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Paper 1 · Set 08

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
75:00 PDF

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1Indices and surds · spec MM1.1 · Paper 1 style · ★☆☆

Simplify \(\dfrac{(2x^3)^4 \times 8x^{-5}}{4x^2}\).

  1. A\(32x^{9}\)
  2. B\(16x^{7}\)
  3. C\(x^{5}\)
  4. D\(16x^{5}\)
  5. E\(4x^{5}\)
  6. F\(32x^{5}\)
Worked solution

\((2x^3)^4 = 16x^{12}\), so the numerator is \(16x^{12}\times 8x^{-5} = 128x^{7}\). Dividing by \(4x^2\) gives \(32x^{5}\).

Full method — how and why
A pure index-laws question. Nothing here needs thought once you treat the numbers and the powers of \(x\) as two separate bookkeeping jobs.
  1. Deal with the bracket first: \((2x^3)^4 = 2^4 x^{12} = 16x^{12}\).
    A power outside a bracket multiplies every factor inside, so the \(2\) is raised to the 4th power as well as the \(x^3\). This is the step people skip.
  2. Multiply the numerator: \(16x^{12} \times 8x^{-5} = 128x^{7}\).
    Multiplying powers of the same base adds the indices: \(12 + (-5) = 7\). The numbers just multiply normally.
  3. Divide by \(4x^2\): \(\dfrac{128x^{7}}{4x^{2}} = 32x^{5}\).
    Dividing subtracts indices: \(7 - 2 = 5\), and \(128 \div 4 = 32\).
Where it goes wrong
  • Forgetting to raise the \(2\) to the 4th power gives \(2x^{12}\) at step 1 and the answer \(4x^5\).
  • Subtracting instead of adding the \(-5\) gives \(x^{9}\) and the answer \(32x^{9}\) - which is offered.
Takeaway Handle the number part and the index part separately, and always distribute an outside power over everything in the bracket.
Answer F · Indices and surds (MM1.1) · question IND-01
2Indices and surds · spec MM1.1 · Paper 1 style · ★☆☆

Solve \(4^{x+1} = 8^{2x-3}\).

  1. A\(x = \dfrac{11}{4}\)
  2. B\(x = \dfrac{5}{4}\)
  3. C\(x = \dfrac{9}{4}\)
  4. D\(x = \dfrac{7}{2}\)
  5. E\(x = 2\)
  6. F\(x = 4\)
Worked solution

Write both sides as powers of 2: \(2^{2x+2} = 2^{6x-9}\). So \(2x + 2 = 6x - 9\), giving \(4x = 11\) and \(x = \frac{11}{4}\).

Full method — how and why
An exponential equation where both sides can be written as powers of the same base. Once the bases match, the indices must be equal.
  1. Spot that \(4 = 2^2\) and \(8 = 2^3\).
    Both numbers are powers of \(2\), so the whole equation can be rewritten in base \(2\) and the exponentials compared directly.
  2. Rewrite: \(4^{x+1} = 2^{2(x+1)} = 2^{2x+2}\) and \(8^{2x-3} = 2^{3(2x-3)} = 2^{6x-9}\).
    \((a^m)^n = a^{mn}\): the outer index multiplies the whole inner index, so brackets are essential.
  3. Equate the indices: \(2x + 2 = 6x - 9\), so \(4x = 11\) and \(x = \frac{11}{4}\).
    \(2^u = 2^v\) forces \(u = v\) because \(2^x\) is a one-to-one function (it never takes the same value twice).
Where it goes wrong
  • Multiplying out as \(3 \times 2x - 3\) instead of \(3(2x - 3)\) loses the bracket and gives a different linear equation.
  • Writing \(8 = 2^4\) (a slip) gives \(x = \frac{7}{2}\), which is offered.
Takeaway Whenever an equation has different bases, look for a common base first; comparing indices is then legitimate because exponentials are one-to-one.
Answer A · Indices and surds (MM1.1) · question IND-03
3Quadratic functions · spec MM1.3 · Paper 1 style · ★☆☆

The quadratic \(x^2 + px + q\) has minimum value \(-4\) and one of its roots is \(0\). Which of the following could be the quadratic?

  1. A\(x^2 - 6x + 8\)
  2. B\(x^2 - 2x - 3\)
  3. C\(x^2 + 2x - 8\)
  4. D\(x^2 - 4x\)
  5. E\(x^2 + 6x + 5\)
Worked solution

Minimum \(-4\) means \(x^2 + px + q = (x-h)^2 - 4\), with roots \(h \pm 2\). One root is \(0\), so \(h = \pm 2\): the quadratic is \(x^2 - 4x\) or \(x^2 + 4x\). Only \(x^2 - 4x\) is offered.

