UPSC Mains 2019 Mathematics Optional Question Papers with Solutions

All 59 questions of the 2019 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 2019 Mathematics Paper I

Question 1 (50 marks)

Q1(a) · Calculus · 10 marks≈ 7 min
Real Numbers, Limits, Continuity and Differentiability
Let f : [0, π/2] → ℝ be a continuous function such that
f(x) = cos²x/(4x² − π²), 0 ≤ x < π/2.
Find the value of f(π/2).
हिंदी में प्रश्न
माना कि f : [0, π/2] → ℝ एक संतत फलन है, जैसा कि
f(x) = cos²x/(4x² − π²), 0 ≤ x < π/2.
f(π/2) का मान ज्ञात कीजिए।
Practise this question Calculus notes
Q1(b) · Calculus · 10 marks≈ 7 min
Functions of Several Variables: Partial Derivatives, Extrema, Lagrange Multipliers and Jacobians
Let f : D (⊆ ℝ²) → ℝ be a function and (a, b) ∈ D. If f(x, y) is continuous at (a, b), then show that the functions f(x, b) and f(a, y) are continuous at x = a and at y = b respectively.
हिंदी में प्रश्न
माना कि f : D (⊆ ℝ²) → ℝ एक फलन है और (a, b) ∈ D. अगर f(x, y) बिंदु (a, b) पर संतत है, तो दर्शाइए कि फलन f(x, b) और f(a, y) क्रमशः x = a और y = b पर संतत हैं।
Practise this question Calculus notes
Q1(c) · Linear Algebra · 10 marks≈ 7 min
Linear Transformations, Rank-Nullity and Matrix Representation
Let T : ℝ² → ℝ² be a linear map such that T(2, 1) = (5, 7) and T(1, 2) = (3, 3). If A is the matrix corresponding to T with respect to the standard bases e₁, e₂, then find Rank (A).
हिंदी में प्रश्न
माना कि T : ℝ² → ℝ² एक रैखिक प्रतिचित्र है, जैसा कि T(2, 1) = (5, 7) एवं T(1, 2) = (3, 3). अगर A मानक आधारों e₁, e₂ के सापेक्ष T के संगत आव्यूह है, तो A की कोटि ज्ञात कीजिए।
Practise this question Linear Algebra notes
Q1(d) · Linear Algebra · 10 marks≈ 7 min
Systems of Linear Equations
If
A = [[1, 2, 1], [1, −4, 1], [3, 0, −3]] and B = [[2, 1, 1], [1, −1, 0], [2, 1, −1]]
then show that AB = 6I₃. Use this result to solve the following system of equations :
2x + y + z = 5
x − y = 0
2x + y − z = 1
हिंदी में प्रश्न
अगर
A = [[1, 2, 1], [1, −4, 1], [3, 0, −3]] और B = [[2, 1, 1], [1, −1, 0], [2, 1, −1]]
है, तो दर्शाइए कि AB = 6I₃. इस परिणाम का उपयोग करते हुए निम्नलिखित समीकरण निकाय को हल कीजिए :
2x + y + z = 5
x − y = 0
2x + y − z = 1
Practise this question Linear Algebra notes
Worked solution
Row 1 of A against the columns of B: (2 + 2 + 2, 1 − 2 + 1, 1 + 0 − 1) = (6, 0, 0); row 2: (2 − 4 + 2, 1 + 4 + 1, 1 + 0 − 1) = (0, 6, 0); row 3: (6 + 0 − 6, 3 + 0 − 3, 3 + 0 + 3) = (0, 0, 6). So AB = 6I₃.
Hence B⁻¹ = A/6. The given system is BX = C with C = (5, 0, 1)^T, since its coefficient matrix is exactly B.
X = B⁻¹C = (1/6)AC = (1/6)(5 + 0 + 1, 5 − 0 + 1, 15 + 0 − 3) = (1, 1, 2). Check: 2 + 1 + 2 = 5, 1 − 1 = 0, 2 + 1 − 2 = 1.
Answer: x = 1, y = 1, z = 2.
Final answer: x = 1, y = 1, z = 2.
Q1(e) · Analytic Geometry · 10 marks≈ 7 min
Straight Lines and Shortest Distance between Skew Lines
Show that the lines
(x + 1)/(−3) = (y − 3)/2 = (z + 2)/1 and x/1 = (y − 7)/(−3) = (z + 7)/2
intersect. Find the coordinates of the point of intersection and the equation of the plane containing them.
हिंदी में प्रश्न
दर्शाइए कि
(x + 1)/(−3) = (y − 3)/2 = (z + 2)/1 और x/1 = (y − 7)/(−3) = (z + 7)/2
प्रतिच्छेदी रेखाएँ हैं। प्रतिच्छेद बिंदु के निर्देशांकों और उस समतल, जिसमें दोनों रेखाएँ हैं, का समीकरण ज्ञात कीजिए।
Practise this question Analytic Geometry notes

Question 2 (50 marks)

