Applied Mathematics (4320001) - Summer 2024 Solution

Q.1 Fill in the blanks [14 marks]

Q1.1 [1 mark]

Order of the matrix $\begin{bmatrix} 1 & 2 & 3 \\ -4 & 5 & 6 \end{bmatrix}$ is = ___________

Answer: (b) $2 \times 3$

Solution: A matrix with 2 rows and 3 columns has order $2 \times 3$.

Q1.2 [1 mark]

If $\begin{bmatrix} x-3 & 2 \\ 4 & 0 \end{bmatrix} = \begin{bmatrix} 5 & 2 \\ 4 & 0 \end{bmatrix}$ then $x$ = ____

Answer: (d) 8

Solution: For matrix equality, corresponding elements must be equal: $x - 3 = 5$ $x = 8$

Q1.3 [1 mark]

The adjoint of $\begin{bmatrix} -3 & 2 \\ 0 & 1 \end{bmatrix}$ = _____________

Answer: (b) $\begin{bmatrix} 1 & -2 \\ 0 & -3 \end{bmatrix}$

Solution: For matrix $A = \begin{bmatrix} a & b \\ c & d \end{bmatrix}$, $\text{adj}(A) = \begin{bmatrix} d & -b \\ -c & a \end{bmatrix}$ $\text{adj}\begin{bmatrix} -3 & 2 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 1 & -2 \\ 0 & -3 \end{bmatrix}$

Q1.4 [1 mark]

For any square matrix $A$, $(A^{-1})^{-1}$ = ____________

Answer: (b) $A$

Solution: By definition of inverse matrices: $(A^{-1})^{-1} = A$

Q1.5 [1 mark]

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

Answer: (b) $\frac{1}{x}$

Solution: The derivative of natural logarithm: $\frac{d}{dx} \log x = \frac{1}{x}$

Q1.6 [1 mark]

$\frac{d}{dx}(\tan^{-1} x + \cot^{-1} x)$ = _______

Answer: (d) 0

Solution: $\tan^{-1} x + \cot^{-1} x = \frac{\pi}{2}$ (constant) Therefore, $\frac{d}{dx}(\tan^{-1} x + \cot^{-1} x) = 0$

Q1.7 [1 mark]

If $x = a \cos \theta$, $y = a \sin \theta$ then $\frac{dy}{dx}$ = __________

Answer: (a) $-\cot \theta$

Solution: $\frac{dx}{d\theta} = -a \sin \theta$, $\frac{dy}{d\theta} = a \cos \theta$ $\frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta} = \frac{a \cos \theta}{-a \sin \theta} = -\cot \theta$

Q1.8 [1 mark]

$\int 5x^4 dx$ = ____________ + $c$

Answer: (d) $x^5$

Solution: $\int 5x^4 dx = 5 \cdot \frac{x^5}{5} = x^5 + c$

Q1.9 [1 mark]

$\int_0^1 e^x dx$ = __________

Answer: (a) $e - 1$

Solution: $\int_0^1 e^x dx = [e^x]_0^1 = e^1 - e^0 = e - 1$

Q1.10 [1 mark]

$\int_{-1}^1 3x^2 - 2x + 1 dx$ = __________

Answer: (c) 4

Solution: $\int_{-1}^1 (3x^2 - 2x + 1) dx = [x^3 - x^2 + x]_{-1}^1$ $= (1 - 1 + 1) - (-1 - 1 - 1) = 1 - (-3) = 4$

Q1.11 [1 mark]

The order of differential equation $(\frac{dy}{dx})^2 + 4y = x$ is ___________

Answer: (d) 1

Solution: Order is the highest derivative present. Here, only first derivative $\frac{dy}{dx}$ appears, so order = 1.

Q1.12 [1 mark]

The integrating factor of $\frac{dy}{dx} + 3y = x$ is _____________

Answer: (d) $e^{3x}$

Solution: For linear DE $\frac{dy}{dx} + Py = Q$, integrating factor = $e^{\int P dx}$ Here $P = 3$, so I.F. = $e^{\int 3 dx} = e^{3x}$

Q1.13 [1 mark]

The mean of first ten natural numbers is_________

Answer: (a) 5.5

Solution: Mean = $\frac{1 + 2 + 3 + ... + 10}{10} = \frac{55}{10} = 5.5$

Q1.14 [1 mark]

The range of the data 17, 15, 25, 34, 32 is _______________

Answer: (d) 19

Solution: Range = Maximum - Minimum = 34 - 15 = 19

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

Q2.1 [3 marks]

If $A = \begin{bmatrix} 1 & -1 \\ 2 & 3 \end{bmatrix}$ then find $A + A^T + I$.

