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Paper 1 · Set 09

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
75:00 PDF

Answer grid

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1Quadratic functions · spec MM1.3 · Paper 1 style · ★☆☆

Which of the following quadratics has its vertex in the second quadrant (where \(x < 0\) and \(y > 0\))?

  1. A\(y = -x^2 - 2x - 3\)
  2. B\(y = -x^2 + 2x + 1\)
  3. C\(y = x^2 - 4x + 5\)
  4. D\(y = x^2 + 6x + 10\)
  5. E\(y = x^2 + 4x + 1\)
Worked solution

Complete the square: A vertex \((-2, -3)\); B \((2, 1)\); C \((-1, -2)\); D \((-3, 1)\), which has \(x < 0\) and \(y > 0\); E \((1, 2)\). So D.

Full method — how and why
A 'which one' question about vertices. Complete the square on each option - but use the sign of the \(x\)-coordinate to eliminate most of them first.
  1. Recall that for \(y = x^2 + bx + c\) the vertex is at \(x = -\frac{b}{2}\).
    You only need the vertex position, not the whole completed square, to test the first condition \(x < 0\).
  2. Second quadrant needs \(x < 0\), so \(-\frac b2 < 0\), i.e. \(b > 0\): only \(y = x^2 + 4x + 1\) and \(y = x^2 + 6x + 10\) survive among the upward parabolas.
    Eliminating on one condition first halves the work.
  3. Compute the \(y\)-coordinates: \(x^2 + 4x + 1 = (x+2)^2 - 3\), vertex \((-2, -3)\): fails \(y > 0\). \(x^2 + 6x + 10 = (x+3)^2 + 1\), vertex \((-3, 1)\): passes.
    Now apply the second condition to the short list.
  4. Check the downward ones quickly: \(-x^2 - 2x - 3\) has vertex \((-1, -2)\) and \(-x^2 + 2x + 1\) has vertex \((1, 2)\) - neither is in the second quadrant.
    Worth 10 seconds of confirmation on a 'which one' question.
Where it goes wrong
  • Reading the vertex of \((x+3)^2 + 1\) as \((3, 1)\) - the sign flips.
  • Mixing up the quadrants: the second quadrant is up and to the left.
Takeaway On 'which of these' questions, test the cheapest condition first and eliminate before doing any heavy algebra.
Answer D · Quadratic functions (MM1.3) · question QUA-13
2Polynomials, factor and remainder theorems · spec MM1.6 · Paper 1 style · ★☆☆

Find the remainder when \(x^{4} + 2x^{3} - 3x + 1\) is divided by \((2x - 1)\).

  1. A\(\dfrac{13}{16}\)
  2. B\(\dfrac{3}{16}\)
  3. C\(1\)
  4. D\(-\dfrac{1}{4}\)
  5. E\(-\dfrac{7}{16}\)
  6. F\(-\dfrac{3}{16}\)
Worked solution

\(f\!\left(\frac12\right) = \frac1{16} + \frac{2}{8} - \frac32 + 1 = \frac{1}{16} + \frac{4}{16} - \frac{24}{16} + \frac{16}{16} = -\frac{3}{16}\).

Full method — how and why
A divisor of the form \(ax - b\). The Remainder Theorem still applies, but the substitution value is \(\frac{b}{a}\), not \(b\).
  1. \(2x - 1 = 0\) at \(x = \frac12\).
    The remainder on dividing by \((ax - b)\) is \(f(b/a)\) - the root of the divisor, whatever the leading coefficient.
  2. Substitute: \(\left(\frac12\right)^4 + 2\left(\frac12\right)^3 - 3\left(\frac12\right) + 1 = \frac{1}{16} + \frac{2}{8} - \frac32 + 1\).
    Work each power out separately: \(\frac1{16}\), \(\frac18\), then the linear and constant terms.
  3. Put everything over \(16\): \(\frac{1}{16} + \frac{4}{16} - \frac{24}{16} + \frac{16}{16} = -\frac{3}{16}\).
    A common denominator prevents fraction slips; \(\frac{2}{8} = \frac{4}{16}\) and \(\frac32 = \frac{24}{16}\).
Where it goes wrong
  • Substituting \(x = 1\) or \(x = 2\) instead of \(\frac12\).
  • Arithmetic sign slips giving \(+\frac{3}{16}\), which is offered.
Takeaway For divisor \(ax - b\), evaluate at \(x = \frac ba\); convert everything to a common denominator before adding.
Answer F · Polynomials, factor and remainder theorems (MM1.6) · question POL-04
3Trigonometric functions and equations · spec MM4.6 · Paper 1 style · ★☆☆

