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Paper 1 · Set 02

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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1Simultaneous equations and inequalities · spec MM1.4 · Paper 1 style · ★☆☆

The simultaneous equations \(y = x^2 + 2x + c\) and \(y = 4x + 1\) have exactly one solution. Find the value of \(c\).

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

\(x^2 - 2x + (c - 1) = 0\) has one solution when \(4 - 4(c-1) = 0\), so \(c = 2\).

Full method — how and why
'Exactly one solution' for a line and a parabola is again the discriminant-zero (tangency) condition.
  1. Set the expressions equal: \(x^2 + 2x + c = 4x + 1\).
    A simultaneous solution is a point where both \(y\)-values agree.
  2. Collect: \(x^2 - 2x + (c - 1) = 0\).
    Move everything to one side; the \(2x - 4x\) gives \(-2x\).
  3. Discriminant zero: \((-2)^2 - 4(1)(c - 1) = 0\).
    One solution means a repeated root.
  4. Solve: \(4 - 4c + 4 = 0 \Rightarrow 4c = 8 \Rightarrow c = 2\).
    Careful with \(-4(c-1) = -4c + 4\).
Where it goes wrong
  • Mishandling \(-4(c-1)\) and getting \(c = 0\).
  • Setting the discriminant \(> 0\) ('has a solution') rather than \(= 0\) ('exactly one').
Takeaway Exactly one intersection between a line and a parabola always means discriminant \(= 0\).
Answer C · Simultaneous equations and inequalities (MM1.4) · question SIM-09
2Angles, polygons, congruence and vectors · spec M5.4 · Paper 1 style · ★☆☆

Which of the following sets of information is NOT sufficient to prove that two triangles are congruent?

  1. Atwo sides and a non-included angle equal (SSA)
  2. Btwo sides and the included angle equal (SAS)
  3. Ctwo angles and a corresponding side equal (ASA)
  4. Dright angle, hypotenuse and one other side equal (RHS)
  5. Ethree sides equal (SSS)
Worked solution

SSA is not a congruence condition (it is the ambiguous case of the sine rule); the other four are the standard criteria.

Full method — how and why
The four congruence criteria are SSS, SAS, ASA and RHS. Anything else - notably SSA - is not sufficient.
  1. Recall the valid criteria: SSS, SAS, ASA (or AAS), RHS.
    These four are the ones that fix a triangle completely.
  2. Consider SSA: two sides and a non-included angle. This is the ambiguous case of the sine rule, where two different triangles can share the same data.
    The side opposite the known angle can 'swing' to two positions - exactly the situation in TRI-07.
  3. So SSA is the odd one out.
    RHS is really a special case of SSA that works, because the right angle removes the ambiguity.
Where it goes wrong
  • Thinking RHS is invalid because it looks like SSA - the right angle makes it unambiguous.
  • Confusing AAA (similar but not congruent) with ASA.
Takeaway SSS, SAS, ASA, RHS prove congruence; SSA does not, and AAA gives only similarity.
Answer A · Angles, polygons, congruence and vectors (M5.4) · question GEO-08
3GCSE algebra · spec M4.19 · Paper 1 style · ★☆☆

A sequence has first term \(2\) and each subsequent term is obtained by squaring the previous term and subtracting \(2\). What is the \(50\)th term?

  1. A\(2^{50}\)
  2. B\(-2\)
  3. C\(14\)
  4. D\(4\)
  5. E\(0\)
  6. F\(2\)
Worked solution

\(2^2 - 2 = 2\): every term is \(2\).

Full method — how and why
A recurrence with a fixed point. Computing one term reveals the sequence never changes.
  1. Apply the rule to the first term: \(u_2 = 2^2 - 2 = 2\).
    'Square then subtract 2' applied to \(2\) returns \(2\).
  2. Since \(u_2 = u_1\), the same calculation repeats forever: every term is \(2\).
    \(2\) is a fixed point of the map \(x \mapsto x^2 - 2\).
  3. So \(u_{50} = 2\).
    No need to iterate fifty times.
Where it goes wrong
  • Assuming the terms must grow because of the squaring, and choosing \(2^{50}\).
  • Computing \(2^2 - 2\) as \(0\) or \(-2\).
Takeaway Always compute the second term: if it repeats the first, the sequence is constant.
Answer F · GCSE algebra (M4.19) · question ALG-12
4Exponentials and logarithms · spec MM5.2 · Paper 1 style · ★☆☆

Solve \(\log_2 x + \log_4 x = 6\). (You may use \(\log_4 x = \dfrac{1}{2}\log_2 x\).)

