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Paper 1 · Set 10

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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Answer grid

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1Binomial expansion · spec MM2.4 · Paper 1 style · ★☆☆

Expand \((1 - 2x)^4\) and hence find the coefficient of \(x^2\) in \((1 + 3x)(1 - 2x)^4\).

  1. A\(24\)
  2. B\(-24\)
  3. C\(48\)
  4. D\(0\)
  5. E\(12\)
  6. F\(-48\)
Worked solution

\((1-2x)^4 = 1 - 8x + 24x^2 - 32x^3 + 16x^4\). Coefficient of \(x^2\) in the product: \(1\cdot 24 + 3\cdot(-8) = 0\).

Full method — how and why
A product of a binomial expansion with a linear factor. Only two products can give \(x^2\), so collect just those.
  1. Expand \((1-2x)^4\): coefficients \(\binom4r(-2)^r\) give \(1 - 8x + 24x^2 - 32x^3 + 16x^4\).
    \(\binom41(-2) = -8\) and \(\binom42(-2)^2 = 6 \times 4 = 24\); remember to raise the \(-2\) to the power as well.
  2. To make \(x^2\) in the product, pair \(1 \times 24x^2\) and \(3x \times (-8x)\).
    Degrees must add to \(2\): \((0, 2)\) and \((1, 1)\) are the only options.
  3. Add: \(24 + (-24) = 0\).
    The two contributions cancel exactly - the answer is \(0\), not 'no term'.
Where it goes wrong
  • Taking the \(x^2\) coefficient of \((1-2x)^4\) as \(+24\) but forgetting the \(3x \times (-8x)\) contribution, giving \(24\).
  • Sign error making \((-2)^2 = -4\).
Takeaway For a coefficient in a product, list the degree pairs that add to the target and sum their contributions.
Answer D · Binomial expansion (MM2.4) · question BIN-07
2Exponentials and logarithms · spec MM5.2 · Paper 1 style · ★☆☆

If \(\log_a b = 2\) and \(\log_b c = 3\), find \(\log_a c\).

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

\(b = a^2\) and \(c = b^3 = a^6\), so \(\log_a c = 6\).

Full method — how and why
A chain of logarithms. Convert each statement to exponential form and substitute.
  1. \(\log_a b = 2\) means \(b = a^2\).
    \(\log_a b = c \Leftrightarrow a^c = b\).
  2. \(\log_b c = 3\) means \(c = b^3\).
    Same conversion with the next pair.
  3. Substitute: \(c = (a^2)^3 = a^6\).
    Powers of powers multiply.
  4. Therefore \(\log_a c = 6\).
    Read the exponent straight off. (In general \(\log_a c = \log_a b \times \log_b c = 2 \times 3\).)
Where it goes wrong
  • Adding the values to get \(5\).
  • Reading \((a^2)^3\) as \(a^5\).
Takeaway Logarithms chain by multiplication: \(\log_a c = \log_a b \times \log_b c\).
Answer B · Exponentials and logarithms (MM5.2) · question LOG-18
3Angles, polygons, congruence and vectors · spec M5.2 · Paper 1 style · ★☆☆

The interior angle of a regular polygon is \(156^\circ\). How many sides does it have?

  1. A\(24\)
  2. B\(20\)
  3. C\(10\)
  4. D\(12\)
  5. E\(18\)
  6. F\(15\)
Worked solution

Exterior angle \(24^\circ\); \(360/24 = 15\) sides.

Full method — how and why
Work with the exterior angle: they always sum to \(360^\circ\), which makes the arithmetic trivial.
  1. Exterior angle \(= 180^\circ - 156^\circ = 24^\circ\).
    Interior and exterior angles at a vertex lie on a straight line.
  2. The exterior angles of any polygon sum to \(360^\circ\), and for a regular polygon they are all equal.
    Walking once around the shape turns you through a full circle.
  3. Number of sides \(= \dfrac{360}{24} = 15\).
    Divide the total turn by the turn per vertex.
Where it goes wrong
  • Using \(\frac{360}{156}\) or \(\frac{156}{24}\).
  • Using the interior-angle-sum formula and then having to solve \(\frac{180(n-2)}{n} = 156\) - correct but slower.
Takeaway For regular polygons, go via the exterior angle: \(n = \frac{360}{\text{exterior}}\).
Answer F · Angles, polygons, congruence and vectors (M5.2) · question GEO-01
4Sequences and series · spec MM2.3 · Paper 1 style · ★☆☆

A geometric series has first term \(2\) and common ratio \(r > 0\). The sum of the first three terms is \(14\). Find the sum of the first five terms.