Find the smallest positive solution of \(\cos\left(x - \dfrac{\pi}{6}\right) = -\dfrac{1}{2}\).

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

\(x - \frac\pi6 = \frac{2\pi}{3}\) gives \(x = \frac{5\pi}{6}\) (the other solution \(x - \frac\pi6 = \frac{4\pi}{3}\) gives \(\frac{3\pi}{2}\), larger).

Full method — how and why
An equation with a shifted argument. Solve for the bracket first, then unshift - and be careful about which solution is smallest.
  1. Let \(u = x - \frac\pi6\) and solve \(\cos u = -\frac12\).
    Substituting keeps the shift out of the way until the end.
  2. \(\cos u = -\frac12\) at \(u = \frac{2\pi}{3}\) and \(u = \frac{4\pi}{3}\) (second and third quadrants).
    \(\cos\) is negative in quadrants 2 and 3; the reference angle is \(\frac\pi3\).
  3. Convert back: \(x = u + \frac\pi6\), giving \(x = \frac{2\pi}{3} + \frac{\pi}{6} = \frac{5\pi}{6}\) and \(x = \frac{4\pi}{3} + \frac{\pi}{6} = \frac{3\pi}{2}\).
    Common denominator \(6\): \(\frac{4\pi}{6} + \frac{\pi}{6} = \frac{5\pi}{6}\).
  4. The smallest positive one is \(\dfrac{5\pi}{6}\).
    Compare the candidates; \(\frac{5\pi}{6} < \frac{3\pi}{2}\).
Where it goes wrong
  • Subtracting the shift instead of adding when converting back.
  • Using the reference angle \(\frac\pi3\) directly, which gives \(\frac\pi2\) - an offered option.
Takeaway Solve for the whole bracket, then undo the shift; adjust the interval for the bracket if the question restricts \(x\).
Answer B · Trigonometric functions and equations (MM4.6) · question TRG-08
4Angles, polygons, congruence and vectors · spec M5.6 · Paper 1 style · ★☆☆

A shape is reflected in the line \(y = x\) and then rotated \(90^\circ\) anticlockwise about the origin. The point \((3, 1)\) ends up at

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

Reflection in \(y = x\): \((3, 1) \to (1, 3)\). Rotation \(90^\circ\) anticlockwise: \((x, y) \to (-y, x)\), so \((1, 3) \to (-3, 1)\).

Full method — how and why
Two transformations in sequence. Apply each rule in the stated order to the coordinates.
  1. Reflection in \(y = x\) swaps the coordinates: \((3,1) \to (1,3)\).
    The line \(y = x\) exchanges the roles of the two axes.
  2. Rotation \(90^\circ\) anticlockwise about the origin sends \((x,y) \to (-y,x)\).
    Worth memorising; check it on \((1,0) \to (0,1)\) ✓.
  3. Apply it: \((1,3) \to (-3,1)\).
    Here \(x = 1\), \(y = 3\), so the image is \((-3, 1)\).
Where it goes wrong
  • Applying the transformations in the wrong order.
  • Using the clockwise rule \((x,y) \to (y,-x)\), giving \((3,-1)\) - an offered option.
Takeaway Anticlockwise \(90^\circ\): \((x,y) \to (-y,x)\). Clockwise: \((x,y) \to (y,-x)\). Order matters.
Answer B · Angles, polygons, congruence and vectors (M5.6) · question GEO-12
5Functions and transformations · spec MM8.2 · Paper 1 style · ★☆☆

The curve \(y = x^2 - 6x + 5\) is reflected in the \(y\)-axis. Find the equation of the reflected curve.