Q2(a) · Calculus · 15 marks≈ 11 min
Real Numbers, Limits, Continuity and Differentiability
Is f(x) = |cos x| + |sin x| differentiable at x = π/2 ? If yes, then find its derivative at x = π/2. If no, then give a proof of it.
हिंदी में प्रश्न
क्या f(x) = |cos x| + |sin x|, x = π/2 पर अवकलनीय है? अगर आपका उत्तर हाँ है, तो f(x) का अवकलज x = π/2 पर ज्ञात कीजिए। अगर आपका उत्तर ना है, तो अपने उत्तर का प्रमाण दीजिए।
Practise this question Calculus notes
Worked solution
Near x = π/2, sin x > 0, so |sin x| = sin x, while cos x > 0 to the left of π/2 and cos x < 0 to the right. Also f(π/2) = 0 + 1 = 1.
With x = π/2 + h: cos x = −sin h and sin x = cos h. For h < 0, |cos x| = −sin h, so (f(π/2 + h) − 1)/h = (cos h − 1)/h − (sin h)/h → 0 − 1 = −1; thus Lf′(π/2) = −1.
For h > 0, |cos x| = sin h, so (f(π/2 + h) − 1)/h = (sin h)/h + (cos h − 1)/h → 1; thus Rf′(π/2) = 1.
Lf′(π/2) ≠ Rf′(π/2), so f is not differentiable at π/2 (it is continuous there, with a corner).
Answer: No: the left derivative is −1 and the right derivative is 1.
Final answer: No: the left derivative is −1 and the right derivative is 1.
Q2(b) · Linear Algebra · 15 marks≈ 11 min
Symmetric, Hermitian, Orthogonal and Unitary Matrices and Their Eigenvalues
Let A and B be two orthogonal matrices of same order and det A + det B = 0. Show that A + B is a singular matrix.
हिंदी में प्रश्न
माना कि A और B समान कोटि के दो लांबिक आव्यूह हैं तथा det A + det B = 0. दर्शाइए कि A + B एक अव्युत्क्रमणीय (सिंगुलर) आव्यूह है।
Practise this question Linear Algebra notes
Worked solution
For an orthogonal matrix AA^T = A^TA = I, so (det A)² = 1 and det A = ±1; similarly det B = ±1. With det A + det B = 0, one is 1 and the other −1, so det A · det B = −1.
A(A^T + B^T)B = (AA^T)B + A(B^TB) = B + A, and A^T + B^T = (A + B)^T.
Taking determinants: det(A + B) = det A · det(A + B)^T · det B = (det A · det B) det(A + B) = −det(A + B).
Hence 2 det(A + B) = 0, so det(A + B) = 0.
Answer: det(A + B) = 0, so A + B is singular.
Final answer: det(A + B) = 0, so A + B is singular.
Q2(c) · Analytic Geometry · 20 marks≈ 14 min
The Sphere
(i) The plane x + 2y + 3z = 12 cuts the axes of coordinates in A, B, C. Find the equations of the circle circumscribing the triangle ABC. (10 marks)
(ii) Prove that the plane z = 0 cuts the enveloping cone of the sphere x² + y² + z² = 11 which has the vertex at (2, 4, 1) in a rectangular hyperbola. (10 marks)
हिंदी में प्रश्न
(i) समतल x + 2y + 3z = 12 निर्देशांक अक्षों को A, B, C पर प्रतिच्छेद करता है। त्रिभुज ABC के परिवृत्त का समीकरण ज्ञात कीजिए। (10 अंक)
(ii) सिद्ध कीजिए कि समतल z = 0 गोलक x² + y² + z² = 11 के अन्वालोपी शंकु, जिसका शीर्ष (2, 4, 1) पर है, को एक समकोणीय अतिपरवलय पर प्रतिच्छेद करता है। (10 अंक)
Practise this question Analytic Geometry notes
Worked solution
(i)
A = (12, 0, 0), B = (0, 6, 0), C = (0, 0, 4). The sphere through O, A, B, C is x² + y² + z² − 12x − 6y − 4z = 0.
The circumcircle of ABC is the section of this sphere by the plane ABC.
Answer: x² + y² + z² − 12x − 6y − 4z = 0, x + 2y + 3z = 12.
(ii)
S = x² + y² + z² − 11, S₁ = 4 + 16 + 1 − 11 = 10, T = 2x + 4y + z − 11. The enveloping cone is SS₁ = T²: 10(x² + y² + z² − 11) = (2x + 4y + z − 11)².
Put z = 0: 10x² + 10y² − 110 = 4x² + 16y² + 16xy − 44x − 88y + 121, that is 6x² − 16xy − 6y² + 44x + 88y − 231 = 0.
Here h² − ab = 64 + 36 > 0, so the conic is a hyperbola (its 3 × 3 discriminant is non-zero, so it is not a line pair), and the coefficients of x² and y² add to 6 − 6 = 0.
Answer: the section is the rectangular hyperbola 6x² − 16xy − 6y² + 44x + 88y − 231 = 0, z = 0; hence proved.
Final answer: (i) x² + y² + z² − 12x − 6y − 4z = 0, x + 2y + 3z = 12; (ii) the section is the rectangular hyperbola 6x² − 16xy − 6y² + 44x + 88y − 231 = 0, z = 0; hence proved

Question 3 (50 marks)

