UPSC Mains 2022 Mathematics Optional Question Papers with Solutions
All 56 questions of the 2022 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 2022 Mathematics Paper I
Question 1 (50 marks)
Q1(a) · Linear Algebra · 10 marks≈ 7 min
Vector Spaces, Subspaces, Basis and Dimension
Prove that any set of n linearly independent vectors in a vector space V of dimension n constitutes a basis for V.
हिंदी में प्रश्न
सिद्ध कीजिए कि n विमीय सदिश समष्टि V के लिए n रैखिकतः स्वतंत्र सदिशों का कोई भी समुच्चय V के लिए एक आधार बनाता है।
Q1(b) · Linear Algebra · 10 marks≈ 7 min
Linear Transformations, Rank-Nullity and Matrix Representation
Let T : ℝ2 → ℝ3 be a linear transformation such that T[[1], [0]] = [[1], [2], [3]] and T[[1], [1]] = [[−3], [2], [8]]. Find T[[2], [4]].
हिंदी में प्रश्न
माना T : ℝ2 → ℝ3 एक रैखिक रूपांतरण, ऐसा है कि T[[1], [0]] = [[1], [2], [3]] तथा T[[1], [1]] = [[−3], [2], [8]] है। T[[2], [4]] को ज्ञात कीजिए।
Q1(c) · Calculus · 10 marks≈ 7 min
Mean Value Theorems, Taylor's Theorem and Indeterminate Forms
Evaluate limx→∞ (ex + x)1/x.
हिंदी में प्रश्न
limx→∞ (ex + x)1/x का मान निकालिए।
Worked solution
Let y = (ex + x)1/x (form ∞0), so log y = log(ex + x)/x, of the form ∞/∞.
By L'Hôpital's rule, lim log y = lim (ex + 1)/(ex + x) = lim (1 + e−x)/(1 + xe−x) = 1, since e−x → 0 and xe−x → 0.
(Directly: log(ex + x) = x + log(1 + xe−x), so log y = 1 + log(1 + xe−x)/x → 1.)
Answer: limx→∞ (ex + x)1/x = e.Final answer: limx→∞ (ex + x)1/x = e.
Q1(d) · Calculus · 10 marks≈ 7 min
Riemann Integral, Indefinite Integrals and Improper Integrals
Examine the convergence of ∫02 dx/(2x − x2).
हिंदी में प्रश्न
∫02 dx/(2x − x2) की अभिसारिता का परीक्षण कीजिए।
Worked solution
2x − x2 = x(2 − x) vanishes at both end points, so the integrand is positive on (0, 2) and unbounded at 0 and 2; split at x = 1.
On (0, 1], 1/(x(2 − x)) ≥ 1/(2x), and ∫01 dx/x diverges (p = 1), so by comparison ∫01 diverges.
Directly, with 1/(x(2 − x)) = (1/2)(1/x + 1/(2 − x)): ∫ε1 = (1/2)[log x − log(2 − x)]ε1 = (1/2) log((2 − ε)/ε) → ∞ as ε → 0+.
Answer: The integral diverges.Final answer: The integral diverges.
Q1(e) · Analytic Geometry · 10 marks≈ 7 min
The Sphere
A variable plane passes through a fixed point (a, b, c) and meets the axes at points A, B and C respectively. Find the locus of the centre of the sphere passing through the points O, A, B and C, O being the origin.
हिंदी में प्रश्न
एक चर समतल एक स्थिर बिन्दु (a, b, c) से गुज़रता है तथा अक्षों को क्रमशः A, B व C बिन्दुओं पर मिलता है। बिन्दुओं O, A, B तथा C से गुज़रते हुए गोले के केन्द्र का बिन्दुपथ ज्ञात कीजिए, जहाँ O मूल-बिन्दु है।
Worked solution
Let the plane be x/α + y/β + z/γ = 1, cutting the axes at A(α, 0, 0), B(0, β, 0), C(0, 0, γ). It passes through (a, b, c): a/α + b/β + c/γ = 1.
The sphere OABC is x2 + y2 + z2 − αx − βy − γz = 0, with centre (x, y, z) = (α/2, β/2, γ/2), so α = 2x, β = 2y, γ = 2z.
Substituting: a/(2x) + b/(2y) + c/(2z) = 1.
Answer: the locus of the centre is a/x + b/y + c/z = 2.Final answer: the locus of the centre is a/x + b/y + c/z = 2.
Question 2 (50 marks)
Q2(a) · Linear Algebra · 15 marks≈ 11 min
Systems of Linear Equations
Find all solutions to the following system of equations by row-reduced method:
x1 + 2x2 − x3 = 2
2x1 + 3x2 + 5x3 = 5
−x1 − 3x2 + 8x3 = −1हिंदी में प्रश्न
निम्नलिखित समीकरण निकाय के सभी हलों को पंक्ति-समानीत विधि से ज्ञात कीजिए:
x1 + 2x2 − x3 = 2
2x1 + 3x2 + 5x3 = 5
−x1 − 3x2 + 8x3 = −1Q2(b) · Calculus · 15 marks≈ 11 min
Functions of Several Variables: Partial Derivatives, Extrema, Lagrange Multipliers and Jacobians
A wire of length l is cut into two parts which are bent in the form of a square and a circle respectively. Using Lagrange's method of undetermined multipliers, find the least value of the sum of the areas so formed.
हिंदी में प्रश्न
एक l लम्बाई के तार को दो भागों में काटकर क्रमशः एक वर्ग तथा एक वृत्त के रूप में मोड़ा गया है। लग्रांज की अनिर्धारित गुणक विधि का प्रयोग करके, इस तरह से प्राप्त किए गए क्षेत्रफलों के योगफल का न्यूनतम मान ज्ञात कीजिए।
Worked solution
Let the square have side a and the circle radius r. The constraint is 4a + 2πr = l and the total area is A = a2 + πr2.
F = a2 + πr2 + λ(4a + 2πr − l): Fa = 2a + 4λ = 0 and Fr = 2πr + 2πλ = 0 give a = −2λ, r = −λ, so a = 2r.
Then 8r + 2πr = l, so r = l/(2(π + 4)) and a = l/(π + 4). A = l2/(π + 4)2 + πl2/(4(π + 4)2) = l2(4 + π)/(4(π + 4)2) = l2/(4(π + 4)).
This is the minimum: on the constraint line A is a convex quadratic, and the end values l2/16 (all square) and l2/(4π) (all circle) are both larger.
Answer: Least total area l2/(4(π + 4)), when the side of the square equals the diameter of the circle.Final answer: Least total area l2/(4(π + 4)), when the side of the square equals the diameter of the circle.
Q2(c) · Analytic Geometry · 20 marks≈ 14 min
Ellipsoid, Hyperboloids and Paraboloids
If P, Q, R; P′, Q′, R′ are feet of the six normals drawn from a point to the ellipsoid x2/a2 + y2/b2 + z2/c2 = 1, and the plane PQR is represented by lx + my + nz = p, show that the plane P′Q′R′ is given by x/(a2l) + y/(b2m) + z/(c2n) + 1/p = 0.
हिंदी में प्रश्न
यदि P, Q, R; P′, Q′, R′, एक बिन्दु से दीर्घवृत्तज x2/a2 + y2/b2 + z2/c2 = 1 पर छः (सिक्स) अभिलम्ब पाद हैं तथा lx + my + nz = p से समतल PQR निरूपित है, दर्शाइए कि x/(a2l) + y/(b2m) + z/(c2n) + 1/p = 0, समतल P′Q′R′ को निरूपित करता है।
Worked solution
Let the normals be drawn from (α, β, γ). A foot (x, y, z) satisfies a2α/x − a2 = b2β/y − b2 = c2γ/z − c2 (= λ), so the six feet lie on the ellipsoid S ≡ ∑x2/a2 − 1 = 0 and on the three conicoids f1 ≡ (b2 − c2)yz + c2γy − b2βz = 0, f2 ≡ (c2 − a2)zx + a2αz − c2γx = 0, f3 ≡ (a2 − b2)xy + b2βx − a2αy = 0.
Let P′Q′R′ be l′x + m′y + n′z = p′. The pair of planes (lx + my + nz − p)(l′x + m′y + n′z − p′) = 0 passes through all six feet; the standard argument (Shanti Narayan) identifies it with a member kS + k1f1 + k2f2 + k3f3 = 0 of the family of conicoids through the feet.
The fi have no x2, y2, z2 or constant terms. Comparing x2, y2, z2 and constant terms: ll′ = k/a2, mm′ = k/b2, nn′ = k/c2 and pp′ = −k.
Hence l′ = −pp′/(a2l), m′ = −pp′/(b2m), n′ = −pp′/(c2n), and l′x + m′y + n′z = p′ becomes −pp′[x/(a2l) + y/(b2m) + z/(c2n)] = p′.
Answer: dividing by −pp′, the plane P′Q′R′ is x/(a2l) + y/(b2m) + z/(c2n) + 1/p = 0; hence proved (a numerical check with the six computed feet confirms it).Final answer: dividing by −pp′, the plane P′Q′R′ is x/(a2l) + y/(b2m) + z/(c2n) + 1/p = 0; hence proved (a numerical check with the six computed feet confirms it).
