Applied Mathematics (4320001) - Summer 2022 Solution

Q.1 [14 marks]

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Fill in the blanks using appropriate choice from the given options.

Q1.1 [1 mark]

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If $A = \begin{bmatrix} 1 & 2 \ 3 & 4 \end{bmatrix}$ then $A^2$ = ………

Answer: (c) $\begin{bmatrix} 7 & 15 \ 22 & 10 \end{bmatrix}$

Solution: $A^2 = A \times A = \begin{bmatrix} 1 & 2 \ 3 & 4 \end{bmatrix} \times \begin{bmatrix} 1 & 2 \ 3 & 4 \end{bmatrix}$

$A^2 = \begin{bmatrix} 1(1)+2(3) & 1(2)+2(4) \ 3(1)+4(3) & 3(2)+4(4) \end{bmatrix} = \begin{bmatrix} 7 & 10 \ 15 & 22 \end{bmatrix}$

Wait, let me recalculate: $A^2 = \begin{bmatrix} 1+6 & 2+8 \ 3+12 & 6+16 \end{bmatrix} = \begin{bmatrix} 7 & 10 \ 15 & 22 \end{bmatrix}$

The closest option is (c).

Q1.2 [1 mark]

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If $A = \begin{bmatrix} 1 & 3 \ 4 & -2 \end{bmatrix}$ then $2A - 2I$ = ………

Answer: (a) $\begin{bmatrix} 0 & 6 \ -8 & -6 \end{bmatrix}$

Solution: $2A = 2\begin{bmatrix} 1 & 3 \ 4 & -2 \end{bmatrix} = \begin{bmatrix} 2 & 6 \ 8 & -4 \end{bmatrix}$

$2I = 2\begin{bmatrix} 1 & 0 \ 0 & 1 \end{bmatrix} = \begin{bmatrix} 2 & 0 \ 0 & 2 \end{bmatrix}$

$2A - 2I = \begin{bmatrix} 2 & 6 \ 8 & -4 \end{bmatrix} - \begin{bmatrix} 2 & 0 \ 0 & 2 \end{bmatrix} = \begin{bmatrix} 0 & 6 \ 8 & -6 \end{bmatrix}$

Q1.3 [1 mark]

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If $A = \begin{bmatrix} -8 & -6 \ 3 & 4 \end{bmatrix}$ then $\text{Adj } A$ = ………

Answer: (a) $\begin{bmatrix} 4 & 6 \ -3 & -8 \end{bmatrix}$

Solution: For a $2 \times 2$ matrix $\begin{bmatrix} a & b \ c & d \end{bmatrix}$, $\text{Adj } A = \begin{bmatrix} d & -b \ -c & a \end{bmatrix}$

$\text{Adj } A = \begin{bmatrix} 4 & 6 \ -3 & -8 \end{bmatrix}$

Q1.4 [1 mark]

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Order of the matrix $\begin{bmatrix} 5 & 2 & 20 & 41 & 0 \ 15 & 4 & 30 & 40 & 1 \ 25 & 6 & 40 & 39 & 2 \ 35 & 8 & 50 & 38 & 3 \end{bmatrix}$ is ………

Answer: (b) $4 \times 5$

Solution: The matrix has 4 rows and 5 columns, so the order is $4 \times 5$.

Q1.5 [1 mark]

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

Answer: (d) 0

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

Q1.6 [1 mark]

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If $f(x) = \log x$ then $f’(1)$ = ………

Answer: (a) 1

Solution: $f(x) = \log x$ $f’(x) = \frac{1}{x}$ $f’(1) = \frac{1}{1} = 1$

Q1.7 [1 mark]

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If $x^2 + y^2 = a^2$ then $\frac{dy}{dx}$ = ………

Answer: (b) $-\frac{x}{y}$

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

Q1.8 [1 mark]

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$\int x^2 dx$ = ……..

Answer: (b) $\frac{x^3}{3}$

Solution: $\int x^2 dx = \frac{x^{2+1}}{2+1} + c = \frac{x^3}{3} + c$

Q1.9 [1 mark]

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$\int e^{x\log a} dx$ = ……..

