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Paper 1 · Set 04

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

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

Find the range of values of \(x\) for which \(x^2 - 7x + 10 < 0\) and \(x^2 - 4 > 0\).

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

\((x-2)(x-5) < 0\) gives \(2 < x < 5\); \(x^2 > 4\) gives \(x < -2\) or \(x > 2\). The intersection is \(2 < x < 5\).

Full method — how and why
Two simultaneous quadratic inequalities. Solve each separately, then intersect on a number line.
  1. First: \(x^2 - 7x + 10 < 0\) factorises as \((x-2)(x-5) < 0\), giving \(2 < x < 5\).
    An upward parabola is below the axis between its roots.
  2. Second: \(x^2 - 4 > 0\) gives \(x < -2\) or \(x > 2\).
    Above the axis means outside the roots - two intervals.
  3. Intersect: \((2, 5)\) overlaps \((2, \infty)\) entirely and misses \((-\infty, -2)\) completely, so the answer is \(2 < x < 5\).
    'And' means intersection: a value must satisfy both, so draw both solution sets on one line and take the common part.
Where it goes wrong
  • Taking the union and including \(x < -2\) - the word is 'and', not 'or'.
  • Getting the first inequality backwards (\(x < 2\) or \(x > 5\)) by forgetting that \(< 0\) means between the roots.
Takeaway \(<0\) between the roots, \(>0\) outside them (for an upward parabola); then combine solution sets on a number line.
Answer A · Quadratic functions (MM1.3) · question QUA-18
2Differentiation · spec MM6.3 · Paper 1 style · ★☆☆

Find the equation of the normal to the curve \(y = \sqrt{x}\) at the point \((4, 2)\).

  1. A\(y = -4x + 16\)
  2. B\(y = -4x + 18\)
  3. C\(y = -4x + 2\)
  4. D\(y = \dfrac14 x + 1\)
  5. E\(y = -\dfrac14 x + 3\)
  6. F\(y = 4x - 14\)
Worked solution

\(y' = \frac{1}{2\sqrt x} = \frac14\) at \(x = 4\), so the normal has gradient \(-4\): \(y - 2 = -4(x - 4)\), i.e. \(y = -4x + 18\).

Full method — how and why
Normal line: the negative reciprocal of the tangent's gradient, through the same point.
  1. Write \(y = x^{1/2}\) and differentiate: \(y' = \frac12 x^{-1/2} = \dfrac{1}{2\sqrt x}\).
    Convert the root to a fractional power before differentiating.
  2. At \(x = 4\): \(y' = \dfrac{1}{2 \times 2} = \dfrac14\).
    \(\sqrt4 = 2\).
  3. Normal gradient \(= -\dfrac{1}{1/4} = -4\).
    Normal is perpendicular to the tangent: negative reciprocal.
  4. Line through \((4,2)\): \(y - 2 = -4(x - 4) \Rightarrow y = -4x + 18\).
    \(-4 \times -4 = +16\), plus the \(2\).
Where it goes wrong
  • Using the tangent gradient \(\frac14\) and choosing \(y = \frac14x + 1\).
  • Arithmetic on the constant: \(y = -4x + 16\) is offered.
Takeaway Normal gradient \(= -1/f'(x_1)\); then substitute the point into the point-gradient form.
Answer B · Differentiation (MM6.3) · question DIF-04
3Number, units and estimation · spec M2.3 · Paper 1 style · ★☆☆

Find the highest common factor of \(2^{3}\times 3^{2}\times 5\) and \(2^{2}\times 3^{3}\times 7\).

  1. A\(18\)
  2. B\(6\)
  3. C\(12\)
  4. D\(36\)
  5. E\(72\)
  6. F\(108\)
Worked solution

Take the lower power of each common prime: \(2^2 \times 3^2 = 36\).

Full method — how and why
HCF from prime factorisations: take each common prime to its lower power.
  1. List the primes in each: \(2^3 \times 3^2 \times 5\) and \(2^2 \times 3^3 \times 7\).
    Both are already factorised, which is the form the rule needs.
  2. Take the common primes only: \(2\) and \(3\) appear in both; \(5\) and \(7\) do not.
    A prime missing from one number cannot divide both.
  3. Take the lower power of each: \(2^{\min(3,2)} = 2^2\) and \(3^{\min(2,3)} = 3^2\).
    The HCF must divide both, so it cannot use more copies of a prime than the smaller number has.
  4. Multiply: \(4 \times 9 = 36\).
    Check: \(36\) divides \(360\) and \(756\) ✓.
Where it goes wrong
  • Taking the higher powers (that gives the LCM contribution, not the HCF).
  • Including \(5\) or \(7\), which gives a number dividing only one of them.
Takeaway HCF: common primes, lowest powers. LCM: all primes, highest powers.
Answer D · Number, units and estimation (M2.3) · question NUM-01
4Integration · spec MM7.6 · Paper 1 style · ★☆☆

The gradient of a curve at \((x, y)\) is \(3\sqrt{x}\) and the curve passes through \((4, 20)\). Find the \(y\)-intercept of the curve.

