Maths I — Crystal Clear Practice Drill

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Maths I — Crystal Clear Practice Drill

How to use this: Cover the answers below each question. Solve on paper first. Then reveal and grade yourself. If you got it wrong, go back to the specific week note and re-read the pattern.

🔵 Tier 1 — Core Mechanics (Try these first, every day)

D1. Sets and Relations (Week 1)

Q: Let

R = \{(x, y) \mid x + y = 0\}

\mathbb{Z}

. Is

R

an equivalence relation?

<details> <summary>Model Answer</summary>

Reflexive: Does $(x, x) \in R$ ? That requires $x + x = 0$ , i.e., $2x = 0$ . This is only true for $x = 0$ . Not all integers satisfy it. Fails.
Symmetric: If $x + y = 0$ , then $y + x = 0$ . Holds.
Transitive: If $x + y = 0$ and $y + z = 0$ , then $x = -y$ and $z = -y$ , so $x = z$ , meaning $x + z = 2x \neq 0$ in general. Fails.
Answer: Not an equivalence relation (fails reflexivity and transitivity).

</details>

D2. Quadratic (Week 2)

Q: For

3x^2 - 7x + 2 = 0

, find

\alpha^2 + \beta^2

without solving for

\alpha, \beta

<details> <summary>Model Answer</summary>

By Vieta's:

\alpha + \\beta = 7/3

\alpha\\beta = 2/3

\alpha^2 + \beta^2 = (\alpha+\beta)^2 - 2\alpha\\beta = \frac{49}{9} - \frac{4}{3} = \frac{49}{9} - \frac{12}{9} = \frac{37}{9}

</details>

D3. Limit via Factoring (Week 3)

Q: Evaluate

\lim_{x \to 3} \frac{x^2 - 9}{x - 3}

<details> <summary>Model Answer</summary>

Direct substitution gives

0/0

. Factor:

\frac{(x-3)(x+3)}{(x-3)} = x + 3

. Substitute

x = 3

: Answer = 6.

</details>

D4. Chain Rule (Week 4)

Q: Differentiate

y = (3x^2 + 1)^4

<details> <summary>Model Answer</summary>

Chain Rule: outer is

u^4

, inner is

3x^2 + 1

\frac{dy}{dx} = 4(3x^2+1)^3 \cdot 6x = 24x(3x^2+1)^3

</details>

D5. Optimization (Week 5)

Q: A farmer has 60 metres of fencing for a rectangular plot (one side is a wall — no fence needed). Maximize the area.

<details> <summary>Model Answer</summary>

Let width =

x

(two of these), length =

60 - 2x

. Area

A = x(60 - 2x) = 60x - 2x^2

A' = 60 - 4x = 0 \implies x = 15

A'' = -4 < 0

(maximum). Area =

15 \times 30 = 450\text{ m}^2

. Dimensions: 15m × 30m.

</details>

D6. Definite Integral (Week 6)

Q: Compute

\int_0^2 (x^3 - 2x)\,dx

<details> <summary>Model Answer</summary>

F(x) = \frac{x^4}{4} - x^2

F(2) = 4 - 4 = 0

F(0) = 0

. Answer = 0.

This is a signed area problem — the region above and below cancel perfectly.

</details>

D7. Integration by Parts (Week 7)

Q: Compute

\int x e^x\,dx

<details> <summary>Model Answer</summary>

ILATE:

u = x

(Algebraic),

dv = e^x dx

du = dx

v = e^x

\int xe^x\,dx = xe^x - \int e^x\,dx = xe^x - e^x + C = e^x(x-1) + C

</details>

D8. Linear Approximation (Week 8)

Q: Find the best linear approximation

L(x)

f(x) = x^3

x = 2

<details> <summary>Model Answer</summary>

f(2) = 8

f'(x) = 3x^2 \implies f'(2) = 12

L(x) = f(2) + f'(2)(x-2) = 8 + 12(x-2) = 12x - 16

</details>

🟡 Tier 2 — Tricky But Crucial

T1. Standard Limit (Week 3 / Week 8)

Q: Find

\lim_{n\to\infty} n\left(e^{1/n} - 1\right)

<details> <summary>Model Answer</summary>

Let

u = 1/n

. As

n\to\infty

u\to 0

n\left(e^{1/n}-1\right) = \frac{e^u - 1}{u} \xrightarrow{u\to 0} 1

Answer: 1. (Standard limit

\lim_{u\to 0}\frac{e^u-1}{u} = 1

)

</details>

T2. Continuity Constant (Week 3)

Q: Find

k

so that

f(x) = \begin{cases} kx + 3 & x \leq 2 \ x^2 - 1 & x > 2 \end{cases}

is continuous everywhere.

<details> <summary>Model Answer</summary>

x = 2

: LHL =

2k + 3

. RHL =

4 - 1 = 3

. Set equal:

2k + 3 = 3 \implies k = 0

</details>

T3. Monotonicity (Week 5)

Q: Find the intervals where

f(x) = x^3 - 6x^2 + 9x + 1

is strictly increasing.

