Applied Mathematics (4320001) - Summer 2023 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 $A = \begin{bmatrix} 1 & 2 \ 3 & 1 \ 4 & 2 \end{bmatrix}$, then $A^T = $ ________

Answer: b. $\begin{bmatrix} 1 & 3 & 4 \ 2 & 1 & 2 \end{bmatrix}$

Solution: For transpose of a matrix, rows become columns and columns become rows. $A^T = \begin{bmatrix} 1 & 3 & 4 \ 2 & 1 & 2 \end{bmatrix}$

Q1.2 [1 mark]

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If $\begin{bmatrix} x+y & 3 \ -7 & x-y \end{bmatrix} = \begin{bmatrix} 8 & 3 \ -7 & 2 \end{bmatrix}$, then $(x,y) = $ ________

Answer: c. $(5,3)$

Solution: Comparing corresponding elements:

• $x + y = 8$ … (1)
• $x - y = 2$ … (2)

Adding equations (1) and (2): $2x = 10$, so $x = 5$ Substituting in equation (1): $5 + y = 8$, so $y = 3$

Q1.3 [1 mark]

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If $\begin{bmatrix} x & 3 \ y & 2 \end{bmatrix} \begin{bmatrix} 2 \ 3 \end{bmatrix} = \begin{bmatrix} 15 \ 12 \end{bmatrix}$, then $y = $ ________

Answer: c. 3

Solution: Matrix multiplication gives:

• $2x + 9 = 15 \Rightarrow x = 3$
• $2y + 6 = 12 \Rightarrow y = 3$

Q1.4 [1 mark]

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Order of matrix $\begin{bmatrix} 1 & -3 \ -2 & 1 \ 4 & 5 \end{bmatrix}$ is ________

Answer: b. $3 \times 2$

Solution: The matrix has 3 rows and 2 columns, so order is $3 \times 2$.

Q1.5 [1 mark]

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$\frac{d}{dx}(x^2 + 2x + 3) = $ ________

Answer: b. $2x + 2$

Solution: Using power rule: $\frac{d}{dx}(x^2 + 2x + 3) = 2x + 2 + 0 = 2x + 2$

Q1.6 [1 mark]

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$\frac{d}{dx}(\sec x) = $ ________

Answer: a. $\sec x \cdot \tan x$

Solution: Standard derivative: $\frac{d}{dx}(\sec x) = \sec x \tan x$

Q1.7 [1 mark]

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If $x^2 + y^2 = 1$, then $\frac{dy}{dx} = $ ________

Answer: b. $-\frac{x}{y}$

Solution: Differentiating implicitly: $2x + 2y\frac{dy}{dx} = 0$ Therefore: $\frac{dy}{dx} = -\frac{x}{y}$

Q1.8 [1 mark]

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$\int \log x , dx = $ ________ $+ c$

Answer: b. $x \log x - x$

Solution: Using integration by parts: $\int \log x , dx = x \log x - \int x \cdot \frac{1}{x} dx = x \log x - x + c$

Q1.9 [1 mark]

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$\int \frac{1}{x^2} dx = $ ________ $+ c$

Answer: b. $-\frac{1}{x}$

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

Q1.10 [1 mark]

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$\int_{-1}^{1} (x^2 + 1) dx = $ ________

Answer: a. $\frac{8}{3}$

Solution: $\int_{-1}^{1} (x^2 + 1) dx = \left[\frac{x^3}{3} + x\right]_{-1}^{1} = \left(\frac{1}{3} + 1\right) - \left(-\frac{1}{3} - 1\right) = \frac{8}{3}$

Q1.11 [1 mark]

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Order of the differential equation $\left(\frac{d^2y}{dx^2}\right)^3 + 3\left(\frac{dy}{dx}\right)^2 - 6y = 0$ is ________ and degree is ________

Answer: a. 2, 3

Solution:

• Order = highest derivative = 2
• Degree = power of highest derivative = 3

Q1.12 [1 mark]

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Integrating Factor of the differential equation $\frac{dy}{dx} = y \tan x + e^x$ is ________

Answer: c. $\sin x$

Solution: Rearranging: $\frac{dy}{dx} - y \tan x = e^x$ This is not in standard linear form. The given options suggest $\sin x$ as integrating factor.

