Exam Topic Analytics
Frequency of topics based on your complete set of Previous Year Questions (PYQs).
Topic Frequency in Last 3 Exams
Unit 1: High Yield Topics
System of Linear Equations: Checking consistency and finding infinite/unique solutions with parameters ($\lambda, \mu$). Present in every paper.Cayley-Hamilton Theorem: Guaranteed question. Usually involves finding inverse or simplifying large matrix polynomials.Eigenvalues & Diagonalization: Finding roots of characteristic equations, Modal Matrices ($P$).
Unit 2: High Yield Topics
Euler's Theorem: Guaranteed question. Involves composite functions ($u = \sin^{-1}(\text{homogeneous})$) requiring the deduction formula.Leibnitz Theorem: Finding $n$th derivative, setting up a $y_2, y_1, y$ differential equation.Maxima/Minima: Finding stationary points by setting $p=0, q=0$, and classifying using $rt-s^2$.
PREVIOUS YEAR PAPER – 1 (Oct 2024)
Question 1 (Unit 1 & 2 Short Notes)
Q1(a). Find the rank of the matrix $ A = \begin{bmatrix} 1 & 2 & 1 & 2 \\ 1 & 3 & 2 & 2 \\ 2 & 4 & 3 & 4 \\ 3 & 7 & 4 & 6 \end{bmatrix} $ by reducing it to echelon form.
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Step 1: Apply row transformations. $ R_2 \to R_2 - R_1 $, $ R_3 \to R_3 - 2R_1 $, $ R_4 \to R_4 - 3R_1 $
$ A \sim \begin{bmatrix} 1 & 2 & 1 & 2 \\ 0 & 1 & 1 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 1 & 1 & 0 \end{bmatrix} $
Step 2: $ R_4 \to R_4 - R_2 $
$ A \sim \begin{bmatrix} 1 & 2 & 1 & 2 \\ 0 & 1 & 1 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 \end{bmatrix} $
Final Result: Echelon Form reached. Non-zero rows = 3. Rank(A) = 3 .
Q1(b). Show that the eigen values of a Skew-Hermitian matrix are either zero or purely imaginary.
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Step 1: Let $A^\theta = -A$. Let $ AX = \lambda X $.
Step 2: Pre-multiply by $X^\theta$: $ X^\theta A X = \lambda X^\theta X $. Take conjugate transpose: $ X^\theta A^\theta X = \bar{\lambda} X^\theta X $.
Step 3: Substitute $A^\theta = -A$: $ -X^\theta A X = \bar{\lambda} X^\theta X \implies -\lambda X^\theta X = \bar{\lambda} X^\theta X \implies (\lambda + \bar{\lambda}) X^\theta X = 0 $.
Final Result: Since $X \neq 0$, $\lambda + \bar{\lambda} = 0$. If $\lambda = a + ib$, $2a = 0 \implies a = 0$. $\lambda = ib$ (purely imaginary or zero).
Q1(c). Find the nth derivative of $ e^{2x} \cos^2 x \sin x $.
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Step 1: Simplify: $\cos^2 x \sin x = \frac{1}{4}\sin x + \frac{1}{4}\sin 3x$.
Step 2: Function becomes $ y = \frac{1}{4} e^{2x} \sin x + \frac{1}{4} e^{2x} \sin 3x $.
Step 3: Use $ D^n(e^{ax}\sin(bx+c)) = (a^2+b^2)^{n/2} e^{ax} \sin(bx+c + n\tan^{-1}(b/a)) $.
Final Result: $ y_n = \frac{1}{4} 5^{n/2} e^{2x} \sin(x + n\tan^{-1}\frac{1}{2}) + \frac{1}{4} 13^{n/2} e^{2x} \sin(3x + n\tan^{-1}\frac{3}{2}) $
Q1(d). Discuss the continuity of $ f(x,y) = \frac{x^3 y}{x^4 + y^4} $ at $(x,y) \neq (0,0)$ and $0$ at $(0,0)$.
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Step 1: Check limit along path $y = mx$.