  1. A\(12\)
  2. B\(20\)
  3. C\(4\)
  4. D\(8\)
  5. E\(-4\)
  6. F\(0\)
Worked solution

\(y = 2x^{3/2} + c\); \(20 = 16 + c \Rightarrow c = 4\). The \(y\)-intercept is \(4\).

Full method — how and why
Integrate the gradient function, use the given point for \(c\), then read the intercept.
  1. Integrate: \(\displaystyle\int 3x^{1/2}dx = 3 \times \frac{x^{3/2}}{3/2} = 2x^{3/2}\), so \(y = 2x^{3/2} + c\).
    \(3 \div \frac32 = 2\).
  2. Substitute \((4, 20)\): \(4^{3/2} = 8\), so \(2(8) + c = 20\), giving \(c = 4\).
    \(4^{3/2} = (\sqrt4)^3 = 8\).
  3. The \(y\)-intercept is the value at \(x = 0\): \(y = 0 + 4 = 4\).
    For this curve the intercept is just \(c\).
Where it goes wrong
  • Using \(4^{3/2} = 6\) and getting \(c = 8\) - an offered option.
  • Reporting \(20\) (the given \(y\)-value) as the intercept.
Takeaway The constant of integration is the \(y\)-intercept when the rest of the expression vanishes at \(x = 0\).
Answer C · Integration (MM7.6) · question INT-17
5Sequences and series · spec MM2.3 · Paper 1 style · ★☆☆

Evaluate \(\displaystyle\sum_{k=1}^{10} 3 \times 2^{k}\).

  1. A\(6141\)
  2. B\(3069\)
  3. C\(3072\)
  4. D\(3066\)
  5. E\(6144\)
  6. F\(6138\)
Worked solution

This is \(6 + 12 + \ldots + 6\cdot2^{9}\): first term \(6\), ratio \(2\), \(10\) terms: \(6(2^{10} - 1) = 6 \times 1023 = 6138\).

Full method — how and why
Sigma notation hiding a geometric series. Write out the first term and the ratio, and count the terms carefully.
  1. Write out the start: \(k = 1\) gives \(3 \times 2 = 6\); \(k = 2\) gives \(12\). So \(a = 6\), \(r = 2\).
    The first term is the value at the lower limit, not \(3\).
  2. Count the terms: \(k\) runs from \(1\) to \(10\), which is \(10\) terms.
    Upper minus lower plus one.
  3. Apply the sum formula: \(S_{10} = 6\dfrac{2^{10} - 1}{2 - 1} = 6 \times 1023\).
    \(2^{10} = 1024\) is worth knowing by heart.
  4. Evaluate: \(6138\).
    \(6 \times 1000 = 6000\) plus \(6 \times 23 = 138\).
Where it goes wrong
  • Taking \(a = 3\) (ignoring the \(k = 1\) factor of \(2\)), giving \(3069\).
  • Using \(2^{10}\) instead of \(2^{10} - 1\), giving \(6144\).
Takeaway In \(\sum a r^k\) the first term is the value at the lower limit; count terms as (upper \(-\) lower \(+ 1\)).
Answer F · Sequences and series (MM2.3) · question SEQ-15
6GCSE algebra · spec M4.18 · Paper 1 style · ★☆☆

The distance-time graph of a journey is a straight line from \((0, 0)\) to \((2, 60)\), then horizontal to \((3, 60)\), then a straight line to \((5, 0)\) (time in hours, distance from home in km). What was the average speed for the whole journey?

  1. A\(0\) km/h
  2. B\(12\) km/h
  3. C\(24\) km/h
  4. D\(20\) km/h
  5. E\(30\) km/h
  6. F\(40\) km/h
Worked solution

Total distance \(60 + 60 = 120\) km in \(5\) hours: \(24\) km/h. (Displacement is \(0\) but distance travelled is not.)