  1. A\(30\)
  2. B\(52\)
  3. C\(46\)
  4. D\(94\)
  5. E\(126\)
  6. F\(62\)
Worked solution

\(2 + 2r + 2r^2 = 14 \Rightarrow r^2 + r - 6 = 0 \Rightarrow r = 2\). \(S_5 = 2(2^5 - 1) = 62\).

Full method — how and why
The first three terms give a quadratic in \(r\); the positivity condition picks one root.
  1. Write the sum of three terms: \(2 + 2r + 2r^2 = 14\).
    Terms are \(a\), \(ar\), \(ar^2\) with \(a = 2\).
  2. Divide by \(2\) and rearrange: \(r^2 + r - 6 = 0\).
    Dividing first keeps the numbers small.
  3. Factorise: \((r+3)(r-2) = 0\), so \(r = 2\) or \(r = -3\). The condition \(r > 0\) gives \(r = 2\).
    The question's restriction exists precisely to discard the negative root.
  4. Sum five terms: \(S_5 = 2\dfrac{2^5 - 1}{2-1} = 2 \times 31 = 62\).
    \(2^5 = 32\), so the bracket is \(31\).
Where it goes wrong
  • Using \(r = -3\) and getting \(S_5 = 2 \times \frac{-243-1}{-4} = 122\).
  • Adding two more terms by hand and slipping: \(2 + 4 + 8 + 16 + 32 = 62\) is a good check.
Takeaway Conditions like '\(r > 0\)' are there to select between roots - apply them before continuing.
Answer F · Sequences and series (MM2.3) · question SEQ-10
5Indices and surds · spec MM1.2 · Paper 1 style · ★☆☆

The number \(\dfrac{\sqrt{50} + \sqrt{18}}{\sqrt{8}}\) is equal to

  1. A\(4\)
  2. B\(\dfrac{8}{\sqrt 2}\)
  3. C\(\sqrt{34}\)
  4. D\(8\)
  5. E\(2\sqrt{2}\)
  6. F\(4\sqrt{2}\)
Worked solution

\(\sqrt{50} = 5\sqrt2\), \(\sqrt{18} = 3\sqrt2\), \(\sqrt8 = 2\sqrt2\). So \(\dfrac{8\sqrt2}{2\sqrt2} = 4\).

Full method — how and why
A quotient of surds that collapses completely. Simplify each root first and the \(\sqrt2\)'s cancel.
  1. Simplify: \(\sqrt{50} = 5\sqrt2\), \(\sqrt{18} = 3\sqrt2\), \(\sqrt8 = 2\sqrt2\).
    Every number is (square) \(\times\, 2\), so every root is a multiple of \(\sqrt2\) - which is why the fraction will be rational.
  2. Add the numerator: \(5\sqrt2 + 3\sqrt2 = 8\sqrt2\).
    Like terms again - \(\sqrt2\) behaves as a common factor.
  3. Divide: \(\dfrac{8\sqrt2}{2\sqrt2} = 4\).
    The \(\sqrt2\) cancels top and bottom, leaving plain arithmetic.
Where it goes wrong
  • Working out \(\sqrt{50} + \sqrt{18} = \sqrt{68}\), which is wrong and leads nowhere.
  • Cancelling only the numbers and leaving a stray \(\sqrt2\), giving \(4\sqrt2\).
Takeaway Simplify every surd to the form \(k\sqrt m\) before adding or dividing; common surd factors then cancel visibly.
Answer A · Indices and surds (MM1.2) · question IND-14
6Integration · spec MM7.2 · Paper 1 style · ★☆☆

Evaluate \(\displaystyle\int_{-1}^{2} |x|\,dx\).

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

Two triangles: \(\frac12\cdot1\cdot1 + \frac12\cdot2\cdot2 = \frac12 + 2 = \frac52\).

