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Paper 1 · Set 06

Paper 1: Applications of Mathematical Knowledge. 20 questions, 75 minutes. Answer on the grid or click the options; nothing is marked until you press Submit.

75 minutes20 marksNo calculator
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1Quadratic functions · spec MM1.3 · Paper 1 style · ★☆☆

For what value of \(c\) does the line \(y = 2x + c\) touch the parabola \(y = x^2 - 4x + 7\)?

  1. A\(7\)
  2. B\(2\)
  3. C\(-2\)
  4. D\(-1\)
  5. E\(0\)
  6. F\(5\)
Worked solution

Equate: \(x^2 - 6x + (7 - c) = 0\). Tangency needs discriminant zero: \(36 - 4(7-c) = 0\), so \(8 + 4c = 0\) and \(c = -2\).

Full method — how and why
'Touches' means exactly one intersection, which after substitution means a repeated root: discriminant zero.
  1. Set the two expressions equal: \(2x + c = x^2 - 4x + 7\).
    Intersections of \(y = f(x)\) and \(y = g(x)\) are the solutions of \(f(x) = g(x)\).
  2. Collect into a standard quadratic: \(x^2 - 6x + (7 - c) = 0\).
    Everything to one side, so the discriminant test can be applied.
  3. Set the discriminant to zero: \((-6)^2 - 4(1)(7 - c) = 0\).
    Touching means a repeated root, i.e. the line meets the curve once - exactly the \(b^2 = 4ac\) case.
  4. Solve: \(36 - 28 + 4c = 0 \Rightarrow 4c = -8 \Rightarrow c = -2\).
    Watch the double negative: \(-4(7 - c) = -28 + 4c\).
Where it goes wrong
  • Sign slip in \(-4(7-c)\) giving \(c = 2\), which is offered.
  • Using \(b^2 - 4ac > 0\) (crossing) rather than \(= 0\) (touching).
Takeaway Tangency between a line and a curve: substitute, then set the discriminant of the resulting quadratic to zero.
Answer C · Quadratic functions (MM1.3) · question QUA-05
2Integration · spec MM7.6 · Paper 1 style · ★☆☆

A curve has \(\dfrac{dy}{dx} = 6x^2 - 4x + 1\) and passes through \((1, 3)\). Find its equation.

  1. A\(y = 12x - 4\)
  2. B\(y = 2x^3 - 4x^2 + x + 4\)
  3. C\(y = 2x^3 - 2x^2 + x + 2\)
  4. D\(y = 2x^3 - 2x^2 + x + 3\)
  5. E\(y = 2x^3 - 2x^2 + x\)
  6. F\(y = 2x^3 - 2x^2 + 2\)
Worked solution

\(y = 2x^3 - 2x^2 + x + c\); at \((1, 3)\): \(2 - 2 + 1 + c = 3 \Rightarrow c = 2\).

Full method — how and why
Reverse differentiation plus a boundary condition to pin down the constant.
  1. Integrate each term: \(\int(6x^2 - 4x + 1)dx = 2x^3 - 2x^2 + x + c\).
    Add one to each index and divide: \(\frac{6x^3}{3} = 2x^3\), \(\frac{-4x^2}{2} = -2x^2\).
  2. Do not forget the constant \(c\) - it is exactly what the given point determines.
    Every antiderivative differs by a constant, so one point selects the right curve.
  3. Substitute \((1,3)\): \(2 - 2 + 1 + c = 3\), so \(c = 2\).
    Evaluate the polynomial at \(x = 1\) and set it equal to \(3\).
  4. The curve is \(y = 2x^3 - 2x^2 + x + 2\).
    Check: differentiating gives back \(6x^2 - 4x + 1\) ✓.
Where it goes wrong
  • Omitting \(c\) altogether and choosing \(y = 2x^3 - 2x^2 + x\).
  • Arithmetic slip giving \(c = 3\) - an offered option.
Takeaway \(\frac{dy}{dx} = f(x)\) plus one point: integrate, then substitute to find \(c\).
Answer C · Integration (MM7.6) · question INT-04
3GCSE algebra · spec M4.8 · Paper 1 style · ★☆☆

Which of the following is an identity (true for all values of \(x\))?

  1. A\((x + 3)^2 - (x - 3)^2 = 12x\)
  2. B\(x^2 + 6x + 9 = 0\)
  3. C\((x + 3)^2 = x^2 + 9\)
  4. D\((x - 3)^2 = x^2 - 6x - 9\)
  5. E\((x + 3)(x - 3) = x^2 + 9\)
Worked solution

\((x+3)^2 - (x-3)^2 = (x^2 + 6x + 9) - (x^2 - 6x + 9) = 12x\) for all \(x\).