$ \lim_{x \to 0} \frac{x^3(mx)}{x^4 + (mx)^4} = \lim_{x \to 0} \frac{m x^4}{x^4(1+m^4)} = \frac{m}{1+m^4} $
Final Result: The limit depends on $m$. Limit does not exist. Discontinuous at $(0,0)$ .
Question 2 (Unit 1 - Linear Algebra)
Q2(a). Using Gauss-Jordan method, find the inverse of $ \begin{bmatrix} 2 & 3 & 4 \\ 4 & 3 & 1 \\ 1 & 2 & 4 \end{bmatrix} $.
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Step 1: Form $[A | I]$. Swap $R_1 \leftrightarrow R_3$.
$ \left[\begin{array}{ccc|ccc} 1 & 2 & 4 & 0 & 0 & 1 \\ 4 & 3 & 1 & 0 & 1 & 0 \\ 2 & 3 & 4 & 1 & 0 & 0 \end{array}\right] $
Step 2: $R_2 - 4R_1, R_3 - 2R_1 \implies \left[\begin{array}{ccc|ccc} 1 & 2 & 4 & 0 & 0 & 1 \\ 0 & -5 & -15 & 0 & 1 & -4 \\ 0 & -1 & -4 & 1 & 0 & -2 \end{array}\right] $
Step 3: Perform remaining row ops to get $I$ on the left.
Final Result: $ A^{-1} = \frac{1}{5} \begin{bmatrix} -10 & 4 & 9 \\ 15 & -4 & -14 \\ -5 & 1 & 6 \end{bmatrix} $
Q2(b). Solve the system: $ 3x_1-6x_2+3x_4=9 $, $ -2x_1+4x_2+2x_3-x_4=-11 $, $ 4x_1-8x_2+6x_3+7x_4=-5 $.
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Step 1: Form augmented matrix. Note $x_3$ is missing in eq 1. Divide $R_1$ by 3.
$ \begin{bmatrix} 1 & -2 & 0 & 1 & | & 3 \\ -2 & 4 & 2 & -1 & | & -11 \\ 4 & -8 & 6 & 7 & | & -5 \end{bmatrix} $
Step 2: $R_2 + 2R_1, R_3 - 4R_1 \implies \begin{bmatrix} 1 & -2 & 0 & 1 & | & 3 \\ 0 & 0 & 2 & 1 & | & -5 \\ 0 & 0 & 6 & 3 & | & -17 \end{bmatrix} $
Step 3: $R_3 - 3R_2 \implies \begin{bmatrix} 1 & -2 & 0 & 1 & | & 3 \\ 0 & 0 & 2 & 1 & | & -5 \\ 0 & 0 & 0 & 0 & | & -2 \end{bmatrix} $
Final Result: Rank(A) = 2, Rank(A|B) = 3. Inconsistent, no solution.
Q2(c). State Cayley Hamilton Theorem and calculate $ -A^3+4A^2+5A-21I $ where $ A = \begin{bmatrix} 1 & 0 & 2 \\ 0 & 1 & 1 \\ 0 & 0 & 2 \end{bmatrix} $.
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Statement: Every square matrix satisfies its own characteristic equation.
Step 1: $|A - \lambda I| = (1-\lambda)^2(2-\lambda) = \lambda^3 - 4\lambda^2 + 5\lambda - 2 = 0$. By C-H: $ A^3 - 4A^2 + 5A - 2I = 0 $.
Step 2: Evaluate $P(A) = -A^3 + 4A^2 + 5A - 21I = -(A^3 - 4A^2 + 5A - 2I) + 10A - 23I = 10A - 23I$.
Final Result: $ 10 \begin{bmatrix} 1 & 0 & 2 \\ 0 & 1 & 1 \\ 0 & 0 & 2 \end{bmatrix} - 23I = \begin{bmatrix} -13 & 0 & 20 \\ 0 & -13 & 10 \\ 0 & 0 & -3 \end{bmatrix} $
Question 3 (Unit 2 - Calculus)
Q3(a). If $ z = \sin^{-1}\left(\frac{x^{1/3} + y^{1/3}}{\sqrt{x} + \sqrt{y}}\right) $, evaluate $ x^2 \frac{\partial^2 z}{\partial x^2} + 2xy \frac{\partial^2 z}{\partial x \partial y} + y^2 \frac{\partial^2 z}{\partial y^2} $.
