Q.1 [14 marks]

Fill in the blanks using appropriate choice from the given options

Q1.1 [1 mark]

Order of the matrix $\begin{bmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \end{bmatrix}$ = ………

Answer: (a) 2 × 3

Solution: Matrix has 2 rows and 3 columns, so order is 2 × 3.

Q1.2 [1 mark]

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

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

Solution: Transpose means rows become columns: $A^T = \begin{bmatrix} 1 & 3 \\ 2 & 4 \end{bmatrix}$

Q1.3 [1 mark]

If $A = \begin{bmatrix} 1 & -1 \\ 2 & 3 \end{bmatrix}$ then $adj(A)$ =…………..

Answer: (d) $\begin{bmatrix} 3 & 1 \\ -2 & 1 \end{bmatrix}$

Solution: For $2×2$ matrix $\begin{bmatrix} a & b \\ c & d \end{bmatrix}$ , $adj = \begin{bmatrix} d & -b \\ -c & a \end{bmatrix}$

Q1.4 [1 mark]

$[1 \; 2 \; 3] \begin{bmatrix} 4 \\ 5 \\ -1 \end{bmatrix}$ =……………….

Answer: (c) 11

Solution: $1×4 + 2×5 + 3×(-1) = 4 + 10 - 3 = 11$

Q1.5 [1 mark]

$\frac{d}{dx}(x^3 + 1)$ =……

Answer: (a) $3x^2$

Solution: $\frac{d}{dx}(x^3 + 1) = 3x^2 + 0 = 3x^2$

Q1.6 [1 mark]

$\frac{d}{dx}(\sec^2 x - \tan^2 x)$ =……

Answer: (b) 0

Solution: Since $\sec^2 x - \tan^2 x = 1$ (constant), derivative = 0

Q1.7 [1 mark]

$\frac{d}{dx}(\log x)$ =……

Answer: (c) $\frac{1}{x}$

Solution: Standard derivative: $\frac{d}{dx}(\log x) = \frac{1}{x}$

Q1.8 [1 mark]

$\int x^2 dx$ =……..+ C

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

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

Q1.9 [1 mark]

$\int_{-\pi/2}^{\pi/2} \sin x \, dx$ =……. + C

Answer: (d) $2$

Solution: $\int_{-\pi/2}^{\pi/2} \sin x \, dx = [-\cos x]_{-\pi/2}^{\pi/2} = -\cos(\pi/2) + \cos(-\pi/2) = 0 + 0 = 2$

Q1.10 [1 mark]

$\int_1^3 \frac{1}{x} dx$ =……….

Answer: (c) $\log 3$

Solution: $\int_1^3 \frac{1}{x} dx = [\log x]_1^3 = \log 3 - \log 1 = \log 3$

Q1.11 [1 mark]

Order and Degree of the differential equation $\left(\frac{d^2y}{dx^2}\right)^3 + \frac{dy}{dx} + 1 = 0$ are .............

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 $\frac{dy}{dx} + y = 1$ is

Answer: (b) $e^x$

Solution: For $\frac{dy}{dx} + Py = Q$ , I.F. = $e^{\int P dx} = e^{\int 1 dx} = e^x$

Q1.13 [1 mark]

Mean of 1,3,5,7,9 is

Answer: (a) 5

Solution: Mean = $\frac{1+3+5+7+9}{5} = \frac{25}{5} = 5$

Q1.14 [1 mark]

If the Mean of 15, 7, 6, a, 3 is 4 then a = ………….

Answer: (c) -11

Solution: $\frac{15+7+6+a+3}{5} = 4$ $31 + a = 20$ $a = -11$

Q.2 [14 marks]

Q.2(A) Attempt any two [6 marks]

Q2(A).1 [3 marks]

If $A = \begin{bmatrix} 3 & 2 \\ -1 & 4 \end{bmatrix}$ , then prove that $A^2 - 7A + 14I_2 = 0$ .

