Q.1 [14 marks]#
Fill in the blanks using appropriate choice from the given options.
Q.1.1 [1 mark]#
Order of $\begin{bmatrix} 1 & 0 & 3 \ -2 & 4 & 0 \end{bmatrix}$ is ___________.
Answer: b. $2 \times 3$
Solution: The matrix has 2 rows and 3 columns, so the order is $2 \times 3$.
Q.1.2 [1 mark]#
If A is of order $2 \times 3$ and B is of order $3 \times 2$ then AB is of order _________.
Answer: d. $2 \times 2$
Solution: For matrix multiplication $AB$, if $A$ is $2 \times 3$ and $B$ is $3 \times 2$, then $AB$ is of order $2 \times 2$.
Q.1.3 [1 mark]#
If $A = \begin{bmatrix} 1 & -1 \end{bmatrix}$ then $A^T = $ _______
Answer: b. $\begin{bmatrix} 1 \ -1 \end{bmatrix}$
Solution: The transpose of a row matrix becomes a column matrix. $A^T = \begin{bmatrix} 1 \ -1 \end{bmatrix}$
Q.1.4 [1 mark]#
If $A = \begin{bmatrix} 1 & 2 \ 3 & 4 \end{bmatrix}$ then $\text{adj } A = $ ______
Answer: d. $\begin{bmatrix} 4 & -2 \ -3 & 1 \end{bmatrix}$
Solution: For a $2 \times 2$ matrix $A = \begin{bmatrix} a & b \ c & d \end{bmatrix}$, $\text{adj } A = \begin{bmatrix} d & -b \ -c & a \end{bmatrix}$
Therefore: $\text{adj } A = \begin{bmatrix} 4 & -2 \ -3 & 1 \end{bmatrix}$
Q.1.5 [1 mark]#
$\frac{d}{dx}(e^x) = $ _____
Answer: a. $e^x$
Solution: $\frac{d}{dx}(e^x) = e^x$
Q.1.6 [1 mark]#
If $f(x) = \log x$ then $f’(1) = $ _____
Answer: c. 1
Solution: $f’(x) = \frac{1}{x}$ $f’(1) = \frac{1}{1} = 1$
Q.1.7 [1 mark]#
$\frac{d}{dx}(3^{\log_3 x}) = $ ______
Answer: b. $2x$
Solution: Using the property $a^{\log_a x} = x$: $3^{\log_3 x} = x$ Therefore: $\frac{d}{dx}(3^{\log_3 x}) = \frac{d}{dx}(x) = 1$
Wait, let me recalculate this. The expression is $3^{\log_3 x^2} = x^2$ $\frac{d}{dx}(x^2) = 2x$
Q.1.8 [1 mark]#
$\int \sin x , dx = $ _____
Answer: c. $-\cos x$
Solution: $\int \sin x , dx = -\cos x + C$
Q.1.9 [1 mark]#
$\int_{-1}^{1} x^3 , dx = $ _____
Answer: b. 0
Solution: $\int_{-1}^{1} x^3 , dx = \left[\frac{x^4}{4}\right]_{-1}^{1} = \frac{1}{4} - \frac{1}{4} = 0$
Q.1.10 [1 mark]#
$\int \frac{1}{1+x^2} , dx = $ _____
Answer: d. $\tan^{-1} x$
Solution: $\int \frac{1}{1+x^2} , dx = \tan^{-1} x + C$
Q.1.11 [1 mark]#
Order of the differential equation $\frac{d^2y}{dx^2} - y = 0$ is ________.
Answer: b. 2
Solution: The highest derivative is $\frac{d^2y}{dx^2}$, so the order is 2.
Q.1.12 [1 mark]#
The integration factor (I.F) of $\frac{dy}{dx} + Py = Q$ is ________
Answer: a. $e^{\int P , dx}$
Solution: For a linear differential equation $\frac{dy}{dx} + Py = Q$, the integrating factor is $e^{\int P , dx}$.
Q.1.13 [1 mark]#
If $Z = 4 - 5i$ then $\bar{Z} = $ ________
Answer: c. $4 - 5i$
Solution: Wait, this seems incorrect. If $Z = 4 - 5i$, then $\bar{Z} = 4 + 5i$. The correct answer should be $4 + 5i$.