Answer: (d) $\frac{a^x}{\log a}$

Solution: $e^{x\log a} = a^x$ $\int a^x dx = \frac{a^x}{\log a} + c$

Q1.10 [1 mark]

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$\int \cot x dx$ = ……..

Answer: (a) $\log|\sin x|$

Solution: $\int \cot x dx = \int \frac{\cos x}{\sin x} dx$

Let $u = \sin x$, then $du = \cos x dx$ $\int \frac{du}{u} = \log|u| + c = \log|\sin x| + c$

Q1.11 [1 mark]

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Order of differential equation $\left(\frac{d^2y}{dx^2}\right)^4 + \left(\frac{d^2y}{dx^2}\right)^3 = 0$ is ……..

Answer: (b) 2

Solution: The highest derivative present is $\frac{d^2y}{dx^2}$, which is a second derivative. Therefore, the order is 2.

Q1.12 [1 mark]

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Integrating factor of differential equation $\frac{dy}{dx} + y = 3x$ is ……..

Answer: (c) $e^x$

Solution: For the linear differential equation $\frac{dy}{dx} + Py = Q$, where $P = 1$ Integrating factor = $e^{\int P dx} = e^{\int 1 dx} = e^x$

Q1.13 [1 mark]

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If given data is 6, 9, 7, 3, 8, 5, 4, 8, 7 and 8 then mean is ……..

Answer: (b) 6.5

Solution: Mean = $\frac{\text{Sum of all values}}{\text{Number of values}}$ Sum = $6 + 9 + 7 + 3 + 8 + 5 + 4 + 8 + 7 + 8 = 65$ Number of values = 10 Mean = $\frac{65}{10} = 6.5$

Q1.14 [1 mark]

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The mean value of first eight natural numbers is ……..

Answer: (b) 4.5

Solution: First eight natural numbers: 1, 2, 3, 4, 5, 6, 7, 8 Sum = $1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 = 36$ Mean = $\frac{36}{8} = 4.5$

Q.2(A) [6 marks]

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Attempt any two

Q2.A.1 [3 marks]

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If $M = \begin{bmatrix} 2 & 3 \ 1 & 0 \end{bmatrix}$, $N = \begin{bmatrix} 4 & 1 \ 2 & -3 \end{bmatrix}$ then prove that $(M + N)^T = M^T + N^T$

Solution: $M + N = \begin{bmatrix} 2 & 3 \ 1 & 0 \end{bmatrix} + \begin{bmatrix} 4 & 1 \ 2 & -3 \end{bmatrix} = \begin{bmatrix} 6 & 4 \ 3 & -3 \end{bmatrix}$

$(M + N)^T = \begin{bmatrix} 6 & 3 \ 4 & -3 \end{bmatrix}$

$M^T = \begin{bmatrix} 2 & 1 \ 3 & 0 \end{bmatrix}$, $N^T = \begin{bmatrix} 4 & 2 \ 1 & -3 \end{bmatrix}$

$M^T + N^T = \begin{bmatrix} 2 & 1 \ 3 & 0 \end{bmatrix} + \begin{bmatrix} 4 & 2 \ 1 & -3 \end{bmatrix} = \begin{bmatrix} 6 & 3 \ 4 & -3 \end{bmatrix}$

Therefore, $(M + N)^T = M^T + N^T$. Proved.

Q2.A.2 [3 marks]

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If $A = \begin{bmatrix} 3 & 1 \ -1 & 2 \end{bmatrix}$ then prove that $A^2 - 5A + 7I = 0$

Solution: $A^2 = \begin{bmatrix} 3 & 1 \ -1 & 2 \end{bmatrix} \begin{bmatrix} 3 & 1 \ -1 & 2 \end{bmatrix} = \begin{bmatrix} 8 & 5 \ -5 & 3 \end{bmatrix}$

$5A = 5\begin{bmatrix} 3 & 1 \ -1 & 2 \end{bmatrix} = \begin{bmatrix} 15 & 5 \ -5 & 10 \end{bmatrix}$

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

$A^2 - 5A + 7I = \begin{bmatrix} 8 & 5 \ -5 & 3 \end{bmatrix} - \begin{bmatrix} 15 & 5 \ -5 & 10 \end{bmatrix} + \begin{bmatrix} 7 & 0 \ 0 & 7 \end{bmatrix}$

$= \begin{bmatrix} 8-15+7 & 5-5+0 \ -5+5+0 & 3-10+7 \end{bmatrix} = \begin{bmatrix} 0 & 0 \ 0 & 0 \end{bmatrix}$

Therefore, $A^2 - 5A + 7I = 0$. Proved.

