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Home Official UAT-UK Maths 2 Guide (pp. 6–159 • SymPy Verified)
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Official UAT-UK Mathematics 2 Guide — 31 Solved Exercises

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6-Card Lesson (ARCH-MM-01)Practice Matching Archetype Drill

Zero Base with Positive, Zero & Negative Exponents (a^m when a = 0)

Official UAT-UK Guide Exercise Box (Page 10):
Something to think about: what happens to ama^m when a=0a = 0?
Official Footnote: 2 The case when m is negative takes a little more care to deal with. Have a think about how it works. 3 A good website to use to help you understand graph sketching is DESMOS GRAPHING
SymPy-Verified Result:
For m>0m > 0, 0m=00^m = 0; for m=0m = 0, 000^0 is indeterminate/undefined in ESAT algebra; for m<0m < 0, 0m=10∣m∣0^m = \frac{1}{0^{|m|}} is strictly undefined (division by zero).
Sub-90 Dry-Wipe Shortcut15s
Line 1 (3-Regime Domain Split): Split mm immediately into m>0m > 0 (0m=00^m = 0), m=0m = 0 (indeterminate), and m<0m < 0 (10∣m∣\frac{1}{0^{|m|}} undefined).
Line 2 (Domain Prune): Whenever ama^m appears with negative or variable mm, immediately cross out any option that includes a=0a = 0 in its domain!
Standard Classroom Derivation~90s
Test individual integer cases m=1,2,3m = 1, 2, 3 to see 0m=00^m = 0, test m=0m = 0 via limit laws lim⁡a→0+a0=1\lim_{a\to 0^+} a^0 = 1 vs lim⁡m→0+0m=0\lim_{m\to 0^+} 0^m = 0, and write 0−k=10k0^{-k} = \frac{1}{0^k} to see division by zero occurs for all negative powers.
Examiner Trap Warning: fence_post_or_domain_trap: Assuming a⁰ = 1 holds at a = 0 without restriction, or forgetting that a^(-2) = 1/a² introduces a vertical asymptote at a = 0.
MM-GUIDE-EX-02Page 12MM1: Algebra and functionsSaves 95s
6-Card Lesson (ARCH-MM-01)Practice Matching Archetype Drill

Squaring and Re-Factorising Binomial Surds (a ± b√c)^2

Official UAT-UK Guide Exercise Box (Page 12):
Start with some random surd expressions (a±bc)2(a \pm b\sqrt{c})^2, square them and simplify them. Then look at the expressions and see if you can factorise them back to their original form.
Official Footnote: 4 The answer is not 2 - 3√5. Why not?
SymPy-Verified Result:
(a±bc)2=(a2+b2c)±2abc(a \pm b\sqrt{c})^2 = (a^2 + b^2 c) \pm 2ab\sqrt{c}; to reverse, write A±2B=∣u±v∣\sqrt{A \pm 2\sqrt{B}} = |\sqrt{u} \pm \sqrt{v}| where u+v=Au + v = A and uv=Buv = B.
Sub-90 Dry-Wipe Shortcut25s
Line 1 (Half-the-Surd-Coefficient Rule): Pull a factor of 22 outside the inner root so it reads A±2B\sqrt{A \pm 2\sqrt{B}}, then factor B=u×vB = u \times v with u+v=Au + v = A.
Line 2 (Principal Root Sign Lock): Write u±v\sqrt{u} \pm \sqrt{v} with the larger surd first so the result is strictly ≥0\ge 0.
Standard Classroom Derivation~120s
Expand (p+qr)2=(p2+q2r)+2pqr(p + q\sqrt{r})^2 = (p^2 + q^2 r) + 2pq\sqrt{r}, then set up simultaneous equations A=p2+q2rA = p^2 + q^2 r and B=2pqB = 2pq, substitute q=B2pq = \frac{B}{2p}, and solve the resulting quartic rp4−Ap2+B24=0r p^4 - A p^2 + \frac{B^2}{4} = 0.
Examiner Trap Warning: sign_or_direction_flip: Writing a negative value for a principal square root sqrt(...) when the second term b*sqrt(c) exceeds a.
MM-GUIDE-EX-03Page 12MM1: Algebra and functionsSaves 110s
6-Card Lesson (ARCH-MM-01)Practice Matching Archetype Drill

Principal Square Root Trap: What is √(49 - 12√5)?

Official UAT-UK Guide Exercise Box (Page 12):
What is 49−125\sqrt{49 - 12\sqrt{5}}? Be very careful — you might get the wrong answer (4^4 The answer is not 2−352 - 3\sqrt{5}. Why not?)!
Official Footnote: 4 The answer is not 2 - 3√5. Why not?
SymPy-Verified Result:
49−125=35−2\sqrt{49 - 12\sqrt{5}} = 3\sqrt{5} - 2 (since 35=45>4=23\sqrt{5} = \sqrt{45} > \sqrt{4} = 2, whereas 2−35<02 - 3\sqrt{5} < 0).
Sub-90 Dry-Wipe Shortcut25s
**Line 1 (Split 49=45+449 = 45 + 4):** Rewrite 125=2×2×35=244512\sqrt{5} = 2 \times 2 \times 3\sqrt{5} = 2\sqrt{4}\sqrt{45}, where 45+4=4945 + 4 = 49.
Line 2 (Positive Principal Root Check): Since 45>4\sqrt{45} > \sqrt{4}, (45−4)2=45−4=35−2>0\sqrt{(\sqrt{45} - \sqrt{4})^2} = \sqrt{45} - \sqrt{4} = 3\sqrt{5} - 2 > 0 (25s).
Standard Classroom Derivation~135s
Assume 49−125=a−b5\sqrt{49 - 12\sqrt{5}} = a - b\sqrt{5}, square both sides to obtain a2+5b2−2ab5=49−125a^2 + 5b^2 - 2ab\sqrt{5} = 49 - 12\sqrt{5}, equate rational and irrational parts (a2+5b2=49a^2 + 5b^2 = 49, ab=6ab = 6), find a=2,b=3a = 2, b = 3, and then check whether 2−352 - 3\sqrt{5} or 35−23\sqrt{5} - 2 is positive.
Examiner Trap Warning: sign_or_direction_flip: Selecting 2 - 3√5, forgetting that the radical symbol √x denotes the non-negative principal square root.
MM-GUIDE-EX-04Page 14MM1: Algebra and functionsSaves 130s
6-Card Lesson (ARCH-MM-01)Practice Matching Archetype Drill

Sum/Difference of Cubes a^3 ± b^3 & Cube-Root Rationalisation

Official UAT-UK Guide Exercise Box (Page 14):
What is the factorisation of a3−b3a^3 - b^3 and of a3+b3a^3 + b^3? Use these to rationalise the denominators in: (a) 7363+63+1\frac{7}{\sqrt[3]{36} + \sqrt[3]{6} + 1} and (b) x−27x3−3\frac{x - 27}{\sqrt[3]{x} - 3}.
SymPy-Verified Result:
a3∓b3=(a∓b)(a2±ab+b2)a^3 \mp b^3 = (a \mp b)(a^2 \pm ab + b^2); (a) 7(63−1)5\frac{7(\sqrt[3]{6} - 1)}{5}; (b) x2/3+3x1/3+9x^{2/3} + 3x^{1/3} + 9 (for xe27x e 27).
Sub-90 Dry-Wipe Shortcut35s
Line 1 (SOAP Identity): a3∓b3=(a∓b)(a2±ab+b2)a^3 \mp b^3 = (a \mp b)(a^2 \pm ab + b^2) (Same, Opposite, Always Positive).
Line 2 (Instant Collapse): In (a), a=63,b=1  ⟹  a = \sqrt[3]{6}, b = 1 \implies multiply by 63−163−1\frac{\sqrt[3]{6}-1}{\sqrt[3]{6}-1} to collapse denominator to (63)3−13=5  ⟹  7(63−1)5(\sqrt[3]{6})^3 - 1^3 = 5 \implies \frac{7(\sqrt[3]{6}-1)}{5}. In (b), a=x1/3,b=3  ⟹  a3−33a−3=a2+3a+9=x2/3+3x1/3+9a = x^{1/3}, b = 3 \implies \frac{a^3 - 3^3}{a - 3} = a^2 + 3a + 9 = x^{2/3} + 3x^{1/3} + 9 (35s).
Standard Classroom Derivation~165s
Perform algebraic long division of a3−b3a^3 - b^3 by a−ba - b and a3+b3a^3 + b^3 by a+ba + b, then multiply numerator and denominator of (a) by (63−1)(\sqrt[3]{6} - 1) and expand all nine terms in the denominator.
Examiner Trap Warning: sign_or_direction_flip: Writing +ab instead of -ab in a³ + b³ = (a + b)(a² - ab + b²), or including 2ab as if expanding a square.
MM-GUIDE-EX-05Page 17MM1: Algebra and functionsSaves 85s
6-Card Lesson (ARCH-MM-02)Practice Matching Archetype Drill