Full method — how and why
An identity holds for every \(x\). Expand each candidate and compare, or test values to eliminate.
  1. Expand the first option: \((x+3)^2 - (x-3)^2 = (x^2 + 6x + 9) - (x^2 - 6x + 9)\).
    Expand both squares fully, including the cross terms.
  2. Simplify: the \(x^2\) and \(9\) cancel, leaving \(12x\) - true for all \(x\). This is the identity.
    Both sides agree identically, which is what 'identity' requires.
  3. Check the others quickly: \((x+3)^2 = x^2 + 6x + 9 \neq x^2 + 9\); \((x+3)(x-3) = x^2 - 9\), not \(+9\); \(x^2 + 6x + 9 = 0\) holds only at \(x = -3\); \((x-3)^2 = x^2 - 6x + 9\), not \(-9\).
    Each wrong option drops or mis-signs a term - a quick test at \(x = 1\) kills them all.
Where it goes wrong
  • Confusing an equation (true sometimes) with an identity (true always).
  • Forgetting the middle term when squaring a bracket.
Takeaway \(a^2 - b^2 = (a-b)(a+b)\) applied to brackets gives quick identities; test \(x = 1\) to eliminate impostors.
Answer A · GCSE algebra (M4.8) · question ALG-04
4Indices and surds · spec MM1.1 · Paper 1 style · ★☆☆

How many real solutions does the equation \(2^{x^2} = 16 \times 2^{-3x}\) have?

  1. A\(3\)
  2. B\(2\)
  3. C\(4\)
  4. D\(1\)
  5. E\(0\)
  6. Finfinitely many
Worked solution

\(2^{x^2} = 2^{4 - 3x}\), so \(x^2 + 3x - 4 = 0\), i.e. \((x+4)(x-1) = 0\): two real solutions.

Full method — how and why
'How many real solutions' with exponentials - convert to a single base and the exponents give a quadratic whose discriminant answers the question.
  1. Write \(16 = 2^4\), so the right-hand side is \(2^{4}\cdot2^{-3x} = 2^{4 - 3x}\).
    Multiplying powers of the same base adds indices; this puts both sides in base \(2\).
  2. Equate indices: \(x^2 = 4 - 3x\), i.e. \(x^2 + 3x - 4 = 0\).
    Legitimate because \(2^u\) is one-to-one.
  3. Factorise: \((x+4)(x-1) = 0\) - two distinct real roots.
    The question asks only for the *number* of solutions, so the discriminant (\(9 + 16 > 0\)) would do; factorising confirms both roots are real.
Where it goes wrong
  • Rejecting \(x = -4\) because it is negative - the exponent may be negative; it is \(2^x\) that must stay positive.
  • Reading \(16 \times 2^{-3x}\) as \((16 \times 2)^{-3x}\).
Takeaway Same base \(\Rightarrow\) equate exponents; then it is an ordinary quadratic and every real root counts.
Answer B · Indices and surds (MM1.1) · question IND-19
5Differentiation · spec MM6.2 · Paper 1 style · ★☆☆

Find \(\dfrac{dy}{dx}\) when \(y = \dfrac{(3x - 2)^2}{x^2}\).

  1. A\(\dfrac{12}{x^2} + \dfrac{8}{x^3}\)
  2. B\(\dfrac{6}{x} - \dfrac{12}{x^2}\)
  3. C\(-\dfrac{12}{x^2} + \dfrac{8}{x^3}\)
  4. D\(\dfrac{12}{x^2} - \dfrac{8}{x^3}\)
  5. E\(\dfrac{6(3x-2)}{x^2}\)
  6. F\(9 - \dfrac{12}{x}\)
Worked solution

\(y = 9 - 12x^{-1} + 4x^{-2}\), so \(y' = 12x^{-2} - 8x^{-3}\).

Full method — how and why
The spec only allows differentiating powers of \(x\), so the expression must first be written as a sum of such powers - expand and divide term by term.
  1. Expand the numerator: \((3x-2)^2 = 9x^2 - 12x + 4\).
    There is no quotient or chain rule available, so the fraction must be dismantled.
  2. Divide each term by \(x^2\): \(y = 9 - 12x^{-1} + 4x^{-2}\).
    \(\frac{9x^2}{x^2} = 9\), \(\frac{-12x}{x^2} = -12x^{-1}\), \(\frac{4}{x^2} = 4x^{-2}\).
  3. Differentiate term by term with \(\frac{d}{dx}x^n = nx^{n-1}\): \(0 + 12x^{-2} - 8x^{-3}\).
    \(-12 \times (-1) = +12\) and \(4 \times (-2) = -8\); the constant \(9\) differentiates to \(0\).
  4. Write in fraction form: \(\dfrac{12}{x^2} - \dfrac{8}{x^3}\).
    Match the form used in the options.
Where it goes wrong
  • Differentiating the bracket directly as \(\frac{6(3x-2)}{x^2}\) - that would need the quotient rule (and is wrong anyway).
  • Sign slips: differentiating \(x^{-1}\) gives \(-x^{-2}\), so \(-12x^{-1}\) gives \(+12x^{-2}\).
Takeaway Rewrite as \(\sum ax^n\) before differentiating; negative powers change sign when differentiated.
Answer D · Differentiation (MM6.2) · question DIF-01
6Straight lines · spec MM3.1 · Paper 1 style · ★☆☆

Find the equation of the line through \((2, -3)\) that is perpendicular to \(3x - 2y + 5 = 0\).

