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Paper 1 · Set 11

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

Given that \(2^{x} = 5\) and \(2^{y} = 20\), find the value of \(2^{2x - y}\).

  1. A\(\dfrac{1}{4}\)
  2. B\(\dfrac{4}{5}\)
  3. C\(\dfrac{5}{4}\)
  4. D\(-10\)
  5. E\(\dfrac{25}{2}\)
  6. F\(5\)
Worked solution

\(2^{2x - y} = \dfrac{(2^x)^2}{2^y} = \dfrac{25}{20} = \dfrac54\).

Full method — how and why
You are given two exponentials and asked for a third. Express the target in terms of the given ones using index laws - you never need to find \(x\) or \(y\).
  1. Split the index: \(2^{2x - y} = \dfrac{2^{2x}}{2^{y}}\).
    Subtraction in the index is division - this is what lets you use the two given facts separately.
  2. Write \(2^{2x} = (2^x)^2 = 5^2 = 25\).
    A coefficient in the index is a power of the whole exponential: \(2^{2x} = (2^x)^2\), not \(2 \times 2^x\).
  3. Substitute: \(\dfrac{25}{20} = \dfrac54\).
    \(2^y = 20\) is given directly.
Where it goes wrong
  • Using \(2^{2x} = 2 \times 2^x = 10\), giving \(\frac12\).
  • Trying to solve for \(x = \log_2 5\) and \(y = \log_2 20\) first - correct but slow, and it invites arithmetic slips.
Takeaway Rewrite the target index as sums and multiples of the given ones; substitution then beats solving for the unknowns.
Answer C · Indices and surds (MM1.1) · question IND-15
2Simultaneous equations and inequalities · spec MM1.5 · Paper 1 style · ★☆☆

The inequality \(x^2 + bx + 9 > 0\) holds for all real \(x\). Find the complete set of values of \(b\).

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

Need discriminant negative: \(b^2 - 36 < 0\), so \(-6 < b < 6\).

Full method — how and why
'True for all \(x\)' for an upward parabola means it never touches the axis: discriminant strictly negative.
  1. Note the leading coefficient is \(1 > 0\), so the parabola opens upwards.
    For \(>0\) everywhere you need both the right shape and no real roots; here the shape is automatic.
  2. Require no real roots: \(b^2 - 4(1)(9) < 0\).
    If there were real roots the curve would touch or cross the axis and the expression would fail to be positive there.
  3. Solve: \(b^2 < 36 \Rightarrow -6 < b < 6\).
    \(b^2 < 36\) means \(|b| < 6\), a single interval.
Where it goes wrong
  • Solving \(b^2 < 36\) as two outer intervals - that is the solution of \(b^2 > 36\).
  • Using \(\leq\): at \(b = \pm6\) the curve touches the axis, where the expression equals \(0\) and is not \(> 0\).
Takeaway 'Positive for all \(x\)' \(\Leftrightarrow\) \(a > 0\) and \(b^2 - 4ac < 0\); \(x^2 < k\) gives one interval, \(x^2 > k\) gives two.
Answer D · Simultaneous equations and inequalities (MM1.5) · question SIM-07
3Circles and circle theorems · spec MM3.2 · Paper 1 style · ★☆☆

The circle \(x^2 + y^2 = r^2\) and the line \(3x + 4y = 20\) have exactly one point in common. Find \(r\).

  1. A\(\dfrac{20}{7}\)
  2. B\(4\)
  3. C\(20\)
  4. D\(5\)
  5. E\(\sqrt{20}\)
  6. F\(2\)
Worked solution

Distance from origin to the line \(= \frac{20}{\sqrt{9 + 16}} = 4\). Tangency means \(r = 4\).

Full method — how and why
One common point means tangency, so the radius equals the distance from the centre to the line.
  1. The circle is centred at the origin, so compute the distance from \((0,0)\) to \(3x + 4y = 20\).
    Centre-at-origin makes the distance formula especially clean.
  2. Distance \(= \dfrac{|3(0) + 4(0) - 20|}{\sqrt{3^2+4^2}} = \dfrac{20}{5} = 4\).
    \(\sqrt{9+16} = 5\) - the familiar 3-4-5 triple.
  3. Tangency requires \(r\) equal to this distance: \(r = 4\).
    If \(r\) were larger the line would cut the circle twice; smaller, and it would miss.
Where it goes wrong
  • Dividing by \(\sqrt{20}\) or by \(7\) instead of \(5\).
  • Answering \(20\) or \(\sqrt{20}\) by stopping too early.
Takeaway Distance from the origin to \(ax+by=c\) is \(\frac{|c|}{\sqrt{a^2+b^2}}\); tangency sets it equal to \(r\).
Answer B · Circles and circle theorems (MM3.2) · question CIR-13
4Integration · spec MM7.3 · Paper 1 style · ★☆☆

Let \(I = \displaystyle\int_{0}^{1} (x^2 + 1)\,dx\) and \(J = \displaystyle\int_{0}^{1} (x + 1)^2\,dx\). Which of the following is true?

