Q.1 Fill in the blanks [14 marks]

Q1.1 [1 mark]

Order of the matrix $\begin{bmatrix} 2 & 5 \\ 7 & 8 \end{bmatrix}$ is _________

Answer: (d) $2 \times 2$

Solution: The matrix has 2 rows and 2 columns, so its order is $2 \times 2$ .

Q1.2 [1 mark]

$\begin{bmatrix} 4 & 3 \\ 6 & 2 \end{bmatrix} + \begin{bmatrix} 1 & 5 \\ 5 & 8 \end{bmatrix} =$ _________

Answer: (a) $\begin{bmatrix} 5 & 8 \\ 11 & 10 \end{bmatrix}$

Solution: $\begin{bmatrix} 4 & 3 \\ 6 & 2 \end{bmatrix} + \begin{bmatrix} 1 & 5 \\ 5 & 8 \end{bmatrix} = \begin{bmatrix} 4+1 & 3+5 \\ 6+5 & 2+8 \end{bmatrix} = \begin{bmatrix} 5 & 8 \\ 11 & 10 \end{bmatrix}$

Q1.3 [1 mark]

Which of the following is a square matrix?

Answer: (c) $\begin{bmatrix} 1 & 3 \\ 5 & 4 \end{bmatrix}$

Solution: A square matrix has equal number of rows and columns. Only option (c) has $2 \times 2$ dimensions.

Q1.4 [1 mark]

If $A = [3]$ and $B = [4]$ then $A \cdot B =$ _________

Answer: (b) 12

Solution: $A \cdot B = [3] \times [4] = [3 \times 4] = [12] = 12$

Q1.5 [1 mark]

$\frac{d}{dx}\sin x =$ _________

Answer: (d) $\cos x$

Solution: The derivative of $\sin x$ is $\cos x$ .

Q1.6 [1 mark]

If $f(x) = xe^x$ then $f'(0) =$ _________

Answer: (b) 1

Solution: Using product rule: $f'(x) = \frac{d}{dx}(xe^x) = e^x + xe^x = e^x(1 + x)$ $f'(0) = e^0(1 + 0) = 1 \times 1 = 1$

Q1.7 [1 mark]

If $y = x^2$ then $\frac{d^2y}{dx^2} =$ _________

Answer: (b) 2

Solution: $y = x^2$ $\frac{dy}{dx} = 2x$ $\frac{d^2y}{dx^2} = 2$

Q1.8 [1 mark]

$\int \cos x dx =$ _________ $+ c$

Answer: (a) $\sin x$

Solution: $\int \cos x dx = \sin x + c$

Q1.9 [1 mark]

$\int_0^1 x dx =$ _________

Answer: (c) $\frac{1}{2}$

Solution: $\int_0^1 x dx = \left[\frac{x^2}{2}\right]_0^1 = \frac{1^2}{2} - \frac{0^2}{2} = \frac{1}{2}$

Q1.10 [1 mark]

$\int \frac{1}{1+x^2} dx =$ _________ $+ c$

Answer: (a) $\tan^{-1} x$

Solution: $\int \frac{1}{1+x^2} dx = \tan^{-1} x + c$

Q1.11 [1 mark]

Order of differential equation $x\sin y + xy = x$ is _________

Answer: (b) 1

Solution: The equation can be written as $\frac{dy}{dx} = \frac{1-xy}{\sin y}$ . The highest order derivative is first order.

