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
Solve each sub-part systematically with clear mathematical reasoning. For (a), verify linear independence and extend to basis; for (b), construct the matrix representation and apply rank-nullity; for (c), set up the volume function and optimize using calculus; for (d), use the functional equation to derive the differential equation; for (e), use the cone generator method. Allocate approximately 20% time per sub-part given equal marks distribution, ensuring each part receives complete treatment with proper justification.
- (a) Verify linear independence of the three vectors by checking no non-trivial combination equals zero; extend to basis by adding a fourth vector not in their span, such as (0,0,1,0) or (1,0,0,0)
- (b) Construct the 3×4 matrix representation of T; find rank by row reduction or determinant of 3×3 minors; determine nullity using rank-nullity theorem (nullity = 4 - rank = 2)
- (c) Express volume V = (6-2h)(2-2h)h = 4h(3-h)(1-h); find dV/dh = 0 yielding 3h² - 8h + 3 = 0; select valid root h = (4-√7)/3 ≈ 0.45m within domain 0 < h < 1
- (d) Use f(x+y) = f(x)f(y) to show [f(x+h)-f(x)]/h = f(x)[f(h)-1]/h; as h→0, f'(x) = f(x)f'(0) = 2f(x); solve to get f(x) = e^(2x)
- (e) For cone with vertex (1,1,0) and guiding curve y=0, x²+z²=4: parametrize generator lines (1+(x₀-1)t, 1-t, z₀t) where x₀²+z₀²=4; eliminate parameters to get (x-1)² + z² = 4(y-1)² with y ≤ 1
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
Solve all four sub-parts systematically, allocating approximately 30% time to part (a) on linear transformation, 30% to part (b) on Mean Value Theorem proof, 20% to part (c)(i) on cylinder equation, and 20% to part (c)(ii) on shortest distance. Begin each part with clear statement of the approach, show complete working with proper mathematical notation, and conclude with boxed final answers.
- For (a): Verify that {(1,1,-1), (4,1,1), (1,-1,2)} forms a basis for ℝ³, then express T as a 2×3 matrix by solving for images of standard basis vectors or using linearity directly
- For (b): Apply MVT to f(x) = sin⁻¹x on [1/2, 3/5], showing f'(c) = 1/√(1-c²) lies between 4/5 and 5/√39, then manipulate to obtain the required bounds
- For (c)(i): Use the condition that for any point (x,y,z) on cylinder, its distance from the axis line x/1=y/2=z/3 equals the radius 4, giving (2x-y)² + (3y-2z)² + (3x-z)² = 16×14 or equivalent simplified form
- For (c)(ii): Verify the lines are skew, then apply SD formula |(a₂-a₁)·(b₁×b₂)|/|b₁×b₂| with correct identification of points and direction vectors, obtaining exact numerical value
- Clear demonstration of linearity properties in (a), proper interval selection in (b), correct generator direction handling in (c)(i), and accurate cross product computation in (c)(ii)
Q3 50M solve Linear algebra, analytical geometry, multivariable calculus
(a) Reduce the following matrix to echelon form:
$$A = \begin{bmatrix} 2 & -2 & 2 & 1 \\ -3 & 6 & 0 & -1 \\ 1 & -7 & 10 & 2 \end{bmatrix}$$ (15 marks)
(b) Find the equations of the spheres which pass through the circle $x^2 + y^2 + z^2 - 2x + 2y + 4z - 3 = 0$, $2x + y + z = 4$ and touch the plane $3x + 4y = 14$. (15 marks)
(c) (i) Evaluate $\displaystyle\iint\limits_R y\, dx\, dy$, where R is the region bounded by $y = x$ and $y = 4x - x^2$. (10 marks)
(ii) If $u(x,y) = x\, f\left(\dfrac{y}{x}\right) + g\left(\dfrac{y}{x}\right)$, where f and g are arbitrary functions, then show that
I. $\quad x\,\dfrac{\partial u}{\partial x} + y\,\dfrac{\partial u}{\partial y} = x\, f\left(\dfrac{y}{x}\right)$,
II. $\quad x^2\,\dfrac{\partial^2 u}{\partial x^2} + 2xy\,\dfrac{\partial^2 u}{\partial x\,\partial y} + y^2\,\dfrac{\partial^2 u}{\partial y^2} = 0$. (10 marks)
हिंदी में पढ़ें
(a) निम्नलिखित आव्यूह को सोपानक (एशेलोन) रूप में समानीत कीजिए :