Answer:

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

$A^T = \begin{bmatrix} 1 & 2 \\ -1 & 3 \end{bmatrix}$

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

$A + A^T + I = \begin{bmatrix} 1 & -1 \\ 2 & 3 \end{bmatrix} + \begin{bmatrix} 1 & 2 \\ -1 & 3 \end{bmatrix} + \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$

$= \begin{bmatrix} 3 & 1 \\ 1 & 7 \end{bmatrix}$

Q2.2 [3 marks]

If $A = \begin{bmatrix} 2 & 3 \\ -1 & 2 \end{bmatrix}$ then prove that $A^2 - 4A + 7I_2 = 0$

Answer: Proved

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

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

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

$7I_2 = \begin{bmatrix} 7 & 0 \\ 0 & 7 \end{bmatrix}$

$A^2 - 4A + 7I_2 = \begin{bmatrix} 1 & 12 \\ -4 & 1 \end{bmatrix} - \begin{bmatrix} 8 & 12 \\ -4 & 8 \end{bmatrix} + \begin{bmatrix} 7 & 0 \\ 0 & 7 \end{bmatrix}$

$= \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix} = 0$ ✓

Q2.3 [3 marks]

Solve differential equation $dy - 3x^2e^{-y}dx = 0$

Answer: $e^y = x^3 + C$

Solution: $dy - 3x^2e^{-y}dx = 0$ $dy = 3x^2e^{-y}dx$ $e^y dy = 3x^2 dx$

Integrating both sides: $\int e^y dy = \int 3x^2 dx$ $e^y = x^3 + C$

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

Q2.1 [4 marks]

Find the inverse of matrix $\begin{bmatrix} 3 & -1 & 2 \\ 4 & 1 & -1 \\ 5 & 0 & 1 \end{bmatrix}$

Answer: $A^{-1} = \begin{bmatrix} 1/14 & 1/14 & -1/14 \\ -9/14 & -7/14 & 11/14 \\ -5/14 & -5/14 & 1/2 \end{bmatrix}$

Solution: Let $A = \begin{bmatrix} 3 & -1 & 2 \\ 4 & 1 & -1 \\ 5 & 0 & 1 \end{bmatrix}$

First, find $\det(A)$: $\det(A) = 3(1 \cdot 1 - (-1) \cdot 0) - (-1)(4 \cdot 1 - (-1) \cdot 5) + 2(4 \cdot 0 - 1 \cdot 5)$ $= 3(1) + 1(9) + 2(-5) = 3 + 9 - 10 = 2$

Since $\det(A) \neq 0$, inverse exists.

Finding cofactors and adjoint matrix: $C_{11} = 1$, $C_{12} = -9$, $C_{13} = -5$ $C_{21} = 1$, $C_{22} = -7$, $C_{23} = -5$
$C_{31} = -1$, $C_{32} = 11$, $C_{33} = 7$

$\text{adj}(A) = \begin{bmatrix} 1 & 1 & -1 \\ -9 & -7 & 11 \\ -5 & -5 & 7 \end{bmatrix}$

$A^{-1} = \frac{1}{\det(A)} \cdot \text{adj}(A) = \frac{1}{2} \begin{bmatrix} 1 & 1 & -1 \\ -9 & -7 & 11 \\ -5 & -5 & 7 \end{bmatrix}$

Q2.2 [4 marks]

If $A + B = \begin{bmatrix} 1 & -1 \\ 3 & 0 \end{bmatrix}$ and $A - B = \begin{bmatrix} 3 & 1 \\ 1 & 4 \end{bmatrix}$ then find $AB$.