  1. A\(x = 8\)
  2. B\(x = 2^{6}\)
  3. C\(x = 16\)
  4. D\(x = 64\)
  5. E\(x = 4\)
  6. F\(x = 32\)
Worked solution

\(\frac32\log_2 x = 6 \Rightarrow \log_2 x = 4 \Rightarrow x = 16\).

Full method — how and why
Two different bases. The hint converts one to the other, after which it is a linear equation in \(\log_2 x\).
  1. Use the given relation \(\log_4 x = \frac12\log_2 x\).
    It holds because \(4 = 2^2\): it takes twice as many factors of \(2\) as of \(4\) to build \(x\).
  2. Substitute: \(\log_2 x + \frac12\log_2 x = 6\), i.e. \(\frac32\log_2 x = 6\).
    Both terms now involve the same quantity, so treat \(\log_2 x\) as a single unknown.
  3. Solve for the log: \(\log_2 x = 4\).
    Multiply by \(\frac23\).
  4. Convert: \(x = 2^4 = 16\).
    \(\log_2 x = 4\) means \(x = 2^4\).
Where it goes wrong
  • Solving \(\log_2 x = 6\) and answering \(64\) - which is offered (twice, as \(64\) and \(2^6\)).
  • Adding the bases or the arguments instead of using the relation.
Takeaway Get every logarithm to the same base first; then it is ordinary algebra in that one quantity.
Answer C · Exponentials and logarithms (MM5.2) · question LOG-11
5Functions and transformations · spec MM8.4 · Paper 1 style · ★☆☆

The graph of \(y = a(x + b)^2 + c\) has vertex \((-2, 5)\) and passes through \((0, -3)\). Find \(a\).

  1. A\(-2\)
  2. B\(2\)
  3. C\(-4\)
  4. D\(-1\)
  5. E\(1\)
  6. F\(-\dfrac12\)
Worked solution

\(b = 2\), \(c = 5\). At \(x = 0\): \(4a + 5 = -3\), so \(a = -2\).

Full method — how and why
The completed-square form is handed to you, so the vertex gives two constants and the extra point gives the third.
  1. Match the vertex: \(y = a(x + b)^2 + c\) has vertex \((-b, c)\), and we are told it is \((-2, 5)\), so \(b = 2\) and \(c = 5\).
    Note the sign: the form uses \(x + b\), so the vertex \(x\)-coordinate is \(-b\), not \(b\).
  2. Substitute the point \((0, -3)\): \(a(0 + 2)^2 + 5 = -3\).
    One point is enough to find the remaining unknown.
  3. Solve: \(4a = -8\), so \(a = -2\).
    Negative, which makes sense: the curve passes below its vertex, so it opens downwards.
Where it goes wrong
  • Taking \(b = -2\) and getting \(a = -\frac12\) from \(a(0-2)^2\)... (the same here, but the sign confusion bites when the point is not symmetric).
  • Answering \(+2\) and ignoring that the parabola must open downwards to reach \(y = -3\).
Takeaway Sanity-check the sign of \(a\) against the picture: if the curve drops below the vertex, \(a < 0\).
Answer A · Functions and transformations (MM8.4) · question FUN-13
6Trigonometric functions and equations · spec MM4.6 · Paper 1 style · ★☆☆

Find the number of solutions of \(\tan^2 x = 3\) for \(0 \leq x < 2\pi\).

