल हैं तथा y2 = y1u है, तो दर्शाइए कि u = 1 + a e−∫(Q/y1) dx है, जहाँ a एक शून्येतर अचर है।
Worked solution
w = y2 − y1 solves the homogeneous equation: w′ + Pw = 0. Also y1′ + Py1 = Q gives P = (Q − y1′)/y1.
With y2 = y1u, w = y1(u − 1). Then y1′(u − 1) + y1u′ + (Q − y1′)(u − 1) = 0, that is y1u′ + Q(u − 1) = 0.
Separating: du/(u − 1) = −(Q/y1) dx, so u − 1 = a e−∫(Q/y1) dx; a ≠ 0 because y2 ≠ y1.
Answer: u = 1 + a e−∫(Q/y1) dx; hence proved.Final answer: u = 1 + a e−∫(Q/y1) dx; hence proved.
Q5(b) · Ordinary Differential Equations · 10 marks≈ 7 min
Laplace Transforms and Initial Value Problems
Apply Laplace transform to solve (D2 + D) y(t) = 2, D ≡ d/dt, subject to the conditions y(0) = 3, y′(0) = 1.
हिंदी में प्रश्न
शर्तों y(0) = 3, y′(0) = 1 के अधीन, लाप्लास रूपांतरण लागू करके (D2 + D) y(t) = 2, D ≡ d/dt को हल कीजिए।
Q5(c) · Dynamics and Statics · 10 marks≈ 7 min
Virtual Work, Stability and Forces in Three Dimensions
A smooth paraboloid of revolution is fixed with its axis vertical and vertex upwards; on it is placed a heavy elastic string of unstretched length 2πb. When the string is in equilibrium, apply the principle of virtual work to show that it rests in the form of a circle of radius 4πabλ/(4πaλ − Wb), where W is the weight of the string, λ is the modulus of elasticity and 4a is the latus rectum of the generating parabola.
हिंदी में प्रश्न
परिक्रमण का एक चिकना परवलयज इस प्रकार स्थित है कि उसका अक्ष ऊर्ध्वाधर है तथा शीर्ष ऊपर की ओर है; इस पर 2πb अतानित लंबाई की एक भारी प्रत्यास्थ डोरी रखी है। जब डोरी संतुलन में हो, तो कल्पित कार्य का सिद्धांत (नियम) लागू करके यह दर्शाइए कि यह 4πabλ/(4πaλ − Wb) त्रिज्या के एक वृत्त के रूप में है, जहाँ W डोरी का भार है, λ प्रत्यास्थता गुणांक है तथा 4a जनक परवलय का नाभिलंब है।
Worked solution
Let the string rest as a horizontal circle of radius x; the generating parabola (vertex up, latus rectum 4a) is x2 = 4ad, d being the depth below the vertex, so the string is at height −x2/(4a).
Its stretched length is 2πx, so its tension is T = λ(2πx − 2πb)/(2πb) = λ(x − b)/b.
Virtual work for a change δx (the smooth surface does no work): −W δ(−x2/(4a)) − T δ(2πx) = 0, that is Wx/(2a) = 2πλ(x − b)/b.
So 4πaλ(x − b) = Wbx, that is x(4πaλ − Wb) = 4πabλ.
Answer: the radius is x = 4πabλ/(4πaλ − Wb), which requires 4πaλ > Wb.Final answer: the radius is x = 4πabλ/(4πaλ − Wb), which requires 4πaλ > Wb.
Q5(d) · Dynamics and Statics · 10 marks≈ 7 min
Constrained Motion, Work and Energy
A particle falls from rest at the vertex of an inverted catenary. Prove that the particle will leave the curve when the path described is √3 times the vertical distance through which the particle has fallen.
हिंदी में प्रश्न
एक कण, एक उल्टी कैटेनरी के शीर्ष से विराम अवस्था से नीचे गिरता है। सिद्ध कीजिए कि कण वक्र को तब छोड़ेगा, जब कण द्वारा तय किया गया पथ उस ऊर्ध्वाधर दूरी का √3 गुना हो, जितनी दूरी तक कण नीचे गिरा है।
Worked solution
For the inverted catenary s = c tan ψ (vertex up), the radius of curvature is ρ = ds/dψ = c sec2ψ and the height above the directrix is y = c sec ψ.
Starting from rest at the vertex (y = c), after a vertical fall z = c sec ψ − c the speed is given by v2 = 2gc(sec ψ − 1).
On the convex side, mg cos ψ − R = mv2/ρ, so R = mg cos ψ (2 cos ψ − 1), which vanishes when cos ψ = 1/2, ψ = 60°.
There s = c tan 60° = √3c and z = c(sec 60° − 1) = c.
Answer: the particle leaves the curve at ψ = 60°, where the arc described, √3c, is √3 times the vertical fall c.Final answer: the particle leaves the curve at ψ = 60°, where the arc described, √3c, is √3 times the vertical fall c.
Q5(e) · Vector Analysis · 10 marks≈ 7 min
Gauss and Stokes' Theorems and Green's Identities
Given F = (2x + 3y) î − 4z ĵ − 5x k̂ and S is the surface 2x + 4y + 4z = 5, bounded by x = 0, x = 2, y = 0 and y = 3. Evaluate ∬S (∇ × F) · n̂ dS.
हिंदी में प्रश्न
दिया गया है कि F = (2x + 3y) î − 4z ĵ − 5x k̂ है तथा S सतह 2x + 4y + 4z = 5 है, जो x = 0, x = 2, y = 0 और y = 3 द्वारा परिबद्ध है। ∬S (∇ × F) · n̂ dS का मान ज्ञात कीजिए।
Worked solution
For F = (2x + 3y, −4z, −5x): ∇ × F = (0 + 4)i + (0 + 5)j + (0 − 3)k = 4i + 5j − 3k.
On 2x + 4y + 4z = 5 the upward unit normal is n = (1, 2, 2)/3, so (∇ × F)·n = (4 + 10 − 6)/3 = 8/3.
dS = dx dy/|n·k| = (3/2) dx dy over the projection 0 ≤ x ≤ 2, 0 ≤ y ≤ 3, of area 6.
∬(∇ × F)·n dS = (8/3)(3/2)(6) = 24; the line integral round the four edges of this piece of the plane also gives 24.
Answer: 24 with the upward normal (−24 with the downward normal).Final answer: 24 with the upward normal (−24 with the downward normal).
Question 6 (50 marks)
Q6(a) · Ordinary Differential Equations · 15 marks≈ 11 min
Second Order Equations with Variable Coefficients
Solve x2 d2y/dx2 − 3x dy/dx + y = ((sin(log x) + 1)/x) log x.
हिंदी में प्रश्न
x2 d2y/dx2 − 3x dy/dx + y = ((sin(log x) + 1)/x) log x को हल कीजिए।
Worked solution
With x = ez: (θ2 − 4θ + 1)y = e−zz(sin z + 1). Roots θ = 2 ± √3, so CF = c1x2 + √3 + c2x2 − √3.
Exponential shift: PI = e−z[1/f(θ)](z sin z + z), where f(θ) = (θ − 1)2 − 4(θ − 1) + 1 = θ2 − 6θ + 6.
[1/f(θ)]z = (1/6)(1 + θ + ...)z = (z + 1)/6.
For z sin z use [1/f]zV = z[1/f]V − [f′/f2]V with V = sin z. With θ2 = −1, f = 5 − 6θ, so [1/f] sin z = (5 + 6θ) sin z/61 = (5 sin z + 6 cos z)/61.
Applying 1/f again: [1/f2] sin z = (5 + 6θ)(5 sin z + 6 cos z)/3721 = (60 cos z − 11 sin z)/3721; then f′ = 2θ − 6 gives [f′/f2] sin z = −(54 sin z + 382 cos z)/3721.
So [1/f](z sin z) = z(5 sin z + 6 cos z)/61 + (54 sin z + 382 cos z)/3721.
Answer: y = c1x2 + √3 + c2x2 − √3 + (1/x)[(log x + 1)/6 + log x (5 sin log x + 6 cos log x)/61 + (54 sin log x + 382 cos log x)/3721] (verified by substitution).Final answer: y = c1x2 + √3 + c2x2 − √3 + (1/x)[(log x + 1)/6 + log x (5 sin log x + 6 cos log x)/61 + (54 sin log x + 382 cos log x)/3721] (verified by substitution).