Q1.12 [1 mark]

Integration factor of $\frac{dy}{dx} + y = x$ is _________

Answer: (d) $e^x$

Solution: For $\frac{dy}{dx} + Py = Q$ , integration factor $= e^{\int P dx} = e^{\int 1 dx} = e^x$

Q1.13 [1 mark]

$i^2 =$ _________

Answer: (b) -1

Solution: By definition, $i^2 = -1$

Q1.14 [1 mark]

$(2+3i)(2-3i) =$ _________

Answer: (c) 13

Solution: $(2+3i)(2-3i) = 2^2 - (3i)^2 = 4 - 9i^2 = 4 - 9(-1) = 4 + 9 = 13$

Q.2(A) Attempt any two [6 marks]

Q2.1(A)(1) [3 marks]

If $A = \begin{bmatrix} 2 & 5 \\ -1 & 3 \end{bmatrix}$ , $B = \begin{bmatrix} 5 & 8 \\ 4 & 6 \end{bmatrix}$ and $C = \begin{bmatrix} 4 & 2 \\ 1 & 5 \end{bmatrix}$ then find $2A + 3B - C$

Solution: $2A = 2\begin{bmatrix} 2 & 5 \\ -1 & 3 \end{bmatrix} = \begin{bmatrix} 4 & 10 \\ -2 & 6 \end{bmatrix}$

$3B = 3\begin{bmatrix} 5 & 8 \\ 4 & 6 \end{bmatrix} = \begin{bmatrix} 15 & 24 \\ 12 & 18 \end{bmatrix}$

$2A + 3B = \begin{bmatrix} 4 & 10 \\ -2 & 6 \end{bmatrix} + \begin{bmatrix} 15 & 24 \\ 12 & 18 \end{bmatrix} = \begin{bmatrix} 19 & 34 \\ 10 & 24 \end{bmatrix}$

$2A + 3B - C = \begin{bmatrix} 19 & 34 \\ 10 & 24 \end{bmatrix} - \begin{bmatrix} 4 & 2 \\ 1 & 5 \end{bmatrix} = \begin{bmatrix} 15 & 32 \\ 9 & 19 \end{bmatrix}$

Q2.1(A)(2) [3 marks]

If $M = \begin{bmatrix} 1 & 4 \\ 3 & 7 \end{bmatrix}$ and $N = \begin{bmatrix} 6 & 9 \\ 0 & 5 \end{bmatrix}$ then prove that $(M+N)^T = M^T + N^T$

Solution: $M + N = \begin{bmatrix} 1 & 4 \\ 3 & 7 \end{bmatrix} + \begin{bmatrix} 6 & 9 \\ 0 & 5 \end{bmatrix} = \begin{bmatrix} 7 & 13 \\ 3 & 12 \end{bmatrix}$

$(M+N)^T = \begin{bmatrix} 7 & 3 \\ 13 & 12 \end{bmatrix}$

$M^T = \begin{bmatrix} 1 & 3 \\ 4 & 7 \end{bmatrix}$ , $N^T = \begin{bmatrix} 6 & 0 \\ 9 & 5 \end{bmatrix}$

$M^T + N^T = \begin{bmatrix} 1 & 3 \\ 4 & 7 \end{bmatrix} + \begin{bmatrix} 6 & 0 \\ 9 & 5 \end{bmatrix} = \begin{bmatrix} 7 & 3 \\ 13 & 12 \end{bmatrix}$

Hence, $(M+N)^T = M^T + N^T$ is proved.

Q2.1(A)(3) [3 marks]

Solve differential equation: $x\frac{dy}{dx} + y = xy$

Solution: $x\frac{dy}{dx} + y = xy$ $\frac{dy}{dx} + \frac{y}{x} = y$ $\frac{dy}{dx} = y - \frac{y}{x} = y\left(1 - \frac{1}{x}\right) = y\left(\frac{x-1}{x}\right)$

Separating variables: $\frac{dy}{y} = \frac{x-1}{x}dx$

Integrating: $\ln|y| = \int\frac{x-1}{x}dx = \int\left(1 - \frac{1}{x}\right)dx = x - \ln|x| + C$

$y = Ae^{x-\ln|x|} = A\frac{e^x}{x}$

Q.2(B) Attempt any two [8 marks]

Q2.1(B)(1) [4 marks]