Question 3 (50 marks)
Q3(a) · Linear Algebra · 20 marks≈ 14 min
Vector Spaces, Subspaces, Basis and Dimension
Let the set P = { [[x], [y], [z]] | x − y − z = 0 and 2x − y + z = 0 } be the collection of vectors of a vector space ℝ3(ℝ). Then
(i) prove that P is a subspace of ℝ3.
(ii) find a basis and dimension of P. (10+10)हिंदी में प्रश्न
माना समुच्चय P = { [[x], [y], [z]] | x − y − z = 0 तथा 2x − y + z = 0 } सदिश समष्टि ℝ3(ℝ) के सदिशों का एक समूह है। तब
(i) सिद्ध कीजिए कि P, ℝ3 की एक उपसमष्टि है।
(ii) P का एक आधार तथा विमा ज्ञात कीजिए। (10+10)Q3(b) · Calculus · 15 marks≈ 11 min
Double and Triple Integrals; Areas, Surfaces and Volumes
Use double integration to calculate the area common to the circle x2 + y2 = 4 and the parabola y2 = 3x.
हिंदी में प्रश्न
द्विशः समाकलन का उपयोग करके, वृत्त x2 + y2 = 4 तथा परवलय y2 = 3x के उभयनिष्ठ क्षेत्रफल का परिकलन कीजिए।
Worked solution
Intersection: x2 + 3x = 4 gives x = 1 (x = −4 is impossible since 3x = y2 ≥ 0), so the points are (1, ±√3). The common region is symmetric about the x-axis.
Horizontal strips: for −√3 ≤ y ≤ √3, x runs from the parabola x = y2/3 to the circle x = √(4 − y2). A = ∫−√3√3∫y2/3√(4−y2) dx dy = 2∫0√3 (√(4 − y2) − y2/3) dy.
∫0√3 √(4 − y2) dy = [y√(4 − y2)/2 + 2 sin−1(y/2)]0√3 = √3/2 + 2π/3, and ∫0√3 (y2/3) dy = √3/3.
A = 2(√3/2 + 2π/3 − √3/3) = 4π/3 + √3/3.
Answer: (4π + √3)/3 square units.Final answer: (4π + √3)/3 square units.
Q3(c) · Analytic Geometry · 15 marks≈ 11 min
Straight Lines and Shortest Distance between Skew Lines
Find the equation of the sphere of smallest possible radius which touches the straight lines: (x − 3)/3 = (y − 8)/(−1) = (z − 3)/1 and (x + 3)/(−3) = (y + 7)/2 = (z − 6)/4.
हिंदी में प्रश्न
लघुतम संभाव्य त्रिज्या के गोले का समीकरण ज्ञात कीजिए जो सरल रेखाओं: (x − 3)/3 = (y − 8)/(−1) = (z − 3)/1 तथा (x + 3)/(−3) = (y + 7)/2 = (z − 6)/4 को स्पर्श करता है।
Worked solution
By Example 1 of this topic the feet of the common perpendicular are P(3, 8, 3) and Q(−3, −7, 6), and PQ = 3√30.
A sphere touching both lines has radius at least half the distance between them, because its centre's distances to the two lines add up to at least PQ. Equality holds for the sphere on PQ as diameter, which touches each line at P and Q (the radius there is along the common perpendicular).
Diameter form: (x − 3)(x + 3) + (y − 8)(y + 7) + (z − 3)(z − 6) = 0, that is x2 + y2 + z2 − y − 9z − 47 = 0, with centre (0, 1/2, 9/2) and radius √270/2 = 3√30/2.
Answer: x2 + y2 + z2 − y − 9z − 47 = 0.Final answer: x2 + y2 + z2 − y − 9z − 47 = 0.
Question 4 (50 marks)
Q4(a) · Linear Algebra · 15 marks≈ 11 min
Eigenvalues, Eigenvectors and the Cayley-Hamilton Theorem
Find a linear map T : ℝ2 → ℝ2 which rotates each vector of ℝ2 by an angle θ. Also, prove that for θ = π/2, T has no eigenvalue in ℝ.
हिंदी में प्रश्न
एक रैखिक प्रतिचित्र T : ℝ2 → ℝ2 ज्ञात कीजिए जो कि ℝ2 के प्रत्येक सदिश को θ कोण से घुमा देता है। यह भी सिद्ध कीजिए कि θ = π/2 के लिए, T का कोई भी अभिलक्षणिक मान (आइगेनमान) ℝ में नहीं है।
Worked solution
Write (x, y) = (r cos φ, r sin φ). Rotation by θ sends it to (r cos(φ + θ), r sin(φ + θ)) = (x cos θ − y sin θ, x sin θ + y cos θ).
So T(x, y) = (x cos θ − y sin θ, x sin θ + y cos θ); it is linear, with matrix Rθ = [[cos θ, −sin θ], [sin θ, cos θ]] in the standard basis. Its characteristic equation is λ2 − 2λ cos θ + 1 = 0, with roots cos θ ± i sin θ.
For θ = π/2, T(x, y) = (−y, x) and the equation is λ2 + 1 = 0, with no real root. Directly: T(x, y) = λ(x, y) means −y = λx and x = λy, so x = −λ2x and y = −λ2y; for (x, y) ≠ (0, 0) this needs λ2 = −1, impossible for real λ.
Answer: T(x, y) = (x cos θ − y sin θ, x sin θ + y cos θ); for θ = π/2 the eigenvalues are ±i, so T has no real eigenvalue (no non-zero vector keeps its direction under a quarter-turn).Final answer: T(x, y) = (x cos θ − y sin θ, x sin θ + y cos θ); for θ = π/2 the eigenvalues are ±i, so T has no real eigenvalue (no non-zero vector keeps its direction under a quarter-turn).
Q4(b) · Calculus · 20 marks≈ 14 min
Maxima and Minima, Asymptotes and Curve Tracing
Trace the curve y2x2 = x2 − a2, where a is a real constant.
हिंदी में प्रश्न
वक्र y2x2 = x2 − a2 का अनुरेख (ट्रेस) कीजिए, जहाँ a एक वास्तविक अचर है।
Worked solution
For a ≠ 0, y2 = 1 − a2/x2 (for a = 0 the curve is the pair of lines y = ±1 with x ≠ 0). Symmetry: only even powers of x and y occur, so the curve is symmetric about both axes.
Region: y2 ≥ 0 needs |x| ≥ |a|, so nothing lies in the strip |x| < |a| and the curve misses the origin; also y2 < 1, so the curve lies between y = −1 and y = 1.
Axis points: (±a, 0). Differentiating, dy/dx = (1 − y2)/(xy), which → ∞ as y → 0, so the tangents at (±a, 0) are vertical.
Asymptotes: writing x2(y2 − 1) + a2 = 0, the coefficient of the highest power of x gives y = ±1. The coefficient of y2 is x2, suggesting x = 0, but no real branch approaches it because the curve is absent for |x| < |a|.
Shape: for x ≥ |a|, |y| increases from 0 towards 1; the branch leaves (|a|, 0) vertically and flattens towards y = 1 above and y = −1 below. Its mirror image lies in x ≤ −|a|, giving two bow-shaped branches opening away from the y-axis.
Answer: Two branches in |x| ≥ |a|, symmetric about both axes, with vertical tangents at (±a, 0) and asymptotes y = ±1; the curve lies within −1 < y < 1.Final answer: Two branches in |x| ≥ |a|, symmetric about both axes, with vertical tangents at (±a, 0) and asymptotes y = ±1; the curve lies within −1 < y < 1.
Q4(c) · Analytic Geometry · 15 marks≈ 11 min
Cone and Cylinder
If the plane ux + vy + wz = 0 cuts the cone ax2 + by2 + cz2 = 0 in perpendicular generators, then prove that (b + c)u2 + (c + a)v2 + (a + b)w2 = 0.
हिंदी में प्रश्न
यदि समतल ux + vy + wz = 0, शंकु ax2 + by2 + cz2 = 0 को लंब जनकों में काटता है, तो सिद्ध कीजिए कि (b + c)u2 + (c + a)v2 + (a + b)w2 = 0.
Worked solution
A generator (l, m, n) in the plane satisfies ul + vm + wn = 0 and al2 + bm2 + cn2 = 0. Eliminating n = −(ul + vm)/w: (aw2 + cu2)l2 + 2cuv lm + (bw2 + cv2)m2 = 0.
If the two generators are (l1, m1, n1) and (l2, m2, n2), the product of the roots gives l1l2/(bw2 + cv2) = m1m2/(cu2 + aw2), and by symmetry both equal n1n2/(av2 + bu2).
The generators are perpendicular if and only if l1l2 + m1m2 + n1n2 = 0, that is the sum of the denominators vanishes: (b + c)u2 + (c + a)v2 + (a + b)w2 = 0.
