Applied Mathematics (4320001) - Summer 2024 Solution

Q.1 Fill in the blanks [14 marks]

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Q1.1 [1 mark]

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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]

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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]

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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]

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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]

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

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

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

Q1.6 [1 mark]

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

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]

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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]

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

Answer: (d) $x^5$

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

Q1.9 [1 mark]

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

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]

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

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]

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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]

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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]

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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]

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

Answer: (d) 19

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

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

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Q2.1 [3 marks]

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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]

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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]

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

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

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Q2.1 [4 marks]

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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]

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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]

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

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

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Q3.1 [3 marks]

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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]

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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]

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

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

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Q3.1 [4 marks]

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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]

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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]

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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.

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

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Q4.1 [3 marks]

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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]

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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]

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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

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

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Q4.1 [4 marks]

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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]

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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]

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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 = $\sqrt{132} = 11.49$

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

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Q5.1 [3 marks]

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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]

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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]

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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 = $\sqrt{170} = 13.04$

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

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Q5.1 [4 marks]

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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]

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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]

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

Formula Cheat Sheet

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Matrix Operations

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• 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

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• $\frac{d}{dx}(x^n) = nx^{n-1}$
• $\frac{d}{dx}(e^x) = e^x$
• $\frac{d}{dx}(\ln x) = \frac{1}{x}$
• $\frac{d}{dx}(\sin x) = \cos x$
• $\frac{d}{dx}(\cos x) = -\sin x$
• $\frac{d}{dx}(\tan x) = \sec^2 x$
• $\frac{d}{dx}(\tan^{-1} x) = \frac{1}{1+x^2}$
• 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

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• $\int x^n , dx = \frac{x^{n+1}}{n+1} + C$ (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

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• 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

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• 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: $\bar{x} = \frac{n_1\bar{x_1} + n_2\bar{x_2}}{n_1 + n_2}$

Problem-Solving Strategies

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Matrix Problems

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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

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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

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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

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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

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

Common Mistakes to Avoid

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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

Exam Tips

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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