Full method — how and why
A modulus integrand. Split at the corner, or - faster - read the areas off as triangles.
  1. Note \(|x|\) changes formula at \(x = 0\), which is inside \([-1, 2]\).
    Split at every point where the modulus switches branch.
  2. On \([-1, 0]\): \(|x| = -x\), and the graph is a triangle with base \(1\) and height \(1\), area \(\frac12\).
    Sketching \(y = |x|\) makes the integral a pair of triangles - no antiderivative needed.
  3. On \([0, 2]\): \(|x| = x\), a triangle with base \(2\) and height \(2\), area \(2\).
    \(\frac12 \times 2 \times 2 = 2\).
  4. Add: \(\frac12 + 2 = \frac52\).
    Both pieces are above the axis, so both count positively.
Where it goes wrong
  • Integrating \(x\) straight through and getting \(\frac{4}{2} - \frac12 = \frac32\) - an offered option.
  • Treating the left piece as negative.
Takeaway \(|x|\) is never negative: split at the corner, and use triangle areas when the graph is piecewise linear.
Answer D · Integration (MM7.2) · question INT-20
7Polynomials, factor and remainder theorems · spec MM1.6 · Paper 1 style · ★☆☆

\((x + 2)\) is a factor of \(2x^3 + ax^2 - 5x - 6\). Find the value of \(a\).

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

Factor Theorem: \(f(-2) = -16 + 4a + 10 - 6 = 4a - 12 = 0\), so \(a = 3\).

Full method — how and why
'Is a factor' means remainder zero, so the Factor Theorem turns the statement into a single equation for \(a\).
  1. \((x + 2)\) vanishes at \(x = -2\).
    Write \(x + 2 = x - (-2)\) if the sign is confusing; the root is the value that makes the bracket zero.
  2. Substitute: \(f(-2) = 2(-8) + a(4) - 5(-2) - 6 = -16 + 4a + 10 - 6\).
    Careful with the odd power: \((-2)^3 = -8\), while \((-2)^2 = +4\).
  3. Simplify: \(4a - 12\).
    \(-16 + 10 - 6 = -12\).
  4. Set to zero (factor \(\Rightarrow\) remainder zero): \(4a = 12\), so \(a = 3\).
    That is precisely the Factor Theorem.
Where it goes wrong
  • Using \(x = 2\) instead of \(x = -2\), which gives \(a = -3\) - an offered option.
  • Sign slips on \((-2)^3\) and \(-5(-2)\); write each term out separately.
Takeaway Factor \((x + k)\) means substitute \(x = -k\) and set the result to zero.
Answer F · Polynomials, factor and remainder theorems (MM1.6) · question POL-02
8Pythagoras, mensuration and 3D · spec M5.16 · Paper 1 style · ★☆☆

A cylindrical tin of radius \(4\) cm and height \(10\) cm is full of water. The water is poured into an empty cylindrical jug of radius \(5\) cm. How deep is the water in the jug?

  1. A\(6\) cm
  2. B\(12.5\) cm
  3. C\(8\) cm
  4. D\(7.2\) cm
  5. E\(5\) cm
  6. F\(6.4\) cm
Worked solution

\(\pi\cdot16\cdot10 = \pi\cdot25\cdot h \Rightarrow h = \frac{160}{25} = 6.4\).

Full method — how and why
Conservation of volume: the water keeps its volume, so equate the two cylinder volumes.
  1. Volume in the tin: \(\pi r^2 h = \pi(16)(10) = 160\pi\).
    \(r = 4\) so \(r^2 = 16\).
  2. In the jug the same volume occupies \(\pi(25)h\).
    Radius \(5\), unknown depth \(h\); the \(\pi\) will cancel.
  3. Equate and solve: \(160\pi = 25\pi h \Rightarrow h = \dfrac{160}{25} = 6.4\) cm.
    The wider jug means a shallower depth - a good sanity check.
Where it goes wrong
  • Using the radii rather than their squares, giving \(8\) cm - an offered option.
  • Expecting the depth to increase because the jug is 'bigger'.
Takeaway Pouring conserves volume: equate \(\pi r_1^2h_1 = \pi r_2^2h_2\); a wider container gives a shallower depth.
Answer F · Pythagoras, mensuration and 3D (M5.16) · question MEN-14
9Sequences and series · spec MM2.2 · Paper 1 style · ★☆☆

How many terms of the arithmetic series \(3 + 7 + 11 + \ldots\) are needed for the sum to first exceed \(500\)?