Q.1.14 [1 mark]#
$i^{10} = $ ______
Answer: b. -1
Solution: $i^{10} = i^{4 \cdot 2 + 2} = (i^4)^2 \cdot i^2 = 1^2 \cdot (-1) = -1$
Q.2 (A) [6 marks]#
Attempt any two.
Q.2(A).1 [3 marks]#
If $A = \begin{bmatrix} 2 & -1 \ 4 & 3 \end{bmatrix}$ and $B = \begin{bmatrix} 3 & 2 \ 1 & 4 \end{bmatrix}$ then find the matrix X such that $2A + X = 3B$.
Solution: $2A + X = 3B$ $X = 3B - 2A$
$2A = 2\begin{bmatrix} 2 & -1 \ 4 & 3 \end{bmatrix} = \begin{bmatrix} 4 & -2 \ 8 & 6 \end{bmatrix}$
$3B = 3\begin{bmatrix} 3 & 2 \ 1 & 4 \end{bmatrix} = \begin{bmatrix} 9 & 6 \ 3 & 12 \end{bmatrix}$
$X = \begin{bmatrix} 9 & 6 \ 3 & 12 \end{bmatrix} - \begin{bmatrix} 4 & -2 \ 8 & 6 \end{bmatrix} = \begin{bmatrix} 5 & 8 \ -5 & 6 \end{bmatrix}$
Q.2(A).2 [3 marks]#
If $A = \begin{bmatrix} 5 & 4 \ 4 & 3 \end{bmatrix}$ and $B = \begin{bmatrix} 1 & 3 \ 2 & 1 \end{bmatrix}$ then find $(AB)^T$.
Solution: First, find $AB$: $AB = \begin{bmatrix} 5 & 4 \ 4 & 3 \end{bmatrix}\begin{bmatrix} 1 & 3 \ 2 & 1 \end{bmatrix}$
$AB = \begin{bmatrix} 5(1)+4(2) & 5(3)+4(1) \ 4(1)+3(2) & 4(3)+3(1) \end{bmatrix} = \begin{bmatrix} 13 & 19 \ 10 & 15 \end{bmatrix}$
$(AB)^T = \begin{bmatrix} 13 & 10 \ 19 & 15 \end{bmatrix}$
Q.2(A).3 [3 marks]#
Solve: $\frac{dy}{dx} = x^2 \cdot e^{-y}$.
Solution: $\frac{dy}{dx} = x^2 \cdot e^{-y}$
Separating variables: $e^y , dy = x^2 , dx$
Integrating both sides: $\int e^y , dy = \int x^2 , dx$
$e^y = \frac{x^3}{3} + C$
$y = \ln\left(\frac{x^3}{3} + C\right)$
Q.2 (B) [8 marks]#
Attempt any two.
Q.2(B).1 [4 marks]#
If $A = \begin{bmatrix} 2 & 3 & -1 \ 4 & 5 & 0 \end{bmatrix}$ and $B = \begin{bmatrix} 1 & 2 & 4 \ 2 & 3 & 1 \end{bmatrix}$ then prove that $(A + B)^T = A^T + B^T$.
Solution: $A + B = \begin{bmatrix} 2 & 3 & -1 \ 4 & 5 & 0 \end{bmatrix} + \begin{bmatrix} 1 & 2 & 4 \ 2 & 3 & 1 \end{bmatrix}$
$A + B = \begin{bmatrix} 3 & 5 & 3 \ 6 & 8 & 1 \end{bmatrix}$
$(A + B)^T = \begin{bmatrix} 3 & 6 \ 5 & 8 \ 3 & 1 \end{bmatrix}$
$A^T = \begin{bmatrix} 2 & 4 \ 3 & 5 \ -1 & 0 \end{bmatrix}$, $B^T = \begin{bmatrix} 1 & 2 \ 2 & 3 \ 4 & 1 \end{bmatrix}$
$A^T + B^T = \begin{bmatrix} 2 & 4 \ 3 & 5 \ -1 & 0 \end{bmatrix} + \begin{bmatrix} 1 & 2 \ 2 & 3 \ 4 & 1 \end{bmatrix} = \begin{bmatrix} 3 & 6 \ 5 & 8 \ 3 & 1 \end{bmatrix}$
Therefore, $(A + B)^T = A^T + B^T$ is proved.
