UPSC Mains 2016 Mathematics Optional Question Papers with Solutions

All 59 questions of the 2016 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 2016 Mathematics Paper I

Question 1 (50 marks)

Q1(a) · Linear Algebra · 10 marks≈ 7 min
Algebra of Matrices, Echelon Form, Rank, Inverse, Congruence and Similarity
(i) Using elementary row operations, find the inverse of A = [[1, 2, 1], [1, 3, 2], [1, 0, 1]]. (6 marks)
(ii) If A = [[1, 1, 3], [5, 2, 6], [−2, −1, −3]], then find A¹⁴ + 3A − 2I. (4 marks)
हिंदी में प्रश्न
(i) यदि A = [[1, 2, 1], [1, 3, 2], [1, 0, 1]] है, तो प्रारम्भिक पंक्ति संक्रिया (elementary row operation) के प्रयोग से A⁻¹ निकालिये। (6 अंक)
(ii) यदि A = [[1, 1, 3], [5, 2, 6], [−2, −1, −3]] है, तो A¹⁴ + 3A − 2I का मान निकालिये। (4 अंक)
Practise this question Linear Algebra notes
Worked solution
(i) [A | I] with rows (1, 2, 1 | 1, 0, 0), (1, 3, 2 | 0, 1, 0), (1, 0, 1 | 0, 0, 1). R₂ − R₁ and R₃ − R₁ give (0, 1, 1 | −1, 1, 0) and (0, −2, 0 | −1, 0, 1).
R₃ + 2R₂ gives (0, 0, 2 | −3, 2, 1), and R₃/2 gives (0, 0, 1 | −3/2, 1, 1/2). Then R₂ − R₃ = (0, 1, 0 | 1/2, 0, −1/2) and R₁ − 2R₂ − R₃ = (1, 0, 0 | 3/2, −1, 1/2).
So A⁻¹ = (1/2)[[3, −2, 1], [1, 0, −1], [−3, 2, 1]]; check: the first row of A against the columns gives (3 + 2 − 3)/2 = 1, (−2 + 0 + 2)/2 = 0, (1 − 2 + 1)/2 = 0.
(ii) For A = [[1, 1, 3], [5, 2, 6], [−2, −1, −3]]: tr A = 0, the sum of principal 2 × 2 minors is −3 + 3 + 0 = 0 and det A = 0, so the characteristic polynomial is λ³ and, by Cayley-Hamilton, A³ = 0.
Directly, A² = [[0, 0, 0], [3, 3, 9], [−1, −1, −3]] and A·A² = 0. Hence A¹⁴ = A¹¹A³ = 0 and A¹⁴ + 3A − 2I = 3A − 2I.
Answer: (i) A⁻¹ = (1/2)[[3, −2, 1], [1, 0, −1], [−3, 2, 1]]; (ii) A¹⁴ + 3A − 2I = [[1, 3, 9], [15, 4, 18], [−6, −3, −11]].
Final answer: (i) A⁻¹ = (1/2)[[3, −2, 1], [1, 0, −1], [−3, 2, 1]]; (ii) A¹⁴ + 3A − 2I = [[1, 3, 9], [15, 4, 18], [−6, −3, −11]].
Q1(b) · Linear Algebra · 10 marks≈ 7 min
Systems of Linear Equations
(i) Using elementary row operations, find the condition that the linear equations
x − 2y + z = a
2x + 7y − 3z = b
3x + 5y − 2z = c
have a solution. (7 marks)
(ii) If
W₁ = {(x, y, z) | x + y − z = 0}
W₂ = {(x, y, z) | 3x + y − 2z = 0}
W₃ = {(x, y, z) | x − 7y + 3z = 0}
then find dim(W₁ ∩ W₂ ∩ W₃) and dim(W₁ + W₂). (3 marks)
हिंदी में प्रश्न
(i) प्रारम्भिक पंक्ति संक्रिया (elementary row operation) के प्रयोग से वह शर्त निकालिये, जिससे कि प्रथम-घातीय समीकरणों (linear equations)
x − 2y + z = a
2x + 7y − 3z = b
3x + 5y − 2z = c
का एक हल हो। (7 अंक)
(ii) यदि
W₁ = {(x, y, z) | x + y − z = 0}
W₂ = {(x, y, z) | 3x + y − 2z = 0}
W₃ = {(x, y, z) | x − 7y + 3z = 0}
तो dim(W₁ ∩ W₂ ∩ W₃) तथा dim(W₁ + W₂) का मान निकालिये। (3 अंक)
Practise this question Linear Algebra notes
Worked solution
(i) In [A | B], R₂ − 2R₁ gives (0, 11, −5 | b − 2a) and R₃ − 3R₁ gives (0, 11, −5 | c − 3a); then R₃ − R₂ gives (0, 0, 0 | c − a − b).
ρ(A) = 2, so the system is consistent ⇔ ρ([A | B]) = 2 ⇔ c = a + b (then it has infinitely many solutions).
