Q.1 Fill in the blanks [14 marks]

Q1.1 [1 mark]

If A = [1 2; 3 -1] then 4A = ...

Answer: (b) [4 8; 12 -4]

Solution: $4A = 4 \begin{bmatrix} 1 & 2 \\ 3 & -1 \end{bmatrix} = \begin{bmatrix} 4 & 8 \\ 12 & -4 \end{bmatrix}$

Q1.2 [1 mark]

Order of the matrix [1 1 2; -3 2 3] is ...

Answer: (a) 2 × 3

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

Q1.3 [1 mark]

If A = [1 1; 1 1] then A² = ...

Answer: (d) [2 2; 2 2]

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

Q1.4 [1 mark]

If A = [2 -1; 3 4] then adjoint of A = ...

Answer: (c) [4 1; -3 2]

Solution: For matrix A = [a b; c d], adj(A) = [d -b; -c a] adj(A) = [4 1; -3 2]

Q1.5 [1 mark]

d/dx(tan x) = ...

Answer: (d) sec²x

Solution: $\frac{d}{dx}(\tan x) = \sec^2 x$

Q1.6 [1 mark]

d/dx(sin 5x) = ...

Answer: (b) 5cos5x

Solution: $\frac{d}{dx}(\sin 5x) = 5\cos 5x$ (using chain rule)

Q1.7 [1 mark]

If function y = f(x) is maximum at x = a then f'(a) = ...

Answer: (c) 0

Solution: At maximum point, first derivative equals zero: f'(a) = 0

Q1.8 [1 mark]

∫sin x dx = ... + C

Answer: (a) -cos x

Solution: $\int \sin x \, dx = -\cos x + C$

Q1.9 [1 mark]

∫1/(x²+4) dx = ... + C

Answer: (d) (1/2)tan⁻¹(x/2)

Solution: $\int \frac{1}{x^2+4} dx = \frac{1}{2}\tan^{-1}\left(\frac{x}{2}\right) + C$

Q1.10 [1 mark]

∫₁² x² dx = ...

Answer: (a) 7/3

Solution: $\int_1^2 x^2 dx = \left[\frac{x^3}{3}\right]_1^2 = \frac{8}{3} - \frac{1}{3} = \frac{7}{3}$

Q1.11 [1 mark]

Order of differential equation (d³y/dx³)⁴ + dy/dx + 5y = 0 is ...

Answer: (c) 3

Solution: Order is the highest derivative present = 3

Q1.12 [1 mark]

Integrating factor of dy/dx + y/x = 1 is ...

Answer: (b) x

Solution: I.F. = $e^{\int \frac{1}{x} dx} = e^{\ln x} = x$

Q1.13 [1 mark]

Mean of 39,23,58,47,50,16,61 is ...

Answer: (b) 42

Solution: Mean = $\frac{39+23+58+47+50+16+61}{7} = \frac{294}{7} = 42$

Q1.14 [1 mark]

Mean of first five natural numbers is ...

Answer: (a) 3

Solution: Mean = $\frac{1+2+3+4+5}{5} = \frac{15}{5} = 3$

Q.2 Attempt any two [14 marks total]

Q2(A).1 [3 marks]

If A = [1 3 5; -1 0 2; 4 3 6], B = [3 4 5; 5 4 3; 3 5 4], C = [1 2 1; 3 3 3; 4 5 6], find 3A+2B-4C

Solution: $3A = \begin{bmatrix} 3 & 9 & 15 \\ -3 & 0 & 6 \\ 12 & 9 & 18 \end{bmatrix}$

$2B = \begin{bmatrix} 6 & 8 & 10 \\ 10 & 8 & 6 \\ 6 & 10 & 8 \end{bmatrix}$

$4C = \begin{bmatrix} 4 & 8 & 4 \\ 12 & 12 & 12 \\ 16 & 20 & 24 \end{bmatrix}$

$3A + 2B - 4C = \begin{bmatrix} 5 & 9 & 21 \\ -5 & -4 & 0 \\ 2 & -1 & 2 \end{bmatrix}$

Q2(A).2 [3 marks]

If A = [7 5; -1 2], B = [1 -1; 3 2], show that (A+B)ᵀ = Aᵀ + Bᵀ

Solution: $A + B = \begin{bmatrix} 8 & 4 \\ 2 & 4 \end{bmatrix}$

$(A + B)^T = \begin{bmatrix} 8 & 2 \\ 4 & 4 \end{bmatrix}$

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

$A^T + B^T = \begin{bmatrix} 8 & 2 \\ 4 & 4 \end{bmatrix}$

Hence proved: $(A + B)^T = A^T + B^T$

Q2(A).3 [3 marks]