Full method — how and why
Average speed uses total distance, not net displacement - and this journey returns home.
  1. Read the outward distance: the traveller goes from \(0\) to \(60\) km, so \(60\) km out.
    The graph shows distance from home against time.
  2. The horizontal section is a stop (no distance covered), and the final section returns to \(0\): another \(60\) km.
    Coming back still counts as distance travelled.
  3. Total distance \(= 60 + 60 = 120\) km over a total time of \(5\) hours.
    Include the stopped hour in the time - it is part of the journey.
  4. Average speed \(= \dfrac{120}{5} = 24\) km/h.
    Total distance divided by total time.
Where it goes wrong
  • Using the net displacement \(0\) and answering \(0\) km/h - an offered option.
  • Excluding the stopped hour, giving \(30\) km/h.
Takeaway Average speed \(=\) total distance \(\div\) total time; stops count in the time, return legs count in the distance.
Answer C · GCSE algebra (M4.18) · question ALG-15
7Pythagoras, mensuration and 3D · spec M5.7 · Paper 1 style · ★☆☆

A ladder of length \(6.5\) m leans against a vertical wall with its foot \(2.5\) m from the wall on horizontal ground. How far up the wall does it reach?

  1. A\(7\) m
  2. B\(5\) m
  3. C\(4\) m
  4. D\(5.5\) m
  5. E\(6\) m
  6. F\(\sqrt{48.5}\) m
Worked solution

\(\sqrt{6.5^2 - 2.5^2} = \sqrt{42.25 - 6.25} = \sqrt{36} = 6\).

Full method — how and why
A ladder against a wall is a right-angled triangle with the ladder as hypotenuse.
  1. Identify the hypotenuse: the ladder, \(6.5\) m - the longest side, opposite the right angle at the base of the wall.
    The wall and ground are perpendicular.
  2. Apply Pythagoras: height \(= \sqrt{6.5^2 - 2.5^2} = \sqrt{42.25 - 6.25}\).
    Subtract, because we want a shorter side.
  3. Compute: \(\sqrt{36} = 6\) m.
    Decimals chosen to give a whole number - a sign the working is right.
Where it goes wrong
  • Adding the squares, giving \(\sqrt{48.5}\) - an offered option.
  • Arithmetic: \(6.5^2 = 42.25\), not \(36.25\).
Takeaway Hypotenuse known \(\Rightarrow\) subtract squares; hypotenuse unknown \(\Rightarrow\) add them.
Answer E · Pythagoras, mensuration and 3D (M5.7) · question MEN-07
8Straight lines · spec MM3.1 · Paper 1 style · ★☆☆

The line \(ax + by = 12\) has \(x\)-intercept \(3\) and \(y\)-intercept \(-4\). Find \(a - b\).

  1. A\(12\)
  2. B\(7\)
  3. C\(-1\)
  4. D\(0\)
  5. E\(-7\)
  6. F\(1\)
Worked solution

\((3, 0)\): \(3a = 12 \Rightarrow a = 4\). \((0, -4)\): \(-4b = 12 \Rightarrow b = -3\). \(a - b = 7\).

Full method — how and why
Intercepts are points on the line. Substitute each one to get an equation for \(a\) and for \(b\) separately.
  1. \(x\)-intercept \(3\) means the point \((3, 0)\) lies on the line: \(3a + 0 = 12\), so \(a = 4\).
    At an \(x\)-intercept, \(y = 0\).
  2. \(y\)-intercept \(-4\) means \((0, -4)\) lies on it: \(0 - 4b = 12\), so \(b = -3\).
    At a \(y\)-intercept, \(x = 0\). The negative intercept makes \(b\) negative.
  3. Compute \(a - b = 4 - (-3) = 7\).
    Subtracting a negative adds.
Where it goes wrong
  • Writing \(b = 3\) and answering \(1\).
  • Mixing up which intercept is which and substituting \((0,3)\) and \((-4,0)\).
Takeaway An intercept is just a point with one coordinate zero - substitute it into the equation.
Answer B · Straight lines (MM3.1) · question LIN-07
9Binomial expansion · spec MM2.4 · Paper 1 style · ★☆☆

In the expansion of \((1 + x)^n\), the coefficients of \(x^4\) and \(x^5\) are equal. Find \(n\).

  1. A\(20\)
  2. B\(5\)
  3. C\(11\)
  4. D\(9\)
  5. E\(10\)
  6. F\(8\)
Worked solution

\(\binom n4 = \binom n5\) requires \(4 + 5 = n\) (by symmetry \(\binom nr = \binom n{n-r}\)), so \(n = 9\).

