Q.1 [14 marks]

Fill in the blanks using appropriate choice from the given options

Q1.1 [1 mark]

If $A_{2×3}$ and $B_{3×4}$ are two matrices then find order of AB =______

Answer: b. $2×4$

Solution: When multiplying matrices, if $A$ is of order $m×n$ and $B$ is of order $n×p$ , then $AB$ is of order $m×p$ . Given: $A_{2×3}$ and $B_{3×4}$ Therefore, $AB$ will be of order $2×4$ .

Q1.2 [1 mark]

If $A = [1\ 3\ 2]$ and $B = \begin{bmatrix} 1 \\ 2 \\ 1 \end{bmatrix}$ then find AB =______

Answer: b. 9

Solution: $AB = [1\ 3\ 2] \begin{bmatrix} 1 \\ 2 \\ 1 \end{bmatrix} = 1(1) + 3(2) + 2(1) = 1 + 6 + 2 = 9$

Q1.3 [1 mark]

$A.I_2 = A$ then $I_2$ =______

Answer: c. $\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$

Solution: $I_2$ is the identity matrix of order 2×2, which has 1's on the main diagonal and 0's elsewhere.

Q1.4 [1 mark]

If $\frac{d}{dx}(\sin^2 x + \cos^2 x) =$ ______

Answer: b. 0

Solution: Since $\sin^2 x + \cos^2 x = 1$ (fundamental trigonometric identity) $\frac{d}{dx}(\sin^2 x + \cos^2 x) = \frac{d}{dx}(1) = 0$

Q1.5 [1 mark]

$\frac{d}{dx}(\cot x) =$ ______

Answer: d. $-\csc^2 x$

Solution: $\frac{d}{dx}(\cot x) = -\csc^2 x$

Q1.6 [1 mark]

$\frac{d}{dx}\log(\sin x)$ then find out $\frac{d^2y}{dx^2} =$ ______

Answer: d. $-\cot^2 x$

Solution: Let $y = \log(\sin x)$ $\frac{dy}{dx} = \frac{1}{\sin x} \cdot \cos x = \cot x$ $\frac{d^2y}{dx^2} = \frac{d}{dx}(\cot x) = -\csc^2 x$

However, since $\csc^2 x = 1 + \cot^2 x$ , the answer is $-\csc^2 x$ .

Q1.7 [1 mark]

$\frac{d}{dx}(\frac{1}{x}) =$ ______

Answer: c. $-\frac{1}{x^2}$

Solution: $\frac{d}{dx}(\frac{1}{x}) = \frac{d}{dx}(x^{-1}) = -1 \cdot x^{-2} = -\frac{1}{x^2}$

Q1.8 [1 mark]

If $\int x^5 dx =$ ______+ c

Answer: a. $\frac{x^6}{6}$

Solution: $\int x^5 dx = \frac{x^{5+1}}{5+1} + c = \frac{x^6}{6} + c$

Q1.9 [1 mark]

$\int_0^{2\pi} (\sin^2 θ + \cos^2 θ)dθ =$ ______+ c

Answer: a. $2π$

Solution: $\int_0^{2\pi} (\sin^2 θ + \cos^2 θ)dθ = \int_0^{2\pi} 1 \, dθ = [θ]_0^{2\pi} = 2\pi - 0 = 2\pi$

Q1.10 [1 mark]

$\int_{-1}^{1} x^3 dx =$ ______+ c

Answer: c. 0

Solution: $\int_{-1}^{1} x^3 dx = \left[\frac{x^4}{4}\right]_{-1}^{1} = \frac{1^4}{4} - \frac{(-1)^4}{4} = \frac{1}{4} - \frac{1}{4} = 0$

Q1.11 [1 mark]

The order and degree of the differential equation $x^2 \frac{d^2y}{dx^2} + 3y^2 = 0$ is =______

Answer: c. 2 and 1

Solution: Order is the highest derivative present = 2 (from $\frac{d^2y}{dx^2}$ ) Degree is the power of the highest derivative = 1

Q1.12 [1 mark]

An integrating factor of the differential equation $\frac{dy}{dx} + py = Q$ is ______

Answer: c. $e^{\int p dx}$

Solution: For a first-order linear differential equation $\frac{dy}{dx} + py = Q$ , the integrating factor is $e^{\int p dx}$ .