Full method — how and why
Two pieces of information about a quadratic - its minimum and one root - pin the whole thing down. Working in completed-square form makes the roots visible.
  1. Minimum \(-4\) and leading coefficient \(1\) means \(x^2 + px + q = (x - h)^2 - 4\) for some \(h\).
    A monic quadratic with minimum \(m\) must be \((x - h)^2 + m\); only the position \(h\) is unknown.
  2. Its roots satisfy \((x - h)^2 = 4\), i.e. \(x = h \pm 2\).
    Solve the completed-square form directly rather than expanding.
  3. One root is \(0\), so \(h + 2 = 0\) or \(h - 2 = 0\): \(h = -2\) or \(h = 2\).
    Both cases are legitimate; the options decide which one is wanted.
  4. That gives \(x^2 + 4x\) or \(x^2 - 4x\); the list offers \(x^2 - 4x\).
    Check it: \(x^2 - 4x = (x-2)^2 - 4\), minimum \(-4\), roots \(0\) and \(4\).
Where it goes wrong
  • Picking \(x^2 - 2x - 3\): its minimum is \(-4\) but its roots are \(-1\) and \(3\), so \(0\) is not a root.
  • Checking only the minimum (several options have minimum \(-4\)) or only the roots.
Takeaway Test every stated condition - an option that satisfies one of them is designed to catch you.
Answer D · Quadratic functions (MM1.3) · question QUA-10
4Quadratic functions · spec MM1.3 · Paper 1 style · ★☆☆

Given that \(x^2 + y^2 = 10\) and \(x + y = 4\), find the value of \(xy\).

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

\((x+y)^2 = x^2 + y^2 + 2xy\), so \(16 = 10 + 2xy\) and \(xy = 3\).

Full method — how and why
A symmetric pair of equations. Expanding \((x+y)^2\) links the two given quantities to \(xy\) directly.
  1. Write the identity \((x + y)^2 = x^2 + y^2 + 2xy\).
    This is the only identity that connects a sum, a sum of squares and a product.
  2. Substitute the given values: \(4^2 = 10 + 2xy\).
    \(x + y = 4\) and \(x^2 + y^2 = 10\) are both given, so only \(xy\) is unknown.
  3. Solve: \(16 - 10 = 2xy \Rightarrow xy = 3\).
    Halve at the end - forgetting to divide by \(2\) gives \(6\), which is offered.
Where it goes wrong
  • Answering \(6\) by omitting the factor \(2\) in \(2xy\).
  • Trying to solve for \(x\) and \(y\) individually (they are \(1\) and \(3\)) - slower and unnecessary.
Takeaway \((x+y)^2 = x^2 + y^2 + 2xy\) converts between the three symmetric quantities; recognise which two you are given.
Answer C · Quadratic functions (MM1.3) · question QUA-16
5Simultaneous equations and inequalities · spec MM1.4 · Paper 1 style · ★☆☆

Two numbers have sum \(10\) and the sum of their squares is \(58\). What is the positive difference between them?

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

\((x - y)^2 = 2(x^2 + y^2) - (x+y)^2 = 116 - 100 = 16\), so the difference is \(4\) (the numbers are \(3\) and \(7\)).

Full method — how and why
Symmetric information again: a sum and a sum of squares. The identity for \((x-y)^2\) gets the difference without finding the numbers.
  1. Use \((x - y)^2 = x^2 + y^2 - 2xy\) and \((x + y)^2 = x^2 + y^2 + 2xy\).
    Adding these gives \((x+y)^2 + (x-y)^2 = 2(x^2 + y^2)\) - the relation that links all three given quantities.
  2. Rearrange: \((x - y)^2 = 2(x^2 + y^2) - (x + y)^2\).
    This avoids having to find \(xy\) as an intermediate step.
  3. Substitute: \(2(58) - 10^2 = 116 - 100 = 16\).
    Careful: it is \(2(x^2+y^2)\), not \(x^2 + y^2\).
  4. Take the positive square root: the difference is \(4\) (the numbers are \(3\) and \(7\)).
    The question asks for the positive difference, so take \(+4\).
Where it goes wrong
  • Forgetting the factor \(2\) and getting \((x-y)^2 = -42\), which is impossible.
  • Giving \(16\) (the square of the difference) as the answer.
Takeaway \((x+y)^2 + (x-y)^2 = 2(x^2+y^2)\): any two of sum, difference and sum-of-squares give the third.
Answer B · Simultaneous equations and inequalities (MM1.4) · question SIM-17
6Probability · spec M7.4 · Paper 1 style · ★☆☆

The probabilities that a train is early, on time or late are \(0.1\), \(x\) and \(2x\) respectively. Find the probability that the train is not late.

