Q1 50M Compulsory solve Linear algebra and calculus
(a) Let H be a subspace of R⁴ spanned by the vectors v₁ = (1, –2, 5, –3), v₂ = (2, 3, 1, –4), v₃ = (3, 8, –3, –5). Then find a basis and dimension of H, and extend the basis of H to a basis of R⁴. (10 marks)
(b) Let T : R³ → R³ be a linear operator and B = {v₁, v₂, v₃} be a basis of R³ over R. Suppose that Tv₁ = (1, 1, 0), Tv₂ = (1, 0, –1), Tv₃ = (2, 1, –1). Find a basis for the range space and null space of T. (10 marks)
(c) Discuss the continuity of the function
f(x) = { 1/(1–e^(–1/x)), x ≠ 0
{ 0, x = 0
for all values of x. (10 marks)
(d) Expand ln(x) in powers of (x–1) by Taylor's theorem and hence find the value of ln(1·1) correct up to four decimal places. (10 marks)
(e) Find the equation of the right circular cylinder which passes through the circle x² + y² + z² = 9, x – y + z = 3. (10 marks)
हिंदी में पढ़ें
(a) माना H, R⁴ की एक उपसमष्टि है, जो कि सदिशों v₁ = (1, –2, 5, –3), v₂ = (2, 3, 1, –4), v₃ = (3, 8, –3, –5) द्वारा जनित है। तब H का एक आधार एवं विमा ज्ञात कीजिए तथा H के इस आधार को R⁴ के एक आधार तक विस्तृत कीजिए। (10 अंक)
(b) माना T : R³ → R³ एक रैखिक संकारक है तथा R पर R³ का एक आधार B = {v₁, v₂, v₃} है। माना कि Tv₁ = (1, 1, 0), Tv₂ = (1, 0, –1), Tv₃ = (2, 1, –1) है। T की परिसर समष्टि तथा शून्य समष्टि के लिए एक आधार ज्ञात कीजिए। (10 अंक)
(c) x के सभी मानों के लिए फलन
f(x) = { 1/(1–e^(–1/x)), x ≠ 0
{ 0, x = 0
के सांतत्य की चर्चा कीजिए। (10 अंक)
(d) टेलर प्रमेय द्वारा ln(x) का (x–1) की घात में प्रसार कीजिए तथा ln(1·1) का दशमलव के चार स्थानों तक सही मान ज्ञात कीजिए। (10 अंक)
(e) वृत्त x² + y² + z² = 9, x – y + z = 3 से होकर जाने वाले लम्ब वृत्तीय बेलन का समीकरण ज्ञात कीजिए। (10 अंक)
Answer approach & key points
Framework: UPSC Mathematics Paper 1. (a) calculate: given > formula > substitution > result with units > interpretation | (b) calculate: given > formula > substitution > result with units > interpretation | (c) discuss: intro > 3-4 dimensions > example > balanced close | (d) calculate: given > formula > substitution > result with units > interpretation | (e) derive: given > assumptions > stepwise derivation > result > check Full marks: All parts solved with full working, correct results, and verification steps.
- Dependence relation v1 − 2v2 + v3 = 0 (or row reduction)
- Basis {v1, v2}, dim H = 2
- Extension with e3, e4 (or other valid vectors)
- Proof that the four vectors are independent
- General image T(av1+bv2+cv3) = (a+b+2c, a+c, −b−c)
- Tv3 = Tv1 + Tv2; range basis {(1,1,0),(1,0,−1)}, rank 2
- Null space: a = b = −c; basis {v3 − v1 − v2}, nullity 1
- Rank–nullity check
- Continuity for x ≠ 0 (denominator never zero)
- Right limit at 0 = 1
- Left limit at 0 = 0 = f(0)
- Conclusion: discontinuous at 0 (jump); left-continuous, not right-continuous
- Derivatives f⁽ⁿ⁾(1) = (−1)^(n−1)(n−1)!