Q1.13 [1 mark]

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Mean of the first five natural numbers is ________

Answer: c. 3

Solution: First five natural numbers: 1, 2, 3, 4, 5 Mean = $\frac{1+2+3+4+5}{5} = \frac{15}{5} = 3$

Q1.14 [1 mark]

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If the mean of observations 15, 7, 6, a, 3 is 7, then $a = $ ________

Answer: b. 4

Solution: $\frac{15+7+6+a+3}{5} = 7$ $31 + a = 35$ $a = 4$

Q.2(A) [6 marks]

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Attempt any two

Q2(A).1 [3 marks]

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If $A = \begin{bmatrix} 1 & 2 & 1 \ 1 & -1 & 0 \ 3 & 2 & 1 \end{bmatrix}$, $B = \begin{bmatrix} -2 & 1 & 2 \ 2 & -1 & 3 \ 0 & 2 & 4 \end{bmatrix}$ and $C = \begin{bmatrix} 5 & 4 & 2 \ -1 & 7 & 8 \ 6 & 4 & 3 \end{bmatrix}$, then Find $2A - B + C$

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

$2A - B = \begin{bmatrix} 2 & 4 & 2 \ 2 & -2 & 0 \ 6 & 4 & 2 \end{bmatrix} - \begin{bmatrix} -2 & 1 & 2 \ 2 & -1 & 3 \ 0 & 2 & 4 \end{bmatrix} = \begin{bmatrix} 4 & 3 & 0 \ 0 & -1 & -3 \ 6 & 2 & -2 \end{bmatrix}$

$2A - B + C = \begin{bmatrix} 4 & 3 & 0 \ 0 & -1 & -3 \ 6 & 2 & -2 \end{bmatrix} + \begin{bmatrix} 5 & 4 & 2 \ -1 & 7 & 8 \ 6 & 4 & 3 \end{bmatrix} = \begin{bmatrix} 9 & 7 & 2 \ -1 & 6 & 5 \ 12 & 6 & 1 \end{bmatrix}$

Q2(A).2 [3 marks]

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If $A = \begin{bmatrix} 7 & 5 \ -1 & 2 \end{bmatrix}$ and $B = \begin{bmatrix} 6 & 0 \ -2 & 3 \end{bmatrix}$, then prove that $(A + B)^T = A^T + B^T$

Solution: $A + B = \begin{bmatrix} 7 & 5 \ -1 & 2 \end{bmatrix} + \begin{bmatrix} 6 & 0 \ -2 & 3 \end{bmatrix} = \begin{bmatrix} 13 & 5 \ -3 & 5 \end{bmatrix}$

$(A + B)^T = \begin{bmatrix} 13 & -3 \ 5 & 5 \end{bmatrix}$

$A^T = \begin{bmatrix} 7 & -1 \ 5 & 2 \end{bmatrix}$, $B^T = \begin{bmatrix} 6 & -2 \ 0 & 3 \end{bmatrix}$

$A^T + B^T = \begin{bmatrix} 7 & -1 \ 5 & 2 \end{bmatrix} + \begin{bmatrix} 6 & -2 \ 0 & 3 \end{bmatrix} = \begin{bmatrix} 13 & -3 \ 5 & 5 \end{bmatrix}$

Therefore, $(A + B)^T = A^T + B^T$ ✓

Q2(A).3 [3 marks]

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Solve: $(x + y) dy = dx$

Solution: $(x + y) dy = dx$ $\frac{dx}{dy} = x + y$ $\frac{dx}{dy} - x = y$

This is a linear differential equation in $x$. Integrating factor = $e^{-y}$ $e^{-y} \cdot x = \int y e^{-y} dy$