(ii) W₁ ∩ W₂ ∩ W₃ is the solution space of the matrix [[1, 1, −1], [3, 1, −2], [1, −7, 3]]; R₂ − 3R₁ = (0, −2, 1) and R₃ − R₁ = (0, −8, 4) = 4(0, −2, 1), so the rank is 2 and the dimension is 3 − 2 = 1 (the line L{(1, 1, 2)}).
W₁, W₂ are distinct planes through 0 (dimension 2 each) with W₁ ∩ W₂ = L{(1, 1, 2)}, so dim(W₁ + W₂) = 2 + 2 − 1 = 3, i.e. W₁ + W₂ = ℝ³.
Answer: (i) c = a + b; (ii) dim(W₁ ∩ W₂ ∩ W₃) = 1 and dim(W₁ + W₂) = 3.
Final answer: (i) c = a + b; (ii) dim(W₁ ∩ W₂ ∩ W₃) = 1 and dim(W₁ + W₂) = 3.
Q1(c) · Calculus · 10 marks≈ 7 min
Riemann Integral, Indefinite Integrals and Improper Integrals
Evaluate:
I = ∫₀¹ ∛(x log(1/x)) dx
हिंदी में प्रश्न
मान निकालिये :
I = ∫₀¹ ∛(x log(1/x)) dx
Practise this question Calculus notes
Worked solution
Put x = e^−t: log(1/x) = t, dx = −e^−t dt, and x from 0 to 1 corresponds to t from ∞ to 0.
I = ∫₀^∞ (te^−t)^1/3e^−t dt = ∫₀^∞ t^1/3e^−4t/3 dt, a convergent Gamma-type integral.
With s = 4t/3: I = (3/4)^4/3∫₀^∞ s^1/3e^−s ds = (3/4)^4/3Γ(4/3) = (1/3)(3/4)^4/3Γ(1/3).
Answer: I = (3/4)^4/3 Γ(4/3) ≈ 0.6085.
Final answer: I = (3/4)^4/3 Γ(4/3) ≈ 0.6085.
Q1(d) · Analytic Geometry · 10 marks≈ 7 min
The Sphere
Find the equation of the sphere which passes through the circle x² + y² = 4; z = 0 and is cut by the plane x + 2y + 2z = 0 in a circle of radius 3.
हिंदी में प्रश्न
उस गोले (sphere) का समीकरण निकालिये, जो वृत्त x² + y² = 4; z = 0 से गुजरता है और जो तल x + 2y + 2z = 0 से एक वृत्त, जिसकी त्रिज्या 3 है, में काटा जाता है।
Practise this question Analytic Geometry notes
Worked solution
Spheres through the circle: x² + y² + z² − 4 + λz = 0, with centre (0, 0, −λ/2) and R² = λ²/4 + 4.
Distance of the centre from x + 2y + 2z = 0: |2(−λ/2)|/3 = |λ|/3. The section has radius 3, so R² − λ²/9 = 9, that is 5λ²/36 = 5 and λ = ±6.
Answer: x² + y² + z² ± 6z − 4 = 0 (two spheres, each of radius √13).
Final answer: x² + y² + z² ± 6z − 4 = 0 (two spheres, each of radius √13).
Q1(e) · Analytic Geometry · 10 marks≈ 7 min
Straight Lines and Shortest Distance between Skew Lines
Find the shortest distance between the lines (x − 1)/2 = (y − 2)/4 = z − 3 and y − mx = z = 0. For what value of m will the two lines intersect?
हिंदी में प्रश्न
रेखाओं (x − 1)/2 = (y − 2)/4 = z − 3 तथा y − mx = z = 0 के बीच लघुतम दूरी (shortest distance) निकालिये। m के किस मान के लिए दोनों रेखाएँ प्रतिच्छेद (intersect) करेंगी?
Practise this question Analytic Geometry notes
Worked solution
Line 1: A(1, 2, 3), b₁ = (2, 4, 1). Line 2 (y = mx, z = 0): through O, b₂ = (1, m, 0).
b₁ × b₂ = (−m, 1, 2m − 4), with |b₁ × b₂|² = m² + 1 + (2m − 4)² = 5m² − 16m + 17, and (O − A)·(b₁ × b₂) = (−1, −2, −3)·(−m, 1, 2m − 4) = 10 − 5m.
SD = |10 − 5m|/√(5m² − 16m + 17). The lines are never parallel, so they intersect exactly when SD = 0, that is m = 2; then line 1 meets z = 0 at (−5, −10, 0), which lies on y = 2x.
Answer: SD = 5|2 − m|/√(5m² − 16m + 17); the lines intersect for m = 2, at (−5, −10, 0).
Final answer: SD = 5|2 − m|/√(5m² − 16m + 17); the lines intersect for m = 2, at (−5, −10, 0).