Q6(b) · Dynamics and Statics · 20 marks≈ 14 min
Motion in a Plane and Projectiles
(i) A battleship is steaming ahead with velocity u. A gun is mounted on the ship so as to point straight backwards and is set at an angle of elevation α. If v be the velocity of projection relative to the gun, show that the range is (2v/g) sin α (v cos α − u) and the angle for maximum range is cos−1((u + √(u2 + 8v2))/(4v)), where g stands for acceleration due to gravity. (10)
(ii) A uniform chain of length l has one end fixed at a height h above a rough table and rests in a vertical plane so that a portion of it lies in a straight line on the table. Prove that if the chain is on the point of slipping, the length on the table is (l + μh) − √((μ2 + 1)h2 + 2μlh), where μ is the coefficient of friction. (10)हिंदी में प्रश्न
(i) एक जंगी जहाज u वेग से आगे बढ़ रहा है। जहाज पर एक बन्दूक इस तरह लगाई गई है कि उसका मुँह ठीक पीछे की ओर है तथा उसे उन्नयन कोण α पर रखा गया है। यदि बन्दूक के सापेक्ष प्रक्षेपण (प्रोजेक्शन) का वेग v हो, तो दर्शाइए कि परास (2v/g) sin α (v cos α − u) है तथा अधिकतम परास के लिए कोण cos−1((u + √(u2 + 8v2))/(4v)) है, जहाँ g गुरुत्वीय त्वरण को दर्शाता है। (10)
(ii) एक एकसमान जंजीर (चेन), जिसकी लंबाई l है, का एक सिरा एक खुरदरी मेज से h ऊँचाई पर स्थिर है और यह एक ऊर्ध्वाधर तल में इस प्रकार रखी है कि इसका एक हिस्सा मेज पर एक सीधी रेखा में है। सिद्ध कीजिए कि यदि चेन फिसलने की कगार पर है, तो मेज पर उसकी लंबाई (l + μh) − √((μ2 + 1)h2 + 2μlh) है, जहाँ μ घर्षण गुणांक है। (10)Worked solution
(i)
Relative to the gun the shell leaves with v cos α backwards and v sin α upwards; adding the ship's velocity u forwards, its actual velocity is (v cos α − u) backwards and v sin α upwards.
The time of flight is 2v sin α/g, so the range (measured on the sea from the point of firing) is R = (2v/g) sin α (v cos α − u).
R(α) = (v/g)(v sin 2α − 2u sin α), and dR/dα = (2v/g)(v cos 2α − u cos α) = 0.
With c = cos α: v(2c2 − 1) − uc = 0, that is 2vc2 − uc − v = 0, whose positive root is c = (u + √(u2 + 8v2))/(4v).
d2R/dα2 = (2v/g)(−2v sin 2α + u sin α) = (2v/g) sin α (u − 4vc) = −(2v/g) sin α √(u2 + 8v2) < 0 there, so this is a maximum.
Answer: R = (2v/g) sin α (v cos α − u), greatest when α = cos−1((u + √(u2 + 8v2))/(4v)).
(ii)
Let x be the length on the table; the rest, l − x, hangs as a catenary which touches the table at its lowest point, where the tension is horizontal and equal to wc.
The part on the table is in limiting equilibrium under this pull and the friction μwx, so c = μx.
For the hanging part, measured from its lowest point, s = l − x and the fixed end is h higher: (c + h)2 = c2 + s2, that is (l − x)2 = h2 + 2μhx.
So x2 − 2(l + μh)x + (l2 − h2) = 0, giving x = (l + μh) ± √((l + μh)2 − l2 + h2).
Since x < l the minus sign is taken, and (l + μh)2 − l2 + h2 = (μ2 + 1)h2 + 2μlh.
Answer: the length on the table is (l + μh) − √((μ2 + 1)h2 + 2μlh).Final answer: (i) R = (2v/g) sin α (v cos α − u), greatest when α = cos−1((u + √(u2 + 8v2))/(4v)); (ii) the length on the table is (l + μh) − √((μ2 + 1)h2 + 2μlh)
Q6(c) · Vector Analysis · 15 marks≈ 11 min
Curves in Space: Curvature, Torsion and Serret-Frenet Formulae
Determine T̂, N̂, B̂, κ and τ for the parabola y2 = 4ax, where a is a constant. Here T̂, N̂, B̂, κ and τ denote unit tangent vector, unit principal normal vector, unit binormal vector, curvature and torsion respectively.
हिंदी में प्रश्न
परवलय y2 = 4ax के लिए T̂, N̂, B̂, κ और τ निर्धारित कीजिए, जहाँ a एक अचर है। यहाँ T̂, N̂, B̂, κ तथा τ क्रमशः एकक स्पर्शी सदिश, एकक मुख्य अभिलम्ब सदिश, एकक उपाभिलम्ब सदिश, वक्रता तथा मरोड़ को निरूपित करते हैं।
Worked solution
Parametrise y2 = 4ax as r = (at2, 2at, 0): r′ = 2a(t, 1, 0), |r′| = 2a√(1 + t2), r″ = (2a, 0, 0) and r′ × r″ = (0, 0, −4a2).
T = (ti + j)/√(1 + t2) and B = (r′ × r″)/|r′ × r″| = −k.
N = B × T = (i − tj)/√(1 + t2), pointing into the concave side (N = i at the vertex).
κ = 4a2/[8a3(1 + t2)3/2] = 1/[2a(1 + t2)3/2]; since 1 + t2 = (a + x)/a, κ = √a/[2(a + x)3/2].
r‴ = 0, so [r′ r″ r‴] = 0 and τ = 0, as for every plane curve.
Answer: T = (ti + j)/√(1 + t2), N = (i − tj)/√(1 + t2), B = −k (for the sense of increasing y), κ = 1/[2a(1 + t2)3/2] = √a/[2(a + x)3/2], τ = 0.Final answer: T = (ti + j)/√(1 + t2), N = (i − tj)/√(1 + t2), B = −k (for the sense of increasing y), κ = 1/[2a(1 + t2)3/2] = √a/[2(a + x)3/2], τ = 0.
Question 7 (50 marks)
Q7(a) · Dynamics and Statics · 15 marks≈ 11 min
Rectilinear Motion and Simple Harmonic Motion
A light elastic string of natural length l has one extremity fixed at a point A and the other attached to a stone (mass m) the weight of which in equilibrium would extend the string to a length b. Show that if the stone be dropped from rest at A, it will come to instantaneous rest at a depth √(b2 − l2) below the equilibrium position and this depth is attained in time √(2l/g) + √((b − l)/g) {π − cos−1 √((b − l)/(b + l))}. Prove also that if the greatest depth below A be l cot2(θ/2), then the modulus of elasticity is (1/2) mg tan2 θ.
हिंदी में प्रश्न
प्राकृतिक लंबाई l की एक हल्की प्रत्यास्थ डोरी का एक सिरा बिंदु A पर स्थिर है तथा दूसरा सिरा एक पत्थर (द्रव्यमान m) से जुड़ा है; संतुलन की स्थिति में इस पत्थर का भार डोरी को खींचकर b लंबाई तक बढ़ा देता है। दर्शाइए कि यदि पत्थर को बिंदु A से विराम अवस्था से गिराया जाए, तो वह √(2l/g) + √((b − l)/g) {π − cos−1 √((b − l)/(b + l))} समय में संतुलन की स्थिति से √(b2 − l2) नीचे तात्क्षणिक विराम अवस्था में आ जाएगा। यह भी सिद्ध कीजिए कि यदि A के नीचे अधिकतम गहराई l cot2(θ/2) है, तो प्रत्यास्थता गुणांक (1/2) mg tan2 θ है।
Worked solution
The equilibrium extension is b − l, so λ = mgl/(b − l); while the string is taut the motion is SHM about the equilibrium point E (depth b below A) with n2 = g/(b − l).
Free fall from A through l takes √(2l/g) and gives speed √(2gl) at depth l, which is b − l above E.
Amplitude: 2gl = n2(A2 − (b − l)2) gives A2 = 2l(b − l) + (b − l)2 = b2 − l2, so the stone comes to rest √(b2 − l2) below E.
With x measured downwards from E, write x = −A cos ψ, where ψ = nt + constant increases; the start x = −(b − l) with x increasing gives ψ0 = cos−1((b − l)/A) = cos−1√((b − l)/(b + l)), and the lowest point is ψ = π.
Time in the SHM = (π − ψ0)/n = √((b − l)/g){π − cos−1√((b − l)/(b + l))}; adding √(2l/g) gives the stated time.
Second part: the greatest depth below A is D = b + √(b2 − l2), so (D − b)2 = b2 − l2 and b = (D2 + l2)/(2D).
With D = lk2, k = cot(θ/2): b − l = (D − l)2/(2D) = l(k2 − 1)2/(2k2), so λ = mgl/(b − l) = 2mgk2/(k2 − 1)2.
Since tan θ = 2 tan(θ/2)/(1 − tan2(θ/2)) = 2k/(k2 − 1), this is λ = (1/2)mg tan2θ.