Full method — how and why
Two equal binomial coefficients from the same row. Symmetry gives the answer without computing factorials.
  1. The condition is \(\binom n4 = \binom n5\).
    In \((1+x)^n\) the coefficient of \(x^r\) is \(\binom nr\).
  2. Within one row of Pascal's triangle, two different entries are equal only when they are symmetric partners: \(\binom nr = \binom ns\) with \(r \neq s\) forces \(r + s = n\).
    The row rises to a single peak and falls symmetrically, so equal values are mirror images.
  3. Hence \(n = 4 + 5 = 9\).
    Check: \(\binom94 = 126 = \binom95\) ✓.
Where it goes wrong
  • Setting up the factorial equation and mis-cancelling to get \(n = 10\) or \(n = 8\).
  • Assuming \(n = 5\) because the coefficients 'stop' there.
Takeaway Equal entries in the same row of Pascal's triangle are symmetric partners: \(r + s = n\).
Answer D · Binomial expansion (MM2.4) · question BIN-06
10Binomial expansion · spec MM2.4 · Paper 1 style · ★☆☆

Find the coefficient of \(x^3\) in the expansion of \((2 + 3x)^5\).

  1. A\(270\)
  2. B\(1080\)
  3. C\(2160\)
  4. D\(720\)
  5. E\(540\)
  6. F\(360\)
Worked solution

\(\binom{5}{3} 2^{2} (3x)^3 = 10 \times 4 \times 27 x^3 = 1080x^3\).

Full method — how and why
A single coefficient from \((a + bx)^n\). Use the general term rather than expanding the whole thing.
  1. Write the general term: \(\binom{5}{r} 2^{5-r}(3x)^r\).
    Every term of \((a+b)^n\) has the form \(\binom nr a^{n-r}b^r\); the powers of the two parts always add to \(n\).
  2. For \(x^3\) take \(r = 3\): \(\binom53 2^{2}(3x)^3\).
    The power of \(x\) comes only from the second bracket, so \(r\) is the power you want.
  3. Evaluate each factor: \(\binom53 = 10\), \(2^2 = 4\), \(3^3 = 27\).
    Keep them separate - combining too early is where slips happen.
  4. Multiply: \(10 \times 4 \times 27 = 1080\).
    \(10 \times 108 = 1080\).
Where it goes wrong
  • Forgetting the \(2^{2}\) and answering \(270\).
  • Using \(3^3 = 9\) or cubing only the \(x\), giving \(360\) - both are offered.
Takeaway The general term \(\binom nr a^{n-r}b^r\) carries three factors: the binomial coefficient and a power of each part.
Answer B · Binomial expansion (MM2.4) · question BIN-01
11Functions and transformations · spec MM8.3 · Paper 1 style · ★☆☆

The line \(y = mx + c\) passes through \((2, 7)\). If \(m\) is increased by \(1\) and \(c\) is decreased by \(2\), the new line passes through \((2, k)\). Find \(k\).

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

New value at \(x = 2\): \((m+1)\cdot 2 + (c - 2) = 2m + c = 7\). So \(k = 7\).

Full method — how and why
The two changes are designed to cancel at \(x = 2\). Substitute and watch the terms collapse.
  1. From the original line: \(2m + c = 7\).
    The line passes through \((2, 7)\), so substituting gives one equation in \(m\) and \(c\).
  2. The new line is \(y = (m + 1)x + (c - 2)\); evaluate it at \(x = 2\): \(2(m+1) + (c - 2) = 2m + 2 + c - 2\).
    Substitute \(x = 2\) and expand - do not try to find \(m\) and \(c\), which are not determined.
  3. Simplify: \(2m + c = 7\), so \(k = 7\).
    The \(+2\) from the gradient change and the \(-2\) from the intercept change cancel exactly.
Where it goes wrong
  • Trying to solve for \(m\) and \(c\) individually - there is not enough information, and none is needed.
  • Forgetting that the gradient change is multiplied by \(x = 2\), and answering \(5\) or \(9\).
Takeaway When a question gives only one equation for two unknowns, look for a combination that is determined - here \(2m + c\).
Answer B · Functions and transformations (MM8.3) · question FUN-17
12Number, units and estimation · spec M2.2 · Paper 1 style · ★☆☆

Evaluate \(\left(2\dfrac{1}{3} - 1\dfrac{3}{4}\right) \div \dfrac{7}{6}\).