Q3(a) · Calculus · 15 marks≈ 11 min
Maxima and Minima, Asymptotes and Curve Tracing
Find the maximum and the minimum value of the function f(x) = 2x³ − 9x² + 12x + 6 on the interval [2, 3].
हिंदी में प्रश्न
फलन f(x) = 2x³ − 9x² + 12x + 6 का अंतराल [2, 3] पर अधिकतम और न्यूनतम मान ज्ञात कीजिए।
Practise this question Calculus notes
Worked solution
f′(x) = 6x² − 18x + 12 = 6(x − 1)(x − 2), which vanishes at x = 1, 2 and is positive for x > 2.
So f is increasing on [2, 3] (x = 2 is an end point, not an interior extremum), and the extremes are at the ends: f(2) = 16 − 36 + 24 + 6 = 10, f(3) = 54 − 81 + 36 + 6 = 15.
Answer: Minimum 10 (at x = 2), maximum 15 (at x = 3).
Final answer: Minimum 10 (at x = 2), maximum 15 (at x = 3).
Q3(b) · Analytic Geometry · 15 marks≈ 11 min
Ellipsoid, Hyperboloids and Paraboloids
Prove that, in general, three normals can be drawn from a given point to the paraboloid x² + y² = 2az, but if the point lies on the surface
27a(x² + y²) + 8(a − z)³ = 0
then two of the three normals coincide.
हिंदी में प्रश्न
सिद्ध कीजिए कि साधारणतः किसी एक बिंदु से परवलयज x² + y² = 2az पर तीन अभिलंब बनाए जा सकते हैं, लेकिन अगर बिंदु सतह 27a(x² + y²) + 8(a − z)³ = 0 पर स्थित है, तो इन तीन अभिलंबों में से दो अभिलंब एक ही हैं।
Practise this question Analytic Geometry notes
Worked solution
The normal at (x₁, y₁, z₁) to x² + y² = 2az is (x − x₁)/x₁ = (y − y₁)/y₁ = (z − z₁)/(−a). If it passes through (α, β, γ), with common ratio t: x₁ = α/(1 + t), y₁ = β/(1 + t), z₁ = γ + at.
The foot lies on the paraboloid: f(t) = 2a(γ + at)(1 + t)² − (α² + β²) = 0, a cubic in t; each root gives one normal, so in general three normals pass through (α, β, γ).
Two normals coincide when f has a double root: f′(t) = 2a(1 + t)[3at + a + 2γ] = 0. The root t = −1 would need α² + β² = 0, so t = −(a + 2γ)/(3a), giving 1 + t = 2(a − γ)/(3a) and γ + at = −(a − γ)/3.
Then f(t) = 0 becomes −2a·(a − γ)/3·4(a − γ)²/(9a²) = α² + β², that is 27a(α² + β²) + 8(a − γ)³ = 0.
Answer: two of the three normals coincide exactly when (α, β, γ) lies on 27a(x² + y²) + 8(a − z)³ = 0; hence proved.
Final answer: two of the three normals coincide exactly when (α, β, γ) lies on 27a(x² + y²) + 8(a − z)³ = 0; hence proved.
Q3(c) · Linear Algebra · 20 marks≈ 14 min
Algebra of Matrices, Echelon Form, Rank, Inverse, Congruence and Similarity
Let
A = [[5, 7, 2, 1], [1, 1, −8, 1], [2, 3, 5, 0], [3, 4, −3, 1]]
(i) Find the rank of matrix A.
(ii) Find the dimension of the subspace
V = {(x₁, x₂, x₃, x₄) ∈ ℝ⁴ | A[[x₁], [x₂], [x₃], [x₄]] = 0}
(15 + 5 = 20 marks)
हिंदी में प्रश्न
माना कि
A = [[5, 7, 2, 1], [1, 1, −8, 1], [2, 3, 5, 0], [3, 4, −3, 1]]
(i) आव्यूह A की कोटि ज्ञात कीजिए।
(ii) उपसमष्टि
V = {(x₁, x₂, x₃, x₄) ∈ ℝ⁴ | A[[x₁], [x₂], [x₃], [x₄]] = 0}
की विमा ज्ञात कीजिए। (15 + 5 = 20 अंक)
Practise this question Linear Algebra notes

Question 4 (50 marks)

