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Paper 1 · Set 07

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

Find the value of \(\left(\dfrac{27}{8}\right)^{-\frac{2}{3}}\).

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

The negative index inverts the fraction: \(\left(\frac{8}{27}\right)^{2/3}\). Cube root first: \(\left(\frac{2}{3}\right)^2 = \frac{4}{9}\).

Full method — how and why
A fractional negative index of a fraction. Three separate instructions are packed into \(-\frac23\): invert, take the cube root, square.
  1. The minus sign inverts the fraction: \(\left(\frac{27}{8}\right)^{-2/3} = \left(\frac{8}{27}\right)^{2/3}\).
    \(a^{-n} = \frac{1}{a^n}\), and the reciprocal of a fraction is the fraction turned upside down. Doing this first keeps the numbers small.
  2. Take the cube root of top and bottom: \(\left(\frac{8}{27}\right)^{1/3} = \frac{2}{3}\).
    \(a^{m/n} = \left(\sqrt[n]{a}\right)^m\); rooting before powering keeps you working with \(2\) and \(3\) rather than \(64\) and \(729\).
  3. Square the result: \(\left(\frac23\right)^2 = \frac49\).
    The numerator \(2\) in the index is the power.
Where it goes wrong
  • Treating the minus sign as making the answer negative gives \(-\frac94\) or \(-\frac49\); a negative index never makes a positive number negative.
  • Forgetting to square gives \(\frac23\); forgetting to invert gives \(\frac94\). All three are offered as options.
Takeaway Read a fractional index as three instructions - sign (invert), denominator (root), numerator (power) - and do them in that order.
Answer C · Indices and surds (MM1.1) · question IND-02
2Statistics · spec M6.3 · Paper 1 style · ★☆☆

A cumulative frequency graph for \(80\) observations passes through \((20, 20)\), \((30, 40)\) and \((45, 60)\). Estimate the interquartile range.

  1. A\(40\)
  2. B\(10\)
  3. C\(20\)
  4. D\(25\)
  5. E\(15\)
  6. F\(30\)
Worked solution

Lower quartile at cumulative frequency \(20\): value \(20\). Upper quartile at \(60\): value \(45\). IQR \(= 25\).

Full method — how and why
On a cumulative frequency graph, quartiles are read by going in from the cumulative frequency axis at \(\frac n4\), \(\frac n2\), \(\frac{3n}{4}\).
  1. With \(n = 80\): the lower quartile is at cumulative frequency \(\frac{80}{4} = 20\), the upper at \(\frac{3 \times 80}{4} = 60\).
    Cumulative frequency graphs use \(\frac n4\) and \(\frac{3n}{4}\) directly (no \(+1\), unlike a raw list).
  2. The point \((20, 20)\) gives \(Q_1 = 20\).
    Read across at CF \(= 20\) and down to the value axis.
  3. The point \((45, 60)\) gives \(Q_3 = 45\).
    Read across at CF \(= 60\).
  4. IQR \(= Q_3 - Q_1 = 45 - 20 = 25\).
    The interquartile range is a difference of values, not of frequencies.
Where it goes wrong
  • Subtracting the cumulative frequencies, \(60 - 20 = 40\) - an offered option.
  • Using the median point \((30, 40)\) as a quartile.
Takeaway Go in from the frequency axis at \(\frac n4\) and \(\frac{3n}{4}\), read off values, then subtract the values.
Answer D · Statistics (M6.3) · question STA-08
3Functions and transformations · spec MM8.2 · Paper 1 style · ★☆☆

The graph of \(y = f(x)\) has a maximum point at \((3, 8)\). The graph of \(y = f(x - a) + b\) has its maximum at \((1, 2)\). Find \(a + b\).

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

The maximum moves from \((3, 8)\) to \((3 + a, 8 + b) = (1, 2)\), so \(a = -2\), \(b = -6\) and \(a + b = -8\).