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

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

Full method — how and why
A squared tangent equation. Two values of \(\tan x\), each giving solutions once per \(\pi\).
  1. Take square roots: \(\tan x = \pm\sqrt3\).
    Both signs are needed.
  2. \(\tan x = \sqrt3\) gives \(x = \frac{\pi}{3}\) and \(\frac{\pi}{3} + \pi = \frac{4\pi}{3}\).
    Period \(\pi\) for \(\tan\).
  3. \(\tan x = -\sqrt3\) gives \(x = \frac{2\pi}{3}\) and \(\frac{2\pi}{3} + \pi = \frac{5\pi}{3}\).
    \(\tan\) is negative in quadrants 2 and 4; the reference angle is still \(\frac\pi3\).
  4. Four solutions in \([0, 2\pi)\).
    Two values \(\times\) two periods.
Where it goes wrong
  • Taking only the positive root and answering \(2\).
  • Using period \(2\pi\) for \(\tan\).
Takeaway \(\tan^2 x = k\) gives \(\tan x = \pm\sqrt k\), and each has one solution per interval of length \(\pi\).
Answer B · Trigonometric functions and equations (MM4.6) · question TRG-12
7Differentiation · spec MM6.3 · Paper 1 style · ★☆☆

The tangent to \(y = x^2 - 4x + 1\) at the point where \(x = 3\) meets the \(y\)-axis at

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

\(y' = 2x - 4 = 2\) at \(x = 3\); the point is \((3, -2)\). Tangent: \(y + 2 = 2(x - 3)\), so at \(x = 0\), \(y = -8\).

Full method — how and why
Tangent line: gradient from the derivative, point from the curve, then find where it meets the axis.
  1. Differentiate: \(y' = 2x - 4\); at \(x = 3\) the gradient is \(2\).
    The derivative evaluated at a point gives the tangent's gradient there.
  2. Find the point: \(y(3) = 9 - 12 + 1 = -2\), so the tangent touches at \((3, -2)\).
    The tangent passes through the curve at that \(x\) - use the original function.
  3. Write the tangent: \(y + 2 = 2(x - 3)\).
    Point-gradient form with the point and gradient just found.
  4. Set \(x = 0\): \(y + 2 = -6\), so \(y = -8\). The tangent meets the \(y\)-axis at \((0, -8)\).
    The \(y\)-axis is \(x = 0\).
Where it goes wrong
  • Using \(y(3) = -2\) as the intercept and answering \((0,-2)\).
  • Sign slip: \(y = 2x - 8\), not \(2x + 8\).
Takeaway Tangent \(=\) gradient from \(f'\) \(+\) point from \(f\); then substitute \(x = 0\) for the intercept.
Answer F · Differentiation (MM6.3) · question DIF-03
8Quadratic functions · spec MM1.3 · Paper 1 style · ★☆☆

The parabola \(y = x^2 - 4x + 3\) is reflected in the \(x\)-axis and then translated \(2\) units in the positive \(y\)-direction. What is the maximum value of the resulting function?

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

\(x^2 - 4x + 3 = (x-2)^2 - 1\), minimum \(-1\). Reflecting gives maximum \(1\); translating up by \(2\) gives maximum \(3\).

Full method — how and why
Transformations applied to a quadratic. Track what happens to the vertex rather than to the whole equation.
  1. Complete the square: \(x^2 - 4x + 3 = (x - 2)^2 - 1\), so the minimum is \(-1\) at \(x = 2\).
    The turning point is the only feature the transformations act on in a way that matters.
  2. Reflect in the \(x\)-axis: \(y \to -y\), so the minimum \(-1\) becomes a maximum of \(1\).
    Reflection in the \(x\)-axis flips the sign of every \(y\)-value, turning a minimum into a maximum.
  3. Translate \(2\) up: the maximum becomes \(1 + 2 = 3\).
    A vertical translation adds to every \(y\)-value, including the extreme one.
Where it goes wrong
  • Forgetting that the reflection converts the minimum into a maximum and reporting \(-1 + 2 = 1\).
  • Reflecting the \(x\)-coordinate as well - the turning point stays at \(x = 2\) throughout.
Takeaway For transformation questions, follow one key point (usually the vertex) through each step in order.
Answer A · Quadratic functions (MM1.3) · question QUA-21
9Triangles, radians and sectors · spec MM4.2 · Paper 1 style · ★☆☆

The minute hand of a clock is \(12\) cm long. How far does its tip travel in \(25\) minutes?

  1. A\(25\pi\) cm
  2. B\(5\pi\) cm
  3. C\(10\pi\) cm
  4. D\(12\pi\) cm
  5. E\(\dfrac{25\pi}{6}\) cm
  6. F\(20\pi\) cm
Worked solution

\(25\) minutes is \(\frac{25}{60} = \frac{5}{12}\) of a revolution: \(\frac{5}{12} \times 2\pi \times 12 = 10\pi\) cm.