Q4(a) · Linear Algebra · 15 marks≈ 11 min
Eigenvalues, Eigenvectors and the Cayley-Hamilton Theorem
State the Cayley-Hamilton theorem. Use this theorem to find A¹⁰⁰, where
A = [[1, 0, 0], [1, 0, 1], [0, 1, 0]]
हिंदी में प्रश्न
कैले-हैमिल्टन प्रमेय का कथन लिखिए। इस प्रमेय का उपयोग करके A¹⁰⁰ का मान ज्ञात कीजिए, जहाँ
A = [[1, 0, 0], [1, 0, 1], [0, 1, 0]]
Practise this question Linear Algebra notes
Worked solution
Cayley-Hamilton theorem: every square matrix satisfies its characteristic equation; if det(λI − A) = λⁿ + c₁λⁿ⁻¹ + ... + c_n, then Aⁿ + c₁Aⁿ⁻¹ + ... + c_nI = 0.
Here det(λI − A) = λ³ − λ² − λ + 1, so A³ = A² + A − I and (multiplying by Aⁿ⁻³ and inducting) Aⁿ = Aⁿ⁻² + A² − I for n ≥ 3.
Hence A^2k = kA² − (k − 1)I, and with k = 50, A¹⁰⁰ = 50A² − 49I.
Answer: A¹⁰⁰ = [[1, 0, 0], [50, 1, 0], [50, 0, 1]].
Final answer: A¹⁰⁰ = [[1, 0, 0], [50, 1, 0], [50, 0, 1]].
Q4(b) · Analytic Geometry · 15 marks≈ 11 min
Ellipsoid, Hyperboloids and Paraboloids
Find the length of the normal chord through a point P of the ellipsoid
x²/a² + y²/b² + z²/c² = 1
and prove that if it is equal to 4PG₃, where G₃ is the point where the normal chord through P meets the xy-plane, then P lies on the cone
(x²/a⁶)(2c² − a²) + (y²/b⁶)(2c² − b²) + z²/c⁴ = 0
हिंदी में प्रश्न
बिंदु P से गुजरने वाली दीर्घवृत्तज
x²/a² + y²/b² + z²/c² = 1
की अभिलंब जीवा की लंबाई ज्ञात कीजिए और सिद्ध कीजिए कि अगर यह 4PG₃ के समान है, जहाँ G₃ वह बिंदु है जहाँ P से गुजरने वाली अभिलंब जीवा xy-तल पर मिलती है, तो P शंकु
(x²/a⁶)(2c² − a²) + (y²/b⁶)(2c² − b²) + z²/c⁴ = 0
पर स्थित है।
Practise this question Analytic Geometry notes
Worked solution
Let the normal at P(x₁, y₁, z₁) have direction cosines (l, m, n) = p(x₁/a², y₁/b², z₁/c²), with 1/p² = ∑x₁²/a⁴. The point at distance r along it is (x₁ + lr, y₁ + mr, z₁ + nr).
It lies on the ellipsoid when 2r∑(lx₁/a²) + r²∑(l²/a²) = 0. Here ∑lx₁/a² = p∑x₁²/a⁴ = 1/p and ∑l²/a² = p²∑x₁²/a⁶.
So the length of the normal chord is |r| = 2/(p³∑x₁²/a⁶).
The normal meets z = 0 where z₁ + nr = 0, that is r = −z₁/n = −c²/p, so PG₃ = c²/p.
Chord = 4PG₃ gives 2/(p³∑x₁²/a⁶) = 4c²/p, that is ∑x₁²/a⁴ = 2c²∑x₁²/a⁶ (using 1/p² = ∑x₁²/a⁴).
Collecting terms: (x₁²/a⁶)(2c² − a²) + (y₁²/b⁶)(2c² − b²) + z₁²(2c² − c²)/c⁶ = 0.
Answer: normal chord = 2/(p³∑x₁²/a⁶), and P lies on the cone (x²/a⁶)(2c² − a²) + (y²/b⁶)(2c² − b²) + z²/c⁴ = 0; hence proved.
Final answer: normal chord = 2/(p³∑x₁²/a⁶), and P lies on the cone (x²/a⁶)(2c² − a²) + (y²/b⁶)(2c² − b²) + z²/c⁴ = 0; hence proved.
Q4(c) · Calculus · 20 marks≈ 14 min
Functions of Several Variables: Partial Derivatives, Extrema, Lagrange Multipliers and Jacobians
(i) If
u = sin⁻¹ √((x^1/3 + y^1/3)/(x^1/2 + y^1/2))
then show that sin²u is a homogeneous function of x and y of degree −1/6. Hence show that
x² ∂²u/∂x² + 2xy ∂²u/∂x∂y + y² ∂²u/∂y² = (tan u/12)(13/12 + tan²u/12) (12 marks)
(ii) Using the Jacobian method, show that if f′(x) = 1/(1 + x²) and f(0) = 0, then
f(x) + f(y) = f((x + y)/(1 − xy)) (8 marks)
हिंदी में प्रश्न
(i) अगर
u = sin⁻¹ √((x^1/3 + y^1/3)/(x^1/2 + y^1/2))
है, तो दर्शाइए कि sin²u, x और y का −1/6 घातविशिष्ट समांगी फलन है। अतएव दर्शाइए कि
x² ∂²u/∂x² + 2xy ∂²u/∂x∂y + y² ∂²u/∂y² = (tan u/12)(13/12 + tan²u/12) (12 अंक)
(ii) जैकोबियन विधि का व्यवहार करते हुए दर्शाइए कि अगर f′(x) = 1/(1 + x²) और f(0) = 0 है, तो
f(x) + f(y) = f((x + y)/(1 − xy)) (8 अंक)
Practise this question Calculus notes
Worked solution
(i) z = sin²u = (x^1/3 + y^1/3)/(x^1/2 + y^1/2); replacing x, y by tx, ty multiplies it by t^1/3/t^1/2 = t^−1/6, so z is homogeneous of degree −1/6.
Euler: xz_x + yz_y = −z/6. Since z_x = 2 sin u cos u · u_x (similarly for y), 2 sin u cos u (xu_x + yu_y) = −sin²u/6, so xu_x + yu_y = −tan u/12 = g(u).
Differentiating xu_x + yu_y = g(u) w.r.t. x and y, multiplying by x and y and adding: x²u_xx + 2xyu_xy + y²u_yy = g(u)(g′(u) − 1) = (−tan u/12)(−sec²u/12 − 1) = (tan u/12)(1 + (1 + tan²u)/12) = (tan u/12)(13/12 + tan²u/12).
(ii) Let u = f(x) + f(y), v = (x + y)/(1 − xy). Then u_x = 1/(1 + x²), u_y = 1/(1 + y²), v_x = (1 + y²)/(1 − xy)², v_y = (1 + x²)/(1 − xy)².
∂(u, v)/∂(x, y) = u_xv_y − u_yv_x = 1/(1 − xy)² − 1/(1 − xy)² = 0, so u and v are functionally related: u = φ(v).
Put y = 0: f(x) + f(0) = φ(x), i.e. φ = f since f(0) = 0. Hence f(x) + f(y) = f((x + y)/(1 − xy)).
Answer: Both identities hold as shown (f is in fact tan⁻¹x).
Final answer: Both identities hold as shown (f is in fact tan⁻¹x).