$$A = \begin{bmatrix} 2 & -2 & 2 & 1 \\ -3 & 6 & 0 & -1 \\ 1 & -7 & 10 & 2 \end{bmatrix}$$ (15 अंक)
(b) उन गोलों के समीकरण ज्ञात कीजिए जो वृत्त $x^2 + y^2 + z^2 - 2x + 2y + 4z - 3 = 0$, $2x + y + z = 4$ से होकर गुजरते हैं और समतल $3x + 4y = 14$ को स्पर्श करते हैं। (15 अंक)
(c) (i) $\displaystyle\iint\limits_R y\, dx\, dy$ का मान ज्ञात कीजिए, जहाँ R, $y = x$ तथा $y = 4x - x^2$ से परिवृत क्षेत्र है। (10 अंक)
(ii) यदि $u(x,y) = x\, f\left(\dfrac{y}{x}\right) + g\left(\dfrac{y}{x}\right)$ है, जहाँ f और g स्वेच्छ फलन हैं, तो दर्शाइए कि
I. $\quad x\,\dfrac{\partial u}{\partial x} + y\,\dfrac{\partial u}{\partial y} = x\, f\left(\dfrac{y}{x}\right)$ है,
II. $\quad x^2\,\dfrac{\partial^2 u}{\partial x^2} + 2xy\,\dfrac{\partial^2 u}{\partial x\,\partial y} + y^2\,\dfrac{\partial^2 u}{\partial y^2} = 0$ है। (10 अंक)
Answer approach & key points
Solve all four sub-parts systematically, allocating approximately 30% time to part (a) matrix reduction, 30% to part (b) sphere equations, 25% to part (c)(i) double integration, and 15% to part (c)(ii) partial derivative proofs. Begin each sub-part with clear identification of the mathematical technique, show complete working with row operations for (a), sphere family parameterization for (b), region sketching and limits for (c)(i), and chain rule applications for (c)(ii). Conclude with boxed final answers for each part.
- For (a): Correct application of elementary row operations to achieve row echelon form with leading entries 1, 2, 0 in successive rows and proper identification of pivot positions
- For (b): Formation of sphere family equation S + λP = 0, correct application of tangency condition using distance from center to plane equals radius, yielding two valid sphere equations
- For (c)(i): Accurate sketch of region R bounded by line y=x and parabola y=4x-x², correct intersection points (0,0) and (3,3), proper order of integration with limits 0 to 3 for x and x to 4x-x² for y
- For (c)(ii): Verification of Euler's homogeneous function theorem for part I using substitution v=y/x and chain rule, and confirmation of part II as second-degree Euler equation for homogeneous functions
- Clear presentation of all row operation steps in (a) with explicit notation (R₂ → R₂ + ³⁄₂R₁ etc.) to enable partial credit tracing
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
Prove the geometric impossibility in (a), calculate mixed partial derivatives in (b), and solve the eigenvalue problem with quotient space verification in (c). Allocate approximately 30% time to part (a) as it requires conceptual rigour, 30% to part (b) for careful limit analysis, 25% to (c)(i) for eigenvalue computation, and 15% to (c)(ii) for the dimension theorem verification. Begin with sphere-line analysis, proceed to partial derivative limits, then matrix characteristic polynomial, and conclude with basis construction for the quotient space.
- For (a): Complete the square to find sphere centre (2,-1,2) and radius 3; verify the given line does not intersect the sphere or lies entirely outside; show the distance from centre to line exceeds radius, making tangent plane impossible
- For (b): Compute f_x(0,y) using limit definition, then f_xy(0,0); compute f_y(x,0) then f_yx(0,0); demonstrate f_xy(0,0) = -1 and f_yx(0,0) = 1, showing inequality of mixed partials
- For (c)(i): Find characteristic equation det(A-λI)=0 yielding eigenvalues λ=3,3,-2; find eigenvectors for λ=3 (geometric multiplicity 1) and λ=-2; handle the defective case for repeated eigenvalue appropriately
- For (c)(ii): Identify basis {1,x,x²,x³,x⁴} for P₄ and {1,x,x²} for P₂; construct coset basis {x³+P₂, x⁴+P₂} for P₄/P₂; verify dim(P₄/P₂)=2=5-3=dim P₄-dim P₂
- Cross-part rigour: Use ε-δ arguments or explicit limit calculations in (b); justify why tangent plane condition fails via distance formula or system inconsistency in (a)