Full method — how and why
Distance travelled by a rotating tip is arc length. Find the fraction of a full turn, then multiply by the circumference.
  1. Find the fraction of a revolution: \(\dfrac{25}{60} = \dfrac{5}{12}\).
    The minute hand takes \(60\) minutes for one full turn.
  2. Compute the full circumference: \(2\pi r = 2\pi(12) = 24\pi\) cm.
    The tip traces a circle of radius equal to the hand's length.
  3. Multiply: \(\dfrac{5}{12} \times 24\pi = 10\pi\) cm.
    Equivalently, \(\theta = \frac{5}{12}\times 2\pi = \frac{5\pi}{6}\) and \(s = r\theta = 12 \times \frac{5\pi}{6} = 10\pi\).
Where it goes wrong
  • Using the radius as the distance, or the area formula instead of arc length.
  • Using \(\frac{25}{12}\) or \(\frac{25}{60}\) of the radius rather than the circumference.
Takeaway Arc length \(=\) (fraction of a turn) \(\times 2\pi r\); a clock hand makes one turn per hour.
Answer C · Triangles, radians and sectors (MM4.2) · question TRI-13
10Probability · spec M7.6 · Paper 1 style · ★☆☆

Two cards are drawn at random from a standard pack of \(52\) without replacement. Which of the following expressions is the probability that both are aces?

  1. A\(\dfrac{4}{52} + \dfrac{3}{51}\)
  2. B\(\dfrac{2}{52}\)
  3. C\(\dfrac{4}{52} \times \dfrac{3}{52}\)
  4. D\(\dfrac{1}{13} \times \dfrac{1}{13}\)
  5. E\(\dfrac{4}{52} \times \dfrac{3}{51}\)
  6. F\(\dfrac{4}{52} \times \dfrac{4}{52}\)
Worked solution

Without replacement the second draw has \(3\) aces among \(51\) cards.

Full method — how and why
The question asks for the expression, so the reasoning is entirely about how the second draw changes.
  1. First card an ace: \(\dfrac{4}{52}\) - four aces in the pack.
    All \(52\) cards are equally likely.
  2. Given that, one ace and one card overall have gone: \(3\) aces remain among \(51\) cards.
    Without replacement, both numbers drop by one.
  3. Multiply: \(\dfrac{4}{52} \times \dfrac{3}{51}\).
    This is \(P(A) \times P(B \mid A)\), the multiplication rule for dependent events.
Where it goes wrong
  • \(\frac{4}{52} \times \frac{4}{52}\) - that is with replacement.
  • \(\frac{4}{52} \times \frac{3}{52}\) - reducing the aces but forgetting the total also falls to \(51\).
  • Adding rather than multiplying: 'and' means multiply.
Takeaway Without replacement, reduce numerator and denominator; 'and' \(\Rightarrow\) multiply.
Answer E · Probability (M7.6) · question PRO-11
11Number, units and estimation · spec M2.5 · Paper 1 style · ★☆☆

How many three-digit positive integers have digits that are all different and all odd?

  1. A\(120\)
  2. B\(24\)
  3. C\(125\)
  4. D\(100\)
  5. E\(30\)
  6. F\(60\)
Worked solution

Odd digits \(1, 3, 5, 7, 9\): \(5 \times 4 \times 3 = 60\).

Full method — how and why
Counting with two restrictions: digits from a fixed set, and all different.
  1. List the allowed digits: \(1, 3, 5, 7, 9\) - five odd digits.
    'All odd' restricts the pool; note \(0\) is even so no leading-zero problem arises.
  2. First digit: \(5\) choices. Second: \(4\) (one used). Third: \(3\).
    'All different' removes one option at each stage.
  3. Multiply: \(5 \times 4 \times 3 = 60\).
    The product rule.
Where it goes wrong
  • Using \(5^3 = 125\) by ignoring 'all different' - an offered option.
  • Worrying about leading zeros, which cannot occur here.
Takeaway Count position by position, reducing the pool by one each time when repetition is banned.
Answer F · Number, units and estimation (M2.5) · question NUM-15
12Functions and transformations · spec MM8.2 · Paper 1 style · ★☆☆

The graph of \(y = f(x)\) is transformed by the sequence: reflect in the \(x\)-axis, then translate \(2\) units in the positive \(y\)-direction, then reflect in the \(y\)-axis. The resulting graph has equation

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

Reflect in \(x\)-axis: \(y = -f(x)\). Translate up \(2\): \(y = -f(x) + 2\). Reflect in \(y\)-axis (replace \(x\) by \(-x\)): \(y = 2 - f(-x)\).