  1. A\(2x - 3y - 13 = 0\)
  2. B\(3x + 2y = 0\)
  3. C\(2x + 3y + 5 = 0\)
  4. D\(2x + 3y - 13 = 0\)
  5. E\(2x + 3y - 5 = 0\)
  6. F\(3x - 2y - 12 = 0\)
Worked solution

The given line has gradient \(\frac32\), so the perpendicular has gradient \(-\frac23\): \(y + 3 = -\frac23(x - 2) \Rightarrow 3y + 9 = -2x + 4 \Rightarrow 2x + 3y + 5 = 0\).

Full method — how and why
Perpendicular line through a point: find the gradient of the given line, take the negative reciprocal, then use the point-gradient form.
  1. Rearrange \(3x - 2y + 5 = 0\) into \(y = \frac32 x + \frac52\), so its gradient is \(\frac32\).
    Alternatively use the shortcut: \(ax + by + c = 0\) has gradient \(-\frac ab = -\frac{3}{-2} = \frac32\).
  2. Perpendicular gradient: \(-\dfrac{1}{3/2} = -\dfrac23\).
    Perpendicular gradients multiply to \(-1\); invert and change sign.
  3. Use \(y - y_1 = m(x - x_1)\) at \((2, -3)\): \(y + 3 = -\frac23(x - 2)\).
    \(y - (-3)\) is \(y + 3\) - the commonest sign slip in this whole topic.
  4. Clear fractions and tidy: \(3y + 9 = -2x + 4 \Rightarrow 2x + 3y + 5 = 0\).
    Multiply through by \(3\) first, then collect everything on one side.
Where it goes wrong
  • Using \(y - 3\) instead of \(y + 3\), which gives \(2x + 3y - 5 = 0\) - an offered option.
  • Taking the perpendicular gradient as \(-\frac32\) or \(\frac23\).
Takeaway Negative reciprocal, and substitute the point with its own sign: \(y - (-3) = y + 3\).
Answer C · Straight lines (MM3.1) · question LIN-01
7Circles and circle theorems · spec MM3.2 · Paper 1 style · ★☆☆

Find the radius of the circle \(x^2 + y^2 - 6x + 4y - 12 = 0\).

  1. A\(4\)
  2. B\(5\)
  3. C\(1\)
  4. D\(\sqrt{12}\)
  5. E\(\sqrt{13}\)
  6. F\(25\)
Worked solution

\((x-3)^2 + (y+2)^2 = 12 + 9 + 4 = 25\), so \(r = 5\).

Full method — how and why
The expanded form of a circle. Complete the square in \(x\) and in \(y\) to recover the centre-radius form.
  1. Group the terms: \((x^2 - 6x) + (y^2 + 4y) = 12\).
    Move the constant to the right first so the completed squares have somewhere to go.
  2. Complete each square: \(x^2 - 6x = (x-3)^2 - 9\) and \(y^2 + 4y = (y+2)^2 - 4\).
    Halve each coefficient (\(-6 \to -3\), \(4 \to 2\)) and subtract its square.
  3. Substitute: \((x-3)^2 + (y+2)^2 - 13 = 12\), so \((x-3)^2 + (y+2)^2 = 25\).
    The \(-9\) and \(-4\) move across to the right, adding to the constant.
  4. Read off \(r^2 = 25\), so \(r = 5\).
    The right-hand side is \(r^2\), not \(r\).
Where it goes wrong
  • Answering \(25\) - that is \(r^2\).
  • Subtracting the \(9\) and \(4\) on the right instead of adding, giving \(r^2 = 12\) or similar.
Takeaway \(x^2+y^2+cx+dy+e=0\) has centre \(\left(-\frac c2, -\frac d2\right)\) and \(r^2 = \frac{c^2}{4}+\frac{d^2}{4}-e\).
Answer B · Circles and circle theorems (MM3.2) · question CIR-01
8Differentiation · spec MM6.3 · Paper 1 style · ★☆☆

Find the sum of the \(y\)-coordinates of the stationary points of the curve \(y = 2x^3 - 9x^2 + 12x - 3\).

  1. A\(2\)
  2. B\(0\)
  3. C\(5\)
  4. D\(4\)
  5. E\(3\)
  6. F\(1\)
Worked solution

\(y' = 6x^2 - 18x + 12 = 6(x - 1)(x - 2)\): stationary points at \(x = 1\) (\(y = 2 - 9 + 12 - 3 = 2\)) and \(x = 2\) (\(y = 16 - 36 + 24 - 3 = 1\)). Sum \(3\).