  1. A\(17\)
  2. B\(16\)
  3. C\(11\)
  4. D\(14\)
  5. E\(18\)
  6. F\(15\)
Worked solution

\(S_n = \frac n2(6 + 4(n-1)) = n(2n + 1)\). \(n = 15\): \(465\); \(n = 16\): \(528 > 500\). So \(16\).

Full method — how and why
An inequality on a sum. Form \(S_n\), then find the first integer \(n\) that works - by trial, since \(n\) must be a whole number.
  1. Identify \(a = 3\), \(d = 4\) and form the sum: \(S_n = \frac n2\left[6 + 4(n-1)\right] = \frac n2(4n + 2) = n(2n+1)\).
    Simplifying to \(n(2n+1)\) makes the trial values much easier to compute.
  2. Test \(n = 15\): \(15 \times 31 = 465\), which is less than \(500\).
    Estimate first: \(2n^2 \approx 500\) gives \(n \approx 15.8\), so start near there.
  3. Test \(n = 16\): \(16 \times 33 = 528 > 500\) ✓.
    The sum is increasing, so the first \(n\) that exceeds \(500\) is the answer.
  4. Answer: \(16\) terms.
    Check the one below to be sure it fails - that is what 'first exceeds' demands.
Where it goes wrong
  • Solving \(2n^2 + n = 500\) and rounding down to \(15\).
  • Forgetting that \(n\) must be a positive integer and quoting a decimal.
Takeaway For 'first exceeds' questions, solve approximately then test the two neighbouring integers.
Answer B · Sequences and series (MM2.2) · question SEQ-09
10Binomial expansion · spec MM2.4 · Paper 1 style · ★☆☆

Evaluate \(\binom{10}{3} - \binom{10}{7} + \binom{9}{2}\).

  1. A\(36\)
  2. B\(240\)
  3. C\(0\)
  4. D\(84\)
  5. E\(120\)
  6. F\(156\)
Worked solution

\(\binom{10}{3} = \binom{10}{7}\), so the first two cancel; \(\binom92 = 36\).

Full method — how and why
A short calculation made trivial by the symmetry of binomial coefficients.
  1. Use the symmetry \(\binom nr = \binom n{n-r}\): since \(3 + 7 = 10\), \(\binom{10}{3} = \binom{10}{7}\).
    Choosing \(3\) objects from \(10\) is the same as choosing which \(7\) to leave behind.
  2. So the first two terms cancel, leaving \(\binom92\).
    Spotting the cancellation avoids computing \(120\) twice.
  3. Evaluate \(\binom92 = \frac{9 \times 8}{2} = 36\).
    \(\binom n2 = \frac{n(n-1)}{2}\).
Where it goes wrong
  • Computing all three coefficients and mis-adding, e.g. \(120 - 120 + 36\) done as \(156\).
  • Thinking \(\binom{10}{3} - \binom{10}{7}\) is \(\binom{10}{-4}\) or similar.
Takeaway \(\binom nr = \binom n{n-r}\): look for pairs whose lower entries add to \(n\).
Answer A · Binomial expansion (MM2.4) · question BIN-05
11Trigonometric functions and equations · spec MM4.6 · Paper 1 style · ★☆☆

Given that \(\sin x + \cos x = 0\) and \(0 \leq x < 2\pi\), find the sum of all possible values of \(x\).

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

\(\tan x = -1\): \(x = \frac{3\pi}{4}\) or \(\frac{7\pi}{4}\). Sum \(= \frac{10\pi}{4} = \frac{5\pi}{2}\).

