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Paper 1 · Set 03

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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1Statistics · spec M6.3 · Paper 1 style · ★☆☆

The mean of five numbers is \(12\). When a sixth number is added the mean becomes \(14\). What is the sixth number?

  1. A\(24\)
  2. B\(84\)
  3. C\(20\)
  4. D\(26\)
  5. E\(16\)
  6. F\(14\)
Worked solution

Total goes from \(60\) to \(84\): the sixth number is \(24\).

Full method — how and why
Means questions are almost always really about totals. Convert each mean into a total before doing anything else.
  1. Original total \(= 5 \times 12 = 60\).
    mean \(=\dfrac{\text{total}}{\text{how many}}\), so total \(=\) mean \(\times\) how many.
  2. New total \(= 6 \times 14 = 84\).
    There are now six numbers, and their mean is \(14\).
  3. The sixth number is the difference: \(84 - 60 = 24\).
    Nothing else changed, so the whole increase in the total is the new number.
Where it goes wrong
  • Answering \(14\) - the new mean, not the new number.
  • Answering \(84\), the new total rather than the added value.
  • Thinking the new number must be \(2\) more than the old mean; the mean rose by \(2\) for all six, so the newcomer must carry \(6 \times 2 = 12\) extra above the old mean of \(12\), giving \(24\).
Takeaway Turn means into totals; the added value is the change in total.
Answer A · Statistics (M6.3) · question STA-01
2Statistics · spec M6.3 · Paper 1 style · ★☆☆

The numbers \(3, 7, x, 12, 18\) (with \(7 \leq x \leq 12\)) have mean equal to median. Find \(x\).

  1. A\(12\)
  2. B\(7\)
  3. C\(10\)
  4. D\(8\)
  5. E\(11\)
  6. F\(9\)
Worked solution

The median is \(x\) and the mean is \(\frac{40 + x}{5}\). Setting them equal: \(40 + x = 5x\), so \(x = 10\).

Full method — how and why
Write expressions for the mean and the median, then equate. The given range for \(x\) tells you where it sits in the order.
  1. Since \(7 \leq x \leq 12\), the sorted list is \(3, 7, x, 12, 18\), so the median is \(x\).
    With five values the median is the third; the constraint is there precisely to fix \(x\)'s position.
  2. Mean \(= \dfrac{3 + 7 + x + 12 + 18}{5} = \dfrac{40 + x}{5}\).
    Sum the known values first: \(3+7+12+18 = 40\).
  3. Set mean \(=\) median: \(\dfrac{40+x}{5} = x\).
    That is the condition stated.
  4. Solve: \(40 + x = 5x \Rightarrow 4x = 40 \Rightarrow x = 10\).
    And \(10\) does lie in \([7, 12]\), so the assumed ordering holds.
Where it goes wrong
  • Taking the median to be \(12\) (the middle of the numbers as originally listed without \(x\)).
  • Forgetting to check that the answer is consistent with the assumed position in the order.
Takeaway Order first, then write the median; always check the solution respects the ordering you assumed.
Answer C · Statistics (M6.3) · question STA-03
3Polynomials, factor and remainder theorems · spec MM1.6 · Paper 1 style · ★☆☆

Fully factorise \(x^3 - 7x + 6\).

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

\(f(1) = 0\), so \((x-1)\) is a factor: \(x^3 - 7x + 6 = (x-1)(x^2 + x - 6) = (x-1)(x+3)(x-2)\).