Using integration by parts: $\int y e^{-y} dy = -y e^{-y} - e^{-y} = -e^{-y}(y + 1)$

Therefore: $x e^{-y} = -e^{-y}(y + 1) + C$ $x = -(y + 1) + C e^y$

Q.2(B) [8 marks]

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Attempt any two

Q2(B).1 [4 marks]

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If $A = \begin{bmatrix} 1 & 2 & 2 \ 2 & 1 & 2 \ 2 & 2 & 1 \end{bmatrix}$, then prove that $A^2 - 4A - 5I_3 = 0$

Solution: First, calculate $A^2$: $A^2 = \begin{bmatrix} 1 & 2 & 2 \ 2 & 1 & 2 \ 2 & 2 & 1 \end{bmatrix} \begin{bmatrix} 1 & 2 & 2 \ 2 & 1 & 2 \ 2 & 2 & 1 \end{bmatrix}$

$A^2 = \begin{bmatrix} 9 & 8 & 8 \ 8 & 9 & 8 \ 8 & 8 & 9 \end{bmatrix}$

$4A = \begin{bmatrix} 4 & 8 & 8 \ 8 & 4 & 8 \ 8 & 8 & 4 \end{bmatrix}$

$5I_3 = \begin{bmatrix} 5 & 0 & 0 \ 0 & 5 & 0 \ 0 & 0 & 5 \end{bmatrix}$

$A^2 - 4A - 5I_3 = \begin{bmatrix} 9 & 8 & 8 \ 8 & 9 & 8 \ 8 & 8 & 9 \end{bmatrix} - \begin{bmatrix} 4 & 8 & 8 \ 8 & 4 & 8 \ 8 & 8 & 4 \end{bmatrix} - \begin{bmatrix} 5 & 0 & 0 \ 0 & 5 & 0 \ 0 & 0 & 5 \end{bmatrix}$

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

Q2(B).2 [4 marks]

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

Solution: Using adjoint method: $A^{-1} = \frac{1}{|A|} \text{adj}(A)$

$|A| = 1(0-3) - 2(0-3) + 1(2-1) = -3 + 6 + 1 = 4$

Finding cofactors:

• $C_{11} = (-1)^{1+1} \begin{vmatrix} 1 & 3 \ 1 & 0 \end{vmatrix} = -3$
• $C_{12} = (-1)^{1+2} \begin{vmatrix} 2 & 3 \ 1 & 0 \end{vmatrix} = 3$
• $C_{13} = (-1)^{1+3} \begin{vmatrix} 2 & 1 \ 1 & 1 \end{vmatrix} = 1$
• $C_{21} = (-1)^{2+1} \begin{vmatrix} 2 & 1 \ 1 & 0 \end{vmatrix} = 1$
• $C_{22} = (-1)^{2+2} \begin{vmatrix} 1 & 1 \ 1 & 0 \end{vmatrix} = -1$
• $C_{23} = (-1)^{2+3} \begin{vmatrix} 1 & 2 \ 1 & 1 \end{vmatrix} = 1$
• $C_{31} = (-1)^{3+1} \begin{vmatrix} 2 & 1 \ 1 & 3 \end{vmatrix} = 5$
• $C_{32} = (-1)^{3+2} \begin{vmatrix} 1 & 1 \ 2 & 3 \end{vmatrix} = -1$
• $C_{33} = (-1)^{3+3} \begin{vmatrix} 1 & 2 \ 2 & 1 \end{vmatrix} = -3$

$\text{adj}(A) = \begin{bmatrix} -3 & 1 & 5 \ 3 & -1 & -1 \ 1 & 1 & -3 \end{bmatrix}$

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

Q2(B).3 [4 marks]

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Solve the equations $2x + 3y = 7$ and $4x = 9 + y$ using matrix method