  1. A\(0.5\)
  2. B\(0.9\)
  3. C\(0.7\)
  4. D\(0.4\)
  5. E\(0.6\)
  6. F\(0.3\)
Worked solution

\(0.1 + 3x = 1 \Rightarrow x = 0.3\); not late \(= 0.1 + 0.3 = 0.4\).

Full method — how and why
The probabilities of an exhaustive set of outcomes sum to \(1\). Use that to find \(x\) first.
  1. Early, on time, late are the only possibilities, so \(0.1 + x + 2x = 1\).
    Exhaustive and mutually exclusive outcomes have total probability \(1\).
  2. Solve: \(3x = 0.9\), so \(x = 0.3\) and late \(= 0.6\).
    Subtract \(0.1\), then divide by \(3\).
  3. Not late \(=\) early \(+\) on time \(= 0.1 + 0.3 = 0.4\).
    Or use the complement: \(1 - 0.6 = 0.4\) - quicker and a useful check.
Where it goes wrong
  • Stopping at \(x = 0.3\) and answering that, or answering \(0.6\), the probability of being late.
  • Forgetting 'early' counts as 'not late'.
Takeaway Probabilities of all outcomes sum to \(1\); 'not late' is the complement of 'late'.
Answer D · Probability (M7.4) · question PRO-09
7Pythagoras, mensuration and 3D · spec M5.18 · Paper 1 style · ★☆☆

From the top of a cliff \(40\) m high, the angle of depression of a boat is \(30^\circ\). How far is the boat from the foot of the cliff?

  1. A\(20\) m
  2. B\(20\sqrt3\) m
  3. C\(40\sqrt{3}\) m
  4. D\(40\) m
  5. E\(80\) m
  6. F\(\dfrac{40}{\sqrt3}\) m
Worked solution

\(\tan30^\circ = \frac{40}{d} \Rightarrow d = \frac{40}{1/\sqrt3} = 40\sqrt3\).

Full method — how and why
An angle of depression from the top equals the angle of elevation from the boat. Then it is right-angled trigonometry.
  1. Draw the triangle: the cliff is vertical (\(40\) m), the sea horizontal, and the angle at the boat is \(30^\circ\).
    The angle of depression from the cliff top equals the angle of elevation from the boat (alternate angles with the horizontal).
  2. Relative to the \(30^\circ\) angle, the cliff is opposite and the distance \(d\) is adjacent: \(\tan 30^\circ = \dfrac{40}{d}\).
    \(\tan = \frac{\text{opposite}}{\text{adjacent}}\).
  3. Solve: \(d = \dfrac{40}{\tan 30^\circ} = \dfrac{40}{1/\sqrt3} = 40\sqrt3\) m.
    Dividing by \(\frac{1}{\sqrt3}\) multiplies by \(\sqrt3\).
Where it goes wrong
  • Multiplying by \(\tan 30^\circ\) instead of dividing, giving \(\frac{40}{\sqrt3}\) - an offered option.
  • Placing the \(30^\circ\) angle at the top of the cliff inside the triangle, which would swap opposite and adjacent.
Takeaway Depression from the top \(=\) elevation from the bottom; a shallow angle means a large horizontal distance.
Answer C · Pythagoras, mensuration and 3D (M5.18) · question MEN-16
8Functions and transformations · spec MM8.2 · Paper 1 style · ★☆☆

The graph of \(y = f(x)\) passes through \((4, 6)\). Which point must lie on the graph of \(y = 3f(2x) - 1\)?

  1. A\((2, 11)\)
  2. B\((2, 17)\)
  3. C\((8, 17)\)
  4. D\((8, 18)\)
  5. E\((4, 17)\)
  6. F\((2, 18)\)
Worked solution

\(f(2x)\) halves \(x\)-coordinates: \((2, 6)\). Then \(3f - 1\): \(y = 18 - 1 = 17\). Point \((2, 17)\).