Full method — how and why
Factorising a cubic: find one root by inspection, divide it out, then factorise the quadratic that remains.
  1. Test small factors of the constant term \(6\): \(f(1) = 1 - 7 + 6 = 0\), so \((x - 1)\) is a factor.
    Any integer root must divide the constant term, so only \(\pm1, \pm2, \pm3, \pm6\) are worth testing - and \(1\) is always the quickest to try.
  2. Divide: \(x^3 - 7x + 6 = (x - 1)(x^2 + x - 6)\).
    Note the missing \(x^2\) term in the original - treat it as \(0x^2\) when dividing or comparing coefficients.
  3. Factorise the quadratic: \(x^2 + x - 6 = (x + 3)(x - 2)\).
    Two numbers multiplying to \(-6\) and adding to \(+1\).
  4. Full factorisation: \((x-1)(x-2)(x+3)\).
    Check by multiplying the constants: \((-1)(-2)(3) = 6\). ✓
Where it goes wrong
  • Forgetting the missing \(x^2\) term and producing the wrong quadratic factor.
  • Sign errors giving \((x-1)(x+2)(x-3)\), whose constant term is \(+6\) too - always check the \(x\) coefficient as well.
Takeaway Root-hunt among the divisors of the constant term, divide out, then factorise what remains; check by multiplying the constants back.
Answer B · Polynomials, factor and remainder theorems (MM1.6) · question POL-03
4Indices and surds · spec MM1.2 · Paper 1 style · ★☆☆

Given that \(\sqrt{a} - \sqrt{b} = 2\) and \(a - b = 12\), find \(\sqrt{a} + \sqrt{b}\).

  1. A\(4\)
  2. B\(8\)
  3. C\(10\)
  4. D\(6\)
  5. E\(24\)
  6. F\(3\)
Worked solution

\(a - b = (\sqrt a - \sqrt b)(\sqrt a + \sqrt b)\), so \(12 = 2(\sqrt a + \sqrt b)\), giving \(6\).

Full method — how and why
You are given a difference of surds and a difference of the numbers. The link is the difference of two squares - no need to find \(a\) and \(b\).
  1. Factorise \(a - b\) as a difference of two squares in \(\sqrt a\) and \(\sqrt b\): \(a - b = (\sqrt a - \sqrt b)(\sqrt a + \sqrt b)\).
    \(a = (\sqrt a)^2\) for \(a \geq 0\), so the whole expression is \(X^2 - Y^2\) with \(X = \sqrt a\), \(Y = \sqrt b\) - this is the only structural idea in the question.
  2. Substitute the given values: \(12 = 2 \times (\sqrt a + \sqrt b)\).
    The first bracket is exactly the quantity given as \(2\).
  3. Divide: \(\sqrt a + \sqrt b = 6\).
    Direct.
Where it goes wrong
  • Solving simultaneously for \(a\) and \(b\) (possible: \(a = 16\), \(b = 4\)) - correct but three times the work and easy to fumble.
  • Answering \(24\) by multiplying rather than dividing.
Takeaway \(a - b = (\sqrt a - \sqrt b)(\sqrt a + \sqrt b)\) converts between sums and differences of surds instantly.
Answer D · Indices and surds (MM1.2) · question IND-20
5Graphs of functions · spec MM8.6 · Paper 1 style · ★☆☆

The curve \(y = x^4 - 5x^2 + 4\) meets the coordinate axes at how many points in total?

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

\(x\)-axis: \((x^2 - 1)(x^2 - 4) = 0\), four points. \(y\)-axis: \((0, 4)\), one point. Total \(5\).

Full method — how and why
Count both axes. The \(x\)-axis meetings come from the roots; the \(y\)-axis always contributes exactly one point.
  1. Factorise: \(x^4 - 5x^2 + 4 = (x^2-1)(x^2-4)\).
    Quadratic in \(x^2\): two numbers multiplying to \(4\) and adding to \(-5\).
  2. Solve: \(x = \pm1\) and \(x = \pm2\) - four \(x\)-axis crossings.
    Each positive value of \(x^2\) gives two values of \(x\).
  3. \(y\)-axis: substitute \(x = 0\) to get \(y = 4\) - one point.
    A function meets the \(y\)-axis exactly once.
  4. Total: \(4 + 1 = 5\).
    The \(y\)-intercept is a separate point (it is not one of the roots here, since \(y(0) \neq 0\)).
Where it goes wrong
  • Forgetting the \(y\)-axis and answering \(4\).
  • Counting \(x^2 = 1, 4\) as only two points.
Takeaway 'Meets the coordinate axes' means both axes: roots plus the single \(y\)-intercept.
Answer A · Graphs of functions (MM8.6) · question GRA-06
6Straight lines · spec MM3.1 · Paper 1 style · ★☆☆

Which of the following lines is parallel to \(4x + 6y = 9\)?