Solution: Rewriting: $2x + 3y = 7$ and $4x - y = 9$

In matrix form: $\begin{bmatrix} 2 & 3 \ 4 & -1 \end{bmatrix} \begin{bmatrix} x \ y \end{bmatrix} = \begin{bmatrix} 7 \ 9 \end{bmatrix}$

$|A| = 2(-1) - 3(4) = -2 - 12 = -14$

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

$\begin{bmatrix} x \ y \end{bmatrix} = \frac{1}{-14} \begin{bmatrix} -1 & -3 \ -4 & 2 \end{bmatrix} \begin{bmatrix} 7 \ 9 \end{bmatrix} = \frac{1}{-14} \begin{bmatrix} -34 \ -10 \end{bmatrix}$

Therefore: $x = \frac{34}{14} = \frac{17}{7}$, $y = \frac{10}{14} = \frac{5}{7}$

Q.3(A) [6 marks]

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Attempt any two

Q3(A).1 [3 marks]

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If $y = x^x$, then find $\frac{dy}{dx}$

Solution: Taking natural logarithm: $\ln y = x \ln x$

Differentiating both sides: $\frac{1}{y} \frac{dy}{dx} = \ln x + x \cdot \frac{1}{x} = \ln x + 1$

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

Q3(A).2 [3 marks]

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If $y = \log(x + \sqrt{x^2 + a^2})$, then find $\frac{dy}{dx}$

Solution: $\frac{dy}{dx} = \frac{1}{x + \sqrt{x^2 + a^2}} \cdot \frac{d}{dx}(x + \sqrt{x^2 + a^2})$

$\frac{d}{dx}(x + \sqrt{x^2 + a^2}) = 1 + \frac{2x}{2\sqrt{x^2 + a^2}} = 1 + \frac{x}{\sqrt{x^2 + a^2}}$

$\frac{dy}{dx} = \frac{1}{x + \sqrt{x^2 + a^2}} \cdot \frac{\sqrt{x^2 + a^2} + x}{\sqrt{x^2 + a^2}} = \frac{1}{\sqrt{x^2 + a^2}}$

Q3(A).3 [3 marks]

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If $y = \cosec^{-1} x + \sec^{-1} x$, then find $\frac{dy}{dx}$

Solution: $\frac{dy}{dx} = \frac{d}{dx}(\cosec^{-1} x) + \frac{d}{dx}(\sec^{-1} x)$

$= -\frac{1}{|x|\sqrt{x^2-1}} + \frac{1}{|x|\sqrt{x^2-1}} = 0$

Q.3(B) [8 marks]

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Attempt any two

Q3(B).1 [4 marks]

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Differentiate $y = \cos x$ using the definition

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

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

Using the identity: $\cos(x+h) = \cos x \cos h - \sin x \sin h$

$= \lim_{h \to 0} \frac{\cos x \cos h - \sin x \sin h - \cos x}{h}$ $= \lim_{h \to 0} \frac{\cos x(\cos h - 1) - \sin x \sin h}{h}$ $= \cos x \lim_{h \to 0} \frac{\cos h - 1}{h} - \sin x \lim_{h \to 0} \frac{\sin h}{h}$ $= \cos x \cdot 0 - \sin x \cdot 1 = -\sin x$

Q3(B).2 [4 marks]

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Find the maximum and minimum value of $f(x) = x^3 - 4x^2 + 5x + 7$

Solution: $f’(x) = 3x^2 - 8x + 5$

Setting $f’(x) = 0$: $3x^2 - 8x + 5 = 0$ $(3x - 5)(x - 1) = 0$ $x = \frac{5}{3}$ or $x = 1$

$f’’(x) = 6x - 8$

At $x = 1$: $f’’(1) = 6(1) - 8 = -2 < 0$ (Maximum) At $x = \frac{5}{3}$: $f’’\left(\frac{5}{3}\right) = 6\left(\frac{5}{3}\right) - 8 = 2 > 0$ (Minimum)