Solve equations $2x + 3y = 8$ , $3x + 4y = 11$ using matrix method

Solution: Writing in matrix form: $AX = B$ $\begin{bmatrix} 2 & 3 \\ 3 & 4 \end{bmatrix}\begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 8 \\ 11 \end{bmatrix}$

Finding $A^{-1}$ : $|A| = 2(4) - 3(3) = 8 - 9 = -1$

$A^{-1} = \frac{1}{|A|}\begin{bmatrix} 4 & -3 \\ -3 & 2 \end{bmatrix} = \begin{bmatrix} -4 & 3 \\ 3 & -2 \end{bmatrix}$

$X = A^{-1}B = \begin{bmatrix} -4 & 3 \\ 3 & -2 \end{bmatrix}\begin{bmatrix} 8 \\ 11 \end{bmatrix} = \begin{bmatrix} -32+33 \\ 24-22 \end{bmatrix} = \begin{bmatrix} 1 \\ 2 \end{bmatrix}$

Therefore: $x = 1, y = 2$

Q2.1(B)(2) [4 marks]

If $A = \begin{bmatrix} 3 & 2 \\ 1 & 4 \end{bmatrix}$ and $B = \begin{bmatrix} 1 & 2 \\ 0 & 1 \end{bmatrix}$ then prove that $(AB)^T = B^T A^T$

Solution: $AB = \begin{bmatrix} 3 & 2 \\ 1 & 4 \end{bmatrix}\begin{bmatrix} 1 & 2 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 3 & 8 \\ 1 & 6 \end{bmatrix}$

$(AB)^T = \begin{bmatrix} 3 & 1 \\ 8 & 6 \end{bmatrix}$

$A^T = \begin{bmatrix} 3 & 1 \\ 2 & 4 \end{bmatrix}$ , $B^T = \begin{bmatrix} 1 & 0 \\ 2 & 1 \end{bmatrix}$

$B^T A^T = \begin{bmatrix} 1 & 0 \\ 2 & 1 \end{bmatrix}\begin{bmatrix} 3 & 1 \\ 2 & 4 \end{bmatrix} = \begin{bmatrix} 3 & 1 \\ 8 & 6 \end{bmatrix}$

Hence, $(AB)^T = B^T A^T$ is proved.

Q2.1(B)(3) [4 marks]

If $A = \begin{bmatrix} 2 & 3 \\ -1 & 2 \end{bmatrix}$ then prove that $A^2 - 4A + 7I = O$

Solution: $A^2 = \begin{bmatrix} 2 & 3 \\ -1 & 2 \end{bmatrix}\begin{bmatrix} 2 & 3 \\ -1 & 2 \end{bmatrix} = \begin{bmatrix} 1 & 12 \\ -4 & 1 \end{bmatrix}$

$4A = 4\begin{bmatrix} 2 & 3 \\ -1 & 2 \end{bmatrix} = \begin{bmatrix} 8 & 12 \\ -4 & 8 \end{bmatrix}$

$7I = 7\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 7 & 0 \\ 0 & 7 \end{bmatrix}$

$A^2 - 4A + 7I = \begin{bmatrix} 1 & 12 \\ -4 & 1 \end{bmatrix} - \begin{bmatrix} 8 & 12 \\ -4 & 8 \end{bmatrix} + \begin{bmatrix} 7 & 0 \\ 0 & 7 \end{bmatrix} = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix} = O$

Hence proved.

Q.3(A) Attempt any two [6 marks]

Q3.1(A)(1) [3 marks]

Find derivative of $f(x) = e^x$ using definition of differentiation

Solution: Using definition: $f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}$

$f'(x) = \lim_{h \to 0} \frac{e^{x+h} - e^x}{h} = \lim_{h \to 0} \frac{e^x \cdot e^h - e^x}{h}$

$= \lim_{h \to 0} \frac{e^x(e^h - 1)}{h} = e^x \lim_{h \to 0} \frac{e^h - 1}{h}$

Since $\lim_{h \to 0} \frac{e^h - 1}{h} = 1$

Therefore: $f'(x) = e^x$

Q3.1(A)(2) [3 marks]