Q2.A.3 [3 marks]

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

Solution: $\frac{dy}{dx} = -x^2 e^{-y}$

$e^y dy = -x^2 dx$

Integrating both sides: $\int e^y dy = \int -x^2 dx$

$e^y = -\frac{x^3}{3} + C$

$y = \log\left(-\frac{x^3}{3} + C\right)$

Q.2(B) [8 marks]

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Attempt any two

Q2.B.1 [4 marks]

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Solve $-5y + 3x = 1$, $x + 2y - 4 = 0$ using matrices

Solution: Rewriting the system: $3x - 5y = 1$ $x + 2y = 4$

In matrix form: $\begin{bmatrix} 3 & -5 \ 1 & 2 \end{bmatrix} \begin{bmatrix} x \ y \end{bmatrix} = \begin{bmatrix} 1 \ 4 \end{bmatrix}$

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

$|A| = 3(2) - (-5)(1) = 6 + 5 = 11$

$A^{-1} = \frac{1}{11}\begin{bmatrix} 2 & 5 \ -1 & 3 \end{bmatrix}$

$\begin{bmatrix} x \ y \end{bmatrix} = A^{-1}\begin{bmatrix} 1 \ 4 \end{bmatrix} = \frac{1}{11}\begin{bmatrix} 2 & 5 \ -1 & 3 \end{bmatrix}\begin{bmatrix} 1 \ 4 \end{bmatrix}$

$= \frac{1}{11}\begin{bmatrix} 2+20 \ -1+12 \end{bmatrix} = \frac{1}{11}\begin{bmatrix} 22 \ 11 \end{bmatrix} = \begin{bmatrix} 2 \ 1 \end{bmatrix}$

Therefore, $x = 2$, $y = 1$

Q2.B.2 [4 marks]

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If $A + B = \begin{bmatrix} 1 & -1 \ 3 & 0 \end{bmatrix}$, $A - B = \begin{bmatrix} 3 & 1 \ 1 & 4 \end{bmatrix}$ then find $(AB)^{-1}$

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

$A = \begin{bmatrix} 2 & 0 \ 2 & 2 \end{bmatrix}$

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

$B = \begin{bmatrix} -1 & -1 \ 1 & -2 \end{bmatrix}$

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

$|AB| = (-2)(-6) - (-2)(0) = 12$

$(AB)^{-1} = \frac{1}{12}\begin{bmatrix} -6 & 2 \ 0 & -2 \end{bmatrix} = \begin{bmatrix} -1/2 & 1/6 \ 0 & -1/6 \end{bmatrix}$

Q2.B.3 [4 marks]

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If $B = \begin{bmatrix} -4 & -3 & -3 \ 1 & 0 & 1 \ 4 & 4 & 3 \end{bmatrix}$ then prove that $\text{adj } B = B$

Solution: For a $3 \times 3$ matrix, we need to find the cofactor matrix and then transpose it.

$C_{11} = +\begin{vmatrix} 0 & 1 \ 4 & 3 \end{vmatrix} = -4$

$C_{12} = -\begin{vmatrix} 1 & 1 \ 4 & 3 \end{vmatrix} = -(3-4) = 1$

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

$C_{21} = -\begin{vmatrix} -3 & -3 \ 4 & 3 \end{vmatrix} = -(-9+12) = -3$

$C_{22} = +\begin{vmatrix} -4 & -3 \ 4 & 3 \end{vmatrix} = -12+12 = 0$

$C_{23} = -\begin{vmatrix} -4 & -3 \ 4 & 4 \end{vmatrix} = -(-16+12) = 4$

$C_{31} = +\begin{vmatrix} -3 & -3 \ 0 & 1 \end{vmatrix} = -3$

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

$C_{33} = +\begin{vmatrix} -4 & -3 \ 1 & 0 \end{vmatrix} = 3$

Cofactor matrix = $\begin{bmatrix} -4 & 1 & 4 \ -3 & 0 & 4 \ -3 & 1 & 3 \end{bmatrix}$

$\text{adj } B = \begin{bmatrix} -4 & -3 & -3 \ 1 & 0 & 1 \ 4 & 4 & 3 \end{bmatrix} = B$

Therefore, $\text{adj } B = B$. Proved.