Question 2 (50 marks)

Q2(a) · Linear Algebra · 16 marks≈ 12 min
Linear Transformations, Rank-Nullity and Matrix Representation
(i) If M₂(R) is space of real matrices of order 2 × 2 and P₂(x) is the space of real polynomials of degree at most 2, then find the matrix representation of T : M₂(R) → P₂(x), such that T([[a, b], [c, d]]) = a + c + (a − d)x + (b + c)x², with respect to the standard bases of M₂(R) and P₂(x). Further find the null space of T. (10 marks)
(ii) If T : P₂(x) → P₃(x) is such that T(f(x)) = f(x) + 5∫₀^x f(t) dt, then choosing {1, 1 + x, 1 − x²} and {1, x, x², x³} as bases of P₂(x) and P₃(x) respectively, find the matrix of T. (6 marks)
हिंदी में प्रश्न
(i) यदि M₂(R), 2 × 2 कोटि (order) के वास्तविक आव्यूहों की समष्टि (space) तथा P₂(x), वास्तविक बहुपदों (polynomials), जिनकी अधिकतम घात (degree) 2 है, की समष्टि (space) हो, तो T : M₂(R) → P₂(x), जहाँ T([[a, b], [c, d]]) = a + c + (a − d)x + (b + c)x², का M₂(R) एवं P₂(x) के मानक आधारों (standard bases) के सापेक्ष आव्यूह निरूपित कीजिये। इसके अलावा T का शून्य समष्टि (null space) प्राप्त कीजिये। (10 अंक)
(ii) यदि T : P₂(x) → P₃(x) इस प्रकार है कि T(f(x)) = f(x) + 5∫₀^x f(t) dt, तो {1, 1 + x, 1 − x²} एवं {1, x, x², x³} को क्रमशः P₂(x) एवं P₃(x) का आधार (bases) लेते हुए T का आव्यूह निकालिये। (6 अंक)
Practise this question Linear Algebra notes
Worked solution
(i) Bases: E₁₁, E₁₂, E₂₁, E₂₂ of M₂(ℝ) and 1, x, x² of P₂(x). T(E₁₁) = 1 + x, T(E₁₂) = x², T(E₂₁) = 1 + x², T(E₂₂) = −x.
Writing the coordinates as columns, [T] = [[1, 0, 1, 0], [1, 0, 0, −1], [0, 1, 1, 0]] (a 3 × 4 matrix).
Null space: a + c = 0, a − d = 0, b + c = 0 give c = −a, d = a, b = a. So N(T) = {a[[1, 1], [−1, 1]]: a ∈ ℝ}, nullity 1 and rank 3 (T is onto P₂(x)).
(ii) T(1) = 1 + 5x; T(1 + x) = 1 + x + 5(x + x²/2) = 1 + 6x + (5/2)x²; T(1 − x²) = 1 − x² + 5(x − x³/3) = 1 + 5x − x² − (5/3)x³.
Their coordinates in {1, x, x², x³} are (1, 5, 0, 0), (1, 6, 5/2, 0) and (1, 5, −1, −5/3), which are the columns of [T].
Answer: (i) [T] = [[1, 0, 1, 0], [1, 0, 0, −1], [0, 1, 1, 0]], N(T) = L{[[1, 1], [−1, 1]]}; (ii) [T] = [[1, 1, 1], [5, 6, 5], [0, 5/2, −1], [0, 0, −5/3]].
Final answer: (i) [T] = [[1, 0, 1, 0], [1, 0, 0, −1], [0, 1, 1, 0]], N(T) = L{[[1, 1], [−1, 1]]}; (ii) [T] = [[1, 1, 1], [5, 6, 5], [0, 5/2, −1], [0, 0, −5/3]].
Q2(b) · Linear Algebra · 16 marks≈ 12 min
Symmetric, Hermitian, Orthogonal and Unitary Matrices and Their Eigenvalues
(i) If A = [[1, 1, 0], [1, 1, 0], [0, 0, 1]], then find the eigenvalues and eigenvectors of A. (8 marks)
(ii) Prove that eigenvalues of a Hermitian matrix are all real. (8 marks)
हिंदी में प्रश्न
(i) यदि A = [[1, 1, 0], [1, 1, 0], [0, 0, 1]] है, तो A के अभिलक्षणिक मान (eigenvalues) तथा अभिलक्षणिक सदिशों (eigenvectors) को निकालिये। (8 अंक)
(ii) सिद्ध कीजिये कि हर्मिटी (Hermitian) आव्यूह के सभी अभिलक्षणिक मान वास्तविक हैं। (8 अंक)
Practise this question Linear Algebra notes
Q2(c) · Linear Algebra · 18 marks≈ 13 min
Linear Transformations, Rank-Nullity and Matrix Representation
If A = [[1, −1, 2], [−2, 1, −1], [1, 2, 3]] is the matrix representation of a linear transformation T : P₂(x) → P₂(x) with respect to the bases {1 − x, x(1 − x), x(1 + x)} and {1, 1 + x, 1 + x²}, then find T.
हिंदी में प्रश्न
यदि आधारों (bases) {1 − x, x(1 − x), x(1 + x)} एवं {1, 1 + x, 1 + x²} के सापेक्ष रैखिक रूपांतरण (linear transformation) T : P₂(x) → P₂(x) के तहत आव्यूह निरूपण A = [[1, −1, 2], [−2, 1, −1], [1, 2, 3]] हो, तो T प्राप्त कीजिये।
Practise this question Linear Algebra notes
Worked solution
By definition T(u_j) = ∑_i a_ijv_i, with u₁ = 1 − x, u₂ = x − x², u₃ = x + x² and v₁ = 1, v₂ = 1 + x, v₃ = 1 + x².
T(u₁) = 1 − 2(1 + x) + (1 + x²) = −2x + x²; T(u₂) = −1 + (1 + x) + 2(1 + x²) = 2 + x + 2x²; T(u₃) = 2 − (1 + x) + 3(1 + x²) = 4 − x + 3x².
The standard basis in terms of the u_j: x = (u₂ + u₃)/2, x² = (u₃ − u₂)/2, 1 = u₁ + x.
Hence T(x) = (6 + 5x²)/2 = 3 + (5/2)x², T(x²) = (2 − 2x + x²)/2 = 1 − x + (1/2)x², and T(1) = T(u₁) + T(x) = 3 − 2x + (7/2)x².
By linearity T(a + bx + cx²) = aT(1) + bT(x) + cT(x²). Check: T(1 − x) = T(1) − T(x) = −2x + x², as required.
Answer: T(a + bx + cx²) = (3a + 3b + c) − (2a + c)x + ((7a + 5b + c)/2)x².
Final answer: T(a + bx + cx²) = (3a + 3b + c) − (2a + c)x + ((7a + 5b + c)/2)x².