Solve the differential equation xy dy = (x+1)(y+1)dx

Solution: Separating variables: $\frac{y}{y+1} dy = \frac{x+1}{x} dx$

$\left(1 - \frac{1}{y+1}\right) dy = \left(1 + \frac{1}{x}\right) dx$

Integrating: $y - \ln|y+1| = x + \ln|x| + C$

Final answer: $y - x = \ln|y+1| + \ln|x| + C$

Q2(B).1 [4 marks]

Find the inverse of matrix [3 1 2; 2 -3 -1; 1 2 1]

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

$|A| = 3(-3-(-2)) - 1(2-(-1)) + 2(4-(-3)) = 3(-1) - 1(3) + 2(7) = -3 - 3 + 14 = 8$

Cofactors:

C₁₁ = -1, C₁₂ = -3, C₁₃ = 7
C₂₁ = 3, C₂₂ = 1, C₂₃ = -5
C₃₁ = 5, C₃₂ = 7, C₃₃ = -11

$adj(A) = \begin{bmatrix} -1 & 3 & 5 \\ -3 & 1 & 7 \\ 7 & -5 & -11 \end{bmatrix}$

$A^{-1} = \frac{1}{8} \begin{bmatrix} -1 & 3 & 5 \\ -3 & 1 & 7 \\ 7 & -5 & -11 \end{bmatrix}$

Q2(B).2 [4 marks]

Solve 3x - 2y = 8, 5x + 4y = 6 using matrix method

Solution: $\begin{bmatrix} 3 & -2 \\ 5 & 4 \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 8 \\ 6 \end{bmatrix}$

$|A| = 3(4) - (-2)(5) = 12 + 10 = 22$

$A^{-1} = \frac{1}{22} \begin{bmatrix} 4 & 2 \\ -5 & 3 \end{bmatrix}$

$\begin{bmatrix} x \\ y \end{bmatrix} = \frac{1}{22} \begin{bmatrix} 4 & 2 \\ -5 & 3 \end{bmatrix} \begin{bmatrix} 8 \\ 6 \end{bmatrix} = \frac{1}{22} \begin{bmatrix} 44 \\ -22 \end{bmatrix}$

Answer: x = 2, y = -1

Q2(B).3 [4 marks]

If A = [1 2 1; 2 3 1; 1 2 2], find A·adj(A)

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

For any matrix A: $A \cdot adj(A) = |A| \cdot I$

$A \cdot adj(A) = (-1) \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} = \begin{bmatrix} -1 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & -1 \end{bmatrix}$

Q.3 Attempt any two [14 marks total]

Q3(A).1 [3 marks]

If y = log(sin x/(1+cos x)), find dy/dx

Solution: $y = \log(\sin x) - \log(1+\cos x)$

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

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

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

Using identity: $\frac{\sin x}{1+\cos x} = \tan(\frac{x}{2})$

Answer: $\frac{dy}{dx} = \cot x + \tan(\frac{x}{2})$

Q3(A).2 [3 marks]

If y = sin(x+y), find dy/dx

Solution: Differentiating both sides: $\frac{dy}{dx} = \cos(x+y) \cdot \left(1 + \frac{dy}{dx}\right)$

$\frac{dy}{dx} = \cos(x+y) + \cos(x+y) \cdot \frac{dy}{dx}$

$\frac{dy}{dx} - \cos(x+y) \cdot \frac{dy}{dx} = \cos(x+y)$

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

Answer: $\frac{dy}{dx} = \frac{\cos(x+y)}{1-\cos(x+y)}$

Q3(A).3 [3 marks]

Obtain ∫x²log x dx

Solution: Using integration by parts: ∫u dv = uv - ∫v du

Let u = log x, dv = x² dx Then du = (1/x) dx, v = x³/3

$\int x^2 \log x \, dx = \log x \cdot \frac{x^3}{3} - \int \frac{x^3}{3} \cdot \frac{1}{x} dx$

$= \frac{x^3 \log x}{3} - \int \frac{x^2}{3} dx$

$= \frac{x^3 \log x}{3} - \frac{x^3}{9} + C$

Answer: $\frac{x^3}{3}(\log x - \frac{1}{3}) + C$

Q3(B).1 [4 marks]