Full method — how and why
Convert a sum of \(\sin\) and \(\cos\) into a single \(\tan\) equation by dividing.
  1. Rearrange: \(\sin x = -\cos x\).
    Move one term across.
  2. Divide by \(\cos x\): \(\tan x = -1\). (Check \(\cos x = 0\) separately: then \(\sin x = 0\) too, impossible.)
    Dividing by \(\cos x\) is safe only after ruling out \(\cos x = 0\) - and here that case gives a contradiction.
  3. Solve on \([0, 2\pi)\): \(\tan x = -1\) at \(x = \frac{3\pi}{4}\) and \(x = \frac{3\pi}{4} + \pi = \frac{7\pi}{4}\).
    Reference angle \(\frac\pi4\), negative tangent in quadrants 2 and 4.
  4. Sum: \(\frac{3\pi}{4} + \frac{7\pi}{4} = \frac{10\pi}{4} = \frac{5\pi}{2}\).
    Add over the common denominator and simplify.
Where it goes wrong
  • Finding only one solution and answering \(\frac{3\pi}{4}\) - an offered option.
  • Forgetting to justify dividing by \(\cos x\) (here it is fine, but the habit matters).
Takeaway \(a\sin x + b\cos x = 0\) becomes \(\tan x = -\frac ba\); check the \(\cos x = 0\) case before dividing.
Answer B · Trigonometric functions and equations (MM4.6) · question TRG-16
12Number, units and estimation · spec M1.1 · Paper 1 style · ★☆☆

Water flows through a pipe at \(1.2\) litres per second. How long does it take to fill a tank of volume \(3.6\) m\(^3\)? (1 m\(^3\) = 1000 litres.)

  1. A\(30\) minutes
  2. B\(50\) minutes
  3. C\(5\) minutes
  4. D\(1\) hour
  5. E\(3\) hours
  6. F\(500\) seconds
Worked solution

\(3600\) litres at \(1.2\) litres/s takes \(3000\) s \(= 50\) minutes.

Full method — how and why
A rate problem with a unit conversion. Convert the volume to litres, divide by the rate, then convert the time.
  1. Convert the volume: \(3.6\) m\(^3 = 3.6 \times 1000 = 3600\) litres.
    The conversion is given in the question.
  2. Divide by the rate: \(\dfrac{3600}{1.2} = 3000\) seconds.
    Volume divided by volume-per-second gives seconds.
  3. Convert to minutes: \(\dfrac{3000}{60} = 50\) minutes.
    The options are in minutes, so finish the conversion.
Where it goes wrong
  • Leaving the answer as \(3000\) seconds and picking \(500\) seconds by a slip.
  • Dividing by \(1.2\) as if it were \(12\), giving \(5\) minutes.
Takeaway Make the units match before dividing, and convert the answer into the units the options use.
Answer B · Number, units and estimation (M1.1) · question NUM-16
13Graphs of functions · spec MM8.7 · Paper 1 style · ★☆☆

The graphs of \(y = x^2 - 2x\) and \(y = k\) meet at exactly two points. Find the complete set of values of \(k\).

  1. A\(k > -1\)
  2. B\(k \geq -1\)
  3. C\(k > 0\)
  4. D\(k = -1\)
  5. E\(k < -1\)
  6. Fall real \(k\)
Worked solution

\(y = (x-1)^2 - 1\) has minimum \(-1\); a horizontal line above the minimum meets the parabola twice.

Full method — how and why
A horizontal line meeting a parabola twice: the line must be above the minimum.
  1. Complete the square: \(x^2 - 2x = (x-1)^2 - 1\), so the minimum value is \(-1\) at \(x = 1\).
    The turning point governs how many times a horizontal line can cut the curve.
  2. A line \(y = k\) meets the parabola twice when \(k\) is strictly above the minimum.
    At \(k = -1\) the line is tangent at the vertex: exactly one point. Below it: none.
  3. So \(k > -1\).
    Strict inequality, because equality gives only one intersection.
Where it goes wrong
  • Including \(k = -1\) and choosing \(k \geq -1\) - offered as a distractor.
  • Using the \(y\)-intercept \(0\) instead of the minimum value.
Takeaway Count intersections with a horizontal line by sliding it against the turning point; equality gives tangency.
Answer A · Graphs of functions (MM8.7) · question GRA-02
14GCSE algebra · spec M4.16 · Paper 1 style · ★☆☆

The sum of two numbers is \(12\) and the difference of their squares is \(48\). Find the larger number.