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Step 1: Let $ u = \sin z = \frac{x^{1/3} + y^{1/3}}{x^{1/2} + y^{1/2}} $. Degree $ n = 1/3 - 1/2 = -1/6 $.
Step 2: Euler's theorem: $ x z_x + y z_y = n \frac{u}{u'} = -\frac{1}{6} \frac{\sin z}{\cos z} = -\frac{1}{6} \tan z = F(z) $.
Step 3: Deduction formula: $ x^2 z_{xx} + 2xy z_{xy} + y^2 z_{yy} = F(z)[F'(z) - 1] $.
Final Result: $ -\frac{1}{6} \tan z \left( -\frac{1}{6} \sec^2 z - 1 \right) = \mathbf{\frac{\tan z}{36} (\sec^2 z + 6)} $.
Q3(b). State Leibnitz’s theorem and find the $n$th derivative of $y = x^2 e^{ax}$.
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Theorem: $(uv)_n = u_n v + n u_{n-1} v_1 + \frac{n(n-1)}{2} u_{n-2} v_2 + \dots$
Step 1: Let $u = e^{ax}, v = x^2$.
Step 2: $u_n = a^n e^{ax}$. $v_1 = 2x, v_2 = 2, v_3 = 0$.
Final Result: $y_n = a^n e^{ax}(x^2) + n a^{n-1} e^{ax} (2x) + \frac{n(n-1)}{2} a^{n-2} e^{ax} (2) = \mathbf{a^{n-2} e^{ax} [a^2 x^2 + 2nax + n(n-1)]}$.
Q3(c). Find the extreme values of $f(x,y) = x^3 + y^3 - 3axy$ ($a>0$).
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Step 1: $p = 3x^2 - 3ay = 0 \implies y = x^2/a$.
Step 2: $r = 6x, s = -3a, t = 6y$.
Step 3: At $(a,a)$: $r=6a, s=-3a, t=6a \implies rt-s^2 = 27a^2 > 0$. Since $a>0$, $r>0$.
Final Result: Minimum occurs at $(a,a)$ with Minimum Value = $-a^3$ . Point $(0,0)$ is a saddle point ($rt-s^2 < 0$).
PREVIOUS YEAR PAPER – 2 (March 2025)
Question 1 (Short Notes)
Q1(a). Express matrix $ A = \begin{bmatrix} 1+i & 2 & 0 \\ 3 & 4-i & 1 \\ -i & 2+i & 5 \end{bmatrix} $ as sum of Hermitian and Skew-Hermitian matrix.
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Formula: $ A = P + Q $ where $ P = \frac{1}{2}(A + A^\theta) $ and $ Q = \frac{1}{2}(A - A^\theta) $.
$ A^\theta = \begin{bmatrix} 1-i & 3 & i \\ 2 & 4+i & 2-i \\ 0 & 1 & 5 \end{bmatrix} $
Final Result:
Q1(b). Find non singular matrices P and Q such that PAQ is in normal form for $ A = \begin{bmatrix} 1 & 2 & 3 \\ 3 & 1 & 2 \end{bmatrix} $.
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Step 1: $ \begin{bmatrix} 1 & 2 & 3 \\ 3 & 1 & 2 \end{bmatrix} = I_2 A I_3 $.
Step 2: Apply $R_2 \to R_2 - 3R_1$, $C_2 \to C_2 - 2C_1$, $C_3 \to C_3 - 3C_1$.
Final Result: Normal form $[I_2 \ 0]$.
Q1(c). Find the nth derivative of $y = \sin^4 x$.
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Step 1: $ y = (\sin^2 x)^2 = \left(\frac{1-\cos 2x}{2}\right)^2 = \frac{1}{4}(1 - 2\cos 2x + \cos^2 2x) $.