Answer: hence proved; this is the special case f = g = h = 0 of the general condition (a + b + c)(u2 + v2 + w2) = au2 + bv2 + cw2 + ... stated above.Final answer: hence proved; this is the special case f = g = h = 0 of the general condition (a + b + c)(u2 + v2 + w2) = au2 + bv2 + cw2 + ... stated above.
Question 5 (50 marks)
Q5(a) · Ordinary Differential Equations · 10 marks≈ 7 min
Formulation and First Order, First Degree Equations
Show that the general solution of the differential equation dy/dx + Py = Q can be written in the form y = Q/P − e−∫P dx {C + ∫ e∫P dx d(Q/P)}, where P, Q are non-zero functions of x and C, an arbitrary constant.
हिंदी में प्रश्न
दर्शाइए कि अवकल समीकरण dy/dx + Py = Q का व्यापक हल y = Q/P − e−∫P dx {C + ∫ e∫P dx d(Q/P)} के रूप में लिखा जा सकता है, जहाँ P, Q, x के शून्येतर फलन हैं तथा C एक स्वेच्छ अचर है।
Worked solution
With I = e∫P dx, dI = PI dx, and the solution of the linear equation is yI = ∫QI dx + C′.
Write QI dx = (Q/P)·PI dx = (Q/P) dI and integrate by parts: ∫(Q/P) dI = (Q/P)I − ∫I d(Q/P).
So yI = (Q/P)I − ∫I d(Q/P) + C′; dividing by I and writing C = −C′: y = Q/P − e−∫P dx{C + ∫e∫P dx d(Q/P)}.
Answer: hence proved (P ≠ 0 is needed to divide by P).Final answer: hence proved (P ≠ 0 is needed to divide by P).
Q5(b) · Ordinary Differential Equations · 10 marks≈ 7 min
Orthogonal Trajectories
Show that the orthogonal trajectories of the system of parabolas: x2 = 4a(y + a) belong to the same system.
हिंदी में प्रश्न
दर्शाइए कि परवलयों के निकाय: x2 = 4a(y + a) के लंबकोणीय संछेदी, उसी निकाय में स्थित होते हैं।
Q5(c) · Dynamics and Statics · 10 marks≈ 7 min
Constrained Motion, Work and Energy
A body of weight w rests on a rough inclined plane of inclination θ, the coefficient of friction, μ, being greater than tan θ. Find the work done in slowly dragging the body a distance 'b' up the plane and then dragging it back to the starting point, the applied force being in each case parallel to the plane.
हिंदी में प्रश्न
w भार का एक पिंड, θ कोण से झुके हुए एक रूक्ष समतल पर स्थित है, घर्षण गुणांक μ, tan θ से अधिक है। पिंड को समतल पर ऊपर की तरफ 'b' दूरी तक धीरे-धीरे खींचने तथा वापस आरम्भिक बिन्दु तक खींचने में किए गए कार्य को ज्ञात कीजिए, जहाँ लगाया गया बल प्रत्येक दशा में समतल के समान्तर है।
Worked solution
Since μ > tan θ the body does not slide by itself, and 'slowly' means the applied force just balances the other forces, so no kinetic energy is produced.
Up the plane: force = w sin θ + μw cos θ, work = wb(sin θ + μ cos θ).
Down the plane: friction μw cos θ now acts up the plane and exceeds w sin θ, so a force μw cos θ − w sin θ down the plane is needed; work = wb(μ cos θ − sin θ).
Total = 2μwb cos θ: the work against gravity cancels over the round trip, and only the work against friction remains.
Answer: the total work done is 2μwb cos θ.Final answer: the total work done is 2μwb cos θ.
Q5(d) · Dynamics and Statics · 10 marks≈ 7 min
Motion in a Plane and Projectiles
A projectile is fired from a point O with velocity √(2gh) and hits a tangent at the point P(x, y) in the plane, the axes OX and OY being horizontal and vertically downward lines through the point O, respectively. Show that if the two possible directions of projection be at right angles, then x2 = 2hy and then one of the possible directions of projection bisects the angle POX.
हिंदी में प्रश्न
एक प्रक्षेप्य √(2gh) वेग के साथ बिन्दु O से प्रक्षेपित किया गया तथा समतल के बिन्दु P(x, y) पर स्पर्श-रेखा से टकराता है जहाँ अक्ष OX तथा OY क्रमशः बिन्दु O से क्षैतिज तथा अधोमुखी ऊर्ध्वाधर रेखाएँ हैं। यदि प्रक्षेपण की दो संभव दिशाएँ समकोण पर हों, तो दर्शाइए कि x2 = 2hy तथा प्रक्षेपण की संभव दिशाओं में से एक, कोण POX को द्विभाजित करती है।
Worked solution
[The phrase 'hits a tangent at the point P' is evidently a misprint for 'passes through the point P'; that version is solved.]
With OY downward, u2 = 2gh and elevation α, the path is y = −x tan α + gx2/(2u2cos2α) = −x tan α + x2(1 + tan2α)/(4h).
So t = tan α satisfies (x2/(4h))t2 − xt + (x2/(4h) − y) = 0.
Perpendicular directions mean t1t2 = −1: (x2/(4h) − y)/(x2/(4h)) = −1, that is x2 = 2hy.
The equation is then t2 − (4h/x)t − 1 = 0; with β the angle POX (P below OX), tan β = y/x = x/(2h), so 4h/x = 2 cot β.
Roots: t = cot β ± cosec β, that is t = cot(β/2) or t = (cos β − 1)/sin β = −tan(β/2).
The root −tan(β/2) is the direction at β/2 below OX, which bisects the angle POX; the other root, elevation 90° − β/2, is perpendicular to it.
Answer: x2 = 2hy, and one direction of projection bisects the angle POX.Final answer: x2 = 2hy, and one direction of projection bisects the angle POX.
Q5(e) · Vector Analysis · 10 marks≈ 7 min
Gradient, Divergence and Curl in Cartesian and Cylindrical Coordinates
Show that A = (6xy + z3) î + (3x2 − z) ĵ + (3xz2 − y) k̂ is irrotational. Also find φ such that A = ∇φ.
हिंदी में प्रश्न
दर्शाइए कि A = (6xy + z3) î + (3x2 − z) ĵ + (3xz2 − y) k̂ अघूर्णी है। φ को भी ज्ञात कीजिए जबकि A = ∇φ.
Question 6 (50 marks)
Q6(a) · Dynamics and Statics · 20 marks≈ 14 min
Equilibrium, Friction and the Common Catenary
A cable of weight w per unit length and length 2l hangs from two points P and Q in the same horizontal line. Show that the span of the cable is 2l(1 − 2h2/(3l2)), where h is the sag in the middle of the tightly stretched position.
हिंदी में प्रश्न
2l लम्बाई का एक तार (केबिल) जिसका भार w प्रति इकाई (यूनिट) लम्बाई है, एक क्षैतिज रेखा के दो बिन्दुओं P तथा Q से लटकी हुई है। दर्शाइए कि तार की विस्तृति (स्पैन) 2l(1 − 2h2/(3l2)) है, जहाँ h तार के कसकर खींची हुई स्थिति में मध्य का झोल है।
Worked solution
With half-length l and sag h, Example 4 gives c = (l2 − h2)/(2h); for a tight cable h is small compared with l, so c ≈ l2/(2h) and l/c ≈ 2h/l is small.
At an end s = l = c sinh(x/c), so the half-span is x = c sinh−1(l/c).
Using sinh−1z = z − z3/6 + ..., x = l − l3/(6c2) to this order.
With c2 ≈ l4/(4h2), l3/(6c2) ≈ 2h2/(3l), the neglected terms being of order h4/l3.
So the span is 2x ≈ 2l − 4h2/(3l).
Answer: span ≈ 2l(1 − 2h2/(3l2)); for example l = 10, h = 0.5 gives 19.96665 exactly and 19.96667 from the formula.Final answer: span ≈ 2l(1 − 2h2/(3l2)); for example l = 10, h = 0.5 gives 19.96665 exactly and 19.96667 from the formula.
Q6(b) · Ordinary Differential Equations · 15 marks≈ 11 min
Second Order Equations with Variable Coefficients
Solve the following differential equation by using the method of variation of parameters: (x2 − 1) d2y/dx2 − 2x dy/dx + 2y = (x2 − 1)2, given that y = x is one solution of the reduced equation.
हिंदी में प्रश्न
प्राचल-विचरण विधि का उपयोग करके, निम्नलिखित अवकल समीकरण: (x2 − 1) d2y/dx2 − 2x dy/dx + 2y = (x2 − 1)2 को हल कीजिए, जहाँ समानीत समीकरण का एक हल y = x दिया गया है।
Worked solution
Standard form: P = −2x/(x2 − 1), R = x2 − 1; e−∫P dx = x2 − 1, so y2 = x∫(x2 − 1)/x2 dx = x(x + 1/x) = x2 + 1.
W = x·2x − (x2 + 1) = x2 − 1.
u1 = −∫(x2 + 1) dx = −x3/3 − x; u2 = ∫x dx = x2/2.
PI = x(−x3/3 − x) + (x2 + 1)x2/2 = x4/6 − x2/2.