How Changing b and c Moves the Parabola y = x^2 + bx + c

Official UAT-UK Guide Exercise Box (Page 17):
How does changing the values of bb and cc in y=x2+bx+cy = x^2 + bx + c affect the position of the graph? Justify your findings by completing the square.
SymPy-Verified Result:
y=(x+b2)2+(c−b24)y = \left(x + \frac{b}{2}\right)^2 + \left(c - \frac{b^2}{4}\right): changing cc translates the graph vertically by Δc\Delta c; changing bb slides the vertex (−b2,c−b24)\left(-\frac{b}{2}, c - \frac{b^2}{4}\right) along the inverted parabola y=c−x2y = c - x^2.
Sub-90 Dry-Wipe Shortcut25s
Line 1 (Completed Square Lock): y=(x+b2)2+c−b24  ⟹  y = \left(x + \frac{b}{2}\right)^2 + c - \frac{b^2}{4} \implies Vertex is (−b2,c−b24)\left(-\frac{b}{2}, c - \frac{b^2}{4}\right).
Line 2 (Locus Elimination): Eliminating b=−2xvb = -2x_v gives yv=c−xv2y_v = c - x_v^2: varying bb moves the minimum along a downward parabola through (0,c)(0, c)!
Standard Classroom Derivation~110s
Differentiate dydx=2x+b=0\frac{dy}{dx} = 2x + b = 0 to find xv=−b2x_v = -\frac{b}{2}, substitute back to get yv=b24−b22+c=c−b24y_v = \frac{b^2}{4} - \frac{b^2}{2} + c = c - \frac{b^2}{4}, and plot multiple curves.
Examiner Trap Warning: sign_or_direction_flip: Thinking increasing b > 0 shifts the parabola to the right (+b/2) instead of to the left (-b/2) and downwards (-b²/4).
MM-GUIDE-EX-06Page 20MM1: Algebra and functionsSaves 110s
6-Card Lesson (ARCH-MM-02)Practice Matching Archetype Drill

Reciprocal Quadratic Formula x = 2c / (-b ± √(b^2 - 4ac)) & Domain Condition c ≠ 0

Official UAT-UK Guide Exercise Box (Page 20):
Start with ax2+bx+c=0ax^2 + bx + c = 0, divide through by x2x^2 to get c(1x)2+b(1x)+a=0c\left(\frac{1}{x}\right)^2 + b\left(\frac{1}{x}\right) + a = 0, and show x=2c−b±b2−4acx = \frac{2c}{-b \pm \sqrt{b^2 - 4ac}}. Are there any conditions on the equivalence (e.g., can we have c=0c = 0)?
SymPy-Verified Result:
Rationalising 2c−b±b2−4ac\frac{2c}{-b \pm \sqrt{b^2 - 4ac}} yields −b∓b2−4ac2a\frac{-b \mp \sqrt{b^2 - 4ac}}{2a}; strictly requires **ae0a e 0 and ce0c e 0** (since c=0  ⟹  x=0c = 0 \implies x = 0 is a root, making 1x\frac{1}{x} and 00\frac{0}{0} undefined).
Sub-90 Dry-Wipe Shortcut30s
Line 1 (Conjugate Difference-of-Squares): Multiply 2c−b±Δ\frac{2c}{-b \pm \sqrt{\Delta}} by −b∓Δ−b∓Δ  ⟹  2c(−b∓Δ)b2−(b2−4ac)=−b∓Δ2a\frac{-b \mp \sqrt{\Delta}}{-b \mp \sqrt{\Delta}} \implies \frac{2c(-b \mp \sqrt{\Delta})}{b^2 - (b^2 - 4ac)} = \frac{-b \mp \sqrt{\Delta}}{2a}.
Line 2 (Zero-Root Trap Check): Dividing by x2x^2 assumes xe0x e 0, which requires **ce0c e 0** (if c=0c = 0, one root yields 00\frac{0}{0}!).
Standard Classroom Derivation~140s
Let u=1xu = \frac{1}{x}, solve cu2+bu+a=0cu^2 + bu + a = 0 via the quadratic formula u=−b±b2−4ac2cu = \frac{-b \pm \sqrt{b^2 - 4ac}}{2c}, invert to get x=1ux = \frac{1}{u}, and rationalise the denominator to compare with the standard formula.
Examiner Trap Warning: fence_post_or_domain_trap: Dividing an equation by x² without checking if x = 0 (which happens precisely when the constant term c = 0).
MM-GUIDE-EX-07Page 41MM1: Algebra and functionsSaves 90s
6-Card Lesson (ARCH-MM-01)Practice Matching Archetype Drill

Polynomial Division Degree Invariants: f(x) = g(x)h(x) + r(x)

Official UAT-UK Guide Exercise Box (Page 41):
Here is a general expression for dividing some polynomial f(x)f(x) by a polynomial h(x)h(x): f(x)=g(x)h(x)+r(x)f(x) = g(x)h(x) + r(x). What can you say about the degree of h(x),g(x)h(x), g(x) and r(x)r(x) in this expression? Be very careful about how you express your answer.
SymPy-Verified Result:
When deg⁡(f)≥deg⁡(h)≥1\deg(f) \ge \deg(h) \ge 1: deg⁡(g)=deg⁡(f)−deg⁡(h)\deg(g) = \deg(f) - \deg(h), and **either r(x)≡0r(x) \equiv 0 (the zero polynomial) or 0≤deg⁡(r)<deg⁡(h)0 \le \deg(r) < \deg(h)** (if deg⁡(f)<deg⁡(h)\deg(f) < \deg(h), then g(x)≡0g(x) \equiv 0 and r(x)=f(x)r(x) = f(x)).
Sub-90 Dry-Wipe Shortcut25s
Line 1 (Quotient Degree Difference): deg⁡(g)=deg⁡(f)−deg⁡(h)\deg(g) = \deg(f) - \deg(h) (for deg⁡(f)≥deg⁡(h)\deg(f) \ge \deg(h)).
Line 2 (Remainder Strict Inequality): Never write deg⁡(r)=deg⁡(h)−1\deg(r) = \deg(h) - 1; always write **deg⁡(r)≤deg⁡(h)−1\deg(r) \le \deg(h) - 1 (or r(x)≡0r(x) \equiv 0)** because intermediate remainder terms can cancel!
Standard Classroom Derivation~115s
Perform long division on several sample polynomials of degrees 3, 4, and 5 and compare the highest powers of xx in the quotient and remainder.
Examiner Trap Warning: multi_statement_linchpin_foil: Claiming that dividing by a cubic h(x) (degree 3) must always leave a quadratic remainder (degree 2), forgetting it can be linear, constant, or zero.
MM-GUIDE-EX-08Page 50MM2: Sequences and seriesSaves 80s
6-Card Lesson (ARCH-MM-05)Practice Matching Archetype Drill

Linear Combinations of Arithmetic Sequences: α u_n + β v_n + γ w_n

Official UAT-UK Guide Exercise Box (Page 50):
We could also look at any linear combination of two arithmetic sequences: αun+βvn\alpha u_n + \beta v_n; verify these are also arithmetic sequences. Does this work for linear sums of more than two sequences (e.g., αun+βvn+γwn\alpha u_n + \beta v_n + \gamma w_n)?
SymPy-Verified Result:
Yes, for any finite number of APs: xn=∑k=1mαkun(k)x_n = \sum_{k=1}^m \alpha_k u_{n}^{(k)} is an AP with first term A=∑αkakA = \sum \alpha_k a_k and constant common difference D=∑αkdkD = \sum \alpha_k d_k.
Sub-90 Dry-Wipe Shortcut20s
**Line 1 (Linearity of Difference Operator Δ\Delta):** Δ(αun+βvn+γwn)=αd1+βd2+γd3=constant\Delta(\alpha u_n + \beta v_n + \gamma w_n) = \alpha d_1 + \beta d_2 + \gamma d_3 = \text{constant}.
Line 2 (Instant Parameters): First term is αu1+βv1+γw1\alpha u_1 + \beta v_1 + \gamma w_1; common difference is αd1+βd2+γd3\alpha d_1 + \beta d_2 + \gamma d_3 (20s).
Standard Classroom Derivation~100s
Write un=a1+(n−1)d1u_n = a_1 + (n-1)d_1 and vn=a2+(n−1)d2v_n = a_2 + (n-1)d_2, expand αun+βvn=(αa1+βa2)+(n−1)(αd1+βd2)\alpha u_n + \beta v_n = (\alpha a_1 + \beta a_2) + (n-1)(\alpha d_1 + \beta d_2), and compute xn+1−xnx_{n+1} - x_n.
Examiner Trap Warning: subsystem_isolation_trap: Confusing linear combinations of APs (which remain APs) with products u_n * v_n (which become quadratic sequences) or sums of GPs (which are generally NOT GPs unless ratios match).
MM-GUIDE-EX-09Page 54MM2: Sequences and seriesSaves 105s
6-Card Lesson (ARCH-MM-05)Practice Matching Archetype Drill