Question 3 (50 marks)

Q3(a) · Calculus · 20 marks≈ 14 min
Functions of Several Variables: Partial Derivatives, Extrema, Lagrange Multipliers and Jacobians
Find the maximum and minimum values of x² + y² + z² subject to the conditions x²/4 + y²/5 + z²/25 = 1 and x + y − z = 0.
हिंदी में प्रश्न
x² + y² + z² का अधिकतम तथा न्यूनतम मान निकालिये, जहाँ x²/4 + y²/5 + z²/25 = 1 तथा x + y − z = 0 हो।
Practise this question Calculus notes
Worked solution
This is the previous example with a = 1/4, b = 1/5, c = 1/25 and l = m = 1, n = −1. The stationary values u satisfy 4/(4 − u) + 5/(5 − u) + 25/(25 − u) = 0.
Clearing denominators: 4(5 − u)(25 − u) + 5(4 − u)(25 − u) + 25(4 − u)(5 − u) = (500 − 120u + 4u²) + (500 − 145u + 5u²) + (500 − 225u + 25u²) = 0.
So 34u² − 490u + 1500 = 0, i.e. 17u² − 245u + 750 = 0, with discriminant 245² − 51000 = 9025 = 95² and roots u = (245 ± 95)/34 = 10, 75/17.
Both are attained: the plane x + y = z cuts the ellipsoid in an ellipse with semi-axes √10 and √(75/17).
Answer: Maximum 10, minimum 75/17.
Final answer: Maximum 10, minimum 75/17.
Q3(b) · Calculus · 15 marks≈ 11 min
Real Numbers, Limits, Continuity and Differentiability
Let
f(x, y) = (2x⁴y − 5x²y² + y⁵)/(x² + y²)², (x, y) ≠ (0, 0)
f(x, y) = 0, (x, y) = (0, 0)
Find a δ > 0 such that |f(x, y) − f(0, 0)| < ·01, whenever √(x² + y²) < δ.
हिंदी में प्रश्न
मान लीजिये
f(x, y) = (2x⁴y − 5x²y² + y⁵)/(x² + y²)², (x, y) ≠ (0, 0)
f(x, y) = 0, (x, y) = (0, 0)
δ > 0 प्राप्त कीजिये इस प्रकार कि |f(x, y) − f(0, 0)| < ·01, जब √(x² + y²) < δ हो।
Practise this question Calculus notes
Worked solution
[Misprint: as printed, the middle term 5x²y² has degree 4, and along y = x, f = (3x⁵ − 5x⁴)/(4x⁴) → −5/4 ≠ 0, so f is not continuous at (0, 0) and no δ exists. The intended term is 5x²y³, which is solved below.]
Put r = √(x² + y²), so |x| ≤ r and |y| ≤ r. Then |2x⁴y| ≤ 2r⁵, |5x²y³| ≤ 5r⁵, |y⁵| ≤ r⁵, and (x² + y²)² = r⁴.
Hence |f(x, y) − f(0, 0)| ≤ 8r⁵/r⁴ = 8r for (x, y) ≠ (0, 0), and 8r < 0.01 whenever r < 0.01/8 = 0.00125.
So δ = 0.00125 (or any smaller positive number) works; the same argument with ε in place of 0.01 proves that f is continuous at (0, 0).
Answer: δ = 0.00125 = 1/800 (for the intended function).
Final answer: δ = 0.00125 = 1/800 (for the intended function).
Q3(c) · Calculus · 15 marks≈ 11 min
Double and Triple Integrals; Areas, Surfaces and Volumes
Find the surface area of the plane x + 2y + 2z = 12 cut off by x = 0, y = 0 and x² + y² = 16.
हिंदी में प्रश्न
तल x + 2y + 2z = 12 के x = 0, y = 0 तथा x² + y² = 16 द्वारा काटे गये क्षेत्र का पृष्ठीय क्षेत्रफल (surface area) निकालिये।
Practise this question Calculus notes
Worked solution
On the plane z = (12 − x − 2y)/2, so z_x = −1/2, z_y = −1 and √(1 + z_x² + z_y²) = √(1 + 1/4 + 1) = 3/2.
The portion cut off by x = 0, y = 0 and the cylinder x² + y² = 16 projects onto the quarter disc D: x, y ≥ 0, x² + y² ≤ 16, of area 4π (z stays positive there).
S = ∬_D (3/2) dx dy = (3/2)(4π). (Equivalently, S = (area of projection)/cos γ with cos γ = 2/3 for the unit normal (1, 2, 2)/3.)
Answer: 6π.
Final answer: 6π.