Full method — how and why
Follow the turning point through the transformation and match coordinates - each coordinate gives one equation.
  1. \(y = f(x - a) + b\) translates the graph \(a\) right and \(b\) up, so the maximum moves from \((3, 8)\) to \((3 + a,\; 8 + b)\).
    A translation moves every point, including the turning point, by the same vector.
  2. Match the \(x\)-coordinates: \(3 + a = 1\), so \(a = -2\).
    A negative \(a\) simply means the shift is to the left.
  3. Match the \(y\)-coordinates: \(8 + b = 2\), so \(b = -6\).
    Same for the vertical shift.
  4. Add: \(a + b = -8\).
    The question asks for the sum, not for either value alone.
Where it goes wrong
  • Using \(3 - a = 1\) and getting \(a = 2\), which flips the sign of the answer.
  • Answering \(-2\) or \(-6\) (one of the two constants) instead of their sum.
Takeaway \(f(x - a) + b\) shifts by \((+a, +b)\); set up one equation per coordinate.
Answer D · Functions and transformations (MM8.2) · question FUN-14
4Quadratic functions · spec MM1.3 · Paper 1 style · ★☆☆

The graph of \(y = x^2 + bx + c\) crosses the \(x\)-axis at \((1, 0)\) and \((5, 0)\). What is the \(y\)-coordinate of the vertex?

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

\(y = (x-1)(x-5)\), vertex at \(x = 3\): \(y = (2)(-2) = -4\).

Full method — how and why
Given the two roots, use the factorised form and the symmetry of the parabola - no expansion needed.
  1. Write the factorised form: \(y = (x - 1)(x - 5)\).
    Roots at \(1\) and \(5\) with leading coefficient \(1\) (given, since the equation is \(x^2 + bx + c\)).
  2. The vertex lies midway between the roots: \(x = \frac{1 + 5}{2} = 3\).
    A parabola is symmetric, so its turning point sits on the axis of symmetry, exactly halfway between the roots.
  3. Evaluate there: \(y = (3-1)(3-5) = 2 \times (-2) = -4\).
    Substituting into the factorised form is far quicker than expanding to \(x^2 - 6x + 5\).
Where it goes wrong
  • Averaging the roots but then substituting into the wrong expression, e.g. getting \(-9\) from \((3)^2 - 6(3) + 5\) miscomputed.
  • Giving the \(x\)-coordinate \(3\) instead of the \(y\)-coordinate.
Takeaway Midpoint of the roots gives the axis of symmetry; substitute into the factorised form for the vertex height.
Answer C · Quadratic functions (MM1.3) · question QUA-15
5Probability · spec M7.6 · Paper 1 style · ★☆☆

Three fair coins are tossed. Find the probability of getting more heads than tails.

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

Need \(2\) or \(3\) heads: \(\frac{3 + 1}{8} = \frac12\) (by symmetry between heads and tails).

Full method — how and why
Either count the favourable outcomes out of \(8\), or exploit the symmetry between heads and tails.
  1. With three coins there are \(2^3 = 8\) equally likely outcomes.
    Each coin independently doubles the number of outcomes.
  2. 'More heads than tails' means \(2\) or \(3\) heads: \(\binom32 + \binom33 = 3 + 1 = 4\) outcomes.
    With an odd number of coins there is no tie, so every outcome has a strict majority.
  3. \(P = \dfrac48 = \dfrac12\).
    Or instantly: heads and tails are symmetric and a tie is impossible, so each must have probability \(\frac12\).
Where it goes wrong
  • Answering \(\frac38\) - the probability of exactly \(2\) heads, forgetting the all-heads case.
  • Trying the symmetry argument with an even number of coins, where ties break it.
Takeaway An odd number of fair coins: \(P(\text{more heads}) = \frac12\) by symmetry, since ties cannot occur.
Answer F · Probability (M7.6) · question PRO-07
6Trigonometric functions and equations · spec MM4.3 · Paper 1 style · ★☆☆

Find the exact value of \(\sin 60^\circ \cos 30^\circ + \cos 60^\circ \sin 30^\circ\).

  1. A\(\dfrac{\sqrt3}{2}\)
  2. B\(\dfrac{3}{4}\)
  3. C\(\dfrac{\sqrt3}{4}\)
  4. D\(1\)
  5. E\(0\)
  6. F\(\dfrac12\)
Worked solution

\(\frac{\sqrt3}{2}\cdot\frac{\sqrt3}{2} + \frac12\cdot\frac12 = \frac34 + \frac14 = 1\).

