Q.1 [14 marks]

Fill in the blanks using appropriate choice from the given options

Q1.1 [1 mark]

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]

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]

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]

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]

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

Answer: b. $2x + 2$

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

Q1.6 [1 mark]

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

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

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

Q1.7 [1 mark]

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]

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

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]

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

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

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

Q1.10 [1 mark]

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

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]

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]

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]

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]

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]

Attempt any two

Q2(A).1 [3 marks]

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]

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]

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]

Attempt any two

Q2(B).1 [4 marks]

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]

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]

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]

Attempt any two

Q3(A).1 [3 marks]

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]

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]

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]

Attempt any two

Q3(B).1 [4 marks]

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]

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]

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]

Attempt any two

Q4(A).1 [3 marks]

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]

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]

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

$x_i$	6	10	14	18	24	28	30
$f_i$	2	4	7	f	8	4	3

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]

Attempt any two

Q4(B).1 [4 marks]

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]

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]

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:

$x_i$	$x_i - \bar{x}$	$(x_i - \bar{x})^2$
10	-8.5	72.25
15	-3.5	12.25
7	-11.5	132.25
19	0.5	0.25
9	-9.5	90.25
21	2.5	6.25
23	4.5	20.25
25	6.5	42.25
26	7.5	56.25
30	11.5	132.25

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

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]

Attempt any two

Q5(A).1 [3 marks]

Find the standard deviation for the following data:

$x_i$	4	8	11	17	20	24	32
$f_i$	3	5	9	5	4	3	1

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:

$x_i$	$f_i$	$x_i - \bar{x}$	$(x_i - \bar{x})^2$	$f_i(x_i - \bar{x})^2$
4	3	-10	100	300
8	5	-6	36	180
11	9	-3	9	81
17	5	3	9	45
20	4	6	36	144
24	3	10	100	300
32	1	18	324	324

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

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]

Find the standard deviation for the following data:

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

Solution: First, find class midpoints and calculate mean:

Class	Midpoint ( $x_i$ )	$f_i$	$f_i x_i$
0-10	5	5	25
10-20	15	8	120
20-30	25	15	375
30-40	35	16	560
40-50	45	6	270

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

Standard Deviation Table:

$x_i$	$f_i$	$x_i - \bar{x}$	$(x_i - \bar{x})^2$	$f_i(x_i - \bar{x})^2$
5	5	-22	484	2420
15	8	-12	144	1152
25	15	-2	4	60
35	16	8	64	1024
45	6	18	324	1944

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

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

Q5(A).3 [3 marks]

Find the mean for the following data:

Class	30-40	40-50	50-60	60-70	70-80	80-90	90-100
Frequency	3	7	12	15	8	3	2

Solution: Using midpoint method:

Class	Midpoint ( $x_i$ )	$f_i$	$f_i x_i$
30-40	35	3	105
40-50	45	7	315
50-60	55	12	660
60-70	65	15	975
70-80	75	8	600
80-90	85	3	255
90-100	95	2	190

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

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

Q.5(B) [8 marks]

Attempt any two

Q5(B).1 [4 marks]

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]

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]

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

Matrix Operations

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

Derivatives

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

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

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

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

For Matrix Problems

Check dimensions for multiplication compatibility
Use cofactor method for finding inverse
Apply transpose properties systematically

For Differentiation

Identify the type of function (composite, implicit, parametric)
Apply appropriate rules (chain rule, product rule, quotient rule)
Simplify the final expression

For Integration

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

For Differential Equations

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

Common Mistakes to Avoid

Matrix Operations

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

Differentiation

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

Integration

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

Statistics

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

Exam Tips

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

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