Answer: the stone is at rest √(b2 − l2) below E after √(2l/g) + √((b − l)/g){π − cos−1√((b − l)/(b + l))}, and the modulus is λ = (1/2)mg tan2θ.Final answer: the stone is at rest √(b2 − l2) below E after √(2l/g) + √((b − l)/g){π − cos−1√((b − l)/(b + l))}, and the modulus is λ = (1/2)mg tan2θ.
Q7(b) · Vector Analysis · 15 marks≈ 11 min
Gauss and Stokes' Theorems and Green's Identities
If F = 4xz î − y2 ĵ + yz k̂, then evaluate ∬S F · n̂ dS, where S is the surface of a unit cube with two opposite corners at (0, 0, 0) and (1, 1, 1) respectively. Hence verify the divergence theorem.
हिंदी में प्रश्न
यदि F = 4xz î − y2 ĵ + yz k̂ है, तो ∬S F · n̂ dS का मान ज्ञात कीजिए, जहाँ S एक इकाई घन का पृष्ठ है जिसके दो विपरीत कोने क्रमशः (0, 0, 0) तथा (1, 1, 1) पर हैं। अतः अपसरण (डाइवर्जेंस) प्रमेय को सत्यापित कीजिए।
Worked solution
x = 1: F·n = 4z, flux 2; x = 0: F·n = −4xz = 0.
y = 1: F·n = −y2 = −1, flux −1; y = 0: F·n = y2 = 0.
z = 1: F·n = yz = y, flux 1/2; z = 0: F·n = −yz = 0.
Answer: ∬F·n dS = 2 − 1 + 1/2 = 3/2, equal to ∭(4z − y) dV = 3/2 (Example 4), which verifies the theorem.Final answer: ∬F·n dS = 2 − 1 + 1/2 = 3/2, equal to ∭(4z − y) dV = 3/2 (Example 4), which verifies the theorem.
Q7(c) · Ordinary Differential Equations · 20 marks≈ 14 min
Formulation and First Order, First Degree Equations
(i) Show that (x + y + 1)−4 is an integrating factor of (2x − y − 1)y dx + (2y − x − 1)x dy = 0. Hence solve it. (10)
(ii) If L{f(t)} = f̄(s), then show that L{f(t)/t} = ∫s∞ f̄(u) du. Hence evaluate L{(cos at − cos bt)/t}. (10)हिंदी में प्रश्न
(i) दर्शाइए कि (2x − y − 1)y dx + (2y − x − 1)x dy = 0 का एक समाकलन गुणक (x + y + 1)−4 है। अतः इसे हल कीजिए। (10)
(ii) यदि L{f(t)} = f̄(s) है, तो दर्शाइए कि L{f(t)/t} = ∫s∞ f̄(u) du है। अतः L{(cos at − cos bt)/t} का मान ज्ञात कीजिए। (10)Worked solution
(i)
M = (2x − y − 1)y, N = (2y − x − 1)x, s = x + y + 1. μ = s−4 is an IF if (μM)y = (μN)x, that is s(My − Nx) = 4(M − N).
My − Nx = (2x − 2y − 1) − (2y − 2x − 1) = 4(x − y), and M − N = x2 − y2 + x − y = (x − y)s. Both sides equal 4(x − y)s, so s−4 is an IF.
Let F = xy/s3: Fx = (ys − 3xy)/s4 = −(2xy − y2 − y)/s4 = −M/s4, and by symmetry Fy = −N/s4. So the exact equation is −dF = 0.
Answer: xy = c(x + y + 1)3.
(ii)
Proof: ∫s∞ f̄(u) du = ∫s∞∫0∞ e−utf(t) dt du = ∫0∞ f(t)[∫s∞ e−ut du] dt = ∫0∞ e−st(f(t)/t) dt = L{f(t)/t} (order changed under absolute convergence; limt→0+ f(t)/t must exist).
For f = cos at − cos bt, f(t)/t → 0 as t → 0, and L{(cos at − cos bt)/t} = ∫s∞[u/(u2 + a2) − u/(u2 + b2)] du = [(1/2) log((u2 + a2)/(u2 + b2))]s∞.
Answer: L{(cos at − cos bt)/t} = (1/2) log((s2 + b2)/(s2 + a2)).Final answer: (i) xy = c(x + y + 1)3; (ii) L{(cos at − cos bt)/t} = (1/2) log((s2 + b2)/(s2 + a2))
Question 8 (50 marks)
Q8(a) · Ordinary Differential Equations · 15 marks≈ 11 min
Second Order Equations with Variable Coefficients
Solve d2y/dx2 + a2y = sec ax by the method of variation of parameters.
हिंदी में प्रश्न
प्राचल-विचरण विधि द्वारा d2y/dx2 + a2y = sec ax को हल कीजिए।
Q8(b) · Dynamics and Statics · 15 marks≈ 11 min
Virtual Work, Stability and Forces in Three Dimensions
AB is a uniform rod of length l and weight W, which can turn freely about a fixed point in its length distant l/3 from A. AC and BC are light strings each of length (5/6)l, attached to a particle C of weight w. Prove that if W is less than 2w, there will be stable equilibrium with AB inclined to the horizontal at an angle tan−1((W + w)/(4w)).
हिंदी में प्रश्न
l लंबाई तथा W भार की एक एकसमान छड़ AB है, जो अपनी लंबाई में बिंदु A से l/3 दूरी पर स्थित एक निश्चित बिंदु के परितः स्वतंत्र रूप से घूम सकती है। AC तथा BC दो हल्की डोरियाँ हैं, जिनमें से प्रत्येक की लंबाई (5/6)l है और जो w भार के एक कण C से जुड़ी है। सिद्ध कीजिए कि यदि W, 2w से कम है, तो स्थायी संतुलन की स्थिति होगी, जिसमें AB क्षैतिज से tan−1((W + w)/(4w)) कोण पर झुकी होगी।
Worked solution
Let O be the fixed point (OA = l/3, OB = 2l/3) and let AB make angle θ with the horizontal, B being higher when θ > 0; the centre G of the rod is l/6 from O towards B.
Triangle ACB has AC = BC = 5l/6 and AB = l, so C is at distance √(25l2/36 − l2/4) = 2l/3 from G, below AB on its perpendicular.
While both strings are taut the system is rigid; heights above O: zG = (l/6) sin θ and zC = (l/6) sin θ − (2l/3) cos θ.
V = (W + w)(l/6) sin θ − (2l/3)w cos θ, and dV/dθ = (W + w)(l/6) cos θ + (2l/3)w sin θ = 0 gives tan θ = −(W + w)/(4w): AB is inclined at α = tan−1((W + w)/(4w)), with B lower.
At θ = −α, d2V/dθ2 = (W + w)(l/6) sin α + (2l/3)w cos α > 0, so this equilibrium is stable.
Both strings stay taut only if the vertical through C meets AB between A and B; C is at horizontal distance (l/6) cos α − (2l/3) sin α from O, and this is not beyond A, at −(l/3) cos α, only if tan α ≤ 3/4.
tan α = (W + w)/(4w) < 3/4 is exactly W < 2w.
Answer: if W < 2w the stable position has AB inclined to the horizontal at tan−1((W + w)/(4w)); if W > 2w one string would slacken.Final answer: if W < 2w the stable position has AB inclined to the horizontal at tan−1((W + w)/(4w)); if W > 2w one string would slacken.
Q8(c) · Vector Analysis · 20 marks≈ 14 min
Vector Identities and Vector Equations
(i) For a scalar point function φ and a vector point function F, prove that ∇ × (φF) = (∇φ) × F + φ(∇ × F). If r = x î + y ĵ + z k̂ and r = |r|, then verify the above identity for φ = 1/r and F = r. (10)
(ii) If r = a cos t î + a sin t ĵ + bt k̂, then show that |r′ × r″|2 = a2(a2 + b2) and [r′ r″ r‴] = a2b. (10)हिंदी में प्रश्न
(i) एक अदिश बिंदु फलन φ तथा एक सदिश बिंदु फलन F के लिए सिद्ध कीजिए कि ∇ × (φF) = (∇φ) × F + φ(∇ × F)। यदि r = x î + y ĵ + z k̂ तथा r = |r| हैं, तो φ = 1/r तथा F = r के लिए उपर्युक्त सर्वसमिका (आइडेन्टिटी) को सत्यापित कीजिए। (10)
(ii) यदि r = a cos t î + a sin t ĵ + bt k̂ है, तो दर्शाइए कि |r′ × r″|2 = a2(a2 + b2) और [r′ r″ r‴] = a2b। (10)Worked solution
(i) Proof: ∇ × (φF) = Σ i × ∂(φF)/∂x = Σ i × [(∂φ/∂x)F + φ ∂F/∂x] = (Σ i ∂φ/∂x) × F + φ Σ i × ∂F/∂x = (∇φ) × F + φ(∇ × F).