Full method — how and why
Pure recall of exact values. (It is also \(\sin(60^\circ + 30^\circ) = \sin 90^\circ\), but the addition formula is off-specification.)
  1. Write down the exact values: \(\sin 60^\circ = \frac{\sqrt3}{2}\), \(\cos 30^\circ = \frac{\sqrt3}{2}\), \(\cos 60^\circ = \frac12\), \(\sin 30^\circ = \frac12\).
    These come from the half-equilateral triangle with sides \(1, \sqrt3, 2\) - worth being able to redraw in seconds.
  2. Substitute: \(\dfrac{\sqrt3}{2}\cdot\dfrac{\sqrt3}{2} + \dfrac12\cdot\dfrac12 = \dfrac34 + \dfrac14\).
    \((\sqrt3)^2 = 3\), so the first product is \(\frac34\).
  3. Add: \(1\).
    A clean answer - the sign that the exact values were right.
Where it goes wrong
  • Using \(\sin 60^\circ = \frac12\) (that is \(\sin 30^\circ\)), giving \(\frac{\sqrt3}{2}\).
  • Writing \(\left(\frac{\sqrt3}{2}\right)^2 = \frac{\sqrt3}{4}\).
Takeaway \(\sin\) and \(\cos\) of \(30^\circ\) and \(60^\circ\) swap over; square a surd fraction by squaring top and bottom.
Answer D · Trigonometric functions and equations (MM4.3) · question TRG-03
7Indices and surds · spec MM1.1 · Paper 1 style · ★☆☆

Solve \(x^{\frac{3}{2}} = 8x^{-\frac{3}{2}}\) for \(x > 0\).

  1. A\(x = 2\sqrt{2}\)
  2. B\(x = 16\)
  3. C\(x = 8\)
  4. D\(x = \sqrt{2}\)
  5. E\(x = 2\)
  6. F\(x = 4\)
Worked solution

Multiply by \(x^{3/2}\): \(x^3 = 8\), so \(x = 2\).

Full method — how and why
An equation with the unknown in two fractional powers. Multiplying by the negative power clears it entirely.
  1. Multiply both sides by \(x^{3/2}\): \(x^{3/2} \cdot x^{3/2} = 8\), i.e. \(x^3 = 8\).
    \(x^{-3/2}\) becomes \(x^0 = 1\) on the right, and the left gains \(\frac32 + \frac32 = 3\). Multiplying by a positive quantity (we are told \(x > 0\)) cannot introduce false solutions.
  2. Take the cube root: \(x = 2\).
    \(x^3 = 8\) has one real root.
Where it goes wrong
  • Cancelling the powers to get \(x = 8\).
  • Dividing by \(x^{3/2}\) instead of multiplying, which leaves \(x^{3} = 8\) on the wrong side and invites sign errors in the index.
Takeaway To clear a negative fractional power, multiply through by that power - stated positivity guarantees the step is reversible.
Answer E · Indices and surds (MM1.1) · question IND-17
8Triangles, radians and sectors · spec MM4.1 · Paper 1 style · ★☆☆

A triangle has sides \(4\), \(5\) and \(6\). Find the cosine of its largest angle.

  1. A\(\dfrac{9}{16}\)
  2. B\(\dfrac{1}{8}\)
  3. C\(\dfrac{5}{8}\)
  4. D\(-\dfrac{1}{8}\)
  5. E\(\dfrac{1}{4}\)
  6. F\(\dfrac{3}{4}\)
Worked solution

The largest angle is opposite the side \(6\): \(\cos\theta = \dfrac{16 + 25 - 36}{2\cdot4\cdot5} = \dfrac{5}{40} = \dfrac18\).

Full method — how and why
Three sides given, an angle wanted: the rearranged cosine rule. The word 'largest' tells you which angle.
  1. Identify the largest angle: it is opposite the longest side, \(6\).
    In any triangle the ordering of sides matches the ordering of opposite angles.
  2. Use \(\cos\theta = \dfrac{a^2 + b^2 - c^2}{2ab}\) with \(a = 4\), \(b = 5\), \(c = 6\).
    The two sides adjacent to the angle go on the bottom; the opposite side is the one subtracted on top.
  3. Substitute: \(\dfrac{16 + 25 - 36}{2(4)(5)} = \dfrac{5}{40} = \dfrac18\).
    Positive, so the largest angle is acute - this triangle has no obtuse angle at all.
Where it goes wrong
  • Putting the longest side on the bottom, giving \(\frac{9}{16}\) or similar.
  • Assuming the largest angle must be obtuse and choosing \(-\frac18\).
Takeaway \(\cos\theta = \frac{a^2+b^2-c^2}{2ab}\) with \(c\) opposite \(\theta\); a positive result means an acute angle.
Answer B · Triangles, radians and sectors (MM4.1) · question TRI-02
9Probability · spec M7.7 · Paper 1 style · ★☆☆

In a class of \(30\), \(18\) study French, \(12\) study German and \(5\) study both. A student is chosen at random. Given that the student studies French, what is the probability that they also study German?