  1. A\(I < J\) and \(J - I = \dfrac{4}{3}\)
  2. B\(I < J\) and \(J - I = \dfrac{1}{2}\)
  3. C\(I = J\)
  4. D\(I < J\) and \(J - I = 1\)
  5. E\(I > J\)
Worked solution

\(I = \frac13 + 1 = \frac43\). \(J = \int_0^1 (x^2 + 2x + 1)\,dx = \frac13 + 1 + 1 = \frac73\). \(J - I = 1\).

Full method — how and why
Two similar-looking integrals. Expand the second - \((x+1)^2 \neq x^2 + 1\) - and compare.
  1. Compute \(I\): \(\displaystyle\int_0^1 (x^2+1)dx = \left[\frac{x^3}{3} + x\right]_0^1 = \frac13 + 1 = \frac43\).
    Term by term.
  2. Expand the second integrand: \((x+1)^2 = x^2 + 2x + 1\).
    The cross term \(2x\) is the whole difference between the two integrals.
  3. Compute \(J\): \(\left[\frac{x^3}{3} + x^2 + x\right]_0^1 = \frac13 + 1 + 1 = \frac73\).
    The extra \(\int_0^1 2x\,dx = 1\).
  4. Compare: \(J - I = \frac73 - \frac43 = 1\), and \(J > I\).
    Equivalently, the difference is exactly \(\int_0^1 2x\,dx = 1\).
Where it goes wrong
  • Assuming \(I = J\) because the integrands 'look the same'.
  • Arithmetic giving \(J - I = \frac43\) - an offered option.
Takeaway \((x+1)^2 = x^2 + 2x + 1\): the difference between the two integrals is the integral of the cross term.
Answer D · Integration (MM7.3) · question INT-18
5Ratio, proportion and percentages · spec M3.11 · Paper 1 style · ★☆☆

A bacteria population doubles every \(3\) hours. By what factor does it grow in \(24\) hours?

  1. A\(16\)
  2. B\(48\)
  3. C\(64\)
  4. D\(128\)
  5. E\(8\)
  6. F\(256\)
Worked solution

\(24/3 = 8\) doublings: \(2^8 = 256\).

Full method — how and why
Repeated doubling. Count the doublings, then raise \(2\) to that power.
  1. Count the doubling periods: \(24 \div 3 = 8\).
    Each \(3\)-hour block doubles the population.
  2. Each doubling multiplies by \(2\), so eight of them multiply by \(2^8\).
    Repeated multiplication is exponentiation - not multiplication by \(8\).
  3. \(2^8 = 256\).
    Worth knowing the powers of \(2\) up to \(2^{10} = 1024\).
Where it goes wrong
  • Multiplying by \(8\) or computing \(2 \times 8 = 16\) - both offered.
  • Using \(2^6 = 64\) by dividing \(24\) by \(4\).
Takeaway Growth factor \(= (\text{factor per period})^{\text{number of periods}}\).
Answer F · Ratio, proportion and percentages (M3.11) · question RAT-05
6Differentiation · spec MM6.3 · Paper 1 style · ★☆☆

The line \(y = 4x + c\) is a tangent to \(y = x^2 + 2x + 5\). Find \(c\).

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

\(2x + 2 = 4 \Rightarrow x = 1\), \(y = 8\). \(8 = 4 + c \Rightarrow c = 4\).

Full method — how and why
Tangency via calculus: the gradients must match, which locates the point of contact.
  1. Differentiate the curve: \(y' = 2x + 2\).
    The tangent's gradient equals the curve's gradient at the contact point.
  2. Set it equal to the line's gradient: \(2x + 2 = 4\), so \(x = 1\).
    This finds where the tangent touches.
  3. Find the \(y\)-value on the curve: \(y(1) = 1 + 2 + 5 = 8\).
    The contact point is on the curve.
  4. The point must also be on the line: \(8 = 4(1) + c\), so \(c = 4\).
    Substituting the contact point into the line's equation gives \(c\).
Where it goes wrong
  • Using the discriminant route but mis-expanding and getting \(c = 1\) or \(5\) (both offered).
  • Substituting into the line the \(y\)-value from the wrong \(x\).
Takeaway Tangency: equal gradients give the \(x\); equal \(y\)-values give the constant.
Answer A · Differentiation (MM6.3) · question DIF-12
7Number, units and estimation · spec M2.11 · Paper 1 style · ★☆☆

Find the exact value of \(\left(\sqrt{5} - 1\right)^2 + \left(\sqrt{5} + 1\right)^2\).