Question 4 (50 marks)

Q4(a) · Analytic Geometry · 10 marks≈ 7 min
Straight Lines and Shortest Distance between Skew Lines
Find the surface generated by a line which intersects the lines y = a = z, x + 3z = a = y + z and parallel to the plane x + y = 0.
हिंदी में प्रश्न
एक रेखा, जो रेखाओं y = a = z, x + 3z = a = y + z को प्रतिच्छेद (intersect) करती है तथा तल x + y = 0 के समानान्तर है, द्वारा जनित सतह (surface generated) निकालिये।
Practise this question Analytic Geometry notes
Worked solution
Any line meeting both given lines is (y − a) + λ(z − a) = 0, (x + 3z − a) + μ(y + z − a) = 0. Its direction is (0, 1, λ) × (1, μ, 3 + μ) = (3 + μ − λμ, λ, −1).
Parallel to x + y = 0: (3 + μ − λμ) + λ = 0. Put λ = −(y − a)/(z − a) and μ = −(x + 3z − a)/(y + z − a) and clear denominators: 3(z − a)(y + z − a) − (x + 3z − a)(z − a) − (y − a)(y + z − a) − (y − a)(x + 3z − a) = 0.
Expanding, the z² terms, the y terms and the constants cancel, leaving −y² − yz − zx − xy + 2ax + 2az = 0, that is y² + yz + zx + xy = 2a(x + z).
Answer: the surface is (x + y)(y + z) = 2a(x + z), a hyperbolic paraboloid.
Final answer: the surface is (x + y)(y + z) = 2a(x + z), a hyperbolic paraboloid.
Q4(b) · Analytic Geometry · 10 marks≈ 7 min
Cone and Cylinder
Show that the cone 3yz − 2zx − 2xy = 0 has an infinite set of three mutually perpendicular generators. If x/1 = y/1 = z/2 is a generator belonging to one such set, find the other two.
हिंदी में प्रश्न
सिद्ध कीजिये कि शंकु (cone) 3yz − 2zx − 2xy = 0 के तीन परस्पर लम्बीय जनकों (generators) का एक अनन्त समुच्चय है। यदि x/1 = y/1 = z/2 ऐसे किसी समुच्चय का एक जनक (generator) हो, तो बाकी दो निकालिये।
Practise this question Analytic Geometry notes
Worked solution
The cone has a = b = c = 0, so a + b + c = 0: by the result above it has infinitely many sets of three mutually perpendicular generators. The line x/1 = y/1 = z/2 is a generator: 3·2 − 2·2 − 2·1 = 0.
Plane perpendicular to it through O: x + y + 2z = 0. With x = −y − 2z: 3yz + 2z(y + 2z) + 2y(y + 2z) = 2y² + 9yz + 4z² = (2y + z)(y + 4z) = 0.
z = −2y gives x = 3y: direction (3, 1, −2). y = −4z gives x = 2z: direction (2, −4, 1). Check: 6 − 4 − 2 = 0, 3 + 1 − 4 = 0, 2 − 4 + 2 = 0.
Answer: x/3 = y/1 = z/(−2) and x/2 = y/(−4) = z/1.
Final answer: x/3 = y/1 = z/(−2) and x/2 = y/(−4) = z/1.
Q4(c) · Calculus · 15 marks≈ 11 min
Double and Triple Integrals; Areas, Surfaces and Volumes
Evaluate ∬_R f(x, y) dx dy over the rectangle R = [0, 1; 0, 1] where
f(x, y) = x + y, if x² < y < 2x²
f(x, y) = 0, elsewhere
हिंदी में प्रश्न
∬_R f(x, y) dx dy का मान निकालिये, जहाँ आयत R = [0, 1; 0, 1] तथा
f(x, y) = x + y, यदि x² < y < 2x²
f(x, y) = 0, अन्यथा
है।
Practise this question Calculus notes
Worked solution
f is non-zero only between the parabolas y = x² and y = 2x² inside the square, and 2x² ≤ 1 for x ≤ 1/√2. f is bounded with discontinuities only on two curves, so it is integrable.
For 0 ≤ x ≤ 1/√2, y runs from x² to 2x²: ∫(x + y) dy = x³ + 3x⁴/2, and ∫₀^1/√2 (x³ + 3x⁴/2) dx = 1/16 + 3/(40√2).
For 1/√2 ≤ x ≤ 1, y runs from x² to 1: ∫(x + y) dy = x − x³ + (1 − x⁴)/2, and ∫_1/√2¹ = [x²/2 − x⁴/4 + x/2 − x⁵/10]_1/√2¹ = 13/20 − 3/16 − 19/(40√2).
Total = 1/16 + 13/20 − 3/16 − 16/(40√2) = 21/40 − √2/5.
Answer: 21/40 − √2/5 ≈ 0.242.
Final answer: 21/40 − √2/5 ≈ 0.242.
Q4(d) · Analytic Geometry · 15 marks≈ 11 min
Ellipsoid, Hyperboloids and Paraboloids
Find the locus of the point of intersection of three mutually perpendicular tangent planes to the conicoid ax² + by² + cz² = 1.
हिंदी में प्रश्न
शांकवज (conicoid) ax² + by² + cz² = 1 के तीन पारस्परिक लम्बीय स्पर्शी तलों के प्रतिच्छेदन बिन्दु का बिन्दुपथ निकालिये।
Practise this question Analytic Geometry notes

Question 5 (50 marks)