If $y = \log(\sin x)$ then find $\frac{dy}{dx}$

Solution: $y = \log(\sin x)$

Using chain rule: $\frac{dy}{dx} = \frac{1}{\sin x} \cdot \frac{d}{dx}(\sin x) = \frac{1}{\sin x} \cdot \cos x = \frac{\cos x}{\sin x} = \cot x$

Q3.1(A)(3) [3 marks]

Evaluate: $\int\left(4x^3 + 3x^2 + \frac{2}{x}\right)dx$

Solution: $\int\left(4x^3 + 3x^2 + \frac{2}{x}\right)dx$

$= \int 4x^3 dx + \int 3x^2 dx + \int \frac{2}{x} dx$

$= 4 \cdot \frac{x^4}{4} + 3 \cdot \frac{x^3}{3} + 2\ln|x| + C$

$= x^4 + x^3 + 2\ln|x| + C$

Q.3(B) Attempt any two [8 marks]

Q3.1(B)(1) [4 marks]

If $y = e^{\tan x} + \log(\sin x)$ then find $\frac{dy}{dx}$

Solution: $y = e^{\tan x} + \log(\sin x)$

$\frac{dy}{dx} = \frac{d}{dx}[e^{\tan x}] + \frac{d}{dx}[\log(\sin x)]$

For first term: $\frac{d}{dx}[e^{\tan x}] = e^{\tan x} \cdot \sec^2 x$

For second term: $\frac{d}{dx}[\log(\sin x)] = \frac{1}{\sin x} \cdot \cos x = \cot x$

Therefore: $\frac{dy}{dx} = e^{\tan x} \sec^2 x + \cot x$

Q3.1(B)(2) [4 marks]

The equation of motion of a particle is $s = t^4 + 3t$ . Find its velocity and acceleration at $t = 2$ sec

Solution: Given: $s = t^4 + 3t$

Velocity: $v = \frac{ds}{dt} = 4t^3 + 3$

At $t = 2$ : $v = 4(2)^3 + 3 = 4(8) + 3 = 32 + 3 = 35$ units/sec

Acceleration: $a = \frac{dv}{dt} = \frac{d^2s}{dt^2} = 12t^2$

At $t = 2$ : $a = 12(2)^2 = 12(4) = 48$ units/sec²

Q3.1(B)(3) [4 marks]

Find the maximum and minimum value of the function $f(x) = 2x^3 - 3x^2 - 12x + 5$

Solution: $f(x) = 2x^3 - 3x^2 - 12x + 5$

$f'(x) = 6x^2 - 6x - 12 = 6(x^2 - x - 2) = 6(x-2)(x+1)$

For critical points: $f'(x) = 0$ $6(x-2)(x+1) = 0$ $x = 2$ or $x = -1$

$f''(x) = 12x - 6$

At $x = -1$ : $f''(-1) = 12(-1) - 6 = -18 < 0$ (Maximum) At $x = 2$ : $f''(2) = 12(2) - 6 = 18 > 0$ (Minimum)

$f(-1) = 2(-1)^3 - 3(-1)^2 - 12(-1) + 5 = -2 - 3 + 12 + 5 = 12$ (Maximum) $f(2) = 2(8) - 3(4) - 12(2) + 5 = 16 - 12 - 24 + 5 = -15$ (Minimum)

Maximum value: 12 at $x = -1$ Minimum value: -15 at $x = 2$

Q.4(A) Attempt any two [6 marks]

Q4.1(A)(1) [3 marks]

Evaluate: $\int xe^x dx$

Solution: Using integration by parts: $\int u dv = uv - \int v du$

Let $u = x$ , $dv = e^x dx$ Then $du = dx$ , $v = e^x$

$\int xe^x dx = x \cdot e^x - \int e^x dx = xe^x - e^x + C = e^x(x-1) + C$

Q4.1(A)(2) [3 marks]