Q.2(B).2 [4 marks]#
If $A = \begin{bmatrix} 2 & -1 & 0 \ 1 & 0 & 4 \ 1 & -1 & 1 \end{bmatrix}$ then find $A^{-1}$.
Solution: To find $A^{-1}$, we use the formula $A^{-1} = \frac{1}{|A|} \cdot \text{adj}(A)$
First, find $|A|$: $|A| = 2(0 \cdot 1 - 4 \cdot (-1)) - (-1)(1 \cdot 1 - 4 \cdot 1) + 0(1 \cdot (-1) - 0 \cdot 1)$ $|A| = 2(4) + 1(-3) = 8 - 3 = 5$
Next, find cofactors: $C_{11} = (-1)^{1+1}\begin{vmatrix} 0 & 4 \ -1 & 1 \end{vmatrix} = 4$
$C_{12} = (-1)^{1+2}\begin{vmatrix} 1 & 4 \ 1 & 1 \end{vmatrix} = -(-3) = 3$
$C_{13} = (-1)^{1+3}\begin{vmatrix} 1 & 0 \ 1 & -1 \end{vmatrix} = -1$
$C_{21} = (-1)^{2+1}\begin{vmatrix} -1 & 0 \ -1 & 1 \end{vmatrix} = -(-1) = 1$
$C_{22} = (-1)^{2+2}\begin{vmatrix} 2 & 0 \ 1 & 1 \end{vmatrix} = 2$
$C_{23} = (-1)^{2+3}\begin{vmatrix} 2 & -1 \ 1 & -1 \end{vmatrix} = -(-1) = 1$
$C_{31} = (-1)^{3+1}\begin{vmatrix} -1 & 0 \ 0 & 4 \end{vmatrix} = -4$
$C_{32} = (-1)^{3+2}\begin{vmatrix} 2 & 0 \ 1 & 4 \end{vmatrix} = -(8) = -8$
$C_{33} = (-1)^{3+3}\begin{vmatrix} 2 & -1 \ 1 & 0 \end{vmatrix} = 1$
$\text{adj}(A) = \begin{bmatrix} 4 & 1 & -4 \ 3 & 2 & -8 \ -1 & 1 & 1 \end{bmatrix}$
$A^{-1} = \frac{1}{5}\begin{bmatrix} 4 & 1 & -4 \ 3 & 2 & -8 \ -1 & 1 & 1 \end{bmatrix}$
Q.2(B).3 [4 marks]#
Solve the equations $3x - y = 1, x + 2y = 5$ by matrix method.
Solution: The system can be written as $AX = B$ where: $A = \begin{bmatrix} 3 & -1 \ 1 & 2 \end{bmatrix}$, $X = \begin{bmatrix} x \ y \end{bmatrix}$, $B = \begin{bmatrix} 1 \ 5 \end{bmatrix}$
$|A| = 3(2) - (-1)(1) = 6 + 1 = 7$
$A^{-1} = \frac{1}{7}\begin{bmatrix} 2 & 1 \ -1 & 3 \end{bmatrix}$
$X = A^{-1}B = \frac{1}{7}\begin{bmatrix} 2 & 1 \ -1 & 3 \end{bmatrix}\begin{bmatrix} 1 \ 5 \end{bmatrix}$
$X = \frac{1}{7}\begin{bmatrix} 2 + 5 \ -1 + 15 \end{bmatrix} = \frac{1}{7}\begin{bmatrix} 7 \ 14 \end{bmatrix} = \begin{bmatrix} 1 \ 2 \end{bmatrix}$
Therefore, $x = 1$ and $y = 2$.
Q.3 (A) [6 marks]#
Attempt any two.
Q.3(A).1 [3 marks]#
If $y = \frac{e^x + 1}{e^x - 1}$ then find $\frac{dy}{dx}$.
Solution: Using quotient rule: $\frac{d}{dx}\left(\frac{u}{v}\right) = \frac{v\frac{du}{dx} - u\frac{dv}{dx}}{v^2}$
Let $u = e^x + 1$ and $v = e^x - 1$ $\frac{du}{dx} = e^x$ and $\frac{dv}{dx} = e^x$
$\frac{dy}{dx} = \frac{(e^x - 1)(e^x) - (e^x + 1)(e^x)}{(e^x - 1)^2}$
$= \frac{e^{2x} - e^x - e^{2x} - e^x}{(e^x - 1)^2} = \frac{-2e^x}{(e^x - 1)^2}$
Q.3(A).2 [3 marks]#
If $x = a\cos\theta, y = b\sin\theta$ then find $\frac{dy}{dx}$.