Full method — how and why
Find both stationary points and add their \(y\)-values - two substitutions after one factorisation.
  1. Differentiate: \(y' = 6x^2 - 18x + 12 = 6(x^2 - 3x + 2) = 6(x-1)(x-2)\).
    Take out the \(6\), then factorise.
  2. Stationary points at \(x = 1\) and \(x = 2\).
    Roots of the derivative.
  3. Compute the \(y\)-values: \(y(1) = 2 - 9 + 12 - 3 = 2\); \(y(2) = 16 - 36 + 24 - 3 = 1\).
    Substitute into the original cubic, carefully: \(2(8) = 16\) and \(9(4) = 36\).
  4. Add: \(2 + 1 = 3\).
    The question asks for the sum of the \(y\)-coordinates.
Where it goes wrong
  • Adding the \(x\)-coordinates (\(1 + 2 = 3\)) - the same answer here by coincidence, but the wrong method.
  • Arithmetic slips in \(y(2)\); recompute \(16 - 36 = -20\), \(-20 + 24 = 4\), \(4 - 3 = 1\).
Takeaway Two stationary points mean two substitutions into the original function - the derivative gives only the positions.
Answer E · Differentiation (MM6.3) · question DIF-20
9Sequences and series · spec MM2.1 · Paper 1 style · ★☆☆

A sequence is defined by \(x_{n+1} = 3x_n - 2\) with \(x_1 = 2\). Find \(x_5\).

  1. A\(244\)
  2. B\(80\)
  3. C\(28\)
  4. D\(82\)
  5. E\(162\)
  6. F\(54\)
Worked solution

\(x_2 = 4\), \(x_3 = 10\), \(x_4 = 28\), \(x_5 = 82\).

Full method — how and why
A recurrence with a small index - just iterate. There is no shortcut worth finding for only four steps.
  1. Apply the rule repeatedly from \(x_1 = 2\): \(x_2 = 3(2) - 2 = 4\).
    \(x_{n+1} = 3x_n - 2\) means 'triple the last term, subtract 2'.
  2. \(x_3 = 3(4) - 2 = 10\); \(x_4 = 3(10) - 2 = 28\); \(x_5 = 3(28) - 2 = 82\).
    Write each step on its own line - the numbers grow quickly and a single slip propagates.
Where it goes wrong
  • Stopping at \(x_4 = 28\) (an offered option) by miscounting the steps: going from \(x_1\) to \(x_5\) takes four applications.
  • Reading the rule as \(3(x_n - 2)\), which gives \(x_5 = 162\) - also offered.
Takeaway Count the applications carefully: reaching \(x_k\) from \(x_1\) takes \(k - 1\) steps.
Answer D · Sequences and series (MM2.1) · question SEQ-03
10Probability · spec M7.6 · Paper 1 style · ★☆☆

Two fair dice are thrown. Find the probability that the product of the two scores is odd.

  1. A\(\dfrac{2}{3}\)
  2. B\(\dfrac{3}{4}\)
  3. C\(\dfrac{1}{4}\)
  4. D\(\dfrac{1}{6}\)
  5. E\(\dfrac{1}{2}\)
  6. F\(\dfrac{1}{3}\)
Worked solution

The product is odd only if both scores are odd: \(\frac12 \times \frac12 = \frac14\).

Full method — how and why
A product is odd only when every factor is odd - so this is an 'and' of two independent events.
  1. Odd \(\times\) odd \(=\) odd; any even factor makes the product even.
    Recognising this converts the question into a much simpler one.
  2. \(P(\text{first die odd}) = \frac36 = \frac12\); same for the second.
    Three odd faces out of six: \(1, 3, 5\).
  3. Independent dice, so multiply: \(\frac12 \times \frac12 = \frac14\).
    One die's outcome has no effect on the other.
Where it goes wrong
  • Answering \(\frac12\) by thinking 'half of all products are odd' - in fact only \(9\) of the \(36\) outcomes are.
  • Treating it as a sum (where odd \(+\) odd \(=\) even), a different and much more common question.
Takeaway Product odd \(\iff\) all factors odd; product even \(\iff\) at least one even.
Answer C · Probability (M7.6) · question PRO-02
11Binomial expansion · spec MM2.4 · Paper 1 style · ★☆☆

Find the coefficient of \(x^5\) in the expansion of \((1 + x)^{3}(1 + x)^{4}\).

  1. A\(35\)
  2. B\(21\)
  3. C\(12\)
  4. D\(1\)
  5. E\(28\)
  6. F\(7\)
Worked solution

\((1+x)^3(1+x)^4 = (1+x)^7\), and \(\binom75 = 21\).