Evaluate: $\int \frac{dx}{\sqrt{9-4x^2}}$

Solution: $\int \frac{dx}{\sqrt{9-4x^2}} = \int \frac{dx}{\sqrt{9(1-\frac{4x^2}{9})}} = \int \frac{dx}{3\sqrt{1-\left(\frac{2x}{3}\right)^2}}$

Let $\frac{2x}{3} = \sin \theta$ , then $x = \frac{3\sin \theta}{2}$ , $dx = \frac{3\cos \theta}{2} d\theta$

$= \int \frac{\frac{3\cos \theta}{2} d\theta}{3\sqrt{1-\sin^2 \theta}} = \int \frac{\frac{3\cos \theta}{2} d\theta}{3\cos \theta} = \int \frac{1}{2} d\theta = \frac{\theta}{2} + C$

$= \frac{1}{2}\sin^{-1}\left(\frac{2x}{3}\right) + C$

Q4.1(A)(3) [3 marks]

Find complex conjugate of $\frac{1-i}{1+i}$

Solution: $\frac{1-i}{1+i} = \frac{(1-i)(1-i)}{(1+i)(1-i)} = \frac{(1-i)^2}{1-i^2} = \frac{1-2i+i^2}{1-(-1)} = \frac{1-2i-1}{2} = \frac{-2i}{2} = -i$

Complex conjugate of $-i$ is $\overline{-i} = i$

Q.4(B) Attempt any two [8 marks]

Q4.1(B)(1) [4 marks]

Evaluate: $\int_0^{\pi/2} \frac{\sqrt{\cos x}}{\sqrt{\cos x} + \sqrt{\sin x}} dx$

Solution: Let $I = \int_0^{\pi/2} \frac{\sqrt{\cos x}}{\sqrt{\cos x} + \sqrt{\sin x}} dx$

Using property: $\int_0^a f(x)dx = \int_0^a f(a-x)dx$

$I = \int_0^{\pi/2} \frac{\sqrt{\cos(\pi/2-x)}}{\sqrt{\cos(\pi/2-x)} + \sqrt{\sin(\pi/2-x)}} dx = \int_0^{\pi/2} \frac{\sqrt{\sin x}}{\sqrt{\sin x} + \sqrt{\cos x}} dx$

Adding both expressions: $2I = \int_0^{\pi/2} \frac{\sqrt{\cos x} + \sqrt{\sin x}}{\sqrt{\cos x} + \sqrt{\sin x}} dx = \int_0^{\pi/2} 1 dx = \frac{\pi}{2}$

Therefore: $I = \frac{\pi}{4}$

Q4.1(B)(2) [4 marks]

Find the area of circle $x^2 + y^2 = a^2$ using integration

Solution: For circle $x^2 + y^2 = a^2$ , we have $y = \pm\sqrt{a^2-x^2}$

Area of circle = $4 \times$ Area in first quadrant $= 4\int_0^a \sqrt{a^2-x^2} dx$

Let $x = a\sin \theta$ , $dx = a\cos \theta d\theta$ When $x = 0$ , $\theta = 0$ ; when $x = a$ , $\theta = \pi/2$

$= 4\int_0^{\pi/2} \sqrt{a^2-a^2\sin^2 \theta} \cdot a\cos \theta d\theta$ $= 4\int_0^{\pi/2} a\cos \theta \cdot a\cos \theta d\theta$ $= 4a^2\int_0^{\pi/2} \cos^2 \theta d\theta$ $= 4a^2 \cdot \frac{\pi}{4} = \pi a^2$

Q4.1(B)(3) [4 marks]

Simplify: $\frac{(\cos 3\theta + i\sin 3\theta)^4 \cdot (\cos \theta - i\sin \theta)^5}{(\cos 2\theta - i\sin 2\theta)^3 \cdot (\cos 12\theta + i\sin 12\theta)}$