For φ = 1/r, F = r: ∇(1/r) = −r/r3 and ∇ × r = 0, so the right side is (−r/r3) × r + 0 = 0.
Directly, the i-component of ∇ × (r/r) is ∂(z/r)/∂y − ∂(y/r)/∂z = −yz/r3 + yz/r3 = 0, and likewise for j and k, so both sides are 0.
(ii) r′ = (−a sin t, a cos t, b), r″ = (−a cos t, −a sin t, 0), r‴ = (a sin t, −a cos t, 0), and r′ × r″ = (ab sin t, −ab cos t, a2).
|r′ × r″|2 = a2b2 + a4 = a2(a2 + b2) and [r′ r″ r‴] = (r′ × r″)·r‴ = a2b sin2t + a2b cos2t = a2b.
Answer: the identity holds (both sides are 0 for φ = 1/r, F = r); |r′ × r″|2 = a2(a2 + b2) and [r′ r″ r‴] = a2b.Final answer: the identity holds (both sides are 0 for φ = 1/r, F = r); |r′ × r″|2 = a2(a2 + b2) and [r′ r″ r‴] = a2b.
UPSC Mains 2026 Mathematics Paper II
Question 1 (50 marks)
Q1(a) · Abstract Algebra · 10 marks≈ 7 min
Normal Subgroups, Quotient Groups, Homomorphisms and Isomorphism Theorems
Let G be a group of order pq, where p and q are primes. Show that either G is an abelian group or no non-identity element commutes with every element of G.
हिंदी में प्रश्न
माना कोटि pq का एक समूह G है, जहाँ p और q अभाज्य हैं। दर्शाइए कि या तो G एक आबेली समूह है या कोई भी अतत्समक (नॉन-आइडेन्टिटी) अवयव, G के प्रत्येक अवयव से क्रमविनिमेय नहीं करता।
Worked solution
Z(G) is a subgroup, so by Lagrange o(Z(G)) ∈ {1, p, q, pq}.
If o(Z(G)) = pq, then G = Z(G) is abelian.
If o(Z(G)) = p or q, then o(G/Z(G)) is q or p, a prime, so G/Z(G) is cyclic. Then G is abelian, so Z(G) = G, a contradiction.
Hence either G is abelian, or Z(G) = {e}, which says exactly that no non-identity element commutes with every element of G.
Answer: G is abelian or has trivial centre.Final answer: G is abelian or has trivial centre.
Q1(b) · Abstract Algebra · 10 marks≈ 7 min
Integral Domains, PID, Euclidean Domains, UFD, Fields and Quotient Fields
Show that every non-zero prime ideal in a Euclidean Domain is maximal.
हिंदी में प्रश्न
दर्शाइए कि यूक्लिडीय प्रांत में प्रत्येक शून्येतर अभाज्य गुणजावली उच्चिष्ठ है।
Worked solution
Let D be a Euclidean domain; it is a PID, so a non-zero prime ideal is P = ⟨p⟩ with p ≠ 0, and p is not a unit because P ≠ D.
Let J be an ideal with P ⊆ J ⊆ D. Then J = ⟨a⟩ and p ∈ ⟨a⟩, so p = ab for some b ∈ D.
ab ∈ P and P is prime, so a ∈ P or b ∈ P.
If a ∈ P, then J = ⟨a⟩ ⊆ P, so J = P. If b ∈ P, write b = pc; then p = apc, and cancelling p ≠ 0 gives ac = 1, so a is a unit and J = D.
Answer: P is maximal. ('Non-zero' is needed: ⟨0⟩ is prime in any integral domain but maximal only in a field.)Final answer: P is maximal. ('Non-zero' is needed: ⟨0⟩ is prime in any integral domain but maximal only in a field.)
Q1(c) · Real Analysis · 10 marks≈ 7 min
Continuity, Uniform Continuity and Continuous Functions on Compact Sets
Find the points of discontinuity of the function f : (0, 1) → ℝ defined by f(x) = limn→∞ [(1 + sin(π/x))n − 1]/[(1 + sin(π/x))n + 1], x ∈ (0, 1).
हिंदी में प्रश्न
f(x) = limn→∞ [(1 + sin(π/x))n − 1]/[(1 + sin(π/x))n + 1], x ∈ (0, 1) द्वारा परिभाषित फलन f : (0, 1) → ℝ के असांतत्य बिंदु ज्ञात कीजिए।
Worked solution
Put t = 1 + sin(π/x) ∈ [0, 2]. If t > 1, (tn − 1)/(tn + 1) = (1 − t−n)/(1 + t−n) → 1; if t = 1 the quotient is 0; if 0 ≤ t < 1, tn → 0 and the quotient tends to −1.
Hence f(x) = 1, 0 or −1 according as sin(π/x) > 0, = 0 or < 0; that is, f(x) = sgn(sin(π/x)).
For x ∈ (0, 1), π/x ∈ (π, ∞), and sin(π/x) = 0 iff π/x = kπ, i.e. x = 1/k with k ≥ 2 an integer.
If sin(π/x0) ≠ 0, then by continuity sin(π/x) keeps its sign near x0, so f is constant near x0 and continuous there.
At x = 1/k, f(1/k) = 0, while sin(π/x) changes sign as π/x crosses kπ, so the one-sided limits of f are 1 and −1: a jump discontinuity.
Answer: f is discontinuous exactly at x = 1/2, 1/3, 1/4, ... (all jump discontinuities) and continuous at every other point of (0, 1).Final answer: f is discontinuous exactly at x = 1/2, 1/3, 1/4, ... (all jump discontinuities) and continuous at every other point of (0, 1).
Q1(d) · Complex Analysis · 10 marks≈ 7 min
Analytic Functions and Cauchy-Riemann Equations
Show that the function f(z) = √|xy| is not analytic at the origin, although Cauchy-Riemann equations are satisfied at that point.
हिंदी में प्रश्न
दर्शाइए कि फलन f(z) = √|xy| मूल-बिंदु पर विश्लेषिक नहीं है, जबकि इस बिंदु पर कौशी-रीमान समीकरण संतुष्ट होते हैं।
Q1(e) · Linear Programming · 10 marks≈ 7 min
Duality in Linear Programming
Solve the dual problem of the following linear programming problem by the graphical method:
Minimize Y = 10y1 + 6y2 + 2y3 + y4
subject to the constraints:
y1 + y2 + y3 + y4 ≥ 2
2y1 + y2 − y3 − 2y4 ≥ 1
y1, y2, y3, y4 ≥ 0हिंदी में प्रश्न
निम्नलिखित रैखिक प्रोग्रामन समस्या की द्वैत समस्या को ग्राफीय विधि से हल कीजिए:
न्यूनतमीकरण कीजिए Y = 10y1 + 6y2 + 2y3 + y4
निम्न व्यवरोधों के अंतर्गत:
y1 + y2 + y3 + y4 ≥ 2
2y1 + y2 − y3 − 2y4 ≥ 1
y1, y2, y3, y4 ≥ 0
Question 2 (50 marks)
Q2(a) · Abstract Algebra · 15 marks≈ 11 min
Normal Subgroups, Quotient Groups, Homomorphisms and Isomorphism Theorems
Let G be a finite group generated by two elements u and v of order 2 each. Show that G is isomorphic to the dihedral group of order 2n, where O(uv) = n.
हिंदी में प्रश्न
माना कोटि 2 के दो अवयवों u और v से जनित एक परिमित समूह G है। दर्शाइए कि G, 2n कोटि के द्वितल समूह के तुल्याकारी है, जहाँ O(uv) = n है।
Worked solution
Put r = uv and s = u. Then o(r) = n and s2 = e; since u−1 = u and v−1 = v, srs−1 = u(uv)u = vu = (uv)−1 = r−1.
v = u(uv) = sr, so G = ⟨u, v⟩ = ⟨r, s⟩. Using sr = r−1s, every word in r, s reduces to ri or ris with 0 ≤ i ≤ n − 1, so G = ⟨r⟩ ∪ ⟨r⟩s and o(G) ≤ 2n.
s ∉ ⟨r⟩: otherwise u and v = sr both lie in the cyclic group ⟨r⟩, which has at most one element of order 2; then u = v, r = e, n = 1, and ⟨r⟩ = {e} cannot contain u. So ⟨r⟩ and ⟨r⟩s are disjoint and o(G) = 2n.