Full method — how and why
Track a single known point through the transformations, remembering that inside changes act inversely on \(x\).
  1. Start with the known point \((4, 6)\) on \(y = f(x)\).
    One point is all you are given, so the answer must be its image.
  2. \(f(2x)\) squashes horizontally by factor \(\frac12\): the point with \(x\)-coordinate \(4\) moves to \(x = 2\), since \(f(2 \times 2) = f(4) = 6\).
    To get the same output from \(f(2x)\) you need half the input - this is the inverse action of an inside change.
  3. Multiply by \(3\): \(y = 3 \times 6 = 18\). Then subtract \(1\): \(y = 17\).
    Outside changes act directly on the output, in the order written.
  4. The point is \((2, 17)\).
    Check: \(3f(2 \times 2) - 1 = 3(6) - 1 = 17\) ✓.
Where it goes wrong
  • Doubling the \(x\)-coordinate to \(8\) instead of halving it.
  • Forgetting the \(-1\) and answering \((2, 18)\).
Takeaway Substitute the point back into the new equation to check: \(f(2x)\) at \(x = 2\) really is \(f(4)\).
Answer B · Functions and transformations (MM8.2) · question FUN-05
9Simultaneous equations and inequalities · spec MM1.4 · Paper 1 style · ★☆☆

Solve \(2x + 3y = 12\) and \(x^2 - y^2 = 0\). How many solutions \((x, y)\) are there?

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

\(x^2 = y^2\) means \(y = x\) or \(y = -x\). \(y = x\): \(5x = 12\), one solution. \(y = -x\): \(-x = 12\), one solution. Two in total.

Full method — how and why
\(x^2 - y^2 = 0\) factorises into two lines. The system is a line meeting a pair of lines, so count the intersections of each case.
  1. Factorise: \(x^2 - y^2 = (x - y)(x + y) = 0\), so \(y = x\) or \(y = -x\).
    A product is zero when either factor is zero - this splits the problem into two straight-line cases.
  2. Case \(y = x\): \(2x + 3x = 12 \Rightarrow 5x = 12\), one solution \(\left(\frac{12}{5}, \frac{12}{5}\right)\).
    Substitute into the linear equation; a linear equation in one unknown has exactly one root.
  3. Case \(y = -x\): \(2x - 3x = 12 \Rightarrow -x = 12\), one solution \((-12, 12)\).
    Same method; the coefficient of \(x\) is non-zero so again exactly one solution.
  4. Total: two solutions.
    The two cases give different points, so no double counting.
Where it goes wrong
  • Taking the square root as \(y = x\) only, halving the answer.
  • Concluding 'a line meets a quadratic in at most 2 points, so 2' without checking that neither case is parallel or degenerate.
Takeaway \(A^2 = B^2\) means \(A = B\) or \(A = -B\); both branches must be followed.
Answer B · Simultaneous equations and inequalities (MM1.4) · question SIM-14
10Pythagoras, mensuration and 3D · spec M5.16 · Paper 1 style · ★☆☆

A trapezium has parallel sides of length \(8\) cm and \(14\) cm and area \(77\) cm\(^2\). Find the perpendicular distance between the parallel sides.

  1. A\(14\) cm
  2. B\(6\) cm
  3. C\(7\) cm
  4. D\(11\) cm
  5. E\(3.5\) cm
  6. F\(5.5\) cm
Worked solution

\(\frac12(8 + 14)h = 11h = 77 \Rightarrow h = 7\).

Full method — how and why
Trapezium area formula, rearranged for the height.
  1. Write the formula: area \(= \frac12(a + b)h\) where \(a, b\) are the parallel sides.
    The height is the perpendicular distance between the parallel sides.
  2. Substitute: \(\frac12(8 + 14)h = 11h\).
    \(\frac12 \times 22 = 11\).
  3. Solve: \(11h = 77\), so \(h = 7\) cm.
    Divide both sides by \(11\).
Where it goes wrong
  • Forgetting the \(\frac12\) and answering \(3.5\) - an offered option.
  • Using the difference of the parallel sides instead of the sum.
Takeaway Trapezium: average the parallel sides, then multiply by the perpendicular height.
Answer C · Pythagoras, mensuration and 3D (M5.16) · question MEN-03
11Polynomials, factor and remainder theorems · spec MM1.6 · Paper 1 style · ★☆☆

\(p(x)\) is a cubic polynomial with \(p(1) = p(2) = p(3) = 0\) and \(p(0) = 12\). Find \(p(4)\).