  1. A\(6\)
  2. B\(10\)
  3. C\(8\)
  4. D\(7\)
  5. E\(4\)
  6. F\(9\)
Worked solution

\(x^2 - y^2 = (x - y)(x + y) = 12(x - y) = 48 \Rightarrow x - y = 4\). So \(x = 8\), \(y = 4\).

Full method — how and why
A sum and a difference of squares. Factorising the difference of squares links them without solving for both numbers.
  1. Factorise: \(x^2 - y^2 = (x-y)(x+y)\).
    This is the identity that connects the two given facts.
  2. Substitute the known sum: \(48 = (x-y)(12)\), so \(x - y = 4\).
    The sum \(x + y = 12\) is given directly.
  3. Solve the pair \(x + y = 12\), \(x - y = 4\): adding gives \(2x = 16\), so \(x = 8\) (and \(y = 4\)).
    Adding the two equations eliminates \(y\).
  4. The larger number is \(8\).
    Check: \(64 - 16 = 48\) ✓.
Where it goes wrong
  • Answering \(4\) (the smaller number or the difference).
  • Trying to solve \(x^2 - y^2 = 48\) by trial.
Takeaway \(x^2 - y^2 = (x-y)(x+y)\) converts a difference of squares into a linear pair.
Answer C · GCSE algebra (M4.16) · question ALG-14
15Exponentials and logarithms · spec MM5.2 · Paper 1 style · ★☆☆

Find the value of \(\log_2 40 - \log_2 5 + \log_2 \dfrac{1}{4}\).

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

\(\log_2\left(\frac{40}{5}\cdot\frac14\right) = \log_2 2 = 1\).

Full method — how and why
Combine everything into a single logarithm using the laws, then evaluate.
  1. Combine the first two: \(\log_2 40 - \log_2 5 = \log_2\dfrac{40}{5} = \log_2 8\).
    A difference of logs (same base) is the log of the quotient.
  2. Add the third: \(\log_2 8 + \log_2\frac14 = \log_2\left(8 \times \frac14\right) = \log_2 2\).
    A sum of logs is the log of the product. Note \(\log_2\frac14\) is negative, so this step reduces the total.
  3. Evaluate: \(\log_2 2 = 1\).
    \(\log_a a = 1\) for any valid base.
Where it goes wrong
  • Treating \(\log_2\frac14\) as \(+2\) rather than \(-2\) and answering \(5\).
  • Dividing the logarithms instead of subtracting them.
Takeaway \(\log\) turns \(\times\) into \(+\) and \(\div\) into \(-\); combine into one log before evaluating.
Answer A · Exponentials and logarithms (MM5.2) · question LOG-01
16Functions and transformations · spec MM8.2 · Paper 1 style · ★☆☆

Which of the following is the equation of the graph obtained by stretching \(y = x^3 - 3x\) parallel to the \(x\)-axis with scale factor \(\dfrac{1}{2}\)?

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

A stretch parallel to the \(x\)-axis with factor \(\frac12\) gives \(y = f(2x) = (2x)^3 - 3(2x) = 8x^3 - 6x\).

Full method — how and why
A horizontal stretch of factor \(\frac12\) replaces \(x\) by \(2x\) - the reciprocal of the factor.
  1. A stretch parallel to the \(x\)-axis with scale factor \(k\) gives \(y = f\!\left(\frac{x}{k}\right)\); with \(k = \frac12\) that is \(y = f(2x)\).
    Inside changes act inversely: to squash the graph to half its width you must double the input.
  2. Substitute: \(y = (2x)^3 - 3(2x)\).
    Replace every \(x\) in the formula, including the one in the linear term.
  3. Simplify: \(8x^3 - 6x\).
    \((2x)^3 = 8x^3\) - the coefficient is cubed too.
Where it goes wrong
  • Replacing only the first \(x\), giving \(8x^3 - 3x\) - an offered option.
  • Using \(y = f(x/2)\) and getting \(\frac{x^3}{8} - \frac{3x}{2}\), which is a stretch of factor \(2\).
Takeaway Stretch factor \(k\) parallel to the \(x\)-axis \(\Rightarrow\) replace \(x\) by \(x/k\); substitute into every occurrence.
Answer E · Functions and transformations (MM8.2) · question FUN-15
17Polynomials, factor and remainder theorems · spec MM1.6 · Paper 1 style · ★★☆

When \(x^3 - 3x^2 + 2x + 7\) is divided by \(x^2 + 1\), what is the remainder?