Q1.13 [1 mark]

$i^4 =$ ______

Answer: a. 1

Solution: $i^4 = (i^2)^2 = (-1)^2 = 1$

Q1.14 [1 mark]

(3+4i)(4-5i) =______

Answer: d. -32+ i

Solution: $(3+4i)(4-5i) = 3(4) + 3(-5i) + 4i(4) + 4i(-5i)$ $= 12 - 15i + 16i - 20i^2$ $= 12 + i - 20(-1)$ $= 12 + i + 20 = 32 + i$

Wait, let me recalculate: $(3+4i)(4-5i) = 12 - 15i + 16i - 20i^2 = 12 + i + 20 = 32 + i$

The correct answer should be b. 32+ i, but option d shows -32+ i. There might be an error in the options.

Q.2 [14 marks]

Q.2(a) [6 marks]

Attempt any two

Q2.1 [3 marks]

If $A = \begin{bmatrix} 1 & -1 & 1 \\ 3 & 2 & 1 \end{bmatrix}$ and $B = \begin{bmatrix} 1 & 2 \\ 4 & 2 \\ 1 & 7 \end{bmatrix}$ then find out AB & BA.

Solution:

AB calculation: $AB = \begin{bmatrix} 1 & -1 & 1 \\ 3 & 2 & 1 \end{bmatrix} \begin{bmatrix} 1 & 2 \\ 4 & 2 \\ 1 & 7 \end{bmatrix}$

$AB = \begin{bmatrix} 1(1) + (-1)(4) + 1(1) & 1(2) + (-1)(2) + 1(7) \\ 3(1) + 2(4) + 1(1) & 3(2) + 2(2) + 1(7) \end{bmatrix}$

$AB = \begin{bmatrix} 1 - 4 + 1 & 2 - 2 + 7 \\ 3 + 8 + 1 & 6 + 4 + 7 \end{bmatrix} = \begin{bmatrix} -2 & 7 \\ 12 & 17 \end{bmatrix}$

BA calculation: $BA = \begin{bmatrix} 1 & 2 \\ 4 & 2 \\ 1 & 7 \end{bmatrix} \begin{bmatrix} 1 & -1 & 1 \\ 3 & 2 & 1 \end{bmatrix}$

$BA = \begin{bmatrix} 1(1) + 2(3) & 1(-1) + 2(2) & 1(1) + 2(1) \\ 4(1) + 2(3) & 4(-1) + 2(2) & 4(1) + 2(1) \\ 1(1) + 7(3) & 1(-1) + 7(2) & 1(1) + 7(1) \end{bmatrix}$

$BA = \begin{bmatrix} 7 & 3 & 3 \\ 10 & 0 & 6 \\ 22 & 13 & 8 \end{bmatrix}$

Q2.2 [3 marks]

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

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

$A^2 = \begin{bmatrix} (-1)(-1) + (2)(3) & (-1)(2) + (2)(1) \\ (3)(-1) + (1)(3) & (3)(2) + (1)(1) \end{bmatrix}$

$A^2 = \begin{bmatrix} 1 + 6 & -2 + 2 \\ -3 + 3 & 6 + 1 \end{bmatrix} = \begin{bmatrix} 7 & 0 \\ 0 & 7 \end{bmatrix}$

$7I_2 = 7\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 7 & 0 \\ 0 & 7 \end{bmatrix}$

Therefore, $A^2 - 7I_2 = \begin{bmatrix} 7 & 0 \\ 0 & 7 \end{bmatrix} - \begin{bmatrix} 7 & 0 \\ 0 & 7 \end{bmatrix} = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix} = 0$

Hence proved.

Q2.3 [3 marks]

Find the inverse complex number of $\frac{2+3i}{4-3i}$

Solution: First, let's find $\frac{2+3i}{4-3i}$ :

$\frac{2+3i}{4-3i} = \frac{(2+3i)(4+3i)}{(4-3i)(4+3i)} = \frac{8 + 6i + 12i + 9i^2}{16 - 9i^2}$

$= \frac{8 + 18i - 9}{16 + 9} = \frac{-1 + 18i}{25} = -\frac{1}{25} + \frac{18}{25}i$

The inverse of a complex number $z = a + bi$ is $\frac{1}{z} = \frac{\bar{z}}{|z|^2}$

Let $z = -\frac{1}{25} + \frac{18}{25}i$

$|z|^2 = \left(-\frac{1}{25}\right)^2 + \left(\frac{18}{25}\right)^2 = \frac{1}{625} + \frac{324}{625} = \frac{325}{625} = \frac{13}{25}$

$\bar{z} = -\frac{1}{25} - \frac{18}{25}i$

$\frac{1}{z} = \frac{-\frac{1}{25} - \frac{18}{25}i}{\frac{13}{25}} = \frac{-1 - 18i}{13}$

Q.2(b) [8 marks]

Attempt any two

Q2.1 [4 marks]

2y+5x-4 =0 and 7x +3y = 5 solve the equations using matrix method.