  1. A\(\dfrac{2}{3}\)
  2. B\(\dfrac{7}{12}\)
  3. C\(\dfrac{49}{72}\)
  4. D\(\dfrac{1}{2}\)
  5. E\(\dfrac{7}{24}\)
  6. F\(1\)
Worked solution

\(\frac73 - \frac74 = \frac{28 - 21}{12} = \frac{7}{12}\); \(\frac{7}{12} \times \frac67 = \frac12\).

Full method — how and why
Mixed numbers and division of fractions. Convert to improper fractions, subtract, then multiply by the reciprocal.
  1. Convert: \(2\frac13 = \frac73\) and \(1\frac34 = \frac74\).
    Whole part times denominator plus numerator.
  2. Subtract with a common denominator \(12\): \(\dfrac{28}{12} - \dfrac{21}{12} = \dfrac{7}{12}\).
    \(\frac73 = \frac{28}{12}\) and \(\frac74 = \frac{21}{12}\).
  3. Divide by \(\frac76\) means multiply by \(\frac67\): \(\dfrac{7}{12} \times \dfrac{6}{7}\).
    Dividing by a fraction is multiplying by its reciprocal.
  4. Cancel and multiply: the \(7\)s cancel and \(\frac{6}{12} = \frac12\).
    Cancel before multiplying to keep numbers small.
Where it goes wrong
  • Multiplying by \(\frac76\) instead of its reciprocal, giving \(\frac{49}{72}\) - an offered option.
  • Subtracting the whole parts and fractions separately and mishandling the borrow.
Takeaway Improper fractions, common denominator, then multiply by the reciprocal - and cancel before multiplying.
Answer D · Number, units and estimation (M2.2) · question NUM-11
13Graphs of functions · spec MM8.7 · Paper 1 style · ★☆☆

The line \(y = mx\) meets the curve \(y = x^3 - x\) at three distinct points. Find the complete set of values of \(m\).

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

\(x^3 - x - mx = x(x^2 - (1 + m)) = 0\): three distinct roots iff \(1 + m > 0\), i.e. \(m > -1\).

Full method — how and why
Intersections of a line through the origin with a cubic. Factor out the common \(x\) and use the discriminant on what remains.
  1. Set them equal: \(mx = x^3 - x\), so \(x^3 - x - mx = 0\).
    Bring everything to one side.
  2. Factor out \(x\): \(x\left(x^2 - (1 + m)\right) = 0\).
    \(x = 0\) is always a solution, because both graphs pass through the origin.
  3. The other roots satisfy \(x^2 = 1 + m\), which has two distinct non-zero solutions exactly when \(1 + m > 0\).
    If \(1 + m = 0\) the root coincides with \(x = 0\); if negative, there are no more real roots.
  4. So \(m > -1\).
    Three distinct points in total.
Where it goes wrong
  • Forgetting the root \(x = 0\) and requiring three roots from the quadratic.
  • Including \(m = -1\), where all three roots coincide at \(x = 0\).
Takeaway A line through the origin always meets \(y = x^3 - x\) there; factor out \(x\) and analyse the quadratic that remains.
Answer E · Graphs of functions (MM8.7) · question GRA-16
14Differentiation · spec MM6.3 · Paper 1 style · ★☆☆

Find the coordinates of the stationary points of \(y = x^3 - 3x^2 - 9x + 5\) and state their nature.

  1. Amaximum at \((1, -6)\), minimum at \((-3, -22)\)
  2. Bminimum at \((-1, 10)\), maximum at \((3, -22)\)
  3. Cmaximum at \((-1, 10)\), minimum at \((3, -22)\)
  4. Dmaximum at \((-1, 0)\), minimum at \((3, -32)\)
  5. Emaximum at \((-1, 10)\) only
Worked solution

\(y' = 3x^2 - 6x - 9 = 3(x - 3)(x + 1)\). \(y'' = 6x - 6\): negative at \(x = -1\) (max, \(y = -1 - 3 + 9 + 5 = 10\)), positive at \(x = 3\) (min, \(y = 27 - 27 - 27 + 5 = -22\)).