Dn = ⟨ρ, σ⟩ has the same description: 2n elements ρiσj with o(ρ) = n, σ2 = e, σρσ−1 = ρ−1. So ψ(risj) = ρiσj (0 ≤ i ≤ n − 1, j = 0, 1) is a bijection, and it preserves products because in both groups products are computed by the same rules rn = s2 = e, sri = r−is.
Answer: G ≅ Dn, the dihedral group of order 2n, with uv as the rotation and u as a reflection.Final answer: G ≅ Dn, the dihedral group of order 2n, with uv as the rotation and u as a reflection.
Q2(b) · Real Analysis · 15 marks≈ 11 min
Real Number System, Sequences, Cauchy Sequences and Completeness
Prove that the sequence (un) defined by the recursion formula un+1 = √(7 + un), u1 = √7, converges to the positive root of x2 − x − 7 = 0.
हिंदी में प्रश्न
सिद्ध कीजिए कि प्रतिवर्तन सूत्र un+1 = √(7 + un), u1 = √7, द्वारा परिभाषित अनुक्रम (un), समीकरण x2 − x − 7 = 0 के धनात्मक मूल पर अभिसरित होता है।
Q2(c) · Complex Analysis · 20 marks≈ 14 min
Contour Integration: Evaluation of Real Integrals
Use the method of contour integration to prove ∫02π sin2θ/(a + b cos θ) dθ = (2π/b2){a − √(a2 − b2)}, if a > b > 0.
हिंदी में प्रश्न
कन्टूर समाकलन विधि के उपयोग से सिद्ध कीजिए कि यदि a > b > 0 है, तो ∫02π sin2θ/(a + b cos θ) dθ = (2π/b2){a − √(a2 − b2)} है।
Worked solution
With z = eiθ: sin2θ = −(z2 − 1)2/(4z2), a + b cos θ = (bz2 + 2az + b)/(2z), dθ = dz/(iz).
So I = −(1/(2ib))∮|z|=1 (z2 − 1)2/(z2(z − α)(z − β)) dz, where α, β = (−a ± √(a2 − b2))/b are the roots of bz2 + 2az + b.
αβ = 1 and |α| = (a − √(a2 − b2))/b = b/(a + √(a2 − b2)) < 1, so α is inside and β outside; z = 0 is a double pole inside.
Res at 0: (z2 − 1)2/(z2 + (2a/b)z + 1) = (1 − 2z2 + ...)(1 − (2a/b)z + ...), whose z-coefficient −2a/b is the residue. Res at α: (α2 − 1)2/(α2(α − β)); since α2 − 1 = α2 − αβ = α(α − β), this equals α − β = 2√(a2 − b2)/b.
I = −(1/(2ib))·2πi·(−2a/b + 2√(a2 − b2)/b) = (2π/b2)(a − √(a2 − b2)).
Answer: ∫02π sin2θ/(a + b cos θ) dθ = (2π/b2){a − √(a2 − b2)}, as required.Final answer: ∫02π sin2θ/(a + b cos θ) dθ = (2π/b2){a − √(a2 − b2)}, as required.
Question 3 (50 marks)
Q3(a) · Complex Analysis · 15 marks≈ 11 min
Power Series, Taylor's Series and Laurent's Series
Prove that cosh(z + 1/z) = a0 + ∑n=1∞ an(zn + 1/zn), where an = (1/(2π)) ∫02π cos nθ cosh(2 cos θ) dθ, for n = 0, 1, 2, ….
हिंदी में प्रश्न
सिद्ध कीजिए कि cosh(z + 1/z) = a0 + ∑n=1∞ an(zn + 1/zn) है, जहाँ n = 0, 1, 2, … के लिए an = (1/(2π)) ∫02π cos nθ cosh(2 cos θ) dθ है।
Worked solution
g(z) = cosh(z + 1/z) is analytic in 0 < |z| < ∞, so it has a Laurent series ∑n=−∞∞ anzn there, with an = (1/2πi)∮|z|=1 g(z)/zn+1 dz.
On |z| = 1 put z = eiθ: z + 1/z = 2cos θ and dz = ieiθ dθ, so an = (1/2π)∫02π cosh(2cos θ)(cos nθ − i sin nθ) dθ.
Under θ → 2π − θ, cosh(2cos θ) is unchanged and sin nθ changes sign, so ∫02π cosh(2cos θ) sin nθ dθ = 0 and an = (1/2π)∫02π cos nθ cosh(2cos θ) dθ.
cos(−nθ) = cos nθ gives a−n = an (equivalently g(1/z) = g(z)); grouping the terms n and −n gives cosh(z + 1/z) = a0 + ∑n=1∞ an(zn + 1/zn).
Answer: the expansion holds for all z ≠ 0, with an = (1/(2π))∫02π cos nθ cosh(2cos θ) dθ, n = 0, 1, 2, ....Final answer: the expansion holds for all z ≠ 0, with an = (1/(2π))∫02π cos nθ cosh(2cos θ) dθ, n = 0, 1, 2, ....
Q3(b) · Real Analysis · 20 marks≈ 14 min
Riemann Integral, Improper Integrals and Fundamental Theorems of Integral Calculus
A function f is defined on [0, 1] by
f(x) = x2, when x is rational;
f(x) = x3, when x is irrational.
Evaluate the lower integral ∫01 f and the upper integral ∫01 f. Does the integral ∫01 f exist?हिंदी में प्रश्न
अंतराल [0, 1] पर एक फलन
f(x) = x2, जब x परिमेय है;
f(x) = x3, जब x अपरिमेय है
द्वारा परिभाषित है। निम्न समाकल ∫01 f और उच्च समाकल ∫01 f का मूल्यांकन कीजिए। क्या समाकल ∫01 f का अस्तित्व है?Worked solution
On [0, 1], x3 ≤ x2, with equality only at 0 and 1; both functions are continuous.
By density, on every sub-interval sup f = sup x2 and inf f = inf x3, so U(P, f) = U(P, x2) and L(P, f) = L(P, x3) for every partition P.
Hence the upper integral is ∫01 x2 dx = 1/3 and the lower integral is ∫01 x3 dx = 1/4.
Answer: lower integral 1/4, upper integral 1/3; since they differ, ∫01 f does not exist.Final answer: lower integral 1/4, upper integral 1/3; since they differ, ∫01 f does not exist.
Q3(c) · Linear Programming · 15 marks≈ 11 min
Assignment Problem
The following table represents the estimated costs of assigning jobs to machines. Solve this assignment problem to minimize the total cost.
[Cost table, rows Jobs J1, J2, J3; columns Machines M1, M2, M3, M4]
J1: 8, 24, 28, 32
J2: 8, 13, 17, 19
J3: 8, 15, 19, 22
Is the assignment unique for the total minimum cost? Justify your answer and obtain an alternate assignment, if it exists.हिंदी में प्रश्न
निम्नलिखित सारणी कार्यों को मशीनें नियत करने पर अनुमानित लागत निरूपित करती है। कुल लागत को न्यूनतम करने के लिए इस नियतन समस्या को हल कीजिए।
[लागत सारणी, पंक्तियाँ कार्य J1, J2, J3; स्तंभ मशीन M1, M2, M3, M4]
J1: 8, 24, 28, 32
J2: 8, 13, 17, 19
J3: 8, 15, 19, 22
क्या कुल न्यूनतम लागत के लिए नियतन अद्वितीय है? अपने उत्तर का तर्क दीजिए और एक वैकल्पिक (alternate) नियतन, यदि इसका अस्तित्व है, प्राप्त कीजिए।
Question 4 (50 marks)
Q4(a) · Abstract Algebra · 15 marks≈ 11 min
Integral Domains, PID, Euclidean Domains, UFD, Fields and Quotient Fields
Let n be a positive integer. Show that the following statements are equivalent:
(i) ℤ/nℤ is an integral domain.
(ii) ℤ/nℤ is a field.
(iii) n is prime.हिंदी में प्रश्न
माना n एक धनात्मक पूर्णांक है। दर्शाइए कि निम्नलिखित कथन तुल्य हैं:
(i) ℤ/nℤ एक पूर्णांकीय प्रांत है।
(ii) ℤ/nℤ एक क्षेत्र है।
(iii) n अभाज्य है।Q4(b) · Real Analysis · 15 marks≈ 11 min
Riemann Integral, Improper Integrals and Fundamental Theorems of Integral Calculus
Prove that the improper integral ∫0∞ dx/(1 + x2 sin2x) is divergent.
हिंदी में प्रश्न
सिद्ध कीजिए कि अनंत समाकल ∫0∞ dx/(1 + x2 sin2x) अपसारी है।
Worked solution
f(x) = 1/(1 + x2sin2x) > 0, so ∫0∞ f converges iff ∑n≥0 In converges, where In = ∫nπ(n+1)π f(x) dx.