Full method — how and why
Three transformations in a stated order. Apply them to the equation one at a time, and be careful that the last one acts on everything built so far.
  1. Reflect in the \(x\)-axis: \(y = -f(x)\).
    Negating the output flips the graph vertically.
  2. Translate \(2\) up: \(y = -f(x) + 2\).
    Add to the whole expression, not just to \(f\).
  3. Reflect in the \(y\)-axis: replace every \(x\) by \(-x\), giving \(y = -f(-x) + 2 = 2 - f(-x)\).
    Only \(x\) is replaced; the constant \(2\) is unaffected because it does not involve \(x\).
Where it goes wrong
  • Writing \(-f(-x) - 2\) by also negating the constant.
  • Applying the transformations in the wrong order - with mixed inside/outside operations, order matters.
Takeaway Rewrite the equation after each step; a \(y\)-axis reflection substitutes \(-x\) for \(x\) everywhere it appears.
Answer D · Functions and transformations (MM8.2) · question FUN-11
13Trigonometric functions and equations · spec MM4.5 · Paper 1 style · ★☆☆

For all \(\theta\) where the expressions are defined, \((\sin\theta + \cos\theta)^2 + (\sin\theta - \cos\theta)^2\) equals

  1. A\(2\sin^2\theta\)
  2. B\(2 + 2\sin\theta\cos\theta\)
  3. C\(0\)
  4. D\(2\)
  5. E\(2\cos^2\theta\)
  6. F\(1\)
Worked solution

Expanding: \((1 + 2\sin\theta\cos\theta) + (1 - 2\sin\theta\cos\theta) = 2\).

Full method — how and why
Expanding two brackets whose cross terms cancel. The Pythagorean identity finishes it.
  1. Expand the first: \((\sin\theta + \cos\theta)^2 = \sin^2\theta + 2\sin\theta\cos\theta + \cos^2\theta\).
    Standard \((a+b)^2\) expansion.
  2. Expand the second: \((\sin\theta - \cos\theta)^2 = \sin^2\theta - 2\sin\theta\cos\theta + \cos^2\theta\).
    Only the middle sign changes.
  3. Add: the \(\pm2\sin\theta\cos\theta\) terms cancel, leaving \(2(\sin^2\theta + \cos^2\theta)\).
    This cancellation is the point of the question.
  4. Apply the identity: \(2 \times 1 = 2\).
    \(\sin^2 + \cos^2 = 1\) for every \(\theta\), so the answer is a constant.
Where it goes wrong
  • Keeping a cross term and choosing \(2 + 2\sin\theta\cos\theta\) - an offered option.
  • Answering \(1\) by forgetting the factor \(2\).
Takeaway \((a+b)^2 + (a-b)^2 = 2(a^2+b^2)\); with \(\sin\) and \(\cos\) that is always \(2\).
Answer D · Trigonometric functions and equations (MM4.5) · question TRG-13
14Simultaneous equations and inequalities · spec MM1.4 · Paper 1 style · ★☆☆

The curves \(y = x^2 - 4\) and \(y = -x^2 + 2x + k\) intersect at exactly two points. Find the complete set of values of \(k\).

  1. A\(k > \dfrac{9}{2}\)
  2. B\(k < -\dfrac{9}{2}\)
  3. C\(k > -\dfrac{9}{2}\)
  4. D\(k \neq -\dfrac92\)
  5. E\(k > -4\)
  6. Fall real \(k\)
Worked solution

\(2x^2 - 2x - 4 - k = 0\): discriminant \(4 + 8(4 + k) > 0 \Rightarrow 36 + 8k > 0 \Rightarrow k > -\frac92\).