Step 2: $ = \frac{1}{4}\left[ 1 - 2\cos 2x + \frac{1+\cos 4x}{2} \right] = \frac{3}{8} - \frac{1}{2}\cos 2x + \frac{1}{8}\cos 4x $.
Final Result: $ y_n = \mathbf{-2^{n-1} \cos(2x + \frac{n\pi}{2}) + \frac{4^n}{8} \cos(4x + \frac{n\pi}{2})} $.
Q1(d). If $ u = \text{cosec}^{-1}\frac{x+2y+3z}{\sqrt{x^8+y^8+z^8}} $, show that $ x u_x + y u_y + z u_z + 3 \tan u = 0 $.
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Step 1: $ v = \text{cosec } u $. Degree $n = 1 - 4 = -3$.
Step 2: $ x u_x + y u_y + z u_z = n \frac{v}{v'} = (-3) \frac{\text{cosec } u}{-\text{cosec } u \cot u} = 3 \tan u $.
Final Result: $ x u_x + y u_y + z u_z - 3 \tan u = 0 $ (Proved, note standard sign logic yields -3).
Question 2 (Unit 1 - Linear Algebra)
Q2(a). Discuss consistency of the system: $ 2x-3y+6z-5w=3 $, $ y-4z+w=1 $, $ 4x-5y+8z-9w=\mu $. If consistent, solve.
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Step 1: Augmented matrix $\implies$ Apply $R_3 \to R_3 - 2R_1 \implies R_3 \to R_3 - R_2$.
$ \begin{bmatrix} 2 & -3 & 6 & -5 & | & 3 \\ 0 & 1 & -4 & 1 & | & 1 \\ 0 & 0 & 0 & 0 & | & \mu - 7 \end{bmatrix} $
Step 2: For consistency, Rank(A) = Rank(A|B) $\implies \mu - 7 = 0 \implies \mu = 7$.
Final Result: Let $z=k_1, w=k_2$. Solutions: $ y = 1+4k_1-k_2 $, $ x = 3+3k_1+k_2 $.
Q2(b). Find eigenvalues and eigenvectors of $ A = \begin{bmatrix} 2 & -2 & 2 \\ 1 & 1 & 1 \\ 1 & 3 & -1 \end{bmatrix} $. Check if diagonalizable.
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Step 1: Characteristic eq: $\lambda^3 - 2\lambda^2 - 4\lambda + 8 = 0 \implies (\lambda-2)^2(\lambda+2)=0$. Roots: $2, 2, -2$.
Step 2: Eigenvector for $\lambda=-2$: $X_1 = [4, 1, -7]^T$.
Step 3: Eigenvector for $\lambda=2$: Matrix $A-2I$ yields equations $-2y+2z=0, x-y+z=0 \implies y=z, x=0$. Only one independent vector $X_2 = [0, 1, 1]^T$.
Final Result: Alg Mult(2)=2, Geom Mult(2)=1. Not Diagonalizable.
Q2(c). State Cayley Hamilton Theorem. Find characteristic equation of $ A = \begin{bmatrix} 2 & -1 & 1 \\ -1 & 2 & -1 \\ 1 & -1 & 2 \end{bmatrix} $. Find $A^6-6A^5+9A^4-2A^3-12A^2+23A-9I$.
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Step 1: Char Eq: $\lambda^3 - 6\lambda^2 + 9\lambda - 4 = 0$. By C-H: $A^3 - 6A^2 + 9A - 4I = 0$.
Step 2: Divide given polynomial $P(A)$ by $(A^3 - 6A^2 + 9A - 4I)$.
Remainder is $5A - I$.
Final Result: $ 5 \begin{bmatrix} 2 & -1 & 1 \\ -1 & 2 & -1 \\ 1 & -1 & 2 \end{bmatrix} - I = \mathbf{\begin{bmatrix} 9 & -5 & 5 \\ -5 & 9 & -5 \\ 5 & -5 & 9 \end{bmatrix}} $.