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

Reflecting in the \(y\)-axis replaces \(x\) with \(-x\): \(y = x^2 + 6x + 5\).

Full method — how and why
Reflection in the \(y\)-axis replaces \(x\) by \(-x\). Only the odd powers change sign.
  1. Replace \(x\) with \(-x\): \(y = (-x)^2 - 6(-x) + 5\).
    Reflecting in the \(y\)-axis maps the point \((x, y)\) to \((-x, y)\), which is achieved by substituting \(-x\).
  2. Simplify: \((-x)^2 = x^2\) and \(-6(-x) = +6x\), so \(y = x^2 + 6x + 5\).
    Even powers are unaffected; odd powers flip sign. The constant is also unchanged.
Where it goes wrong
  • Changing the sign of the constant too, giving \(x^2 + 6x - 5\).
  • Confusing it with reflection in the \(x\)-axis, which would negate the whole right-hand side.
Takeaway \(y\)-axis reflection: \(x \to -x\), so only odd-degree terms change sign. \(x\)-axis reflection: \(y \to -y\), so everything does.
Answer A · Functions and transformations (MM8.2) · question FUN-06
6Ratio, proportion and percentages · spec M3.4 · Paper 1 style · ★☆☆

\(\pounds 600\) is divided between \(A\), \(B\) and \(C\) in the ratio \(2 : 3 : 7\). How much more does \(C\) receive than \(A\)?

  1. A\(\pounds 300\)
  2. B\(\pounds 250\)
  3. C\(\pounds 150\)
  4. D\(\pounds 200\)
  5. E\(\pounds 350\)
  6. F\(\pounds 100\)
Worked solution

Each part is \(\pounds 50\); \(C - A = 5\) parts \(= \pounds 250\).

Full method — how and why
Sharing in a ratio: find the value of one part, then answer the question actually asked.
  1. Add the ratio parts: \(2 + 3 + 7 = 12\).
    The total is divided into this many equal parts.
  2. One part: \(\pounds 600 \div 12 = \pounds 50\).
    Divide the total by the number of parts.
  3. The difference between \(C\) and \(A\) is \(7 - 2 = 5\) parts.
    Work with the difference in parts rather than computing both shares.
  4. So the difference is \(5 \times \pounds 50 = \pounds 250\).
    Check: \(A\) gets \(\pounds100\), \(C\) gets \(\pounds350\) ✓.
Where it goes wrong
  • Giving \(C\)'s share (\(\pounds350\)) instead of the difference - an offered option.
  • Dividing by \(3\) (the number of people) instead of \(12\) (the number of parts).
Takeaway Find the value of one part first; then read carefully whether a share, a difference or a total is wanted.
Answer B · Ratio, proportion and percentages (M3.4) · question RAT-03
7Sequences and series · spec MM2.3 · Paper 1 style · ★☆☆

Find the sum to infinity of the geometric series \(12 - 4 + \dfrac{4}{3} - \dfrac{4}{9} + \ldots\)

  1. A\(16\)
  2. B\(6\)
  3. C\(9\)
  4. D\(18\)
  5. E\(24\)
  6. F\(8\)
Worked solution

\(a = 12\), \(r = -\frac13\). \(S_\infty = \dfrac{12}{1 + \frac13} = \dfrac{12}{4/3} = 9\).

Full method — how and why
A sum to infinity. Find the ratio (including its sign) and check \(|r| < 1\) before using the formula.
  1. Find \(r\) by dividing consecutive terms: \(\frac{-4}{12} = -\frac13\).
    Always divide a term by the one before it; the alternating signs mean \(r\) is negative.
  2. Check convergence: \(\left|-\frac13\right| < 1\) ✓.
    The sum to infinity only exists when \(|r| < 1\); otherwise the formula is meaningless.
  3. Apply \(S_\infty = \dfrac{a}{1 - r} = \dfrac{12}{1 - (-\frac13)} = \dfrac{12}{\frac43}\).
    \(1 - r\) with a negative \(r\) becomes \(1 + \frac13\) - the double negative is the main trap.
  4. Divide: \(12 \times \frac34 = 9\).
    Dividing by a fraction is multiplying by its reciprocal.
Where it goes wrong
  • Using \(r = +\frac13\) and getting \(18\).
  • Computing \(\frac{12}{\frac43}\) as \(16\) by multiplying instead of dividing.
Takeaway \(S_\infty = \frac{a}{1-r}\); if \(r\) is negative the denominator grows, so the sum is smaller than \(a/(1-|r|)\).
Answer C · Sequences and series (MM2.3) · question SEQ-02
8Integration · spec MM7.1 · Paper 1 style · ★☆☆

