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

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

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

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

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.

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:

Age	20-24	25-29	30-34	35-39	40-44	45-49	50-54	55-59
Staff	5	7	9	11	10	8	6	4

Answer: Mean = 37.5 years

Solution:

Class	Midpoint (x)	Frequency (f)	fx
20-24	22	5	110
25-29	27	7	189
30-34	32	9	288
35-39	37	11	407
40-44	42	10	420
45-49	47	8	376
50-54	52	6	312
55-59	57	4	228
Total		60	2330

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

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:

Class	0-10	10-20	20-30	30-40	40-50
Frequency	5	8	15	16	6

Answer: Standard deviation = 10.95

Solution:

Class	Midpoint (x)	f	fx	$x^2$	$fx^2$
0-10	5	5	25	25	125
10-20	15	8	120	225	1800
20-30	25	15	375	625	9375
30-40	35	16	560	1225	19600
40-50	45	6	270	2025	12150
Total		50	1350		43050

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]

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:

$x_i$	3	4	5	6	7	8
$f_i$	1	3	7	5	2	2

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:

$x$	$x - \bar{x}$	$(x - \bar{x})^2$
120	-19.5	380.25
132	-7.5	56.25
148	8.5	72.25
136	-3.5	12.25
142	2.5	6.25
140	0.5	0.25
165	25.5	650.25
153	13.5	182.25
Total	0	1360

$n = 8$ , $\sum x = 1116$ 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]

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

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

Problem-Solving Strategies

Matrix Problems

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

Differentiation Problems

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

Integration Problems

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

Differential Equations

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

Statistics Problems

Organize data in tabular form
Calculate systematically using formulas
Use class midpoints for grouped data
Double-check calculations

Common Mistakes to Avoid

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

Exam Tips

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