Question 5 (50 marks)

Q5(a) · Ordinary Differential Equations · 10 marks≈ 7 min
Formulation and First Order, First Degree Equations
Solve the differential equation
(2y sin x + 3y⁴ sin x cos x) dx − (4y³ cos²x + cos x) dy = 0
हिंदी में प्रश्न
अवकल समीकरण
(2y sin x + 3y⁴ sin x cos x) dx − (4y³ cos²x + cos x) dy = 0
को हल कीजिए।
Practise this question Ordinary Differential Equations notes
Q5(b) · Ordinary Differential Equations · 10 marks≈ 7 min
Linear Equations with Constant Coefficients
Determine the complete solution of the differential equation
d²y/dx² − 4 dy/dx + 4y = 3x²e^2x sin 2x
हिंदी में प्रश्न
अवकल समीकरण
d²y/dx² − 4 dy/dx + 4y = 3x²e^2x sin 2x
का पूर्ण हल ज्ञात कीजिए।
Practise this question Ordinary Differential Equations notes
Q5(c) · Dynamics and Statics · 10 marks≈ 7 min
Equilibrium, Friction and the Common Catenary
One end of a heavy uniform rod AB can slide along a rough horizontal rod AC, to which it is attached by a ring. B and C are joined by a string. When the rod is on the point of sliding, then AC² − AB² = BC². If θ is the angle between AB and the horizontal line, then prove that the coefficient of friction is cot θ/(2 + cot²θ).
हिंदी में प्रश्न
एक भारी एकसमान छड़ AB का एक सिरा एक रूक्ष क्षैतिज छड़ AC, जिसके साथ वह वलय (रिंग) के द्वारा जुड़ी हुई है, पर सरक सकती है। B एवं C एक रस्सी से जुड़े हैं। जब छड़ सर्पण बिंदु पर है, तब AC² − AB² = BC² है। यदि क्षैतिज रेखा व AB के बीच का कोण θ है, तो सिद्ध कीजिए कि घर्षण गुणांक cot θ/(2 + cot²θ) है।
Practise this question Dynamics and Statics notes
Worked solution
AC² = AB² + BC² means ∠ABC = 90°, so the string BC is perpendicular to the rod.
Let AB = 2a and its weight be W; at A act the normal reaction R (vertical) and friction F (along AC), and T acts at B along BC.
Moments about A: T × 2a = W × a cos θ, so T = (W/2) cos θ.
T is perpendicular to AB, so it makes angle θ with the vertical: resolving, F = T sin θ = (W/2) sin θ cos θ and R = W − T cos θ = W(1 − (1/2)cos²θ).
μ = F/R = sin θ cos θ/(2 − cos²θ); dividing above and below by sin²θ, μ = cot θ/(2 cosec²θ − cot²θ) = cot θ/(2 + cot²θ).
Answer: μ = cot θ/(2 + cot²θ).
Final answer: μ = cot θ/(2 + cot²θ).
Q5(d) · Dynamics and Statics · 10 marks≈ 7 min
Constrained Motion, Work and Energy
The force of attraction of a particle by the earth is inversely proportional to the square of its distance from the earth's centre. A particle, whose weight on the surface of the earth is W, falls to the surface of the earth from a height 3h above it. Show that the magnitude of work done by the earth's attraction force is (3/4)hW, where h is the radius of the earth.
हिंदी में प्रश्न
एक कण का पृथ्वी द्वारा आकर्षण बल उस कण के पृथ्वी के केन्द्र से दूरी के वर्ग के व्युत्क्रमानुपाती है। एक कण, जिसका भार पृथ्वी की सतह पर W है, सतह से 3h ऊँचाई से पृथ्वी की सतह पर गिरता है। दर्शाइए कि पृथ्वी के आकर्षण बल द्वारा किए गए कार्य का परिमाण (3/4)hW है, जहाँ h पृथ्वी की त्रिज्या है।
Practise this question Dynamics and Statics notes
Worked solution
At distance x from the centre the attraction is Wh²/x², since it equals W at the surface x = h.
The particle falls from x = h + 3h = 4h to x = h, and the force acts along the motion.
Work = ∫_h^4h Wh²/x² dx = Wh²(1/h − 1/(4h)) = (3/4)Wh.
Answer: the work done by the earth's attraction is (3/4)hW (it would be 3hW if gravity were uniform).
Final answer: the work done by the earth's attraction is (3/4)hW (it would be 3hW if gravity were uniform).
Q5(e) · Vector Analysis · 10 marks≈ 7 min
Gradient, Divergence and Curl in Cartesian and Cylindrical Coordinates
Find the directional derivative of the function xy² + yz² + zx² along the tangent to the curve x = t, y = t², z = t³ at the point (1, 1, 1).
हिंदी में प्रश्न
वक्र x = t, y = t², z = t³ के बिंदु (1, 1, 1) पर स्पर्श-रेखा की दिशा में फलन xy² + yz² + zx² का दिशात्मक अवकलज ज्ञात कीजिए।
Practise this question Vector Analysis notes
Worked solution
∇φ = (y² + 2zx, 2xy + z², 2yz + x²) = (3, 3, 3) at (1, 1, 1).
The point is t = 1, where the tangent r′ = (1, 2t, 3t²) = (1, 2, 3) has length √14.
Directional derivative = (3, 3, 3)·(1, 2, 3)/√14 = 18/√14.
Answer: 18/√14 = 9√14/7 ≈ 4.81.
Final answer: 18/√14 = 9√14/7 ≈ 4.81.