Full method — how and why
Two binomials with the same bracket: combine the powers first, then use a single coefficient.
  1. Combine: \((1+x)^3(1+x)^4 = (1+x)^7\).
    Same base, so add the indices - this converts a product of expansions into one expansion.
  2. The coefficient of \(x^5\) is \(\binom75\).
    Standard general term with \(r = 5\).
  3. Evaluate using symmetry: \(\binom75 = \binom72 = \frac{7 \times 6}{2} = 21\).
    Computing \(\binom72\) is quicker than \(\binom75\) and gives the same value.
Where it goes wrong
  • Multiplying the separate expansions term by term and mis-collecting.
  • Multiplying the coefficients \(\binom35\) and \(\binom45\) - both of which are zero, and neither is relevant.
Takeaway Combine equal brackets before expanding; then one binomial coefficient does the whole job.
Answer B · Binomial expansion (MM2.4) · question BIN-10
12Circles and circle theorems · spec MM3.3 · Paper 1 style · ★☆☆

\(A\), \(B\) and \(C\) lie on a circle with centre \(O\). Angle \(AOB = 100^\circ\) and \(C\) lies on the major arc \(AB\). Find angle \(ACB\).

  1. A\(200^\circ\)
  2. B\(100^\circ\)
  3. C\(40^\circ\)
  4. D\(50^\circ\)
  5. E\(130^\circ\)
  6. F\(80^\circ\)
Worked solution

The angle at the centre is twice the angle at the circumference subtended by the same arc: \(ACB = 50^\circ\).

Full method — how and why
The angle at the centre is twice the angle at the circumference on the same arc - provided you use the right arc.
  1. Identify the arc: angle \(AOB = 100^\circ\) at the centre stands on the minor arc \(AB\).
    The reflex angle at the centre (\(260^\circ\)) would correspond to the other arc - which arc you use decides the answer.
  2. \(C\) is on the major arc, so it sees the same minor arc \(AB\).
    The circumference angle must be on the opposite arc from the one it stands on.
  3. Apply the theorem: angle \(ACB = \frac12 \times 100^\circ = 50^\circ\).
    Centre angle is double the circumference angle on the same arc.
Where it goes wrong
  • Doubling instead of halving, giving \(200^\circ\) - impossible for an angle in a triangle.
  • Using the reflex angle \(260^\circ\) and answering \(130^\circ\) (which would be right if \(C\) were on the minor arc).
Takeaway Angle at centre \(= 2 \times\) angle at circumference on the same arc; check which side of the chord the point lies.
Answer D · Circles and circle theorems (MM3.3) · question CIR-06
13Functions and transformations · spec MM8.2 · Paper 1 style · ★☆☆

The graph of \(y = x^2\) is translated by \(3\) units in the positive \(x\)-direction and then stretched parallel to the \(y\)-axis with scale factor \(2\). What is the equation of the resulting graph?

  1. A\(y = 2(x-3)^2\)
  2. B\(y = (x-3)^2 + 2\)
  3. C\(y = 2(x+3)^2\)
  4. D\(y = 2x^2 - 3\)
  5. E\(y = (2x-3)^2\)
  6. F\(y = \frac12 (x-3)^2\)
Worked solution

Translation: \(y = (x-3)^2\). Stretch by factor \(2\) in the \(y\)-direction: \(y = 2(x-3)^2\).

Full method — how and why
Two transformations applied in a stated order. Apply them one at a time to the equation, in the order given.
  1. Translate \(3\) in the positive \(x\)-direction: replace \(x\) by \(x - 3\), giving \(y = (x - 3)^2\).
    A shift right by \(a\) replaces \(x\) with \(x - a\) - the sign is opposite to what it looks like, because the new graph reaches a given height \(3\) later.
  2. Stretch parallel to the \(y\)-axis with factor \(2\): multiply the whole function by \(2\), giving \(y = 2(x-3)^2\).
    A vertical stretch multiplies every \(y\)-value, so it acts on the outside of the function.
Where it goes wrong
  • Writing \((x+3)^2\) - shifting right uses \(x - 3\).
  • Putting the \(2\) inside as \((2x - 3)^2\), which is a horizontal stretch, not a vertical one.
Takeaway Changes inside the function affect \(x\) and act in the opposite sense; changes outside affect \(y\) and act as they read.
Answer A · Functions and transformations (MM8.2) · question FUN-01
14Exponentials and logarithms · spec MM5.3 · Paper 1 style · ★☆☆

Solve \(2^{2x} - 3\cdot 2^{x+2} + 32 = 0\). Find the sum of the solutions.

  1. A\(6\)
  2. B\(12\)
  3. C\(4\)
  4. D\(3\)
  5. E\(8\)
  6. F\(5\)
Worked solution

Let \(y = 2^x\): \(y^2 - 12y + 32 = (y - 4)(y - 8) = 0\), so \(2^x = 4\) or \(8\): \(x = 2\) or \(3\). Sum \(5\).