Answer:

Solution: First calculate $A^2$ : $A^2 = \begin{bmatrix} 3 & 2 \\ -1 & 4 \end{bmatrix} \begin{bmatrix} 3 & 2 \\ -1 & 4 \end{bmatrix} = \begin{bmatrix} 7 & 14 \\ -7 & 14 \end{bmatrix}$

Calculate $7A$ : $7A = 7\begin{bmatrix} 3 & 2 \\ -1 & 4 \end{bmatrix} = \begin{bmatrix} 21 & 14 \\ -7 & 28 \end{bmatrix}$

Calculate $14I_2$ : $14I_2 = 14\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 14 & 0 \\ 0 & 14 \end{bmatrix}$

Now: $A^2 - 7A + 14I_2 = \begin{bmatrix} 7 & 14 \\ -7 & 14 \end{bmatrix} - \begin{bmatrix} 21 & 14 \\ -7 & 28 \end{bmatrix} + \begin{bmatrix} 14 & 0 \\ 0 & 14 \end{bmatrix} = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix}$

Hence proved.

Q2(A).2 [3 marks]

Using matrix, solve the following system: $3x - y = 1$ , $2x + y = 4$ .

Answer:

Solution: System in matrix form: $\begin{bmatrix} 3 & -1 \\ 2 & 1 \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 1 \\ 4 \end{bmatrix}$

Find determinant: $|A| = 3(1) - (-1)(2) = 3 + 2 = 5$

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

Solution: $\begin{bmatrix} x \\ y \end{bmatrix} = A^{-1}B = \frac{1}{5}\begin{bmatrix} 1 & 1 \\ -2 & 3 \end{bmatrix}\begin{bmatrix} 1 \\ 4 \end{bmatrix} = \frac{1}{5}\begin{bmatrix} 5 \\ 10 \end{bmatrix} = \begin{bmatrix} 1 \\ 2 \end{bmatrix}$

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

Q2(A).3 [3 marks]

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

Answer:

Solution: Rewrite as: $\frac{dy}{dx} + \frac{2xy}{x^2+1} = \frac{e^x}{x^2+1}$

This is linear form with $P = \frac{2x}{x^2+1}$ , $Q = \frac{e^x}{x^2+1}$

I.F. = $e^{\int \frac{2x}{x^2+1}dx} = e^{\ln(x^2+1)} = x^2+1$

Solution: $y(x^2+1) = \int e^x dx = e^x + C$

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

Q.2(B) Attempt any two [8 marks]

Q2(B).1 [4 marks]

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

Answer:

Solution: Calculate determinant: $|A| = 1(-2-2) - 2(3-4) + 3(6+8) = -4 + 2 + 42 = 40$

Find cofactor matrix: $C_{11} = -4$ , $C_{12} = 1$ , $C_{13} = 14$ $C_{21} = 4$ , $C_{22} = -11$ , $C_{23} = 6$ $C_{31} = 8$ , $C_{32} = 8$ , $C_{33} = -8$

$adj(A) = \begin{bmatrix} -4 & 4 & 8 \\ 1 & -11 & 8 \\ 14 & 6 & -8 \end{bmatrix}$

$A^{-1} = \frac{1}{40}\begin{bmatrix} -4 & 4 & 8 \\ 1 & -11 & 8 \\ 14 & 6 & -8 \end{bmatrix}$

Q2(B).2 [4 marks]

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

Answer:

Solution: Calculate $AB = \begin{bmatrix} 1 & -3 \\ 2 & 4 \end{bmatrix}\begin{bmatrix} 3 & 2 \\ 1 & 5 \end{bmatrix} = \begin{bmatrix} 0 & -13 \\ 10 & 24 \end{bmatrix}$

$|AB| = 0(24) - (-13)(10) = 130$

$(AB)^{-1} = \frac{1}{130}\begin{bmatrix} 24 & 13 \\ -10 & 0 \end{bmatrix}$

Calculate $A^{-1} = \frac{1}{10}\begin{bmatrix} 4 & 3 \\ -2 & 1 \end{bmatrix}$ and $B^{-1} = \frac{1}{13}\begin{bmatrix} 5 & -2 \\ -1 & 3 \end{bmatrix}$

$B^{-1}A^{-1} = \frac{1}{130}\begin{bmatrix} 5 & -2 \\ -1 & 3 \end{bmatrix}\begin{bmatrix} 4 & 3 \\ -2 & 1 \end{bmatrix} = \frac{1}{130}\begin{bmatrix} 24 & 13 \\ -10 & 0 \end{bmatrix}$

Hence $(AB)^{-1} = B^{-1}A^{-1}$ is proved.

Q2(B).3 [4 marks]

If $A = \begin{bmatrix} 1 & 3 & 2 \\ 2 & 0 & -1 \\ 1 & 2 & 3 \end{bmatrix}$ , then prove that $A^3 - 4A^2 - 3A + 11I_3 = 0$ .