  1. A\(24\)
  2. B\(-24\)
  3. C\(12\)
  4. D\(0\)
  5. E\(-12\)
  6. F\(6\)
Worked solution

\(p(x) = a(x-1)(x-2)(x-3)\); \(p(0) = -6a = 12 \Rightarrow a = -2\). \(p(4) = -2 \cdot 3\cdot2\cdot1 = -12\).

Full method — how and why
Three roots determine a cubic up to a constant multiple. One extra value fixes the multiple.
  1. Write \(p(x) = a(x-1)(x-2)(x-3)\).
    Given roots, start from the factorised form; the leading coefficient \(a\) is the only freedom left.
  2. Use \(p(0) = 12\): \(a(-1)(-2)(-3) = -6a = 12\), so \(a = -2\).
    Three negative factors give a negative product - a very common sign slip.
  3. Evaluate \(p(4) = -2(3)(2)(1) = -12\).
    Substitute directly into the factorised form; no expansion needed.
Where it goes wrong
  • Taking \(a = 2\) from \(6a = 12\) and answering \(+12\).
  • Assuming \(a = 1\) because the question says 'a cubic' - the leading coefficient is not given.
Takeaway Roots give the shape, one extra point gives the scale; keep everything in factorised form.
Answer E · Polynomials, factor and remainder theorems (MM1.6) · question POL-14
12GCSE algebra · spec M4.6 · Paper 1 style · ★☆☆

Simplify \(\dfrac{3}{x + 2} - \dfrac{2}{x - 1}\).

  1. A\(\dfrac{5x - 1}{(x + 2)(x - 1)}\)
  2. B\(\dfrac{x + 1}{(x + 2)(x - 1)}\)
  3. C\(\dfrac{1}{x + 3}\)
  4. D\(\dfrac{x - 7}{(x + 2)(x - 1)}\)
  5. E\(\dfrac{x - 1}{(x + 2)(x - 1)}\)
  6. F\(\dfrac{1}{(x + 2)(x - 1)}\)
Worked solution

\(\dfrac{3(x - 1) - 2(x + 2)}{(x + 2)(x - 1)} = \dfrac{x - 7}{(x + 2)(x - 1)}\).

Full method — how and why
Subtracting algebraic fractions: common denominator, then expand the numerator carefully - especially the minus sign.
  1. Common denominator: \((x+2)(x-1)\).
    The denominators share no factor, so their product is the lowest common denominator.
  2. Rewrite each fraction: \(\dfrac{3(x-1)}{(x+2)(x-1)} - \dfrac{2(x+2)}{(x+2)(x-1)}\).
    Multiply each numerator by whatever its denominator was missing.
  3. Combine and expand: \(3x - 3 - (2x + 4) = 3x - 3 - 2x - 4 = x - 7\).
    The minus sign applies to both terms of the second bracket - the classic error.
  4. So the answer is \(\dfrac{x-7}{(x+2)(x-1)}\).
    Numerator does not factorise further.
Where it goes wrong
  • Getting \(x + 1\) by only negating the \(2x\) - an offered option.
  • Subtracting the denominators as well.
Takeaway Bracket the whole second numerator before subtracting, then distribute the minus sign.
Answer D · GCSE algebra (M4.6) · question ALG-07
13Circles and circle theorems · spec MM3.3 · Paper 1 style · ★☆☆

\(PQRS\) is a cyclic quadrilateral with angle \(P = 3x\), angle \(R = x + 40^\circ\). Find \(x\).

  1. A\(45^\circ\)
  2. B\(50^\circ\)
  3. C\(35^\circ\)
  4. D\(30^\circ\)
  5. E\(40^\circ\)
  6. F\(25^\circ\)
Worked solution

Opposite angles of a cyclic quadrilateral sum to \(180^\circ\): \(4x + 40 = 180\), \(x = 35^\circ\).