Full method — how and why
A hidden quadratic in \(2^x\). The middle term needs rewriting before the substitution is visible.
  1. Rewrite the middle term: \(3 \cdot 2^{x+2} = 3 \cdot 2^2 \cdot 2^x = 12 \cdot 2^x\).
    \(2^{x+2} = 2^x \times 4\) - separating the constant part is essential before substituting.
  2. Substitute \(y = 2^x\), noting \(2^{2x} = y^2\): \(y^2 - 12y + 32 = 0\).
    \(2^{2x} = (2^x)^2\); that is what makes it quadratic.
  3. Factorise: \((y-4)(y-8) = 0\), so \(y = 4\) or \(y = 8\).
    Both are positive, so both give solutions for \(x\).
  4. Convert back: \(2^x = 4 \Rightarrow x = 2\); \(2^x = 8 \Rightarrow x = 3\). Sum \(= 5\).
    The question asks for the sum of the \(x\) values, not the \(y\) values.
Where it goes wrong
  • Summing the \(y\) values (\(4 + 8 = 12\)) - which is offered.
  • Reading \(2^{x+2}\) as \(2^x + 2\).
Takeaway \(a^{x+k} = a^k a^x\): split off the constant before substituting, then convert the roots back to \(x\).
Answer F · Exponentials and logarithms (MM5.3) · question LOG-03
15Angles, polygons, congruence and vectors · spec M5.2 · Paper 1 style · ★★☆

A regular hexagon and a regular octagon share a common side. What is the size of the angle between their two other sides that meet at one end of the common side (the angle outside both polygons)?

  1. A\(120^\circ\)
  2. B\(95^\circ\)
  3. C\(105^\circ\)
  4. D\(135^\circ\)
  5. E\(115^\circ\)
  6. F\(75^\circ\)
Worked solution

Angles at the shared vertex: \(120^\circ\) (hexagon) \(+ 135^\circ\) (octagon) \(+ x = 360^\circ\), so \(x = 105^\circ\).

Full method — how and why
Angles at a point sum to \(360^\circ\). Find each polygon's interior angle and subtract.
  1. Hexagon interior angle: exterior \(= \frac{360}{6} = 60^\circ\), so interior \(= 120^\circ\).
    Regular polygon: interior \(= 180 - \frac{360}{n}\).
  2. Octagon interior angle: exterior \(= \frac{360}{8} = 45^\circ\), so interior \(= 135^\circ\).
    Same formula with \(n = 8\).
  3. At the shared vertex three angles meet and fill a full turn: \(120 + 135 + x = 360\).
    The two interior angles plus the gap between the outer sides make a complete revolution.
  4. Solve: \(x = 360 - 255 = 105^\circ\).
    That gap is the angle asked for.
Where it goes wrong
  • Using \(180^\circ\) instead of \(360^\circ\) at the vertex.
  • Using exterior angles in the sum, giving \(255^\circ\) or similar.
Takeaway Regular \(n\)-gon interior angle \(= 180 - \frac{360}{n}\); angles round a point total \(360^\circ\).
Answer C · Angles, polygons, congruence and vectors (M5.2) · question GEO-15
16Integration · spec MM7.1 · Paper 1 style · ★★☆

The curve \(y = x(x - 1)(x - 2)\) and the \(x\)-axis enclose two regions. What is the ratio of the area of the region for \(0 \leq x \leq 1\) to the area for \(1 \leq x \leq 2\)?

  1. A\(1 : 2\)
  2. B\(2 : 1\)
  3. C\(1 : 3\)
  4. D\(1 : 4\)
  5. E\(1 : 1\)
  6. F\(3 : 1\)
Worked solution

The cubic has rotational symmetry about \((1, 0)\) (it is \(u^3 - u\) with \(u = x - 1\), an odd function), so the two regions are congruent: ratio \(1 : 1\).

Full method — how and why
A symmetry question. Shifting the cubic reveals it is odd about \((1,0)\), so the two regions are congruent - no integration needed.
  1. Substitute \(u = x - 1\) to centre the cubic: \(y = (u+1)u(u-1) = u(u^2 - 1) = u^3 - u\).
    The roots \(0, 1, 2\) become \(-1, 0, 1\) - symmetric about the origin, which is the hint to try this.
  2. \(u^3 - u\) is an odd function: replacing \(u\) by \(-u\) negates it.
    Odd functions have rotational symmetry of order 2 about the origin.
  3. So the piece on \(-1 \leq u \leq 0\) is the \(180^\circ\) rotation of the piece on \(0 \leq u \leq 1\): the regions are congruent and have equal areas.
    Rotation preserves area.
  4. The ratio is \(1 : 1\).
    You could integrate both (each has area \(\frac14\)), but the symmetry argument is instant.
Where it goes wrong
  • Assuming the regions differ because one is above the axis and one below - sign differs, area does not.
  • Integrating carelessly and comparing signed values.
Takeaway Shift a cubic so its middle root is at the origin: if it becomes odd, the two enclosed regions are congruent.
Answer E · Integration (MM7.1) · question INT-19
17Polynomials, factor and remainder theorems · spec MM1.6 · Paper 1 style · ★★☆

The expression \(x^4 + 4\) can be factorised into two quadratics with integer coefficients. What is the sum of the coefficients of \(x\) in these two quadratics?