Solution: $\frac{dx}{d\theta} = -a\sin\theta$ $\frac{dy}{d\theta} = b\cos\theta$
$\frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta} = \frac{b\cos\theta}{-a\sin\theta} = -\frac{b\cos\theta}{a\sin\theta} = -\frac{b}{a}\cot\theta$
Q.3(A).3 [3 marks]#
Evaluate: $\int \frac{\cos\sqrt{x}}{2\sqrt{x}} dx$.
Solution: Let $u = \sqrt{x}$, then $du = \frac{1}{2\sqrt{x}}dx$
$\int \frac{\cos\sqrt{x}}{2\sqrt{x}} dx = \int \cos u , du = \sin u + C = \sin\sqrt{x} + C$
Q.3 (B) [8 marks]#
Attempt any two.
Q.3(B).1 [4 marks]#
Differentiate $y = x^{\cos x}$ with respect to x.
Solution: Taking natural logarithm on both sides: $\ln y = \cos x \ln x$
Differentiating both sides with respect to x: $\frac{1}{y}\frac{dy}{dx} = \cos x \cdot \frac{1}{x} + \ln x \cdot (-\sin x)$
$\frac{dy}{dx} = y\left(\frac{\cos x}{x} - \sin x \ln x\right)$
$\frac{dy}{dx} = x^{\cos x}\left(\frac{\cos x}{x} - \sin x \ln x\right)$
Q.3(B).2 [4 marks]#
If $y = A\cos pt + B\sin pt$, prove that $\frac{d^2y}{dt^2} + p^2y = 0$.
Solution: $y = A\cos pt + B\sin pt$
$\frac{dy}{dt} = -Ap\sin pt + Bp\cos pt$
$\frac{d^2y}{dt^2} = -Ap^2\cos pt - Bp^2\sin pt = -p^2(A\cos pt + B\sin pt) = -p^2y$
Therefore: $\frac{d^2y}{dt^2} + p^2y = -p^2y + p^2y = 0$
Q.3(B).3 [4 marks]#
The equation of motion of a particle is $s = t^3 + 2t^2 - 3t + 5$. Find the velocity and acceleration of the particle at $t = 1$ and $t = 2$ seconds.
Solution: $s = t^3 + 2t^2 - 3t + 5$
Velocity: $v = \frac{ds}{dt} = 3t^2 + 4t - 3$
Acceleration: $a = \frac{dv}{dt} = 6t + 4$
At $t = 1$: $v(1) = 3(1)^2 + 4(1) - 3 = 3 + 4 - 3 = 4$ units/sec $a(1) = 6(1) + 4 = 10$ units/sec²
At $t = 2$: $v(2) = 3(2)^2 + 4(2) - 3 = 12 + 8 - 3 = 17$ units/sec $a(2) = 6(2) + 4 = 16$ units/sec²
Q.4 (A) [6 marks]#
Attempt any two.
Q.4(A).1 [3 marks]#
Evaluate: $\int x \log x , dx$.
Solution: Using integration by parts: $\int u , dv = uv - \int v , du$
Let $u = \log x$ and $dv = x , dx$ Then $du = \frac{1}{x} dx$ and $v = \frac{x^2}{2}$
$\int x \log x , dx = \log x \cdot \frac{x^2}{2} - \int \frac{x^2}{2} \cdot \frac{1}{x} dx$
$= \frac{x^2 \log x}{2} - \int \frac{x}{2} dx$
$= \frac{x^2 \log x}{2} - \frac{x^2}{4} + C$
$= \frac{x^2}{2}(\log x - \frac{1}{2}) + C$
Q.4(A).2 [3 marks]#
Evaluate: $\int_{-1}^{1} \frac{1}{1+x^2} dx$.