Three Equivalent Methods for Summing an Arithmetic Progression (Fence-Post Check)

Official UAT-UK Guide Exercise Box (Page 54):
For each of the three summation methods in the guide (Reverse-and-Add, Index Shifting ∑r=mnur\sum_{r=m}^n u_r, and Subtracting Prefix Sums Sn−Sm−1S_n - S_{m-1}), work out the sum and verify all three methods give you the same answer.
SymPy-Verified Result:
∑r=mnur=n−m+12(um+un)=Sn−Sm−1\sum_{r=m}^{n} u_r = \frac{n - m + 1}{2}(u_m + u_n) = S_n - S_{m-1}, where the inclusive term count is strictly N=n−m+1N = n - m + 1.
Sub-90 Dry-Wipe Shortcut25s
Line 1 (Fence-Post Term Count): Inclusive terms from r=mr = m to r=nr = n is **N=n−m+1N = n - m + 1** (never n−mn - m!).
**Line 2 (Average ×\times Count):** Sum=(First Term um+Last Term un2)×(n−m+1)\text{Sum} = \left(\frac{\text{First Term } u_m + \text{Last Term } u_n}{2}\right) \times (n - m + 1) (25s).
Standard Classroom Derivation~130s
Evaluate Sn=n2(2a+(n−1)d)S_n = \frac{n}{2}(2a + (n-1)d), subtract Sm−1=m−12(2a+(m−2)d)S_{m-1} = \frac{m-1}{2}(2a + (m-2)d), and factor the resulting quadratic in nn and mm.
Examiner Trap Warning: fence_post_or_domain_trap: Subtracting S_n - S_m instead of S_n - S_{m-1}, or using (n - m) terms instead of (n - m + 1) terms.
MM-GUIDE-EX-10Page 65MM3: Coordinate geometry in the (x, y) planeSaves 110s
6-Card Lesson (ARCH-MM-07)Practice Matching Archetype Drill

Perpendicular Gradients m_1 m_2 = -1 & Condition for Lines Meeting at Angle φ (45° or 60°)

Official UAT-UK Guide Exercise Box (Page 65):
Use m=tan⁡θm = \tan\theta to explain why m1m2=−1m_1 m_2 = -1 if and only if the lines are perpendicular (excluding vertical lines). What is the condition on m1m_1 and m2m_2 when two lines meet at 45∘45^\circ or 60∘60^\circ?
SymPy-Verified Result:
For perpendicular lines (θ2=θ1±90∘\theta_2 = \theta_1 \pm 90^\circ), m2=tan⁡(θ1+90∘)=−cot⁡θ1=−1m1  ⟺  m1m2=−1m_2 = \tan(\theta_1 + 90^\circ) = -\cot\theta_1 = -\frac{1}{m_1} \iff m_1 m_2 = -1. For acute angle ϕ\phi, ∣m2−m11+m1m2∣=tan⁡ϕ\left|\frac{m_2 - m_1}{1 + m_1 m_2}\right| = \tan\phi (=1= 1 for 45∘45^\circ, =3= \sqrt{3} for 60∘60^\circ).
Sub-90 Dry-Wipe Shortcut30s
Line 1 (90° Triangle Rotation): Rotating displacement (Δx,Δy)=(1,m1)(\Delta x, \Delta y) = (1, m_1) by 90∘90^\circ gives (−m1,1)  ⟹  m2=1−m1  ⟹  m1m2=−1(-m_1, 1) \implies m_2 = \frac{1}{-m_1} \implies m_1 m_2 = -1.
**Line 2 (General Angle ϕ\phi Formula):** tan⁡ϕ=∣m2−m11+m1m2∣  ⟹  ∣m2−m1∣=∣1+m1m2∣\tan\phi = \left|\frac{m_2 - m_1}{1 + m_1 m_2}\right| \implies |m_2 - m_1| = |1 + m_1 m_2| for 45∘45^\circ, and ∣m2−m1∣=3∣1+m1m2∣|m_2 - m_1| = \sqrt{3}|1 + m_1 m_2| for 60∘60^\circ.
Standard Classroom Derivation~140s
Construct right-angled triangles on the coordinate axes, rotate (1,m1)(1, m_1) by 90∘90^\circ to (−m1,1)(-m_1, 1), and compute the ratio of rise over run.
Examiner Trap Warning: sign_or_direction_flip: Writing m₁ * m₂ = +1 (reciprocal without negation) or forgetting the modulus bars yielding both acute and obtuse intersections.
MM-GUIDE-EX-11Page 66MM3: Coordinate geometry in the (x, y) planeSaves 80s
6-Card Lesson (ARCH-MM-07)Practice Matching Archetype Drill

Deriving y - y_1 = m(x - x_1) via Sequential Graph Transformations of y = x

Official UAT-UK Guide Exercise Box (Page 66):
Describe what each transformation does to the graph of y=xy = x: (a) y=mxy = mx; (b) y−y1=xy - y_1 = x; (c) y=x−x1y = x - x_1; (d) y−y1=mxy - y_1 = mx; (e) y=m(x−x1)y = m(x - x_1); (f) y−y1=m(x−x1)y - y_1 = m(x - x_1).
SymPy-Verified Result:
(a) Vertical stretch by mm; (b) Shift up by y1y_1; (c) Shift right by x1x_1; (d) Stretch by mm then shift up by y1y_1; (e) Stretch by mm then shift right by x1x_1; (f) Stretch by mm then translate (0,0)→(x1,y1)(0,0) \to (x_1, y_1).
Sub-90 Dry-Wipe Shortcut20s
Line 1 (Anchor Point Tracking): y=mxy = mx passes through (0,0)(0,0) with gradient mm.
**Line 2 (Translation Vector [x1,y1]T[x_1, y_1]^T):** Replacing x→(x−x1)x \to (x - x_1) and y→(y−y1)y \to (y - y_1) translates the anchor point (0,0)→(x1,y1)(0,0) \to (x_1, y_1) without changing gradient mm, giving y−y1=m(x−x1)y - y_1 = m(x - x_1) directly!
Standard Classroom Derivation~100s
Rearrange each equation into y=mx+cy = mx + c and calculate the xx- and yy-intercepts separately.
Examiner Trap Warning: sign_or_direction_flip: Expanding m(x - x₁) as mx - x₁ (forgetting to multiply -x₁ by m in the y-intercept c = y₁ - m*x₁).
MM-GUIDE-EX-12Page 67MM3: Coordinate geometry in the (x, y) planeSaves 75s
6-Card Lesson (ARCH-MM-07)Practice Matching Archetype Drill

Finding Equations of Lines from Gradient & Point: 3 Worked Cases

Official UAT-UK Guide Exercise Box (Page 67):
Find the equation of the line with the following gradients and points: (a) m=4m = 4, point (3,2)(3, 2); (b) m=−5m = -5, point (5,3)(5, 3); (c) m=−2m = -2, point (−2,−4)(-2, -4).
Official Footnote: 42 With line questions it is very easy to make sign errors when putting coordinates into equations. Make sure you are careful!
SymPy-Verified Result:
(a) y=4x−10y = 4x - 10; (b) y=−5x+28y = -5x + 28 (or 5x+y=285x + y = 28); (c) y=−2x−8y = -2x - 8 (or 2x+y+8=02x + y + 8 = 0).
Sub-90 Dry-Wipe Shortcut20s
**Line 1 (Direct Standard Form mx−y=mx1−y1mx - y = mx_1 - y_1):** Plug (x1,y1)(x_1, y_1) directly into c=y1−mx1c = y_1 - mx_1:
- (a) c=2−4(3)=−10  ⟹  y=4x−10c = 2 - 4(3) = -10 \implies y = 4x - 10.
- (b) c=3−(−5)(5)=28  ⟹  y=−5x+28c = 3 - (-5)(5) = 28 \implies y = -5x + 28.
- (c) c=−4−(−2)(−2)=−8  ⟹  y=−2x−8c = -4 - (-2)(-2) = -8 \implies y = -2x - 8 (20s).
Standard Classroom Derivation~95s
Substitute into y−y1=m(x−x1)y - y_1 = m(x - x_1) for each part: (a) y−2=4(x−3)y - 2 = 4(x - 3); (b) y−3=−5(x−5)y - 3 = -5(x - 5); (c) y−(−4)=−2(x−(−2))y - (-4) = -2(x - (-2)), then expand brackets and collect constants.
Examiner Trap Warning: sign_or_direction_flip: In part (c), writing -4 - (-2)(-2) = 0 instead of -4 - 4 = -8 due to triple negative signs.
MM-GUIDE-EX-13Page 68MM3: Coordinate geometry in the (x, y) planeSaves 90s
6-Card Lesson (ARCH-MM-07)Practice Matching Archetype Drill