  1. A\(2x + 3y = 1\)
  2. B\(3x - 2y = 4\)
  3. C\(y = \frac{3}{2}x + 1\)
  4. D\(6x + 4y = 9\)
  5. E\(4x - 6y = 9\)
Worked solution

\(4x + 6y = 9\) has gradient \(-\frac23\); so does \(2x + 3y = 1\).

Full method — how and why
Parallel means equal gradient. Convert each option to gradient form - or compare the coefficient ratios directly.
  1. Gradient of \(4x + 6y = 9\) is \(-\frac{4}{6} = -\frac23\).
    For \(ax + by = c\) the gradient is \(-\frac ab\).
  2. Test the options: \(2x + 3y = 1\) has gradient \(-\frac23\) ✓.
    Equivalently, the coefficients \((2,3)\) are proportional to \((4,6)\) - the constant term is irrelevant to direction.
  3. Reject the rest: \(3x - 2y = 4\) gives \(\frac32\) (perpendicular), \(4x - 6y = 9\) gives \(\frac23\), \(6x + 4y = 9\) gives \(-\frac32\), and \(y = \frac32x+1\) gives \(\frac32\).
    Note how many distractors are the perpendicular gradient - a deliberate trap.
Where it goes wrong
  • Choosing \(6x + 4y = 9\) because the numbers \(4\) and \(6\) reappear - but swapped, so the gradient changes.
  • Forgetting the minus sign in \(-\frac ab\).
Takeaway Parallel \(\Leftrightarrow\) the coefficients of \(x\) and \(y\) are in the same ratio; the constant plays no part.
Answer A · Straight lines (MM3.1) · question LIN-05
7Integration · spec MM7.4 · Paper 1 style · ★☆☆

Let \(F(x) = \displaystyle\int_{2}^{x} (t^2 - 3t)\,dt\). Find \(F'(3)\).

  1. A\(9\)
  2. B\(\dfrac{9}{2}\)
  3. C\(-3\)
  4. D\(-\dfrac{7}{6}\)
  5. E\(0\)
  6. F\(3\)
Worked solution

By the Fundamental Theorem of Calculus \(F'(x) = x^2 - 3x\), so \(F'(3) = 9 - 9 = 0\).

Full method — how and why
A direct test of the Fundamental Theorem of Calculus: differentiating an integral with a variable upper limit gives back the integrand.
  1. Apply the FTC: \(\dfrac{d}{dx}\displaystyle\int_2^x g(t)\,dt = g(x)\), so \(F'(x) = x^2 - 3x\).
    The lower limit is a constant, so it contributes nothing to the derivative.
  2. Substitute \(x = 3\): \(F'(3) = 9 - 9 = 0\).
    Just evaluate the integrand at \(3\) - no integration is needed at all.
Where it goes wrong
  • Actually integrating and then differentiating - correct but slow, and error-prone.
  • Substituting \(t = 3\) into an antiderivative and answering \(-\frac{7}{6}\) or \(\frac92\), both offered.
Takeaway \(\frac{d}{dx}\int_a^x f(t)\,dt = f(x)\): differentiation undoes integration, so just evaluate the integrand.
Answer E · Integration (MM7.4) · question INT-08
8Trigonometric functions and equations · spec MM4.6 · Paper 1 style · ★☆☆

Solve \(\sin^2 x = \dfrac{3}{4}\) for \(0 \leq x < 2\pi\). How many solutions are there?

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

\(\sin x = \pm\frac{\sqrt3}{2}\): \(x = \frac\pi3, \frac{2\pi}{3}, \frac{4\pi}{3}, \frac{5\pi}{3}\). Four solutions.