  1. A\(12\)
  2. B\(8\)
  3. C\(10\)
  4. D\(6\)
  5. E\(12 + 4\sqrt5\)
  6. F\(4\sqrt5\)
Worked solution

\((6 - 2\sqrt5) + (6 + 2\sqrt5) = 12\).

Full method — how and why
A sum of conjugate squares. The surd terms cancel, so the answer is rational.
  1. Expand the first: \((\sqrt5 - 1)^2 = 5 - 2\sqrt5 + 1 = 6 - 2\sqrt5\).
    \((a-b)^2 = a^2 - 2ab + b^2\) with \((\sqrt5)^2 = 5\).
  2. Expand the second: \((\sqrt5 + 1)^2 = 6 + 2\sqrt5\).
    Only the middle sign differs.
  3. Add: the \(\mp2\sqrt5\) terms cancel, leaving \(12\).
    Conjugates always cancel their surd parts when added.
Where it goes wrong
  • Dropping the cross terms: here it happens to give \(12\) anyway, but the same slip fails whenever the two brackets are not conjugates.
  • Adding the surd terms instead of cancelling, giving \(12 + 4\sqrt5\) - an offered option.
Takeaway \((a+b)^2 + (a-b)^2 = 2(a^2+b^2)\): here \(2(5 + 1) = 12\), with no surds surviving.
Answer A · Number, units and estimation (M2.11) · question NUM-20
8Sequences and series · spec MM2.1 · Paper 1 style · ★☆☆

A sequence is defined by \(u_1 = 1\) and \(u_{n+1} = u_n + 2n + 1\). Which of the following is an expression for \(u_n\)?

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

\(u_2 = 4\), \(u_3 = 9\), \(u_4 = 16\): \(u_n = n^2\). Check: \((n+1)^2 = n^2 + 2n + 1\).

Full method — how and why
A recurrence that generates a familiar sequence. Compute terms, recognise the pattern, then verify it satisfies the rule.
  1. Generate terms: \(u_1 = 1\), \(u_2 = 1 + 3 = 4\), \(u_3 = 4 + 5 = 9\), \(u_4 = 9 + 7 = 16\).
    The increments \(3, 5, 7\) are the odd numbers - a strong hint.
  2. Recognise \(1, 4, 9, 16\) as the squares: conjecture \(u_n = n^2\).
    Pattern-spotting is legitimate here because the options let you verify.
  3. Verify against the recurrence: if \(u_n = n^2\) then \(u_n + 2n + 1 = n^2 + 2n + 1 = (n+1)^2 = u_{n+1}\) ✓.
    This is the crucial step: checking the formula satisfies the rule turns a guess into a proof.
Where it goes wrong
  • Choosing \(2n - 1\) (the increments) instead of the terms themselves.
  • Testing only \(u_1\), which several options match.
Takeaway Spot the pattern from a few terms, then prove it by substituting into the recurrence.
Answer F · Sequences and series (MM2.1) · question SEQ-18
9Quadratic functions · spec MM1.3 · Paper 1 style · ★☆☆

The line \(y = mx\) meets the curve \(y = x^2 + 9\) at two distinct points. Find the complete set of values of \(m\).

  1. A\(m < -3\) or \(m > 3\)
  2. B\(m < -6\) or \(m > 6\)
  3. C\(-6 < m < 6\)
  4. D\(-3 < m < 3\)
  5. E\(m > 6\)
  6. Fall real \(m\)
Worked solution

\(x^2 - mx + 9 = 0\) needs \(m^2 - 36 > 0\), so \(m < -6\) or \(m > 6\).

Full method — how and why
A line meeting a curve twice: substitute and demand a positive discriminant.
  1. Substitute \(y = mx\) into \(y = x^2 + 9\): \(mx = x^2 + 9\).
    Intersections are the common solutions, found by equating the two expressions for \(y\).
  2. Rearrange: \(x^2 - mx + 9 = 0\).
    Standard form is needed before the discriminant can be read off.
  3. Two distinct intersections \(\Rightarrow\) discriminant \(> 0\): \(m^2 - 36 > 0\).
    Here \(b = -m\), and \(b^2 = m^2\) regardless of sign.
  4. Solve \(m^2 > 36\): \(m < -6\) or \(m > 6\).
    Two intervals, since \(m^2 > 36\) means \(|m| > 6\); geometrically, steep lines in either direction cut the parabola twice.
Where it goes wrong
  • Answering \(-6 < m < 6\) (the region where the line misses the parabola entirely).
  • Writing the discriminant as \(-m^2 - 36\) by mishandling the sign of \(b\).
Takeaway \(b^2\) kills the sign of \(b\): conditions like this are almost always symmetric in \(\pm m\).
Answer B · Quadratic functions (MM1.3) · question QUA-23
10Indices and surds · spec MM1.2 · Paper 1 style · ★☆☆

Simplify \(\sqrt{75} - \sqrt{27} + \sqrt{12}\).