Two-Point Line Equations & Zero-Work Origin/Horizontal Shortcuts

Official UAT-UK Guide Exercise Box (Page 68):
Find the equation of the line through the following pairs of points: (a) (0,0)(0, 0) and (2,3)(2, 3) [is there a shortcut here?]; (b) (−2,5)(-2, 5) and (−4,−7)(-4, -7); (c) (3,−7)(3, -7) and (8,−7)(8, -7).
SymPy-Verified Result:
(a) y=32xy = \frac{3}{2}x (or 3x−2y=03x - 2y = 0); (b) y=6x+17y = 6x + 17; (c) y=−7y = -7 (horizontal line, m=0m = 0).
Sub-90 Dry-Wipe Shortcut20s
Line 1 (Inspect Coordinates Before Calculating):
- (a) Passes through (0,0)  ⟹  c=0(0,0) \implies c = 0 immediately: y=32xy = \frac{3}{2}x (2s).
- (b) m=−7−5−4−(−2)=−12−2=6  ⟹  c=5−6(−2)=17  ⟹  y=6x+17m = \frac{-7 - 5}{-4 - (-2)} = \frac{-12}{-2} = 6 \implies c = 5 - 6(-2) = 17 \implies y = 6x + 17.
- (c) Both points have y=−7  ⟹  y = -7 \implies horizontal line y=−7y = -7 (2s)!
Standard Classroom Derivation~110s
Compute m=y2−y1x2−x1m = \frac{y_2 - y_1}{x_2 - x_1} and substitute into y−y1=m(x−x1)y - y_1 = m(x - x_1) for all three pairs of coordinates.
Examiner Trap Warning: intermediate_stop_trap: Spending 30 seconds substituting (3, -7) and (8, -7) into the gradient formula when identical y-coordinates immediately give y = -7.
MM-GUIDE-EX-14Page 73MM3: Coordinate geometry in the (x, y) planeSaves 105s
6-Card Lesson (ARCH-MM-08)Practice Matching Archetype Drill

Why Equations Fail to Be Circles & Necessary/Sufficient Conditions on px^2 + qy^2 + ax + by + c = 0

Official UAT-UK Guide Exercise Box (Page 73):
Why are (1) 2x2+3y2−4x−8y−19=02x^2 + 3y^2 - 4x - 8y - 19 = 0, (2) 2x2−2y2−4x−8y−19=02x^2 - 2y^2 - 4x - 8y - 19 = 0, and (3) x2+2y+5=y2+4x+7x^2 + 2y + 5 = y^2 + 4x + 7 not equations of circles? Given px2+qy2+ax+by+c=0px^2 + qy^2 + ax + by + c = 0, what conditions on p,q,a,b,cp, q, a, b, c make it a circle?
SymPy-Verified Result:
(1) Unequal coefficients 2e32 e 3 (ellipse); (2) & (3) Opposite signs p=−qp = -q (hyperbolas). General conditions: **p=qe0p = q e 0**, no xyxy term, and **a2+b2−4pc>0a^2 + b^2 - 4pc > 0** (ensuring strictly positive radius R2>0R^2 > 0).
Sub-90 Dry-Wipe Shortcut30s
Line 1 (Coefficient Match): Bring all terms to LHS: must have p=qe0p = q e 0 (kills (1) 2e32 e 3, (2) 2e−22 e -2, and (3) x2−y2x^2 - y^2).
Line 2 (Strict Positive Radius Discriminant): Completing the square gives R2=a2+b2−4pc4p2>0  ⟺  R^2 = \frac{a^2 + b^2 - 4pc}{4p^2} > 0 \iff **a2+b2−4pc>0a^2 + b^2 - 4pc > 0**!
Standard Classroom Derivation~135s
Divide px2+py2+ax+by+c=0px^2 + py^2 + ax + by + c = 0 by pp and complete the square in both xx and yy to find the centre and radius.
Examiner Trap Warning: multi_statement_linchpin_foil: Stating only p = q != 0 and forgetting that if a² + b² - 4pc <= 0, the locus is a single point (R = 0) or the empty set (R² < 0), NOT a circle!
MM-GUIDE-EX-15Page 76MM3: Coordinate geometry in the (x, y) planeSaves 130s
6-Card Lesson (ARCH-MM-08)Practice Matching Archetype Drill

Shortest Distance Between Line y = 2x + 15 and Circle (x + 2)^2 + (y - 3)^2 = 9

Official UAT-UK Guide Exercise Box (Page 76):
Find the shortest length between the line y=2x+15y = 2x + 15 and the circle (x+2)2+(y−3)2=9(x + 2)^2 + (y - 3)^2 = 9. Try to solve it using a number of different methods. Which method do you think is best?
SymPy-Verified Result:
dmin⁡=855−3≈0.578d_{\min} = \frac{8\sqrt{5}}{5} - 3 \approx 0.578 (Centre (−2,3)(-2, 3) to line 2x−y+15=02x - y + 15 = 0 is 85\frac{8}{\sqrt{5}}, minus radius R=3R = 3).
Sub-90 Dry-Wipe Shortcut35s
Line 1 (Point-to-Line Distance Formula): Distance from centre (−2,3)(-2, 3) to 2x−y+15=02x - y + 15 = 0 is D=∣2(−2)−1(3)+15∣22+(−1)2=85=855D = \frac{|2(-2) - 1(3) + 15|}{\sqrt{2^2 + (-1)^2}} = \frac{8}{\sqrt{5}} = \frac{8\sqrt{5}}{5}.
**Line 2 (Subtract Radius R=9=3R = \sqrt{9} = 3):** dmin⁡=855−3d_{\min} = \frac{8\sqrt{5}}{5} - 3 (saves 130s!).
Standard Classroom Derivation~165s
Write the normal line through (−2,3)(-2, 3) with gradient −12-\frac{1}{2}: y−3=−12(x+2)  ⟹  y=−12x+2y - 3 = -\frac{1}{2}(x + 2) \implies y = -\frac{1}{2}x + 2. Intersect with y=2x+15  ⟹  52x=−13  ⟹  x=−265,y=235y = 2x + 15 \implies \frac{5}{2}x = -13 \implies x = -\frac{26}{5}, y = \frac{23}{5}. Compute distance (−165)2+(85)2=855\sqrt{(-\frac{16}{5})^2 + (\frac{8}{5})^2} = \frac{8\sqrt{5}}{5} and subtract R=3R = 3.
Examiner Trap Warning: intermediate_stop_trap: Outputting the centre-to-line distance 8*sqrt(5)/5 without subtracting the circle radius R = 3 (or subtracting R² = 9 instead of R = 3).
MM-GUIDE-EX-16Page 82MM4: TrigonometrySaves 100s
6-Card Lesson (ARCH-MM-10)Practice Matching Archetype Drill

Ambiguous Case of the Sine Rule (SSA): Conditions for 0, 1, or 2 Triangles

Official UAT-UK Guide Exercise Box (Page 82):
Explore when the Sine Rule ambiguity arises: what are the conditions for there to be one answer, two answers, and no answers to a question giving two sides a,ba, b and non-included angle AA?
SymPy-Verified Result:
For acute AA with altitude h=bsin⁡Ah = b\sin A: (1) a<bsin⁡A  ⟹  a < b\sin A \implies 0 triangles; (2) a=bsin⁡Aa = b\sin A or a≥b  ⟹  a \ge b \implies 1 triangle; (3) bsin⁡A<a<b  ⟹  b\sin A < a < b \implies 2 triangles (B1B_1 acute and B2=180∘−B1B_2 = 180^\circ - B_1 obtuse). For A≥90∘A \ge 90^\circ: a>b  ⟹  a > b \implies 1 triangle, a≤b  ⟹  a \le b \implies 0 triangles.
Sub-90 Dry-Wipe Shortcut30s
**Line 1 (Perpendicular Altitude h=bsin⁡Ah = b\sin A):** Compare opposite side aa against h=bsin⁡Ah = b\sin A and adjacent side bb.
Line 2 (Swing-Arc Test): If bsin⁡A<a<bb\sin A < a < b, the arc of radius aa cuts the base twice (2 solutions); if a≥ba \ge b or a=bsin⁡Aa = b\sin A, 1 solution; if a<bsin⁡Aa < b\sin A, 0 solutions.
Standard Classroom Derivation~130s
Apply Cosine Rule a2=b2+c2−2bccos⁡Aa^2 = b^2 + c^2 - 2bc\cos A as a quadratic in cc: c2−(2bcos⁡A)c+(b2−a2)=0c^2 - (2b\cos A)c + (b^2 - a^2) = 0 and examine the discriminant Δ\Delta and sum/product of roots.
Examiner Trap Warning: fence_post_or_domain_trap: Forgetting the obtuse supplement B₂ = 180° - B₁ when b*sin(A) < a < b, or including B₂ when a >= b (where A + B₂ >= 180° is impossible).
MM-GUIDE-EX-17Page 83MM4: TrigonometrySaves 95s
6-Card Lesson (ARCH-MM-10)Practice Matching Archetype Drill