Full method — how and why
A squared trig equation. Taking the square root produces two values, each with its own solutions.
  1. Take square roots: \(\sin x = \pm\dfrac{\sqrt3}{2}\).
    \(X^2 = k\) has two roots - forgetting the negative one halves the answer.
  2. \(\sin x = \frac{\sqrt3}{2}\) gives \(x = \frac{\pi}{3}\) and \(\frac{2\pi}{3}\).
    Second-quadrant partner: \(\pi - \frac\pi3\).
  3. \(\sin x = -\frac{\sqrt3}{2}\) gives \(x = \frac{4\pi}{3}\) and \(\frac{5\pi}{3}\).
    Third and fourth quadrants, by symmetry below the axis.
  4. Total: \(4\) solutions in \([0, 2\pi)\).
    Count them; symmetry says there should be an even number here.
Where it goes wrong
  • Taking only the positive root and answering \(2\).
  • Reading \(\sin^2 x\) as \(\sin(x^2)\).
Takeaway \(\sin^2 x = c\) gives \(\sin x = \pm\sqrt c\), and each sign contributes its own solutions.
Answer E · Trigonometric functions and equations (MM4.6) · question TRG-07
9Integration · spec MM7.4 · Paper 1 style · ★☆☆

Given that \(\displaystyle\int_{0}^{4} f(x)\,dx = 10\), find \(\displaystyle\int_{0}^{4} \left(f(x) + 3x\right)dx - \int_{4}^{0} f(x)\,dx\).

  1. A\(20\)
  2. B\(0\)
  3. C\(14\)
  4. D\(44\)
  5. E\(24\)
  6. F\(34\)
Worked solution

\(10 + 24 - (-10) = 44\).

Full method — how and why
Three integral rules in one: linearity, the integral of \(3x\), and the effect of swapping limits.
  1. Split the first integral: \(\int_0^4 (f(x) + 3x)dx = \int_0^4 f(x)dx + \int_0^4 3x\,dx = 10 + \int_0^4 3x\,dx\).
    Sums split into separate integrals.
  2. Compute \(\int_0^4 3x\,dx = \left[\frac{3x^2}{2}\right]_0^4 = 24\).
    \(\frac{3 \times 16}{2} = 24\).
  3. Handle the second term: \(\int_4^0 f(x)dx = -\int_0^4 f(x)dx = -10\).
    Reversing the limits reverses the sign - the key fact here.
  4. Combine: \((10 + 24) - (-10) = 44\).
    Subtracting a negative adds.
Where it goes wrong
  • Treating \(\int_4^0 f\) as \(+10\) and answering \(24\).
  • Forgetting the \(3x\) term, giving \(20\).
Takeaway \(\int_b^a = -\int_a^b\); watch for reversed limits used as a deliberate trap.
Answer D · Integration (MM7.4) · question INT-16
10Sequences and series · spec MM2.2 · Paper 1 style · ★☆☆

The sum of the first \(n\) terms of an arithmetic series is \(S_n\). Given that \(S_{10} = 145\) and \(S_{20} = 490\), find the common difference.

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

\(S_{10} = 5(2a + 9d) = 145 \Rightarrow 2a + 9d = 29\); \(S_{20} = 10(2a + 19d) = 490 \Rightarrow 2a + 19d = 49\). Subtract: \(10d = 20\), \(d = 2\).

Full method — how and why
Two sums give two equations in \(a\) and \(d\). Subtracting eliminates \(a\) - and only \(d\) is wanted.
  1. \(S_{10} = \frac{10}{2}\left[2a + 9d\right] = 145\), so \(2a + 9d = 29\).
    Divide out the \(\frac n2\) factor straight away to keep the equations clean.
  2. \(S_{20} = \frac{20}{2}\left[2a + 19d\right] = 490\), so \(2a + 19d = 49\).
    Same formula with \(n = 20\).
  3. Subtract: \(10d = 20\), so \(d = 2\).
    The \(2a\) terms are identical in both equations, so subtraction removes them at once.
Where it goes wrong
  • Using \((n-1)d\) as \(10d\) and \(20d\) rather than \(9d\) and \(19d\).
  • Dividing \(490\) by \(20\) instead of by \(10\).
Takeaway Two sums, two unknowns: write both in the \(2a + (n-1)d\) form and subtract to eliminate \(a\).
Answer A · Sequences and series (MM2.2) · question SEQ-16
11Binomial expansion · spec MM2.4 · Paper 1 style · ★☆☆

Use the binomial expansion to find the value of \(1.01^{5}\) correct to \(4\) decimal places.