Solution: The system can be written as: $5x + 2y = 4$ $7x + 3y = 5$

In matrix form: $\begin{bmatrix} 5 & 2 \\ 7 & 3 \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 4 \\ 5 \end{bmatrix}$

Let $A = \begin{bmatrix} 5 & 2 \\ 7 & 3 \end{bmatrix}$

$|A| = 5(3) - 2(7) = 15 - 14 = 1$

$A^{-1} = \frac{1}{|A|} \begin{bmatrix} 3 & -2 \\ -7 & 5 \end{bmatrix} = \begin{bmatrix} 3 & -2 \\ -7 & 5 \end{bmatrix}$

$\begin{bmatrix} x \\ y \end{bmatrix} = A^{-1} \begin{bmatrix} 4 \\ 5 \end{bmatrix} = \begin{bmatrix} 3 & -2 \\ -7 & 5 \end{bmatrix} \begin{bmatrix} 4 \\ 5 \end{bmatrix}$

$\begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 3(4) + (-2)(5) \\ -7(4) + 5(5) \end{bmatrix} = \begin{bmatrix} 12 - 10 \\ -28 + 25 \end{bmatrix} = \begin{bmatrix} 2 \\ -3 \end{bmatrix}$

Therefore, $x = 2$ and $y = -3$ .

Q2.2 [4 marks]

If $A = \begin{bmatrix} 2 & -2 \\ 3 & 1 \end{bmatrix}$ and $B = \begin{bmatrix} -1 & 5 \\ 4 & -3 \end{bmatrix}$ then Prove that $(AB)^T = B^T.A^T$

Solution: First, let's find $AB$ : $AB = \begin{bmatrix} 2 & -2 \\ 3 & 1 \end{bmatrix} \begin{bmatrix} -1 & 5 \\ 4 & -3 \end{bmatrix}$

$AB = \begin{bmatrix} 2(-1) + (-2)(4) & 2(5) + (-2)(-3) \\ 3(-1) + 1(4) & 3(5) + 1(-3) \end{bmatrix}$

$AB = \begin{bmatrix} -2 - 8 & 10 + 6 \\ -3 + 4 & 15 - 3 \end{bmatrix} = \begin{bmatrix} -10 & 16 \\ 1 & 12 \end{bmatrix}$

$(AB)^T = \begin{bmatrix} -10 & 1 \\ 16 & 12 \end{bmatrix}$

Now, let's find $B^T$ and $A^T$ : $A^T = \begin{bmatrix} 2 & 3 \\ -2 & 1 \end{bmatrix}$ , $B^T = \begin{bmatrix} -1 & 4 \\ 5 & -3 \end{bmatrix}$

$B^T \cdot A^T = \begin{bmatrix} -1 & 4 \\ 5 & -3 \end{bmatrix} \begin{bmatrix} 2 & 3 \\ -2 & 1 \end{bmatrix}$

$B^T \cdot A^T = \begin{bmatrix} -1(2) + 4(-2) & -1(3) + 4(1) \\ 5(2) + (-3)(-2) & 5(3) + (-3)(1) \end{bmatrix}$

$B^T \cdot A^T = \begin{bmatrix} -2 - 8 & -3 + 4 \\ 10 + 6 & 15 - 3 \end{bmatrix} = \begin{bmatrix} -10 & 1 \\ 16 & 12 \end{bmatrix}$

Since $(AB)^T = B^T \cdot A^T$ , the property is proved.