Circumcircle Proof of the Extended Sine Rule: a / sin A = b / sin B = c / sin C = 2R

Official UAT-UK Guide Exercise Box (Page 83):
Prove using the circumcircle of radius RR and circle theorems that 2Rsin⁡B=b2R\sin B = b, and hence 2R=asin⁡A=bsin⁡B=csin⁡C2R = \frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}. Also explain why the perpendicular bisectors of the sides meet at the circumcentre.
SymPy-Verified Result:
Perpendicular bisectors are the loci of points equidistant from vertices (OA=OBOA = OB and OB=OC  ⟹  OA=OB=OC=ROB = OC \implies OA = OB = OC = R). Drawing diameter A′C=2RA'C = 2R gives ∠A′BC=90∘\angle A'BC = 90^\circ (angle in semicircle) and ∠BA′C=B\angle BA'C = B (same segment), so sin⁡B=b2R  ⟹  bsin⁡B=2R\sin B = \frac{b}{2R} \implies \frac{b}{\sin B} = 2R.
Sub-90 Dry-Wipe Shortcut25s
Line 1 (Diameter Construction): Replace vertex AA with diameter endpoint A′A' so A′C=2RA'C = 2R and ∠A′BC=90∘\angle A'BC = 90^\circ (Thales' Theorem) while ∠A′=∠B\angle A' = \angle B.
Line 2 (Right-Triangle Readout): In △A′BC\triangle A'BC, sin⁡B=opphyp=b2R  ⟹  bsin⁡B=2R\sin B = \frac{\text{opp}}{\text{hyp}} = \frac{b}{2R} \implies \frac{b}{\sin B} = 2R directly!
Standard Classroom Derivation~120s
Drop perpendiculars from the circumcentre OO to each chord a,b,ca, b, c, use central angles 2A,2B,2C2A, 2B, 2C, and solve right-angled half-chord triangles.
Examiner Trap Warning: factor_of_two_or_stoichiometry_trap: Writing a / sin(A) = R (circumradius) instead of 2R (circumdiameter).
MM-GUIDE-EX-18Page 91MM4: TrigonometrySaves 110s
6-Card Lesson (ARCH-MM-09)Practice Matching Archetype Drill

Trig Graph Transformations, cos x = ±sin x Roots, and Phase Shift sin(2x ± π/6) vs sin 2(x ± π/6)

Official UAT-UK Guide Exercise Box (Page 91):
Sketch y=sin⁡x,2sin⁡x,sin⁡2xy = \sin x, 2\sin x, \sin 2x (and cos⁡,tan⁡\cos, \tan) on −720∘<x≤720∘-720^\circ < x \le 720^\circ and −2π<x≤2π-2\pi < x \le 2\pi. Find all x∈(−2π,2π]x \in (-2\pi, 2\pi] where cos⁡x=sin⁡x\cos x = \sin x and cos⁡x=−sin⁡x\cos x = -\sin x. Compare y=sin⁡(2x±π6)y = \sin(2x \pm \frac{\pi}{6}) with y=sin⁡2(x±π6)y = \sin 2(x \pm \frac{\pi}{6}).
Official Footnote: y = sin 2(x - π/6)
SymPy-Verified Result:
cos⁡x=sin⁡x  ⟹  x∈{−7π4,−3π4,π4,5π4}\cos x = \sin x \implies x \in \left\{-\frac{7\pi}{4}, -\frac{3\pi}{4}, \frac{\pi}{4}, \frac{5\pi}{4}\right\}; cos⁡x=−sin⁡x  ⟹  x∈{−5π4,−π4,3π4,7π4}\cos x = -\sin x \implies x \in \left\{-\frac{5\pi}{4}, -\frac{\pi}{4}, \frac{3\pi}{4}, \frac{7\pi}{4}\right\}. In sin⁡(2x±π6)=sin⁡2(x±π12)\sin(2x \pm \frac{\pi}{6}) = \sin 2(x \pm \frac{\pi}{12}), the horizontal phase shift is ∓π12\mp\frac{\pi}{12}, whereas in sin⁡2(x±π6)\sin 2(x \pm \frac{\pi}{6}) the shift is ∓π6\mp\frac{\pi}{6}.
Sub-90 Dry-Wipe Shortcut25s
Line 1 (Range Trap): −720∘<x≤720∘-720^\circ < x \le 720^\circ is **44 full periods**, whereas −2π<x≤2π-2\pi < x \le 2\pi is only **22 full periods**!
Line 2 (Factor Frequency Coefficient First): Always rewrite sin⁡(2x+π6)=sin⁡(2(x+π12))\sin(2x + \frac{\pi}{6}) = \sin\left(2\left(x + \frac{\pi}{12}\right)\right) so you see the horizontal translation is **−π12-\frac{\pi}{12}** (half of −π6-\frac{\pi}{6}!).
Standard Classroom Derivation~135s
Expand sin⁡(2x+π/6)\sin(2x + \pi/6) using addition formulas or plot point-by-point tables across [−2π,2π][-2\pi, 2\pi].
Examiner Trap Warning: factor_of_two_or_stoichiometry_trap: Reading the horizontal shift of sin(2x + pi/6) as -pi/6 instead of factoring out 2 to get -pi/12.
MM-GUIDE-EX-19Page 93MM4: TrigonometrySaves 80s
6-Card Lesson (ARCH-MM-09)Practice Matching Archetype Drill

Quadrant Sign Invariant of x tan θ Across All 4 Unit-Circle Quadrants

Official UAT-UK Guide Exercise Box (Page 93):
What is the sign of xtan⁡θx \tan\theta in each of the four quadrant diagrams? How does your answer fit with the sign of tan⁡θ\tan\theta when 90∘<θ<180∘90^\circ < \theta < 180^\circ and when 270∘<θ<360∘270^\circ < \theta < 360^\circ?
SymPy-Verified Result:
Since tan⁡θ=yx\tan\theta = \frac{y}{x}, xtan⁡θ=yx\tan\theta = y. Thus xtan⁡θ>0x\tan\theta > 0 in Quadrants I & II (0∘<θ<180∘,θe90∘0^\circ < \theta < 180^\circ, \theta e 90^\circ) and xtan⁡θ<0x\tan\theta < 0 in Quadrants III & IV (180∘<θ<360∘,θe270∘180^\circ < \theta < 360^\circ, \theta e 270^\circ).
Sub-90 Dry-Wipe Shortcut15s
Line 1 (Algebraic Cancellation): Substitute tan⁡θ=yx  ⟹  xtan⁡θ=x(yx)=y\tan\theta = \frac{y}{x} \implies x\tan\theta = x\left(\frac{y}{x}\right) = y (for xe0x e 0).
**Line 2 (Sign of yy):** In Q2 (90∘<θ<180∘90^\circ < \theta < 180^\circ), x<0x < 0 and tan⁡θ<0\tan\theta < 0, so (−)(−)=+y>0(-)(-) = +y > 0; in Q4 (270∘<θ<360∘270^\circ < \theta < 360^\circ), x>0x > 0 and tan⁡θ<0\tan\theta < 0, so (+)(−)=+y<0(+)(-) = +y < 0 (15s)!
Standard Classroom Derivation~95s
Make a 4-row CAST table listing the sign of xx, the sign of yy, the sign of tan⁡θ=y/x\tan\theta = y/x, and the product x⋅tan⁡θx \cdot \tan\theta in each quadrant.
Examiner Trap Warning: sign_or_direction_flip: Confusing the sign of tan(theta) with the sign of x*tan(theta) = y in Quadrant II and Quadrant III.
MM-GUIDE-EX-20Page 98MM4: TrigonometrySaves 125s
6-Card Lesson (ARCH-MM-09)Practice Matching Archetype Drill