Answer: $AB = \begin{bmatrix} 0 & -1 \\ 4 & -2 \end{bmatrix}$

Solution: Adding the equations: $(A + B) + (A - B) = 2A$ $2A = \begin{bmatrix} 1 & -1 \\ 3 & 0 \end{bmatrix} + \begin{bmatrix} 3 & 1 \\ 1 & 4 \end{bmatrix} = \begin{bmatrix} 4 & 0 \\ 4 & 4 \end{bmatrix}$ $A = \begin{bmatrix} 2 & 0 \\ 2 & 2 \end{bmatrix}$

Subtracting the equations: $(A + B) - (A - B) = 2B$ $2B = \begin{bmatrix} 1 & -1 \\ 3 & 0 \end{bmatrix} - \begin{bmatrix} 3 & 1 \\ 1 & 4 \end{bmatrix} = \begin{bmatrix} -2 & -2 \\ 2 & -4 \end{bmatrix}$ $B = \begin{bmatrix} -1 & -1 \\ 1 & -2 \end{bmatrix}$

$AB = \begin{bmatrix} 2 & 0 \\ 2 & 2 \end{bmatrix} \begin{bmatrix} -1 & -1 \\ 1 & -2 \end{bmatrix} = \begin{bmatrix} -2 & -2 \\ 0 & -6 \end{bmatrix}$

Q2.3 [4 marks]

Solve the system of linear equation $2x + 3y = 1$, $y - 4x = 2$ using matrices.

Answer: $x = -\frac{1}{11}$, $y = \frac{13}{11}$

Solution: The system can be written as: $AX = B$ $\begin{bmatrix} 2 & 3 \\ -4 & 1 \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 1 \\ 2 \end{bmatrix}$

$\det(A) = 2(1) - 3(-4) = 2 + 12 = 14$

$A^{-1} = \frac{1}{14} \begin{bmatrix} 1 & -3 \\ 4 & 2 \end{bmatrix}$

$X = A^{-1}B = \frac{1}{14} \begin{bmatrix} 1 & -3 \\ 4 & 2 \end{bmatrix} \begin{bmatrix} 1 \\ 2 \end{bmatrix} = \frac{1}{14} \begin{bmatrix} -5 \\ 8 \end{bmatrix}$

Therefore: $x = -\frac{5}{14}$, $y = \frac{8}{14} = \frac{4}{7}$

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

Q3.1 [3 marks]

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

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

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

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

Q3.2 [3 marks]

If $\sqrt{x} + \sqrt{y} = \sqrt{a}$ then prove that $\frac{dy}{dx} = -\sqrt{\frac{y}{x}}$

Answer: Proved

Solution: $\sqrt{x} + \sqrt{y} = \sqrt{a}$

Differentiating both sides with respect to $x$: $\frac{1}{2\sqrt{x}} + \frac{1}{2\sqrt{y}} \cdot \frac{dy}{dx} = 0$

$\frac{1}{2\sqrt{y}} \cdot \frac{dy}{dx} = -\frac{1}{2\sqrt{x}}$

$\frac{dy}{dx} = -\frac{\sqrt{y}}{\sqrt{x}} = -\sqrt{\frac{y}{x}}$ ✓

Q3.3 [3 marks]

Evaluate $\int \frac{\tan x}{\sec x + \tan x} dx$

Answer: $x - \ln|\sec x + \tan x| + C$

Solution: Let $I = \int \frac{\tan x}{\sec x + \tan x} dx$

Multiply numerator and denominator by $(\sec x - \tan x)$: $I = \int \frac{\tan x(\sec x - \tan x)}{(\sec x + \tan x)(\sec x - \tan x)} dx$ $= \int \frac{\tan x(\sec x - \tan x)}{\sec^2 x - \tan^2 x} dx$ $= \int \frac{\tan x(\sec x - \tan x)}{1} dx$ $= \int (\tan x \sec x - \tan^2 x) dx$ $= \int \tan x \sec x dx - \int (\sec^2 x - 1) dx$ $= \sec x - \tan x + x + C$

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

Q3.1 [4 marks]

If $e^x + e^y = e^{x+y}$ then find $\frac{dy}{dx}$.

Answer: $\frac{dy}{dx} = \frac{e^x(e^y - 1)}{e^y(e^x - 1)}$

Solution: $e^x + e^y = e^{x+y}$

Differentiating both sides with respect to $x$: $e^x + e^y \frac{dy}{dx} = e^{x+y}(1 + \frac{dy}{dx})$ $e^x + e^y \frac{dy}{dx} = e^{x+y} + e^{x+y} \frac{dy}{dx}$

Rearranging: $e^x - e^{x+y} = e^{x+y} \frac{dy}{dx} - e^y \frac{dy}{dx}$ $e^x - e^{x+y} = \frac{dy}{dx}(e^{x+y} - e^y)$

$\frac{dy}{dx} = \frac{e^x - e^{x+y}}{e^{x+y} - e^y} = \frac{e^x(1 - e^y)}{e^y(e^x - 1)} = \frac{e^x(e^y - 1)}{e^y(e^x - 1)}$

Q3.2 [4 marks]

For $y = 2e^{3x} + 3e^{-2x}$, prove that $\frac{d^2y}{dx^2} - \frac{dy}{dx} - 6y = 0$.