Q2.3 [4 marks]

Simplify: $\frac{(cos2θ+isin2θ)^{-3}.(cos3θ-isin3θ)^2}{(cos2θ+isin2θ)^{-7}.(cos5θ-isin5θ)^3}$

Solution: Using De Moivre's theorem: $(\cos\theta + i\sin\theta)^n = \cos(n\theta) + i\sin(n\theta)$

$(\cos2θ+i\sin2θ)^{-3} = \cos(-6θ) + i\sin(-6θ) = \cos(6θ) - i\sin(6θ)$

$(\cos3θ-i\sin3θ)^2 = (\cos(-3θ) + i\sin(-3θ))^2 = \cos(-6θ) + i\sin(-6θ) = \cos(6θ) - i\sin(6θ)$

$(\cos2θ+i\sin2θ)^{-7} = \cos(-14θ) + i\sin(-14θ) = \cos(14θ) - i\sin(14θ)$

$(\cos5θ-i\sin5θ)^3 = (\cos(-5θ) + i\sin(-5θ))^3 = \cos(-15θ) + i\sin(-15θ) = \cos(15θ) - i\sin(15θ)$

The expression becomes: $\frac{[\cos(6θ) - i\sin(6θ)][\cos(6θ) - i\sin(6θ)]}{[\cos(14θ) - i\sin(14θ)][\cos(15θ) - i\sin(15θ)]}$

$= \frac{[\cos(6θ) - i\sin(6θ)]^2}{[\cos(14θ) - i\sin(14θ)][\cos(15θ) - i\sin(15θ)]}$

$= \frac{\cos(12θ) - i\sin(12θ)}{\cos(29θ) - i\sin(29θ)}$

$= \cos(12θ - 29θ) + i\sin(12θ - 29θ) = \cos(-17θ) + i\sin(-17θ) = \cos(17θ) - i\sin(17θ)$

Q.3 [14 marks]

Q.3(a) [6 marks]

Attempt any two

Q3.1 [3 marks]

If $y = \frac{1+\tan x}{1-\tan x}$ 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 $u = 1+\tan x$ and $v = 1-\tan x$

$\frac{du}{dx} = \sec^2 x$ and $\frac{dv}{dx} = -\sec^2 x$

$\frac{dy}{dx} = \frac{(1-\tan x)(\sec^2 x) - (1+\tan x)(-\sec^2 x)}{(1-\tan x)^2}$

$= \frac{(1-\tan x)\sec^2 x + (1+\tan x)\sec^2 x}{(1-\tan x)^2}$

$= \frac{\sec^2 x[(1-\tan x) + (1+\tan x)]}{(1-\tan x)^2}$

$= \frac{2\sec^2 x}{(1-\tan x)^2}$

Q3.2 [3 marks]

Using Definition of differentiation differentiate $x^3$ with respect to $x$ .

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

For $f(x) = x^3$ :

$\frac{d}{dx}(x^3) = \lim_{h \to 0} \frac{(x+h)^3 - x^3}{h}$

$= \lim_{h \to 0} \frac{x^3 + 3x^2h + 3xh^2 + h^3 - x^3}{h}$

$= \lim_{h \to 0} \frac{3x^2h + 3xh^2 + h^3}{h}$

$= \lim_{h \to 0} \frac{h(3x^2 + 3xh + h^2)}{h}$

$= \lim_{h \to 0} (3x^2 + 3xh + h^2)$

$= 3x^2 + 0 + 0 = 3x^2$

Q3.3 [3 marks]

Simplify: $\int \frac{4+3\cos x}{\sin^2 x} dx$

Solution: $\int \frac{4+3\cos x}{\sin^2 x} dx = \int \frac{4}{\sin^2 x} dx + \int \frac{3\cos x}{\sin^2 x} dx$

$= 4\int \csc^2 x \, dx + 3\int \frac{\cos x}{\sin^2 x} dx$

For the first integral: $\int \csc^2 x \, dx = -\cot x$

For the second integral, let $u = \sin x$ , then $du = \cos x \, dx$ : $\int \frac{\cos x}{\sin^2 x} dx = \int \frac{1}{u^2} du = -\frac{1}{u} = -\frac{1}{\sin x} = -\csc x$

Therefore: $\int \frac{4+3\cos x}{\sin^2 x} dx = 4(-\cot x) + 3(-\csc x) + C = -4\cot x - 3\csc x + C$

Q.3(b) [8 marks]

Attempt any two

Q3.1 [4 marks]

If $y = \log\left(\frac{\cos x}{1+\sin x}\right)$ then find $\frac{dy}{dx}$

Solution: $y = \log\left(\frac{\cos x}{1+\sin x}\right) = \log(\cos x) - \log(1+\sin x)$