On [nπ, (n + 1)π], x ≤ (n + 1)π, and with x = nπ + t, sin2x = sin2t. So In ≥ ∫0π dt/(1 + A sin2t) with A = (n + 1)2π2.
∫0π dt/(1 + A sin2t) = 2∫0π/2 sec2t dt/(1 + (1 + A)tan2t) = π/√(1 + A) (put u = tan t).
Hence In ≥ π/√(1 + (n + 1)2π2) ≥ π/√(2(n + 1)2π2) = 1/(√2(n + 1)), since 1 ≤ (n + 1)2π2.
∑1/(n + 1) diverges, so ∑In diverges by comparison. (Note that the integrand does not tend to 0: it equals 1 at every x = nπ.)
Answer: ∫0∞ dx/(1 + x2sin2x) diverges.Final answer: ∫0∞ dx/(1 + x2sin2x) diverges.
Q4(c) · Linear Programming · 20 marks≈ 14 min
The Simplex Method
Solve the following linear programming problem by the Simplex method:
Maximize Z = 4x1 + 10x2
subject to the constraints
2x1 + x2 ≤ 10
2x1 + 5x2 ≤ 20
2x1 + 3x2 ≤ 18
x1, x2 ≥ 0
Obtain an alternate optimal solution, if it exists, with explanation.हिंदी में प्रश्न
निम्नलिखित रैखिक प्रोग्रामन समस्या को एकधा विधि से हल कीजिए:
अधिकतमीकरण कीजिए Z = 4x1 + 10x2
निम्न व्यवरोधों के अंतर्गत
2x1 + x2 ≤ 10
2x1 + 5x2 ≤ 20
2x1 + 3x2 ≤ 18
x1, x2 ≥ 0
व्याख्या सहित, एक वैकल्पिक (alternate) इष्टतम हल, यदि इसका अस्तित्व है, प्राप्त कीजिए।
Question 5 (50 marks)
Q5(a) · Partial Differential Equations · 10 marks≈ 7 min
Lagrange's Method for Quasilinear First-Order PDEs
Find the integral surface of the equations dx/(xz − y) = dy/(yz − x) = dz/(1 − z2).
हिंदी में प्रश्न
समीकरणों dx/(xz − y) = dy/(yz − x) = dz/(1 − z2) का समाकल पृष्ठ ज्ञात कीजिए।
Worked solution
Adding the first two: d(x + y)/((x + y)(z − 1)) = dz/(1 − z2), so d(x + y)/(x + y) = −dz/(1 + z) and (x + y)(1 + z) = c1.
Subtracting: d(x − y)/((x − y)(z + 1)) = dz/((1 − z)(1 + z)), so d(x − y)/(x − y) = dz/(1 − z) and (x − y)(1 − z) = c2.
The two integrals are independent, and the system is the auxiliary system of (xz − y)p + (yz − x)q = 1 − z2.
Answer: φ((x + y)(1 + z), (x − y)(1 − z)) = 0, φ arbitrary.Final answer: φ((x + y)(1 + z), (x − y)(1 − z)) = 0, φ arbitrary.
Q5(b) · Numerical Analysis and Computer Programming · 10 marks≈ 7 min
Roots of Algebraic and Transcendental Equations
By using the Newton-Raphson iterative formula, establish the formula xi+1 = (1/3)(2xi + N/xi2) to find the cube root of N. Use this formula, if established, to find the cube root of 63 correct to four decimal places with the initial approximation 3·9.
हिंदी में प्रश्न
न्यूटन-रैफ्सन पुनरावृत्ति सूत्र के उपयोग से, N का घनमूल ज्ञात करने के लिए सूत्र xi+1 = (1/3)(2xi + N/xi2) को प्रमाणित कीजिए। प्रमाणित होने पर, इस सूत्र के उपयोग से, प्रारंभिक सन्निकटन 3·9 लेकर, 63 का घनमूल दशमलव के चार स्थानों तक सही ज्ञात कीजिए।
Worked solution
f(x) = x3 − N, f′(x) = 3x2: xi+1 = xi − (xi3 − N)/(3xi2) = (2xi3 + N)/(3xi2) = (1/3)(2xi + N/xi2), which establishes the formula.
N = 63, x0 = 3.9: x1 = (1/3)(7.8 + 63/15.21) = (1/3)(7.8 + 4.142012) = 3.980671.
x2 = (1/3)(7.961341 + 63/15.845741) = (1/3)(7.961341 + 3.975833) = 3.979058, and x3 = 3.979057, which agrees with x2 to four decimals.
Answer: ∛63 = 3.9791 (to four decimal places).Final answer: ∛63 = 3.9791 (to four decimal places).
Q5(c) · Numerical Analysis and Computer Programming · 10 marks≈ 7 min
Computer Systems, Logic Gates, Boolean Algebra and Algorithms
Consider the logic circuit L:
[Figure: inputs A, B, C feed three AND gates whose outputs go to one OR gate with output Y. First AND gate: inputs A, B, C. Second AND gate: inputs A, NOT B (through an inverter), C. Third AND gate: inputs NOT A (through an inverter), B.]
(i) Express Y as a Boolean expression.
(ii) Write 8-bit special sequences for A, B and C.
(iii) Find the truth table of L using 8-bit special sequences.हिंदी में प्रश्न
निम्न तर्क परिपथ L का विचार कीजिए:
[चित्र: निवेश A, B, C तीन AND गेटों को जाते हैं जिनके निर्गम एक OR गेट में जाते हैं जिसका निर्गम Y है। पहला AND गेट: निवेश A, B, C। दूसरा AND गेट: निवेश A, NOT B (इन्वर्टर द्वारा), C। तीसरा AND गेट: निवेश NOT A (इन्वर्टर द्वारा), B।]
(i) Y को बूलीय व्यंजक के रूप में अभिव्यक्त कीजिए।
(ii) A, B और C के लिए 8-बिट विशिष्ट अनुक्रम लिखिए।
(iii) 8-बिट विशिष्ट अनुक्रमों के उपयोग से L की सत्यमान सारणी ज्ञात कीजिए।Worked solution
(i) Y = ABC + AB′C + A′B = AC(B + B′) + A′B = AC + A′B.
(ii) Special sequences (rows 000 to 111 in order): A = 00001111, B = 00110011, C = 01010101.
(iii) ABC = 00000001, AB′C = 00001111 AND 11001100 AND 01010101 = 00000100, A′B = 11110000 AND 00110011 = 00110000; OR-ing gives Y = 00110101.
Answer: Y = AC + A′B; truth-table column 00110101, i.e. Y = 1 exactly for ABC = 010, 011, 101, 111.Final answer: Y = AC + A′B; truth-table column 00110101, i.e. Y = 1 exactly for ABC = 010, 011, 101, 111.
Q5(d) · Mechanics and Fluid Dynamics · 10 marks≈ 7 min
Two-Dimensional and Axisymmetric Motion: Sources, Sinks and Vortices
A simple source of strength m is fixed at the origin O in a uniform stream of incompressible fluid moving with velocity U î. Find out the velocity potential φ at any point P of the stream where OP = r and θ is the angle OP makes with the direction î. Find the differential equation of the stream lines and show that they lie on the surface Ur2 sin2θ − 2m cos θ = constant.
हिंदी में प्रश्न
असंपीड्य तरल का वेग U î की एक एकसमान धारा में मूल-बिंदु O पर एक m प्रबलता का सरल स्रोत स्थित है। धारा के किसी बिंदु P पर वेग विभव φ ज्ञात कीजिए, जहाँ OP = r और θ वह कोण है जो OP, î की दिशा से बनाता है। धारा रेखाओं के अवकल समीकरण ज्ञात कीजिए और दर्शाइए कि वे पृष्ठ Ur2 sin2θ − 2m cos θ = अचर पर स्थित हैं।
Q5(e) · Mechanics and Fluid Dynamics · 10 marks≈ 7 min
Moment of Inertia and Motion of Rigid Bodies in Two Dimensions
A uniform heavy solid hemisphere of radius 'a' is held at rest with its base vertical and its curved surface in contact with a horizontal plane. If the hemisphere is released when the plane is rough enough to prevent slipping, show that the angle θ that the base makes with the horizontal at time t, is such that (dθ/dt)2 = 15g cos θ/(a(28 − 15 cos θ)).