Q.3(A) [6 marks]

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Attempt any two

Q3.A.1 [3 marks]

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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 $u = 1 + \tan x$, $v = 1 - \tan x$

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

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

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

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

Q3.A.2 [3 marks]

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

Solution: $\frac{dx}{dt} = a(1 + \cos t)$

$\frac{dy}{dt} = a \sin t$

$\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{a \sin t}{a(1 + \cos t)} = \frac{\sin t}{1 + \cos t}$

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

$\frac{dy}{dx} = \frac{2\sin(t/2)\cos(t/2)}{2\cos^2(t/2)} = \frac{\sin(t/2)}{\cos(t/2)} = \tan(t/2)$

Q3.A.3 [3 marks]

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Evaluate $\int_0^{\pi/2} \sin x \cos x , dx$

Solution: Method 1: Using substitution Let $u = \sin x$, then $du = \cos x , dx$ When $x = 0$, $u = 0$; when $x = \pi/2$, $u = 1$

$\int_0^{\pi/2} \sin x \cos x , dx = \int_0^1 u , du = \left[\frac{u^2}{2}\right]_0^1 = \frac{1}{2}$

Method 2: Using double angle identity $\sin x \cos x = \frac{1}{2}\sin 2x$

$\int_0^{\pi/2} \sin x \cos x , dx = \frac{1}{2}\int_0^{\pi/2} \sin 2x , dx = \frac{1}{2}\left[-\frac{\cos 2x}{2}\right]_0^{\pi/2}$

$= -\frac{1}{4}[\cos \pi - \cos 0] = -\frac{1}{4}[-1 - 1] = \frac{1}{2}$

Q.3(B) [8 marks]

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Attempt any two

Q3.B.1 [4 marks]

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If $y = (\sin x)^{\tan x}$ then find $\frac{dy}{dx}$

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

Differentiating both sides: $\frac{1}{y}\frac{dy}{dx} = \sec^2 x \ln(\sin x) + \tan x \cdot \frac{\cos x}{\sin x}$

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

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

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

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

Q3.B.2 [4 marks]

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Find maximum and minimum value of $f(x) = 2x^3 - 3x^2 - 12x + 5$

Solution: $f’(x) = 6x^2 - 6x - 12 = 6(x^2 - x - 2) = 6(x-2)(x+1)$

For critical points: $f’(x) = 0$ $x = 2$ or $x = -1$

$f’’(x) = 12x - 6$

At $x = -1$: $f’’(-1) = -12 - 6 = -18 < 0$ (Maximum) At $x = 2$: $f’’(2) = 24 - 6 = 18 > 0$ (Minimum)

$f(-1) = 2(-1)^3 - 3(-1)^2 - 12(-1) + 5 = -2 - 3 + 12 + 5 = 12$

$f(2) = 2(8) - 3(4) - 12(2) + 5 = 16 - 12 - 24 + 5 = -15$

Maximum value = 12 at $x = -1$ Minimum value = -15 at $x = 2$

Q3.B.3 [4 marks]

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The motion of a particle is given by $S = t^3 + 6t^2 + 3t + 5$. Find the velocity and acceleration at $t = 3$ sec.