Question 3 (Unit 2 - Calculus)
Q3(a). If $ u = \log(x^3+y^3+z^3 - 3xyz) $, show that $ \frac{\partial u}{\partial x} + \frac{\partial u}{\partial y} + \frac{\partial u}{\partial z} = \frac{3}{x+y+z} $.
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Step 1: $ e^u = x^3+y^3+z^3 - 3xyz = (x+y+z)(x^2+y^2+z^2-xy-yz-zx) $.
Step 2: Differentiate directly: $ u_x = \frac{3x^2 - 3yz}{x^3+y^3+z^3-3xyz} $.
Step 3: Summing them up: $ u_x + u_y + u_z = \frac{3(x^2+y^2+z^2-xy-yz-zx)}{(x+y+z)(x^2+y^2+z^2-xy-yz-zx)} $.
Final Result: Cancelling terms gives $\mathbf{\frac{3}{x+y+z}}$. Proved.
Q3(b). If $y = (x + \sqrt{1+x^2})^m$, prove $(1+x^2)y_{n+2} + (2n+1)x y_{n+1} + (n^2-m^2)y_n = 0$.
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Step 1: $ y_1 = m(x+\sqrt{1+x^2})^{m-1} \left(1 + \frac{x}{\sqrt{1+x^2}}\right) = \frac{m y}{\sqrt{1+x^2}} $.
Step 2: Square and cross-multiply: $ (1+x^2)y_1^2 = m^2 y^2 $.
Step 3: Differentiate again: $ (1+x^2)2y_1 y_2 + 2x y_1^2 = m^2 2y y_1 \implies (1+x^2)y_2 + x y_1 - m^2 y = 0 $.
Final Result: Apply Leibnitz $n$ times. Yields $(1+x^2)y_{n+2} + (2n+1)x y_{n+1} + (n^2-m^2)y_n = 0$.
Q3(c). Expand $\log(1+x)$ up to $x^4$ using Maclaurin's series.
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Formula: $ f(x) = f(0) + x f'(0) + \frac{x^2}{2!} f''(0) + \dots $
Step 1: $ f(x) = \log(1+x) \implies f(0) = 0 $.
$ f'(x) = (1+x)^{-1} \implies f'(0) = 1 $.
$ f''(x) = -(1+x)^{-2} \implies f''(0) = -1 $.
$ f'''(x) = 2(1+x)^{-3} \implies f'''(0) = 2 $.
$ f^{iv}(x) = -6(1+x)^{-4} \implies f^{iv}(0) = -6 $.
Final Result: $ \log(1+x) = 0 + x(1) + \frac{x^2}{2}(-1) + \frac{x^3}{6}(2) + \frac{x^4}{24}(-6) = \mathbf{x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4}} $.
PREVIOUS YEAR PAPER – 3 (October 2025)
Question 1 (Short Notes)
Q1(a). Using Cayley-Hamilton Theorem, find $ A^6 $ if $ A = \begin{bmatrix} 2 & 1 \\ 5 & -2 \end{bmatrix} $.
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Step 1: Char Eq: $\lambda^2 - 0\lambda + (-4 - 5) = 0 \implies \lambda^2 - 9 = 0$. By C-H, $A^2 = 9I$.
Step 2: Cube both sides: $ (A^2)^3 = (9I)^3 \implies A^6 = 729I $.
Final Result: $ A^6 = \mathbf{\begin{bmatrix} 729 & 0 \\ 0 & 729 \end{bmatrix}} $
Q1(b). For what values of x, the rank of matrix $ \begin{bmatrix} 3-x & 2 & 2 \\ 2 & 4-x & 1 \\ -2 & -4 & -1-x \end{bmatrix} $ is not equal to 3.
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Step 1: Rank $\neq 3 \implies |A| = 0$.
Step 2: $R_2 \to R_2 + R_3 \implies$ Row 2 becomes $[0, -x, -x]$. Take $-x$ common $\implies x=0$ is a root.
Step 3: Expanding remaining yields $(3-x)^2 = 0 \implies x=3$.
Final Result: Rank is not 3 for $x = 0$ and $x = 3$ .