Q5(a) · Ordinary Differential Equations · 10 marks≈ 7 min
Linear Equations with Constant Coefficients
Find a particular integral of d²y/dx² + y = e^x/2 sin(x√3/2).
हिंदी में प्रश्न
d²y/dx² + y = e^x/2 sin(x√3/2) का विशेष समाकल (particular integral) निकालिये।
Practise this question Ordinary Differential Equations notes
Q5(b) · Vector Analysis · 10 marks≈ 7 min
Vector Identities and Vector Equations
Prove that the vectors a⃗ = 3î + ĵ − 2k̂, b⃗ = −î + 3ĵ + 4k̂, c⃗ = 4î − 2ĵ − 6k̂ can form the sides of a triangle. Find the lengths of the medians of the triangle.
हिंदी में प्रश्न
सिद्ध कीजिये कि सदिश a⃗ = 3î + ĵ − 2k̂, b⃗ = −î + 3ĵ + 4k̂, c⃗ = 4î − 2ĵ − 6k̂ एक त्रिभुज की भुजाएँ बना सकते हैं। इस त्रिभुज की माध्यिकाओं की लम्बाई निकालिये।
Practise this question Vector Analysis notes
Worked solution
b + c = (−1 + 4, 3 − 2, 4 − 6) = (3, 1, −2) = a, and no two of a, b, c are parallel (their components are not proportional).
So with AB = b and BC = c we get AC = a: the vectors are the sides of a triangle ABC, with |a| = √14, |b| = √26, |c| = √56.
Median from A: b + c/2 = (1, 2, 1), length √6.
Median from B: −b + a/2 = (5/2, −5/2, −5), length √(75/2) = 5√6/2.
Median from C: −a + b/2 = (−7/2, 1/2, 4), length √(57/2) = √114/2.
Check (Apollonius): 6 + 75/2 + 57/2 = 72 = (3/4)(14 + 26 + 56).
Answer: the medians have lengths √6, 5√6/2 and √114/2.
Final answer: the medians have lengths √6, 5√6/2 and √114/2.
Q5(c) · Ordinary Differential Equations · 10 marks≈ 7 min
Formulation and First Order, First Degree Equations
Solve:
dy/dx = (1/(1 + x²))(e^tan⁻¹x − y)
हिंदी में प्रश्न
हल कीजिये :
dy/dx = (1/(1 + x²))(e^tan⁻¹x − y)
Practise this question Ordinary Differential Equations notes
Q5(d) · Ordinary Differential Equations · 10 marks≈ 7 min
Orthogonal Trajectories
Show that the family of parabolas y² = 4cx + 4c² is self-orthogonal.
हिंदी में प्रश्न
दर्शाइये कि परवलय-कुल y² = 4cx + 4c² स्वलांबिक (self-orthogonal) है।
Practise this question Ordinary Differential Equations notes
Q5(e) · Dynamics and Statics · 10 marks≈ 7 min
Central Orbits and Kepler's Laws
A particle moves with a central acceleration which varies inversely as the cube of the distance. If it is projected from an apse at a distance a from the origin with a velocity which is √2 times the velocity for a circle of radius a, then find the equation to the path.
हिंदी में प्रश्न
एक कण केन्द्रीय त्वरण (central acceleration), जो दूरी के त्रिघात के व्युत्क्रमानुपाती (inversely proportional) है, के तहत चलायमान है। यदि इसे मूलबिन्दु (origin) से दूरी a पर एक स्तब्धिका (apse) से उस वेग से प्रक्षेपित किया जाता है, जो कि a त्रिज्या के वृत्त के वेग का √2 गुना है, तो पथ का समीकरण निकालिये।
Practise this question Dynamics and Statics notes

Question 6 (50 marks)

Q6(a) · Ordinary Differential Equations · 10 marks≈ 7 min
Formulation and First Order, First Degree Equations
Solve:
{y(1 − x tan x) + x² cos x} dx − x dy = 0
हिंदी में प्रश्न
हल कीजिये :
{y(1 − x tan x) + x² cos x} dx − x dy = 0
Practise this question Ordinary Differential Equations notes
Q6(b) · Ordinary Differential Equations · 15 marks≈ 11 min
Second Order Equations with Variable Coefficients
Using the method of variation of parameters, solve the differential equation
(D² + 2D + 1)y = e^−x log(x), [D ≡ d/dx]
हिंदी में प्रश्न
अवकल समीकरण (differential equation)
(D² + 2D + 1)y = e^−x log(x), [D ≡ d/dx]
को प्राचल-विचरण (variation of parameters) विधि से हल कीजिये।
Practise this question Ordinary Differential Equations notes
Q6(c) · Ordinary Differential Equations · 15 marks≈ 11 min
Second Order Equations with Variable Coefficients
Find the general solution of the equation x² d³y/dx³ − 4x d²y/dx² + 6 dy/dx = 4.
हिंदी में प्रश्न
समीकरण x² d³y/dx³ − 4x d²y/dx² + 6 dy/dx = 4 का व्यापक हल (general solution) निकालिये।
Practise this question Ordinary Differential Equations notes
Worked solution
y does not occur, so put p = y′: x²p″ − 4xp′ + 6p = 4, an Euler-Cauchy equation in p.
With x = e^z: [θ(θ − 1) − 4θ + 6]p = (θ − 2)(θ − 3)p = 4, so p = c₁x² + c₂x³ + 4/6.
Answer: y = Ax³ + Bx⁴ + 2x/3 + C (A = c₁/3, B = c₂/4).
Final answer: y = Ax³ + Bx⁴ + 2x/3 + C (A = c₁/3, B = c₂/4).
Q6(d) · Ordinary Differential Equations · 10 marks≈ 7 min
Laplace Transforms and Initial Value Problems
Using Laplace transformation, solve the following:
y″ − 2y′ − 8y = 0, y(0) = 3, y′(0) = 6
हिंदी में प्रश्न
लाप्लास रूपांतरण (Laplace transformation) की मदद से निम्न का हल निकालिये :
y″ − 2y′ − 8y = 0, y(0) = 3, y′(0) = 6
Practise this question Ordinary Differential Equations notes