Motion equation s = 2t³ - 3t² - 12t + 7. Find s and t when acceleration is zero

Solution: $s = 2t^3 - 3t^2 - 12t + 7$

Velocity: $v = \frac{ds}{dt} = 6t^2 - 6t - 12$

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

When acceleration = 0: $12t - 6 = 0$ $t = \frac{1}{2}$

At t = 1/2: $s = 2(\frac{1}{2})^3 - 3(\frac{1}{2})^2 - 12(\frac{1}{2}) + 7 = \frac{1}{4} - \frac{3}{4} - 6 + 7 = \frac{1}{2}$

Answer: t = 1/2, s = 1/2

Q3(B).2 [4 marks]

If y = 2e³ˣ + 3e⁻²ˣ, prove d²y/dx² - dy/dx - 6y = 0

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

$\frac{dy}{dx} = 6e^{3x} - 6e^{-2x}$

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

Now: $\frac{d^2y}{dx^2} - \frac{dy}{dx} - 6y$

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

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

$= (18-6-12)e^{3x} + (12+6-18)e^{-2x} = 0$

Hence proved

Q3(B).3 [4 marks]

Find maximum and minimum values of f(x) = x³ - 3x + 11

Solution: $f(x) = x^3 - 3x + 11$

$f'(x) = 3x^2 - 3 = 3(x^2 - 1) = 3(x-1)(x+1)$

Critical points: x = 1, x = -1

$f''(x) = 6x$

At x = 1: f''(1) = 6 > 0 → Local minimum At x = -1: f''(-1) = -6 < 0 → Local maximum

$f(1) = 1 - 3 + 11 = 9$ (minimum) $f(-1) = -1 + 3 + 11 = 13$ (maximum)

Answer: Maximum = 13 at x = -1, Minimum = 9 at x = 1

Q.4 Attempt any two [14 marks total]

Q4(A).1 [3 marks]

Obtain ∫sin 5x sin 6x dx

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

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

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

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

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

Answer: $\frac{1}{2}\sin x - \frac{\sin(11x)}{22} + C$

Q4(A).2 [3 marks]

Obtain ∫(1+x)eˣ/cos²(xeˣ) dx

Solution: Let $u = xe^x$ , then $du = (1+x)e^x dx$

The integral becomes: $\int \frac{du}{\cos^2 u} = \int \sec^2 u \, du = \tan u + C$

Substituting back: $= \tan(xe^x) + C$

Answer: $\tan(xe^x) + C$

Q4(A).3 [3 marks]

Find standard deviation for data: 6,7,10,12,13,4,8,12

Solution: Data: 6, 7, 10, 12, 13, 4, 8, 12 n = 8

Mean = $\frac{6+7+10+12+13+4+8+12}{8} = \frac{72}{8} = 9$

x	x-9	(x-9)²
6	-3	9
7	-2	4
10	1	1
12	3	9
13	4	16
4	-5	25
8	-1	1
12	3	9

Σ(x-9)² = 74

Standard deviation = $\sqrt{\frac{\sum(x-\bar{x})^2}{n}} = \sqrt{\frac{74}{8}} = \sqrt{9.25} = 3.04$

Answer: σ = 3.04

Q4(B).1 [4 marks]

Obtain ∫(2x+1)/[(x+1)(x-3)] dx

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

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

When x = -1: $2(-1)+1 = A(-4) \Rightarrow -1 = -4A \Rightarrow A = \frac{1}{4}$

When x = 3: $2(3)+1 = B(4) \Rightarrow 7 = 4B \Rightarrow B = \frac{7}{4}$

$\int \frac{2x+1}{(x+1)(x-3)} dx = \frac{1}{4}\int \frac{1}{x+1} dx + \frac{7}{4}\int \frac{1}{x-3} dx$

$= \frac{1}{4}\ln|x+1| + \frac{7}{4}\ln|x-3| + C$

Answer: $\frac{1}{4}\ln|x+1| + \frac{7}{4}\ln|x-3| + C$

Q4(B).2 [4 marks]

Obtain ∫₀^(π/2) √(cot x)/(√(cot x) + √(tan x)) dx

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

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

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

Since $\cot(\pi/2-x) = \tan x$ and $\tan(\pi/2-x) = \cot x$ :

$I = \int_0^{\pi/2} \frac{\sqrt{\tan x}}{\sqrt{\tan x} + \sqrt{\cot x}} dx$

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

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

Q4(B).3 [4 marks]

Find mean deviation for grouped data

xᵢ	4	8	11	17	20	24	32
fᵢ	3	5	9	5	4	3	1

Solution: N = Σfᵢ = 3+5+9+5+4+3+1 = 30

Mean = $\frac{\sum f_i x_i}{N} = \frac{3(4)+5(8)+9(11)+5(17)+4(20)+3(24)+1(32)}{30}$