  1. A\(1.0515\)
  2. B\(1.0510\)
  3. C\(1.0505\)
  4. D\(1.0500\)
  5. E\(1.0410\)
  6. F\(1.0501\)
Worked solution

\((1 + 0.01)^5 = 1 + 5(0.01) + 10(0.0001) + 10(0.000001) + \ldots = 1 + 0.05 + 0.001 + 0.00001 + \ldots \approx 1.0510\).

Full method — how and why
An approximation with a small quantity. The binomial expansion converges fast because \(0.01\) is tiny.
  1. Write \(1.01^5 = (1 + 0.01)^5\) and expand: \(1 + 5(0.01) + 10(0.01)^2 + 10(0.01)^3 + \ldots\)
    Coefficients from row 5 of Pascal's triangle: \(1, 5, 10, 10, 5, 1\).
  2. Evaluate the terms: \(1 + 0.05 + 0.001 + 0.00001 + \ldots\)
    \((0.01)^2 = 0.0001\), so the third term is \(10 \times 0.0001 = 0.001\).
  3. Add as far as the 4-decimal-place accuracy requires: \(1.05101\ldots \approx 1.0510\).
    The fourth term only affects the fifth decimal place, so three terms suffice.
Where it goes wrong
  • Stopping after two terms and answering \(1.0500\).
  • Mis-scaling \((0.01)^2\) as \(0.001\), which gives \(1.0600\)-style errors.
Takeaway With a small \(x\), each binomial term is much smaller than the last - take just enough terms for the required accuracy.
Answer B · Binomial expansion (MM2.4) · question BIN-13
12Differentiation · spec MM6.3 · Paper 1 style · ★☆☆

A rectangle has perimeter \(40\). What is its maximum possible area?

  1. A\(80\)
  2. B\(50\)
  3. C\(160\)
  4. D\(400\)
  5. E\(100\)
  6. F\(200\)
Worked solution

Sides \(x\) and \(20 - x\): \(A = 20x - x^2\), \(A' = 20 - 2x = 0\) at \(x = 10\), giving \(A = 100\) (a square).

Full method — how and why
A standard optimisation: express the area in one variable using the constraint, then differentiate.
  1. Let one side be \(x\); the perimeter condition \(2x + 2y = 40\) gives \(y = 20 - x\).
    Use the constraint to eliminate the second variable - optimisation needs a single variable.
  2. Write the area: \(A = x(20 - x) = 20x - x^2\).
    A quadratic in \(x\).
  3. Differentiate and solve: \(A' = 20 - 2x = 0\) at \(x = 10\).
    Or complete the square: \(A = 100 - (x-10)^2\), which shows the maximum immediately.
  4. Compute the area: \(A = 10 \times 10 = 100\) (a square).
    \(A'' = -2 < 0\) confirms a maximum.
Where it goes wrong
  • Using \(x + y = 40\) instead of \(2x + 2y = 40\), giving \(400\) - an offered option.
  • Reporting the side length \(10\) rather than the area.
Takeaway For fixed perimeter, the rectangle of greatest area is the square - but show it via the constraint and the derivative.
Answer E · Differentiation (MM6.3) · question DIF-09
13Indices and surds · spec MM1.1 · Paper 1 style · ★☆☆

The value of \(\dfrac{2^{n+2} + 2^{n}}{2^{n+1}}\) is

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

Factor \(2^n\): numerator \(= 2^n(4 + 1) = 5\cdot 2^n\); denominator \(= 2\cdot 2^n\). Quotient \(= \frac52\).