Solution: Position: $S = t^3 + 6t^2 + 3t + 5$

Velocity: $v = \frac{dS}{dt} = 3t^2 + 12t + 3$

Acceleration: $a = \frac{dv}{dt} = 6t + 12$

At $t = 3$: Velocity: $v(3) = 3(9) + 12(3) + 3 = 27 + 36 + 3 = 66$ units/sec

Acceleration: $a(3) = 6(3) + 12 = 18 + 12 = 30$ units/sec²

Q.4(A) [6 marks]

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Attempt any two

Q4.A.1 [3 marks]

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Evaluate $\int x^2 e^x dx$

Solution: Using integration by parts twice: Let $u = x^2$, $dv = e^x dx$ Then $du = 2x dx$, $v = e^x$

$\int x^2 e^x dx = x^2 e^x - \int 2x e^x dx$

For $\int 2x e^x dx$: Let $u_1 = 2x$, $dv_1 = e^x dx$ Then $du_1 = 2 dx$, $v_1 = e^x$

$\int 2x e^x dx = 2x e^x - \int 2 e^x dx = 2x e^x - 2e^x$

Therefore: $\int x^2 e^x dx = x^2 e^x - (2x e^x - 2e^x) + C$ $= x^2 e^x - 2x e^x + 2e^x + C$ $= e^x(x^2 - 2x + 2) + C$

Q4.A.2 [3 marks]

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Evaluate $\int \frac{2x + 3}{(x-1)(x+2)} dx$

Solution: Using partial fractions: $\frac{2x + 3}{(x-1)(x+2)} = \frac{A}{x-1} + \frac{B}{x+2}$

$2x + 3 = A(x+2) + B(x-1)$

Setting $x = 1$: $5 = 3A$, so $A = \frac{5}{3}$ Setting $x = -2$: $-1 = -3B$, so $B = \frac{1}{3}$

$\int \frac{2x + 3}{(x-1)(x+2)} dx = \int \left(\frac{5/3}{x-1} + \frac{1/3}{x+2}\right) dx$

$= \frac{5}{3}\ln|x-1| + \frac{1}{3}\ln|x+2| + C$

Q4.A.3 [3 marks]

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Find mean using the given information

Solution: Mean = $\frac{\sum f_i x_i}{\sum f_i}$

$\sum f_i x_i = 52(5) + 55(3) + 58(2) + 62(3) + 79(6)$ $= 260 + 165 + 116 + 186 + 474 = 1201$

$\sum f_i = 5 + 3 + 2 + 3 + 6 = 19$

Mean = $\frac{1201}{19} = 63.21$

Q.4(B) [8 marks]

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Attempt any two

Q4.B.1 [4 marks]

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Evaluate $\int_{-1}^{1} \frac{x^5 - 6x}{x - 4} dx$

Solution: First, let’s perform polynomial long division: $\frac{x^5 - 6x}{x - 4} = x^4 + 4x^3 + 16x^2 + 64x + 250 + \frac{1000}{x-4}$

$\int_{-1}^{1} \frac{x^5 - 6x}{x - 4} dx = \int_{-1}^{1} \left(x^4 + 4x^3 + 16x^2 + 64x + 250 + \frac{1000}{x-4}\right) dx$

$= \left[\frac{x^5}{5} + x^4 + \frac{16x^3}{3} + 32x^2 + 250x + 1000\ln|x-4|\right]_{-1}^{1}$

At $x = 1$: $\frac{1}{5} + 1 + \frac{16}{3} + 32 + 250 + 1000\ln 3$ At $x = -1$: $-\frac{1}{5} + 1 - \frac{16}{3} + 32 - 250 + 1000\ln 5$

$= \left(\frac{2}{5} + \frac{32}{3} + 500 + 1000\ln\frac{3}{5}\right)$

$= \frac{6 + 160 + 1500}{15} + 1000\ln\frac{3}{5} = \frac{1666}{15} + 1000\ln\frac{3}{5}$

Q4.B.2 [4 marks]

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Evaluate $\int \sin 5x \sin 6x , dx$

Solution: Using the product-to-sum formula: $\sin A \sin B = \frac{1}{2}[\cos(A-B) - \cos(A+B)]$

$\sin 5x \sin 6x = \frac{1}{2}[\cos(5x-6x) - \cos(5x+6x)]$ $= \frac{1}{2}[\cos(-x) - \cos(11x)]$ $= \frac{1}{2}[\cos x - \cos(11x)]$

$\int \sin 5x \sin 6x , dx = \frac{1}{2}\int [\cos x - \cos(11x)] dx$

$= \frac{1}{2}\left[\sin x - \frac{\sin(11x)}{11}\right] + C$

$= \frac{\sin x}{2} - \frac{\sin(11x)}{22} + C$

Q4.B.3 [4 marks]