Solution: Using De Moivre's theorem: $(\cos \theta + i\sin \theta)^n = \cos n\theta + i\sin n\theta$

Numerator: $(\cos 3\theta + i\sin 3\theta)^4 \cdot (\cos \theta - i\sin \theta)^5$ $= (\cos 12\theta + i\sin 12\theta) \cdot (\cos(-5\theta) + i\sin(-5\theta))$ $= \cos(12\theta - 5\theta) + i\sin(12\theta - 5\theta)$ $= \cos 7\theta + i\sin 7\theta$

Denominator: $(\cos 2\theta - i\sin 2\theta)^3 \cdot (\cos 12\theta + i\sin 12\theta)$ $= (\cos(-6\theta) + i\sin(-6\theta)) \cdot (\cos 12\theta + i\sin 12\theta)$ $= \cos(-6\theta + 12\theta) + i\sin(-6\theta + 12\theta)$ $= \cos 6\theta + i\sin 6\theta$

Result: $\frac{\cos 7\theta + i\sin 7\theta}{\cos 6\theta + i\sin 6\theta} = \cos(7\theta - 6\theta) + i\sin(7\theta - 6\theta) = \cos \theta + i\sin \theta$

Q.5(A) Attempt any two [6 marks]

Q5.1(A)(1) [3 marks]

If $(3x - 7) + 2iy = 5y + (5 + x)i$ then find value of x and y

Solution: $(3x - 7) + 2iy = 5y + (5 + x)i$

Comparing real and imaginary parts: Real parts: $3x - 7 = 5y$ ... (1) Imaginary parts: $2y = 5 + x$ ... (2)

From equation (2): $x = 2y - 5$ ... (3)

Substituting (3) in (1): $3(2y - 5) - 7 = 5y$ $6y - 15 - 7 = 5y$ $6y - 22 = 5y$ $y = 22$

From (3): $x = 2(22) - 5 = 44 - 5 = 39$

Therefore: $x = 39, y = 22$

Q5.1(A)(2) [3 marks]

Convert $z = 1 + \sqrt{3}i$ into polar form

Solution: $z = 1 + \sqrt{3}i$

Modulus: $|z| = \sqrt{1^2 + (\sqrt{3})^2} = \sqrt{1 + 3} = \sqrt{4} = 2$

Argument: $\arg(z) = \tan^{-1}\left(\frac{\sqrt{3}}{1}\right) = \tan^{-1}(\sqrt{3}) = \frac{\pi}{3}$

Polar form: $z = |z|(\cos \theta + i\sin \theta) = 2\left(\cos \frac{\pi}{3} + i\sin \frac{\pi}{3}\right)$

Q5.1(A)(3) [3 marks]

Express $\frac{4 + 2i}{(3 + 2i)(5 - 3i)}$ in $a + ib$ form

Solution: First, simplify denominator: $(3 + 2i)(5 - 3i) = 15 - 9i + 10i - 6i^2 = 15 + i - 6(-1) = 15 + i + 6 = 21 + i$

$\frac{4 + 2i}{21 + i} = \frac{(4 + 2i)(21 - i)}{(21 + i)(21 - i)} = \frac{84 - 4i + 42i - 2i^2}{21^2 - i^2} = \frac{84 + 38i + 2}{441 + 1} = \frac{86 + 38i}{442}$

$= \frac{86}{442} + \frac{38}{442}i = \frac{43}{221} + \frac{19}{221}i$

Q.5(B) Attempt any two [8 marks]

Q5.1(B)(1) [4 marks]

Solve differential equation: $\frac{dy}{dx} + 2y = 3e^x$

Solution: This is a first-order linear differential equation of the form $\frac{dy}{dx} + Py = Q$

Here: $P = 2$ , $Q = 3e^x$

Integration factor: $\mu = e^{\int P dx} = e^{\int 2 dx} = e^{2x}$

Multiplying equation by $\mu$ : $e^{2x}\frac{dy}{dx} + 2e^{2x}y = 3e^{2x} \cdot e^x = 3e^{3x}$