Full method — how and why
Powers with an unknown exponent. Factorising out the smallest power turns the expression into a number.
  1. Write every term with the common factor \(2^n\): \(2^{n+2} = 4\cdot 2^n\), \(2^{n+1} = 2\cdot 2^n\).
    \(2^{n+k} = 2^k \cdot 2^n\). Choosing the *smallest* power (\(2^n\)) as the common factor makes every remaining factor a whole number.
  2. Substitute: \(\dfrac{4\cdot2^n + 2^n}{2\cdot 2^n} = \dfrac{5 \cdot 2^n}{2\cdot 2^n}\).
    Factorising the numerator is legitimate because both terms contain \(2^n\).
  3. Cancel \(2^n\): the value is \(\frac52\), independent of \(n\).
    \(2^n \neq 0\) for every \(n\), so cancelling is always valid here.
Where it goes wrong
  • 'Cancelling' the \(n\)'s in the indices, e.g. claiming \(\frac{2^{n+2}}{2^{n+1}} = 2^{2}/2^{1}\) term by term across a sum - you cannot split a fraction over a sum in the denominator.
  • Answering \(2^n\) because the \(n\) 'has to appear somewhere'.
Takeaway With unknown exponents, factor out the lowest power; the \(n\) almost always cancels and the answer is a plain number.
Answer E · Indices and surds (MM1.1) · question IND-10
14GCSE algebra · spec M4.5 · Paper 1 style · ★☆☆

Factorise \(6x^2 - 7x - 20\).

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

\((3x + 4)(2x - 5) = 6x^2 - 15x + 8x - 20 = 6x^2 - 7x - 20\).

Full method — how and why
Factorising \(ax^2 + bx + c\) with \(a \neq 1\). Use the 'split the middle term' method, or expand the options.
  1. Multiply \(a\) and \(c\): \(6 \times (-20) = -120\). Find two numbers with product \(-120\) and sum \(-7\): \(-15\) and \(8\).
    This is the standard method; the pair always exists when the quadratic factorises over the integers.
  2. Split the middle term: \(6x^2 - 15x + 8x - 20\).
    Order does not matter.
  3. Factor in pairs: \(3x(2x - 5) + 4(2x - 5) = (3x+4)(2x-5)\).
    The bracket must match in both halves - that is the check that the split was right.
  4. Verify by expanding: \(6x^2 - 15x + 8x - 20 = 6x^2 - 7x - 20\) ✓.
    Expanding the chosen option is the fastest check in a multiple-choice setting.
Where it goes wrong
  • Choosing \((3x-5)(2x+4)\), which has a common factor \(2\) in the second bracket - a sign it cannot be right for this expression.
  • Getting the signs the wrong way round: \((3x-4)(2x+5)\) gives \(+7x\).
Takeaway Split the middle term using two numbers with product \(ac\) and sum \(b\); then expand your answer to check.
Answer C · GCSE algebra (M4.5) · question ALG-13
15Probability · spec M7.2 · Paper 1 style · ★☆☆

A spinner has probability \(0.15\) of landing on red. It is spun \(200\) times. What is the expected number of times it lands on red?

  1. A\(30\)
  2. B\(40\)
  3. C\(15\)
  4. D\(35\)
  5. E\(25\)
  6. F\(20\)
Worked solution

\(0.15 \times 200 = 30\).

Full method — how and why
Expected number of successes in repeated independent trials is simply \(n \times p\).
  1. Identify \(n = 200\) spins and \(p = 0.15\).
    Each spin is an independent trial with the same probability.
  2. Expected number \(= 200 \times 0.15 = 30\).
    \(0.15 = \frac{15}{100}\), and \(15\%\) of \(200\) is \(30\).
Where it goes wrong
  • Answering \(15\) by reading the percentage as the answer.
  • Expecting the answer to be a whole number in general - expectations often are not, and that is fine.
Takeaway Expected frequency \(= n \times p\).
Answer A · Probability (M7.2) · question PRO-05
16Binomial expansion · spec MM2.4 · Paper 1 style · ★☆☆

What is the sum of the coefficients in the expansion of \((3x - 2)^{7}\)?

  1. A\(2187\)
  2. B\(-128\)
  3. C\(0\)
  4. D\(1\)
  5. E\(-1\)
  6. F\(128\)
Worked solution

Put \(x = 1\): \((3 - 2)^7 = 1\).