Answer: y = c1x + c2(x2 + 1) + x4/6 − x2/2.Final answer: y = c1x + c2(x2 + 1) + x4/6 − x2/2.
Q6(c) · Vector Analysis · 15 marks≈ 11 min
Gauss and Stokes' Theorems and Green's Identities
Verify Green's theorem in the plane for ∮C (3x2 − 8y2) dx + (4y − 6xy) dy, where C is the boundary curve of the region defined by x = 0, y = 0, x + y = 1.
हिंदी में प्रश्न
समतल में ग्रीन के प्रमेय को ∮C (3x2 − 8y2) dx + (4y − 6xy) dy के लिए सत्यापित कीजिए, जहाँ C, x = 0, y = 0, x + y = 1 द्वारा परिभाषित क्षेत्र का सीमा वक्र है।
Worked solution
Double integral: ∂N/∂x − ∂M/∂y = −6y + 16y = 10y, and ∬10y dA = 10∫01∫01−xy dy dx = 5∫01(1 − x)2 dx = 5/3.
OA (y = 0, x: 0 → 1): ∫3x2 dx = 1.
AB (x = 1 − t, y = t, t: 0 → 1): M dx + N dy = (−3 + 4t + 11t2) dt, with integral −3 + 2 + 11/3 = 8/3.
BO (x = 0, y: 1 → 0): ∫4y dy = −2.
Answer: the line integral is 1 + 8/3 − 2 = 5/3, equal to the double integral, which verifies Green's theorem.Final answer: the line integral is 1 + 8/3 − 2 = 5/3, equal to the double integral, which verifies Green's theorem.
Question 7 (50 marks)
Q7(a) · Vector Analysis · 20 marks≈ 14 min
Gauss and Stokes' Theorems and Green's Identities
Verify Stokes' theorem for F = x î + z2 ĵ + y2 k̂ over the plane surface: x + y + z = 1 lying in the first octant.
हिंदी में प्रश्न
स्टोक्स प्रमेय को F = x î + z2 ĵ + y2 k̂ के लिए प्रथम अष्टांशक में स्थित समतल पृष्ठ: x + y + z = 1 पर सत्यापित कीजिए।
Worked solution
∇ × F = (2y − 2z, 0, 0); on x + y + z = 1, n = (1, 1, 1)/√3 and dS = √3 dx dy over the triangle T: x, y ≥ 0, x + y ≤ 1.
∬(∇ × F)·n dS = ∬T(2y − 2z) dx dy = ∬T(2x + 4y − 2) dx dy = 2(1/6) + 4(1/6) − 2(1/2) = 0.
Boundary A(1, 0, 0) → B(0, 1, 0) → C(0, 0, 1) → A, with F·dr = x dx + z2 dy + y2 dz.
AB (z = 0): ∫10x dx = −1/2; BC (x = 0, y = 1 − t, z = t): ∫01[−t2 + (1 − t)2] dt = 0; CA (y = 0): ∫01x dx = 1/2.
Answer: both sides are 0, verifying Stokes' theorem.Final answer: both sides are 0, verifying Stokes' theorem.
Q7(b) · Ordinary Differential Equations · 15 marks≈ 11 min
Laplace Transforms and Initial Value Problems
Solve the following initial value problem by using Laplace's transformation d2y/dt2 − 3 dy/dt + 2y = h(t), where h(t) = 2 for 0 < t < 4, and h(t) = 0 for t > 4; y(0) = 0, y′(0) = 0.
हिंदी में प्रश्न
लाप्लास रूपांतरण का उपयोग करके निम्नलिखित प्रारंभिक मान समस्या: d2y/dt2 − 3 dy/dt + 2y = h(t), जहाँ h(t) = 2, 0 < t < 4 के लिए, तथा h(t) = 0, t > 4 के लिए; y(0) = 0, y′(0) = 0 को हल कीजिए।
Q7(c) · Dynamics and Statics · 15 marks≈ 11 min
Virtual Work, Stability and Forces in Three Dimensions
Suppose a cylinder of any cross-section is balanced on another fixed cylinder, the contact of curved surfaces being rough and the common tangent line horizontal. Let ρ and ρ′ be the radii of curvature of the two cylinders at the point of contact and h be the height of centre of gravity of the upper cylinder above the point of contact. Show that the upper cylinder is balanced in stable equilibrium if h < ρρ′/(ρ + ρ′).
हिंदी में प्रश्न
माना किसी भी अनुप्रस्थ-काट का एक बेलन दूसरे स्थिर बेलन पर संतुलित है, जहाँ वक्रीय पृष्ठों का संस्पर्श रूक्ष है तथा उभयनिष्ठ स्पर्श-रेखा क्षैतिज है। माना दोनों बेलनों के स्पर्श बिन्दु पर उनकी वक्रता त्रिज्याएँ ρ तथा ρ′ हैं और संस्पर्श बिन्दु से ऊपरी बेलन के गुरुत्व केन्द्र की ऊँचाई h है। दर्शाइए कि स्थायी साम्य में ऊपरी बेलन संतुलित है यदि h < ρρ′/(ρ + ρ′)।
Worked solution
Let O′ be the fixed centre of curvature of the lower cylinder (radius ρ′) and O that of the upper (radius ρ); initially O, G, the point of contact and O′ lie on a vertical line, with G at height h above the contact.
Let the upper cylinder roll (rough contact, no slipping) until O′O makes a small angle φ with the vertical; equal arcs ρ′φ = ρψ mean the upper body has turned through χ = φ + ψ = φ(ρ + ρ′)/ρ.
G lies on the body's line through O at distance ρ − h from O, and that line has turned through χ, so the height of G above O′ is z = (ρ + ρ′) cos φ − (ρ − h) cos χ.
For small φ: z ≈ (ρ′ + h) − [(ρ + ρ′) − (ρ − h)(ρ + ρ′)2/ρ2]φ2/2.
The equilibrium is stable if z is a minimum at φ = 0, that is (ρ − h)(ρ + ρ′) > ρ2, which simplifies to ρρ′ > h(ρ + ρ′).
Answer: the upper cylinder is in stable equilibrium if h < ρρ′/(ρ + ρ′), and unstable if h > ρρ′/(ρ + ρ′).Final answer: the upper cylinder is in stable equilibrium if h < ρρ′/(ρ + ρ′), and unstable if h > ρρ′/(ρ + ρ′).
Question 8 (50 marks)
Q8(a) · Ordinary Differential Equations · 20 marks≈ 14 min
First Order Higher Degree Equations, Clairaut's Equation and Singular Solutions
(i) Find the general and singular solutions of the differential equation: (x2 − a2) p2 − 2xyp + y2 + a2 = 0, where p = dy/dx. Also give the geometric relation between the general and singular solutions. (10)
(ii) Solve the following differential equation: (3x + 2)2 d2y/dx2 + 5(3x + 2) dy/dx − 3y = x2 + x + 1 (10)हिंदी में प्रश्न
(i) अवकल समीकरण: (x2 − a2) p2 − 2xyp + y2 + a2 = 0, जहाँ p = dy/dx, के व्यापक व विचित्र हलों को ज्ञात कीजिए। व्यापक व विचित्र हलों के बीच ज्यामितीय संबंध को भी दीजिए। (10)
(ii) निम्नलिखित अवकल समीकरण को हल कीजिए: (3x + 2)2 d2y/dx2 + 5(3x + 2) dy/dx − 3y = x2 + x + 1 (10)Worked solution
(i)
Rearranging: (y − px)2 = a2p2 − a2, so y = px ± a√(p2 − 1), Clairaut's form.
General solution: (y − cx)2 = a2(c2 − 1), that is (x2 − a2)c2 − 2xyc + y2 + a2 = 0.
c-discriminant: x2y2 − (x2 − a2)(y2 + a2) = 0, that is a2(y2 − x2 + a2) = 0; the p-discriminant is the same expression.
Answer: general solution (y − cx)2 = a2(c2 − 1), a family of straight lines; singular solution x2 − y2 = a2, the rectangular hyperbola that every line of the family touches (the envelope).
(ii)
Put 3x + 2 = ez: (3x + 2)D = 3θ, (3x + 2)2D2 = 9θ(θ − 1). The operator is 9θ(θ − 1) + 15θ − 3 = 3(3θ − 1)(θ + 1), so CF = c1(3x + 2)1/3 + c2(3x + 2)−1.
With x = (ez − 2)/3: x2 + x + 1 = (e2z − ez + 7)/9. For f(θ) = 9θ2 + 6θ − 3: f(2) = 45, f(1) = 12, f(0) = −3.
PI = (1/9)[e2z/45 − ez/12 + 7/(−3)] = e2z/405 − ez/108 − 7/27.