Full method — how and why
Two parabolas meeting twice. Subtracting gives one quadratic, and 'two distinct points' is a positive discriminant.
  1. Set them equal: \(x^2 - 4 = -x^2 + 2x + k\).
    Intersections satisfy both equations, so their right-hand sides agree.
  2. Collect everything on one side: \(2x^2 - 2x - 4 - k = 0\).
    Note the \(-x^2\) becomes \(+x^2\) when moved across, giving \(2x^2\).
  3. Discriminant \(> 0\): \((-2)^2 - 4(2)(-4 - k) > 0\).
    Here \(c = -4 - k\); keeping the whole thing in brackets avoids sign errors.
  4. Simplify: \(4 + 32 + 8k > 0 \Rightarrow 8k > -36 \Rightarrow k > -\frac92\).
    \(-4 \times 2 \times (-4-k) = 32 + 8k\) - two negatives make a positive.
Where it goes wrong
  • Losing a sign in \(c = -4 - k\) and getting \(k > \frac92\) or \(k > -4\), both of which are offered.
  • Forgetting to combine \(x^2\) with \(-(-x^2)\) and working with \(x^2\) instead of \(2x^2\).
Takeaway Bring both curves to one side before using the discriminant, and bracket the constant term to keep signs straight.
Answer C · Simultaneous equations and inequalities (MM1.4) · question SIM-12
15Sequences and series · spec MM2.3 · Paper 1 style · ★☆☆

The second term of a geometric sequence is \(6\) and the fifth term is \(48\). Find the sum of the first \(6\) terms.

  1. A\(93\)
  2. B\(186\)
  3. C\(192\)
  4. D\(381\)
  5. E\(126\)
  6. F\(189\)
Worked solution

\(ar = 6\), \(ar^4 = 48\), so \(r^3 = 8\), \(r = 2\), \(a = 3\). \(S_6 = 3\cdot\dfrac{2^6 - 1}{2 - 1} = 3 \times 63 = 189\).

Full method — how and why
Two terms of a geometric sequence determine \(a\) and \(r\). Dividing the two equations isolates \(r\).
  1. Write the facts: \(ar = 6\) and \(ar^4 = 48\).
    The \(k\)th term is \(ar^{k-1}\): the second term has \(r^1\), the fifth has \(r^4\).
  2. Divide: \(\dfrac{ar^4}{ar} = r^3 = \dfrac{48}{6} = 8\), so \(r = 2\).
    Division cancels \(a\) - always the first move with two geometric terms.
  3. Back-substitute: \(a(2) = 6\), so \(a = 3\).
    Use the simpler of the two equations.
  4. Sum: \(S_6 = a\dfrac{r^6 - 1}{r - 1} = 3 \times \dfrac{64 - 1}{1} = 189\).
    With \(r = 2\) the denominator is \(1\), making the arithmetic easy.
Where it goes wrong
  • Using \(r^4/r^1 = r^4\) instead of \(r^3\) and getting a non-integer ratio.
  • Computing \(S_6\) with \(2^6 = 32\) instead of \(64\).
Takeaway Divide two geometric terms to find \(r\); the index difference tells you which power of \(r\) you get.
Answer F · Sequences and series (MM2.3) · question SEQ-08
16Polynomials, factor and remainder theorems · spec MM1.6 · Paper 1 style · ★☆☆

Which of the following is a factor of \(x^4 - 5x^2 + 4\)?

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

\(x^4 - 5x^2 + 4 = (x^2 - 1)(x^2 - 4) = (x-1)(x+1)(x-2)(x+2)\).

