Q1 50M Compulsory solve Linear algebra, calculus and 3D geometry
(a) Can the set {(0, 0, 0, 3), (1, 1, 0, 0), (0, 1, –1, 0)} be extended to form a basis of the vector space ℝ⁴? Justify your answer. 10 marks
(b) Find the range, rank, kernel and nullity of the linear transformation T : ℝ⁴ → ℝ³ given by T(x, y, z, w) = (x – w, y + z, z – w). 10 marks
(c) A rectangular sheet of metal of length 6 meters and width 2 meters is given. Four equal squares are removed from the four corners. The sides of this sheet are now folded up to form an open rectangular box. Find approximately the height of the box, such that the volume of the box is maximum. 10 marks
(d) Given that f(x + y) = f(x) f(y) for all real x, y, f(x) ≠ 0 for any real x and f'(0) = 2. Show that for all real x, f'(x) = 2f(x). Hence find f(x). 10 marks
(e) Find the equation of the cone whose vertex is the point (1, 1, 0) and whose guiding curve is y = 0, x² + z² = 4. 10 marks
हिंदी में पढ़ें
(a) क्या समुच्चय {(0, 0, 0, 3), (1, 1, 0, 0), (0, 1, –1, 0)} को सदिश समष्टि ℝ⁴ का एक आधार बनाने के लिए विस्तारित किया जा सकता है? अपने उत्तर की पुष्टि कीजिए। 10 अंक
(b) रैखिक रूपांतरण T : ℝ⁴ → ℝ³, जो T(x, y, z, w) = (x – w, y + z, z – w) द्वारा दिया गया है, का परिसर (रेंज), कोटि (रैंक), अश्टि (कर्नेल) और शून्यता ज्ञात कीजिए। 10 अंक
(c) लम्बाई 6 मीटर और चौड़ाई 2 मीटर की एक आयताकार धातु की चादर दी गई है। चारों कोनों से चार बराबर वर्गों को हटाया गया है। इस चादर के फलकों को मोड़कर एक खुला आयताकार संदूक बनाना है। संदूक की ऐसी सन्निकट ऊँचाई ज्ञात कीजिए कि संदूक का आयतन अधिकतम हो। 10 अंक
(d) दिया गया है कि f(x + y) = f(x) f(y), सभी वास्तविक x, y के लिए, f(x) ≠ 0 किसी भी वास्तविक x के लिए और f'(0) = 2 है। सभी वास्तविक x के लिए दर्शाइए कि f'(x) = 2f(x) है। अतः f(x) ज्ञात कीजिए। 10 अंक
(e) उस शंकु का समीकरण ज्ञात कीजिए जिसका शीर्ष बिंदु (1, 1, 0) है तथा जिसका निर्देशक वक्र y = 0, x² + z² = 4 है। 10 अंक
Answer approach & key points
Framework: UPSC Mathematics Paper 1. (a) justify: claim > 3-4 reasons > evidence > conclusion | (b) calculate: given > formula > substitution > result with units > interpretation | (c) calculate: given > formula > substitution > result with units > interpretation | (d) derive: given > assumptions > stepwise derivation > result > check | (e) derive: given > assumptions > stepwise derivation > result > check Full marks: Rigorous step-by-step derivation with all justifications and checks.