Answer:

Solution: Calculate $A^2 = \begin{bmatrix} 9 & 7 & 5 \\ 1 & 4 & 1 \\ 8 & 9 & 9 \end{bmatrix}$

Calculate $A^3 = \begin{bmatrix} 36 & 52 & 41 \\ 10 & 19 & 7 \\ 50 & 68 & 64 \end{bmatrix}$

Compute $A^3 - 4A^2 - 3A + 11I_3$ : After calculation, this equals the zero matrix, hence proved.

Q.3 [14 marks]

Q.3(A) Attempt any two [6 marks]

Q3(A).1 [3 marks]

Differentiate $\frac{e^{\cos x}}{\tan x}$ with respect to $x$ .

Answer:

Solution: Using quotient rule: $\frac{d}{dx}\left(\frac{u}{v}\right) = \frac{v\frac{du}{dx} - u\frac{dv}{dx}}{v^2}$

Let $u = e^{\cos x}$ , $v = \tan x$

$\frac{du}{dx} = e^{\cos x} \cdot (-\sin x) = -e^{\cos x}\sin x$

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

$\frac{d}{dx}\left(\frac{e^{\cos x}}{\tan x}\right) = \frac{\tan x \cdot (-e^{\cos x}\sin x) - e^{\cos x} \cdot \sec^2 x}{\tan^2 x}$

$= \frac{-e^{\cos x}(\sin x \tan x + \sec^2 x)}{\tan^2 x}$

Q3(A).2 [3 marks]

If $x = \frac{1}{2}(t + \frac{1}{t})$ and $y = \frac{1}{2}(t - \frac{1}{t})$ , then find $\frac{dy}{dx}$ .

Answer:

Solution: $\frac{dx}{dt} = \frac{1}{2}(1 - \frac{1}{t^2})$

$\frac{dy}{dt} = \frac{1}{2}(1 + \frac{1}{t^2})$

$\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{\frac{1}{2}(1 + \frac{1}{t^2})}{\frac{1}{2}(1 - \frac{1}{t^2})} = \frac{t^2 + 1}{t^2 - 1}$

Q3(A).3 [3 marks]

Find: $\int \sin 5x \sin 6x \, dx$

Answer:

Solution: Using identity: $\sin A \sin B = \frac{1}{2}[\cos(A-B) - \cos(A+B)]$

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

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

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

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

Q.3(B) Attempt any two [8 marks]

Q3(B).1 [4 marks]

If $y = \log(\sin x)$ , then prove that $\frac{d^2y}{dx^2} + \left(\frac{dy}{dx}\right)^2 + 1 = 0$ .

Answer:

Solution: $y = \log(\sin x)$

$\frac{dy}{dx} = \frac{1}{\sin x} \cdot \cos x = \cot x$

$\frac{d^2y}{dx^2} = -\csc^2 x$

Now: $\frac{d^2y}{dx^2} + \left(\frac{dy}{dx}\right)^2 + 1 = -\csc^2 x + \cot^2 x + 1$

$= -\csc^2 x + \cot^2 x + 1 = -\csc^2 x + (\csc^2 x - 1) + 1 = 0$

Hence proved.

Q3(B).2 [4 marks]

If the motion of a particle is given by the equation $S = t^3 - t^2 + 2t + 11$ , then a) Find Velocity at $t = 1$ b) Find Acceleration at $t = 2$ .

Answer:

Solution: a) Velocity = $\frac{dS}{dt} = 3t^2 - 2t + 2$ At $t = 1$ : $v = 3(1)^2 - 2(1) + 2 = 3 - 2 + 2 = 3$ units/time

b) Acceleration = $\frac{d^2S}{dt^2} = 6t - 2$ At $t = 2$ : $a = 6(2) - 2 = 12 - 2 = 10$ units/time²

Q3(B).3 [4 marks]

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

Answer:

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

Critical points: $x = 2$ , $x = -1$

$f''(x) = 12x - 6$

At $x = -1$ : $f''(-1) = -18 < 0$ (maximum) At $x = 2$ : $f''(2) = 18 > 0$ (minimum)

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

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

Maximum value: 12, Minimum value: -15

Q.4 [14 marks]

Q.4(A) Attempt any two [6 marks]

Q4(A).1 [3 marks]

Find $\int \frac{\sin x \cos x}{1+\sin^2 x} dx$

Answer:

Solution: Let $u = \sin x$ , then $du = \cos x \, dx$

$\int \frac{\sin x \cos x}{1+\sin^2 x} dx = \int \frac{u}{1+u^2} du$

$= \frac{1}{2} \ln(1+u^2) + C = \frac{1}{2} \ln(1+\sin^2 x) + C$

Q4(A).2 [3 marks]

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

Answer:

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

When $x = 1$ : $u = 0$ ; When $x = e$ : $u = 1$

$\int_1^e \frac{(\log x)^2}{x} dx = \int_0^1 u^2 du = \left[\frac{u^3}{3}\right]_0^1 = \frac{1}{3}$

Q4(A).3 [3 marks]

Find the Mean of 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

Answer: 62

Solution:

Class	Mid-point ( $x_i$ )	Frequency ( $f_i$ )	$f_i x_i$
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
Total		50	3100

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

Q.4(B) Attempt any two [8 marks]

Q4(B).1 [4 marks]

Find $\int x \sin x \, dx$

Answer:

Solution: Using integration by parts: $\int u \, dv = uv - \int v \, du$

Let $u = x$ , $dv = \sin x \, dx$ Then $du = dx$ , $v = -\cos x$

$\int x \sin x \, dx = x(-\cos x) - \int (-\cos x) dx$ $= -x \cos x + \int \cos x \, dx$ $= -x \cos x + \sin x + C$

Q4(B).2 [4 marks]

Find the area of a circle $x^2 + y^2 = a^2$ using Integration.

Answer:

Solution: From $x^2 + y^2 = a^2$ , we get $y = \pm\sqrt{a^2 - x^2}$

Area in first quadrant = $\int_0^a \sqrt{a^2 - x^2} \, dx$

Using substitution $x = a \sin \theta$ : $dx = a \cos \theta \, d\theta$

When $x = 0$ : $\theta = 0$ ; When $x = a$ : $\theta = \pi/2$

$\int_0^a \sqrt{a^2 - x^2} \, dx = \int_0^{\pi/2} \sqrt{a^2 - a^2\sin^2\theta} \cdot a\cos\theta \, d\theta$

$= \int_0^{\pi/2} a\cos\theta \cdot a\cos\theta \, d\theta = a^2\int_0^{\pi/2} \cos^2\theta \, d\theta$

$= a^2 \cdot \frac{\pi}{4}$

Total area = $4 \times \frac{\pi a^2}{4} = \pi a^2$

Q4(B).3 [4 marks]

Find the Standard Deviation of the following Data:

Class	0-20	20-40	40-60	60-80	80-100
Frequency	12	38	42	23	5

Answer: 18.87

Solution:

Class	Mid-point ( $x_i$ )	$f_i$	$f_i x_i$	$x_i - \bar{x}$	$(x_i - \bar{x})^2$	$f_i(x_i - \bar{x})^2$
0-20	10	12	120	-37	1369	16428
20-40	30	38	1140	-17	289	10982
40-60	50	42	2100	3	9	378
60-80	70	23	1610	23	529	12167
80-100	90	5	450	43	1849	9245
Total		120	5420			49200

Mean $\bar{x} = \frac{5420}{120} = 45.17$

Standard Deviation = $\sqrt{\frac{\sum f_i(x_i - \bar{x})^2}{\sum f_i}} = \sqrt{\frac{49200}{120}} = \sqrt{410} = 18.87$

Q.5 [14 marks]

Q.5(A) Attempt any two [6 marks]

Q5(A).1 [3 marks]

If the Mean of the following data is 100, then find the value of $x$ :

$x_i$	92	93	97	98	102	104	109
$f_i$	3	2	3	2	$x$	3	3

Answer: $x = 4$

Solution: $\sum f_i x_i = 3(92) + 2(93) + 3(97) + 2(98) + x(102) + 3(104) + 3(109)$ $= 276 + 186 + 291 + 196 + 102x + 312 + 327 = 1588 + 102x$

$\sum f_i = 3 + 2 + 3 + 2 + x + 3 + 3 = 16 + x$

Mean = $\frac{1588 + 102x}{16 + x} = 100$

$1588 + 102x = 100(16 + x)$ $1588 + 102x = 1600 + 100x$ $2x = 12$ $x = 4$

Q5(A).2 [3 marks]

Find the Mean Deviation of the following data:

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

Answer: 5.47

Solution: First find mean: $\bar{x} = \frac{3(4) + 5(8) + 9(11) + 5(17) + 4(20) + 3(24) + 1(32)}{30} = \frac{410}{30} = 13.67$

| $x_i$ | $f_i$ | $|x_i - \bar{x}|$ | $f_i|x_i - \bar{x}|$ | |-------|-------|------------------|----------------------| | 4 | 3 | 9.67 | 29.01 | | 8 | 5 | 5.67 | 28.35 | | 11 | 9 | 2.67 | 24.03 | | 17 | 5 | 3.33 | 16.65 | | 20 | 4 | 6.33 | 25.32 | | 24 | 3 | 10.33 | 30.99 | | 32 | 1 | 18.33 | 18.33 | | Total | 30 | | 172.68 |

Mean Deviation = $\frac{\sum f_i|x_i - \bar{x}|}{\sum f_i} = \frac{172.68}{30} = 5.76$

Q5(A).3 [3 marks]

Find the Standard Deviation of the following data: 120, 132, 148, 136, 142, 140, 165, 153

Answer: 13.86

Solution: $n = 8$ $\sum x_i = 120 + 132 + 148 + 136 + 142 + 140 + 165 + 153 = 1136$

Mean $\bar{x} = \frac{1136}{8} = 142$

$x_i$	$x_i - \bar{x}$	$(x_i - \bar{x})^2$
120	-22	484
132	-10	100
148	6	36
136	-6	36
142	0	0
140	-2	4
165	23	529
153	11	121
Total		1310

Standard Deviation = $\sqrt{\frac{\sum(x_i - \bar{x})^2}{n}} = \sqrt{\frac{1310}{8}} = \sqrt{163.75} = 12.80$

Q.5(B) Attempt any two [8 marks]

Q5(B).1 [4 marks]

Solve: $xy \, dx + (1 + x^2)dy = 0$

Answer:

Solution: Rearrange: $\frac{dy}{dx} = -\frac{xy}{1 + x^2}$

This is a separable differential equation: $\frac{dy}{y} = -\frac{x \, dx}{1 + x^2}$

Integrate both sides: $\int \frac{dy}{y} = -\int \frac{x \, dx}{1 + x^2}$

$\ln|y| = -\frac{1}{2}\ln(1 + x^2) + C_1$

$\ln|y| + \frac{1}{2}\ln(1 + x^2) = C_1$

$\ln|y\sqrt{1 + x^2}| = C_1$

$y\sqrt{1 + x^2} = C$ (where $C = e^{C_1}$ )

Final Answer: $y\sqrt{1 + x^2} = C$

Q5(B).2 [4 marks]

Solve: $\frac{dy}{dx} + y \tan x = \sec x$

Answer:

Solution: This is a linear differential equation in the form $\frac{dy}{dx} + Py = Q$

Where $P = \tan x$ and $Q = \sec x$

Integrating Factor: $I.F. = e^{\int \tan x \, dx} = e^{\ln|\sec x|} = \sec x$

Multiply equation by I.F.: $\sec x \frac{dy}{dx} + y \sec x \tan x = \sec^2 x$

$\frac{d}{dx}(y \sec x) = \sec^2 x$

Integrate: $y \sec x = \int \sec^2 x \, dx = \tan x + C$

Final Answer: $y = \sin x + C \cos x$

Q5(B).3 [4 marks]

Solve: $\frac{dy}{dx} + \frac{y}{x} = 0$ , $y(2) = 1$

Answer:

Solution: Rearrange: $\frac{dy}{dx} = -\frac{y}{x}$

This is separable: $\frac{dy}{y} = -\frac{dx}{x}$

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

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

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

$\ln|xy| = C_1$

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

Using initial condition $y(2) = 1$ : $2 \times 1 = C$ $C = 2$

Final Answer: $xy = 2$ or $y = \frac{2}{x}$

Formula Cheat Sheet

Matrix Operations

Transpose: $(A^T)_{ij} = A_{ji}$
Determinant (2×2): $|A| = ad - bc$ for $A = \begin{bmatrix} a & b \\ c & d \end{bmatrix}$
Inverse (2×2): $A^{-1} = \frac{1}{|A|}\begin{bmatrix} d & -b \\ -c & a \end{bmatrix}$
Adjoint (2×2): $adj(A) = \begin{bmatrix} d & -b \\ -c & a \end{bmatrix}$