Find the area enclosed between the curve \(y = x^2 - 4x\) and the \(x\)-axis.

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

Roots \(0\) and \(4\); the curve is below the axis between them. \(\int_0^4 (x^2 - 4x)\,dx = \frac{64}{3} - 32 = -\frac{32}{3}\), so the area is \(\frac{32}{3}\).

Full method — how and why
Area between a curve and the axis. The integral is negative here because the region is below the axis - the area is its magnitude.
  1. Find the limits: \(x^2 - 4x = x(x-4) = 0\) at \(x = 0\) and \(x = 4\).
    The enclosed region runs between consecutive roots.
  2. Note the curve is below the axis there: at \(x = 2\), \(y = -4\).
    Checking one interior value tells you the sign of the integral in advance.
  3. Integrate: \(\displaystyle\int_0^4 (x^2 - 4x)dx = \left[\frac{x^3}{3} - 2x^2\right]_0^4 = \frac{64}{3} - 32 = -\frac{32}{3}\).
    \(32 = \frac{96}{3}\), so \(\frac{64 - 96}{3} = -\frac{32}{3}\).
  4. Take the magnitude: the area is \(\dfrac{32}{3}\).
    Areas are positive; the negative sign only records that the region lies below the axis.
Where it goes wrong
  • Giving \(-\frac{32}{3}\) as the area - it is offered precisely to catch this.
  • Using limits \(0\) and \(2\) (the vertex) instead of the two roots.
Takeaway Integrate between the roots, then take the absolute value: a definite integral is a signed area.
Answer F · Integration (MM7.1) · question INT-02
9Exponentials and logarithms · spec MM5.1 · Paper 1 style · ★☆☆

The curve \(y = a^{x}\) passes through \((2, 12)\) and \((4, 108)\). Find \(a\).

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

\(\frac{a^4}{a^2} = a^2 = \frac{108}{12} = 9\), so \(a = 3\) (as \(a > 0\)).

Full method — how and why
Two points on an exponential curve. Dividing eliminates any constant and isolates a power of \(a\).
  1. Write both facts: \(a^2 = 12\) and \(a^4 = 108\).
    Substitute each point into \(y = a^x\).
  2. Divide: \(\dfrac{a^4}{a^2} = a^2 = \dfrac{108}{12} = 9\).
    Dividing subtracts indices, giving a single power of \(a\) - the standard move with two points.
  3. Solve: \(a = 3\) (bases of exponentials are positive).
    Reject \(a = -3\): \(y = a^x\) requires \(a > 0\).
Where it goes wrong
  • Answering \(9\) - that is \(a^2\), not \(a\).
  • Subtracting the \(y\)-values instead of dividing.
Takeaway Divide two points on an exponential curve; the index difference gives a clean power of the base.
Answer B · Exponentials and logarithms (MM5.1) · question LOG-12
10Integration · spec MM7.3 · Paper 1 style · ★☆☆

Given that \(\displaystyle\int_{1}^{5} f(x)\,dx = 7\) and \(\displaystyle\int_{3}^{5} f(x)\,dx = 2\), find \(\displaystyle\int_{1}^{3} \left(2f(x) - 1\right)dx\).

  1. A\(12\)
  2. B\(16\)
  3. C\(8\)
  4. D\(9\)
  5. E\(10\)
  6. F\(3\)
Worked solution

\(\int_1^3 f = 7 - 2 = 5\). So \(2 \times 5 - (3 - 1) = 8\).