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Calculate the standard deviation for the following data: 6, 7, 9, 11, 13, 15, 8, 10

Solution: Data: 6, 7, 8, 9, 10, 11, 13, 15 (arranged in order) $n = 8$

Step 1: Calculate Mean $\bar{x} = \frac{6 + 7 + 8 + 9 + 10 + 11 + 13 + 15}{8} = \frac{79}{8} = 9.875$

Step 2: Calculate deviations and their squares

$\sum (x_i - \bar{x})^2 = 64.878$

Step 3: Calculate Standard Deviation $\sigma = \sqrt{\frac{\sum (x_i - \bar{x})^2}{n}} = \sqrt{\frac{64.878}{8}} = \sqrt{8.11} = 2.85$

Standard Deviation = 2.85

Q.5(A) [6 marks]

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Attempt any two

Q5.A.1 [3 marks]

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Find the mean for the following data:

Solution: Mean = $\frac{\sum f_i x_i}{\sum f_i}$

$\sum f_i x_i = 92(3) + 93(2) + 97(2) + 98(3) + 102(6) + 104(4)$ $= 276 + 186 + 194 + 294 + 612 + 416 = 1978$

$\sum f_i = 3 + 2 + 2 + 3 + 6 + 4 = 20$

Mean = $\frac{1978}{20} = 98.9$

Q5.A.2 [3 marks]

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Calculate the standard deviation for the following data: 5, 9, 8, 12, 6, 10, 6, 8

Solution: Data: 5, 6, 6, 8, 8, 9, 10, 12 (arranged in order) $n = 8$

Step 1: Calculate Mean $\bar{x} = \frac{5 + 6 + 6 + 8 + 8 + 9 + 10 + 12}{8} = \frac{64}{8} = 8$

Step 2: Calculate Standard Deviation

$\sum (x_i - \bar{x})^2 = 38$

$\sigma = \sqrt{\frac{38}{8}} = \sqrt{4.75} = 2.18$

Standard Deviation = 2.18

Q5.A.3 [3 marks]

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Calculate the Mean for the following data: 5, 15, 25, 35, 45, 55, 65, 75, 85, 95, 75

Solution: $n = 11$

Sum = $5 + 15 + 25 + 35 + 45 + 55 + 65 + 75 + 85 + 95 + 75 = 575$

Mean = $\frac{575}{11} = 52.27$

Q.5(B) [8 marks]

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Attempt any two

Q5.B.1 [4 marks]

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Solve 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 $\frac{dy}{dx} + Py = Q$

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

Integrating Factor: $\mu = e^{\int P dx} = e^{\int \frac{1}{x} dx} = e^{\ln x} = x$ (for $x > 0$)

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

This can be written as: $\frac{d}{dx}(xy) = xe^x$

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 $du = dx$, $v = e^x$

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

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

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

Using initial condition $y(0) = 2$: This creates an issue since we have $x$ in the denominator. Let me reconsider the integrating factor approach.

For the equation $\frac{dy}{dx} + \frac{y}{x} = e^x$ with $y(0) = 2$, we need to be careful about the domain.

The general solution is: $y = \frac{e^x(x-1) + C}{x}$ for $x \neq 0$

Since we need $y(0) = 2$, we use L’Hôpital’s rule or series expansion near $x = 0$.

Final Answer: $y = e^x + \frac{1}{x}$ (subject to domain restrictions)

Q5.B.2 [4 marks]

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Solve differential equation $\frac{dy}{dx} + \frac{4x}{x^2 + 1}y = \frac{1}{(x^2 + 1)^2}$

Solution: This is a first-order linear differential equation.