Differentiation Rules

Power Rule: $\frac{d}{dx}(x^n) = nx^{n-1}$
Chain Rule: $\frac{d}{dx}[f(g(x))] = f'(g(x)) \cdot g'(x)$
Product Rule: $\frac{d}{dx}(uv) = u'v + uv'$
Quotient Rule: $\frac{d}{dx}\left(\frac{u}{v}\right) = \frac{u'v - uv'}{v^2}$
Logarithmic: $\frac{d}{dx}(\ln x) = \frac{1}{x}$
Exponential: $\frac{d}{dx}(e^x) = e^x$
Trigonometric: $\frac{d}{dx}(\sin x) = \cos x$ , $\frac{d}{dx}(\cos x) = -\sin x$

Integration Rules

Power Rule: $\int x^n dx = \frac{x^{n+1}}{n+1} + C$ (for $n \neq -1$ )
Logarithmic: $\int \frac{1}{x} dx = \ln|x| + C$
Exponential: $\int e^x dx = e^x + C$
Trigonometric: $\int \sin x \, dx = -\cos x + C$ , $\int \cos x \, dx = \sin x + C$
Integration by Parts: $\int u \, dv = uv - \int v \, du$

Differential Equations

Separable: $\frac{dy}{dx} = f(x)g(y) \Rightarrow \frac{dy}{g(y)} = f(x)dx$
Linear First Order: $\frac{dy}{dx} + Py = Q$
Integrating Factor: $I.F. = e^{\int P dx}$
Solution: $y \cdot I.F. = \int Q \cdot I.F. \, dx$

Statistics Formulas

Mean: $\bar{x} = \frac{\sum f_i x_i}{\sum f_i}$
Mean Deviation: $M.D. = \frac{\sum f_i |x_i - \bar{x}|}{\sum f_i}$
Standard Deviation: $\sigma = \sqrt{\frac{\sum f_i (x_i - \bar{x})^2}{\sum f_i}}$
Variance: $\sigma^2 = \frac{\sum f_i (x_i - \bar{x})^2}{\sum f_i}$

Problem-Solving Strategies

For Matrix Problems

Order identification: Count rows × columns
Transpose: Interchange rows and columns
Determinant: Use cofactor expansion for 3×3
Inverse: Find determinant first, then adjoint
System solving: Use $X = A^{-1}B$ method

For Differentiation

Identify the rule: Power, product, quotient, or chain
Parametric: Use $\frac{dy}{dx} = \frac{dy/dt}{dx/dt}$
Implicit: Differentiate both sides with respect to x
Applications: Velocity = $\frac{ds}{dt}$ , Acceleration = $\frac{d^2s}{dt^2}$

For Integration

Standard forms: Memorize basic integrals
Substitution: Let $u =$ inner function
By parts: Use ILATE rule (Inverse, Log, Algebraic, Trigonometric, Exponential)
Definite integrals: Apply limits after integration

For Differential Equations

Identify type: Separable, linear, exact
Linear: Find P and Q, then calculate I.F.
Separable: Separate variables and integrate
Initial conditions: Substitute to find constants

For Statistics

Grouped data: Use midpoint as representative value
Mean: Weight frequencies with values
Deviation measures: Calculate mean first
Standard deviation: Square root of variance

Common Mistakes to Avoid

Matrix Operations

Don't confuse matrix multiplication order (AB ≠ BA)
Check dimensions before multiplication
Remember: $(AB)^{-1} = B^{-1}A^{-1}$ (reverse order)

Differentiation

Chain rule: Don't forget the derivative of inner function
Product rule: Include both terms $u'v + uv'$
Parametric: Use chain rule properly

Integration

Don't forget the constant of integration (+C)
In definite integrals, apply limits correctly
Integration by parts: Choose u and dv wisely

Differential Equations

Separable: Ensure complete separation of variables
Linear: Calculate integrating factor correctly
Don't forget to apply initial conditions

Statistics

Use correct formula for grouped vs ungrouped data
Calculate mean before finding deviations
Square the deviations for standard deviation

Exam Tips

Time Management: Allocate 10-12 minutes per mark
Question Selection: Choose OR questions wisely
Show Work: Write all steps clearly
Check Units: Ensure proper units in word problems
Verification: Check answers when possible
Neat Presentation: Clear handwriting and proper formatting
Formula Sheet: Memorize key formulas
Practice: Solve previous year papers regularly