  1. A\(\dfrac{1}{6}\)
  2. B\(\dfrac{5}{25}\)
  3. C\(\dfrac{5}{18}\)
  4. D\(\dfrac{5}{30}\)
  5. E\(\dfrac{13}{18}\)
  6. F\(\dfrac{5}{12}\)
Worked solution

Conditional on French (\(18\) students), \(5\) also study German: \(\frac{5}{18}\).

Full method — how and why
The words 'given that' signal conditional probability: the sample space shrinks to the condition.
  1. Restrict attention to the \(18\) French students - they are now the whole universe.
    'Given that the student studies French' means we only consider those students.
  2. Of those \(18\), the number who also study German is \(5\).
    The overlap is stated directly in the question.
  3. \(P(\text{German} \mid \text{French}) = \dfrac{5}{18}\).
    Equivalently \(\dfrac{P(F \cap G)}{P(F)} = \dfrac{5/30}{18/30} = \dfrac{5}{18}\) - the \(30\)s cancel.
Where it goes wrong
  • Using \(\frac{5}{30}\) - the unconditional probability of studying both, ignoring the condition.
  • Using \(\frac{5}{12}\), which conditions on German instead - the wrong way round.
Takeaway Conditioning replaces the denominator: divide by the size of the group you are told you are in.
Answer C · Probability (M7.7) · question PRO-06
10Quadratic functions · spec MM1.3 · Paper 1 style · ★☆☆

The equation \(x^2 + (k-2)x + (k+1) = 0\) has two distinct real roots. Find the complete set of possible values of \(k\).

  1. A\(0 < k < 8\)
  2. B\(-2 < k < 4\)
  3. C\(k < -2\) or \(k > 4\)
  4. D\(k < 0\) or \(k > 4\)
  5. E\(k < 0\) or \(k > 8\)
  6. Fall real \(k\)
Worked solution

Discriminant \((k-2)^2 - 4(k+1) = k^2 - 8k = k(k-8) > 0\), so \(k < 0\) or \(k > 8\).

Full method — how and why
'Two distinct real roots' is always a discriminant condition. The unknown is in the coefficients, so the discriminant becomes a quadratic in \(k\) and you finish with an inequality.
  1. Identify \(a = 1\), \(b = k - 2\), \(c = k + 1\) and write the discriminant: \((k-2)^2 - 4(k+1)\).
    'Two distinct real roots' \(\Leftrightarrow\) \(b^2 - 4ac > 0\). Everything else in the question is decoration.
  2. Expand and simplify: \(k^2 - 4k + 4 - 4k - 4 = k^2 - 8k\).
    The constants cancel, which is the sign you have expanded correctly.
  3. Factorise and solve \(k(k - 8) > 0\).
    A quadratic inequality needs its roots and its shape: roots \(0\) and \(8\), opening upwards.
  4. The upward parabola is positive outside its roots: \(k < 0\) or \(k > 8\).
    Sketch it if in doubt - 'outside the roots' for \(> 0\), 'between the roots' for \(< 0\).
Where it goes wrong
  • Expanding \((k-2)^2\) as \(k^2 - 4\), which gives \(k^2 - 4k - 8\) and a different answer set.
  • Reading the sign chart backwards and choosing \(0 < k < 8\) - that is where the discriminant is negative and there are no real roots.
Takeaway Conditions on the number of roots are conditions on \(b^2 - 4ac\); if the coefficients contain a parameter, you end up solving a quadratic inequality in that parameter.
Answer E · Quadratic functions (MM1.3) · question QUA-01
11Functions and transformations · spec MM8.2 · Paper 1 style · ★☆☆

\(f(x) = x^2 - 4x\). The graph of \(y = f(x + 2)\) has its minimum point at

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

\(f(x) = (x-2)^2 - 4\) has minimum at \((2, -4)\). Replacing \(x\) by \(x + 2\) translates the graph \(2\) units to the left: \((0, -4)\).