Question 7 (50 marks)

Q7(a) · Dynamics and Statics · 15 marks≈ 11 min
Equilibrium, Friction and the Common Catenary
A uniform rod AB of length 2a movable about a hinge at A rests with other end against a smooth vertical wall. If α is the inclination of the rod to the vertical, prove that the magnitude of reaction of the hinge is
(1/2) W √(4 + tan²α)
where W is the weight of the rod.
हिंदी में प्रश्न
2a लम्बाई की एकसमान एक छड़ (rod) AB, A पर हिन्ज (hinge) के परितः चलायमान (movable) है, जिसका दूसरा छोर एक चिकनी ऊर्ध्व दीवार पर स्थित है। यदि छड़ का ऊर्ध्वाधर से झुकाव α हो, तो सिद्ध कीजिये कि हिन्ज (hinge) पर प्रतिक्रिया (reaction) का परिमाण (1/2) W √(4 + tan²α) है, जहाँ W छड़ का भार है।
Practise this question Dynamics and Statics notes
Worked solution
The wall is smooth, so its reaction S at B is horizontal; the hinge reaction at A has components X (horizontal) and Y (vertical).
Moments about A: S × 2a cos α = W × a sin α, so S = (W/2) tan α.
Resolving: X = S = (W/2) tan α and Y = W.
Reaction = √(X² + Y²) = W√(1 + tan²α/4) = (1/2)W√(4 + tan²α).
Answer: the hinge reaction is (1/2)W√(4 + tan²α), inclined to the vertical at tan⁻¹((1/2) tan α).
Final answer: the hinge reaction is (1/2)W√(4 + tan²α), inclined to the vertical at tan⁻¹((1/2) tan α).
Q7(b) · Dynamics and Statics · 15 marks≈ 11 min
Equilibrium, Friction and the Common Catenary
Two weights P and Q are suspended from a fixed point O by strings OA, OB and are kept apart by a light rod AB. If the strings OA and OB make angles α and β with the rod AB, show that the angle θ which the rod makes with the vertical is given by
tan θ = (P + Q)/(P cot α − Q cot β)
हिंदी में प्रश्न
दो भार P तथा Q एक स्थिर (fixed) बिन्दु O से धागे (strings) OA एवं OB से लटके हैं तथा एक हल्के छड़ (rod) AB द्वारा एक-दूसरे से अलग किये गये हैं। यदि धागे OA एवं OB छड़ (rod) AB से क्रमशः α तथा β कोण बनाते हों, तो सिद्ध कीजिये कि छड़ (rod) ऊर्ध्व दिशा से θ कोण बनाती है, जहाँ
tan θ = (P + Q)/(P cot α − Q cot β)
Practise this question Dynamics and Statics notes
Worked solution
Treat the rod AB with the two weights as one body; the external forces are P at A, Q at B and the two tensions, which act along AO and BO and so meet at O.
P and Q have a vertical resultant P + Q through the point G of AB with AG : GB = Q : P; for equilibrium it must pass through O, so OG is vertical and θ is the angle between GO and GB.
Let N be the foot of the perpendicular from O to AB and ON = d; then AN = d cot α, BN = d cot β and GN = d cot θ, with N between G and B (say).
AG = AN − GN = d(cot α − cot θ) and GB = BN + GN = d(cot β + cot θ).
AG : GB = Q : P gives P(cot α − cot θ) = Q(cot β + cot θ), that is (P + Q) cot θ = P cot α − Q cot β.
Answer: tan θ = (P + Q)/(P cot α − Q cot β); this is the m-n theorem of §4 applied to triangle OAB.
Final answer: tan θ = (P + Q)/(P cot α − Q cot β); this is the m-n theorem of §4 applied to triangle OAB.
Q7(c) · Dynamics and Statics · 20 marks≈ 14 min
Virtual Work, Stability and Forces in Three Dimensions
A square ABCD, the length of whose sides is a, is fixed in a vertical plane with two of its sides horizontal. An endless string of length l (> 4a) passes over four pegs at the angles of the board and through a ring of weight W which is hanging vertically. Show that the tension of the string is W(l − 3a)/(2√(l² − 6la + 8a²)).
हिंदी में प्रश्न
एक वर्ग ABCD, जिसके प्रत्येक भुजा की लम्बाई a है, को एक ऊर्ध्व तल (vertical plane) में स्थिर (fixed) किया जाता है, जिसकी दो भुजाएँ क्षैतिज हैं। एक अन्तहीन धागा (endless string), जिसकी लंबाई l (> 4a) है, बोर्ड के कोणों पर स्थित चार खूँटियों के ऊपर से एक रिंग (ring) के द्वारा गुजरता है। रिंग का भार W है और यह ऊर्ध्व दिशा में लटक रहा है। दर्शाइये कि धागे का तनाव W(l − 3a)/(2√(l² − 6la + 8a²)) है।
Practise this question Dynamics and Statics notes
Worked solution
The string passes over the pegs A, B at the top, down the vertical sides to D, C, and from the lower pegs to the ring; so the length below DC is l − 3a, and each slant part has length s = (l − 3a)/2.
Let the ring be at depth x below DC; then s² = x² + a²/4.
Virtual work, allowing the string to stretch: W δx − T δ(2s) = 0, and δs = (x/s) δx, so T = Ws/(2x).
Here x = √(s² − a²/4) = (1/2)√((l − 3a)² − a²) = (1/2)√(l² − 6la + 8a²).
Answer: T = W(l − 3a)/(2√(l² − 6la + 8a²)); resolving vertically at the ring, 2T(x/s) = W, confirms it.
Final answer: T = W(l − 3a)/(2√(l² − 6la + 8a²)); resolving vertically at the ring, 2T(x/s) = W, confirms it.

