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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)∣x+y=0}R = \{(x, y) \mid x + y = 0\} on Z\mathbb{Z}. Is RR an equivalence relation?
<details> <summary>Model Answer</summary>
  • Reflexive: Does (x,x)∈R(x, x) \in R? That requires x+x=0x + x = 0, i.e., 2x=02x = 0. This is only true for x=0x = 0. Not all integers satisfy it. Fails.
  • Symmetric: If x+y=0x + y = 0, then y+x=0y + x = 0. Holds.
  • Transitive: If x+y=0x + y = 0 and y+z=0y + z = 0, then x=−yx = -y and z=−yz = -y, so x=zx = z, meaning x+z=2x≠0x + z = 2x \neq 0 in general. Fails.
  • Answer: Not an equivalence relation (fails reflexivity and transitivity).
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D2. Quadratic (Week 2)

Q: For 3x2−7x+2=03x^2 - 7x + 2 = 0, find α2+β2\alpha^2 + \beta^2 without solving for α,β\alpha, \beta.
<details> <summary>Model Answer</summary>
By Vieta's: α+beta=7/3\alpha + \\beta = 7/3, αbeta=2/3\alpha\\beta = 2/3. α2+β2=(α+β)2−2αbeta=499−43=499−129=379\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}
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D3. Limit via Factoring (Week 3)

Q: Evaluate lim⁡x→3x2−9x−3\lim_{x \to 3} \frac{x^2 - 9}{x - 3}.
<details> <summary>Model Answer</summary>
Direct substitution gives 0/00/0. Factor: (x−3)(x+3)(x−3)=x+3\frac{(x-3)(x+3)}{(x-3)} = x + 3. Substitute x=3x = 3: Answer = 6.
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D4. Chain Rule (Week 4)

Q: Differentiate y=(3x2+1)4y = (3x^2 + 1)^4.
<details> <summary>Model Answer</summary>
Chain Rule: outer is u4u^4, inner is 3x2+13x^2 + 1. dydx=4(3x2+1)3⋅6x=24x(3x2+1)3\frac{dy}{dx} = 4(3x^2+1)^3 \cdot 6x = 24x(3x^2+1)^3
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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 = xx (two of these), length = 60−2x60 - 2x. Area A=x(60−2x)=60x−2x2A = x(60 - 2x) = 60x - 2x^2. A′=60−4x=0  ⟹  x=15A' = 60 - 4x = 0 \implies x = 15. A′′=−4<0A'' = -4 < 0 (maximum). Area = 15×30=450 m215 \times 30 = 450\text{ m}^2. Dimensions: 15m × 30m.
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D6. Definite Integral (Week 6)

Q: Compute ∫02(x3−2x) dx\int_0^2 (x^3 - 2x)\,dx.
<details> <summary>Model Answer</summary>
F(x)=x44−x2F(x) = \frac{x^4}{4} - x^2. F(2)=4−4=0F(2) = 4 - 4 = 0. F(0)=0F(0) = 0. Answer = 0.
This is a signed area problem — the region above and below cancel perfectly.
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D7. Integration by Parts (Week 7)

Q: Compute ∫xex dx\int x e^x\,dx.
<details> <summary>Model Answer</summary>
ILATE: u=xu = x (Algebraic), dv=exdxdv = e^x dx. du=dxdu = dx, v=exv = e^x. ∫xex dx=xex−∫ex dx=xex−ex+C=ex(x−1)+C\int xe^x\,dx = xe^x - \int e^x\,dx = xe^x - e^x + C = e^x(x-1) + C
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D8. Linear Approximation (Week 8)

Q: Find the best linear approximation L(x)L(x) of f(x)=x3f(x) = x^3 at x=2x = 2.
<details> <summary>Model Answer</summary>
f(2)=8f(2) = 8. f′(x)=3x2  ⟹  f′(2)=12f'(x) = 3x^2 \implies f'(2) = 12. L(x)=f(2)+f′(2)(x−2)=8+12(x−2)=12x−16L(x) = f(2) + f'(2)(x-2) = 8 + 12(x-2) = 12x - 16
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🟡 Tier 2 — Tricky But Crucial

T1. Standard Limit (Week 3 / Week 8)

Q: Find lim⁡n→∞n(e1/n−1)\lim_{n\to\infty} n\left(e^{1/n} - 1\right).
<details> <summary>Model Answer</summary>
Let u=1/nu = 1/n. As n→∞n\to\infty, u→0u\to 0. n(e1/n−1)=eu−1u→u→01n\left(e^{1/n}-1\right) = \frac{e^u - 1}{u} \xrightarrow{u\to 0} 1 Answer: 1. (Standard limit lim⁡u→0eu−1u=1\lim_{u\to 0}\frac{e^u-1}{u} = 1)
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T2. Continuity Constant (Week 3)

Q: Find kk so that f(x)={kx+3x≤2 x2−1x>2f(x) = \begin{cases} kx + 3 & x \leq 2 \ x^2 - 1 & x > 2 \end{cases} is continuous everywhere.
<details> <summary>Model Answer</summary>
At x=2x = 2: LHL = 2k+32k + 3. RHL = 4−1=34 - 1 = 3. Set equal: 2k+3=3  ⟹  k=02k + 3 = 3 \implies k = 0.
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T3. Monotonicity (Week 5)