Answer: Proved

Solution: $y = 2e^{3x} + 3e^{-2x}$

$\frac{dy}{dx} = 6e^{3x} - 6e^{-2x}$

$\frac{d^2y}{dx^2} = 18e^{3x} + 12e^{-2x}$

Now checking the equation: $\frac{d^2y}{dx^2} - \frac{dy}{dx} - 6y$ $= (18e^{3x} + 12e^{-2x}) - (6e^{3x} - 6e^{-2x}) - 6(2e^{3x} + 3e^{-2x})$ $= 18e^{3x} + 12e^{-2x} - 6e^{3x} + 6e^{-2x} - 12e^{3x} - 18e^{-2x}$ $= (18 - 6 - 12)e^{3x} + (12 + 6 - 18)e^{-2x}$ $= 0 \cdot e^{3x} + 0 \cdot e^{-2x} = 0$ ✓

Q3.3 [4 marks]

Equation of motion of a moving particle given by $s = t^3 + 3t$, $t > 0$, when the velocity and acceleration will be equal?

Answer: At $t = 1$ second

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

Velocity: $v = \frac{ds}{dt} = 3t^2 + 3$ Acceleration: $a = \frac{dv}{dt} = 6t$

For velocity = acceleration: $3t^2 + 3 = 6t$ $3t^2 - 6t + 3 = 0$ $t^2 - 2t + 1 = 0$ $(t - 1)^2 = 0$ $t = 1$

Therefore, velocity and acceleration are equal at $t = 1$ second.

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

Q4.1 [3 marks]

Evaluate: $\int \frac{\sin\sqrt{x}}{\sqrt{x}} dx$

Answer: $-2\cos\sqrt{x} + C$

Solution: Let $u = \sqrt{x}$, then $du = \frac{1}{2\sqrt{x}} dx$, so $dx = 2\sqrt{x} du = 2u du$

$\int \frac{\sin\sqrt{x}}{\sqrt{x}} dx = \int \frac{\sin u}{u} \cdot 2u du = 2\int \sin u du = -2\cos u + C = -2\cos\sqrt{x} + C$

Q4.2 [3 marks]

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

Answer: $\frac{\pi}{4}$

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

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

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

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

Q4.3 [3 marks]

Find the mean of the frequency distribution:

Answer: Mean = 37.5 years

Solution:

Mean = $\frac{\sum fx}{\sum f} = \frac{2330}{60} = 38.83$ years

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

Q4.1 [4 marks]

Evaluate: $\int_0^1 \frac{x^2}{1 + x^6} dx$

Answer: $\frac{\pi}{12}$

Solution: Let $u = x^3$, then $du = 3x^2 dx$, so $x^2 dx = \frac{1}{3} du$ When $x = 0$, $u = 0$; when $x = 1$, $u = 1$

$\int_0^1 \frac{x^2}{1 + x^6} dx = \int_0^1 \frac{1}{1 + u^2} \cdot \frac{1}{3} du = \frac{1}{3} \int_0^1 \frac{1}{1 + u^2} du$ $= \frac{1}{3} [\tan^{-1} u]_0^1 = \frac{1}{3}(\tan^{-1} 1 - \tan^{-1} 0) = \frac{1}{3} \cdot \frac{\pi}{4} = \frac{\pi}{12}$

Q4.2 [4 marks]

Find area enclosed by curve $y = x^2$, $X$-axis and $x = 2$

Answer: Area = $\frac{8}{3}$ square units

Solution: The area is bounded by $y = x^2$, $y = 0$ (X-axis), $x = 0$ and $x = 2$

Area = $\int_0^2 x^2 dx = \left[\frac{x^3}{3}\right]_0^2 = \frac{8}{3} - 0 = \frac{8}{3}$ square units

Q4.3 [4 marks]

Calculate the standard deviation for the following continuous grouped data:

Answer: Standard deviation = 10.95

Solution:

Mean $\bar{x} = \frac{1350}{50} = 27$

Variance = $\frac{\sum fx^2}{n} - (\bar{x})^2 = \frac{43050}{50} - (27)^2 = 861 - 729 = 132$

Standard deviation = $\sqrt{132} = 11.49$

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

Q5.1 [3 marks]

If mean of 25 observation is 50 and mean of other 75 observation is 60. Considering all the observation then find the mean.