Full method — how and why
Find the minimum of \(f\) first, then move it. Tracking the single turning point is quicker than re-completing the square.
  1. Complete the square: \(f(x) = x^2 - 4x = (x-2)^2 - 4\), so \(f\) has its minimum at \((2, -4)\).
    Half of \(-4\) is \(-2\); the \(-4\) outside is the minimum value.
  2. \(y = f(x + 2)\) translates the graph \(2\) units to the left.
    Replacing \(x\) by \(x + a\) shifts left by \(a\): the graph now reaches each value \(2\) earlier.
  3. Move the turning point: \((2, -4) \to (0, -4)\).
    Only the \(x\)-coordinate changes; a horizontal translation never alters \(y\).
Where it goes wrong
  • Shifting right and answering \((4, -4)\).
  • Changing the \(y\)-coordinate as well, giving \((0, -2)\) or similar.
Takeaway \(f(x + a)\) moves the graph \(a\) to the left; horizontal shifts leave \(y\)-values untouched.
Answer A · Functions and transformations (MM8.2) · question FUN-02
12Statistics · spec M6.3 · Paper 1 style · ★☆☆

For which of the following data sets is the mean greater than the median?

  1. A\(1, 2, 3, 4, 5\)
  2. B\(5, 5, 5, 5, 5\)
  3. C\(2, 4, 6, 8, 10\)
  4. D\(1, 2, 3, 4, 20\)
  5. E\(1, 10, 11, 12, 13\)
Worked solution

A large outlier pulls the mean above the median: mean \(6 >\) median \(3\). In B, D, E the mean equals the median; in C the mean \(9.4 <\) median \(11\).

Full method — how and why
The mean is pulled by extreme values; the median is not. Look for skew rather than computing everything.
  1. Check \(1, 2, 3, 4, 20\): median \(= 3\), mean \(= \frac{30}{5} = 6\). Mean \(>\) median. Yes.
    The single large outlier \(20\) drags the mean far above the middle.
  2. \(1,2,3,4,5\) and \(2,4,6,8,10\) and \(5,5,5,5,5\) are symmetric, so mean \(=\) median.
    For symmetric data the two coincide - no need to compute.
  3. \(1, 10, 11, 12, 13\): median \(= 11\), mean \(= \frac{47}{5} = 9.4\), so mean \(<\) median.
    Here the outlier is small, so it drags the mean down - the mirror image.
  4. So only \(1, 2, 3, 4, 20\) works.
    One high outlier gives positive skew.
Where it goes wrong
  • Assuming any outlier raises the mean - a low outlier lowers it, as in \(1, 10, 11, 12, 13\).
  • Computing all five means when symmetry settles three of them instantly.
Takeaway Mean \(>\) median signals a tail to the right (a high outlier); mean \(<\) median signals a tail to the left.
Answer D · Statistics (M6.3) · question STA-10
13Differentiation · spec MM6.2 · Paper 1 style · ★☆☆

Find \(\dfrac{d^2y}{dx^2}\) when \(y = \dfrac{1}{x^2} + 4\sqrt{x}\).

  1. A\(\dfrac{3}{x^4} - x^{-\frac32}\)
  2. B\(\dfrac{6}{x^4} + x^{-\frac32}\)
  3. C\(\dfrac{6}{x^4} - 2x^{-\frac32}\)
  4. D\(\dfrac{6}{x^4} - x^{-\frac32}\)
  5. E\(-\dfrac{6}{x^4} - x^{-\frac32}\)
  6. F\(\dfrac{2}{x^3} + 2x^{-\frac12}\)
Worked solution

\(y' = -2x^{-3} + 2x^{-1/2}\); \(y'' = 6x^{-4} - x^{-3/2}\).

Full method — how and why
Two differentiations of powers, with negative and fractional indices. Rewrite first.
  1. Rewrite: \(y = x^{-2} + 4x^{1/2}\).
    Roots and reciprocals must become powers before the rule applies.
  2. First derivative: \(y' = -2x^{-3} + 4 \times \frac12 x^{-1/2} = -2x^{-3} + 2x^{-1/2}\).
    \(-2 \times x^{-2-1}\) and \(\frac12 \times 4 = 2\).
  3. Second derivative: \(y'' = -2(-3)x^{-4} + 2\left(-\frac12\right)x^{-3/2} = 6x^{-4} - x^{-3/2}\).
    Differentiate again, watching the signs: \(-3\) times \(-2\) gives \(+6\).
  4. Write it as \(\dfrac{6}{x^4} - x^{-3/2}\).
    Matching the option format.
Where it goes wrong
  • Losing a sign and answering \(\frac{6}{x^4} + x^{-3/2}\).
  • Stopping after the first derivative.
Takeaway Differentiate twice, carefully: each negative index gains one more negative factor.
Answer D · Differentiation (MM6.2) · question DIF-14
14Sequences and series · spec MM2.2 · Paper 1 style · ★☆☆

The \(n\)th term of an arithmetic sequence is \(u_n = 7 - 4n\). Find the sum of the first \(30\) terms.