  1. A\(1\)
  2. B\(0\)
  3. C\(2\)
  4. D\(4\)
  5. E\(-4\)
  6. FIt cannot be factorised.
Worked solution

\(x^4 + 4 = (x^2 + 2)^2 - 4x^2 = (x^2 + 2x + 2)(x^2 - 2x + 2)\). The \(x\)-coefficients are \(2\) and \(-2\), summing to \(0\).

Full method — how and why
\(x^4 + 4\) has no real roots, so it cannot be split into linear factors - but it can be split into two quadratics via a difference of two squares.
  1. Complete the square in \(x^2\): \(x^4 + 4 = (x^2 + 2)^2 - 4x^2\).
    \((x^2+2)^2 = x^4 + 4x^2 + 4\), which is \(4x^2\) too big - so subtract it back. This manufactured square is the standard trick (the Sophie Germain identity).
  2. Recognise a difference of two squares: \(A^2 - B^2\) with \(A = x^2 + 2\), \(B = 2x\).
    \(4x^2 = (2x)^2\), so the expression is now in the form that factorises.
  3. Factorise: \((x^2 + 2 - 2x)(x^2 + 2 + 2x) = (x^2 - 2x + 2)(x^2 + 2x + 2)\).
    \(A^2 - B^2 = (A-B)(A+B)\).
  4. Add the \(x\)-coefficients: \(-2 + 2 = 0\).
    The question asks only for this sum, which the symmetry makes obvious.
Where it goes wrong
  • Concluding it cannot be factorised because it has no real roots - a quartic with no real roots still factorises into two quadratics.
  • Adding the constant terms (\(2 + 2 = 4\)) instead of the \(x\)-coefficients.
Takeaway \(x^4 + 4a^4 = (x^2 - 2ax + 2a^2)(x^2 + 2ax + 2a^2)\): manufacture a square, then use the difference of two squares.
Answer B · Polynomials, factor and remainder theorems (MM1.6) · question POL-09
18Angles, polygons, congruence and vectors · spec M5.19 · Paper 1 style · ★★☆

\(\overrightarrow{AB} = 3\mathbf{p} + 2\mathbf{q}\) and \(\overrightarrow{BC} = k\mathbf{p} - 4\mathbf{q}\), where \(\mathbf{p}\) and \(\mathbf{q}\) are not parallel. Given that \(A\), \(B\) and \(C\) are collinear, find \(k\).

  1. A\(2\)
  2. B\(6\)
  3. C\(-3\)
  4. D\(\dfrac{3}{2}\)
  5. E\(-6\)
  6. F\(-4\)
Worked solution

Collinear means \(\overrightarrow{BC} = \lambda\overrightarrow{AB}\): \(-4 = 2\lambda \Rightarrow \lambda = -2\), so \(k = 3\lambda = -6\).

Full method — how and why
Collinear points mean one vector is a scalar multiple of the other. Compare coefficients of the two independent directions.
  1. Collinear \(A\), \(B\), \(C\) means \(\overrightarrow{BC} = \lambda\overrightarrow{AB}\) for some scalar \(\lambda\).
    They share the point \(B\), so the two vectors must point along the same line.
  2. Write it out: \(k\mathbf p - 4\mathbf q = \lambda(3\mathbf p + 2\mathbf q)\).
    Because \(\mathbf p\) and \(\mathbf q\) are not parallel, coefficients can be compared independently.
  3. Compare \(\mathbf q\): \(-4 = 2\lambda\), so \(\lambda = -2\).
    Use the equation that contains no unknown besides \(\lambda\).
  4. Compare \(\mathbf p\): \(k = 3\lambda = -6\).
    Substitute the value of \(\lambda\).
Where it goes wrong
  • Taking \(\lambda = 2\) and answering \(6\) - an offered option.
  • Comparing \(\mathbf p\) with \(\mathbf q\) coefficients.
Takeaway Non-parallel vectors are independent: equate their coefficients separately to solve for scalars.
Answer E · Angles, polygons, congruence and vectors (M5.19) · question GEO-11
19Triangles, radians and sectors · spec MM4.1 · Paper 1 style · ★★☆

In triangle \(ABC\), \(AB = 6\), \(BC = 4\) and angle \(BAC = 30^\circ\). How many possible triangles are there?