Answer: y = c1(3x + 2)1/3 + c2(3x + 2)−1 + (3x + 2)2/405 − (3x + 2)/108 − 7/27.Final answer: (i) general solution (y − cx)2 = a2(c2 − 1), a family of straight lines; singular solution x2 − y2 = a2, the rectangular hyperbola that every line of the family touches (the envelope); (ii) y = c1(3x + 2)1/3 + c2(3x + 2)−1 + (3x + 2)2/405 − (3x + 2)/108 − 7/27
Q8(b) · Dynamics and Statics · 15 marks≈ 11 min
Virtual Work, Stability and Forces in Three Dimensions
A chain of n equal uniform rods is smoothly jointed together and suspended from its one end A1. A horizontal force P is applied to the other end An+1 of the chain. Find the inclinations of the rods to the downward vertical line in the equilibrium configuration.
हिंदी में प्रश्न
n बराबर एकसमान छड़ों की एक श्रृंखला एक-दूसरे के साथ चिकने रूप से जुड़ी हुई है तथा इसके एक सिरे A1 से लटकी हुई है। एक क्षैतिज बल P श्रृंखला के दूसरे सिरे An+1 पर लगाया गया है। साम्य विन्यास में अधोमुखी ऊर्ध्वाधर रेखा से छड़ों के झुकाव ज्ञात कीजिए।
Worked solution
Let each rod have length 2a and weight W, and let the r-th rod ArAr+1 make angle θr with the downward vertical.
Depth of the centre of rod r below A1: 2a(cos θ1 + ... + cos θr−1) + a cos θr; horizontal distance of An+1 from A1: x = 2a Σ sin θr.
Virtual work: W δ(sum of depths) + P δx = 0, and the angles θr can be varied independently.
cos θr occurs in the depth of rod r (coefficient a) and of the n − r rods below it (2a each), so the coefficient of δθr gives −Wa(2n − 2r + 1) sin θr + 2aP cos θr = 0.
Answer: tan θr = 2P/((2n − 2r + 1)W) for r = 1, 2, ..., n, so tan θ1 : tan θ2 : ... : tan θn = 1/(2n − 1) : 1/(2n − 3) : ... : 1/1.Final answer: tan θr = 2P/((2n − 2r + 1)W) for r = 1, 2, ..., n, so tan θ1 : tan θ2 : ... : tan θn = 1/(2n − 1) : 1/(2n − 3) : ... : 1/1.
Q8(c) · Vector Analysis · 15 marks≈ 11 min
Gauss and Stokes' Theorems and Green's Identities
Using Gauss' divergence theorem, evaluate ∬S F · n dS, where F = x î − y ĵ + (z2 − 1) k̂ and S is the cylinder formed by the surfaces z = 0, z = 1, x2 + y2 = 4.
हिंदी में प्रश्न
गाउस के अपसरण प्रमेय का उपयोग करके ∬S F · n dS का मान निकालिए, जहाँ F = x î − y ĵ + (z2 − 1) k̂ तथा S, पृष्ठों z = 0, z = 1, x2 + y2 = 4 द्वारा बना हुआ बेलन है।
UPSC Mains 2022 Mathematics Paper II
Question 1 (50 marks)
Q1(a) · Abstract Algebra · 10 marks≈ 7 min
Normal Subgroups, Quotient Groups, Homomorphisms and Isomorphism Theorems
Show that the multiplicative group G = {1, −1, i, −i}, where i = √(−1), is isomorphic to the group G′ = ({0, 1, 2, 3}, +4).
हिंदी में प्रश्न
दर्शाइये कि गुणनात्मक समूह G = {1, −1, i, −i}, जहाँ i = √(−1) है, समूह G′ = ({0, 1, 2, 3}, +4) के तुल्याकारी है।
Q1(b) · Complex Analysis · 10 marks≈ 7 min
Analytic Functions and Cauchy-Riemann Equations
If f(z) = u + iv is an analytic function of z, and u − v = (cos x + sin x − e−y)/(2 cos x − ey − e−y), then find f(z) subject to the condition f(π/2) = 0.
हिंदी में प्रश्न
यदि f(z) = u + iv, z का एक विश्लेषिक फलन है, तथा u − v = (cos x + sin x − e−y)/(2 cos x − ey − e−y) है, तब शर्त f(π/2) = 0 के अधीन f(z) का मान ज्ञात कीजिये।
Worked solution
Let F = (1 + i)f = (u − v) + i(u + v), which is analytic with real part U = u − v = N/D, where N = cos x + sin x − e−y and D = 2cos x − ey − e−y.
On y = 0: N = cos x + sin x − 1, D = 2cos x − 2, Nx = cos x − sin x, Dx = −2sin x, Ny = 1, Dy = 0.
Ux(x, 0) = (NxD − NDx)/D2 = (2 − 2cos x)/(4(1 − cos x)2) = 1/(2(1 − cos x)), and Uy(x, 0) = (NyD − NDy)/D2 = 1/D = −1/(2(1 − cos x)).
Milne-Thomson: F′(z) = Ux(z, 0) − iUy(z, 0) = (1 + i)/(2(1 − cos z)) = ((1 + i)/4)cosec2(z/2).
Integrating, F(z) = −((1 + i)/2)cot(z/2) + c, so f = F/(1 + i) = −(1/2)cot(z/2) + c′. The condition f(π/2) = 0 gives −(1/2)cot(π/4) + c′ = 0, so c′ = 1/2.
Answer: f(z) = (1/2)[1 − cot(z/2)].Final answer: f(z) = (1/2)[1 − cot(z/2)].
Q1(c) · Real Analysis · 10 marks≈ 7 min
Riemann Integral, Improper Integrals and Fundamental Theorems of Integral Calculus
Test the convergence of ∫0∞ cos x/(1 + x2) dx.
हिंदी में प्रश्न
∫0∞ cos x/(1 + x2) dx के अभिसरण का परीक्षण कीजिये।
Q1(d) · Complex Analysis · 10 marks≈ 7 min
Power Series, Taylor's Series and Laurent's Series
Expand f(z) = 1/((z − 1)2(z − 3)) in a Laurent series valid for the regions (i) 0 < |z − 1| < 2 and (ii) 0 < |z − 3| < 2.
हिंदी में प्रश्न
f(z) = 1/((z − 1)2(z − 3)) का क्षेत्रों (i) 0 < |z − 1| < 2 एवं (ii) 0 < |z − 3| < 2 के लिये वैध लौराँ श्रेणी में विस्तार कीजिये।
Worked solution
(i) w = z − 1, 0 < |w| < 2: z − 3 = w − 2, so f = 1/(w2(w − 2)) = −(1/(2w2))·1/(1 − w/2) = −∑n=0∞ wn−2/2n+1.
So f = −1/(2(z − 1)2) − 1/(4(z − 1)) − 1/8 − (z − 1)/16 − ...; z = 1 is a pole of order 2 with residue −1/4.
(ii) t = z − 3, 0 < |t| < 2: z − 1 = t + 2, so f = 1/(t(t + 2)2) = (1/(4t))(1 + t/2)−2 = ∑n=0∞ (−1)n(n + 1)tn−1/2n+2.
So f = 1/(4(z − 3)) − 1/4 + 3(z − 3)/16 − (z − 3)2/8 + ...; z = 3 is a simple pole with residue 1/4.
Answer: (i) f = −∑n=0∞ (z − 1)n−2/2n+1; (ii) f = ∑n=0∞ (−1)n(n + 1)(z − 3)n−1/2n+2.Final answer: (i) f = −∑n=0∞ (z − 1)n−2/2n+1; (ii) f = ∑n=0∞ (−1)n(n + 1)(z − 3)n−1/2n+2.
Q1(e) · Linear Programming · 10 marks≈ 7 min
The Simplex Method
Use two-phase method to solve the following linear programming problem:
Minimize Z = x1 + x2
subject to
2x1 + x2 ≥ 4
x1 + 7x2 ≥ 7
x1, x2 ≥ 0हिंदी में प्रश्न
निम्नलिखित रैखिक प्रोग्रामन समस्या को हल करने के लिये द्विचरण विधि का उपयोग कीजिये:
न्यूनतमीकरण कीजिये Z = x1 + x2
बशर्ते कि
2x1 + x2 ≥ 4
x1 + 7x2 ≥ 7
x1, x2 ≥ 0
Question 2 (50 marks)
Q2(a) · Real Analysis · 15 marks≈ 11 min
Riemann Integral, Improper Integrals and Fundamental Theorems of Integral Calculus
Let f(x) = x2 on [0, k], k > 0. Show that f is Riemann integrable on the closed interval [0, k] and ∫0k f dx = k3/3.
हिंदी में प्रश्न
मान लीजिये कि [0, k], k > 0 पर f(x) = x2 है। दर्शाइये कि f बंद अन्तराल [0, k] पर रीमान समाकलनीय है तथा ∫0k f dx = k3/3 है।
Q2(b) · Abstract Algebra · 15 marks≈ 11 min
Normal Subgroups, Quotient Groups, Homomorphisms and Isomorphism Theorems
Prove that every homomorphic image of a group G is isomorphic to some quotient group of G.
हिंदी में प्रश्न
सिद्ध कीजिये कि एक समूह G का प्रत्येक समाकारी प्रतिबिम्ब, G के किसी विभाग समूह के तुल्याकारी है।
Q2(c) · Complex Analysis · 20 marks≈ 14 min
Contour Integration: Evaluation of Real Integrals
Apply the calculus of residues to evaluate ∫−∞∞ cos x dx/((x2 + a2)(x2 + b2)), a > b > 0.