  1. A\(2\sqrt{3}\)
  2. B\(6\sqrt{3}\)
  3. C\(\sqrt{60}\)
  4. D\(10\sqrt{3}\)
  5. E\(4\sqrt{3}\)
  6. F\(5\sqrt{3}\)
Worked solution

\(\sqrt{75} = 5\sqrt{3}\), \(\sqrt{27} = 3\sqrt{3}\), \(\sqrt{12} = 2\sqrt{3}\), so the total is \((5 - 3 + 2)\sqrt{3} = 4\sqrt{3}\).

Full method — how and why
Simplifying a sum of surds. They can only be combined once each is written as a multiple of the same surd.
  1. Pull out the largest square factor from each: \(\sqrt{75} = \sqrt{25 \times 3} = 5\sqrt3\), \(\sqrt{27} = \sqrt{9\times3} = 3\sqrt3\), \(\sqrt{12} = \sqrt{4\times3} = 2\sqrt3\).
    \(\sqrt{ab} = \sqrt a\sqrt b\), so a perfect-square factor comes out as a whole number. All three reduce to multiples of \(\sqrt3\), which is why the sum simplifies at all.
  2. Collect like terms: \(5\sqrt3 - 3\sqrt3 + 2\sqrt3 = 4\sqrt3\).
    Once every term has the same surd part, \(\sqrt3\) behaves exactly like an algebraic \(x\).
Where it goes wrong
  • Adding under one root sign: \(\sqrt{75} - \sqrt{27} \neq \sqrt{48}\). Roots do not distribute over \(+\) or \(-\).
  • Using \(\sqrt{75} = 3\sqrt5\) (wrong square factor) - always take out the largest square.
Takeaway Surds add only when their irrational parts match, so simplify each one fully before trying to combine.
Answer E · Indices and surds (MM1.2) · question IND-05
11Circles and circle theorems · spec MM3.2 · Paper 1 style · ★☆☆

The circle \(C\) has equation \(x^2 + y^2 - 4x - 6y + 9 = 0\). Which of the following lines does not meet \(C\)?

  1. A\(x = 4\)
  2. B\(x = 2\)
  3. C\(x = 0\)
  4. D\(y = 0\)
  5. E\(y = 5\)
Worked solution

Centre \((2, 3)\), radius \(2\), so the circle spans \(0 \leq x \leq 4\) and \(1 \leq y \leq 5\). The line \(y = 0\) is \(3\) from the centre, more than the radius, so it misses; \(x = 0\), \(x = 4\) and \(y = 5\) touch and \(x = 2\) passes through the centre.

Full method — how and why
Which line misses the circle: compare each line's distance from the centre with the radius. Horizontal and vertical lines make this trivial.
  1. Complete the square: \((x-2)^2 + (y-3)^2 = -9 + 4 + 9 = 4\), so centre \((2,3)\) and radius \(2\).
    Always convert to centre-radius form first.
  2. The circle therefore spans \(x\) from \(0\) to \(4\) and \(y\) from \(1\) to \(5\).
    Centre \(\pm\) radius in each direction gives the bounding box, whose edges are tangent lines.
  3. Check each line: \(x = 0\), \(x = 4\) and \(y = 5\) are tangents (distance exactly \(2\)); \(x = 2\) passes through the centre.
    For a vertical line \(x = k\), the distance from the centre is \(|k - 2|\).
  4. \(y = 0\) is at distance \(|0 - 3| = 3 > 2\), so it misses the circle.
    Distance greater than the radius means no intersection.
Where it goes wrong
  • Assuming a tangent line 'does not meet' the circle - touching counts as meeting.
  • Getting \(r^2 = 9\) by mishandling the \(+9\) constant.
Takeaway For horizontal/vertical lines just compare the centre coordinate with the radius; tangency still counts as meeting.
Answer D · Circles and circle theorems (MM3.2) · question CIR-12
12Sequences and series · spec MM2.2 · Paper 1 style · ★☆☆

An arithmetic sequence has \(u_3 = 10\) and \(u_8 = 25\). Find the smallest value of \(n\) for which \(u_n > 100\).