Solving sin^2(2x + π/3) = 1/4 in Radians for -2π < x < 2π

Official UAT-UK Guide Exercise Box (Page 98):
Repeat the guide's degree solution in radians: solve sin⁡2(2x+π3)=14\sin^2\left(2x + \frac{\pi}{3}\right) = \frac{1}{4} for −2π<x<2π-2\pi < x < 2\pi.
SymPy-Verified Result:
16 solutions given by x=−π12+kπ2x = -\frac{\pi}{12} + \frac{k\pi}{2} and x=−π4+kπ2x = -\frac{\pi}{4} + \frac{k\pi}{2} for k∈{−3,−2,−1,0,1,2,3,4}k \in \{-3, -2, -1, 0, 1, 2, 3, 4\}, i.e., x∈{−7π4,−19π12,−5π4,−13π12,−3π4,−7π12,−π4,−π12,π4,5π12,3π4,11π12,5π4,17π12,7π4,23π12}x \in \left\{-\frac{7\pi}{4}, -\frac{19\pi}{12}, -\frac{5\pi}{4}, -\frac{13\pi}{12}, -\frac{3\pi}{4}, -\frac{7\pi}{12}, -\frac{\pi}{4}, -\frac{\pi}{12}, \frac{\pi}{4}, \frac{5\pi}{12}, \frac{3\pi}{4}, \frac{11\pi}{12}, \frac{5\pi}{4}, \frac{17\pi}{12}, \frac{7\pi}{4}, \frac{23\pi}{12}\right\}.
Sub-90 Dry-Wipe Shortcut35s
Line 1 (Root-Count Shortcut): sin⁡2u=14\sin^2 u = \frac{1}{4} has 22 roots per π\pi period of uu. Since u=2x+π3u = 2x + \frac{\pi}{3} spans 8π8\pi over x∈(−2π,2π)x \in (-2\pi, 2\pi), there are immediately **8×2=168 \times 2 = 16 roots**!
**Line 2 (Base Pair + π2\frac{\pi}{2} Step):** In u∈[0,π)u \in [0, \pi), u=π6,5π6  ⟹  2x=−π6,π2  ⟹  x0=−π12,π4u = \frac{\pi}{6}, \frac{5\pi}{6} \implies 2x = -\frac{\pi}{6}, \frac{\pi}{2} \implies x_0 = -\frac{\pi}{12}, \frac{\pi}{4}, then add multiples of π2\frac{\pi}{2} across (−2π,2π)(-2\pi, 2\pi).
Standard Classroom Derivation~160s
Take sin⁡(2x+π3)=±12\sin(2x + \frac{\pi}{3}) = \pm\frac{1}{2}, list all 16 values of u=2x+π3u = 2x + \frac{\pi}{3} in (−4π+π3,4π+π3)(-4\pi + \frac{\pi}{3}, 4\pi + \frac{\pi}{3}), subtract π3\frac{\pi}{3}, and divide each by 22.
Examiner Trap Warning: factor_of_two_or_stoichiometry_trap: Forgetting the negative root sin(2x + pi/3) = -1/2 when taking the square root of sin²(...) = 1/4, halving the root count from 16 to 8.
MM-GUIDE-EX-21Page 102MM5: Exponentials and logarithmsSaves 35s
6-Card Lesson (ARCH-MM-06)Practice Matching Archetype Drill

Exact Logarithm Evaluation: log_5(5^2), log_3(3^5), and log_7(√7)

Official UAT-UK Guide Exercise Box (Page 102):
Work out each of the following: (a) log⁡5(52)\log_5(5^2); (b) log⁡3(35)\log_3(3^5); (c) log⁡7(7)\log_7(\sqrt{7}).
Official Footnote: 53 Use a graph package to draw y = 2^x and y = 10^x to see how fast they grow [that is exponential growth]
SymPy-Verified Result:
(a) log⁡5(52)=2\log_5(5^2) = 2; (b) log⁡3(35)=5\log_3(3^5) = 5; (c) log⁡7(71/2)=12\log_7(7^{1/2}) = \frac{1}{2}.
Sub-90 Dry-Wipe Shortcut10s
**Line 1 (Inverse Cancellation log⁡b(bk)=k\log_b(b^k) = k):** Rewrite radicals as fractional powers (7=71/2\sqrt{7} = 7^{1/2}) and read off the exponent directly: 22, 55, and 12\frac{1}{2} (10s).
Standard Classroom Derivation~45s
Let y=log⁡a(ak)y = \log_a(a^k), convert to exponential form ay=aka^y = a^k, and equate powers.
Examiner Trap Warning: unit_or_prefix_mismatch: Writing log_7(sqrt(7)) = 2 instead of 1/2 by confusing 7^(1/2) with 7².
MM-GUIDE-EX-22Page 125MM7: IntegrationSaves 130s
6-Card Lesson (ARCH-MM-11)Practice Matching Archetype Drill

Definite Integrals via Odd/Even & Translational Symmetry (Plus the tan x Singularity Trap!)

Official UAT-UK Guide Exercise Box (Page 125):
Using symmetry ideas and your knowledge of integration, explain why: (1) ∫02πcos⁡x dx=0\int_0^{2\pi} \cos x\,dx = 0; (2) ∫02πsin⁡x dx=0\int_0^{2\pi} \sin x\,dx = 0; (3) ∫−ππsin⁡x dx=0\int_{-\pi}^{\pi} \sin x\,dx = 0; (4) ∫−ππtan⁡x dx=0\int_{-\pi}^{\pi} \tan x\,dx = 0; (5) ∫−1010x3 dx=0\int_{-10}^{10} x^3\,dx = 0; (6) ∫−ππx2 dx=2∫0πx2 dx\int_{-\pi}^{\pi} x^2\,dx = 2\int_0^{\pi} x^2\,dx; (7) ∫0πsin⁡x dx=−∫π2πsin⁡x dx\int_0^{\pi} \sin x\,dx = -\int_{\pi}^{2\pi} \sin x\,dx.
Official Footnote: 69 Although trigonometric integration is not on the TMUA/ESAT specification, all these examples are expressions we could expect you to deduce within the TMUA/ESAT specification.
SymPy-Verified Result:
(1) & (2) Equal positive and negative lobes over one full 2π2\pi period cancel to 00; (3) & (5) Odd functions (f(−x)=−f(x)f(-x) = -f(x)) over symmetric intervals [−a,a][-a, a] integrate to 00; (6) Even function (x2x^2) over [−π,π][-\pi, \pi] doubles [0,π][0, \pi]; (7) sin⁡(x+π)=−sin⁡x\sin(x + \pi) = -\sin x flips the lobe sign; (4) Caveat: tan⁡x\tan x is odd, so its Cauchy Principal Value is 00, though it has vertical asymptotes at x=±π2x = \pm\frac{\pi}{2}.
Sub-90 Dry-Wipe Shortcut20s
Line 1 (Parity & Periodicity Laws — Zero Calculus Needed):
- **Odd (f(−x)=−f(x)f(-x) = -f(x)) on [−a,a][-a,a]:** ∫−aaf(x) dx=0\int_{-a}^a f(x)\,dx = 0 immediately ((3), (4), (5)).
- **Even (f(−x)=f(x)f(-x) = f(x)) on [−a,a][-a,a]:** ∫−aaf(x) dx=2∫0af(x) dx\int_{-a}^a f(x)\,dx = 2\int_0^a f(x)\,dx ((6)).
- **Half-Period Anti-Symmetry (f(x+π)=−f(x)f(x+\pi) = -f(x)):** Lobe on [π,2π][\pi, 2\pi] is the exact negative of [0,π][0, \pi] ((1), (2), (7)).
Standard Classroom Derivation~150s
Attempt to anti-differentiate each integrand and substitute the upper and lower limits.
Examiner Trap Warning: sign_or_direction_flip: Confusing signed definite integral (where area below the x-axis is negative and cancels) with total geometric area bounded by the curve.
MM-GUIDE-EX-23Page 144MM8: Graphs of functionsSaves 65s
6-Card Lesson (ARCH-MM-04)Practice Matching Archetype Drill

Vertical Scaling & Reflection: y = a f(x) for x^2 and cos x

Official UAT-UK Guide Exercise Box (Page 144):
Explore y=f(x)y = f(x) and y=af(x)y = af(x) for f(x)=x2f(x) = x^2 and f(x)=cos⁡xf(x) = \cos x with a=2,3,−1,−2,12,−12a = 2, 3, -1, -2, \frac{1}{2}, -\frac{1}{2}. What do you notice when 0<a<10 < a < 1 and when a<0a < 0?
SymPy-Verified Result:
When 0<a<10 < a < 1, the graph is compressed vertically towards the xx-axis by scale factor aa (all xx-intercepts remain invariant). When a<0a < 0, the graph is reflected in the xx-axis and scaled vertically by ∣a∣|a|.
Sub-90 Dry-Wipe Shortcut15s
Line 1 (Outside Multiplier = Vertical Stretch + Fixed Roots): y=af(x)y = af(x) multiplies every yy-coordinate by aa while keeping every root f(x)=0f(x) = 0 pinned on the xx-axis!
Line 2 (Sign & Magnitude): 0<a<10 < a < 1 squashes vertically; a<0a < 0 flips across the xx-axis (turning maxima into minima).
Standard Classroom Derivation~80s
Compute tables of (x,af(x))(x, af(x)) values and plot each curve individually.
Examiner Trap Warning: sign_or_direction_flip: Thinking y = a*f(x) changes the period or x-intercepts of cos(x); roots stay fixed at x = +-pi/2, +-3pi/2.
MM-GUIDE-EX-24Page 145MM8: Graphs of functionsSaves 70s
6-Card Lesson (ARCH-MM-04)Practice Matching Archetype Drill