  1. A\(1\)
  2. B\(3\)
  3. C\(2\)
  4. D\(0\)
  5. Einfinitely many
Worked solution

Sine rule: \(\dfrac{\sin C}{6} = \dfrac{\sin 30^\circ}{4}\), so \(\sin C = \frac34\). Since \(BC = 4 < AB = 6\) and \(4 > 6\sin30^\circ = 3\), both the acute and obtuse values of \(C\) give valid triangles: the ambiguous case, two triangles.

Full method — how and why
Side-side-angle data: the ambiguous case. Compute \(\sin C\), then decide whether both the acute and obtuse angles give a valid triangle.
  1. Apply the sine rule: \(\dfrac{\sin C}{AB} = \dfrac{\sin A}{BC}\), i.e. \(\dfrac{\sin C}{6} = \dfrac{\sin 30^\circ}{4}\).
    Pair each angle with the side opposite it: \(C\) is opposite \(AB = 6\), and \(A\) is opposite \(BC = 4\).
  2. Solve: \(\sin C = \dfrac{6 \times 0.5}{4} = \dfrac34\), which is less than \(1\), so solutions exist.
    If \(\sin C\) came out greater than \(1\), no triangle would exist at all.
  3. Both \(C = \arcsin\frac34\) and \(180^\circ - \arcsin\frac34\) are possible in principle; check whether the obtuse one still leaves a positive third angle.
    \(\arcsin\frac34 \approx 48.6^\circ\), so the obtuse option is about \(131.4^\circ\); with \(A = 30^\circ\) that totals under \(180^\circ\) ✓.
  4. Equivalently: the side opposite the given angle (\(4\)) is shorter than the other given side (\(6\)) but longer than \(6\sin30^\circ = 3\) - the classic two-triangle condition. So there are 2 triangles.
    The condition \(b\sin A < a < b\) is the standard test for the ambiguous case.
Where it goes wrong
  • Assuming one triangle because the sine rule 'gives one answer'.
  • Assuming two triangles always - if \(a \geq b\) the obtuse option would make the angles exceed \(180^\circ\).
Takeaway SSA data: with \(b\sin A < a < b\) there are two triangles; with \(a \geq b\) exactly one; with \(a < b\sin A\) none.
Answer C · Triangles, radians and sectors (MM4.1) · question TRI-07
20Indices and surds · spec MM1.2 · Paper 1 style · ★★☆

Which one of the following is the largest?

  1. A\(2\sqrt{4.5}\)
  2. B\(\sqrt{2} + \sqrt{7}\)
  3. C\(\sqrt{3} + \sqrt{6}\)
  4. D\(\sqrt{4} + \sqrt{5}\)
  5. E\(\sqrt{1} + \sqrt{8}\)
Worked solution

Each option is \(\sqrt a + \sqrt b\) with \(a + b = 9\). Squaring gives \(9 + 2\sqrt{ab}\), which is largest when the product \(ab\) is largest, i.e. when \(a\) and \(b\) are as close as possible. Option E has \(a = b = 4.5\) (product \(20.25\)), beating \(4 \times 5 = 20\) for option C. Numerically: E \(= \sqrt{18} \approx 4.243\), C \(= 2 + \sqrt5 \approx 4.236\).

Full method — how and why
A comparison question. Every option has the form \(\sqrt a + \sqrt b\) with \(a + b = 9\) - spotting that shared structure turns guesswork into a one-line argument.
  1. Notice that in every option the two numbers under the roots add to \(9\) (including \(2\sqrt{4.5} = \sqrt{4.5} + \sqrt{4.5}\)).
    When all the candidates share a constraint, compare them through that constraint rather than by estimating each one.
  2. Square a general option: \((\sqrt a + \sqrt b)^2 = a + b + 2\sqrt{ab} = 9 + 2\sqrt{ab}\).
    All the options are positive, so the largest option is the one with the largest square - and only the \(ab\) term varies.
  3. With a fixed sum, \(ab\) is largest when \(a\) and \(b\) are closest together. The closest pair here is \(a = b = 4.5\), giving \(ab = 20.25\), beating \(4 \times 5 = 20\).
    This is the standard 'fixed perimeter, maximum area' fact; it decides the question without any decimals.
  4. So \(2\sqrt{4.5} = \sqrt{18} \approx 4.243\) is the largest (option C, \(2 + \sqrt5 \approx 4.236\), is a close second).
    Worth a sanity check because the top two are so close.
Where it goes wrong
  • Estimating roots to one decimal place - \(4.24\) versus \(4.24\) does not separate the top two options.
  • Assuming \(\sqrt1 + \sqrt8\) is biggest because it contains the largest single number; the spread-out pair is actually the smallest.
Takeaway For a fixed sum \(a + b\), the product \(ab\) - and hence \(\sqrt a + \sqrt b\) - is maximised when \(a\) and \(b\) are as close as possible.
Answer A · Indices and surds (MM1.2) · question IND-22
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