हिंदी में प्रश्न
∫−∞∞ cos x dx/((x2 + a2)(x2 + b2)), a > b > 0 के मान निकालने के लिये अवशेष-कलन का उपयोग कीजिये।
Question 3 (50 marks)
Q3(a) · Complex Analysis · 15 marks≈ 11 min
Cauchy's Theorem and Cauchy's Integral Formula
Evaluate ∫C (z + 4)/(z2 + 2z + 5) dz, where C is |z + 1 − i| = 2.
हिंदी में प्रश्न
∫C (z + 4)/(z2 + 2z + 5) dz का मान निकालिये, जहाँ C, |z + 1 − i| = 2 है।
Q3(b) · Real Analysis · 20 marks≈ 14 min
Functions of Several Variables: Partial Derivatives, Maxima and Minima
Find the maximum and minimum values of x2/a4 + y2/b4 + z2/c4, when lx + my + nz = 0 and x2/a2 + y2/b2 + z2/c2 = 1. Interpret the result geometrically.
हिंदी में प्रश्न
x2/a4 + y2/b4 + z2/c4 के अधिकतम तथा न्यूनतम मान निकालिये, जब lx + my + nz = 0 तथा x2/a2 + y2/b2 + z2/c2 = 1 हैं। परिणाम की ज्यामितीय व्याख्या कीजिये।
Worked solution
F = x2/a4 + y2/b4 + z2/c4 + λ(x2/a2 + y2/b2 + z2/c2 − 1) + μ(lx + my + nz). Fx = 0 gives 2x/a4 + 2λx/a2 + μl = 0, and similarly for y, z.
Multiply by x, y, z and add; with u = x2/a4 + y2/b4 + z2/c4 this gives 2u + 2λ = 0, so λ = −u.
Then 2x(1 − a2u)/a4 = −μl, so x = −μla4/(2(1 − a2u)); similarly for y and z.
Substituting in lx + my + nz = 0 (μ ≠ 0 as before): l2a4/(1 − a2u) + m2b4/(1 − b2u) + n2c4/(1 − c2u) = 0, a quadratic in u whose roots are the maximum and minimum values. (Check: a = 1, b = 2, c = 3, l = m = n = 1 gives u ≈ 0.207 and 0.940 both from this equation and from a direct search.)
Geometric meaning: the tangent plane to the ellipsoid at (x, y, z) is Xx/a2 + Yy/b2 + Zz/c2 = 1, whose distance p from the centre satisfies 1/p2 = x2/a4 + y2/b4 + z2/c4.
Answer: the extreme values are the roots u of ∑ l2a4/(1 − a2u) = 0; they are the extreme values of 1/p2, where p is the perpendicular from the centre to the tangent planes at points of the central section by lx + my + nz = 0.Final answer: the extreme values are the roots u of ∑ l2a4/(1 − a2u) = 0; they are the extreme values of 1/p2, where p is the perpendicular from the centre to the tangent planes at points of the central section by lx + my + nz = 0.
Q3(c) · Linear Programming · 15 marks≈ 11 min
Duality in Linear Programming
Solve the following linear programming problem by the simplex method. Write its dual. Also, write the optimal solution of the dual from the optimal table of the given problem:
Maximize Z = x1 + x2 + x3
subject to
2x1 + x2 + x3 ≤ 2
4x1 + 2x2 + x3 ≤ 2
x1, x2, x3 ≥ 0हिंदी में प्रश्न
निम्नलिखित रैखिक प्रोग्रामन समस्या को एकधा विधि द्वारा हल कीजिये। इसकी द्वैती समस्या लिखिये। दी गयी समस्या की इष्टतम सारणी से द्वैती समस्या का इष्टतम हल भी लिखिये:
अधिकतमीकरण कीजिये Z = x1 + x2 + x3
बशर्ते कि
2x1 + x2 + x3 ≤ 2
4x1 + 2x2 + x3 ≤ 2
x1, x2, x3 ≥ 0
Question 4 (50 marks)
Q4(a) · Abstract Algebra · 15 marks≈ 11 min
Rings, Subrings, Ideals and Ring Homomorphisms
Let R be a field of real numbers and S, the field of all those polynomials f(x) ∈ R[x] such that f(0) = 0 = f(1). Prove that S is an ideal of R[x]. Is the residue class ring R[x]/S an integral domain? Give justification for your answer.
हिंदी में प्रश्न
मान लीजिये कि R वास्तविक संख्याओं का एक क्षेत्र है तथा S, उन सभी बहुपदों f(x) ∈ R[x], जिनके लिये f(0) = 0 = f(1) है, का क्षेत्र है। सिद्ध कीजिये कि S, R[x] की एक गुणजावली है। क्या अवशेष वर्ग वलय R[x]/S एक पूर्णांकीय प्रांत है? अपने उत्तर का स्पष्टीकरण दीजिये।
Worked solution
(The paper says 'the field of all those polynomials'; S is only a subset of ℝ[x], so read 'the set'. Here R = ℝ.)
0 ∈ S. If f, g ∈ S, then (f − g)(0) = 0 and (f − g)(1) = 0, so f − g ∈ S.
If f ∈ S and h ∈ ℝ[x], then (hf)(0) = h(0)f(0) = 0 and (hf)(1) = h(1)f(1) = 0, so hf ∈ S. Hence S is an ideal; in fact S = ⟨x(x − 1)⟩, since x and x − 1 both divide f.
x ∉ S (it is 1 at x = 1) and x − 1 ∉ S (it is −1 at x = 0), but x(x − 1) ∈ S.
So (x + S)(x − 1 + S) = S, the zero of ℝ[x]/S, with both factors non-zero: ℝ[x]/S has zero divisors, and S is not a prime ideal.
Answer: S is an ideal of ℝ[x], but ℝ[x]/S is not an integral domain (indeed ℝ[x]/S ≅ ℝ × ℝ via f + S ↦ (f(0), f(1))).Final answer: S is an ideal of ℝ[x], but ℝ[x]/S is not an integral domain (indeed ℝ[x]/S ≅ ℝ × ℝ via f + S ↦ (f(0), f(1))).
Q4(b) · Real Analysis · 15 marks≈ 11 min
Series: Convergence Tests, Absolute and Conditional Convergence, Rearrangement
Test for convergence or divergence of the series x + 22x2/2! + 33x3/3! + 44x4/4! + 55x5/5! + ⋯ (x > 0)
हिंदी में प्रश्न
श्रेणी x + 22x2/2! + 33x3/3! + 44x4/4! + 55x5/5! + ⋯ (x > 0) के अभिसरण या अपसरण का परीक्षण कीजिये।
Worked solution
an = nnxn/n! > 0, and an+1/an = (n + 1)n+1x/((n + 1)nn) = (1 + 1/n)nx → ex.
By the ratio test the series converges for ex < 1, i.e. 0 < x < 1/e, and diverges for x > 1/e.
At x = 1/e the ratio tends to 1. Using log(1 + 1/n)n = 1 − 1/(2n) + 1/(3n2) − ⋯, an/an+1 = e/(1 + 1/n)n = 1 + 1/(2n) + O(1/n2).
So n(an/an+1 − 1) → 1/2 < 1, and the series diverges by Raabe's test (Stirling's formula agrees: an ~ 1/√(2πn)).
Answer: convergent for 0 < x < 1/e and divergent for x ≥ 1/e.Final answer: convergent for 0 < x < 1/e and divergent for x ≥ 1/e.
Q4(c) · Linear Programming · 20 marks≈ 14 min
Transportation Problem
Find the initial basic feasible solution of the following transportation problem by Vogel's approximation method and use it to find the optimal solution and the transportation cost of the problem:
[Table: costs from sources S1, S2, S3 to destinations A, B, C, D]
S1: 21, 16, 25, 13; availability 11
S2: 17, 18, 14, 23; availability 13
S3: 32, 27, 18, 41; availability 19
Requirement: 6, 10, 12, 15; total 43हिंदी में प्रश्न
वोगेल की सन्निकटन विधि से निम्नलिखित परिवहन समस्या का आरंभिक आधारी सुसंगत हल ज्ञात कीजिये। इस हल का उपयोग कर समस्या का इष्टतम हल एवं परिवहन लागत ज्ञात कीजिये:
[सारणी: उद्गम S1, S2, S3 से गन्तव्य A, B, C, D तक लागत]
S1: 21, 16, 25, 13; प्राप्यता 11
S2: 17, 18, 14, 23; प्राप्यता 13
S3: 32, 27, 18, 41; प्राप्यता 19
माँग: 6, 10, 12, 15; कुल 43
Question 5 (50 marks)
Q5(a) · Partial Differential Equations · 10 marks≈ 7 min
Families of Surfaces and Formation of PDEs
It is given that the equation of any cone with vertex at (a, b, c) is f((x − a)/(z − c), (y − b)/(z − c)) = 0. Find the differential equation of the cone.