  1. A\(x + 4\)
  2. B\(x + 10\)
  3. C\(10\)
  4. D\(-x + 10\)
  5. E\(x + 7\)
  6. F\(3x + 4\)
Worked solution

\(x^3 - 3x^2 + 2x + 7 = (x^2 + 1)(x - 3) + (x + 10)\): expanding gives \(x^3 - 3x^2 + x - 3 + x + 10\). Remainder \(x + 10\).

Full method — how and why
Division by a quadratic again: remainder is linear. Comparing coefficients is fastest because the divisor has no \(x\) term.
  1. Write \(x^3 - 3x^2 + 2x + 7 = (x^2 + 1)(x + c) + (ax + b)\).
    Cubic divided by quadratic gives a linear quotient and a linear remainder.
  2. Expand the right: \(x^3 + cx^2 + x + c + ax + b\).
    \((x^2+1)(x+c) = x^3 + cx^2 + x + c\).
  3. Compare \(x^2\): \(c = -3\).
    Only the quotient contributes to \(x^2\).
  4. Compare \(x\): \(1 + a = 2\), so \(a = 1\). Compare constants: \(c + b = 7\), so \(b = 7 + 3 = 10\).
    Work down the powers; each comparison introduces exactly one new unknown.
  5. Remainder: \(x + 10\).
    Degree \(1 <\) degree \(2\) ✓.
Where it goes wrong
  • Forgetting the \(+x\) that \((x^2+1)(x)\) contributes, giving \(a = 2\) and the remainder \(2x + 10\).
  • Using the Remainder Theorem with \(x^2 + 1\) - it has no real roots, so that route needs complex numbers and is off-specification.
Takeaway Comparing coefficients handles any divisor; only use root substitution when the divisor has real roots.
Answer B · Polynomials, factor and remainder theorems (MM1.6) · question POL-12
18Simultaneous equations and inequalities · spec MM1.5 · Paper 1 style · ★★☆

Find the complete set of values of \(x\) for which \(\dfrac{x-2}{x+1} \leq 0\).

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

A fraction is \(\leq 0\) when numerator and denominator have opposite signs (or the numerator is 0): \(x - 2 \leq 0\) and \(x + 1 > 0\) gives \(-1 < x \leq 2\); \(x - 2 \geq 0\) and \(x + 1 < 0\) is impossible. \(x = -1\) is excluded as the fraction is undefined.

Full method — how and why
A rational inequality. Never multiply by \(x + 1\) (its sign is unknown) - analyse the sign of the fraction instead.
  1. Find the critical values: numerator zero at \(x = 2\), denominator zero at \(x = -1\).
    The fraction can only change sign where the numerator or denominator vanishes.
  2. Test the three regions. For \(x < -1\) (say \(x = -2\)): \(\frac{-4}{-1} = 4 > 0\). For \(-1 < x < 2\) (say \(x = 0\)): \(\frac{-2}{1} = -2 < 0\). For \(x > 2\) (say \(x = 3\)): \(\frac{1}{4} > 0\).
    One test value per region settles the sign throughout that region.
  3. Include \(x = 2\) (the fraction is \(0\), and \(\leq 0\) allows it) but exclude \(x = -1\) (undefined).
    The two endpoints behave differently: a zero numerator is allowed by \(\leq\); a zero denominator never is.
  4. Answer: \(-1 < x \leq 2\).
    Note the mixed brackets - strict at \(-1\), inclusive at \(2\).
Where it goes wrong
  • Multiplying by \(x + 1\) to get \(x - 2 \leq 0\), i.e. \(x \leq 2\), which wrongly includes everything below \(-1\).
  • Writing \(-1 \leq x \leq 2\) and including a value where the expression is undefined.
Takeaway For \(\frac{A}{B} \leq 0\), use a sign chart on \(A\) and \(B\); a zero denominator is always excluded, a zero numerator is included when the inequality is not strict.
Answer C · Simultaneous equations and inequalities (MM1.5) · question SIM-16
19Differentiation · spec MM6.1 · Paper 1 style · ★★☆

\(f(x)\) is a polynomial with \(f'(x) = (x - 2)^2(x + 1)\). Which of the following is true?