Full method — how and why
Stationary points from \(f' = 0\), their nature from the sign of \(f''\), and the \(y\)-values by substitution.
  1. Differentiate: \(y' = 3x^2 - 6x - 9 = 3(x^2 - 2x - 3) = 3(x-3)(x+1)\).
    Factor out the \(3\) first - it makes the factorisation obvious.
  2. Solve \(y' = 0\): \(x = 3\) and \(x = -1\).
    Stationary points are where the gradient vanishes.
  3. Second derivative: \(y'' = 6x - 6\). At \(x = -1\), \(y'' = -12 < 0\) (maximum); at \(x = 3\), \(y'' = 12 > 0\) (minimum).
    Negative second derivative means the curve bends downwards - a maximum.
  4. Find the \(y\)-values: \(y(-1) = -1 - 3 + 9 + 5 = 10\); \(y(3) = 27 - 27 - 27 + 5 = -22\).
    Substitute into the original function, not the derivative.
Where it goes wrong
  • Swapping the natures - for a positive cubic the maximum always comes first (smaller \(x\)).
  • Substituting into \(y'\) instead of \(y\) for the coordinates.
Takeaway \(f' = 0\) locates stationary points; the sign of \(f''\) classifies them; substitute into \(f\) for the heights.
Answer C · Differentiation (MM6.3) · question DIF-02
15Triangles, radians and sectors · spec MM4.1 · Paper 1 style · ★★☆

A vertical pole \(BT\) stands on horizontal ground. From a point \(A\) on the ground the angle of elevation of \(T\) is \(30^\circ\); from a point \(C\) on the ground, \(20\) m closer to the pole than \(A\) and in line with \(A\) and \(B\), the angle of elevation is \(60^\circ\). Find the height of the pole.

  1. A\(10\) m
  2. B\(20\sqrt{3}\) m
  3. C\(15\) m
  4. D\(10\sqrt{3}\) m
  5. E\(\dfrac{20}{\sqrt3}\) m
  6. F\(30\) m
Worked solution

Let the height be \(h\). \(AB = h\sqrt3\) and \(CB = \frac{h}{\sqrt3}\). \(AB - CB = 20 \Rightarrow h\left(\sqrt3 - \frac1{\sqrt3}\right) = \frac{2h}{\sqrt3} = 20 \Rightarrow h = 10\sqrt3\).

Full method — how and why
Two right-angled triangles sharing the same height. Express both base lengths in terms of \(h\) and use the given difference.
  1. From \(A\): \(\tan 30^\circ = \dfrac{h}{AB}\), so \(AB = \dfrac{h}{\tan 30^\circ} = h\sqrt3\).
    \(\tan 30^\circ = \frac{1}{\sqrt3}\), so dividing by it multiplies by \(\sqrt3\).
  2. From \(C\): \(\tan 60^\circ = \dfrac{h}{CB}\), so \(CB = \dfrac{h}{\sqrt3}\).
    \(\tan 60^\circ = \sqrt3\). The steeper angle means the shorter base - a good sanity check.
  3. Use the \(20\) m gap: \(AB - CB = h\sqrt3 - \dfrac{h}{\sqrt3} = h\left(\dfrac{3 - 1}{\sqrt3}\right) = \dfrac{2h}{\sqrt3} = 20\).
    Common denominator \(\sqrt3\): \(\sqrt3 = \frac{3}{\sqrt3}\).
  4. Solve: \(h = \dfrac{20\sqrt3}{2} = 10\sqrt3\) m.
    Multiply both sides by \(\sqrt3\) and halve.
Where it goes wrong
  • Adding the bases instead of subtracting (the closer point has the shorter base).
  • Using \(\tan 30^\circ = \sqrt3\), which swaps the two triangles.
Takeaway Two elevation angles to the same object: write each horizontal distance in terms of the height and use their difference.
Answer D · Triangles, radians and sectors (MM4.1) · question TRI-14
16Polynomials, factor and remainder theorems · spec MM1.6 · Paper 1 style · ★★☆

For which value of \(k\) does \(x^2 - x + k\) divide exactly into \(x^3 - 2x^2 + 3x - 4\)? (Assume such a value exists.)

  1. A\(-2\)
  2. B\(3\)
  3. C\(4\)
  4. D\(2\)
  5. E\(1\)
  6. Fno such value exists
Worked solution

Try \(x^3 - 2x^2 + 3x - 4 = (x^2 - x + k)(x + m)\): comparing \(x^2\): \(m - 1 = -2 \Rightarrow m = -1\). Then \(x\): \(k - m = k + 1 = 3 \Rightarrow k = 2\). Constant: \(km = -2 \neq -4\). No value of \(k\) works.