  1. A\(-1650\)
  2. B\(-1530\)
  3. C\(-1740\)
  4. D\(-1590\)
  5. E\(-1770\)
  6. F\(-1860\)
Worked solution

\(u_1 = 3\), \(u_{30} = 7 - 120 = -113\). \(S_{30} = \frac{30}{2}(3 - 113) = 15(-110) = -1650\).

Full method — how and why
You are given the \(n\)th term rather than \(a\) and \(d\). Generate the first and last terms and use the \(\frac n2(a + l)\) form.
  1. First term: \(u_1 = 7 - 4 = 3\). Common difference: the coefficient of \(n\) is \(-4\), so \(d = -4\).
    In \(u_n = An + B\) the gradient \(A\) is the common difference - the sequence is linear in \(n\).
  2. Last term: \(u_{30} = 7 - 120 = -113\).
    Substitute \(n = 30\) directly.
  3. Use \(S_n = \frac n2(a + l) = 15(3 + (-113)) = 15 \times (-110)\).
    This version is quickest when the first and last terms are easy to compute.
  4. Evaluate: \(-1650\).
    Negative, as expected - most terms are negative.
Where it goes wrong
  • Taking \(d = 7\) or \(a = 7\) from the formula rather than computing \(u_1\).
  • Sign slips when adding \(3 + (-113)\).
Takeaway For \(u_n = An + B\): \(d = A\) and \(u_1 = A + B\); then either sum formula works.
Answer A · Sequences and series (MM2.2) · question SEQ-04
15Graphs of functions · spec MM8.7 · Paper 1 style · ★★☆

How many real solutions does the equation \(x^3 = 3x + 1\) have?

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

Sketch \(y = x^3 - 3x - 1\): turning points at \(x = \pm1\) with \(y(-1) = 1 > 0\) and \(y(1) = -3 < 0\). A cubic whose local maximum is positive and local minimum is negative crosses the axis three times.

Full method — how and why
Counting roots of a cubic. Find the turning values and see whether they straddle the axis.
  1. Rearrange to \(f(x) = x^3 - 3x - 1 = 0\).
    Everything on one side, so the question becomes 'how many roots?'.
  2. Differentiate: \(f'(x) = 3x^2 - 3 = 3(x-1)(x+1)\), so the turning points are at \(x = \pm1\).
    A cubic with two turning points can have one, two or three roots.
  3. Evaluate: \(f(-1) = -1 + 3 - 1 = 1 > 0\) (local maximum) and \(f(1) = 1 - 3 - 1 = -3 < 0\) (local minimum).
    The values at the turning points decide the count.
  4. The local maximum is above the axis and the local minimum below, so the curve crosses three times.
    It comes up from \(-\infty\), crosses, comes down past the axis, crosses, then rises again.
Where it goes wrong
  • Trying to factorise - this cubic has no rational roots, so factorising wastes time.
  • Answering \(1\) because 'a cubic always has at least one root' without checking the turning values.
Takeaway Cubic root count: find the turning values; three roots iff the local max and min lie on opposite sides of the axis.
Answer D · Graphs of functions (MM8.7) · question GRA-08
16Simultaneous equations and inequalities · spec MM1.5 · Paper 1 style · ★★☆

How many integers \(x\) satisfy \(x^2 < 5x + 24\) and \(x^2 > 3x\)?

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

\(x^2 - 5x - 24 < 0 \Rightarrow (x-8)(x+3) < 0 \Rightarrow -3 < x < 8\). \(x(x-3) > 0 \Rightarrow x < 0\) or \(x > 3\). Integers: \(-2, -1\) and \(4, 5, 6, 7\): six.

Full method — how and why
Two quadratic inequalities, then an integer count. The second one factorises without a constant term, which changes its shape.
  1. First: \(x^2 - 5x - 24 < 0 \Rightarrow (x - 8)(x + 3) < 0 \Rightarrow -3 < x < 8\).
    Between the roots for \(< 0\).
  2. Second: \(x^2 - 3x > 0 \Rightarrow x(x - 3) > 0 \Rightarrow x < 0\) or \(x > 3\).
    Roots \(0\) and \(3\); outside the roots for \(> 0\). Do not divide by \(x\) - that would lose the \(x<0\) branch.
  3. Intersect: from \((-3, 8)\) keep \((-3, 0)\) and \((3, 8)\).
    Overlap both solution sets on one number line.
  4. Count integers: \(-2, -1\) from the first piece and \(4, 5, 6, 7\) from the second: six.
    Endpoints \(-3, 0, 3, 8\) are all excluded by strict inequalities.
Where it goes wrong
  • Dividing \(x^2 > 3x\) by \(x\) to get \(x > 3\), losing all the negative solutions.
  • Including \(0\) or \(3\) in the count.
Takeaway Never divide an inequality by the variable; factorise instead, then count integers explicitly.
Answer E · Simultaneous equations and inequalities (MM1.5) · question SIM-13
17Graphs of functions · spec MM8.7 · Paper 1 style · ★★☆