Question 6 (50 marks)

Q6(a) · Dynamics and Statics · 15 marks≈ 11 min
Virtual Work, Stability and Forces in Three Dimensions
A body consists of a cone and underlying hemisphere. The base of the cone and the top of the hemisphere have same radius a. The whole body rests on a rough horizontal table with hemisphere in contact with the table. Show that the greatest height of the cone, so that the equilibrium may be stable, is √3 a.
हिंदी में प्रश्न
एक पिण्ड एक शंकु और उसके नीचे अर्धगोले से बना है। शंकु के आधार तथा अर्धगोले के शिखर का अर्धव्यास a है। पूरा पिण्ड एक रूक्ष क्षैतिज मेज पर रखा है, जिसका अर्धगोला मेज को स्पर्श करता है। दर्शाइए कि शंकु की अधिकतम ऊँचाई, जिससे कि साम्यावस्था स्थिर बनी रहे, √3 a है।
Practise this question Dynamics and Statics notes
Q6(b) · Vector Analysis · 15 marks≈ 11 min
Gauss and Stokes' Theorems and Green's Identities
Find the circulation of F round the curve C, where F = (2x + y²)i + (3y − 4x)j and C is the curve y = x² from (0, 0) to (1, 1) and the curve y² = x from (1, 1) to (0, 0).
हिंदी में प्रश्न
वक्र C के चारों तरफ F का परिसंचरण ज्ञात कीजिए, जहाँ F = (2x + y²)i + (3y − 4x)j और C, बिंदु (0, 0) से बिंदु (1, 1) तक वक्र y = x² के द्वारा तथा बिंदु (1, 1) से बिंदु (0, 0) तक वक्र y² = x के द्वारा परिभाषित है।
Practise this question Vector Analysis notes
Worked solution
C goes out along y = x² and back along y² = x, so the region x² ≤ y ≤ √x is on the left: C is anticlockwise.
∂N/∂x − ∂M/∂y = −4 − 2y, so the circulation = ∫₀¹[−4(√x − x²) − (x − x⁴)] dx = −4/3 − 3/10 = −49/30.
Direct check: on y = x², F·dr = (2x + x⁴ + 6x³ − 8x²) dx with integral 1/30; on x = y² (y: 1 → 0), F·dr = (6y³ − 4y² + 3y) dy with integral −5/3; total −49/30.
Answer: −49/30.
Final answer: −49/30.
Q6(c) · Ordinary Differential Equations · 20 marks≈ 14 min
Second Order Equations with Variable Coefficients
(i) Solve the differential equation
d²y/dx² + (3 sin x − cot x) dy/dx + 2y sin²x = e^{−cos x} sin²x (10 marks)
(ii) Find the Laplace transforms of t^−1/2 and t^1/2. Prove that the Laplace transform of t^{n + 1/2}, where n ∈ ℕ, is
Γ(n + 1 + 1/2)/s^{n + 1 + 1/2} (10 marks)
हिंदी में प्रश्न
(i) अवकल समीकरण
d²y/dx² + (3 sin x − cot x) dy/dx + 2y sin²x = e^{−cos x} sin²x
को हल कीजिए। (10 अंक)
(ii) t^−1/2 तथा t^1/2 का लाप्लास रूपांतर ज्ञात कीजिए। सिद्ध कीजिए कि t^{n + 1/2} का लाप्लास रूपांतर
Γ(n + 1 + 1/2)/s^{n + 1 + 1/2}
होता है, जहाँ n ∈ ℕ. (10 अंक)
Practise this question Ordinary Differential Equations notes
Worked solution
(i)
Change the independent variable to z with (z′)² = sin²x, that is z = −cos x: z′ = sin x, z″ = cos x.
P₁ = (z″ + Pz′)/(z′)² = (cos x + 3 sin²x − cos x)/sin²x = 3, Q₁ = 2 sin²x/sin²x = 2, R₁ = e^{−cos x} = e^z.
d²y/dz² + 3dy/dz + 2y = e^z: CF = c₁e^−z + c₂e^−2z, PI = e^z/6.
Answer: y = c₁e^{cos x} + c₂e^{2 cos x} + e^{−cos x}/6.
(ii)
For a > −1: L{t^a} = ∫₀^∞ e^−stt^a dt; with st = x (s > 0) this is (1/s^a+1)∫₀^∞ e^−xx^a dx = Γ(a + 1)/s^a+1.
a = −1/2: L{t^−1/2} = Γ(1/2)/s^1/2 = √(π/s). a = 1/2: L{t^1/2} = Γ(3/2)/s^3/2 = √π/(2s^3/2).
a = n + 1/2 (n ∈ ℕ): L{t^{n + 1/2}} = Γ(n + 1 + 1/2)/s^{n + 1 + 1/2}. (By induction instead: L{t·t^{n − 1/2}} = −d/ds[Γ(n + 1/2)s^{−(n + 1/2)}] = (n + 1/2)Γ(n + 1/2)s^{−(n + 3/2)}, and (n + 1/2)Γ(n + 1/2) = Γ(n + 3/2).)
Answer: √(π/s), √π/(2s^3/2), and Γ(n + 3/2)/s^{n + 3/2} as required.
Final answer: (i) y = c₁e^{cos x} + c₂e^{2 cos x} + e^{−cos x}/6; (ii) √(π/s), √π/(2s^3/2), and Γ(n + 3/2)/s^{n + 3/2} as required