<details> <summary>Model Answer</summary>

f'(x) = 3x^2 - 12x + 9 = 3(x^2 - 4x + 3) = 3(x-1)(x-3)

$f' > 0$ when $x < 1$ or $x > 3$ .
Strictly increasing on $(-\infty, 1)$ and $(3, \infty)$ .

</details>

T4. Partial Fractions (Week 7)

Q: Integrate

\int \frac{3}{x^2 - 1}\,dx

<details> <summary>Model Answer</summary>

Factor:

x^2 - 1 = (x-1)(x+1)

. Decompose:

\frac{3}{(x-1)(x+1)} = \frac{A}{x-1} + \frac{B}{x+1}

x=1

3 = 2A \implies A = 3/2

. At

x=-1

3 = -2B \implies B = -3/2

\int \frac{3/2}{x-1}\,dx - \int \frac{3/2}{x+1}\,dx = \frac{3}{2}\ln|x-1| - \frac{3}{2}\ln|x+1| + C

= \frac{3}{2}\ln\left|\frac{x-1}{x+1}\right| + C

</details>

T5. Matrix Inverse (Week 8)

Q: Find the inverse of

A = \begin{bmatrix} 4 & 7 \\ 2 & 6 \end{bmatrix}

<details> <summary>Model Answer</summary>

\det(A) = (4)(6) - (7)(2) = 24 - 14 = 10

A^{-1} = \frac{1}{10}\begin{bmatrix} 6 & -7 \ -2 & 4 \end{bmatrix} = \begin{bmatrix} 0.6 & -0.7 \ -0.2 & 0.4 \end{bmatrix}

</details>

T6. Functional Equation (Week 8)

f(x+y) = f(x)f(y)

for all reals.

f(1) = 5

and

f'(0) = \ln 5

. Find

f'(2)

<details> <summary>Model Answer</summary>

This functional equation gives

f(x) = a^x

where

a = 5

(since

f(1) = 5

). Key property:

f'(x) = f(x) \cdot f'(0)

f'(2) = f(2) \cdot f'(0) = 5^2 \cdot \ln 5 = 25\ln 5

</details>

🔴 Tier 3 — Exam-Grade Problems

E1. Mixed Limits (Week 7)

Q: Find

\lim_{n\to\infty} \left(\frac{12n^2}{3n+27} - \frac{4n^2+23}{n+5}\right)

<details> <summary>Model Answer</summary>

Combine over common denominator

(3n+27)(n+5) = (n+9) \times 3 \times (n+5)

: After combining: Numerator

= -16n^2 - 23n - 207

, Denominator

\sim n^2

. Degrees match

\implies

Limit

= -16/1 = \mathbf{-16}

</details>

E2. Optimization with Constraint (Week 5)

Q: A box with a square base and no top must have a volume of 32 m³. Find dimensions that minimize surface area.

<details> <summary>Model Answer</summary>

Let base side

= x

, height

= h

. Volume:

x^2 h = 32 \implies h = 32/x^2

. Surface area:

S = x^2 + 4xh = x^2 + 4x \cdot \frac{32}{x^2} = x^2 + \frac{128}{x}

S' = 2x - \frac{128}{x^2} = 0 \implies 2x^3 = 128 \implies x^3 = 64 \implies x = 4

h = 32/16 = 2

. Answer: Base 4m × 4m, Height 2m.

</details>

E3. L'Hôpital (Week 3 / Week 8)

Q: Evaluate

\lim_{x\to 0} \frac{e^x - 1 - x}{x^2}

<details> <summary>Model Answer</summary>

x=0

\frac{0}{0}

. Apply L'Hôpital:

\frac{e^x - 1}{2x} \xrightarrow{x\to 0} \frac{0}{0}

Apply again:

\frac{e^x}{2} \xrightarrow{x\to 0} \frac{1}{2}

Answer: $1/2$ .

</details>

📊 Score Yourself

Tier	Questions	Target
Tier 1 (Core)	D1–D8	7/8 minimum
Tier 2 (Tricky)	T1–T6	4/6 minimum
Tier 3 (Exam)	E1–E3	2/3 minimum

If you scored below target on any tier, revisit that week's note file and then attempt the corresponding graded assignment questions again.

Maths I — Crystal Clear Practice Drill

🔵 Tier 1 — Core Mechanics (Try these first, every day)

D1. Sets and Relations (Week 1)

D2. Quadratic (Week 2)

D3. Limit via Factoring (Week 3)

D4. Chain Rule (Week 4)

D5. Optimization (Week 5)

D6. Definite Integral (Week 6)

D7. Integration by Parts (Week 7)

D8. Linear Approximation (Week 8)

🟡 Tier 2 — Tricky But Crucial

T1. Standard Limit (Week 3 / Week 8)

T2. Continuity Constant (Week 3)

T3. Monotonicity (Week 5)

T4. Partial Fractions (Week 7)

T5. Matrix Inverse (Week 8)

T6. Functional Equation (Week 8)

🔴 Tier 3 — Exam-Grade Problems

E1. Mixed Limits (Week 7)

E2. Optimization with Constraint (Week 5)

E3. L'Hôpital (Week 3 / Week 8)

📊 Score Yourself

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