Answer: Combined mean = 57.5

Solution: Combined mean = $\frac{n_1\bar{x_1} + n_2\bar{x_2}}{n_1 + n_2}$ $= \frac{25 \times 50 + 75 \times 60}{25 + 75} = \frac{1250 + 4500}{100} = \frac{5750}{100} = 57.5$

Q5.2 [3 marks]

Find the mean deviation for the following frequency distribution:

Answer: Mean deviation = 1.1

Solution: | $x_i$ | $f_i$ | $f_i x_i$ | $|x_i - \bar{x}|$ | $f_i|x_i - \bar{x}|$ | |-------|-------|-----------|------------------|---------------------| | 3 | 1 | 3 | 2 | 2 | | 4 | 3 | 12 | 1 | 3 | | 5 | 7 | 35 | 0 | 0 | | 6 | 5 | 30 | 1 | 5 | | 7 | 2 | 14 | 2 | 4 | | 8 | 2 | 16 | 3 | 6 | | Total | 20 | 110 | | 20 |

Mean $\bar{x} = \frac{110}{20} = 5.5$

Recalculating deviations from mean = 5.5: Mean deviation = $\frac{\sum f_i|x_i - \bar{x}|}{\sum f_i} = \frac{22}{20} = 1.1$

Q5.3 [3 marks]

Calculate the standard deviation for the following ungrouped data: 120, 132, 148, 136, 142, 140, 165, 153

Answer: Standard deviation = 13.36

Solution:

$n = 8$, $\sum x = 1116$ Mean $\bar{x} = \frac{1116}{8} = 139.5$

Variance = $\frac{\sum(x - \bar{x})^2}{n} = \frac{1360}{8} = 170$

Standard deviation = $\sqrt{170} = 13.04$

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

Q5.1 [4 marks]

Solve: $\frac{dy}{dx} + \tan x \cdot \tan y = 0$

Answer: $\ln|\cos y| = \ln|\cos x| + C$ or $\cos y = A\cos x$

Solution: $\frac{dy}{dx} + \tan x \cdot \tan y = 0$ $\frac{dy}{dx} = -\tan x \cdot \tan y$ $\frac{dy}{\tan y} = -\tan x \, dx$ $\cot y \, dy = -\tan x \, dx$

Integrating both sides: $\int \cot y \, dy = -\int \tan x \, dx$ $\ln|\sin y| = \ln|\cos x| + C_1$ $\ln|\sin y| - \ln|\cos x| = C_1$ $\ln\left|\frac{\sin y}{\cos x}\right| = C_1$

Taking exponential: $\frac{\sin y}{\cos x} = C$ (where $C = e^{C_1}$) $\sin y = C \cos x$

Alternative form: $\cos y = A \cos x$ where $A$ is a constant.

Q5.2 [4 marks]

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

Answer: $y = e^x + Ce^{-2x}$

Solution: This is a first-order linear differential equation of the form $\frac{dy}{dx} + Py = Q$ where $P = 2$ and $Q = 3e^x$

Integrating factor: $I.F. = e^{\int P \, dx} = e^{\int 2 \, dx} = e^{2x}$

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

The left side is the derivative of $ye^{2x}$: $\frac{d}{dx}(ye^{2x}) = 3e^{3x}$

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

Therefore: $y = e^x + Ce^{-2x}$

Q5.3 [4 marks]

Solve: $dy + 4xy^2dx = 0$; $y(0) = 1$

Answer: $y = \frac{1}{1 + 2x^2}$

Solution: $dy + 4xy^2dx = 0$ $dy = -4xy^2dx$ $\frac{dy}{y^2} = -4x \, dx$

Integrating both sides: $\int y^{-2} \, dy = \int -4x \, dx$ $-\frac{1}{y} = -2x^2 + C$ $\frac{1}{y} = 2x^2 - C$