Full method — how and why
Cyclic quadrilateral: opposite angles are supplementary. That gives a linear equation immediately.
  1. Identify the opposite pair: in \(PQRS\) the vertices go round in order, so \(P\) and \(R\) are opposite.
    Opposite means separated by one vertex on each side - reading the labelling order matters.
  2. Apply the theorem: \(3x + (x + 40) = 180\).
    Opposite angles of a cyclic quadrilateral sum to \(180^\circ\).
  3. Solve: \(4x = 140\), so \(x = 35^\circ\).
    Check: angles \(105^\circ\) and \(75^\circ\) sum to \(180^\circ\) ✓.
Where it goes wrong
  • Setting the angles equal (that is a parallelogram property, not a cyclic one).
  • Using \(360^\circ\) and getting \(x = 80^\circ\).
Takeaway Cyclic quadrilateral: opposite angles sum to \(180^\circ\); adjacent ones have no such relation.
Answer C · Circles and circle theorems (MM3.3) · question CIR-07
14Trigonometric functions and equations · spec MM4.6 · Paper 1 style · ★☆☆

How many solutions does \(\sin 2x = \dfrac{1}{2}\) have for \(0 \leq x < 2\pi\)?

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

\(2x\) runs over \([0, 4\pi)\), which contains two full periods of \(\sin\), each giving two solutions: \(4\) in total.

Full method — how and why
Counting solutions of \(\sin(kx) = c\). Transform the interval for the inside expression and count there.
  1. Let \(u = 2x\). As \(x\) runs over \([0, 2\pi)\), \(u\) runs over \([0, 4\pi)\).
    Multiplying the variable by \(2\) doubles the range of the inside expression - this is the whole trick.
  2. \([0, 4\pi)\) contains exactly two full periods of \(\sin\).
    \(\sin\) has period \(2\pi\), so a \(4\pi\)-long interval holds two complete waves.
  3. In each period, \(\sin u = \frac12\) has two solutions (one on the way up, one on the way down).
    A horizontal line strictly between \(-1\) and \(1\) cuts each wave twice.
  4. Total: \(2 \times 2 = 4\) solutions.
    Each \(u\) gives exactly one \(x\) via \(x = u/2\), so no solutions are lost or duplicated.
Where it goes wrong
  • Answering \(2\) by counting solutions for \(x\) as if the \(2\) were not there.
  • Answering \(8\) by doubling twice.
Takeaway For \(\sin(kx) = c\) on an interval of length \(L\), the inside runs over length \(kL\) - count solutions there.
Answer C · Trigonometric functions and equations (MM4.6) · question TRG-01
15Straight lines · spec MM3.1 · Paper 1 style · ★☆☆

The perpendicular bisector of the line segment joining \((1, 5)\) and \((7, 1)\) has equation

  1. A\(3x - 2y = 6\)
  2. B\(2x - 3y = -1\)
  3. C\(3x + 2y = 18\)
  4. D\(2x + 3y = 17\)
  5. E\(y = \frac32 x - 3\)
  6. F\(3x - 2y = 18\)
Worked solution

Midpoint \((4, 3)\); gradient of segment \(\frac{-4}{6} = -\frac23\), so the bisector has gradient \(\frac32\): \(y - 3 = \frac32(x - 4) \Rightarrow 2y - 6 = 3x - 12 \Rightarrow 3x - 2y = 6\).

Full method — how and why
Perpendicular bisector: through the midpoint, perpendicular to the segment. Two small calculations, then the point-gradient form.
  1. Midpoint: \(\left(\frac{1+7}{2}, \frac{5+1}{2}\right) = (4, 3)\).
    The bisector must pass through the middle of the segment.
  2. Gradient of the segment: \(\dfrac{1 - 5}{7 - 1} = -\dfrac{4}{6} = -\dfrac23\).
    Keep the order of subtraction consistent in numerator and denominator.
  3. Perpendicular gradient: \(\dfrac32\).
    Negative reciprocal of \(-\frac23\).
  4. Line through \((4,3)\): \(y - 3 = \frac32(x - 4) \Rightarrow 2y - 6 = 3x - 12 \Rightarrow 3x - 2y = 6\).
    Multiply by \(2\) to clear the fraction, then collect.
Where it goes wrong
  • Rearranging to \(3x - 2y = 18\) by a sign slip on the constant.
  • Using the segment's gradient instead of the perpendicular one.
Takeaway Perpendicular bisector \(=\) midpoint \(+\) negative reciprocal gradient; it is the locus of points equidistant from the two ends.
Answer A · Straight lines (MM3.1) · question LIN-06
16Exponentials and logarithms · spec MM5.2 · Paper 1 style · ★☆☆

Solve \(\log_3(x + 6) + \log_3(x - 2) = 2\).

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

\((x+6)(x-2) = 9 \Rightarrow x^2 + 4x - 21 = 0 \Rightarrow (x+7)(x-3) = 0\). \(x = -7\) makes both logs undefined, so \(x = 3\) only.