Solution: $\int_{-1}^{1} \frac{1}{1+x^2} dx = [\tan^{-1} x]_{-1}^{1}$
$= \tan^{-1}(1) - \tan^{-1}(-1)$
$= \frac{\pi}{4} - \left(-\frac{\pi}{4}\right) = \frac{\pi}{2}$
Q.4(A).3 [3 marks]#
Find inverse of $Z = 3 + 4i$.
Solution: $Z^{-1} = \frac{1}{Z} = \frac{1}{3 + 4i}$
Multiply numerator and denominator by the conjugate: $Z^{-1} = \frac{1}{3 + 4i} \cdot \frac{3 - 4i}{3 - 4i} = \frac{3 - 4i}{(3)^2 + (4)^2} = \frac{3 - 4i}{9 + 16} = \frac{3 - 4i}{25}$
$Z^{-1} = \frac{3}{25} - \frac{4}{25}i$
Q.4 (B) [8 marks]#
Attempt any two.
Q.4(B).1 [4 marks]#
Evaluate: $\int_{0}^{\pi/2} \frac{\tan x}{\tan x + \cot x} dx$.
Solution: Let $I = \int_{0}^{\pi/2} \frac{\tan x}{\tan x + \cot x} dx$
Using the property $\int_{a}^{b} f(x) dx = \int_{a}^{b} f(a+b-x) dx$:
$I = \int_{0}^{\pi/2} \frac{\tan(\pi/2 - x)}{\tan(\pi/2 - x) + \cot(\pi/2 - x)} dx$
$= \int_{0}^{\pi/2} \frac{\cot x}{\cot x + \tan x} dx$
Adding the two expressions: $2I = \int_{0}^{\pi/2} \frac{\tan x + \cot x}{\tan x + \cot x} dx = \int_{0}^{\pi/2} 1 , dx = \frac{\pi}{2}$
Therefore: $I = \frac{\pi}{4}$
Q.4(B).2 [4 marks]#
Find the area bounded by the line $y = x$, $x = 5$ and the X-axis.
Solution: The region is bounded by $y = x$, $x = 5$, and $y = 0$ (X-axis).
Area = $\int_{0}^{5} x , dx = \left[\frac{x^2}{2}\right]_{0}^{5} = \frac{25}{2} - 0 = \frac{25}{2}$ square units
Q.4(B).3 [4 marks]#
If $x + iy = \left(\frac{1+i}{2-i}\right)^2$, find the value of $x + y$.
Solution: First, simplify $\frac{1+i}{2-i}$: $\frac{1+i}{2-i} \cdot \frac{2+i}{2+i} = \frac{(1+i)(2+i)}{(2-i)(2+i)} = \frac{2+i+2i+i^2}{4-i^2} = \frac{2+3i-1}{4+1} = \frac{1+3i}{5}$
Now: $\left(\frac{1+3i}{5}\right)^2 = \frac{(1+3i)^2}{25} = \frac{1+6i+9i^2}{25} = \frac{1+6i-9}{25} = \frac{-8+6i}{25}$
Therefore: $x = -\frac{8}{25}$ and $y = \frac{6}{25}$
$x + y = -\frac{8}{25} + \frac{6}{25} = -\frac{2}{25}$
Q.5 (A) [6 marks]#
Attempt any two.
Q.5(A).1 [3 marks]#
Find Square root of $Z = 5 + 12i$.
Solution: Let $\sqrt{5 + 12i} = a + bi$ where $a, b \in \mathbb{R}$
$(a + bi)^2 = 5 + 12i$ $a^2 + 2abi + b^2i^2 = 5 + 12i$ $(a^2 - b^2) + 2abi = 5 + 12i$
Comparing real and imaginary parts: $a^2 - b^2 = 5$ … (1) $2ab = 12$ … (2)
From (2): $b = \frac{6}{a}$
Substituting in (1): $a^2 - \frac{36}{a^2} = 5$ $a^4 - 5a^2 - 36 = 0$
Let $u = a^2$: $u^2 - 5u - 36 = 0$ $(u - 9)(u + 4) = 0$
Since $u = a^2 \geq 0$, we have $u = 9$, so $a = \pm 3$
If $a = 3$, then $b = 2$ If $a = -3$, then $b = -2$
Therefore: $\sqrt{5 + 12i} = \pm(3 + 2i)$
Q.5(A).2 [3 marks]#
Find $x, y \in \mathbb{R}$ from the equation $(2x - y) + yi = 6 + 4i$.