Full method — how and why
Sum of all coefficients: substitute \(x = 1\). No expansion needed at all.
  1. Note that setting \(x = 1\) makes every power of \(x\) equal to \(1\), so the expansion collapses to the sum of its coefficients.
    If \((3x-2)^7 = \sum c_k x^k\) then putting \(x = 1\) gives \(\sum c_k\).
  2. Evaluate the original expression at \(x = 1\): \((3 - 2)^7 = 1^7 = 1\).
    You never need a single coefficient.
Where it goes wrong
  • Expanding and trying to add eight terms under time pressure.
  • Substituting \(x = 0\), which gives only the constant term \((-2)^7 = -128\) - an offered option.
Takeaway \(f(1)\) is the sum of the coefficients; \(f(0)\) is the constant term; \(f(-1)\) is the alternating sum.
Answer D · Binomial expansion (MM2.4) · question BIN-04
17Functions and transformations · spec MM1.7 · Paper 1 style · ★★☆

Which of the following functions, each with domain the set of all real numbers, is one-to-one?

  1. A\(f(x) = x^3 - x\)
  2. B\(f(x) = x^2 + 2x\)
  3. C\(f(x) = \cos x\)
  4. D\(f(x) = x^3 + x\)
  5. E\(f(x) = |x|\)
Worked solution

\(x^3 + x\) is strictly increasing (its derivative \(3x^2 + 1 > 0\)), so different inputs give different outputs. \(x^3 - x\) takes the value \(0\) at \(x = -1, 0, 1\); the others are even or periodic.

Full method — how and why
'One-to-one' means no horizontal line meets the graph twice. The quickest test is whether the function is always increasing (or always decreasing).
  1. Rule out the even and periodic functions: \(|x|\) and \(x^2 + 2x\) are symmetric (e.g. \(|1| = |-1|\)), and \(\cos x\) repeats every \(2\pi\).
    Any symmetry or repetition immediately produces two inputs with the same output.
  2. Test \(x^3 - x\): it equals \(0\) at \(x = -1, 0, 1\) - three inputs with the same output.
    A cubic with three distinct roots is certainly not one-to-one.
  3. Test \(x^3 + x\): its derivative \(3x^2 + 1\) is always positive, so the function is strictly increasing.
    A strictly increasing function can never take the same value twice, which is exactly one-to-one.
  4. So \(f(x) = x^3 + x\) is the one-to-one function.
    Alternatively, \(x^3\) and \(x\) both increase, so their sum does.
Where it goes wrong
  • Assuming every cubic is one-to-one - \(x^3 - x\) is a counterexample.
  • Thinking \(|x|\) is one-to-one because it has a 'simple' formula.
Takeaway One-to-one \(\Leftrightarrow\) strictly monotonic; check for symmetry or repeated values first, then look at the derivative's sign.
Answer D · Functions and transformations (MM1.7) · question FUN-04
18Simultaneous equations and inequalities · spec MM1.5 · Paper 1 style · ★★☆

Find the complete set of values of \(x\) for which \((x-1)(x-3)(x-5) > 0\).

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

Sign chart: the product is positive for \(x > 5\), negative for \(3 < x < 5\), positive for \(1 < x < 3\), negative for \(x < 1\).

Full method — how and why
A cubic inequality. With three distinct linear factors, use a sign chart - the sign alternates as you cross each root.
  1. The roots are \(x = 1, 3, 5\); mark them on a number line, splitting it into four regions.
    A product of linear factors can only change sign at a root.
  2. Test the far-right region, e.g. \(x = 6\): \((5)(3)(1) > 0\).
    One test value fixes the whole chart; the far right is easiest because every factor is positive.
  3. Alternate leftwards: positive for \(x > 5\), negative on \((3,5)\), positive on \((1,3)\), negative for \(x < 1\).
    Each simple (unrepeated) root flips the sign exactly once.
  4. Collect the positive regions: \(1 < x < 3\) or \(x > 5\).
    The answer is a union of two intervals.
Where it goes wrong
  • Answering \(x > 5\) only, forgetting the middle positive region.
  • Assuming the expression is positive whenever \(x\) is bigger than all the roots - true here, but the alternation is what you must use.
Takeaway Sign chart: find the roots, test one region, then alternate; repeated factors are the only ones that do not flip the sign.
Answer A · Simultaneous equations and inequalities (MM1.5) · question SIM-10
19Triangles, radians and sectors · spec MM4.1 · Paper 1 style · ★★☆

In triangle \(ABC\), \(\sin A : \sin B : \sin C = 3 : 5 : 7\). Find angle \(C\).