Q1(c). If $ x^x y^y z^z = k $, show that $ \frac{\partial^2 z}{\partial x \partial y} = -(x \log(ex))^{-1} $ at $x=y=z$.
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Step 1: Log diff gives $ \frac{\partial z}{\partial x} = -\frac{1+\log x}{1+\log z} $ and $ \frac{\partial z}{\partial y} = -\frac{1+\log y}{1+\log z} $.
Step 2: Differentiating $z_y$ wrt $x$ gives $ z_{xy} = -\frac{(1+\log x)(1+\log y)}{z(1+\log z)^3} $.
Final Result: At $x=y=z$, it becomes $ -\frac{1}{x(1+\log x)} = \mathbf{-(x \log(ex))^{-1}} $.
Q1(d). Find all the stationary points for $ f(x,y) = (x^2+y^2)e^{-(x^2+y^2)} $.
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Step 1: Let $u = x^2+y^2$. $g(u) = ue^{-u} \implies g'(u) = e^{-u}(1-u)$.
Step 2: $f_x = g'(u)(2x) = 2x e^{-(x^2+y^2)}(1-x^2-y^2) = 0$.
Final Result: $(x,y)=(0,0)$ or $1-x^2-y^2=0$. Points are $(0,0)$ and all points on circle $x^2+y^2=1$ .
Question 2 (Unit 1 - Linear Algebra)
Q2(a). For what values of $\lambda, \mu$, the system $2x+3y+5z=9$, $7x+3y-2z=8$, $2x+3y+\lambda z=\mu$ has No/Unique/Infinite sol.
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Step 1: Augmented matrix $\xrightarrow{R_3 - R_1} \begin{bmatrix} 2 & 3 & 5 & | & 9 \\ 7 & 3 & -2 & | & 8 \\ 0 & 0 & \lambda-5 & | & \mu-9 \end{bmatrix} $
Final Result: Unique: $\lambda \neq 5$. No Sol: $\lambda=5, \mu \neq 9$. Infinite: $\lambda=5, \mu=9$.
Q2(b). Find the modal matrix for $ A = \begin{bmatrix} 1 & 1 & 3 \\ 1 & 5 & 1 \\ 3 & 1 & 1 \end{bmatrix} $.
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Step 1: Char Eq: $\lambda^3 - 7\lambda^2 - 0\lambda + 36 = 0$. Roots: $-2, 3, 6$.
Step 2: Eigenvectors: For $\lambda=-2$: $[1, 0, -1]^T$. For $\lambda=3$: $[1, -1, 1]^T$. For $\lambda=6$: $[1, 2, 1]^T$.
Final Result: Modal matrix $ P = \mathbf{\begin{bmatrix} 1 & 1 & 1 \\ 0 & -1 & 2 \\ -1 & 1 & 1 \end{bmatrix}} $.
Q2(c). Are the vectors $v_1=(1, 2, 3)$, $v_2=(2, 3, 4)$, $v_3=(3, 4, 5)$ linearly dependent?
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Step 1: Form matrix $ A = \begin{bmatrix} 1 & 2 & 3 \\ 2 & 3 & 4 \\ 3 & 4 & 5 \end{bmatrix} $.
Step 2: Reduce to Echelon form. $R_2-2R_1, R_3-3R_1 \implies R_3-2R_2$. Matrix becomes $\begin{bmatrix} 1 & 2 & 3 \\ 0 & -1 & -2 \\ 0 & 0 & 0 \end{bmatrix}$.
Final Result: Rank = 2. Since Rank < Number of Vectors (3), they are Linearly Dependent .
Question 3 (Unit 2 - Calculus)
Q3(a). Verify Euler's theorem for $ u = \frac{x^{1/4} + y^{1/4}}{x^{1/5} + y^{1/5}} $.
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Step 1: Function is homogeneous of degree $n = 1/4 - 1/5 = 1/20$. Euler's states $x u_x + y u_y = \frac{1}{20} u$.
Step 2: Verify by direct differentiation (Quotient Rule). Numerator simplifies heavily when computing $x u_x + y u_y$.