- Series Σ (−1)^(n+1)(x−1)ⁿ/n with validity 0 < x ≤ 2
- Substitution x − 1 = 0.1 with enough terms
- ln 1.1 ≈ 0.0953
- Axis direction = plane normal (1, −1, 1)
- Centre of circle (1, −1, 1) and radius √6
- Distance-from-axis formula |(X − c) × n|²/|n|² = 6
- Equation x² + y² + z² + xy + yz − zx = 9
Q2 50M solve Linear operators and multivariable calculus
(a) Consider a linear operator T on R³ over R defined by T(x, y, z) = (2x, 4x – y, 2x + 3y – z). Is T invertible? If yes, justify your answer and find T⁻¹. (15 marks)
(b) If u = (x+y)/(1-xy) and v = tan⁻¹x + tan⁻¹y, then find ∂(u, v)/∂(x, y). Are u and v functionally related? If yes, find the relationship. (15 marks)
(c) Find the image of the line x = 3-6t, y = 2t, z = 3+2t in the plane 3x+4y-5z+26 = 0. (20 marks)
हिंदी में पढ़ें
(a) माना R के ऊपर R³ पर एक रैखिक संकारक T, T(x, y, z) = (2x, 4x – y, 2x + 3y – z) द्वारा परिभाषित है। क्या T व्युत्क्रमणीय है? यदि हाँ, तो अपने उत्तर का तर्क प्रस्तुत कीजिए तथा T⁻¹ ज्ञात कीजिए। (15 अंक)
(b) यदि u = (x+y)/(1-xy) तथा v = tan⁻¹x + tan⁻¹y है, तब ∂(u, v)/∂(x, y) ज्ञात कीजिए। क्या u तथा v फलनतः सम्बन्धित हैं? यदि हाँ, तो सम्बन्ध ज्ञात कीजिए। (15 अंक)
(c) रेखा x = 3-6t, y = 2t, z = 3+2t का समतल 3x+4y-5z+26 = 0 में प्रतिबिम्ब ज्ञात कीजिए। (20 अंक)
Answer approach & key points
(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 Full marks: Complete working with all steps, correct results, and verification.
- Matrix of T and det = 2 ≠ 0 (or kernel trivial)
- Conclusion: T invertible
- Solving for x, y, z
- T⁻¹(X,Y,Z) = (X/2, 2X − Y, 7X − 3Y − Z)
- Partial derivatives of u and v
- Jacobian = 0
- Inference of functional dependence
- Relation u = tan v (x = tan α, y = tan β)
- Intersection point (−3, 2, 5) at t = 1
- Reflection of a second point, e.g., (3,0,3) → (3/5, −16/5, 7), or reflection of direction vector
- Direction of image (−9, 13, −5)
- Equation (x + 3)/9 = (y − 2)/(−13) = (z − 5)/5
Q3 50M solve Linear algebra, optimization and 3D geometry
(a) Let V = M₂ₓ₂(ℝ) denote a vector space over the field of real numbers. Find the matrix of the linear mapping φ: V → V given by φ(v) = 1 & 2
3 & -1 v with respect to standard basis of M₂ₓ₂(ℝ), and hence find the rank of φ. Is φ invertible? Justify your answer. (15 marks)
(b) Find the volume of the greatest cylinder which can be inscribed in a cone of height h and semi-vertical angle α. (20 marks)
(c) Find the vertex of the cone 4x² - y² + 2z² + 2xy - 3yz + 12x - 11y + 6z + 4 = 0. (15 marks)
हिंदी में पढ़ें
(a) माना V = M₂ₓ₂(ℝ) वास्तविक संख्याओं के क्षेत्र पर एक सदिश समष्टि दर्शाता है। M₂ₓ₂(ℝ) के मानक आधार के सन्दर्भ में φ(v) = 1 & 2
3 & -1 v द्वारा दिए गए रैखिक प्रतिचित्रण φ: V → V का आव्यूह ज्ञात कीजिए और तब φ की कोटि (रैंक) ज्ञात कीजिए। क्या φ व्युत्क्रमणीय है? अपने उत्तर का तर्क प्रस्तुत कीजिए। (15 अंक)
(b) ऊँचाई h तथा अर्ध-शीर्ष कोण α वाले एक शंकु के अंतर्गत सबसे बड़े बेलन का आयतन ज्ञात कीजिए। (20 अंक)
(c) शंकु 4x² - y² + 2z² + 2xy - 3yz + 12x - 11y + 6z + 4 = 0 का शीर्ष ज्ञात कीजिए। (15 अंक)