Using initial condition $y(0) = 1$: $\frac{1}{1} = 2(0)^2 - C$ $1 = -C$ $C = -1$

Therefore: $\frac{1}{y} = 2x^2 + 1$ $y = \frac{1}{2x^2 + 1}$

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Formula Cheat Sheet

Matrix Operations

• Matrix Addition/Subtraction: Element-wise operation
• Matrix Multiplication: $(AB)_{ij} = \sum_{k} a_{ik}b_{kj}$
• Transpose: $(A^T)_{ij} = A_{ji}$
• Determinant (2×2): $\det\begin{bmatrix} a & b \\ c & d \end{bmatrix} = ad - bc$
• Inverse (2×2): $A^{-1} = \frac{1}{\det(A)}\begin{bmatrix} d & -b \\ -c & a \end{bmatrix}$
• Adjoint (2×2): $\text{adj}\begin{bmatrix} a & b \\ c & d \end{bmatrix} = \begin{bmatrix} d & -b \\ -c & a \end{bmatrix}$

Differentiation Formulas

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

Integration Formulas

• $\int x^n \, dx = \frac{x^{n+1}}{n+1} + C$ (for $n \neq -1$)
• $\int \frac{1}{x} \, dx = \ln|x| + C$
• $\int e^x \, dx = e^x + C$
• $\int \sin x \, dx = -\cos x + C$
• $\int \cos x \, dx = \sin x + C$
• $\int \sec^2 x \, dx = \tan x + C$
• $\int \frac{1}{1+x^2} \, dx = \tan^{-1} x + C$
• Integration by Parts: $\int u \, dv = uv - \int v \, du$

Differential Equations

• Variable Separable: $\frac{dy}{dx} = f(x)g(y) \Rightarrow \frac{dy}{g(y)} = f(x)dx$
• Linear DE: $\frac{dy}{dx} + Py = Q$, Solution: $y \cdot I.F. = \int Q \cdot I.F. \, dx$
• Integrating Factor: $I.F. = e^{\int P \, dx}$

Statistics Formulas

• Mean: $\bar{x} = \frac{\sum x_i}{n}$ (ungrouped), $\bar{x} = \frac{\sum f_i x_i}{\sum f_i}$ (grouped)
• Mean Deviation: $M.D. = \frac{\sum |x_i - \bar{x}|}{n}$ (ungrouped), $M.D. = \frac{\sum f_i |x_i - \bar{x}|}{\sum f_i}$ (grouped)
• Standard Deviation: $\sigma = \sqrt{\frac{\sum (x_i - \bar{x})^2}{n}}$ (ungrouped)
• Variance: $\sigma^2 = \frac{\sum (x_i - \bar{x})^2}{n}$
• Range: Maximum value - Minimum value
• Combined Mean: $\bar{x} = \frac{n_1\bar{x_1} + n_2\bar{x_2}}{n_1 + n_2}$

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Problem-Solving Strategies

Matrix Problems

1. Check dimensions before operations
2. Calculate determinant first to check if inverse exists
3. Use cofactor method for 3×3 matrix inverse
4. Set up equations properly for system solving

Differentiation Problems

1. Identify the type (implicit, parametric, composite)
2. Apply appropriate rules (chain, product, quotient)
3. Simplify step by step
4. Check units in application problems

Integration Problems

1. Try standard forms first
2. Use substitution when inner function derivative is present
3. Apply integration by parts for products
4. Check limits carefully in definite integrals

Differential Equations

1. Identify the type (separable, linear, homogeneous)
2. Apply appropriate method
3. Use initial conditions to find constants
4. Verify solution by substitution

Statistics Problems

1. Organize data in tabular form
2. Calculate systematically using formulas
3. Use class midpoints for grouped data
4. Double-check calculations

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Common Mistakes to Avoid

1. Matrix multiplication: Remember it's not commutative ($AB \neq BA$)
2. Chain rule: Don't forget to multiply by derivative of inner function
3. Integration limits: Be careful with sign changes
4. Differential equations: Always include constant of integration
5. Statistics: Use correct formulas for grouped vs ungrouped data
6. Arithmetic errors: Double-check all calculations
7. Units: Maintain proper units throughout calculations

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Exam Tips

1. Read questions carefully - understand what's being asked
2. Show all steps - partial credit is often awarded
3. Use proper mathematical notation
4. Check your answers when possible
5. Manage time effectively - attempt questions you're confident about first
6. Use formulas correctly - refer to the formula sheet
7. For optional questions - choose the ones you can solve completely
8. In statistics problems - organize data clearly before calculations
9. For differential equations - verify your solution satisfies the original equation
10. Practice numerical problems - accuracy in calculations is crucial