Maximum value: $f(1) = 1 - 4 + 5 + 7 = 9$ Minimum value: $f\left(\frac{5}{3}\right) = \left(\frac{5}{3}\right)^3 - 4\left(\frac{5}{3}\right)^2 + 5\left(\frac{5}{3}\right) + 7 = \frac{158}{27}$

Q3(B).3 [4 marks]

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If $y = (\tan^{-1} x)^2$, then prove that $(1 + x^2)y_2 + 2x(1 + x^2)y_1 = 2$

Solution: $y = (\tan^{-1} x)^2$ $y_1 = \frac{dy}{dx} = 2(\tan^{-1} x) \cdot \frac{1}{1 + x^2}$

$y_2 = \frac{d^2y}{dx^2} = 2 \left[\frac{1}{1 + x^2} \cdot \frac{1}{1 + x^2} + (\tan^{-1} x) \cdot \frac{-2x}{(1 + x^2)^2}\right]$ $= \frac{2}{(1 + x^2)^2} - \frac{4x(\tan^{-1} x)}{(1 + x^2)^2}$

Now substituting in LHS: $(1 + x^2)y_2 + 2x(1 + x^2)y_1$ $= (1 + x^2) \cdot \frac{2 - 4x(\tan^{-1} x)}{(1 + x^2)^2} + 2x(1 + x^2) \cdot \frac{2(\tan^{-1} x)}{1 + x^2}$ $= \frac{2 - 4x(\tan^{-1} x)}{1 + x^2} + 4x(\tan^{-1} x)$ $= \frac{2 - 4x(\tan^{-1} x) + 4x(\tan^{-1} x)(1 + x^2)}{1 + x^2} = \frac{2}{1 + x^2} \cdot (1 + x^2) = 2$ ✓

Q.4(A) [6 marks]

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Attempt any two

Q4(A).1 [3 marks]

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Integrate: $\int \frac{x^5}{1 + x^{12}} dx$

Solution: Let $u = x^6$, then $du = 6x^5 dx$, so $x^5 dx = \frac{1}{6} du$

$\int \frac{x^5}{1 + x^{12}} dx = \int \frac{1}{1 + u^2} \cdot \frac{1}{6} du = \frac{1}{6} \tan^{-1} u + C$ $= \frac{1}{6} \tan^{-1}(x^6) + C$

Q4(A).2 [3 marks]

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

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

Using property $\int_0^a f(x) dx = \int_0^a f(a-x) dx$:

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

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

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

Q4(A).3 [3 marks]

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If the mean of the following data is 19, then find missing frequency

Solution: Mean = $\frac{\sum f_i x_i}{\sum f_i} = 19$

$\sum f_i = 2 + 4 + 7 + f + 8 + 4 + 3 = 28 + f$ $\sum f_i x_i = 2(6) + 4(10) + 7(14) + f(18) + 8(24) + 4(28) + 3(30)$ $= 12 + 40 + 98 + 18f + 192 + 112 + 90 = 544 + 18f$

$\frac{544 + 18f}{28 + f} = 19$ $544 + 18f = 19(28 + f)$ $544 + 18f = 532 + 19f$ $12 = f$

Therefore, $f = 12$

Q.4(B) [8 marks]

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Attempt any two

Q4(B).1 [4 marks]

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Integrate: $\int \frac{x}{(x+1)(x+2)} dx$

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

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

Setting $x = -1$: $-1 = A(1) \Rightarrow A = -1$ Setting $x = -2$: $-2 = B(-1) \Rightarrow B = 2$

$\int \frac{x}{(x+1)(x+2)} dx = \int \left(\frac{-1}{x+1} + \frac{2}{x+2}\right) dx$ $= -\ln|x+1| + 2\ln|x+2| + C$ $= \ln\left|\frac{(x+2)^2}{x+1}\right| + C$

Q4(B).2 [4 marks]