Answer approach & key points
(a) calculate: given > formula > substitution > result with units > interpretation | (b) calculate: given > formula > substitution > result with units > interpretation | (c) calculate: given > formula > substitution > result with units > interpretation Full marks: Complete method with all steps, verification, and clear justification
- Images of E11, E12, E21, E22
- 4×4 matrix with columns (1,0,3,0), (0,1,0,3), (2,0,−1,0), (0,2,0,−1)
- det M = 49 ≠ 0 (or det A = −7) ⇒ rank 4
- φ invertible with φ⁻¹(w) = A⁻¹w, A⁻¹ = (1/7)(1 2; 3 −1)
- Relation r = (h − y) tan α from similar triangles
- V(y) = π tan²α · y(h − y)²
- dV/dy = 0 ⇒ y = h/3; second-derivative or endpoint check
- Vmax = 4πh³tan²α/27
- Partial derivatives Fx, Fy, Fz set to zero
- Solving to x = −1, y = −2, z = −3
- Check that the vertex satisfies F = 0 (or constant term vanishes after shift)
- Vertex (−1, −2, −3)
Q4 50M solve Eigenvalues, multiple integration and sphere geometry
(a) Let A = 3 & 2 & 4
2 & 0 & 2
4 & 2 & 3 be a 3×3 matrix. Find the eigenvalues and the corresponding eigenvectors of A. Hence find the eigenvalues and the corresponding eigenvectors of A⁻¹⁵, where A⁻¹⁵ = (A⁻¹)¹⁵. (20 marks)
(b) Using double integration, find the area lying inside the cardioid r = a(1+cos θ) and outside the circle r = a. (15 marks)
(c) Find the equation of the sphere which touches the plane 3x+2y-z+2=0 at the point (1, -2, 1) and cuts orthogonally the sphere x²+y²+z²-4x+6y+4=0. (15 marks)
हिंदी में पढ़ें
(a) माना A = 3 & 2 & 4
2 & 0 & 2
4 & 2 & 3 एक 3×3 आव्यूह है। A का अभिलक्षणिक मान तथा संगत अभिलक्षणिक सदिश ज्ञात कीजिए। अतः A⁻¹⁵ का अभिलक्षणिक मान तथा संगत अभिलक्षणिक सदिश ज्ञात कीजिए, जहाँ A⁻¹⁵ = (A⁻¹)¹⁵ है। (20 अंक)
(b) दिशा: समाकलन का प्रयोग करते हुए हृदयाक्ष (कार्डिओइड) r = a(1+cos θ) के अंदर तथा वृत्त r = a के बाह्य स्थित क्षेत्र का क्षेत्रफल ज्ञात कीजिए। (15 अंक)
(c) उस गोले का समीकरण ज्ञात कीजिए, जो समतल 3x+2y-z+2=0 को बिंदु (1, -2, 1) पर स्पर्श करता है और गोले x²+y²+z²-4x+6y+4=0 को लंबिकतः काटता है। (15 अंक)
Answer approach & key points
(a) calculate: given > formula > substitution > result with units > interpretation | (b) calculate: given > formula > substitution > result with units > interpretation | (c) calculate: given > formula > substitution > result with units > interpretation Full marks: Complete derivations, correct final answers, clear notation
- Characteristic polynomial −(λ + 1)²(λ − 8); eigenvalues −1, −1, 8
- Eigenvector for 8: (2, 1, 2)
- Two independent eigenvectors for −1, e.g., (1, −2, 0), (1, 0, −1)
- A⁻¹⁵: eigenvalues −1, −1, 8⁻¹⁵ with the same eigenvectors (justified via Av = λv ⇒ A⁻¹v = λ⁻¹v)
- Limits a ≤ r ≤ a(1 + cos θ), −π/2 ≤ θ ≤ π/2
- Double integral set-up with r dr dθ
- Inner integration (a²/2)(2cos θ + cos²θ)
- Area = a²(2 + π/4)
- Centre on the normal through P: (1 + 3t, −2 + 2t, 1 − t)
- Given sphere centre (2, −3, 0), radius 3
- Orthogonality d² = R² + r² ⇒ t = −3/2
- Equation x² + y² + z² + 7x + 10y − 5z + 12 = 0