Full method — how and why
Combining integrals over adjacent intervals, plus the linearity rules. No function is ever needed.
  1. Use \(\int_1^5 = \int_1^3 + \int_3^5\): so \(\int_1^3 f = 7 - 2 = 5\).
    Integrals over adjacent ranges add; this is the only way to get at the interval you need.
  2. Split the required integral: \(\int_1^3 (2f(x) - 1)dx = 2\int_1^3 f(x)dx - \int_1^3 1\,dx\).
    Constants come out; sums split - both are properties of the integral.
  3. Evaluate the second piece: \(\int_1^3 1\,dx = 3 - 1 = 2\).
    Integrating \(1\) over an interval gives its length.
  4. Combine: \(2(5) - 2 = 8\).
    Careful with the order of operations.
Where it goes wrong
  • Forgetting the \(-1\) term and answering \(10\).
  • Using \(\int_1^3 f = 7 + 2 = 9\) instead of subtracting.
Takeaway \(\int_a^c = \int_a^b + \int_b^c\), constants factor out, and \(\int_a^b 1\,dx = b - a\).
Answer C · Integration (MM7.3) · question INT-03
11Sequences and series · spec MM2.2 · Paper 1 style · ★☆☆

The sum of the first \(n\) terms of a series is \(S_n = 2n^2 + 3n\). Find the \(10\)th term.

  1. A\(43\)
  2. B\(189\)
  3. C\(21\)
  4. D\(230\)
  5. E\(39\)
  6. F\(41\)
Worked solution

\(u_{10} = S_{10} - S_9 = (200 + 30) - (162 + 27) = 230 - 189 = 41\).

Full method — how and why
Given a formula for \(S_n\), individual terms come from the difference of consecutive sums.
  1. Use \(u_{10} = S_{10} - S_9\).
    The 10th term is what the total gains when you add it - this works for any series, arithmetic or not.
  2. \(S_{10} = 2(100) + 30 = 230\).
    Substitute \(n = 10\).
  3. \(S_9 = 2(81) + 27 = 189\).
    Substitute \(n = 9\); \(2 \times 81 = 162\).
  4. Subtract: \(230 - 189 = 41\).
    Both wrong intermediate values (\(230\) and \(189\)) are offered as distractors.
Where it goes wrong
  • Answering \(230\) (that is the sum of ten terms, not the tenth term).
  • Substituting \(n = 10\) into a 'term' formula that was never given.
Takeaway \(u_n = S_n - S_{n-1}\) recovers terms from a sum formula.
Answer F · Sequences and series (MM2.2) · question SEQ-06
12Trigonometric functions and equations · spec MM4.6 · Paper 1 style · ★☆☆

Find the sum of all solutions of \(\tan x = \sqrt{3}\) in the interval \(0 \leq x \leq 2\pi\).

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

\(x = \frac\pi3\) and \(x = \frac\pi3 + \pi = \frac{4\pi}{3}\). Sum \(\frac{5\pi}{3}\).

Full method — how and why
\(\tan\) has period \(\pi\), so solutions come one per period. Find the first, add \(\pi\), then check the interval.
  1. The principal solution: \(\tan x = \sqrt3\) at \(x = \dfrac{\pi}{3}\).
    From the exact values: \(\tan 60^\circ = \sqrt3\).
  2. \(\tan\) repeats every \(\pi\), so the next solution is \(\dfrac{\pi}{3} + \pi = \dfrac{4\pi}{3}\).
    Unlike \(\sin\) and \(\cos\), \(\tan\) has period \(\pi\), not \(2\pi\).
  3. Check the interval \([0, 2\pi]\): \(\frac{\pi}{3}\) and \(\frac{4\pi}{3}\) are in it; the next, \(\frac{7\pi}{3}\), is not.
    \(\frac{7\pi}{3} > 2\pi\), so stop there.
  4. Sum: \(\dfrac{\pi}{3} + \dfrac{4\pi}{3} = \dfrac{5\pi}{3}\).
    Common denominator \(3\).
Where it goes wrong
  • Using period \(2\pi\) and finding only \(\frac\pi3\).
  • Including \(\frac{7\pi}{3}\) and answering \(\frac{7\pi}{3}\) or more - which is offered.
Takeaway \(\tan\) repeats every \(\pi\): solutions are \(x_0 + k\pi\), so there are usually two in a \(2\pi\) interval.
Answer D · Trigonometric functions and equations (MM4.6) · question TRG-04
13Differentiation · spec MM6.3 · Paper 1 style · ★☆☆

The curve \(y = ax^2 + bx\) has a stationary point at \((2, -8)\). Find \(a + b\).