The line \(y = x + c\) and the curve \(y = \dfrac{1}{x}\) have no points of intersection. Find the complete set of values of \(c\).

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

\(x^2 + cx - 1 = 0\) always has discriminant \(c^2 + 4 > 0\): two intersections for every \(c\). So there are no values of \(c\) with no intersection.

Full method — how and why
A 'find the values' question whose honest answer is 'none'. Substituting shows the discriminant can never be negative.
  1. Set the expressions equal: \(x + c = \dfrac1x\).
    Intersections satisfy both equations.
  2. Multiply by \(x\) (noting \(x \neq 0\), since \(\frac1x\) is undefined there): \(x^2 + cx - 1 = 0\).
    Clearing the fraction gives a quadratic whose roots are the intersection \(x\)-values.
  3. Compute the discriminant: \(c^2 - 4(1)(-1) = c^2 + 4\).
    The \(-1\) constant is what makes the discriminant unavoidably positive.
  4. \(c^2 + 4 > 0\) for every real \(c\), so there are always two intersections: no value of \(c\) avoids them.
    Also check the roots are non-zero: their product is \(-1 \neq 0\), so neither is the excluded \(x = 0\).
Where it goes wrong
  • Assuming a line and a hyperbola must sometimes miss, and choosing an interval.
  • Sign error making the discriminant \(c^2 - 4\), which would give \(-2 < c < 2\) - an offered option.
Takeaway Do the algebra even when the answer 'feels' like an interval; \(b^2 - 4ac\) with \(ac < 0\) is always positive.
Answer F · Graphs of functions (MM8.7) · question GRA-04
18Sequences and series · spec MM2.2 · Paper 1 style · ★★☆

The integers from \(1\) to \(100\) inclusive are written down. What is the sum of those that are multiples of \(3\) or \(5\) (or both)?

  1. A\(2500\)
  2. B\(2350\)
  3. C\(2733\)
  4. D\(2103\)
  5. E\(2418\)
  6. F\(2318\)
Worked solution

Multiples of \(3\): \(3 + \ldots + 99 = 3(1 + \ldots + 33) = 3 \times 561 = 1683\). Multiples of \(5\): \(5(1 + \ldots + 20) = 1050\). Multiples of \(15\): \(15(1 + \ldots + 6) = 315\). Total \(1683 + 1050 - 315 = 2418\).

Full method — how and why
A counting-and-summing problem needing inclusion-exclusion: multiples of \(15\) belong to both lists and would otherwise be counted twice.
  1. Multiples of \(3\) up to \(100\): \(3 + 6 + \cdots + 99 = 3(1 + 2 + \cdots + 33) = 3 \times \frac{33 \times 34}{2} = 3 \times 561 = 1683\).
    Factor out the \(3\) so the bracket is a plain triangular number.
  2. Multiples of \(5\): \(5(1 + \cdots + 20) = 5 \times 210 = 1050\).
    \(\lfloor 100/5 \rfloor = 20\) terms.
  3. Multiples of both, i.e. of \(15\): \(15(1 + \cdots + 6) = 15 \times 21 = 315\).
    'Or' in mathematics includes 'both', and multiples of \(15\) have been counted in each of the first two sums.
  4. Inclusion-exclusion: \(1683 + 1050 - 315 = 2418\).
    Add the two sets, subtract the overlap once.
Where it goes wrong
  • Forgetting to subtract the multiples of \(15\), giving \(2733\) - which is offered.
  • Using \(\text{lcm}(3,5) = 8\) or \(30\) instead of \(15\).
Takeaway For 'multiples of \(a\) or \(b\)', subtract the multiples of \(\text{lcm}(a,b)\) once.
Answer E · Sequences and series (MM2.2) · question SEQ-20
19Simultaneous equations and inequalities · spec MM1.5 · Paper 1 style · ★★☆

Which of the following inequalities has exactly the same solution set as \(\dfrac{1}{x} > 2\)?