Full method — how and why
An 'assume it exists' question where the honest answer is that it does not. Comparing coefficients over-determines \(k\), and the system is inconsistent.
  1. If \(x^2 - x + k\) divides exactly, the quotient is linear: \(x^3 - 2x^2 + 3x - 4 = (x^2 - x + k)(x + m)\).
    Cubic \(\div\) quadratic \(=\) linear, and 'divides exactly' means no remainder term.
  2. Compare \(x^2\): \(m - 1 = -2\), so \(m = -1\).
    The \(x^2\) terms are \(mx^2\) and \(-x^2\).
  3. Compare \(x\): \(k - m = 3\), so \(k = 2\).
    The \(x\) terms are \(kx\) and \(-mx\).
  4. Check the constant term: \(km = 2 \times (-1) = -2\), but the cubic has \(-4\). Contradiction.
    With three equations and two unknowns the system is over-determined - the last equation is a genuine test, not a formality.
  5. So no value of \(k\) works: the answer is 'no such value exists'.
    The bracketed 'assume such a value exists' is a deliberate misdirection.
Where it goes wrong
  • Stopping at \(k = 2\) and never checking the constant term.
  • Trusting the question's hint that a value exists - always verify all the coefficients.
Takeaway When comparing coefficients gives more equations than unknowns, the leftover equations must be checked; they can reveal that no solution exists.
Answer F · Polynomials, factor and remainder theorems (MM1.6) · question POL-16
17Sequences and series · spec MM2.3 · Paper 1 style · ★★☆

The first three terms of a geometric sequence are \(x\), \(x + 6\) and \(x + 15\). Find the fourth term.

  1. A\(\dfrac{81}{2}\)
  2. B\(\dfrac{125}{2}\)
  3. C\(\dfrac{75}{2}\)
  4. D\(\dfrac{225}{4}\)
  5. E\(\dfrac{125}{4}\)
  6. F\(45\)
Worked solution

Geometric: \((x+6)^2 = x(x+15) \Rightarrow x^2 + 12x + 36 = x^2 + 15x \Rightarrow x = 12\). The terms are \(12, 18, 27\) with ratio \(\frac32\), so the fourth term is \(27 \times \frac32 = \frac{81}{2}\).

Full method — how and why
Three consecutive geometric terms: the middle one squared equals the product of its neighbours. That single equation finds \(x\).
  1. Use the geometric condition \(\dfrac{x+6}{x} = \dfrac{x+15}{x+6}\), i.e. \((x+6)^2 = x(x+15)\).
    Consecutive terms have a constant ratio, so equating the two ratios gives the condition \(b^2 = ac\).
  2. Expand: \(x^2 + 12x + 36 = x^2 + 15x\).
    The \(x^2\) terms cancel, leaving a linear equation - which is why this method is quick.
  3. Solve: \(36 = 3x\), so \(x = 12\).
    The terms are \(12, 18, 27\).
  4. Find \(r = \frac{18}{12} = \frac32\) and the fourth term \(27 \times \frac32 = \frac{81}{2}\).
    The ratio need not be an integer; leave the answer as an improper fraction.
Where it goes wrong
  • Assuming the common ratio is \(\frac{x+6}{x+15}\) (upside down).
  • Stopping at the third term \(27\) or giving the ratio instead of the fourth term.
Takeaway Three terms in geometric progression satisfy \(b^2 = ac\); the \(x^2\) terms always cancel, leaving a linear equation.
Answer A · Sequences and series (MM2.3) · question SEQ-19
18Quadratic functions · spec MM1.3 · Paper 1 style · ★★☆

The equation \(x^2 + ax + b = 0\) has roots \(3\) and \(-5\). The equation \(x^2 + bx + a = 0\) has roots \(p\) and \(q\). Find \(p^2 + q^2\).

  1. A\(229\)
  2. B\(195\)
  3. C\(4\)
  4. D\(221\)
  5. E\(225\)
  6. F\(255\)
Worked solution

Sum of roots \(3 + (-5) = -a\), so \(a = 2\); product \(-15 = b\). Then \(x^2 - 15x + 2 = 0\) has \(p + q = 15\) and \(pq = 2\), so \(p^2 + q^2 = 15^2 - 2(2) = 221\).

Full method — how and why
Two linked quadratics: the roots of the first give the coefficients of the second. Sum-and-product relations do all the work.
  1. For \(x^2 + ax + b = 0\) with roots \(3\) and \(-5\): sum \(= 3 + (-5) = -2 = -a\), so \(a = 2\).
    Sum of roots \(= -b/a\); with a monic quadratic that is just \(-a\) - mind the double negative.
  2. Product \(= 3 \times (-5) = -15 = b\).
    Product of roots \(= c/a = b\) here.
  3. The second equation is \(x^2 - 15x + 2 = 0\), so \(p + q = 15\) and \(pq = 2\).
    Substitute the values found, being careful that the roles of \(a\) and \(b\) swap between the two equations.
  4. Use \(p^2 + q^2 = (p+q)^2 - 2pq = 225 - 4 = 221\).
    The same identity as in QUA-04.
Where it goes wrong
  • Getting \(a = -2\) by forgetting the minus in 'sum \(= -a\)', which turns the second equation into \(x^2 - 15x - 2\) and gives \(229\).
  • Computing \((p+q)^2\) and forgetting to subtract \(2pq\), giving \(225\).
Takeaway Read coefficients from roots with sum \(= -b/a\) and product \(= c/a\), and keep track of which equation you are in.
Answer D · Quadratic functions (MM1.3) · question QUA-22
19Indices and surds · spec MM1.1 · Paper 1 style · ★★☆