  1. A\(f\) has local minima at \(x = -1\) and \(x = 2\)
  2. B\(f\) has no turning points
  3. C\(f\) has a local minimum at \(x = -1\) and no turning point at \(x = 2\)
  4. D\(f\) has a local maximum at \(x = -1\) and a local minimum at \(x = 2\)
  5. E\(f\) has a local minimum at \(x = -1\) and a local maximum at \(x = 2\)
Worked solution

\(f'\) changes sign from negative to positive at \(x = -1\) (local minimum). At \(x = 2\) the factor \((x-2)^2\) does not change sign, so \(f'\) stays positive: a stationary point but not a turning point (a point of inflexion).

Full method — how and why
Reading a factorised derivative. A squared factor does not change the sign, so it gives a stationary point that is not a turning point.
  1. The stationary points are the roots of \(f'\): \(x = -1\) and \(x = 2\).
    \(f'(x) = 0\) at both.
  2. Near \(x = -1\): \((x-2)^2 > 0\) always, so the sign of \(f'\) is the sign of \((x+1)\) - negative then positive.
    A sign change from \(-\) to \(+\) means the function falls then rises: a local minimum.
  3. Near \(x = 2\): \((x+1) > 0\) there, and \((x-2)^2 \geq 0\), so \(f' \geq 0\) on both sides - no sign change.
    A repeated (even-power) factor touches zero without crossing, so the gradient does not change sign.
  4. Conclusion: local minimum at \(x = -1\); stationary but not a turning point at \(x = 2\).
    That is a stationary point of inflexion.
Where it goes wrong
  • Assuming every root of \(f'\) is a turning point.
  • Using \(f''(2) = 0\) to conclude 'maximum' or 'minimum' - a zero second derivative is inconclusive.
Takeaway Odd-power factors of \(f'\) change sign (turning points); even-power factors do not (inflexions).
Answer C · Differentiation (MM6.1) · question DIF-13
20Indices and surds · spec MM1.2 · Paper 1 style · ★★☆

The expression \(\dfrac{3}{7 - 2\sqrt{3}} - \dfrac{3}{7 + 2\sqrt{3}}\) simplifies to

  1. A\(\dfrac{12\sqrt{3}}{61}\)
  2. B\(\dfrac{6\sqrt{3}}{37}\)
  3. C\(\dfrac{42}{61}\)
  4. D\(\dfrac{42}{37}\)
  5. E\(0\)
  6. F\(\dfrac{12\sqrt{3}}{37}\)
Worked solution

Common denominator \((7-2\sqrt3)(7+2\sqrt3) = 49 - 12 = 37\). Numerator: \(3(7 + 2\sqrt3) - 3(7 - 2\sqrt3) = 12\sqrt3\).

Full method — how and why
A difference of two rationalisable fractions. Using a single common denominator is faster than rationalising each one.
  1. Use the common denominator \((7 - 2\sqrt3)(7 + 2\sqrt3) = 49 - 12 = 37\).
    The two denominators are conjugates, so their product is automatically rational - no separate rationalising needed.
  2. Numerator: \(3(7 + 2\sqrt3) - 3(7 - 2\sqrt3) = 3 \times 4\sqrt3 = 12\sqrt3\).
    When you subtract conjugates the rational parts cancel and the surd parts double - so only the surd survives.
  3. Result: \(\dfrac{12\sqrt3}{37}\).
    Already rationalised, since the denominator is an integer.
Where it goes wrong
  • \((2\sqrt3)^2 = 4 \times 3 = 12\), not \(6\); using \(6\) gives denominator \(43\).
  • Adding the numerators instead of subtracting, giving \(\frac{42}{37}\) - which is offered.
Takeaway A difference of conjugate fractions leaves only the surd part; their product denominator is rational from the start.
Answer F · Indices and surds (MM1.2) · question IND-18
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