Q: Find the intervals where f(x)=x3−6x2+9x+1f(x) = x^3 - 6x^2 + 9x + 1 is strictly increasing.
<details> <summary>Model Answer</summary>
f′(x)=3x2−12x+9=3(x2−4x+3)=3(x−1)(x−3)f'(x) = 3x^2 - 12x + 9 = 3(x^2 - 4x + 3) = 3(x-1)(x-3).
  • f′>0f' > 0 when x<1x < 1 or x>3x > 3.
  • Strictly increasing on (−∞,1)(-\infty, 1) and (3,∞)(3, \infty).
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T4. Partial Fractions (Week 7)

Q: Integrate ∫3x2−1 dx\int \frac{3}{x^2 - 1}\,dx.
<details> <summary>Model Answer</summary>
Factor: x2−1=(x−1)(x+1)x^2 - 1 = (x-1)(x+1). Decompose: 3(x−1)(x+1)=Ax−1+Bx+1\frac{3}{(x-1)(x+1)} = \frac{A}{x-1} + \frac{B}{x+1} At x=1x=1: 3=2A  ⟹  A=3/23 = 2A \implies A = 3/2. At x=−1x=-1: 3=−2B  ⟹  B=−3/23 = -2B \implies B = -3/2. ∫3/2x−1 dx−∫3/2x+1 dx=32ln⁡∣x−1∣−32ln⁡∣x+1∣+C\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 =32ln⁡∣x−1x+1∣+C= \frac{3}{2}\ln\left|\frac{x-1}{x+1}\right| + C
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T5. Matrix Inverse (Week 8)

Q: Find the inverse of A=[4726]A = \begin{bmatrix} 4 & 7 \\ 2 & 6 \end{bmatrix}.
<details> <summary>Model Answer</summary>
det⁡(A)=(4)(6)−(7)(2)=24−14=10\det(A) = (4)(6) - (7)(2) = 24 - 14 = 10. A−1=110[6−7 −24]=[0.6−0.7 −0.20.4]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}
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T6. Functional Equation (Week 8)

Q: f(x+y)=f(x)f(y)f(x+y) = f(x)f(y) for all reals. f(1)=5f(1) = 5 and f′(0)=ln⁡5f'(0) = \ln 5. Find f′(2)f'(2).
<details> <summary>Model Answer</summary>
This functional equation gives f(x)=axf(x) = a^x where a=5a = 5 (since f(1)=5f(1) = 5). Key property: f′(x)=f(x)⋅f′(0)f'(x) = f(x) \cdot f'(0). f′(2)=f(2)⋅f′(0)=52⋅ln⁡5=25ln⁡5f'(2) = f(2) \cdot f'(0) = 5^2 \cdot \ln 5 = 25\ln 5
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🔴 Tier 3 — Exam-Grade Problems

E1. Mixed Limits (Week 7)

Q: Find lim⁡n→∞(12n23n+27−4n2+23n+5)\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)×3×(n+5)(3n+27)(n+5) = (n+9) \times 3 \times (n+5): After combining: Numerator =−16n2−23n−207= -16n^2 - 23n - 207, Denominator ∼n2\sim n^2. Degrees match   ⟹  \implies Limit =−16/1=−16= -16/1 = \mathbf{-16}.
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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= x, height =h= h. Volume: x2h=32  ⟹  h=32/x2x^2 h = 32 \implies h = 32/x^2. Surface area: S=x2+4xh=x2+4x⋅32x2=x2+128xS = x^2 + 4xh = x^2 + 4x \cdot \frac{32}{x^2} = x^2 + \frac{128}{x}. S′=2x−128x2=0  ⟹  2x3=128  ⟹  x3=64  ⟹  x=4S' = 2x - \frac{128}{x^2} = 0 \implies 2x^3 = 128 \implies x^3 = 64 \implies x = 4. h=32/16=2h = 32/16 = 2. Answer: Base 4m × 4m, Height 2m.
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E3. L'Hôpital (Week 3 / Week 8)

Q: Evaluate lim⁡x→0ex−1−xx2\lim_{x\to 0} \frac{e^x - 1 - x}{x^2}.
<details> <summary>Model Answer</summary>
At x=0x=0: 00\frac{0}{0}. Apply L'Hôpital: ex−12x→x→000\frac{e^x - 1}{2x} \xrightarrow{x\to 0} \frac{0}{0} Apply again: ex2→x→012\frac{e^x}{2} \xrightarrow{x\to 0} \frac{1}{2} Answer: 1/21/2.
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📊 Score Yourself

TierQuestionsTarget
Tier 1 (Core)D1–D87/8 minimum
Tier 2 (Tricky)T1–T64/6 minimum
Tier 3 (Exam)E1–E32/3 minimum
If you scored below target on any tier, revisit that week's note file and then attempt the corresponding graded assignment questions again.
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