  1. A\(36\)
  2. B\(32\)
  3. C\(34\)
  4. D\(31\)
  5. E\(33\)
  6. F\(35\)
Worked solution

\(5d = 15 \Rightarrow d = 3\), \(a = 4\). \(u_n = 4 + 3(n-1) = 3n + 1 > 100 \Rightarrow n > 33\), so \(n = 34\).

Full method — how and why
Two terms give \(a\) and \(d\); then solve an inequality for \(n\) and round to the correct integer.
  1. Subtract the two facts: \(u_8 - u_3 = 5d = 25 - 10 = 15\), so \(d = 3\).
    From \(u_3\) to \(u_8\) is five steps, so the difference is \(5d\) - count the gaps, not the terms.
  2. Find \(a\): \(u_3 = a + 2d = 10\), so \(a = 10 - 6 = 4\).
    The third term is \(a + 2d\), not \(a + 3d\).
  3. Write the general term: \(u_n = 4 + 3(n-1) = 3n + 1\).
    Simplify before solving - it avoids bracket errors.
  4. Solve \(3n + 1 > 100 \Rightarrow n > 33\), so the smallest integer is \(n = 34\).
    Strict inequality: \(n = 33\) gives exactly \(100\), which is not greater than \(100\).
Where it goes wrong
  • Answering \(33\), where \(u_{33} = 100\) exactly.
  • Using \(u_8 - u_3 = 8d - 3d\) correctly but then \(u_3 = a + 3d\) incorrectly.
Takeaway \(u_p\) to \(u_q\) spans \(q - p\) common differences; and 'greater than' excludes the boundary case.
Answer C · Sequences and series (MM2.2) · question SEQ-13
13Quadratic functions · spec MM1.3 · Paper 1 style · ★☆☆

The equation \(kx^2 + 4x + k = 0\) has no real roots. Find the complete set of values of \(k\).

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

Need \(16 - 4k^2 < 0\), i.e. \(k^2 > 4\), so \(k < -2\) or \(k > 2\). (If \(k = 0\) the equation is linear with one root, so \(k = 0\) is excluded anyway.)

Full method — how and why
'No real roots' is the discriminant \(< 0\). The extra subtlety is the coefficient \(k\) appearing in the \(x^2\) term, so \(k = 0\) needs a thought.
  1. Write the discriminant: \(4^2 - 4(k)(k) = 16 - 4k^2\).
    Here \(a = k\) and \(c = k\) - both the first and last coefficients are the parameter.
  2. Set it negative: \(16 - 4k^2 < 0 \Rightarrow k^2 > 4\).
    No real roots \(\Leftrightarrow b^2 - 4ac < 0\).
  3. Solve \(k^2 > 4\): \(k < -2\) or \(k > 2\).
    \(k^2 > 4\) means \(|k| > 2\), which is two separate intervals, not \(-2 < k < 2\).
  4. Check \(k = 0\) separately: the equation becomes \(4x = 0\), which has the root \(x = 0\).
    If \(k = 0\) it is not a quadratic at all, so the discriminant test does not apply - but \(k = 0\) is excluded by the answer anyway.
Where it goes wrong
  • Solving \(k^2 > 4\) as \(-2 < k < 2\) - that is the solution of \(k^2 < 4\).
  • Ignoring the possibility \(k = 0\) entirely (harmless here, fatal in questions where \(0\) lies inside the answer set).
Takeaway \(k^2 > c\) gives two outer intervals; and whenever the leading coefficient contains the parameter, check the value that makes it zero.
Answer C · Quadratic functions (MM1.3) · question QUA-08
14Angles, polygons, congruence and vectors · spec M5.5 · Paper 1 style · ★☆☆

Two similar triangles have corresponding sides of length \(6\) cm and \(9\) cm. The smaller triangle has area \(20\) cm\(^2\). Find the area of the larger triangle.

  1. A\(30\) cm\(^2\)
  2. B\(67.5\) cm\(^2\)
  3. C\(40\) cm\(^2\)
  4. D\(45\) cm\(^2\)
  5. E\(27\) cm\(^2\)
  6. F\(60\) cm\(^2\)
Worked solution

Area scale factor \(\left(\frac96\right)^2 = \frac94\): \(20 \times \frac94 = 45\).

Full method — how and why
Similar shapes: areas scale as the square of the length ratio.
  1. Length scale factor: \(\dfrac96 = \dfrac32\).
    From the smaller to the larger, so the factor is greater than \(1\).
  2. Area scale factor: \(\left(\dfrac32\right)^2 = \dfrac94\).
    Both dimensions scale, so the area scales by the square.
  3. Multiply: \(20 \times \dfrac94 = 45\) cm\(^2\).
    \(20 \div 4 = 5\), then \(\times 9\).
Where it goes wrong
  • Using the length factor directly: \(20 \times \frac32 = 30\) - an offered option.
  • Cubing (that would be for volumes), giving \(67.5\).
Takeaway Area ratio \(=\) (length ratio)\(^2\); check the direction of the scale factor.
Answer D · Angles, polygons, congruence and vectors (M5.5) · question GEO-05
15Binomial expansion · spec MM2.4 · Paper 1 style · ★★☆

The first three terms in the expansion of \((1 + ax)^n\) in ascending powers of \(x\) are \(1 + 12x + 54x^2\). Find \(n\).