$= \frac{12+40+99+85+80+72+32}{30} = \frac{420}{30} = 14$

| xᵢ | fᵢ | |xᵢ-14| | fᵢ|xᵢ-14| | |----|----|---------|-----------| | 4 | 3 | 10 | 30 | | 8 | 5 | 6 | 30 | | 11 | 9 | 3 | 27 | | 17 | 5 | 3 | 15 | | 20 | 4 | 6 | 24 | | 24 | 3 | 10 | 30 | | 32 | 1 | 18 | 18 |

Σfᵢ|xᵢ-14| = 174

Mean deviation = $\frac{\sum f_i |x_i - \bar{x}|}{N} = \frac{174}{30} = 5.8$

Answer: Mean deviation = 5.8

Q.5 Attempt any two [14 marks total]

Q5(A).1 [3 marks]

Find mean deviation for grouped data

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

Solution:

Class	Mid-value	fᵢ	fᵢxᵢ
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 = 50, Σfᵢxᵢ = 3100

Mean = 3100/50 = 62

| Class | xᵢ | fᵢ | |xᵢ-62| | fᵢ|xᵢ-62| | |-------|----|----|---------|-----------| | 30-40 | 35 | 3 | 27 | 81 | | 40-50 | 45 | 7 | 17 | 119 | | 50-60 | 55 | 12 | 7 | 84 | | 60-70 | 65 | 15 | 3 | 45 | | 70-80 | 75 | 8 | 13 | 104 | | 80-90 | 85 | 3 | 23 | 69 | | 90-100| 95 | 2 | 33 | 66 |

Mean deviation = 568/50 = 11.36

Answer: Mean deviation = 11.36

Q5(A).2 [3 marks]

Find standard deviation for given data

Class	60	61	62	63	64	65	66	67	68
Freq	2	1	12	29	25	12	10	4	5

Solution: N = 100, Mean = (2×60 + 1×61 + ... + 5×68)/100 = 6380/100 = 63.8

xᵢ	fᵢ	(xᵢ-63.8)	(xᵢ-63.8)²	fᵢ(xᵢ-63.8)²
60	2	-3.8	14.44	28.88
61	1	-2.8	7.84	7.84
62	12	-1.8	3.24	38.88
63	29	-0.8	0.64	18.56
64	25	0.2	0.04	1.00
65	12	1.2	1.44	17.28
66	10	2.2	4.84	48.40
67	4	3.2	10.24	40.96
68	5	4.2	17.64	88.20

Σfᵢ(xᵢ-x̄)² = 290

Standard deviation = √(290/100) = √2.9 = 1.70

Answer: σ = 1.70

Q5(A).3 [3 marks]

Find mean for grouped data

Class	0-20	20-40	40-60	60-80	80-100	100-120
Freq	26	31	35	42	82	71

Solution:

Class	Mid-value	fᵢ	fᵢxᵢ
0-20	10	26	260
20-40	30	31	930
40-60	50	35	1750
60-80	70	42	2940
80-100	90	82	7380
100-120	110	71	7810

N = 287, Σfᵢxᵢ = 21070

Mean = $\frac{\sum f_i x_i}{N} = \frac{21070}{287} = 73.42$

Answer: Mean = 73.42

Q5(B).1 [4 marks]

Solve differential equation (x + y + 1)² dy/dx = 1

Solution: Let z = x + y + 1, then dz/dx = 1 + dy/dx So dy/dx = dz/dx - 1

Substituting: $z^2(dz/dx - 1) = 1$ $z^2 dz/dx - z^2 = 1$ $z^2 dz/dx = 1 + z^2$ $\frac{z^2}{1 + z^2} dz = dx$

Integrating: $\int \frac{z^2}{1 + z^2} dz = \int dx$

$\int \left(1 - \frac{1}{1 + z^2}\right) dz = x + C$

$z - \tan^{-1}z = x + C$

Substituting back z = x + y + 1: $(x + y + 1) - \tan^{-1}(x + y + 1) = x + C$

Answer: $y + 1 = \tan^{-1}(x + y + 1) + C$

Q5(B).2 [4 marks]

Solve dy/dx + y/x = eˣ, y(0) = 2

Solution: This is a linear differential equation of the form dy/dx + P(x)y = Q(x)

Here P(x) = 1/x, Q(x) = eˣ

Integrating factor: $I.F. = e^{\int \frac{1}{x} dx} = e^{\ln|x|} = |x| = x$ (for x > 0)

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

$\frac{d}{dx}(xy) = xe^x$

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

Using integration by parts for ∫xeˣ dx: Let u = x, dv = eˣ dx Then du = dx, v = eˣ

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

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

Using initial condition y(0) = 2: As x → 0, we need to use L'Hôpital's rule or series expansion.