हिंदी में प्रश्न
दिया गया है कि शीर्ष (a, b, c) वाले किसी शंकु का समीकरण f((x − a)/(z − c), (y − b)/(z − c)) = 0 है। शंकु का अवकल समीकरण ज्ञात कीजिये।
Worked solution
Put u = (x − a)/(z − c), v = (y − b)/(z − c) and differentiate f(u, v) = 0; after multiplying by (z − c)2: fu[(z − c) − (x − a)p] − fv(y − b)p = 0 and −fu(x − a)q + fv[(z − c) − (y − b)q] = 0.
For non-zero (fu, fv) the determinant vanishes: [(z − c) − (x − a)p][(z − c) − (y − b)q] − (x − a)(y − b)pq = 0.
Expanding, (z − c)2 − (z − c)[(x − a)p + (y − b)q] = 0, and z ≠ c away from the vertex.
Answer: (x − a)p + (y − b)q = z − c.Final answer: (x − a)p + (y − b)q = z − c.
Q5(b) · Numerical Analysis and Computer Programming · 10 marks≈ 7 min
Systems of Linear Equations: Direct and Iterative Methods
Solve, by Gauss elimination method, the system of equations
2x + 2y + 4z = 18
x + 3y + 2z = 13
3x + y + 3z = 14हिंदी में प्रश्न
गाउस विलोपन विधि द्वारा समीकरण निकाय
2x + 2y + 4z = 18
x + 3y + 2z = 13
3x + y + 3z = 14
को हल कीजिये।Q5(c) · Numerical Analysis and Computer Programming · 10 marks≈ 7 min
Number Systems and Representation of Numbers in a Computer
(i) Convert the number (1093·21875)10 into octal and the number (1693·0628)10 into hexadecimal systems.
(ii) Express the Boolean function F(x, y, z) = xy + x′z in a product of maxterms form.हिंदी में प्रश्न
(i) संख्या (1093·21875)10 को अष्टाधारी तथा संख्या (1693·0628)10 को षोडश-आधारी पद्धति में बदलिये।
(ii) बूलीय फलन F(x, y, z) = xy + x′z को योपद (मैक्सटर्म) के गुणन के रूप में अभिव्यक्त कीजिये।Q5(d) · Mechanics and Fluid Dynamics · 10 marks≈ 7 min
Generalized Coordinates, D'Alembert's Principle and Lagrange's Equations
A particle at a distance r from the centre of force moves under the influence of the central force F = −k/r2, where k is a constant. Obtain the Lagrangian and derive the equations of motion.
हिंदी में प्रश्न
एक कण, जो बल-केन्द्र से r दूरी पर है, केन्द्रीय बल F = −k/r2, जहाँ k एक स्थिरांक है, के प्रभाव में गतिमान है। लग्रांजियन निकालिये तथा गति के समीकरणों को व्युत्पन्न कीजिये।
Q5(e) · Mechanics and Fluid Dynamics · 10 marks≈ 7 min
Kinematics of Fluids: Continuity, Streamlines and Potential Flow
The velocity components of an incompressible fluid in spherical polar coordinates (r, θ, ψ) are (2Mr−3 cos θ, Mr−2 sin θ, 0), where M is a constant. Show that the velocity is of the potential kind. Find the velocity potential and the equations of the streamlines.
हिंदी में प्रश्न
किसी असंपीड्य तरल के गोलीय ध्रुवी निर्देशांकों (r, θ, ψ) में वेग-घटक (2Mr−3 cos θ, Mr−2 sin θ, 0) हैं, जहाँ M एक स्थिरांक है। दर्शाइये कि वेग, विभव प्रकार का है। वेग विभव तथा धारारेखाओं के समीकरण ज्ञात कीजिये।
Worked solution
[As printed, qθ = Mr−2sin θ is dimensionally inconsistent with qr and fails continuity; the intended field (2Mr−3cos θ, Mr−3sin θ, 0) is solved.]
Continuity: (1/r2)∂(r2qr)/∂r + (1/(r sin θ))∂(qθsin θ)/∂θ = −2M cos θ/r4 + 2M cos θ/r4 = 0.
Curl: (1/r)[∂(rqθ)/∂r − ∂qr/∂θ] = (1/r)[−2M sin θ/r3 + 2M sin θ/r3] = 0, so the velocity is of potential kind; −∂φ/∂r = 2M cos θ/r3, −(1/r)∂φ/∂θ = M sin θ/r3 give φ = M cos θ/r2.
Streamlines: dr/(2cos θ) = r dθ/sin θ, so dr/r = 2cot θ dθ and r = c sin2θ, lying in the planes ψ = constant.
Answer: φ = M cos θ/r2 (a three-dimensional doublet); streamlines r = c sin2θ, ψ = constant.Final answer: φ = M cos θ/r2 (a three-dimensional doublet); streamlines r = c sin2θ, ψ = constant.
Question 6 (50 marks)
Q6(a) · Partial Differential Equations · 20 marks≈ 14 min
Vibrating String, Heat Equation and Laplace Equation
Solve the heat equation ∂u/∂t = ∂2u/∂x2, 0 < x < l, t > 0 subject to the conditions
u(0, t) = u(l, t) = 0
u(x, 0) = x(l − x), 0 ≤ x ≤ lहिंदी में प्रश्न
ऊष्मा समीकरण ∂u/∂t = ∂2u/∂x2, 0 < x < l, t > 0 का शर्तों
u(0, t) = u(l, t) = 0
u(x, 0) = x(l − x), 0 ≤ x ≤ l
से प्रतिबन्धित हल ज्ञात कीजिये।Q6(b) · Numerical Analysis and Computer Programming · 15 marks≈ 11 min
Computer Systems, Logic Gates, Boolean Algebra and Algorithms
Find a combinatorial circuit corresponding to the Boolean function f(x, y, z) = [x · (ȳ + z)] + y and write the input/output table for the circuit.
हिंदी में प्रश्न
बूलीय फलन f(x, y, z) = [x · (ȳ + z)] + y का संगत संचयविन्यास परिपथ (कॉम्बिनेटोरियल सर्किट) ज्ञात कीजिये तथा परिपथ के लिये निवेश/निर्गत (इनपुट/आउटपुट) सारणी लिखिये।
Q6(c) · Mechanics and Fluid Dynamics · 15 marks≈ 11 min
Moment of Inertia and Motion of Rigid Bodies in Two Dimensions
Find the moment of inertia of a right circular solid cone about one of its slant sides (generator) in terms of its mass M, height h and the radius of base as a.
हिंदी में प्रश्न
एक लम्ब वृत्तीय ठोस शंकु का उसकी संहति M, ऊँचाई h तथा आधार की त्रिज्या a के रूप में उसकी एक तिर्यक् रेखा (जनक रेखा) के सापेक्ष जड़त्व-आघूर्ण ज्ञात कीजिये।
Worked solution
Vertex O as origin, axis Oz; discs at height z have radius az/h, and for the solid cone ∫(x2 + y2) dm = (3/10)Ma2, ∫x2 dm = (3/20)Ma2, ∫z2 dm = (3/5)Mh2.
At O: C = (3/10)Ma2 about the axis, A = B = (3/20)Ma2 + (3/5)Mh2 = (3/20)M(a2 + 4h2), and the products of inertia vanish by symmetry.
A generator through O makes angle α with the axis, sin α = a/√(a2 + h2), cos α = h/√(a2 + h2); taking it in the xz-plane, I = A sin2α + C cos2α.
I = [(3/20)M(a2 + 4h2)a2 + (3/10)Ma2h2]/(a2 + h2).
Answer: I = 3Ma2(a2 + 6h2)/(20(a2 + h2)).Final answer: I = 3Ma2(a2 + 6h2)/(20(a2 + h2)).
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 + DD′ − 6D′2) z = x2 sin(x + y) where D ≡ ∂/∂x and D′ ≡ ∂/∂y.
हिंदी में प्रश्न
आंशिक अवकल समीकरण (D2 + DD′ − 6D′2) z = x2 sin(x + y) जहाँ D ≡ ∂/∂x तथा D′ ≡ ∂/∂y, का व्यापक हल ज्ञात कीजिये।
Worked solution
F = (D − 2D′)(D + 3D′), so CF = φ1(y + 2x) + φ2(y − 3x).
(D + 3D′)−1x2sin(x + y) = ∫x2sin(4x + c) dx with c = y − 3x, which gives g = −(x2/4)cos(x + y) + (x/8)sin(x + y) + (1/32)cos(x + y).
(D − 2D′)−1g = ∫g(x, c − 2x) dx with c = y + 2x, where x + y becomes c − x.
Using ∫x2cos(c − x) dx = −x2sin(c − x) + 2x cos(c − x) + 2 sin(c − x) and ∫x sin(c − x) dx = x cos(c − x) + sin(c − x), the integral is (x2/4)sin(c − x) − (3x/8)cos(c − x) − (13/32)sin(c − x).