Full method — how and why
Combine the logs, convert to an equation, then check the domain - that is what this question is really testing.
  1. Combine: \(\log_3\left[(x+6)(x-2)\right] = 2\).
    Sum of logs is the log of the product.
  2. Convert to exponential form: \((x+6)(x-2) = 3^2 = 9\).
    \(\log_a N = c\) means \(N = a^c\).
  3. Expand and solve: \(x^2 + 4x - 12 = 9 \Rightarrow x^2 + 4x - 21 = 0 \Rightarrow (x+7)(x-3) = 0\), so \(x = -7\) or \(x = 3\).
    Ordinary quadratic work.
  4. Check both in the original equation: \(x = 3\) gives \(\log_3 9 + \log_3 1\) ✓; \(x = -7\) gives \(\log_3(-1)\), which is undefined. So \(x = 3\) only.
    Logarithms of negative numbers (or zero) do not exist, so roots that break the original expression must be rejected.
Where it goes wrong
  • Giving both roots - the option '\(x = -7\) or \(x = 3\)' exists precisely to catch this.
  • Checking only one of the two logarithms.
Takeaway Always substitute your solutions back: combining logs can create solutions the original equation never had.
Answer C · Exponentials and logarithms (MM5.2) · question LOG-05
17Exponentials and logarithms · spec MM5.3 · Paper 1 style · ★★☆

Given that \(5^{x} = 2\) and \(2^{y} = 125\), find the value of \(xy\).

  1. A\(1\)
  2. B\(3\)
  3. C\(5\)
  4. D\(2\)
  5. E\(6\)
  6. F\(\log_5 2 \cdot \log_2 125\)
Worked solution

\(5^{xy} = (5^x)^y = 2^y = 125 = 5^3\), so \(xy = 3\).

Full method — how and why
Two exponential facts that chain together. Raising the first to the power \(y\) links them.
  1. Raise the first equation to the power \(y\): \((5^x)^y = 2^y\), i.e. \(5^{xy} = 2^y\).
    \((a^x)^y = a^{xy}\) - this is what connects the two statements.
  2. Substitute the second fact: \(2^y = 125\).
    Given directly.
  3. Write \(125\) as a power of \(5\): \(5^{xy} = 5^3\).
    Both sides are now powers of \(5\).
  4. Equate the indices: \(xy = 3\).
    Valid because \(5^u\) is one-to-one.
Where it goes wrong
  • Trying to compute \(x\) and \(y\) separately as logarithms and getting stuck without a calculator.
  • Multiplying the two equations to get \(10^{x+y} = 250\), which leads nowhere.
Takeaway To find a product of exponents, raise one equation to the other's exponent and chain the results.
Answer B · Exponentials and logarithms (MM5.3) · question LOG-08
18Straight lines · spec MM3.1 · Paper 1 style · ★★☆

The lines \(y = 2x + 1\), \(y = -x + 7\) and the \(x\)-axis enclose a triangle. Find its area.

  1. A\(\dfrac{25}{2}\)
  2. B\(\dfrac{75}{4}\)
  3. C\(\dfrac{15}{2}\)
  4. D\(\dfrac{35}{2}\)
  5. E\(15\)
  6. F\(\dfrac{45}{4}\)
Worked solution

Intersection: \(2x + 1 = 7 - x \Rightarrow x = 2\), \(y = 5\). \(x\)-intercepts: \(-\frac12\) and \(7\), base \(= \frac{15}{2}\). Area \(= \frac12 \cdot \frac{15}{2} \cdot 5 = \frac{75}{4}\).

Full method — how and why
A triangle bounded by two lines and an axis. Find the three vertices, then use the axis segment as the base.
  1. Find where the two lines meet: \(2x + 1 = -x + 7 \Rightarrow 3x = 6 \Rightarrow x = 2\), \(y = 5\). Apex \((2,5)\).
    Equate the two expressions for \(y\).
  2. Find the \(x\)-intercepts: \(2x + 1 = 0\) gives \(x = -\frac12\); \(-x + 7 = 0\) gives \(x = 7\).
    The third side is along the \(x\)-axis, so the other two vertices are the \(x\)-intercepts.
  3. Base length: from \(-\frac12\) to \(7\) is \(7 - (-\frac12) = \frac{15}{2}\).
    Subtract, do not add; the negative intercept makes the base longer, not shorter.
  4. Height is the \(y\)-coordinate of the apex, \(5\): area \(= \frac12 \times \frac{15}{2} \times 5 = \frac{75}{4}\).
    With a horizontal base, the height is just the apex's distance from the axis.
Where it goes wrong
  • Using base \(7 - \frac12 = \frac{13}{2}\) by mishandling the negative intercept.
  • Using the apex \(x\)-coordinate \(2\) as the height.
Takeaway Put the base on an axis whenever you can: the height is then simply a coordinate.
Answer B · Straight lines (MM3.1) · question LIN-04
19Circles and circle theorems · spec MM3.2 · Paper 1 style · ★★☆