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

\(na = 12\) and \(\frac{n(n-1)}{2}a^2 = 54\). Dividing the second by the square of the first: \(\frac{n-1}{2n} = \frac{54}{144} = \frac38\), so \(8n - 8 = 6n\) and \(n = 4\) (then \(a = 3\); check \(\binom42 \cdot 9 = 54\)).

Full method — how and why
Two coefficients, two unknowns (\(a\) and \(n\)). Dividing the equations cleverly eliminates \(a\).
  1. Write the first two coefficients: \(\binom n1 a = na = 12\) and \(\binom n2 a^2 = \frac{n(n-1)}{2}a^2 = 54\).
    The expansion of \((1+ax)^n\) starts \(1 + nax + \frac{n(n-1)}{2}a^2x^2\).
  2. Square the first equation: \(n^2a^2 = 144\).
    Squaring makes the power of \(a\) match the second equation, so dividing will cancel it.
  3. Divide the second by this: \(\dfrac{n(n-1)a^2/2}{n^2a^2} = \dfrac{n-1}{2n} = \dfrac{54}{144} = \dfrac38\).
    Both \(a^2\) and one factor of \(n\) cancel.
  4. Cross-multiply: \(8(n-1) = 6n \Rightarrow 8n - 8 = 6n \Rightarrow n = 4\).
    Then \(a = 3\); check \(\binom42 \times 9 = 6 \times 9 = 54\) ✓.
Where it goes wrong
  • Dividing the second equation by the first (not its square), which leaves an \(a\) behind and stalls.
  • Using \(\binom n2 = \frac{n^2}{2}\).
Takeaway To eliminate a constant that appears to different powers, raise one equation to the matching power first, then divide.
Answer D · Binomial expansion (MM2.4) · question BIN-08
16Polynomials, factor and remainder theorems · spec MM1.6 · Paper 1 style · ★★☆

Given that \(x^3 + px + q\) is divisible by \((x - 2)^2\), find the value of \(q\).

  1. A\(16\)
  2. B\(12\)
  3. C\(8\)
  4. D\(-8\)
  5. E\(-16\)
  6. F\(-12\)
Worked solution

\(x^3 + px + q = (x-2)^2(x - r)\). Comparing \(x^2\) coefficients: \(-4 - r = 0 \Rightarrow r = -4\). Constant term: \((4)(4) = 16 = q\). (Also \(p = 4 + 4r = -12\).)

Full method — how and why
A repeated factor \((x-2)^2\) means \(x = 2\) is a double root. Comparing coefficients in the factorised form gives \(q\) without any calculus.
  1. Write \(x^3 + px + q = (x - 2)^2(x - r)\).
    A cubic with a repeated root \(2\) and one other root \(r\) must take this form (leading coefficient \(1\)).
  2. Compare \(x^2\) coefficients. The left-hand side has none, so the coefficient is \(0\); expanding the right gives \(-(4 + r)\,\)... more precisely \((x^2 - 4x + 4)(x - r)\) has \(x^2\) coefficient \(-r - 4\).
    The absence of a term is information: its coefficient is zero.
  3. So \(-r - 4 = 0\), giving \(r = -4\).
    One comparison is enough to find the third root.
  4. Constant term: \(4 \times (-r) = 4 \times 4 = 16\), so \(q = 16\).
    The constant of \((x-2)^2(x-r)\) is \(4 \times (-r)\); with \(r = -4\) that is \(16\).
Where it goes wrong
  • Forgetting that the missing \(x^2\) term forces a condition, and trying to solve with only the constant term.
  • Sign slip on \(-r\), giving \(q = -16\).
Takeaway A missing power in a polynomial is a coefficient equal to zero - often the most useful equation available.
Answer A · Polynomials, factor and remainder theorems (MM1.6) · question POL-10
17Triangles, radians and sectors · spec MM4.1 · Paper 1 style · ★★☆

Triangle \(ABC\) has \(AB = 7\), \(BC = 8\), \(CA = 9\). Find the area of the triangle.