$\frac{dy}{dx} = \frac{d}{dx}[\log(\cos x)] - \frac{d}{dx}[\log(1+\sin x)]$

$= \frac{1}{\cos x} \cdot (-\sin x) - \frac{1}{1+\sin x} \cdot \cos x$

$= -\frac{\sin x}{\cos x} - \frac{\cos x}{1+\sin x}$

$= -\tan x - \frac{\cos x}{1+\sin x}$

To simplify further: $= -\frac{\sin x(1+\sin x) + \cos^2 x}{\cos x(1+\sin x)}$

$= -\frac{\sin x + \sin^2 x + \cos^2 x}{\cos x(1+\sin x)}$

$= -\frac{\sin x + 1}{\cos x(1+\sin x)} = -\frac{1}{\cos x} = -\sec x$

Q3.2 [4 marks]

Find maximum and minimum value of function $f(x) = 2x^3 - 15x^2 + 36x + 10$ .

Solution: To find extrema, we find where $f'(x) = 0$ :

$f'(x) = 6x^2 - 30x + 36 = 6(x^2 - 5x + 6) = 6(x-2)(x-3)$

Setting $f'(x) = 0$ : $x = 2$ or $x = 3$

To determine nature of critical points, we use the second derivative test: $f''(x) = 12x - 30$

At $x = 2$ : $f''(2) = 24 - 30 = -6 < 0$ → Local maximum At $x = 3$ : $f''(3) = 36 - 30 = 6 > 0$ → Local minimum

Values: $f(2) = 2(8) - 15(4) + 36(2) + 10 = 16 - 60 + 72 + 10 = 38$ $f(3) = 2(27) - 15(9) + 36(3) + 10 = 54 - 135 + 108 + 10 = 37$

Therefore:

Local maximum value: 38 at $x = 2$
Local minimum value: 37 at $x = 3$

Q3.3 [4 marks]

If $y = 2e^{-3x} + 3e^{2x}$ then prove that $y_2 + y_1 - 6y = 0$ .

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

$y_1 = \frac{dy}{dx} = 2(-3)e^{-3x} + 3(2)e^{2x} = -6e^{-3x} + 6e^{2x}$

$y_2 = \frac{d^2y}{dx^2} = -6(-3)e^{-3x} + 6(2)e^{2x} = 18e^{-3x} + 12e^{2x}$

Now let's verify $y_2 + y_1 - 6y = 0$ :

$y_2 + y_1 - 6y = (18e^{-3x} + 12e^{2x}) + (-6e^{-3x} + 6e^{2x}) - 6(2e^{-3x} + 3e^{2x})$

$= 18e^{-3x} + 12e^{2x} - 6e^{-3x} + 6e^{2x} - 12e^{-3x} - 18e^{2x}$

$= (18 - 6 - 12)e^{-3x} + (12 + 6 - 18)e^{2x}$

$= 0 \cdot e^{-3x} + 0 \cdot e^{2x} = 0$

Hence proved.

Q.4 [14 marks]

Q.4(a) [6 marks]

Attempt any two

Q4.1 [3 marks]

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

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

$\int \frac{x^2}{1+x^6} dx = \int \frac{1}{1+(x^3)^2} \cdot x^2 dx = \int \frac{1}{1+u^2} \cdot \frac{1}{3} du$

$= \frac{1}{3} \int \frac{1}{1+u^2} du = \frac{1}{3} \tan^{-1}(u) + C$

$= \frac{1}{3} \tan^{-1}(x^3) + C$

Q4.2 [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 $v = \frac{x^2}{2}$

$\int x \log x \, dx = \log x \cdot \frac{x^2}{2} - \int \frac{x^2}{2} \cdot \frac{1}{x} dx$

$= \frac{x^2 \log x}{2} - \int \frac{x}{2} dx$

$= \frac{x^2 \log x}{2} - \frac{x^2}{4} + C$

$= \frac{x^2}{2}(\log x - \frac{1}{2}) + C$

Q4.3 [3 marks]

Solve the differential equation $x dy + y dx = 0$ .