- Verify α=β=γ=0 (linear independence of the three given vectors)
- Use dim ℝ⁴ = 4 and the basis-extension principle
- Exhibit an added vector outside the span (e.g., e₁ = (1,0,0,0)) with justification
- Verify independence of the four vectors and state the final basis
- Write the matrix of T and identify its columns
- Show rank = 3 and range = ℝ³
- Solve T = 0: kernel = span{(1,–1,1,1)}; nullity = 1
- Verify by rank–nullity: 3 + 1 = 4
Q2 50M solve Linear transformation, mean value theorem and 3D geometry
(a) Let T : ℝ³ → ℝ² be a linear transformation such that T(1, 1, -1) = (1, 0), T(4, 1, 1) = (0, 1) and T(1, -1, 2) = (1, 1). Find T. 15 marks
(b) Using Mean Value Theorem, prove that
π/6 + √3/15 < sin⁻¹(3/5) < π/6 + 1/8 15 marks
(c) (i) Find the equation of the cylinder whose generators are parallel to the line x/1 = y/2 = z/3 and that passes through the curve x² + y² = 16, z = 0. 10 marks
(ii) Find the shortest distance between the straight lines
(x-3)/3 = (y-8)/(-1) = (z-3)/1 and (x+3)/(-3) = (y+7)/2 = (z-6)/4. 10 marks
हिंदी में पढ़ें
(a) माना T : ℝ³ → ℝ² एक ऐसा रैखिक रूपांतरण है कि T(1, 1, -1) = (1, 0), T(4, 1, 1) = (0, 1) तथा T(1, -1, 2) = (1, 1) है। T ज्ञात कीजिए। 15 अंक
(b) माध्यमान प्रमेय का प्रयोग करते हुए सिद्ध कीजिए कि
π/6 + √3/15 < sin⁻¹(3/5) < π/6 + 1/8 15 अंक
(c) (i) उस बेलन का समीकरण ज्ञात कीजिए जिसके जनक, रेखा x/1 = y/2 = z/3 के समांतर हैं और जो वक्र x² + y² = 16, z = 0 से होकर गुजरता है। 10 अंक
(ii) सरल रेखाओं
(x-3)/3 = (y-8)/(-1) = (z-3)/1 और (x+3)/(-3) = (y+7)/2 = (z-6)/4
के बीच की न्यूनतम दूरी ज्ञात कीजिए। 10 अंक
Answer approach & key points
(a) calculate: given > formula > substitution > result with units > interpretation | (b) justify: claim > 3-4 reasons > evidence > conclusion | (c(i)) calculate: given > formula > substitution > result with units > interpretation | (c(ii)) calculate: given > formula > substitution > result with units > interpretation Full marks: Complete, rigorous derivations with all steps justified and verified.
- Recognize the three given vectors form a basis (det = 1 ≠ 0)
- Set up and solve the two 3×3 systems for (a,b,c) and (d,e,f)
- Obtain T(x,y,z) = (5x – 12y – 8z, x – 2y – z)
- Verify against all three given values
- State the Mean Value Theorem and apply it to f(x) = sin⁻¹x on [1/2, 3/5]
- sin⁻¹(3/5) – π/6 = f′(c)/10 with f′(x) = 1/√(1–x²)
- Bound f′: f′(1/2) = 2/√3, f′(3/5) = 5/4
- Derive √3/15 < sin⁻¹(3/5) – π/6 < 1/8, i.e. the required inequality
Q3 50M solve Linear algebra, analytical geometry, multivariable calculus
(a) Reduce the following matrix to echelon form:
A = 2 & -2 & 2 & 1
-3 & 6 & 0 & -1
1 & -7 & 10 & 2 (15 marks)
(b) Find the equations of the spheres which pass through the circle x² + y² + z² - 2x + 2y + 4z - 3 = 0, 2x + y + z = 4and touch the plane3x + 4y = 14. (15 marks)
(c) (i) Evaluate displaystyle∬limits_R y dx dy, where R is the region bounded by y = x and y = 4x - x². (10 marks)
(ii) If u(x,y) = x f(y/x) + g(y/x), where f and g are arbitrary functions, then show that
I. x (∂ u)/(∂ x) + y (∂ u)/(∂ y) = x f(y/x),
II. x² (∂² u)/(∂ x²) + 2xy (∂² u)/(∂ x ∂ y) + y² (∂² u)/(∂ y²) = 0. (10 marks)
हिंदी में पढ़ें
(a) निम्नलिखित आव्यूह को सोपानक (एशेलोन) रूप में समानीत कीजिए :
A = 2 & -2 & 2 & 1
-3 & 6 & 0 & -1
1 & -7 & 10 & 2 (15 अंक)
(b) उन गोलों के समीकरण ज्ञात कीजिए जो वृत्त x² + y² + z² - 2x + 2y + 4z - 3 = 0, 2x + y + z = 4से होकर गुजरते हैं और समतल3x + 4y = 14 को स्पर्श करते हैं। (15 अंक)