Question 7 (50 marks)

Q7(a) · Ordinary Differential Equations · 15 marks≈ 11 min
Second Order Equations with Variable Coefficients
Find the linearly independent solutions of the corresponding homogeneous differential equation of the equation x²y″ − 2xy′ + 2y = x³ sin x and then find the general solution of the given equation by the method of variation of parameters.
हिंदी में प्रश्न
समीकरण x²y″ − 2xy′ + 2y = x³ sin x के संगत समांगी अवकल समीकरण का रेखीय स्वतंत्र हल निकालिए और तब दिए गए समीकरण का प्राचल-विचरण विधि द्वारा सामान्य हल निकालिए।
Practise this question Ordinary Differential Equations notes
Worked solution
The homogeneous equation is Euler-Cauchy: θ(θ − 1) − 2θ + 2 = (θ − 1)(θ − 2), so y₁ = x, y₂ = x², W = x·2x − x²·1 = x².
Standard form: R = x³ sin x/x² = x sin x.
u₁ = −∫x²·x sin x/x² dx = −∫x sin x dx = x cos x − sin x; u₂ = ∫x·x sin x/x² dx = −cos x.
PI = x(x cos x − sin x) − x² cos x = −x sin x.
Answer: y = c₁x + c₂x² − x sin x.
Final answer: y = c₁x + c₂x² − x sin x.
Q7(b) · Vector Analysis · 15 marks≈ 11 min
Curves in Space: Curvature, Torsion and Serret-Frenet Formulae
Find the radius of curvature and radius of torsion of the helix x = a cos u, y = a sin u, z = au tan α.
हिंदी में प्रश्न
कुंडलिनी x = a cos u, y = a sin u, z = au tan α के लिए वक्रता की त्रिज्या तथा विमोटन की त्रिज्या ज्ञात कीजिए।
Practise this question Vector Analysis notes
Q7(c) · Dynamics and Statics · 20 marks≈ 14 min
Rectilinear Motion and Simple Harmonic Motion
A particle moving along the y-axis has an acceleration Fy towards the origin, where F is a positive and even function of y. The periodic time, when the particle vibrates between y = −a and y = a, is T. Show that
2π/√F₁ < T < 2π/√F₂
where F₁ and F₂ are the greatest and the least values of F within the range [−a, a]. Further, show that when a simple pendulum of length l oscillates through 30° on either side of the vertical line, T lies between 2π√(l/g) and 2π√(l/g) √(π/3).
हिंदी में प्रश्न
y-अक्ष की दिशा में गतिमान एक कण का मूलबिंदु की ओर त्वरण Fy है, जहाँ F, y का एक धनात्मक एवं सम फलन है। जब कण y = −a तथा y = a के बीच में कंपन करता है, तब उसका आवर्तकाल T है। दर्शाइए कि
2π/√F₁ < T < 2π/√F₂
जहाँ F₁ एवं F₂ परास [−a, a] में F के अधिकतम एवं न्यूनतम मान हैं। आगे दर्शाइए कि जब लंबाई l का एक सरल लोलक ऊर्ध्वाधर रेखा के किसी भी ओर 30° तक दोलन करता है, तब T, 2π√(l/g) तथा 2π√(l/g) √(π/3) के बीच में रहता है।
Practise this question Dynamics and Statics notes
Worked solution
Here d²y/dt² = −F(y)y with F even and positive, so the motion is symmetric about O and T = 4 × (time from y = 0 to y = a).
Energy: v² = 2∫_y^a F(s)s ds for 0 ≤ y ≤ a.
Since F₂ ≤ F(s) ≤ F₁ and s ≥ 0, F₂(a² − y²) ≤ v² ≤ F₁(a² − y²), with strict inequality on a set of positive length when F is not constant.
Hence T = 4∫₀^a dy/v lies between 4∫₀^a dy/(√F₁ √(a² − y²)) = 2π/√F₁ and 4∫₀^a dy/(√F₂ √(a² − y²)) = 2π/√F₂.
Pendulum: d²θ/dt² = −(g/l) sin θ = −F(θ)θ with F(θ) = (g/l)(sin θ)/θ, which is even and positive on [−π/6, π/6].
(sin θ)/θ decreases as |θ| increases, so F₁ = F(0) = g/l and F₂ = F(π/6) = (g/l)(1/2)/(π/6) = 3g/(πl).
Hence 2π√(l/g) < T < 2π√(πl/(3g)) = 2π√(l/g) √(π/3).
Answer: 2π/√F₁ < T < 2π/√F₂; for a 30° swing 2π√(l/g) < T < 2π√(l/g)√(π/3) (the true period is about 1.017 × 2π√(l/g)).
Final answer: 2π/√F₁ < T < 2π/√F₂; for a 30° swing 2π√(l/g) < T < 2π√(l/g)√(π/3) (the true period is about 1.017 × 2π√(l/g)).

