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!