Which of the following is equal to \(\dfrac{x^2 \sqrt{x}}{\sqrt[3]{x}}\) for \(x > 0\)?

  1. A\(x^{3}\)
  2. B\(x^{\frac{5}{2}}\)
  3. C\(x^{\frac{7}{3}}\)
  4. D\(x^{\frac{11}{6}}\)
  5. E\(x^{\frac{17}{6}}\)
  6. F\(x^{\frac{13}{6}}\)
Worked solution

The index is \(2 + \frac12 - \frac13 = \frac{12 + 3 - 2}{6} = \frac{13}{6}\).

Full method — how and why
Roots disguised as indices. Convert every root to a fractional power and the question becomes adding and subtracting fractions.
  1. Rewrite the roots: \(\sqrt x = x^{1/2}\) and \(\sqrt[3]{x} = x^{1/3}\).
    \(\sqrt[n]{x} = x^{1/n}\). In index form the laws apply; in root form they do not.
  2. Combine: \(\dfrac{x^2 \cdot x^{1/2}}{x^{1/3}} = x^{2 + \frac12 - \frac13}\).
    Multiplication adds indices, division subtracts them - the three operations become one sum.
  3. Add the fractions over a common denominator \(6\): \(2 + \frac12 - \frac13 = \frac{12 + 3 - 2}{6} = \frac{13}{6}\).
    \(6\) is the lowest common denominator of \(1\), \(2\) and \(3\); converting \(2\) to \(\frac{12}{6}\) is where slips happen.
Where it goes wrong
  • Adding the \(\frac13\) instead of subtracting gives \(x^{17/6}\), which is offered.
  • Forgetting to convert the whole number \(2\) into sixths.
Takeaway Turn every root into a fractional index immediately; then all you are doing is arithmetic with fractions.
Answer F · Indices and surds (MM1.1) · question IND-07
20Triangles, radians and sectors · spec MM4.2 · Paper 1 style · ★★☆

The perimeter of a sector of radius \(r\) and angle \(\theta\) radians is \(20\), and its area is \(16\). Find the possible values of \(r\).

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

\(2r + r\theta = 20\) and \(\frac12 r^2\theta = 16\). From the first \(r\theta = 20 - 2r\); then \(\frac12 r(20 - 2r) = 16 \Rightarrow r^2 - 10r + 16 = 0 \Rightarrow r = 2\) or \(8\).

Full method — how and why
Two sector facts, two unknowns. Eliminate \(\theta\) by substitution and solve the resulting quadratic in \(r\).
  1. Perimeter of a sector \(=\) two radii \(+\) arc: \(2r + r\theta = 20\).
    Students often forget the two straight edges - the perimeter is not just the arc.
  2. Rearrange for the arc: \(r\theta = 20 - 2r\).
    Keeping the product \(r\theta\) together is the key, because the area formula contains it.
  3. Area: \(\frac12 r^2\theta = \frac12 r(r\theta) = \frac12 r(20 - 2r) = 16\).
    Writing \(r^2\theta\) as \(r \times (r\theta)\) lets you substitute directly - no need to find \(\theta\).
  4. Simplify and solve: \(10r - r^2 = 16 \Rightarrow r^2 - 10r + 16 = 0 \Rightarrow (r-2)(r-8) = 0\), so \(r = 2\) or \(8\).
    Both are geometrically valid: \(r = 2\) gives \(\theta = 8\) radians... which exceeds \(2\pi\), but the question asks only for the values of \(r\) from the equations.
Where it goes wrong
  • Using perimeter \(= r\theta\) alone, which gives a different quadratic.
  • Finding only one root of the quadratic.
Takeaway Substitute the product \(r\theta\) as a block; sector perimeter is \(2r + r\theta\).
Answer D · Triangles, radians and sectors (MM4.2) · question TRI-08
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