Question 8 (50 marks)

Q8(a) · Ordinary Differential Equations · 15 marks≈ 11 min
First Order Higher Degree Equations, Clairaut's Equation and Singular Solutions
Obtain the singular solution of the differential equation
(dy/dx)² (y/x)² cot²α − 2(dy/dx)(y/x) + (y/x)² cosec²α = 1
Also find the complete primitive of the given differential equation. Give the geometrical interpretations of the complete primitive and singular solution.
हिंदी में प्रश्न
अवकल समीकरण
(dy/dx)² (y/x)² cot²α − 2(dy/dx)(y/x) + (y/x)² cosec²α = 1
का विचित्र हल प्राप्त कीजिए। दिए हुए अवकल समीकरण का पूर्ण पूर्वग भी ज्ञात कीजिए। पूर्ण पूर्वग तथा विचित्र हल की ज्यामितीय व्याख्या कीजिए।
Practise this question Ordinary Differential Equations notes
Worked solution
Multiply by x²: p²y² cot²α − 2pxy + y² cosec²α = x². Put X = x², Y = y², P = dY/dX = yp/x: X(P² cot²α − 2P − 1) + Y cosec²α = 0, that is Y = X g(P) with g(P) = sin²α(1 + 2P − P² cot²α).
Differentiating with respect to X: P − g(P) = Xg′(P) dP/dX, with P − g(P) = cos²α(P − tan²α)(P + 1) and g′(P) = −2cos²α(P − tan²α).
Either P = tan²α, giving Y = X tan²α, or dX/X = −2 dP/(P + 1), giving P + 1 = c/√X = c/x.
In the second case Y = X g(c/x − 1) simplifies to y² = c² sin²α − (x − c)²: the complete primitive is (x − c)² + y² = c² sin²α.
c-discriminant: −2(x − c) = 2c sin²α gives c = x sec²α, and then y² = x² tan²α, the same as the first branch; it satisfies the equation and is not a member of the family.
Answer: complete primitive (x − c)² + y² = c² sin²α: circles with centres on the x-axis whose radius is sin α times the distance of the centre from O; singular solution y = ±x tan α, the pair of lines through O inclined at α to the x-axis, which every circle touches (their envelope).
Final answer: complete primitive (x − c)² + y² = c² sin²α: circles with centres on the x-axis whose radius is sin α times the distance of the centre from O; singular solution y = ±x tan α, the pair of lines through O inclined at α to the x-axis, which every circle touches (their envelope).
Q8(b) · Dynamics and Statics · 15 marks≈ 11 min
Central Orbits and Kepler's Laws
Prove that the path of a planet, which is moving so that its acceleration is always directed to a fixed point (star) and is equal to μ/(distance)², is a conic section. Find the conditions under which the path becomes (i) ellipse, (ii) parabola and (iii) hyperbola.
हिंदी में प्रश्न
एक गतिमान ग्रह का त्वरण μ/(दूरी)² के बराबर है और त्वरण की दिशा हमेशा एक स्थिर बिंदु (तारा) की ओर है। सिद्ध कीजिए कि उस ग्रह का पथ एक शंकु-परिच्छेद है। वे प्रतिबंध ज्ञात कीजिए, जिनके अन्तर्गत पथ (i) दीर्घवृत्त, (ii) परवलय और (iii) अतिपरवलय बन जाता है।
Practise this question Dynamics and Statics notes
Q8(c) · Vector Analysis · 20 marks≈ 14 min
Gauss and Stokes' Theorems and Green's Identities
(i) State Gauss divergence theorem. Verify this theorem for F = 4xi − 2y²j + z²k, taken over the region bounded by x² + y² = 4, z = 0 and z = 3. (15 marks)
(ii) Evaluate by Stokes' theorem ∮_C e^x dx + 2y dy − dz, where C is the curve x² + y² = 4, z = 2. (5 marks)
हिंदी में प्रश्न
(i) गाउस के अपसरण प्रमेय का कथन लिखिए। इस प्रमेय को F = 4xi − 2y²j + z²k के लिए x² + y² = 4, z = 0 और z = 3 से घिरे हुए क्षेत्र में सत्यापित कीजिए। (15 अंक)
(ii) स्टोक्स प्रमेय के द्वारा ∮_C e^x dx + 2y dy − dz का मान ज्ञात कीजिए, जहाँ C, वक्र x² + y² = 4, z = 2 है। (5 अंक)
Practise this question Vector Analysis notes
Worked solution
(i) Statement: as in sub-section 3. Volume side (Example 5): ∭(4 − 4y + 2z) dV = 84π.
Base z = 0: n = −k and F·n = −z² = 0, flux 0; top z = 3: n = k and F·n = 9, flux 9 × 4π = 36π.
Curved surface x = 2 cos φ, y = 2 sin φ: n = (cos φ, sin φ, 0), dS = 2 dφ dz and F·n = 8 cos²φ − 8 sin³φ, so the flux is ∫₀³∫₀^2π(16 cos²φ − 16 sin³φ) dφ dz = 3 × 16π = 48π.
Total flux 0 + 36π + 48π = 84π, equal to the volume integral.
(ii) For F = (e^x, 2y, −1), ∇ × F = 0 (each component depends only on its own variable), so by Stokes' theorem over the disc x² + y² ≤ 4, z = 2, the integral is 0.
Answer: (i) both sides equal 84π; (ii) 0.
Final answer: (i) both sides equal 84π; (ii) 0.

UPSC Mains 2019 Mathematics Paper II