This gives: $\frac{d}{dx}(ye^{2x}) = 3e^{3x}$

Integrating both sides: $ye^{2x} = \int 3e^{3x} dx = 3 \cdot \frac{e^{3x}}{3} + C = e^{3x} + C$

Therefore: $y = \frac{e^{3x} + C}{e^{2x}} = e^x + Ce^{-2x}$

Q5.1(B)(2) [4 marks]

Solve differential equation: $\frac{dy}{dx} = (x + y)^2$

Solution: Let $v = x + y$ , then $\frac{dv}{dx} = 1 + \frac{dy}{dx}$

So $\frac{dy}{dx} = \frac{dv}{dx} - 1$

Substituting in the original equation: $\frac{dv}{dx} - 1 = v^2$ $\frac{dv}{dx} = v^2 + 1$

Separating variables: $\frac{dv}{v^2 + 1} = dx$

Integrating both sides: $\int \frac{dv}{v^2 + 1} = \int dx$ $\tan^{-1}(v) = x + C$ $v = \tan(x + C)$

Substituting back: $x + y = \tan(x + C)$ Therefore: $y = \tan(x + C) - x$

Q5.1(B)(3) [4 marks]

Solve differential equation: $\frac{dy}{dx} + \frac{y}{x} = e^x$ , $y(0) = 2$

Solution: This is a first-order linear differential equation: $\frac{dy}{dx} + \frac{y}{x} = e^x$

Here: $P = \frac{1}{x}$ , $Q = e^x$

Integration factor: $\mu = e^{\int \frac{1}{x} dx} = e^{\ln|x|} = |x| = x$ (for $x > 0$ )

Multiplying equation by $\mu = x$ : $x\frac{dy}{dx} + y = xe^x$

This gives: $\frac{d}{dx}(xy) = xe^x$

Integrating both sides using integration by parts: $xy = \int xe^x dx$

For $\int xe^x dx$ : Let $u = x$ , $dv = e^x dx$ Then $du = dx$ , $v = e^x$ $\int xe^x dx = xe^x - \int e^x dx = xe^x - e^x = e^x(x-1)$

So: $xy = e^x(x-1) + C$ $y = \frac{e^x(x-1) + C}{x}$

Using initial condition $y(0) = 2$ : This presents a problem as we have division by zero. The equation needs to be solved more carefully near $x = 0$ .

For the general solution: $y = e^x\left(1 - \frac{1}{x}\right) + \frac{C}{x}$

Formula Cheat Sheet

Matrix Operations

Matrix addition: $(A + B)_{ij} = A_{ij} + B_{ij}$
Matrix multiplication: $(AB)_{ij} = \sum_{k} A_{ik}B_{kj}$
Transpose: $(A^T)_{ij} = A_{ji}$
Inverse of 2×2 matrix: $A^{-1} = \frac{1}{|A|}\begin{bmatrix} d & -b \\ -c & a \end{bmatrix}$ where $A = \begin{bmatrix} a & b \\ c & d \end{bmatrix}$

Differentiation Formulas

$\frac{d}{dx}(x^n) = nx^{n-1}$
$\frac{d}{dx}(e^x) = e^x$
$\frac{d}{dx}(\sin x) = \cos x$
$\frac{d}{dx}(\cos x) = -\sin x$
$\frac{d}{dx}(\tan x) = \sec^2 x$
$\frac{d}{dx}(\ln x) = \frac{1}{x}$
Product rule: $(uv)' = u'v + uv'$
Chain rule: $\frac{d}{dx}f(g(x)) = f'(g(x)) \cdot g'(x)$