Vertical Translation y = f(x) + a & Trigonometric Power/Inverse Notation

Official UAT-UK Guide Exercise Box (Page 145):
Explore y=f(x)+ay = f(x) + a for f(x)=x2f(x) = x^2 and f(x)=cos⁡xf(x) = \cos x with a=2,3,−1,−2,12,−12a = 2, 3, -1, -2, \frac{1}{2}, -\frac{1}{2}. What happens when a<0a < 0? Contrast the notations a+cos⁡xa + \cos x, cos⁡(x+a)\cos(x + a), cos⁡2x\cos^2 x, cos⁡(x2)\cos(x^2), and cos⁡−1x\cos^{-1} x vs sec⁡x\sec x.
SymPy-Verified Result:
When a<0a < 0, y=f(x)+ay = f(x) + a translates the graph vertically downwards by ∣a∣|a| units. Notation rules: a+cos⁡x=(cos⁡x)+aa + \cos x = (\cos x) + a (vertical shift), cos⁡2x=(cos⁡x)2\cos^2 x = (\cos x)^2, cos⁡(x2)\cos(x^2) squares the angle first, and cos⁡−1x=arccos⁡xe1cos⁡x=sec⁡x\cos^{-1} x = \arccos x e \frac{1}{\cos x} = \sec x.
Sub-90 Dry-Wipe Shortcut15s
Line 1 (Vertical Shift Range Check): For y=a+cos⁡xy = a + \cos x, the range [−1,1][-1, 1] shifts immediately to **[a−1,a+1][a - 1, a + 1]**.
Line 2 (Exponent Convention): cos⁡nx=(cos⁡x)n\cos^n x = (\cos x)^n for n≥2n \ge 2, but n=−1n = -1 is reserved for the inverse function arccos⁡x\arccos x (while (cos⁡x)−1=sec⁡x(\cos x)^{-1} = \sec x).
Standard Classroom Derivation~85s
Plot shifted curves and evaluate sample angles to compare the notations.
Examiner Trap Warning: unit_or_prefix_mismatch: Treating cos^(-1)(x) as 1/cos(x) instead of arccos(x), or reading cos(x) + a as cos(x + a).
MM-GUIDE-EX-25Page 149MM8: Graphs of functionsSaves 80s
6-Card Lesson (ARCH-MM-04)Practice Matching Archetype Drill

Horizontal Translations y = f(x + a): Parabolas & Cosine Phase Shifts

Official UAT-UK Guide Exercise Box (Page 149):
Sketch without a graphing package: (a) y=x2y = x^2 and y=(x+2)2y = (x + 2)^2; (b) y=x2y = x^2 and y=(x−3)2y = (x - 3)^2; (c) y=cos⁡xy = \cos x and y=cos⁡(x+π3)y = \cos(x + \frac{\pi}{3}); (d) y=cos⁡xy = \cos x and y=cos⁡(x−2π3)y = \cos(x - \frac{2\pi}{3}).
Official Footnote: 75 See footnote 59.
SymPy-Verified Result:
(a) Shift left by 22, vertex at (−2,0)(-2, 0); (b) Shift right by 33, vertex at (3,0)(3, 0); (c) Shift left by π3\frac{\pi}{3}, peak moves from 0→−π30 \to -\frac{\pi}{3}; (d) Shift right by 2π3\frac{2\pi}{3}, peak moves from 0→2π30 \to \frac{2\pi}{3}.
Sub-90 Dry-Wipe Shortcut15s
Line 1 (Zero-the-Bracket Rule): Set the inside argument equal to the old feature coordinate xold=0x_{\text{old}} = 0:
- x+2=0  ⟹  x=−2x + 2 = 0 \implies x = -2 (left 22);
- x−3=0  ⟹  x=3x - 3 = 0 \implies x = 3 (right 33);
- x+π3=0  ⟹  x + \frac{\pi}{3} = 0 \implies peak at x=−π3x = -\frac{\pi}{3}; x−2π3=0  ⟹  x - \frac{2\pi}{3} = 0 \implies peak at x=2π3x = \frac{2\pi}{3} (15s).
Standard Classroom Derivation~95s
Expand the quadratics or use cosine addition formulas cos⁡(x+α)=cos⁡xcos⁡α−sin⁡xsin⁡α\cos(x + \alpha) = \cos x\cos\alpha - \sin x\sin\alpha to find intercepts.
Examiner Trap Warning: sign_or_direction_flip: Shifting y = f(x + a) to the right (+a) instead of to the left (-a).
MM-GUIDE-EX-26Page 153MM8: Graphs of functionsSaves 95s
6-Card Lesson (ARCH-MM-04)Practice Matching Archetype Drill

Horizontal Scaling & Reflection: f(2x), f(x/2), and f(-2x) on x^3 - 3x^2 + 2

Official UAT-UK Guide Exercise Box (Page 153):
For f(x)=x3−3x2+2f(x) = x^3 - 3x^2 + 2, compare the graphs of f(2x)f(2x), f(12x)f(\frac{1}{2}x), and f(−2x)f(-2x). What does 0<a<10 < a < 1 do in y=f(ax)y = f(ax)? What do the minus sign and the 22 do in f(−2x)f(-2x)?
SymPy-Verified Result:
For f(x)=x3−3x2+2f(x) = x^3 - 3x^2 + 2 (turning points at (0,2)(0,2) and (2,−2)(2,-2)): f(2x)=8x3−12x2+2f(2x) = 8x^3 - 12x^2 + 2 compresses horizontally by 12\frac{1}{2} (turning points (0,2),(1,−2)(0,2), (1,-2)); f(12x)=18x3−34x2+2f(\frac{1}{2}x) = \frac{1}{8}x^3 - \frac{3}{4}x^2 + 2 stretches horizontally by 22 (turning points (0,2),(4,−2)(0,2), (4,-2)); f(−2x)=−8x3−12x2+2f(-2x) = -8x^3 - 12x^2 + 2 reflects in the yy-axis AND compresses horizontally by 12\frac{1}{2} (turning points (0,2),(−1,−2)(0,2), (-1,-2)).
Sub-90 Dry-Wipe Shortcut20s
**Line 1 (Divide Every xx-Coordinate by aa):** In y=f(ax)y = f(ax), every point (x0,y0)(x_0, y_0) moves to **(x0/a,y0)(x_0 / a, y_0)** while the yy-intercept (0,2)(0, 2) stays pinned!
Line 2 (Instant Stationary Point Map): Old turning points (0,2)(0,2) and (2,−2)(2,-2) map under a=−2a = -2 to (0/−2,2)=(0,2)(0/-2, 2) = (0,2) and (2/−2,−2)=(−1,−2)(2/-2, -2) = (-1, -2) in 5 seconds!
Standard Classroom Derivation~115s
Substitute 2x,12x,−2x2x, \frac{1}{2}x, -2x into x3−3x2+2x^3 - 3x^2 + 2, expand cubics, and differentiate each to find the new stationary points.
Examiner Trap Warning: factor_of_two_or_stoichiometry_trap: Multiplying x-coordinates by a instead of dividing by a in y = f(ax).
MM-GUIDE-EX-27Page 153MM8: Graphs of functionsSaves 95s
6-Card Lesson (ARCH-MM-04)Practice Matching Archetype Drill

Modulus Transformations y = |f(x)| vs y = f(|x|) for sin x and cos x

Official UAT-UK Guide Exercise Box (Page 153):
Sketch y=∣f(x)∣y = |f(x)| and y=f(∣x∣)y = f(|x|) for f(x)=sin⁡xf(x) = \sin x and for f(x)=cos⁡xf(x) = \cos x without using any graphing packages.
SymPy-Verified Result:
∣sin⁡x∣|\sin x| and ∣cos⁡x∣|\cos x| reflect all negative lobes (y<0y < 0) upwards across the xx-axis, halving the period from 2π2\pi to π\pi with range [0,1][0, 1]. For f(∣x∣)f(|x|): cos⁡(∣x∣)=cos⁡x\cos(|x|) = \cos x (unchanged since cos⁡x\cos x is even!), whereas sin⁡(∣x∣)\sin(|x|) mirrors the x≥0x \ge 0 sine wave across the yy-axis, creating a V-shaped cusp at (0,0)(0,0) and making it non-periodic.
Sub-90 Dry-Wipe Shortcut25s
**Line 1 (∣f(x)∣|f(x)| = Bottom-Up Fold):** Fold everything below the xx-axis upwards   ⟹  \implies period of ∣sin⁡x∣|\sin x| and ∣cos⁡x∣|\cos x| halves to π\pi.
**Line 2 (f(∣x∣)f(|x|) = Right-to-Left Mirror):** Erase x<0x < 0 and mirror x≥0x \ge 0 across the yy-axis. Since cos⁡(−x)=cos⁡x\cos(-x) = \cos x, **cos⁡(∣x∣)=cos⁡x\cos(|x|) = \cos x is a classic ESAT invariant**!
Standard Classroom Derivation~120s
Break each function into piecewise intervals where f(x)≥0f(x) \ge 0, f(x)<0f(x) < 0, x≥0x \ge 0, and x<0x < 0.
Examiner Trap Warning: multi_statement_linchpin_foil: Thinking sin(|x|) is equal to |sin(x)|, or failing to spot that cos(|x|) is identical to cos(x) for all real x.
MM-GUIDE-EX-28Page 155MM8: Graphs of functionsSaves 85s
6-Card Lesson (ARCH-MM-06)Practice Matching Archetype Drill