  1. A\(120^\circ\)
  2. B\(150^\circ\)
  3. C\(100^\circ\)
  4. D\(135^\circ\)
  5. E\(90^\circ\)
  6. F\(60^\circ\)
Worked solution

By the sine rule the sides are in ratio \(3 : 5 : 7\). \(\cos C = \dfrac{9 + 25 - 49}{2\cdot3\cdot5} = -\dfrac{15}{30} = -\dfrac12\), so \(C = 120^\circ\).

Full method — how and why
A ratio of sines is a ratio of sides (sine rule). Then the cosine rule finds the angle.
  1. By the sine rule, \(a : b : c = \sin A : \sin B : \sin C = 3 : 5 : 7\).
    \(\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}\) means sides are proportional to the sines of opposite angles.
  2. Take the sides as \(3k, 5k, 7k\); the \(k\) will cancel, so use \(3, 5, 7\).
    Only the shape matters for angles, so the scale factor is irrelevant.
  3. Apply the cosine rule for the angle opposite \(7\): \(\cos C = \dfrac{9 + 25 - 49}{2(3)(5)} = \dfrac{-15}{30} = -\dfrac12\).
    \(C\) is opposite the largest side, so expect the largest angle.
  4. \(\cos C = -\frac12\) gives \(C = 120^\circ\).
    One of the exact values: \(\cos 120^\circ = -\cos 60^\circ = -\frac12\).
Where it goes wrong
  • Treating \(3:5:7\) as the angles themselves, giving \(84^\circ\).
  • Reading \(\cos C = -\frac12\) as \(60^\circ\) and ignoring the sign.
Takeaway Sines in ratio \(\Rightarrow\) sides in the same ratio; then use the cosine rule with any convenient multiple.
Answer A · Triangles, radians and sectors (MM4.1) · question TRI-10
20Differentiation · spec MM6.3 · Paper 1 style · ★★☆

The tangent to \(y = x^3\) at the point \((a, a^3)\), where \(a \neq 0\), meets the curve again at the point with \(x\)-coordinate

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

Tangent: \(y = 3a^2 x - 2a^3\). Solve \(x^3 - 3a^2 x + 2a^3 = 0\); \(x = a\) is a double root, so \((x - a)^2(x + 2a) = 0\): the other intersection is \(x = -2a\).

Full method — how and why
A tangent meeting the curve again. The tangency shows up as a repeated root, which lets you factorise the cubic without solving it.
  1. Find the tangent at \((a, a^3)\): \(y' = 3x^2\), so the gradient is \(3a^2\) and the line is \(y = 3a^2(x - a) + a^3 = 3a^2x - 2a^3\).
    Point-gradient form, then simplify.
  2. Set the tangent equal to the curve: \(x^3 = 3a^2 x - 2a^3\), i.e. \(x^3 - 3a^2x + 2a^3 = 0\).
    Intersections satisfy both equations.
  3. \(x = a\) is a repeated root (that is what tangency means), so factor out \((x-a)^2\): \(x^3 - 3a^2x + 2a^3 = (x-a)^2(x + 2a)\).
    Check by comparing the constant term: \((-a)^2(2a) = 2a^3\) ✓.
  4. The remaining root is \(x = -2a\).
    That is where the tangent cuts the curve again.
Where it goes wrong
  • Forgetting that tangency gives a double root and trying to factor out only one \((x-a)\).
  • Sign error giving \(x = 2a\).
Takeaway A tangent meets the curve in a repeated root; factor \((x - a)^2\) out and read off the third root.
Answer B · Differentiation (MM6.3) · question DIF-16
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