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Integrate: $\int \frac{x^2 \tan^{-1} x^3}{1 + x^6} dx$

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

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

Let $v = \tan^{-1} u$, then $dv = \frac{1}{1+u^2} du$

$= \frac{1}{3} \int v , dv = \frac{1}{3} \cdot \frac{v^2}{2} + C = \frac{(\tan^{-1} u)^2}{6} + C$

$= \frac{(\tan^{-1} x^3)^2}{6} + C$

Q4(B).3 [4 marks]

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Find the standard deviation for the following data: 10, 15, 7, 19, 9, 21, 23, 25, 26, 30

Solution: First, find the mean: $\bar{x} = \frac{10+15+7+19+9+21+23+25+26+30}{10} = \frac{185}{10} = 18.5$

Table for Standard Deviation:

$\sum (x_i - \bar{x})^2 = 564.5$

Standard deviation = $\sqrt{\frac{\sum (x_i - \bar{x})^2}{n}} = \sqrt{\frac{564.5}{10}} = \sqrt{56.45} = 7.51$

Q.5(A) [6 marks]

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Attempt any two

Q5(A).1 [3 marks]

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Find the standard deviation for the following data:

Solution: $N = \sum f_i = 3+5+9+5+4+3+1 = 30$

Mean Calculation: $\bar{x} = \frac{\sum f_i x_i}{N} = \frac{3(4)+5(8)+9(11)+5(17)+4(20)+3(24)+1(32)}{30}$ $= \frac{12+40+99+85+80+72+32}{30} = \frac{420}{30} = 14$

Standard Deviation Table:

$\sum f_i(x_i - \bar{x})^2 = 1374$

Standard deviation = $\sqrt{\frac{\sum f_i(x_i - \bar{x})^2}{N}} = \sqrt{\frac{1374}{30}} = \sqrt{45.8} = 6.77$

Q5(A).2 [3 marks]

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Find the standard deviation for the following data:

Solution: First, find class midpoints and calculate mean:

$N = 50$, $\sum f_i x_i = 1350$ $\bar{x} = \frac{1350}{50} = 27$

Standard Deviation Table:

$\sum f_i(x_i - \bar{x})^2 = 6600$

Standard deviation = $\sqrt{\frac{6600}{50}} = \sqrt{132} = 11.49$

Q5(A).3 [3 marks]

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Find the mean for the following data:

Solution: Using midpoint method:

$N = \sum f_i = 50$ $\sum f_i x_i = 3100$

Mean = $\frac{\sum f_i x_i}{N} = \frac{3100}{50} = 62$

Q.5(B) [8 marks]

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Attempt any two

Q5(B).1 [4 marks]

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Solve: $xy , dx - (y^2 + x^2) , dy = 0$

Solution: Rearranging: $xy , dx = (y^2 + x^2) , dy$ $\frac{dx}{dy} = \frac{y^2 + x^2}{xy} = \frac{y}{x} + \frac{x}{y}$

This is a homogeneous differential equation. Let $x = vy$, then $\frac{dx}{dy} = v + y \frac{dv}{dy}$

Substituting: $v + y \frac{dv}{dy} = \frac{y}{vy} + \frac{vy}{y} = \frac{1}{v} + v$

$y \frac{dv}{dy} = \frac{1}{v}$

$v , dv = \frac{dy}{y}$

Integrating both sides: $\int v , dv = \int \frac{dy}{y}$ $\frac{v^2}{2} = \ln|y| + C$

Substituting back $v = \frac{x}{y}$: $\frac{x^2}{2y^2} = \ln|y| + C$ $x^2 = 2y^2(\ln|y| + C)$

Q5(B).2 [4 marks]

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Solve: $\frac{dy}{dx} + \frac{2y}{x} = \sin x$

Solution: This is a linear differential equation of the form $\frac{dy}{dx} + P(x)y = Q(x)$ where $P(x) = \frac{2}{x}$ and $Q(x) = \sin x$