Answer: z = φ1(y + 2x) + φ2(y − 3x) + (x2/4)sin(x + y) − (3x/8)cos(x + y) − (13/32)sin(x + y) (verified by substitution).Final answer: z = φ1(y + 2x) + φ2(y − 3x) + (x2/4)sin(x + y) − (3x/8)cos(x + y) − (13/32)sin(x + y) (verified by substitution).
Q7(b) · Numerical Analysis and Computer Programming · 15 marks≈ 11 min
Numerical Integration and Numerical Solution of ODEs
The velocity of a train which starts from rest is given by the following table, the time being reckoned in minutes from the start and the velocity in km/hour:
t (minutes): 2, 4, 6, 8, 10, 12, 14, 16, 18, 20
v (km/hour): 16, 28·8, 40, 46·4, 51·2, 32, 17·6, 8, 3·2, 0
Using Simpson's 1/3rd rule, estimate approximately in km the total distance run in 20 minutes.हिंदी में प्रश्न
एक रेलगाड़ी, जो कि विश्राम से चलना प्रारंभ करती है, का वेग निम्नलिखित सारणी द्वारा दिया गया है। प्रस्थान से समय की गणना मिनट में तथा वेग की कि॰ मी॰/घण्टा में की गयी है:
t (मिनट): 2, 4, 6, 8, 10, 12, 14, 16, 18, 20
v (कि॰ मी॰/घण्टा): 16, 28·8, 40, 46·4, 51·2, 32, 17·6, 8, 3·2, 0
सिम्पसन के 1/3 नियम का उपयोग करके 20 मिनट में तय की गयी कुल दूरी (लगभग) का आकलन कि॰ मी॰ में कीजिये।Worked solution
v = 0 at t = 0 (start from rest) and at t = 20; h = 2 min = 1/30 hour and n = 10.
Odd ordinates: 16 + 40 + 51.2 + 17.6 + 3.2 = 128; even ordinates: 28.8 + 46.4 + 32 + 8 = 115.2.
Answer: distance = (1/90)[0 + 0 + 4(128) + 2(115.2)] = 742.4/90 ≈ 8.25 km.Final answer: distance = (1/90)[0 + 0 + 4(128) + 2(115.2)] = 742.4/90 ≈ 8.25 km.
Q7(c) · Mechanics and Fluid Dynamics · 20 marks≈ 14 min
Two-Dimensional and Axisymmetric Motion: Sources, Sinks and Vortices
Two point vortices each of strength k are situated at (±a, 0) and a point vortex of strength −k/2 is situated at the origin. Show that the fluid motion is stationary and also find the equations of streamlines. If the streamlines, which pass through the stagnation points, meet the x-axis at (±b, 0), then show that 3√3 (b2 − a2)2 = 16a3b.
हिंदी में प्रश्न
दो बिन्दु भ्रमिल, जहाँ प्रत्येक का सामर्थ्य k है, (±a, 0) पर स्थित हैं तथा −k/2 सामर्थ्य का एक बिन्दु भ्रमिल, मूलबिन्दु पर स्थित है। दर्शाइये कि तरल गति अचल है तथा धारारेखाओं के समीकरण भी ज्ञात कीजिये। यदि धारारेखाएँ, जो कि प्रगतिरोध बिन्दुओं (स्टैगनेशन पॉइन्ट) से गुजरती हैं, x-अक्ष पर (±b, 0) पर मिलती हैं, तब दर्शाइये कि 3√3 (b2 − a2)2 = 16a3b.
Worked solution
w = ik[log(z − a) + log(z + a) − (1/2)log z]. The vortex at a moves with the velocity of the others: −u + iv = ik/(2a) − (ik/2)/a = 0; by symmetry the vortex at −a is also at rest, and at O the effects of ±a cancel, so the motion is stationary.
ψ = k[log r1 + log r2 − (1/2)log r], so the streamlines are r12r22 = cr, i.e. [(x2 + y2 + a2)2 − 4a2x2]2 = c2(x2 + y2).
Stagnation points: dw/dz = ik(3z2 + a2)/(2z(z2 − a2)) = 0 at z = ±ia/√3, where r12r22 = |z2 − a2|2 = 16a4/9 and r = a/√3, so c = 16√3a3/9.
At (b, 0), r12r22 = (b2 − a2)2 and r = b, so (b2 − a2)2 = 16√3a3b/9.
Answer: the motion is stationary; streamlines r12r22 = cr; multiplying the last relation by 3√3 gives 3√3(b2 − a2)2 = 16a3b.Final answer: the motion is stationary; streamlines r12r22 = cr; multiplying the last relation by 3√3 gives 3√3(b2 − a2)2 = 16a3b.
Question 8 (50 marks)
Q8(a) · Partial Differential Equations · 15 marks≈ 11 min
Classification and Canonical Forms of Second-Order PDEs
Reduce the following partial differential equation to a canonical form and hence solve it: yuxx + (x + y)uxy + xuyy = 0
हिंदी में प्रश्न
निम्नलिखित आंशिक अवकल समीकरण yuxx + (x + y)uxy + xuyy = 0 को विहित रूप में समानीत कीजिये और अतएव इसको हल कीजिये।
Worked solution
A = y, B = x + y, C = x: B2 − 4AC = (x − y)2 > 0 for x ≠ y, hyperbolic; y(dy/dx)2 − (x + y)dy/dx + x = (y dy/dx − x)(dy/dx − 1) = 0 gives ξ = y − x, η = y2 − x2.
ux = −uξ − 2xuη, uy = uξ + 2yuη; uxx = uξξ + 4xuξη + 4x2uηη − 2uη, uxy = −uξξ − 2(x + y)uξη − 4xyuηη, uyy = uξξ + 4yuξη + 4y2uηη + 2uη.
Substituting, the uξξ and uηη coefficients vanish, uξη gets 8xy − 2(x + y)2 = −2(x − y)2 and uη gets 2(x − y).
With x − y = −ξ: canonical form uξη + (1/ξ)uη = 0, i.e. ∂(ξuη)/∂ξ = 0, so ξuη = F′(η) and u = F(η)/ξ + G(ξ).
Answer: uξη = −uη/ξ (ξ = y − x, η = y2 − x2); u = F(y2 − x2)/(y − x) + G(y − x). UPSC 2026 (Q6a) set the same equation.Final answer: uξη = −uη/ξ (ξ = y − x, η = y2 − x2); u = F(y2 − x2)/(y − x) + G(y − x). UPSC 2026 (Q6a) set the same equation.
Q8(b) · Numerical Analysis and Computer Programming · 15 marks≈ 11 min
Numerical Integration and Numerical Solution of ODEs
Using Runge-Kutta method of fourth order, solve the differential equation dy/dx = x + y2 with y(0) = 1, at x = 0·2. Use four decimal places for calculation and step length 0·1.
हिंदी में प्रश्न
चतुर्थ कोटि की रुन्गे-कुट्टा विधि का उपयोग करके अवकल समीकरण dy/dx = x + y2, जबकि y(0) = 1 है, को x = 0·2 पर हल कीजिये। परिकलन में दशमलव के चार स्थानों तक तथा पग लम्बाई (स्टेप लेंथ) 0·1 का उपयोग कीजिये।
Q8(c) · Mechanics and Fluid Dynamics · 20 marks≈ 14 min
Two-Dimensional and Axisymmetric Motion: Sources, Sinks and Vortices
Verify that w = ik log{(z − ia)/(z + ia)} is the complex potential of a steady flow of fluid about a circular cylinder, where the plane y = 0 is a rigid boundary. Find also the force exerted by the fluid on unit length of the cylinder.
हिंदी में प्रश्न
सत्यापित कीजिये कि एक वृत्ताकार बेलन के इर्द-गिर्द एक तरल के अपरिवर्ती प्रवाह का सम्मिश्र विभव w = ik log{(z − ia)/(z + ia)} है, जहाँ समतल y = 0 एक दृढ़ सीमा है। बेलन की एकक लम्बाई (यूनिट लेंथ) पर तरल द्वारा लगाये गये बल को भी ज्ञात कीजिये।
Worked solution
With z − ia = r1eiθ1, z + ia = r2eiθ2: ψ = k log(r1/r2); on y = 0, r1 = r2 and ψ = 0, so the plane y = 0 is a streamline (rigid boundary).
Each circle r1/r2 = λ (0 < λ < 1) is a streamline round z = ia; taking one as the fixed cylinder, w is a steady flow outside it in y > 0: circulation 2πk round the cylinder, with its image −k at −ia in the wall.
Blasius (the contour encloses only z = ia): (dw/dz)2 = −k2[1/(z − ia)2 + 1/(z + ia)2 − 2/((z − ia)(z + ia))], whose residue at ia is −k2(−2/(2ia)) = −ik2/a.
So ∮(dw/dz)2dz = 2πi(−ik2/a) = 2πk2/a and X − iY = (iρ/2)(2πk2/a) = iπρk2/a: X = 0, Y = −πρk2/a.
Answer: w is the stated steady flow; the force per unit length is πρk2/a, perpendicular to the wall and towards it (the cylinder is attracted to the wall).Final answer: w is the stated steady flow; the force per unit length is πρk2/a, perpendicular to the wall and towards it (the cylinder is attracted to the wall).