Applied Mathematics (4320001) - Winter 2023 Solution

Q.1 Fill in the blanks [14 marks]

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Q1.1 [1 mark]

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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]

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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]

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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]

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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]

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d/dx(tan x) = …

Answer: (d) sec²x

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

Q1.6 [1 mark]

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d/dx(sin 5x) = …

Answer: (b) 5cos5x

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

Q1.7 [1 mark]

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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]

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∫sin x dx = … + C

Answer: (a) -cos x

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

Q1.9 [1 mark]

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∫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]

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∫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]

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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]

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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]

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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]

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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]

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Q2(A).1 [3 marks]

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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]

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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]

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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]

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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]

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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]

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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]

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Q3(A).1 [3 marks]

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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]

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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]

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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]

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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]

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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]

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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]

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Q4(A).1 [3 marks]

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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]

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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]

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

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]

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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]

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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]

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Find mean deviation for grouped data

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]

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Q5(A).1 [3 marks]

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Find mean deviation for grouped data

Solution:

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]

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Find standard deviation for given data

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

Σfᵢ(xᵢ-x̄)² = 290

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

Answer: σ = 1.70

Q5(A).3 [3 marks]

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Find mean for grouped data

Solution:

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]

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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]

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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]

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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

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Matrix Operations

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

Differentiation Formulas

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

Integration Formulas

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

Differential Equations

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• 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

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• 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

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Matrix Problems

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

Differentiation Problems

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

Integration Problems

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

Differential Equations

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

Statistics Problems

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

Common Mistakes to Avoid

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

Exam Tips

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