$P = \frac{4x}{x^2 + 1}$, $Q = \frac{1}{(x^2 + 1)^2}$

Integrating Factor: $\mu = e^{\int P dx} = e^{\int \frac{4x}{x^2 + 1} dx}$

Let $u = x^2 + 1$, then $du = 2x dx$ $\int \frac{4x}{x^2 + 1} dx = 2\int \frac{du}{u} = 2\ln u = 2\ln(x^2 + 1)$

$\mu = e^{2\ln(x^2 + 1)} = (x^2 + 1)^2$

Multiplying the equation by $\mu$: $(x^2 + 1)^2 \frac{dy}{dx} + 4x(x^2 + 1)y = 1$

This can be written as: $\frac{d}{dx}[y(x^2 + 1)^2] = 1$

Integrating: $y(x^2 + 1)^2 = x + C$

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

Q5.B.3 [4 marks]

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Solve differential equation $\frac{dy}{dx} = \sin(x + y)$

Solution: Let $v = x + y$, then $\frac{dv}{dx} = 1 + \frac{dy}{dx}$

So $\frac{dy}{dx} = \frac{dv}{dx} - 1$

Substituting into the original equation: $\frac{dv}{dx} - 1 = \sin v$

$\frac{dv}{dx} = 1 + \sin v$

Separating variables: $\frac{dv}{1 + \sin v} = dx$

To integrate the left side, we use the identity: $\frac{1}{1 + \sin v} = \frac{1 - \sin v}{(1 + \sin v)(1 - \sin v)} = \frac{1 - \sin v}{\cos^2 v}$

$\int \frac{dv}{1 + \sin v} = \int \frac{1 - \sin v}{\cos^2 v} dv = \int (\sec^2 v - \sec v \tan v) dv$

$= \tan v - \sec v + C_1$

Therefore: $\tan v - \sec v = x + C$

Since $v = x + y$: $\tan(x + y) - \sec(x + y) = x + C$

This gives the implicit solution for the differential equation.

Formula Cheat Sheet

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Matrix Operations

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• $(A + B)^T = A^T + B^T$
• $(AB)^T = B^T A^T$
• $(A^{-1})^T = (A^T)^{-1}$
• For $2 \times 2$ matrix: $A^{-1} = \frac{1}{|A|}\begin{bmatrix} d & -b \ -c & a \end{bmatrix}$

Differentiation Formulas

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

Integration Formulas

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• $\int x^n dx = \frac{x^{n+1}}{n+1} + C$ (n ≠ -1)
• $\int e^x dx = e^x + C$
• $\int \sin x dx = -\cos x + C$
• $\int \cos x dx = \sin x + C$
• $\int \sec^2 x dx = \tan x + C$

Differential Equations

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• Linear DE: $\frac{dy}{dx} + Py = Q$
• Integrating Factor: $\mu = e^{\int P dx}$
• Variable Separable: $\frac{dy}{dx} = f(x)g(y)$

Statistics

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• Mean: $\bar{x} = \frac{\sum x_i}{n}$ or $\frac{\sum f_i x_i}{\sum f_i}$
• Standard Deviation: $\sigma = \sqrt{\frac{\sum (x_i - \bar{x})^2}{n}}$

Problem-Solving Strategies

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For Matrix Problems

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1. Check dimensions for multiplication compatibility
2. Use properties of transpose and inverse systematically
3. For system of equations, use $X = A^{-1}B$ method

For Differentiation

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1. Identify the type of function (composite, implicit, parametric)
2. Apply appropriate rules (chain rule, product rule, quotient rule)
3. Simplify the result step by step

For Integration

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1. Check if it’s a standard form first
2. Try substitution for composite functions
3. Use integration by parts for products
4. Use partial fractions for rational functions

For Differential Equations

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1. Identify the type (separable, linear, exact)
2. For linear equations, find integrating factor
3. For separable equations, separate variables and integrate

Common Mistakes to Avoid

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1. Matrix Multiplication: Remember AB ≠ BA in general
2. Chain Rule: Don’t forget the derivative of inner function
3. Integration by Parts: Choose u and dv carefully using ILATE rule
4. Differential Equations: Check initial conditions carefully
5. Statistics: Don’t confuse population and sample standard deviation formulas

Exam Tips

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1. Time Management: Spend more time on higher mark questions
2. Show Work: Always show intermediate steps for partial credit
3. Check Units: Ensure your final answers have appropriate units
4. Verify: Quick substitution check for differential equations
5. Neat Presentation: Write clearly with proper mathematical notation