Full method — how and why
A quartic that is quadratic in \(x^2\). Factorise fully and read off which of the offered factors appears.
  1. Substitute \(u = x^2\): \(u^2 - 5u + 4 = (u - 1)(u - 4)\).
    Only even powers appear, so the substitution works.
  2. Undo it: \(x^4 - 5x^2 + 4 = (x^2 - 1)(x^2 - 4)\).
    Replace \(u\) by \(x^2\) in each bracket.
  3. Factorise each as a difference of two squares: \((x-1)(x+1)(x-2)(x+2)\).
    \(x^2 - a^2 = (x-a)(x+a)\) - do not stop at the quadratics.
  4. Compare with the options: \((x + 2)\) appears.
    Equivalently, a factor \((x - a)\) exists iff \(f(a) = 0\): \(f(-2) = 16 - 20 + 4 = 0\) ✓.
Where it goes wrong
  • Choosing \(x^2 + 1\) or \(x^2 + 4\) - these are sums, not differences, and are not factors.
  • Testing only positive values when checking \(f(a) = 0\).
Takeaway Factorise as far as possible; a quick check of \(f(a) = 0\) confirms any candidate factor \((x-a)\).
Answer E · Polynomials, factor and remainder theorems (MM1.6) · question POL-13
17Integration · spec MM7.1 · Paper 1 style · ★★☆

The area between the curve \(y = x^2\) and the line \(y = k\) (where \(k > 0\)) is \(\dfrac{32}{3}\). Find \(k\).

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

Area \(= \int_{-\sqrt k}^{\sqrt k}(k - x^2)\,dx = 2\left(k^{3/2} - \frac{k^{3/2}}{3}\right) = \frac43 k^{3/2} = \frac{32}{3} \Rightarrow k^{3/2} = 8 \Rightarrow k = 4\).

Full method — how and why
Work backwards from a given area to the parameter. Symmetry halves the work.
  1. Find the intersections: \(x^2 = k\) at \(x = \pm\sqrt k\).
    The horizontal line cuts the parabola at two symmetric points.
  2. Set up the area: \(\displaystyle\int_{-\sqrt k}^{\sqrt k}\left(k - x^2\right)dx\), with the line above the curve between the roots.
    Upper minus lower.
  3. Use symmetry: \(= 2\displaystyle\int_0^{\sqrt k}(k - x^2)dx = 2\left[kx - \frac{x^3}{3}\right]_0^{\sqrt k} = 2\left(k^{3/2} - \frac{k^{3/2}}{3}\right) = \frac43 k^{3/2}\).
    The integrand is even, so integrate over half the range and double. Note \(k\sqrt k = k^{3/2}\).
  4. Solve \(\frac43 k^{3/2} = \frac{32}{3}\): \(k^{3/2} = 8\), so \(k = 8^{2/3} = 4\).
    Undo the \(\frac32\) power by raising to the \(\frac23\): \(8^{2/3} = (\sqrt[3]8)^2 = 4\).
Where it goes wrong
  • Solving \(k^{3/2} = 8\) as \(k = 8^{3/2} = 22.6\) or \(k = \sqrt8\) - both offered in spirit.
  • Forgetting to double after using symmetry.
Takeaway Use even symmetry to halve the integration; undo a \(\frac32\) power by raising to \(\frac23\).
Answer D · Integration (MM7.1) · question INT-15
18Exponentials and logarithms · spec MM5.3 · Paper 1 style · ★★☆

The number of integers \(n\) satisfying \(2^{n} < 1000 < 2^{n+3}\) is

  1. A\(3\)
  2. B\(4\)
  3. C\(0\)
  4. D\(1\)
  5. E\(10\)
  6. F\(2\)
Worked solution

\(2^9 = 512 < 1000 < 1024 = 2^{10}\). Need \(n \leq 9\) and \(n + 3 \geq 10\), so \(n = 7, 8, 9\): three integers.

Full method — how and why
A double inequality in powers of \(2\). Locate \(1000\) between consecutive powers, then translate into conditions on \(n\).
  1. Find where \(1000\) sits: \(2^9 = 512\) and \(2^{10} = 1024\), so \(2^9 < 1000 < 2^{10}\).
    Powers of \(2\) up to \(1024\) are worth knowing by heart.
  2. The left condition \(2^n < 1000\) requires \(n \leq 9\).
    \(2^{10} = 1024 > 1000\), so \(n = 10\) fails.
  3. The right condition \(1000 < 2^{n+3}\) requires \(n + 3 \geq 10\), i.e. \(n \geq 7\).
    \(2^{9} = 512 < 1000\), so \(n + 3 = 9\) (i.e. \(n = 6\)) fails.
  4. Combine: \(7 \leq n \leq 9\), giving \(n = 7, 8, 9\) - three integers.
    Count the integers in the range, inclusive.
Where it goes wrong
  • Counting \(9 - 7 = 2\) instead of \(3\).
  • Using strict inequalities on \(n\) where the powers allow equality.
Takeaway Convert conditions on \(2^n\) into conditions on \(n\) by comparing with the nearest powers, then count inclusively.
Answer A · Exponentials and logarithms (MM5.3) · question LOG-14
19Quadratic functions · spec MM1.3 · Paper 1 style · ★★☆

Find the sum of all real values of \(x\) satisfying \((x^2 - 3x)^2 - 2(x^2 - 3x) - 8 = 0\).