Find the length of the tangent from the point \((7, 1)\) to the circle \(x^2 + y^2 - 2x - 4y - 4 = 0\).

  1. A\(3\)
  2. B\(\sqrt{28}\)
  3. C\(6\)
  4. D\(\sqrt{26}\)
  5. E\(\sqrt{37}\)
  6. F\(5\)
Worked solution

Centre \((1, 2)\), radius \(3\). Distance from \((7, 1)\) to centre: \(\sqrt{36 + 1} = \sqrt{37}\). Tangent length \(= \sqrt{37 - 9} = \sqrt{28}\).

Full method — how and why
The tangent, radius and line to the external point form a right-angled triangle. Pythagoras gives the tangent length.
  1. Put the circle in centre-radius form: \((x-1)^2 + (y-2)^2 = 4 + 1 + 4 = 9\), so centre \((1,2)\), radius \(3\).
    Complete the square in \(x\) and \(y\); the constant \(-4\) moves across.
  2. Distance from \((7,1)\) to the centre: \(\sqrt{(7-1)^2 + (1-2)^2} = \sqrt{36 + 1} = \sqrt{37}\).
    This is the hypotenuse of the right-angled triangle (the tangent meets the radius at \(90^\circ\)).
  3. Tangent length \(= \sqrt{PC^2 - r^2} = \sqrt{37 - 9} = \sqrt{28}\).
    Subtract the squares, not the lengths - the right angle is at the point of contact.
Where it goes wrong
  • Answering \(\sqrt{37}\) (that is the distance to the centre) or \(\sqrt{37} - 3\).
  • Getting \(r = 9\) instead of \(3\) from the completed square.
Takeaway Tangent length from \(P\): \(\sqrt{PC^2 - r^2}\), because tangent \(\perp\) radius makes a right-angled triangle.
Answer B · Circles and circle theorems (MM3.2) · question CIR-09
20Binomial expansion · spec MM2.4 · Paper 1 style · ★★★

How many terms in the expansion of \((x + y + z)^{4}\) (after collecting like terms) are there?

  1. A\(16\)
  2. B\(15\)
  3. C\(10\)
  4. D\(5\)
  5. E\(12\)
  6. F\(81\)
Worked solution

Terms are \(x^ay^bz^c\) with \(a + b + c = 4\), \(a, b, c \geq 0\). Counting: for \(a = 0\) there are \(5\) choices of \((b, c)\), \(a = 1\): \(4\), ..., \(a = 4\): \(1\). Total \(5 + 4 + 3 + 2 + 1 = 15\).

Full method — how and why
A trinomial expansion. Counting the distinct terms is a counting problem: how many ways can the powers add to \(4\)?
  1. Every term has the form \(x^ay^bz^c\) with \(a + b + c = 4\) and \(a, b, c \geq 0\).
    Each of the four brackets contributes exactly one of \(x\), \(y\) or \(z\), so the powers add to \(4\).
  2. Count systematically by the value of \(a\): if \(a = 0\) then \(b + c = 4\), giving \(5\) pairs; \(a = 1\) gives \(4\); \(a=2\) gives \(3\); \(a=3\) gives \(2\); \(a=4\) gives \(1\).
    For a fixed \(a\), the number of \((b,c)\) pairs is \((4 - a) + 1\).
  3. Total: \(5 + 4 + 3 + 2 + 1 = 15\).
    A triangular number - the standard result \(\binom{n+2}{2}\) for three variables, here \(\binom62 = 15\).
Where it goes wrong
  • Answering \(3^4 = 81\) - that counts the terms before collecting like terms.
  • Answering \(5\) by counting only the powers of \(x\).
Takeaway Distinct terms of \((x+y+z)^n\) correspond to solutions of \(a+b+c=n\) in non-negative integers: \(\binom{n+2}{2}\) of them.
Answer B · Binomial expansion (MM2.4) · question BIN-14
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