Final Result: Summing yields exactly $\frac{1}{20} u$, verifying the theorem.
Q3(b). Expand $e^x \cos y$ near $(1, \pi/4)$ using Taylor's theorem.
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Formula: $f(x,y) = f(a,b) + [(x-a)f_x + (y-b)f_y] + \frac{1}{2!}[(x-a)^2 f_{xx} + 2(x-a)(y-b)f_{xy} + (y-b)^2 f_{yy}] \dots$
Step 1: Evaluate at $a=1, b=\pi/4$. $f(1,\pi/4) = \frac{e}{\sqrt{2}}$.
Step 2: Derivatives: $f_x = e^x\cos y \implies \frac{e}{\sqrt{2}}$. $f_y = -e^x\sin y \implies -\frac{e}{\sqrt{2}}$.
Final Result: $ \frac{e}{\sqrt{2}} \left[ 1 + (x-1) - (y-\frac{\pi}{4}) + \frac{1}{2}(x-1)^2 - (x-1)(y-\frac{\pi}{4}) - \frac{1}{2}(y-\frac{\pi}{4})^2 \right] $.
Q3(c). Find the maxima and minima of $x^2+y^2+6x+12$.
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Step 1: $p = 2x+6 = 0 \implies x = -3$.
Step 2: $r = f_{xx} = 2$, $s = f_{xy} = 0$, $t = f_{yy} = 2$.
Step 3: $rt-s^2 = (2)(2)-0 = 4 > 0$. Since $r > 0$, it is a Minimum.
Final Result: Minimum at $(-3,0)$ . Value $= 9 + 0 - 18 + 12 = 3$.
COMMONLY USED METHODS FROM NOTES
1. Important Formulas
Euler's Theorem Deduction: If $u = f(\text{homogeneous fn of degree } n)$, let $v = f^{-1}(u)$. Then $ x u_x + y u_y = n \frac{v}{v'} $.Leibnitz Theorem: $ (uv)_n = u_n v + n u_{n-1} v_1 + \frac{n(n-1)}{2} u_{n-2} v_2 + \dots $ (Usually $v = x^2$ or $1-x^2$ so terms vanish after $v_2$).Characteristic Equation ($3 \times 3$): $ \lambda^3 - \text{Trace}(A)\lambda^2 + (\text{Sum of minor diagonals})\lambda - |A| = 0 $.
2. Standard Solving Methods
Maxima/Minima: Find $p = f_x, q = f_y$. Solve $p=0, q=0$ for points $(a,b)$. Calculate $r = f_{xx}, s = f_{xy}, t = f_{yy}$ at $(a,b)$. Gauss-Jordan Inverse: Write $[A | I]$. Use elementary row ops to get $[I | A^{-1}]$.
BAS 101 – MOCK MID SEM EXAM Time: 1.5 Hours | Based on PYQ Analysis
Section A – Short Questions (2.5 marks each)
Find the rank of the matrix $ A = \begin{bmatrix} 1 & 2 & 3 \\ 2 & 3 & 4 \\ 3 & 4 & 5 \end{bmatrix} $.
Verify the Cayley-Hamilton theorem for $ A = \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix} $.
If $ u = \sin^{-1}\left(\frac{x+y}{\sqrt{x}+\sqrt{y}}\right) $, find the value of $ x \frac{\partial u}{\partial x} + y \frac{\partial u}{\partial y} $.
Find the nth derivative of $ y = x^2 e^{3x} $.
Section B – Long Problems Unit 1 (5 marks each)
Test for consistency and solve: $ x+y+z=6, 2x+y-3z=-9, x-2y+3z=6 $.
Find eigenvalues & eigenvectors of $ A = \begin{bmatrix} 8 & -6 & 2 \\ -6 & 7 & -4 \\ 2 & -4 & 3 \end{bmatrix} $.
Using Gauss-Jordan method, find inverse of $ A = \begin{bmatrix} 1 & 1 & 3 \\ 1 & 3 & -3 \\ -2 & -4 & -4 \end{bmatrix} $.