  1. A\(x > \dfrac{1}{2}\)
  2. B\(0 < x < \dfrac{1}{2}\)
  3. C\(x < \dfrac{1}{2}\)
  4. D\(x < 0\) or \(x > \dfrac12\)
  5. E\(x > 2\)
  6. F\(0 < x < 2\)
Worked solution

If \(x > 0\): \(1 > 2x\), so \(0 < x < \frac12\). If \(x < 0\): \(\frac1x\) is negative and cannot exceed \(2\). So \(0 < x < \frac12\). (Multiplying by \(x\) without considering its sign is the classic error.)

Full method — how and why
An inequality with the variable in the denominator. You must not multiply by \(x\) without knowing its sign - split into cases instead.
  1. Case \(x > 0\): multiplying by \(x\) keeps the direction, giving \(1 > 2x\), i.e. \(x < \frac12\). Combined with \(x > 0\): \(0 < x < \frac12\).
    Multiplying an inequality by a positive quantity is safe; the case assumption must be carried into the answer.
  2. Case \(x < 0\): then \(\frac1x < 0 < 2\), so the inequality can never hold.
    Rather than multiplying by a negative (and flipping), just observe the sign - a negative number is never greater than \(2\).
  3. Case \(x = 0\): \(\frac1x\) is undefined, so \(x = 0\) is excluded.
    Always check the value that breaks the expression.
  4. Combine: the solution set is exactly \(0 < x < \frac12\).
    Only the first case contributes.
Where it goes wrong
  • Multiplying through by \(x\) blindly to get \(x < \frac12\), which wrongly includes every negative number.
  • Answering \(x > 2\) by 'inverting' both sides without justification.
Takeaway Never multiply an inequality by an expression of unknown sign: split into cases, or bring everything to one side and analyse the sign of the fraction.
Answer B · Simultaneous equations and inequalities (MM1.5) · question SIM-08
20Integration · spec MM7.1 · Paper 1 style · ★★☆

Find the area of the finite region enclosed by the curve \(y = x^2\) and the line \(y = 2x + 3\).

  1. A\(\dfrac{16}{3}\)
  2. B\(\dfrac{28}{3}\)
  3. C\(12\)
  4. D\(\dfrac{20}{3}\)
  5. E\(\dfrac{32}{3}\)
  6. F\(\dfrac{64}{3}\)
Worked solution

Intersections: \(x^2 - 2x - 3 = (x - 3)(x + 1) = 0\). Area \(= \int_{-1}^{3} (2x + 3 - x^2)\,dx = \left[x^2 + 3x - \frac{x^3}{3}\right]_{-1}^{3} = (9 + 9 - 9) - \left(1 - 3 + \frac13\right) = 9 + \frac53 = \frac{32}{3}\).

Full method — how and why
Area between a line and a curve: integrate (upper \(-\) lower) between the intersection points.
  1. Find the intersections: \(x^2 = 2x + 3 \Rightarrow x^2 - 2x - 3 = (x-3)(x+1) = 0\), so \(x = -1\) and \(x = 3\).
    These are the limits of the enclosed region.
  2. Decide which is on top: at \(x = 0\) the line gives \(3\) and the curve gives \(0\), so the line is above.
    Testing one interior point settles the order - guessing wrong gives the negative of the answer.
  3. Integrate the difference: \(\displaystyle\int_{-1}^{3}\left(2x + 3 - x^2\right)dx = \left[x^2 + 3x - \frac{x^3}{3}\right]_{-1}^{3}\).
    Subtracting first, then integrating once, is quicker and less error-prone than two separate integrals.
  4. Evaluate: at \(3\), \(9 + 9 - 9 = 9\); at \(-1\), \(1 - 3 + \frac13 = -\frac53\). Difference: \(9 + \frac53 = \frac{32}{3}\).
    \(-\frac{(-1)^3}{3} = +\frac13\) - the double negative is the classic slip.
Where it goes wrong
  • Integrating (curve \(-\) line) and giving \(-\frac{32}{3}\), or taking the magnitude of the wrong piece.
  • Sign error at the lower limit, giving \(\frac{16}{3}\) or \(\frac{28}{3}\).
Takeaway Area between curves \(= \int(\text{upper} - \text{lower})\) between intersections; check which is upper with one test value.
Answer E · Integration (MM7.1) · question INT-05
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