Neural Sync Active

Maths I — Crystal Clear Practice Drill

Registry Synced

Maths I — Crystal Clear Practice Drill

1169 words

6 min read

Reading compass

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

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.