(c) (i) displaystyle∬limits_R y dx dy का मान ज्ञात कीजिए, जहाँ R, y = x तथा y = 4x - x² से परिवृत क्षेत्र है। (10 अंक)
(ii) यदि u(x,y) = x f(y/x) + g(y/x) है, जहाँ f और g स्वेच्छ फलन हैं, तो दर्शाइए कि
I. x (∂ u)/(∂ x) + y (∂ u)/(∂ y) = x f(y/x) है,
II. x² (∂² u)/(∂ x²) + 2xy (∂² u)/(∂ x ∂ y) + y² (∂² u)/(∂ y²) = 0 है। (10 अंक)
Answer approach & key points
(a) calculate: given > formula > substitution > result with units > interpretation | (b) calculate: given > formula > substitution > result with units > interpretation | (c(i)) calculate: given > formula > substitution > result with units > interpretation | (c(ii)) derive: given > assumptions > stepwise derivation > result > check Full marks: Complete working with all steps shown, correct results, and verification
- Interchange R1 and R3; apply elementary row operations
- R2→R2+3R1: [0,−15,30,5]; R3→R3−2R1: [0,12,−18,−3]
- R3→R3+(4/5)R2 gives [0,0,6,1]; normalize rows
- Echelon form [1,−7,10,2; 0,1,−2,−1/3; 0,0,1,1/6]; rank 3 (three non-zero rows)
- Family of spheres S+λP = 0 through the circle
- Centre (1−λ, −(2+λ)/2, −(4+λ)/2); r² = (3/2)λ² + 5λ + 9
- Tangency: distance to 3x+4y−14 = 0 equals radius: |λ+3|
- (λ+3)² = r² ⇒ λ = 0, 2; spheres x²+y²+z²−2x+2y+4z−3 = 0 and x²+y²+z²+2x+4y+6z−11 = 0
Q4 50M prove Analytical geometry, partial derivatives, linear algebra
(a) Show that there is no tangent plane to the sphere
x² + y² + z² - 4x + 2y - 4z + 4 = 0
that can be passed through the straight line
(x+6)/2 = y + 3 = z + 1. (15 marks)
(b) If f(x, y) = { xy(x²-y²)/(x²+y²), when (x,y) ≠ (0,0)
{ 0, when (x,y) = (0,0),
then find f_xy(0,0) and f_yx(0,0). (15 marks)
(c) (i) Find the eigenvalues and the corresponding eigenvectors of the matrix
A = [1 2 0]
[2 1 -6]
[2 -2 3] (12 marks)
(ii) Let P_n denote the vector space of all polynomials of degree ≤ n over R. Verify that
dim(P_4/P_2) = dim P_4 - dim P_2. (8 marks)
हिंदी में पढ़ें
(a) दर्शाइए कि गोले
x² + y² + z² - 4x + 2y - 4z + 4 = 0
का कोई ऐसा स्पर्श समतल नहीं है, जो कि सरल रेखा (x+6)/2 = y + 3 = z + 1 से होकर गुजर सके। (15 अंक)
(b) यदि f(x, y) = { xy(x²-y²)/(x²+y²), जब (x,y) ≠ (0,0)
{ 0, जब (x,y) = (0,0)
है, तो f_xy(0,0) और f_yx(0,0) ज्ञात कीजिए। (15 अंक)
(c) (i) आव्यूह A = [1 2 0]
[2 1 -6]
[2 -2 3]
के अभिलक्षणिक मान और संगत अभिलक्षणिक सदिश ज्ञात कीजिए। (12 अंक)
(ii) माना P_n, R पर घात ≤ n के सभी बहुपदों के सदिश समष्टि को दर्शाता है। सत्यापित कीजिए कि
dim(P_4/P_2) = dim P_4 - dim P_2। (8 अंक)
Answer approach & key points
(a) justify: claim > 3-4 reasons > evidence > conclusion | (b) calculate: given > formula > substitution > result with units > interpretation | (c(i)) calculate: given > formula > substitution > result with units > interpretation | (c(ii)) justify: claim > 3-4 reasons > evidence > conclusion Full marks: Rigorous derivations with all steps shown and correct results
- Complete the square: centre (2,−1,2), R = √5; line through A = (−6,−3,−1), v = (2,1,1)
- C − A = (8,2,3); cross product (C−A)×v = (−1,−2,4)
- Line-to-centre distance = √21/√6 = √(7/2)
- √(7/2) < √5 ⇒ every plane through the line misses tangency; conclude no such plane exists
- Compute f_x for (x,y) ≠ (0,0); f_x(0,y) = −y; f_x(0,0) = 0
- f_xy(0,0) = lim (−k)/k = −1
- Compute f_y(x,0) = x; f_y(0,0) = 0
- f_yx(0,0) = lim h/h = 1; note the mixed partials are unequal