Section C – Long Problems Unit 2 (5 marks each)
If $ y = \cos(m \sin^{-1} x) $, show that $ (1-x^2)y_{n+2} - (2n+1)xy_{n+1} - (n^2-m^2)y_n = 0 $.
Find the Maxima and Minima of $ f(x,y) = x^3 + y^3 - 3axy $ ($a>0$).
If $ u = f(r) $ where $ r = \sqrt{x^2+y^2} $, prove that $ \frac{\partial^2 u}{\partial x^2} + \frac{\partial^2 u}{\partial y^2} = f''(r) + \frac{1}{r} f'(r) $.
Show Detailed Solutions for Mock Paper
DETAILED SOLUTIONS – MOCK PAPER
Q1. Rank of $ A = \begin{bmatrix} 1 & 2 & 3 \\ 2 & 3 & 4 \\ 3 & 4 & 5 \end{bmatrix} $. $R_2 \to R_2 - 2R_1$, $R_3 \to R_3 - 3R_1 \implies \begin{bmatrix} 1 & 2 & 3 \\ 0 & -1 & -2 \\ 0 & -2 & -4 \end{bmatrix}$. Rank = 2.
Q2. Verify C-H for $ A = \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix} $. Char Eq: $\lambda^2 - 5\lambda - 2 = 0$.
Q3. If $ u = \sin^{-1}\left(\frac{x+y}{\sqrt{x}+\sqrt{y}}\right) $, find $ x u_x + y u_y $. Let $ v = \sin u $. Degree = $1/2$. $ \frac{1}{2} \tan u $ .
Q4. nth derivative of $ y = x^2 e^{3x} $. Use Leibnitz theorem. $ u = e^{3x}, v = x^2 $. $ 3^{n-2} e^{3x} [ 9x^2 + 6nx + n(n-1) ] $ .
Q5. Consistency: $ x+y+z=6, 2x+y-3z=-9, x-2y+3z=6 $. Rank(A) = 3, Rank(A|B) = 3. Consistent (unique sol). $ x = -3/17, y = 42/17, z = 63/17 $ .
Q6. Eigenvalues/vectors of $ A = \begin{bmatrix} 8 & -6 & 2 \\ -6 & 7 & -4 \\ 2 & -4 & 3 \end{bmatrix} $. Roots: $0, 3, 15$. Eigenvectors: $X_1 = [1, 2, 2]^T$, $X_2 = [2, 1, -2]^T$, $X_3 = [2, -2, 1]^T$.
Q7. Gauss-Jordan inverse of $ A = \begin{bmatrix} 1 & 1 & 3 \\ 1 & 3 & -3 \\ -2 & -4 & -4 \end{bmatrix} $. Apply ops on $[A|I]$ until $[I|A^{-1}]$. $ A^{-1} = \begin{bmatrix} 3 & 1 & 3/2 \\ -5/4 & -1/4 & -3/4 \\ -1/4 & -1/4 & -1/4 \end{bmatrix} $ .
Q8. $ y = \cos(m \sin^{-1} x) $, show Leibnitz eq. Diff twice $\implies (1-x^2)y_2 - x y_1 + m^2 y = 0$. Apply Leibnitz $n$ times $\implies \mathbf{(1-x^2)y_{n+2} - (2n+1)x y_{n+1} - (n^2-m^2)y_n = 0}$.
Q9. Maxima/Minima $ f(x,y) = x^3 + y^3 - 3axy $. Points: $(0,0)$ and $(a,a)$. At $(a,a)$, $rt-s^2 = 27a^2 > 0$. $r=6a>0$. Minimum at $(a,a)$. Value = $-a^3$.
Q10. Prove $\Delta u = f''(r) + \frac{1}{r} f'(r)$. $\frac{\partial r}{\partial x} = \frac{x}{r}$. $u_x = f'(r)\frac{x}{r}$. $u_{xx} = \frac{f'}{r} + \frac{x^2}{r^2} f'' - \frac{x^2}{r^3} f'$. Add to $u_{yy}$ $\implies$ $ f''(r) + \frac{1}{r}f'(r) $. Verified.