  1. A\(28\)
  2. B\(\sqrt{720}\)
  3. C\(24\sqrt5\)
  4. D\(12\sqrt{5}\)
  5. E\(30\)
  6. F\(6\sqrt{5}\)
Worked solution

\(\cos A = \dfrac{49 + 81 - 64}{2\cdot7\cdot9} = \dfrac{66}{126} = \dfrac{11}{21}\), so \(\sin A = \dfrac{\sqrt{441 - 121}}{21} = \dfrac{\sqrt{320}}{21} = \dfrac{8\sqrt5}{21}\). Area \(= \frac12\cdot7\cdot9\cdot\frac{8\sqrt5}{21} = 12\sqrt5\).

Full method — how and why
Three sides, area wanted. Find a cosine, convert to a sine, then use \(\frac12 ab\sin C\) - all without a calculator.
  1. Find \(\cos A\) where \(A\) is between the sides \(7\) and \(9\): \(\cos A = \dfrac{49 + 81 - 64}{2(7)(9)} = \dfrac{66}{126} = \dfrac{11}{21}\).
    The side opposite \(A\) is \(BC = 8\), so that is the one subtracted.
  2. Convert to \(\sin A\) using \(\sin^2 + \cos^2 = 1\): \(\sin A = \sqrt{1 - \frac{121}{441}} = \dfrac{\sqrt{320}}{21}\).
    Keep the denominator \(21\) throughout: \(1 = \frac{441}{441}\).
  3. Simplify the surd: \(\sqrt{320} = \sqrt{64 \times 5} = 8\sqrt5\), so \(\sin A = \dfrac{8\sqrt5}{21}\).
    Pull out the largest square factor.
  4. Area \(= \frac12(7)(9)\sin A = \frac{63}{2} \times \dfrac{8\sqrt5}{21} = 12\sqrt5\).
    \(\frac{63 \times 8}{2 \times 21} = \frac{504}{42} = 12\).
Where it goes wrong
  • Taking \(\sin A\) positive is right (angles in a triangle are between \(0^\circ\) and \(180^\circ\)), but forgetting to square \(\cos A\) before subtracting.
  • Arithmetic slips in \(\sqrt{320}\), e.g. \(16\sqrt5\), doubling the area.
Takeaway Three sides \(\to\) \(\cos\) from the cosine rule \(\to\) \(\sin\) from the Pythagorean identity \(\to\) area from \(\frac12 ab\sin C\).
Answer D · Triangles, radians and sectors (MM4.1) · question TRI-15
18Trigonometric functions and equations · spec MM4.6 · Paper 1 style · ★★☆

Solve \(12\cos^2 x + 6\sin x - 10 = 2\) for \(0^\circ < x < 360^\circ\). Which of the following is the complete set of solutions?

  1. A\(0^\circ, 30^\circ, 150^\circ, 180^\circ\)
  2. B\(30^\circ, 150^\circ, 210^\circ, 330^\circ\)
  3. C\(30^\circ, 150^\circ, 270^\circ\)
  4. D\(30^\circ, 150^\circ\)
  5. E\(210^\circ, 330^\circ\)
  6. F\(30^\circ, 150^\circ, 180^\circ\)
Worked solution

\(12(1 - \sin^2x) + 6\sin x - 12 = 0 \Rightarrow 12\sin^2 x - 6\sin x = 0 \Rightarrow 6\sin x(2\sin x - 1) = 0\). \(\sin x = 0\) gives \(180^\circ\) (the endpoints \(0^\circ\) and \(360^\circ\) are excluded); \(\sin x = \frac12\) gives \(30^\circ\) and \(150^\circ\).

Full method — how and why
Another quadratic in \(\sin x\), but this one has no constant term, so factorising gives \(\sin x = 0\) as well - and the interval is open.
  1. Replace \(\cos^2 x\): \(12(1 - \sin^2 x) + 6\sin x - 10 = 2\).
    Convert to a single trig function.
  2. Tidy: \(12 - 12\sin^2x + 6\sin x - 12 = 0 \Rightarrow -12\sin^2 x + 6\sin x = 0 \Rightarrow 6\sin x(2\sin x - 1) = 0\).
    The constants cancel completely, leaving a factorisable expression with no constant term - do not divide by \(\sin x\), or you lose solutions.
  3. \(\sin x = 0\) gives \(x = 180^\circ\) within the open interval (\(0^\circ\) and \(360^\circ\) are excluded).
    The interval is \(0^\circ < x < 360^\circ\), strict at both ends - so the endpoints are out but \(180^\circ\) is in.
  4. \(\sin x = \frac12\) gives \(x = 30^\circ\) and \(150^\circ\).
    First and second quadrants.
  5. Complete set: \(30^\circ, 150^\circ, 180^\circ\).
    Three solutions - the \(180^\circ\) is what distinguishes the correct option.
Where it goes wrong
  • Dividing by \(\sin x\) and losing \(180^\circ\), giving just \(30^\circ, 150^\circ\) - an offered option.
  • Including \(0^\circ\) and \(360^\circ\), which the strict inequalities exclude.
Takeaway Never divide by a trig function: factorise, so the zero solutions survive; then read the interval's endpoints carefully.
Answer F · Trigonometric functions and equations (MM4.6) · question TRG-14
19Integration · spec MM7.2 · Paper 1 style · ★★☆

Evaluate \(\displaystyle\int_{1}^{4} \dfrac{3x^2 - 5}{\sqrt{x}}\,dx\).

