Applied Mathematics (4320001) - Winter 2023 Solution

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

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

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

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]

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]

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]

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

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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

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

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

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

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

Statistics Problems

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

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

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