Solution: The given equation is: $x dy + y dx = 0$

This can be written as: $x dy = -y dx$

Separating variables: $\frac{dy}{y} = -\frac{dx}{x}$

Integrating both sides: $\int \frac{dy}{y} = \int -\frac{dx}{x}$

$\log|y| = -\log|x| + C_1$

$\log|y| + \log|x| = C_1$

$\log|xy| = C_1$

$|xy| = e^{C_1} = C$ (where $C = e^{C_1}$ )

Therefore: $xy = \pm C$

The general solution is: $xy = k$ (where $k$ is an arbitrary constant)

Q.4(b) [8 marks]

Attempt any two

Q4.1 [4 marks]

Evaluate: $\int_1^e \frac{(\log x)^2}{x} dx$

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

When $x = 1$ : $u = \log 1 = 0$ When $x = e$ : $u = \log e = 1$

$\int_1^e \frac{(\log x)^2}{x} dx = \int_0^1 u^2 du$

$= \left[\frac{u^3}{3}\right]_0^1 = \frac{1^3}{3} - \frac{0^3}{3} = \frac{1}{3}$

Q4.2 [4 marks]

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

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

First, let's simplify the integrand: $\frac{\sec x}{\sec x + \cos x} = \frac{\frac{1}{\cos x}}{\frac{1}{\cos x} + \cos x} = \frac{\frac{1}{\cos x}}{\frac{1 + \cos^2 x}{\cos x}} = \frac{1}{1 + \cos^2 x}$

So $I = \int_0^{\pi/2} \frac{1}{1 + \cos^2 x} dx$

Using the substitution $\tan(x/2) = t$ : $\cos x = \frac{1-t^2}{1+t^2}$ , $dx = \frac{2dt}{1+t^2}$

When $x = 0$ : $t = 0$ When $x = \pi/2$ : $t = 1$

$I = \int_0^1 \frac{1}{1 + \left(\frac{1-t^2}{1+t^2}\right)^2} \cdot \frac{2dt}{1+t^2}$

After simplification (which involves significant algebra), this evaluates to: $I = \frac{\pi}{2\sqrt{2}}$

Q4.3 [4 marks]

Solve the differential equation $\frac{dy}{dx} + \frac{y}{x} = e^x$ , $y(0) = 2$ .

Solution: This is a first-order linear differential equation of the form $\frac{dy}{dx} + P(x)y = Q(x)$

Here, $P(x) = \frac{1}{x}$ and $Q(x) = e^x$

The integrating factor is: $\mu(x) = e^{\int P(x) dx} = e^{\int \frac{1}{x} dx} = e^{\log|x|} = |x| = x$ (for $x > 0$ )

Multiplying the equation by the integrating factor: $x\frac{dy}{dx} + y = xe^x$

The left side is $\frac{d}{dx}(xy)$ , so: $\frac{d}{dx}(xy) = xe^x$

Integrating both sides: $xy = \int xe^x dx$

Using integration by parts for $\int xe^x dx$ : Let $u = x$ , $dv = e^x dx$ Then $du = dx$ , $v = e^x$

$\int xe^x dx = xe^x - \int e^x dx = xe^x - e^x + C = e^x(x-1) + C$

Therefore: $xy = e^x(x-1) + C$

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

Using the initial condition $y(0) = 2$ : This presents a problem as the solution is undefined at $x = 0$ . Let me reconsider the problem.

Actually, let's solve this more carefully. The equation should be valid for $x \neq 0$ .

If we assume the initial condition is at $x = 1$ instead (as $x = 0$ makes the equation singular), and $y(1) = 2$ :

$2 = \frac{e^1(1-1) + C}{1} = \frac{0 + C}{1} = C$

So $C = 2$ , and the solution is: $y = \frac{e^x(x-1) + 2}{x}$

Q.5 [14 marks]

Q.5(a) [6 marks]

Attempt any two

Q5.1 [3 marks]

Find the conjugate complex number and modulus of $\frac{3+7i}{1-i}$ .

Solution: First, let's simplify $\frac{3+7i}{1-i}$ :

$\frac{3+7i}{1-i} = \frac{(3+7i)(1+i)}{(1-i)(1+i)} = \frac{3 + 3i + 7i + 7i^2}{1 - i^2}$

$= \frac{3 + 10i - 7}{1 + 1} = \frac{-4 + 10i}{2} = -2 + 5i$

Conjugate: The conjugate of $-2 + 5i$ is $-2 - 5i$

Modulus: $|{-2 + 5i}| = \sqrt{(-2)^2 + (5)^2} = \sqrt{4 + 25} = \sqrt{29}$

Q5.2 [3 marks]

Find the square root of complex number $3-4i$ .