हिंदी में प्रश्न
त्रिज्या 'a' का एक एकसमान भारी ठोस अर्धगोला विरामावस्था की स्थिति में रखा है जबकि इसका आधार ऊर्ध्वाधर है और इसका वक्र पृष्ठ एक क्षैतिज समतल के संपर्क में है। यदि अर्धगोले को मुक्त (रिलीज़) किया जाता है, जब समतल इतना रूक्ष है कि फिसलन को रोक सके, तो दर्शाइए कि t समय पर कोण θ, जो अर्धगोले का आधार क्षैतिज से बनाता है, इस प्रकार है कि (dθ/dt)2 = 15g cos θ/(a(28 − 15 cos θ)) है।
Worked solution
Let O be the centre of the base and G the centre of mass, OG = c = 3a/8; θ is the angle of the base with the horizontal, which equals the angle of OG with the vertical. O stays at height a, and the contact point C is vertically below O.
CG2 = a2 + c2 − 2ac cos θ and IG = (2/5)Ma2 − Mc2, so IC = M[(7/5)a2 − 2ac cos θ] = Ma2(28 − 15 cos θ)/20.
G is at height a − c cos θ; energy from rest at θ = π/2: (1/2)ICθ̇2 = Mgc cos θ = (3/8)Mga cos θ.
Answer: θ̇2 = (3/4)ga cos θ · 20/(a2(28 − 15 cos θ)) = 15g cos θ/(a(28 − 15 cos θ)).Final answer: θ̇2 = (3/4)ga cos θ · 20/(a2(28 − 15 cos θ)) = 15g cos θ/(a(28 − 15 cos θ)).
Question 6 (50 marks)
Q6(a) · Partial Differential Equations · 20 marks≈ 14 min
Classification and Canonical Forms of Second-Order PDEs
Reduce the partial differential equation y ∂2z/∂x2 + (x + y) ∂2z/∂x∂y + x ∂2z/∂y2 = 0 to canonical form and hence solve it.
हिंदी में प्रश्न
आंशिक अवकल समीकरण y ∂2z/∂x2 + (x + y) ∂2z/∂x∂y + x ∂2z/∂y2 = 0 को विहित रूप में समानीत कीजिए और तब इसे हल कीजिए।
Q6(b) · Numerical Analysis and Computer Programming · 15 marks≈ 11 min
Number Systems and Representation of Numbers in a Computer
(i) Find the decimal equivalent of the following floating point 10-bit numbers with 5 bits as fractional part: 1010111011 and 0100101010. (8)
(ii) Evaluate (11001.1011)10 + (33.24)8 + (101101)2 and convert the final result in hexadecimal system. (7)हिंदी में प्रश्न
(i) निम्नलिखित चल (फ्लोटिंग) बिंदु 10-द्वयंक (बिट) संख्याओं के दशमलव तुल्यांक ज्ञात कीजिए, जिनमें 5 द्वयंक भिन्न भाग हों: 1010111011 और 0100101010। (8)
(ii) (11001.1011)10 + (33.24)8 + (101101)2 का मूल्यांकन कीजिए और अंतिम परिणाम को षोडश आधारी (हैक्साडेसिमल) पद्धति में बदलिए। (7)Worked solution
(i) With 5 integer and 5 fractional bits: 10101.11011 = 21 + 27/32 = 21.84375 and 01001.01010 = 9 + 10/32 = 9.3125. (If the leftmost bit is a sign bit, the first is −(0101.11011)2 = −5.84375 and the second stays 9.3125.)
(ii) [The digits of (11001.1011)10 are all 0 and 1, so base 2 is evidently intended.] (11001.1011)2 = 25 + 11/16 = 25.6875, (33.24)8 = 27 + 2/8 + 4/64 = 27.3125 and (101101)2 = 45.
Sum = 25.6875 + 27.3125 + 45 = 98 = 6 × 16 + 2. Read literally as decimal, 11001.1011 + 27.3125 + 45 = 11073.4136 = (2B41.69E1...)16, since 11073 = 2 × 4096 + 11 × 256 + 4 × 16 + 1.
Answer: (i) 21.84375 and 9.3125; (ii) (62)16 (literal reading: (2B41.69E1...)16).Final answer: (i) 21.84375 and 9.3125; (ii) (62)16 (literal reading: (2B41.69E1...)16).
Q6(c) · Mechanics and Fluid Dynamics · 15 marks≈ 11 min
Two-Dimensional and Axisymmetric Motion: Sources, Sinks and Vortices
Suppose that there is an infinitely long cylinder of radius 'a' placed in a uniform stream having velocity −V î. Then using Milne-Thomson's circle theorem, what will be its complex velocity potential? If in addition, a circulation round the cylinder of k is produced, find out the stagnation points. Do they produce a lifting tendency in the vertical direction? Explain how.
हिंदी में प्रश्न
माना त्रिज्या 'a' के एक अनंत लंबे बेलन को वेग −V î की एक एकसमान धारा में रखा गया है। मिल्ने-थॉमसन के वृत्त प्रमेय के उपयोग से इसका सम्मिश्र वेग विभव क्या होगा? इसके अतिरिक्त, यदि बेलन के परितः k का एक परिसंचरण उत्पन्न किया जाता है, तो प्रगतिरोध बिंदु ज्ञात कीजिए। क्या वे ऊर्ध्वाधर दिशा में उत्थापन प्रवृत्ति उत्पन्न करते हैं? व्याख्या कीजिए कि कैसे।
Worked solution
A stream U along Ox has potential −Uz, so the stream −Vî has f(z) = Vz; the circle theorem gives w = Vz + Va2/z = V(z + a2/z), and with anticlockwise circulation k, w = V(z + a2/z) + (ik/(2π))log z.
dw/dz = V(1 − a2/z2) + ik/(2πz) = 0 gives Vz2 + (ik/(2π))z − Va2 = 0, z = [−ik/(2π) ± √(4V2a2 − k2/(4π2))]/(2V).
If k < 4πaV both lie on r = a with sin θ = −k/(4πaV) (lower half); if k = 4πaV they coincide at z = −ia; if k > 4πaV one lies in the fluid on the negative y-axis.
Blasius: the 1/z term of (dw/dz)2 is iVk/(πz), so ∮(dw/dz)2dz = −2Vk and X − iY = (iρ/2)(−2Vk) = −iρVk: X = 0, Y = ρVk.
Answer: w = V(z + a2/z) + (ik/(2π))log z; stagnation points as above; yes, a lift ρVk per unit length acts vertically upward: the circulation adds to the stream speed above the cylinder and reduces it below, so by Bernoulli the pressure is lower above (Kutta–Joukowski; the Magnus effect).Final answer: w = V(z + a2/z) + (ik/(2π))log z; stagnation points as above; yes, a lift ρVk per unit length acts vertically upward: the circulation adds to the stream speed above the cylinder and reduces it below, so by Bernoulli the pressure is lower above (Kutta–Joukowski; the Magnus effect).
Question 7 (50 marks)
Q7(a) · Partial Differential Equations · 15 marks≈ 11 min
Linear Second-Order PDEs with Constant Coefficients
Solve the partial differential equation (2D2 − 5DD′ + 2D′2) z = 24(y − x) + sin(y − x), where D ≡ ∂/∂x, D′ ≡ ∂/∂y, z = z(x, y).
हिंदी में प्रश्न
आंशिक अवकल समीकरण (2D2 − 5DD′ + 2D′2) z = 24(y − x) + sin(y − x) को हल कीजिए, जहाँ D ≡ ∂/∂x, D′ ≡ ∂/∂y, z = z(x, y) हैं।
Worked solution
The CF and the PI of 24(y − x) are as in the previous example.
For sin(y − x): a = −1, b = 1, so D2 → −1, DD′ → −ab = 1, D′2 → −1 and F → −2 − 5 − 2 = −9; PI = −(1/9)sin(y − x).
Answer: z = φ1(y + 2x) + φ2(2y + x) + (4/9)(y − x)3 − (1/9)sin(y − x).Final answer: z = φ1(y + 2x) + φ2(2y + x) + (4/9)(y − x)3 − (1/9)sin(y − x).
Q7(b) · Numerical Analysis and Computer Programming · 15 marks≈ 11 min
Numerical Integration and Numerical Solution of ODEs
Evaluate the integral ∫00·8 f(x) dx, where f(x) = 0·2 + 25x − 200x2 + 675x3 − 900x4 + 400x5 by using single application of Simpson's 3/8 rule.
हिंदी में प्रश्न
सिम्प्सन के 3/8 नियम को एक बार उपयोग कर समाकल ∫00·8 f(x) dx का मूल्यांकन कीजिए, जहाँ f(x) = 0·2 + 25x − 200x2 + 675x3 − 900x4 + 400x5 है।
Q7(c) · Mechanics and Fluid Dynamics · 20 marks≈ 14 min
Generalized Coordinates, D'Alembert's Principle and Lagrange's Equations
A smooth uniform rod, say OA, of length 2a and mass m is pivoted at one end to a fixed point O. The rod is inclined at an angle θ with the downward vertical line OZ and the plane OAZ makes an angle φ with a fixed vertical plane. A bead of mass λm slides smoothly on the rod and is connected to O by a light elastic string of modulus nmg and natural length a. Show that the kinetic energy T of the system is given by 2T = (4/3)ma2(θ̇2 + φ̇2 sin θ) + λm(ẋ2 + x2θ̇2 + x2θ̇2 sin2θ), where x is the stretched length of the string, and derive the equations of motion.