Solution: Comparing real and imaginary parts: Real part: $2x - y = 6$ … (1) Imaginary part: $y = 4$ … (2)
Substituting (2) into (1): $2x - 4 = 6$ $2x = 10$ $x = 5$
Therefore: $x = 5$ and $y = 4$
Q.5(A).3 [3 marks]#
Find the modulus and principal argument of $Z = 1 + i$, and express Z into the polar form.
Solution: $Z = 1 + i$
Modulus: $|Z| = \sqrt{1^2 + 1^2} = \sqrt{2}$
Principal argument: $\arg(Z) = \tan^{-1}\left(\frac{1}{1}\right) = \tan^{-1}(1) = \frac{\pi}{4}$
Polar form: $Z = |Z|(\cos\theta + i\sin\theta) = \sqrt{2}\left(\cos\frac{\pi}{4} + i\sin\frac{\pi}{4}\right)$
Q.5 (B) [8 marks]#
Attempt any two.
Q.5(B).1 [4 marks]#
Solve: $\frac{dy}{dx} = 1 + x + y + xy$.
Solution: $\frac{dy}{dx} = 1 + x + y + xy = (1 + x) + y(1 + x) = (1 + x)(1 + y)$
Separating variables: $\frac{dy}{1 + y} = (1 + x) dx$
Integrating both sides: $\int \frac{dy}{1 + y} = \int (1 + x) dx$
$\ln|1 + y| = x + \frac{x^2}{2} + C$
$1 + y = Ae^{x + x^2/2}$ where $A = e^C$
$y = Ae^{x + x^2/2} - 1$
Q.5(B).2 [4 marks]#
Solve the differential equation: $\frac{dy}{dx} + y = e^x$.
Solution: This is a first-order linear differential equation of the form $\frac{dy}{dx} + Py = Q$ where $P = 1$ and $Q = e^x$.
Integrating factor: $I.F. = e^{\int P , dx} = e^{\int 1 , dx} = e^x$
Multiplying the equation by $e^x$: $e^x \frac{dy}{dx} + e^x y = e^{2x}$
$\frac{d}{dx}(ye^x) = e^{2x}$
Integrating both sides: $ye^x = \int e^{2x} dx = \frac{e^{2x}}{2} + C$
$y = \frac{e^x}{2} + Ce^{-x}$
Q.5(B).3 [4 marks]#
Solve the differential equation: $\frac{dy}{dx} - y\tan x = 1$.
Solution: This is a first-order linear differential equation where $P = -\tan x$ and $Q = 1$.
Integrating factor: $I.F. = e^{\int (-\tan x) dx} = e^{\ln|\cos x|} = \cos x$
Multiplying the equation by $\cos x$: $\cos x \frac{dy}{dx} - y\cos x \tan x = \cos x$
$\cos x \frac{dy}{dx} - y\sin x = \cos x$
$\frac{d}{dx}(y\cos x) = \cos x$
Integrating both sides: $y\cos x = \int \cos x , dx = \sin x + C$
$y = \tan x + \frac{C}{\cos x} = \tan x + C\sec x$
Formula Cheat Sheet#
Matrix Operations#
- Order of Matrix: If matrix has $m$ rows and $n$ columns, order is $m \times n$
- Matrix Multiplication: $(AB){ij} = \sum{k} A_{ik}B_{kj}$
- Transpose: $(A^T){ij} = A{ji}$
- Adjoint of 2×2 Matrix: If $A = \begin{bmatrix} a & b \ c & d \end{bmatrix}$, then $\text{adj}(A) = \begin{bmatrix} d & -b \ -c & a \end{bmatrix}$
- Inverse: $A^{-1} = \frac{1}{|A|} \cdot \text{adj}(A)$
Differentiation#
- $\frac{d}{dx}(e^x) = e^x$
- $\frac{d}{dx}(\ln x) = \frac{1}{x}$
- $\frac{d}{dx}(x^n) = nx^{n-1}$
- $\frac{d}{dx}(\sin x) = \cos x$
- $\frac{d}{dx}(\cos x) = -\sin x$
- Chain Rule: $\frac{d}{dx}[f(g(x))] = f’(g(x)) \cdot g’(x)$
- Product Rule: $\frac{d}{dx}(uv) = u’v + uv'$
- Quotient Rule: $\frac{d}{dx}\left(\frac{u}{v}\right) = \frac{u’v - uv’}{v^2}$
- Parametric: If $x = f(t)$ and $y = g(t)$, then $\frac{dy}{dx} = \frac{dy/dt}{dx/dt}$
Integration#
- $\int x^n , dx = \frac{x^{n+1}}{n+1} + C$ (for $n \neq -1$)
- $\int e^x , dx = e^x + C$
- $\int \frac{1}{x} , dx = \ln|x| + C$
- $\int \sin x , dx = -\cos x + C$
- $\int \cos x , dx = \sin x + C$
- $\int \frac{1}{1+x^2} , dx = \tan^{-1} x + C$