Shift-Stretch Equivalence: 3^(x+2) = 9(3^x) and log_10(10x) = 1 + log_10(x)

Official UAT-UK Guide Exercise Box (Page 155):
Consider y=f(x)=3xy = f(x) = 3^x: sketch both y=f(x+2)y = f(x + 2) and y=9f(x)y = 9f(x). Now consider y=f(x)=log⁡10xy = f(x) = \log_{10} x: sketch both y=f(10x)y = f(10x) and y=1+f(x)y = 1 + f(x). What do you notice? Explain your answer.
SymPy-Verified Result:
Both pairs produce identical graphs: (1) f(x+2)=3x+2=32⋅3x=9f(x)f(x+2) = 3^{x+2} = 3^2 \cdot 3^x = 9f(x) (a horizontal shift left by 22 equals a vertical stretch by 99); (2) f(10x)=log⁡10(10x)=log⁡1010+log⁡10x=1+f(x)f(10x) = \log_{10}(10x) = \log_{10} 10 + \log_{10} x = 1 + f(x) for x>0x > 0 (a horizontal compression by 110\frac{1}{10} equals a vertical shift up by 11).
Sub-90 Dry-Wipe Shortcut20s
**Line 1 (Exponential Shift ≡\equiv Stretch):** bx+k=bk⋅bx  ⟹  b^{x+k} = b^k \cdot b^x \implies shifting bxb^x left by kk is identical to stretching vertically by bkb^k.
**Line 2 (Log Stretch ≡\equiv Shift):** log⁡b(cx)=log⁡bc+log⁡bx  ⟹  \log_b(cx) = \log_b c + \log_b x \implies scaling xx by cc inside a log is identical to shifting vertically by +log⁡bc+\log_b c!
Standard Classroom Derivation~105s
Plot coordinate tables for 3x+23^{x+2}, 9(3x)9(3^x), log⁡10(10x)\log_{10}(10x), and 1+log⁡10x1 + \log_{10} x and compare points.
Examiner Trap Warning: multi_statement_linchpin_foil: Thinking that a horizontal translation and a vertical stretch can never produce the exact same curve.
MM-GUIDE-EX-29Page 156MM8: Graphs of functionsSaves 100s
6-Card Lesson (ARCH-MM-04)Practice Matching Archetype Drill

Composite Functions f(g(x)), g(f(x)), g(h(x)), h(g(x)) & Non-Commutativity

Official UAT-UK Guide Exercise Box (Page 156):
Given f(x)=x2f(x) = x^2, g(x)=x3−3g(x) = x^3 - 3, and h(x)=2x+3x−4h(x) = \frac{2x + 3}{x - 4} (xe4x e 4), find simplified expressions for: (a) f(g(x))f(g(x)); (b) g(f(x))g(f(x)); (c) g(h(x))g(h(x)); (d) h(g(x))h(g(x)). Is it true that f(g(x))=g(f(x))f(g(x)) = g(f(x))?
SymPy-Verified Result:
(a) f(g(x))=(x3−3)2=x6−6x3+9f(g(x)) = (x^3 - 3)^2 = x^6 - 6x^3 + 9; (b) g(f(x))=x6−3g(f(x)) = x^6 - 3; (c) g(h(x))=(2x+3x−4)3−3=5x3+180x2−405x+219(x−4)3g(h(x)) = \left(\frac{2x+3}{x-4}\right)^3 - 3 = \frac{5x^3 + 180x^2 - 405x + 219}{(x-4)^3}; (d) h(g(x))=2x3−3x3−7h(g(x)) = \frac{2x^3 - 3}{x^3 - 7} (xe73x e \sqrt[3]{7}). Clearly f(g(x))eg(f(x))f(g(x)) e g(f(x)).
Sub-90 Dry-Wipe Shortcut25s
Line 1 (Inside-Out Substitution): Replace every xx in the outer template with (inner)( \text{inner} ): f(g(x))=(x3−3)2=x6−6x3+9f(g(x)) = (x^3-3)^2 = x^6 - 6x^3 + 9 vs g(f(x))=(x2)3−3=x6−3g(f(x)) = (x^2)^3 - 3 = x^6 - 3.
**Line 2 (x=0x = 0 Commutativity Check):** At x=0x = 0, f(g(0))=(−3)2=9f(g(0)) = (-3)^2 = 9 whereas g(f(0))=0−3=−3  ⟹  f∘geg∘fg(f(0)) = 0 - 3 = -3 \implies f \circ g e g \circ f in 2 seconds!
Standard Classroom Derivation~125s
Substitute each inner function into the outer function and expand all binomial powers.
Examiner Trap Warning: intermediate_stop_trap: Expanding (x³ - 3)² as x⁶ + 9 or x⁶ - 9 (omitting the middle cross-term -6x³).
MM-GUIDE-EX-30Page 157MM8: Graphs of functionsSaves 75s
6-Card Lesson (ARCH-MM-04)Practice Matching Archetype Drill

Combined Linear Transformations y = mx + c: Stretch-Then-Shift vs Shift-Then-Stretch

Official UAT-UK Guide Exercise Box (Page 157):
How did you deal with y=x→y=mx+cy = x \to y = mx + c? Did you use f(x)+cf(x) + c or f(x+c)f(x + c)? Pick pairs of values for mm and cc and follow through the graph transformations step by step.
SymPy-Verified Result:
Starting from y=xy = x: Method 1 (Vertical): stretch vertically by mm (y=mxy = mx), then translate up by cc (y=mx+cy = mx + c). Method 2 (Horizontal): rewrite y=m(x+cm)y = m\left(x + \frac{c}{m}\right) — stretch vertically by mm, then translate left by cm\frac{c}{m} (or translate left by cc first to get y=x+cy = x + c, then stretch horizontally by 1m\frac{1}{m}).
Sub-90 Dry-Wipe Shortcut20s
Line 1 (Outside = BIDMAS Order): For outside operations mf(x)+cm f(x) + c, multiply by mm first, then add +c+c.
**Line 2 (Inside = Factor mm First):** If translating horizontally after stretching, factor mx+c=m(x+cm)  ⟹  mx + c = m\left(x + \frac{c}{m}\right) \implies horizontal shift is −cm-\frac{c}{m} (not −c-c!).
Standard Classroom Derivation~95s
Test both orders of transformations on specific numbers like m=2,c=6m = 2, c = 6 and sketch intermediate lines.
Examiner Trap Warning: factor_of_two_or_stoichiometry_trap: Shifting y = mx left by c instead of c/m, which produces y = m(x + c) = mx + mc instead of mx + c.
MM-GUIDE-EX-31Page 158MM8: Graphs of functionsSaves 80s
6-Card Lesson (ARCH-MM-04)Practice Matching Archetype Drill

3-Step Transformation Chain for General Quadratics y = a(x + b)^2 + c

Official UAT-UK Guide Exercise Box (Page 158):
Pick sets of values for a,ba, b and cc and follow through the graph transformations from y=x2y = x^2 to y=a(x+b)2+cy = a(x + b)^2 + c step by step.
SymPy-Verified Result:
From y=x2y = x^2 with vertex (0,0)(0,0): (1) y=(x+b)2y = (x + b)^2 translates horizontally by −b-b to vertex (−b,0)(-b, 0); (2) y=a(x+b)2y = a(x + b)^2 stretches vertically by ∣a∣|a| (and reflects in xx-axis if a<0a < 0), keeping vertex at (−b,0)(-b, 0); (3) y=a(x+b)2+cy = a(x + b)^2 + c translates vertically by +c+c to final vertex (−b,c)(-b, c) and yy-intercept (0,ab2+c)(0, ab^2 + c).
Sub-90 Dry-Wipe Shortcut20s
Line 1 (Vertex Readout): From y=a(x+b)2+cy = a(x + b)^2 + c, read the vertex **(−b,c)(-b, c)** and concavity sign(a)\text{sign}(a) in 2 seconds.
Line 2 (Number of Real Roots Check): Real roots exist iff aa and cc have opposite signs (−ca>0-\frac{c}{a} > 0, or 11 root if c=0c = 0)!
Standard Classroom Derivation~100s
Expand a(x+b)2+c=ax2+2abx+ab2+ca(x + b)^2 + c = ax^2 + 2abx + ab^2 + c and compute turning points and intercepts from the expanded coefficients.
Examiner Trap Warning: sign_or_direction_flip: Writing the vertex of y = a(x + b)² + c as (b, c) instead of (-b, c).