Question 8 (50 marks)

Q8(a) · Vector Analysis · 10 marks≈ 7 min
Gradient, Divergence and Curl in Cartesian and Cylindrical Coordinates
Find f(r) such that ∇f = r⃗/r⁵ and f(1) = 0.
हिंदी में प्रश्न
f(r) निकालिये, इस प्रकार कि ∇f = r⃗/r⁵ तथा f(1) = 0 हो।
Practise this question Vector Analysis notes
Worked solution
∇f(r) = f′(r) r/r, so f′(r)/r = 1/r⁵ and f′(r) = r⁻⁴.
Integrating, f(r) = −1/(3r³) + c, and f(1) = 0 gives c = 1/3.
Answer: f(r) = (1/3)(1 − 1/r³), r ≠ 0.
Final answer: f(r) = (1/3)(1 − 1/r³), r ≠ 0.
Q8(b) · Vector Analysis · 10 marks≈ 7 min
Gauss and Stokes' Theorems and Green's Identities
Prove that
∮_C f dr⃗ = ∬_S dS⃗ × ∇f
हिंदी में प्रश्न
सिद्ध कीजिये कि
∮_C f dr⃗ = ∬_S dS⃗ × ∇f
Practise this question Vector Analysis notes
Q8(c) · Dynamics and Statics · 15 marks≈ 11 min
Rectilinear Motion and Simple Harmonic Motion
A particle moves in a straight line. Its acceleration is directed towards a fixed point O in the line and is always equal to μ(a⁵/x²)^1/3 when it is at a distance x from O. If it starts from rest at a distance a from O, then find the time, the particle will arrive at O.
हिंदी में प्रश्न
एक कण एक सरल रेखा में गतिमान है। इसका त्वरण सरल रेखा पर एक स्थिर (fixed) बिन्दु O की तरफ निर्देशित है तथा हमेशा μ(a⁵/x²)^1/3 के बराबर है, जब यह बिन्दु O से x दूरी पर है। यदि यह बिन्दु O से a दूरी पर शून्य वेग से गतिमान होता है, तो वह समय निकालिये जब कण O पर आएगा।
Practise this question Dynamics and Statics notes
Worked solution
v dv/dx = −μa^5/3x^−2/3; integrating with v = 0 at x = a gives v² = 6μa^5/3(a^1/3 − x^1/3).
The particle moves inwards, so t = ∫₀^a dx/√(6μa^5/3(a^1/3 − x^1/3)).
Put x = as³: dx = 3as² ds and a^1/3 − x^1/3 = a^1/3(1 − s), so the quantity under the root becomes 6μa²(1 − s).
t = (3/√(6μ)) ∫₀¹ s²(1 − s)^−1/2 ds = (3/√(6μ)) B(3, 1/2).
B(3, 1/2) = Γ(3)Γ(1/2)/Γ(7/2) = 2√π/((15/8)√π) = 16/15.
Answer: t = 16/(5√(6μ)) = (8/15)√(6/μ), independent of a.
Final answer: t = 16/(5√(6μ)) = (8/15)√(6/μ), independent of a.
Q8(d) · Vector Analysis · 15 marks≈ 11 min
Curves in Space: Curvature, Torsion and Serret-Frenet Formulae
For the cardioid r = a(1 + cos θ), show that the square of the radius of curvature at any point (r, θ) is proportional to r. Also find the radius of curvature if θ = 0, π/4, π/2.
हिंदी में प्रश्न
दर्शाइये कि कार्डिऑइड (cardioid) r = a(1 + cos θ) के किसी बिन्दु (r, θ) पर वक्रता-त्रिज्या (radius of curvature) का वर्ग r के समानुपाती है। θ = 0, π/4, π/2 पर वक्रता-त्रिज्या भी ज्ञात कीजिये।
Practise this question Vector Analysis notes
Worked solution
r = a(1 + cos θ), r₁ = −a sin θ, r₂ = −a cos θ.
r² + r₁² = a²(2 + 2 cos θ) = 2ar.
r² + 2r₁² − rr₂ = a²[(1 + cos θ)² + 2 sin²θ + cos θ(1 + cos θ)] = 3a²(1 + cos θ) = 3ar.
ρ = (2ar)^3/2/(3ar) = (2√2/3)√(ar), so ρ² = (8a/9)r, proportional to r.
Since r = 2a cos²(θ/2), ρ = (4a/3) cos(θ/2) for −π < θ < π.
Answer: ρ = 4a/3 at θ = 0; ρ = (4a/3) cos(π/8) = (2a/3)√(2 + √2) ≈ 1.232a at θ = π/4; ρ = 2√2a/3 ≈ 0.943a at θ = π/2.
Final answer: ρ = 4a/3 at θ = 0; ρ = (4a/3) cos(π/8) = (2a/3)√(2 + √2) ≈ 1.232a at θ = π/4; ρ = 2√2a/3 ≈ 0.943a at θ = π/2.

UPSC Mains 2016 Mathematics Paper II