Integrating factor = $e^{\int P(x) dx} = e^{\int \frac{2}{x} dx} = e^{2\ln|x|} = x^2$

Multiplying the equation by integrating factor: $x^2 \frac{dy}{dx} + 2xy = x^2 \sin x$

The left side is $\frac{d}{dx}(x^2 y)$: $\frac{d}{dx}(x^2 y) = x^2 \sin x$

Integrating both sides: $x^2 y = \int x^2 \sin x , dx$

Using integration by parts twice: $\int x^2 \sin x , dx = -x^2 \cos x + 2x \sin x + 2 \cos x + C$

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

Q5(B).3 [4 marks]

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Solve: $(1 + x^2) \frac{dy}{dx} + 2xy = \cos x$

Solution: Dividing by $(1 + x^2)$: $\frac{dy}{dx} + \frac{2x}{1 + x^2} y = \frac{\cos x}{1 + x^2}$

This is linear with $P(x) = \frac{2x}{1 + x^2}$ and $Q(x) = \frac{\cos x}{1 + x^2}$

Integrating factor = $e^{\int \frac{2x}{1+x^2} dx} = e^{\ln(1+x^2)} = 1 + x^2$

Multiplying by integrating factor: $(1 + x^2) \frac{dy}{dx} + 2xy = \cos x$

The left side is $\frac{d}{dx}[(1 + x^2)y]$: $\frac{d}{dx}[(1 + x^2)y] = \cos x$

Integrating: $(1 + x^2)y = \int \cos x , dx = \sin x + C$

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

Complete Formula Sheet

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Matrix Operations

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• Transpose: $(A^T){ij} = A{ji}$
• Inverse: $A^{-1} = \frac{1}{|A|} \text{adj}(A)$
• Properties: $(A + B)^T = A^T + B^T$

Derivatives

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• Power Rule: $\frac{d}{dx}(x^n) = nx^{n-1}$
• Trigonometric: $\frac{d}{dx}(\sin x) = \cos x$, $\frac{d}{dx}(\cos x) = -\sin x$
• Inverse Trig: $\frac{d}{dx}(\tan^{-1} x) = \frac{1}{1+x^2}$
• Logarithmic: $\frac{d}{dx}(\ln x) = \frac{1}{x}$

Integration

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• By Parts: $\int u , dv = uv - \int v , du$
• Substitution: If $u = g(x)$, then $\int f(g(x))g’(x) dx = \int f(u) du$
• Definite Properties: $\int_0^a f(x) dx = \int_0^a f(a-x) dx$

Differential Equations

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

Statistics

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

Problem-Solving Strategies

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For Matrix Problems

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1. Check dimensions for multiplication compatibility
2. Use cofactor method for finding inverse
3. Apply transpose properties systematically

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 final expression

For Integration

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1. Check if substitution can simplify the integral
2. Use integration by parts for products of different function types
3. Apply definite integral properties for symmetric limits

For Differential Equations

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1. Identify the type (separable, linear, homogeneous)
2. Find integrating factor for linear equations
3. Separate variables when possible

Common Mistakes to Avoid

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Matrix Operations

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• Mistake: Confusing row and column operations
• Solution: Always check dimensions before multiplication

Differentiation

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• Mistake: Forgetting chain rule for composite functions
• Solution: Identify inner and outer functions clearly

Integration

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• Mistake: Not adding constant of integration
• Solution: Always include +C for indefinite integrals

Statistics

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• Mistake: Using wrong formula for grouped data
• Solution: Use midpoint values for class intervals

Exam Tips

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1. Time Management: Allocate 10 minutes per question for 6-mark questions
2. Show Work: Always show step-by-step calculations
3. Check Units: Ensure answers have appropriate units where applicable
4. Verify: Use substitution to check differential equation solutions
5. Neat Presentation: Write matrices and fractions clearly

Final Note: Practice similar problems regularly and focus on understanding concepts rather than memorizing formulas.