हिंदी में प्रश्न
2a लंबी और m द्रव्यमान की एक चिकनी एकसमान छड़, माना OA, का एक सिरा एक निश्चित बिंदु O पर धुराग्रस्थ (पिवटिड) है। छड़ नीचे की ओर ऊर्ध्वाधर रेखा OZ से θ कोण पर झुकी हुई है और समतल OAZ, एक निश्चित ऊर्ध्वाधर समतल से कोण φ बनाती है। λm द्रव्यमान का एक मनका छड़ पर निर्विघ्न (स्मूथली) फिसलता है और nmg मापांक (मॉड्यूल्स) और प्राकृतिक लंबाई a की एक हल्की प्रत्यास्थ डोरी के द्वारा O से जुड़ा है। दर्शाइए कि इस तंत्र (सिस्टम) की गतिज ऊर्जा T, 2T = (4/3)ma2(θ̇2 + φ̇2 sin θ) + λm(ẋ2 + x2θ̇2 + x2θ̇2 sin2θ) द्वारा दी जाती है, जहाँ x डोरी की तनी हुई (स्ट्रेच्ड) लंबाई है, और गति के समीकरण व्युत्पन्न कीजिए।
Worked solution
[The printed 2T has two misprints: the rod term should be θ̇2 + φ̇2sin2θ and the last bead term x2φ̇2sin2θ; the corrected form is proved.]
Rod: it turns about O, the square of its angular velocity is θ̇2 + sin2θ φ̇2, and its moment about every line through O perpendicular to it is 4ma2/3, so 2Trod = (4/3)ma2(θ̇2 + φ̇2sin2θ).
Bead at distance x along the rod, in spherical coordinates (x, θ, φ): 2Tbead = λm(ẋ2 + x2θ̇2 + x2φ̇2sin2θ).
V = −mga cos θ − λmgx cos θ + nmg(x − a)2/(2a), the string energy being modulus × extension2/(2 × natural length), for x ≥ a.
Put I = (4/3)ma2 + λmx2. φ is cyclic: I sin2θ φ̇ = constant.
θ-equation: d/dt(Iθ̇) − I sin θ cos θ φ̇2 + mg(a + λx)sin θ = 0.
x-equation: λmẍ − λmx(θ̇2 + sin2θ φ̇2) − λmg cos θ + nmg(x − a)/a = 0.
Answer: 2T = (4/3)ma2(θ̇2 + φ̇2sin2θ) + λm(ẋ2 + x2θ̇2 + x2φ̇2sin2θ), with the three equations of motion above.Final answer: 2T = (4/3)ma2(θ̇2 + φ̇2sin2θ) + λm(ẋ2 + x2θ̇2 + x2φ̇2sin2θ), with the three equations of motion above.
Question 8 (50 marks)
Q8(a) · Partial Differential Equations · 15 marks≈ 11 min
Vibrating String, Heat Equation and Laplace Equation
Two-dimensional harmonic equation in plane polar coordinates (r, θ) takes the form:
∂2V/∂r2 + (1/r) ∂V/∂r + (1/r2) ∂2V/∂θ2 = 0
Deduce that it has solutions of the form (Arn + Br−n) e±inθ, where A, B, n are constants. Determine V if it satisfies the two-dimensional harmonic equation in the region 0 ≤ r ≤ a, 0 ≤ θ ≤ 2π and satisfies the conditions:
(i) V remains finite as r → 0,
(ii) V = ∑n cn cos(nθ) on r = a.हिंदी में प्रश्न
समतलीय ध्रुवीय निर्देशांकों (r, θ) में द्विविमीय प्रसंवादी (हार्मोनिक) समीकरण निम्न रूप लेता है:
∂2V/∂r2 + (1/r) ∂V/∂r + (1/r2) ∂2V/∂θ2 = 0
निगमन कीजिए कि इसके हल का रूप (Arn + Br−n) e±inθ है, जहाँ A, B, n अचर हैं। V ज्ञात कीजिए यदि यह क्षेत्र 0 ≤ r ≤ a, 0 ≤ θ ≤ 2π में द्विविमीय प्रसंवादी समीकरण को संतुष्ट करता है और शर्तों:
(i) r → 0 पर V परिमित रहता है,
(ii) r = a पर V = ∑n cn cos(nθ) है
को संतुष्ट करता है।Worked solution
Put V = R(r)Θ(θ) and multiply by r2/(RΘ): (r2R″ + rR′)/R = −Θ″/Θ = n2 (a separation constant).
Θ″ + n2Θ = 0 gives Θ = e±inθ; r2R″ + rR′ − n2R = 0 is an Euler equation, and R = rm gives m2 = n2, so R = Arn + Br−n.
Hence V = (Arn + Br−n)e±inθ; single-valuedness makes n an integer (n = 0 gives A + B log r).
In 0 ≤ r ≤ a, finiteness at r = 0 forces B = 0 and removes log r, so V = ∑ rn(Ancos nθ + Bn′sin nθ); at r = a, Anan = cn and Bn′ = 0.
Answer: V(r, θ) = ∑n cn(r/a)ncos nθ.Final answer: V(r, θ) = ∑n cn(r/a)ncos nθ.
Q8(b) · Numerical Analysis and Computer Programming · 15 marks≈ 11 min
Numerical Integration and Numerical Solution of ODEs
Consider the initial value problem: dy/dx = 1 + y/x, y(1) = 1, h = 1 (Step-size)
(i) Apply the classical fourth order Runge-Kutta method to approximate the solution y(2).
(ii) Use Euler's method to find y(2) and y(3).हिंदी में प्रश्न
प्रारंभिक मान समस्या dy/dx = 1 + y/x, y(1) = 1, h = 1 (पग-माप) का विचार कीजिए।
(i) y(2) का सन्निकट मान प्राप्त करने के लिए चिरप्रतिष्ठित चतुष्कोटि रुंगे-कुट्टा विधि का उपयोग कीजिए।
(ii) y(2) और y(3) के मान ज्ञात करने के लिए ऑयलर विधि का उपयोग कीजिए।Q8(c) · Mechanics and Fluid Dynamics · 20 marks≈ 14 min
Euler's Equation, Bernoulli's Theorem and the Navier–Stokes Equation
A viscous incompressible fluid is filled between two concentric cylinders of radius a, b (b > a). The flow is steady and no body forces are taken into consideration. Discuss and formulate the velocity of the fluid if:
(i) the inner cylinder is given angular velocity Ω while the outer one is held at rest;
(ii) the outer cylinder is rotated with angular velocity Ω while the inner one is held at rest.हिंदी में प्रश्न
त्रिज्या a, b (b > a) के दो संकेंद्री बेलनों के बीच एक श्यान असंपीड्य तरल भरा है। प्रवाह अपरिवर्ती है और कोई भी पिंड बल विचाराधीन नहीं है। तरल के वेग की विवेचना और इसे सूत्रित कीजिए यदि:
(i) आंतरिक बेलन को कोणीय वेग Ω दिया जाता है जबकि बाहरी बेलन को विरामावस्था में रखा जाता है;
(ii) बाहरी बेलन को कोणीय वेग Ω से घुमाया जाता है जबकि आंतरिक बेलन को विरामावस्था में रखा जाता है।Worked solution
Take q = v(r)eθ; continuity holds, and with ∂p/∂θ = 0 the θ-equation gives d/dr[(1/r)d(rv)/dr] = 0, so v = Ar + B/r.
(i) No slip: Aa + B/a = Ωa and Ab + B/b = 0, so A = −Ωa2/(b2 − a2), B = Ωa2b2/(b2 − a2).
(ii) Aa + B/a = 0 and Ab + B/b = Ωb, so A = Ωb2/(b2 − a2), B = −Ωa2b2/(b2 − a2).
In both cases the pressure follows from dp/dr = ρv2/r, and the shear stress is μr d(v/r)/dr = −2μB/r2.
Answer: (i) v = Ωa2(b2 − r2)/((b2 − a2)r); (ii) v = Ωb2(r2 − a2)/((b2 − a2)r) (circular Couette flow).Final answer: (i) v = Ωa2(b2 − r2)/((b2 − a2)r); (ii) v = Ωb2(r2 − a2)/((b2 − a2)r) (circular Couette flow).