  1. A\(\dfrac{146}{5}\)
  2. B\(\dfrac{186}{5}\)
  3. C\(28\)
  4. D\(\dfrac{86}{5}\)
  5. E\(\dfrac{126}{5}\)
  6. F\(\dfrac{136}{5}\)
Worked solution

Integrand \(= 3x^{3/2} - 5x^{-1/2}\); integral \(= \left[\frac65 x^{5/2} - 10x^{1/2}\right]_1^4 = \left(\frac{6\cdot32}{5} - 20\right) - \left(\frac65 - 10\right) = \frac{192}{5} - 20 - \frac65 + 10 = \frac{186}{5} - 10 = \frac{136}{5}\).

Full method — how and why
You can only integrate powers of \(x\), so split the fraction first. Then the limits are straightforward - if the fractions are handled carefully.
  1. Split the fraction: \(\dfrac{3x^2 - 5}{\sqrt x} = 3x^2 x^{-1/2} - 5x^{-1/2} = 3x^{3/2} - 5x^{-1/2}\).
    Dividing by \(\sqrt x = x^{1/2}\) subtracts \(\frac12\) from each index.
  2. Integrate: \(\displaystyle\int 3x^{3/2}\,dx = \frac{3x^{5/2}}{5/2} = \frac65 x^{5/2}\) and \(\displaystyle\int 5x^{-1/2}\,dx = 10x^{1/2}\).
    Add one to the index and divide by the new index: \(\frac32 + 1 = \frac52\), and \(-\frac12 + 1 = \frac12\) so dividing by \(\frac12\) doubles.
  3. Evaluate at \(x = 4\): \(4^{5/2} = 32\) and \(4^{1/2} = 2\), giving \(\frac{192}{5} - 20\).
    \(4^{5/2} = (\sqrt4)^5 = 2^5\).
  4. Evaluate at \(x = 1\): \(\frac65 - 10\). Subtract: \(\left(\frac{192}{5} - 20\right) - \left(\frac65 - 10\right) = \frac{186}{5} - 10 = \frac{136}{5}\).
    \(\frac{192 - 6}{5} = \frac{186}{5}\), then \(-20 + 10 = -10 = -\frac{50}{5}\).
Where it goes wrong
  • Forgetting the lower limit entirely, giving \(\frac{186}{5}\) - an offered option.
  • Dividing by \(\frac52\) incorrectly: \(\frac{3}{5/2} = \frac65\), not \(\frac{15}{2}\).
Takeaway Split fractions into powers of \(x\) before integrating; \(4^{5/2} = 32\) type evaluations are worth doing slowly.
Answer F · Integration (MM7.2) · question INT-01
20Number, units and estimation · spec M2.3 · Paper 1 style · ★★☆

How many positive divisors does \(360\) have?

  1. A\(24\)
  2. B\(20\)
  3. C\(18\)
  4. D\(36\)
  5. E\(12\)
  6. F\(16\)
Worked solution

\(360 = 2^3\cdot3^2\cdot5\), so the number of divisors is \((3+1)(2+1)(1+1) = 24\).

Full method — how and why
Counting divisors from the prime factorisation: each prime's exponent can be chosen independently.
  1. Factorise: \(360 = 8 \times 45 = 2^3 \times 3^2 \times 5\).
    Split into convenient factors and factorise each.
  2. A divisor has the form \(2^a 3^b 5^c\) with \(0 \leq a \leq 3\), \(0 \leq b \leq 2\), \(0 \leq c \leq 1\).
    Any such combination divides \(360\), and every divisor has this form uniquely.
  3. Count the choices: \(4 \times 3 \times 2 = 24\).
    Each exponent has (power \(+ 1\)) possible values, because \(0\) is allowed.
Where it goes wrong
  • Multiplying the exponents (\(3 \times 2 \times 1 = 6\)) instead of (exponent \(+1\)).
  • Forgetting that \(1\) and \(360\) are themselves divisors - the '\(+1\)'s account for them.
Takeaway Number of divisors of \(p^aq^br^c\) is \((a+1)(b+1)(c+1)\).
Answer A · Number, units and estimation (M2.3) · question NUM-09
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