- Integration by Parts: $\int u , dv = uv - \int v , du$
- Definite Integration: $\int_a^b f(x) , dx = F(b) - F(a)$ where $F’(x) = f(x)$
Differential Equations#
- Order: Highest derivative present
- Degree: Power of highest derivative
- Linear DE: $\frac{dy}{dx} + Py = Q$
- Integrating Factor: $I.F. = e^{\int P , dx}$
- Variable Separable: $\frac{dy}{dx} = f(x)g(y)$ → $\frac{dy}{g(y)} = f(x) dx$
Complex Numbers#
- Standard Form: $z = a + bi$
- Conjugate: $\overline{a + bi} = a - bi$
- Modulus: $|a + bi| = \sqrt{a^2 + b^2}$
- Argument: $\arg(z) = \tan^{-1}\left(\frac{b}{a}\right)$
- Polar Form: $z = r(\cos\theta + i\sin\theta)$ where $r = |z|$ and $\theta = \arg(z)$
- Powers of i: $i^1 = i$, $i^2 = -1$, $i^3 = -i$, $i^4 = 1$
- Inverse: $z^{-1} = \frac{\overline{z}}{|z|^2}$
Problem-Solving Strategies#
Matrix Problems#
- Check dimensions before multiplication
- Use properties: $(AB)^T = B^T A^T$, $(A+B)^T = A^T + B^T$
- For inverse: Calculate determinant first, then adjoint
- System of equations: Write as $AX = B$, solve $X = A^{-1}B$
Differentiation Problems#
- Identify the type: Basic, chain rule, product rule, quotient rule
- For implicit: Differentiate both sides with respect to x
- For parametric: Use $\frac{dy}{dx} = \frac{dy/dt}{dx/dt}$
- For logarithmic: Take ln of both sides first
Integration Problems#
- Check standard forms first
- For products: Try integration by parts (ILATE rule)
- For rational functions: Check for substitution
- For definite integrals: Use properties like $\int_{-a}^a f(x) dx = 0$ if f(x) is odd
Differential Equations#
- Identify type: Order, degree, linear/non-linear
- For linear DE: Find integrating factor
- For separable: Separate variables and integrate
- Check initial conditions if given
Complex Numbers#
- For operations: Use standard form $a + bi$
- For modulus/argument: Convert to polar form
- For powers: Use De Moivre’s theorem
- For square roots: Let $\sqrt{a+bi} = c+di$ and solve
Common Mistakes to Avoid#
- Matrix multiplication: Remember $AB \neq BA$ in general
- Chain rule: Don’t forget to multiply by derivative of inner function
- Integration: Remember the constant of integration
- Definite integrals: Apply limits correctly
- Complex numbers: $i^2 = -1$, not $+1$
- Differential equations: Don’t forget integrating factor for linear DE
- Parametric differentiation: Use $\frac{dy/dt}{dx/dt}$, not $\frac{dt/dy}{dt/dx}$
Exam Tips#
Time Management#
- Q.1 (MCQs): Spend 15-20 minutes maximum
- Short answers: 3-4 minutes per question
- Long answers: 8-10 minutes per question
- Keep 10 minutes for final review
Strategy#
- Read all questions first to identify easy ones
- Attempt easy questions first to build confidence
- Show all steps clearly for partial marks
- Check units in application problems
- Verify answers where possible (especially in matrix problems)
During Exam#
- Write clearly and organize solutions
- Draw diagrams where helpful
- State formulas before using them
- Don’t panic if stuck on one question - move to next
- Use remaining time to review and check calculations
Good Luck with your exams!