From the original equation at x = 0: dy/dx = eˣ - y/x This suggests we need to be more careful with the initial condition.

Alternative approach: Since the equation has a singularity at x = 0, we solve in the neighborhood where x ≠ 0.

Answer: $y = \frac{e^x(x-1) + C}{x}$ where C is determined by boundary conditions.

Q5(B).3 [4 marks]

Solve y dy/dx = √(1 + x² + y² + x²y²)

Solution: $y \frac{dy}{dx} = \sqrt{1 + x^2 + y^2 + x^2y^2}$

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

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

Integrating both sides: $\int \frac{y dy}{\sqrt{1 + y^2}} = \int \sqrt{1 + x^2} dx$

For the left side, let u = 1 + y², then du = 2y dy: $\int \frac{y dy}{\sqrt{1 + y^2}} = \frac{1}{2} \int \frac{du}{\sqrt{u}} = \sqrt{u} = \sqrt{1 + y^2}$

For the right side: $\int \sqrt{1 + x^2} dx = \frac{x\sqrt{1 + x^2}}{2} + \frac{1}{2}\ln|x + \sqrt{1 + x^2}| + C$

Therefore: $\sqrt{1 + y^2} = \frac{x\sqrt{1 + x^2}}{2} + \frac{1}{2}\ln|x + \sqrt{1 + x^2}| + C$

Answer: $\sqrt{1 + y^2} = \frac{x\sqrt{1 + x^2}}{2} + \frac{1}{2}\ln|x + \sqrt{1 + x^2}| + C$

Formula Cheat Sheet

Matrix Operations

$(A + B)^T = A^T + B^T$
$(AB)^T = B^T A^T$
$A \cdot adj(A) = |A| \cdot I$
For 2×2 matrix $[a \; b; c \; d]$ : $adj = [d \; -b; -c \; a]$

Differentiation Formulas

$\frac{d}{dx}(\sin x) = \cos x$
$\frac{d}{dx}(\cos x) = -\sin x$
$\frac{d}{dx}(\tan x) = \sec^2 x$
$\frac{d}{dx}(\log x) = \frac{1}{x}$
$\frac{d}{dx}(e^x) = e^x$
Chain rule: $\frac{d}{dx}f(g(x)) = f'(g(x)) \cdot g'(x)$

Integration Formulas

$\int \sin x \, dx = -\cos x + C$
$\int \cos x \, dx = \sin x + C$
$\int \sec^2 x \, dx = \tan x + C$
$\int \frac{1}{x} dx = \ln|x| + C$
$\int e^x dx = e^x + C$
$\int \frac{1}{x^2 + a^2} dx = \frac{1}{a}\tan^{-1}(\frac{x}{a}) + C$

Differential Equations

Linear DE: $\frac{dy}{dx} + P(x)y = Q(x)$
Integrating Factor: $I.F. = e^{\int P(x) dx}$
Variable Separable: $\frac{dy}{dx} = f(x)g(y) \Rightarrow \frac{dy}{g(y)} = f(x)dx$

Statistics

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

Problem-Solving Strategies

Matrix Problems

Always check dimensions before operations
For inverse: Calculate determinant first, then adjoint
For system of equations: Use $X = A^{-1}B$ where $AX = B$

Differentiation Problems

Identify the type: Chain rule, product rule, quotient rule
For implicit differentiation: Differentiate both sides, collect dy/dx terms
For parametric: Use $\frac{dy}{dx} = \frac{dy/dt}{dx/dt}$

Integration Problems

Try substitution if you see function and its derivative
Use integration by parts for products (LIATE rule)
For definite integrals: Check for symmetry properties

Differential Equations

Identify type: Separable, linear, exact
For linear DE: Find integrating factor first
Always verify your solution by substitution

Statistics Problems

Find mean first for deviation calculations
Use grouped data formulas when data is in classes
Create frequency table to organize calculations

Common Mistakes to Avoid

Matrix multiplication: Order matters (AB ≠ BA generally)
Chain rule: Don't forget to multiply by derivative of inner function
Integration by parts: Choose u and dv carefully using LIATE
Differential equations: Don't forget the constant of integration
Statistics: Use correct formula for grouped vs ungrouped data

Exam Tips

Read questions carefully - especially for OR questions
Show all steps - partial marks are awarded
Check units and signs in your final answers
Verify solutions when possible by substitution
Manage time wisely - attempt questions you're confident about first
Use standard formulas - memorize the formula sheet content
For fill-in-blanks: Eliminate obviously wrong options first