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

Let \(u = x^2 - 3x\): \((u-4)(u+2) = 0\). \(x^2 - 3x - 4 = 0\) gives \(x = 4, -1\) (sum 3). \(x^2 - 3x + 2 = 0\) gives \(x = 1, 2\) (sum 3). Total \(6\).

Full method — how and why
A quadratic in the expression \(x^2 - 3x\). After substituting back you get two quadratics, and the question only wants the total of all four roots.
  1. Let \(u = x^2 - 3x\): the equation becomes \(u^2 - 2u - 8 = 0\).
    The same expression appears squared and linearly, which is the signal for a substitution.
  2. Factorise: \((u - 4)(u + 2) = 0\), so \(u = 4\) or \(u = -2\).
    Both values are usable - neither is ruled out, because \(x^2 - 3x\) can take any value \(\geq -\frac94\).
  3. Case \(u = 4\): \(x^2 - 3x - 4 = 0\). Its roots sum to \(3\) (sum \(= -b/a\)). Discriminant \(9 + 16 > 0\), so both are real.
    You never need the roots themselves - the sum of roots comes straight from the coefficients.
  4. Case \(u = -2\): \(x^2 - 3x + 2 = 0\). Roots sum to \(3\); discriminant \(9 - 8 > 0\), so both are real.
    Check the discriminant in each case: a case with no real roots would contribute nothing to the total.
  5. Total: \(3 + 3 = 6\).
    Four real values of \(x\), summing to \(6\).
Where it goes wrong
  • Answering \(3\) by solving only one of the two cases.
  • Adding the \(u\) values (\(4 + (-2) = 2\)) instead of the \(x\) values.
Takeaway With a substitution, each value of \(u\) gives its own equation in \(x\); check each for real roots and use 'sum of roots \(= -b/a\)' to avoid solving.
Answer E · Quadratic functions (MM1.3) · question QUA-09
20Graphs of functions · spec MM8.7 · Paper 1 style · ★★☆

The curves \(y = 2^{x}\) and \(y = x^2\) intersect at how many points?

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

They meet at \(x = 2\) and \(x = 4\) (\(4 = 4\), \(16 = 16\)), and once for negative \(x\) (at \(x = 0\): \(1 > 0\); at \(x = -1\): \(\frac12 < 1\)). For \(x > 4\), \(2^x\) grows faster. Three intersections.

Full method — how and why
Exponential versus quadratic. Test the obvious integer points and use end behaviour to be sure nothing is missed.
  1. Try small integers: \(x = 2\) gives \(4 = 4\) ✓ and \(x = 4\) gives \(16 = 16\) ✓ - two intersections already.
    \(2^x = x^2\) has these two well-known solutions.
  2. Check negative \(x\): at \(x = 0\), \(2^0 = 1 > 0 = x^2\); at \(x = -1\), \(2^{-1} = \frac12 < 1 = x^2\). The order swaps, so there is a crossing between \(-1\) and \(0\).
    A sign change in the difference \(2^x - x^2\) guarantees a crossing.
  3. Check beyond \(x = 4\): \(2^x\) grows faster than \(x^2\) (e.g. at \(x = 5\), \(32 > 25\)), so they never meet again.
    Exponential growth eventually outpaces any polynomial.
  4. Total: three intersections.
    One negative, and \(x = 2\), \(x = 4\).
Where it goes wrong
  • Finding only \(x = 2\) and \(x = 4\) and answering \(2\).
  • Forgetting that \(x^2\) is positive for negative \(x\) while \(2^x\) stays small but positive.
Takeaway Test integer points, then check the two ends: a sign change in the difference means an extra crossing.
Answer A · Graphs of functions (MM8.7) · question GRA-11
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