Solution: Let $\sqrt{3-4i} = a + bi$ where $a, b \in \mathbb{R}$

Then $(a + bi)^2 = 3 - 4i$

$a^2 + 2abi + (bi)^2 = 3 - 4i$

$a^2 - b^2 + 2abi = 3 - 4i$

Comparing real and imaginary parts: $a^2 - b^2 = 3$ ... (1) $2ab = -4$ ... (2)

From equation (2): $b = -\frac{2}{a}$

Substituting in equation (1): $a^2 - \left(-\frac{2}{a}\right)^2 = 3$

$a^2 - \frac{4}{a^2} = 3$

$a^4 - 3a^2 - 4 = 0$

Let $u = a^2$ : $u^2 - 3u - 4 = 0$

$(u-4)(u+1) = 0$

So $u = 4$ or $u = -1$

Since $u = a^2 \geq 0$ , we have $u = 4$ , so $a^2 = 4$

Therefore $a = \pm 2$

If $a = 2$ : $b = -\frac{2}{2} = -1$ If $a = -2$ : $b = -\frac{2}{-2} = 1$

The two square roots are: $2 - i$ and $-2 + i$

Q5.3 [3 marks]

Find $\frac{dy}{dx}$ for $y = (\sin x)^{\tan x}$

Solution: Taking logarithm of both sides: $\log y = \tan x \log(\sin x)$

Differentiating both sides with respect to $x$ : $\frac{1}{y} \frac{dy}{dx} = \frac{d}{dx}[\tan x \log(\sin x)]$

Using product rule on the right side: $\frac{1}{y} \frac{dy}{dx} = \sec^2 x \log(\sin x) + \tan x \cdot \frac{\cos x}{\sin x}$

$\frac{1}{y} \frac{dy}{dx} = \sec^2 x \log(\sin x) + \tan x \cdot \cot x$

$\frac{1}{y} \frac{dy}{dx} = \sec^2 x \log(\sin x) + 1$

Therefore: $\frac{dy}{dx} = y[\sec^2 x \log(\sin x) + 1]$

$\frac{dy}{dx} = (\sin x)^{\tan x}[\sec^2 x \log(\sin x) + 1]$

Q.5(b) [8 marks]

Attempt any two

Q5.1 [4 marks]

Find solution of the differential equation $\tan y \, dx + \tan x \sec^2 y \, dy = 0$ .

Solution: The given equation is: $\tan y \, dx + \tan x \sec^2 y \, dy = 0$

Rearranging: $\tan y \, dx = -\tan x \sec^2 y \, dy$

$\frac{\tan y}{\sec^2 y} dy = -\tan x \, dx$

$\frac{\sin y / \cos y}{1/\cos^2 y} dy = -\tan x \, dx$

$\frac{\sin y}{\cos y} \cdot \cos^2 y \, dy = -\tan x \, dx$

$\sin y \cos y \, dy = -\tan x \, dx$

Integrating both sides: $\int \sin y \cos y \, dy = -\int \tan x \, dx$

For the left side, let $u = \sin y$ , then $du = \cos y \, dy$ : $\int \sin y \cos y \, dy = \int u \, du = \frac{u^2}{2} = \frac{\sin^2 y}{2}$

For the right side: $-\int \tan x \, dx = -\int \frac{\sin x}{\cos x} dx = \log|\cos x| + C_1$

Therefore: $\frac{\sin^2 y}{2} = \log|\cos x| + C$

$\sin^2 y = 2\log|\cos x| + K$ (where $K = 2C$ )

Q5.2 [4 marks]

If $A = \begin{bmatrix} 3 & -1 & 2 \\ 4 & 1 & -1 \\ 5 & 0 & 1 \end{bmatrix}$ then find $A^{-1}$ .

Solution: To find $A^{-1}$ , we use the formula $A^{-1} = \frac{1}{|A|} \text{adj}(A)$

First, let's find $|A|$ : $|A| = 3\begin{vmatrix} 1 & -1 \\ 0 & 1 \end{vmatrix} - (-1)\begin{vmatrix} 4 & -1 \\ 5 & 1 \end{vmatrix} + 2\begin{vmatrix} 4 & 1 \\ 5 & 0 \end{vmatrix}$

$= 3(1 \cdot 1 - (-1) \cdot 0) + 1(4 \cdot 1 - (-1) \cdot 5) + 2(4 \cdot 0 - 1 \cdot 5)$

$= 3(1) + 1(4 + 5) + 2(0 - 5) = 3 + 9 - 10 = 2$

Now we find the cofactor matrix:

$C_{11} = +\begin{vmatrix} 1 & -1 \\ 0 & 1 \end{vmatrix} = 1$

$C_{12} = -\begin{vmatrix} 4 & -1 \\ 5 & 1 \end{vmatrix} = -(4-(-5)) = -9$

$C_{13} = +\begin{vmatrix} 4 & 1 \\ 5 & 0 \end{vmatrix} = 0-5 = -5$

$C_{21} = -\begin{vmatrix} -1 & 2 \\ 0 & 1 \end{vmatrix} = -(-1-0) = 1$

$C_{22} = +\begin{vmatrix} 3 & 2 \\ 5 & 1 \end{vmatrix} = 3-10 = -7$

$C_{23} = -\begin{vmatrix} 3 & -1 \\ 5 & 0 \end{vmatrix} = -(0-(-5)) = -5$

$C_{31} = +\begin{vmatrix} -1 & 2 \\ 1 & -1 \end{vmatrix} = 1-2 = -1$

$C_{32} = -\begin{vmatrix} 3 & 2 \\ 4 & -1 \end{vmatrix} = -(-3-8) = 11$

$C_{33} = +\begin{vmatrix} 3 & -1 \\ 4 & 1 \end{vmatrix} = 3-(-4) = 7$

The cofactor matrix is: $C = \begin{bmatrix} 1 & -9 & -5 \\ 1 & -7 & -5 \\ -1 & 11 & 7 \end{bmatrix}$

The adjugate is the transpose of the cofactor matrix: $\text{adj}(A) = \begin{bmatrix} 1 & 1 & -1 \\ -9 & -7 & 11 \\ -5 & -5 & 7 \end{bmatrix}$

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

Q5.3 [4 marks]

$x = a(\theta - \sin\theta)$ , $y = a(1 - \cos\theta)$ then find $\frac{dy}{dx}$ .

Solution: These are parametric equations. To find $\frac{dy}{dx}$ , we use: $\frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta}$

First, let's find $\frac{dx}{d\theta}$ : $x = a(\theta - \sin\theta)$ $\frac{dx}{d\theta} = a(1 - \cos\theta)$

Next, let's find $\frac{dy}{d\theta}$ : $y = a(1 - \cos\theta)$ $\frac{dy}{d\theta} = a\sin\theta$

Therefore: $\frac{dy}{dx} = \frac{a\sin\theta}{a(1 - \cos\theta)} = \frac{\sin\theta}{1 - \cos\theta}$

Using the identity $1 - \cos\theta = 2\sin^2(\theta/2)$ and $\sin\theta = 2\sin(\theta/2)\cos(\theta/2)$ :

$\frac{dy}{dx} = \frac{2\sin(\theta/2)\cos(\theta/2)}{2\sin^2(\theta/2)} = \frac{\cos(\theta/2)}{\sin(\theta/2)} = \cot(\theta/2)$

Formula Cheat Sheet

Differentiation Formulas

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

Integration Formulas

$\int x^n dx = \frac{x^{n+1}}{n+1} + C$ (for $n \neq -1$ )
$\int \frac{1}{x} dx = \log|x| + C$
$\int e^x dx = e^x + C$
$\int \sin x dx = -\cos x + C$
$\int \cos x dx = \sin x + C$
$\int \sec^2 x dx = \tan x + C$

Matrix Operations

$(AB)^T = B^T A^T$
$A^{-1} = \frac{1}{|A|} \text{adj}(A)$
For 2×2 matrix: $\begin{bmatrix} a & b \\ c & d \end{bmatrix}^{-1} = \frac{1}{ad-bc} \begin{bmatrix} d & -b \\ -c & a \end{bmatrix}$

Complex Numbers

$i^2 = -1$ , $i^3 = -i$ , $i^4 = 1$
$|a + bi| = \sqrt{a^2 + b^2}$
De Moivre's Theorem: $(\cos\theta + i\sin\theta)^n = \cos(n\theta) + i\sin(n\theta)$

Problem-Solving Strategies

For Matrix Problems: Always check dimensions before multiplication
For Differentiation: Use appropriate rules (product, quotient, chain)
For Integration: Look for substitutions or integration by parts
For Differential Equations: Identify type (separable, linear, etc.)
For Complex Numbers: Convert to standard form before operations

Common Mistakes to Avoid

Sign errors in differentiation and integration
Forgetting constant of integration
Matrix dimension mismatch
Not simplifying complex fractions
Missing absolute value signs in logarithms

Exam Tips

Show all steps clearly
Double-check calculations
Use proper mathematical notation
Manage time effectively
Attempt easier questions first