Integration Formulas

$\int x^n dx = \frac{x^{n+1}}{n+1} + C$ (for $n \neq -1$ )
$\int \frac{1}{x} dx = \ln|x| + C$
$\int e^x dx = e^x + C$
$\int \sin x dx = -\cos x + C$
$\int \cos x dx = \sin x + C$
$\int \sec^2 x dx = \tan x + C$
$\int \frac{1}{1+x^2} dx = \tan^{-1} x + C$
$\int \frac{1}{\sqrt{1-x^2}} dx = \sin^{-1} x + C$

Differential Equations

First-order linear: $\frac{dy}{dx} + Py = Q$
Integration factor: $\mu = e^{\int P dx}$
Solution: $y = \frac{1}{\mu}\left[\int \mu Q dx + C\right]$
Variable separable: $\frac{dy}{dx} = f(x)g(y)$ → $\frac{dy}{g(y)} = f(x)dx$

Complex Numbers

$i^2 = -1$ , $i^3 = -i$ , $i^4 = 1$
Modulus: $|a + bi| = \sqrt{a^2 + b^2}$
Argument: $\arg(a + bi) = \tan^{-1}\left(\frac{b}{a}\right)$
Polar form: $z = r(\cos \theta + i\sin \theta)$
De Moivre's theorem: $(\cos \theta + i\sin \theta)^n = \cos n\theta + i\sin n\theta$

Problem-Solving Strategies

Matrix Problems

Always check dimensions before performing operations
For matrix equations: Use inverse method $X = A^{-1}B$
For transpose properties: Use $(AB)^T = B^T A^T$
For matrix powers: Calculate step by step, look for patterns

Differentiation Problems

Identify the type: Product, quotient, chain rule, or implicit
For complex functions: Break down using appropriate rules
For applications: Remember $v = \frac{ds}{dt}$ , $a = \frac{dv}{dt}$
For maxima/minima: Find critical points where $f'(x) = 0$

Integration Problems

Recognize standard forms first
For substitution: Look for $f'(x)$ when $f(x)$ appears
For integration by parts: Choose $u$ as LIATE (Log, Inverse trig, Algebraic, Trig, Exponential)
For definite integrals: Use fundamental theorem or properties

Differential Equations

Identify the type: Linear, separable, or exact
For linear equations: Find integration factor systematically
For separable equations: Separate variables completely before integrating
Always check initial conditions if given

Complex Numbers

For operations: Convert to $a + bi$ form first
For polar form: Calculate modulus and argument carefully
For powers: Use De Moivre's theorem
For division: Multiply by conjugate of denominator

Common Mistakes to Avoid

Matrix Operations

❌ Don't assume $AB = BA$ (matrix multiplication is not commutative)
❌ Don't forget to check if matrices can be multiplied (inner dimensions must match)
❌ Don't confuse transpose with inverse

Differentiation

❌ Don't forget the chain rule for composite functions
❌ Don't mix up $\frac{d}{dx}(\sin x) = \cos x$ and $\frac{d}{dx}(\cos x) = -\sin x$
❌ Don't forget to use product rule when multiplying functions

Integration

❌ Don't forget the constant of integration $+C$
❌ Don't confuse indefinite and definite integrals
❌ Don't forget to substitute limits properly in definite integrals

Complex Numbers

❌ Don't forget $i^2 = -1$ when expanding
❌ Don't confuse modulus with real part
❌ Don't forget to rationalize denominators with complex numbers

Exam Tips

Time Management

Spend 2-3 minutes reading the entire paper first
Attempt easier questions first to build confidence
Reserve 15 minutes at the end for review

Writing Strategy

Show all steps clearly - partial marks are often awarded
Draw diagrams where helpful - especially for geometry problems
Write final answers clearly and box them if possible

Calculation Tips

Double-check arithmetic - many marks are lost due to calculation errors
Use calculator efficiently but don't become dependent on it
Cross-verify answers using different methods when possible

Question Selection

In OR questions, choose the one you're most confident about
Don't spend too much time on any single question
If stuck, move on and return later with fresh perspective

Good luck with your exam preparation!