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

\(y' = 2ax + b = 0\) at \(x = 2\): \(b = -4a\). \(y(2) = 4a + 2b = 4a - 8a = -4a = -8 \Rightarrow a = 2\), \(b = -8\). \(a + b = -6\).

Full method — how and why
A stationary point gives two conditions: the derivative vanishes there, and the point is on the curve.
  1. Differentiate: \(y' = 2ax + b\). At \(x = 2\): \(4a + b = 0\), so \(b = -4a\).
    Stationary means the gradient is zero at that \(x\).
  2. The point is on the curve: \(y(2) = 4a + 2b = -8\).
    Substitute \(x = 2\) into the original function and set it equal to the given \(y\).
  3. Substitute \(b = -4a\): \(4a - 8a = -4a = -8\), so \(a = 2\) and \(b = -8\).
    One equation in one unknown after substitution.
  4. Compute \(a + b = 2 + (-8) = -6\).
    The question asks for the sum.
Where it goes wrong
  • Using only one of the two conditions.
  • Sign error giving \(a = -2\) and the answer \(6\).
Takeaway 'Stationary point at \((p,q)\)' is two equations: \(f'(p) = 0\) and \(f(p) = q\).
Answer A · Differentiation (MM6.3) · question DIF-07
14Probability · spec M7.7 · Paper 1 style · ★☆☆

A fair die is thrown repeatedly until a six appears. Find the probability that the first six appears on the third throw.

  1. A\(\dfrac{25}{216}\)
  2. B\(\dfrac{5}{36}\)
  3. C\(\dfrac{1}{216}\)
  4. D\(\dfrac{1}{6}\)
  5. E\(\dfrac{125}{216}\)
  6. F\(\dfrac{91}{216}\)
Worked solution

\(\left(\frac56\right)^2 \cdot \frac16 = \frac{25}{216}\).

Full method — how and why
'First success on the \(k\)th trial' means failure, failure, ..., then success - a chain of independent events.
  1. The first two throws must be non-sixes: probability \(\frac56\) each.
    'First six on the third' forbids a six on throws one and two.
  2. The third throw must be a six: probability \(\frac16\).
    Exactly one way to succeed on that throw.
  3. Multiply: \(\left(\frac56\right)^2 \times \frac16 = \dfrac{25}{216}\).
    Independent throws, so probabilities multiply.
Where it goes wrong
  • Answering \(\frac16\) by ignoring the first two throws.
  • Using \(\left(\frac16\right)^3 = \frac{1}{216}\), which is the probability of three sixes.
Takeaway First success on trial \(k\): \((1-p)^{k-1}p\) - failures first, then the one success.
Answer A · Probability (M7.7) · question PRO-10
15Exponentials and logarithms · spec MM5.3 · Paper 1 style · ★★☆

The equation \(9^{x} - k\cdot 3^{x} + 9 = 0\) has exactly one real solution. Find the value of \(k\).

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

With \(y = 3^x > 0\): \(y^2 - ky + 9 = 0\). Exactly one solution for \(x\) means exactly one positive root: a repeated root (\(k^2 = 36\), product \(9 > 0\) so the roots have the sign of \(k\)): \(k = 6\) (repeated root \(y = 3\), \(x = 1\)). \(k = -6\) gives \(y = -3\), no solutions. (Two distinct positive roots would give two solutions.)

Full method — how and why
A quadratic in \(3^x\) with a parameter. 'Exactly one real solution' is a condition on the positive roots, not simply on the discriminant.
  1. Substitute \(y = 3^x\), noting \(y > 0\): \(y^2 - ky + 9 = 0\).
    \(9^x = (3^x)^2\). The restriction \(y > 0\) is the heart of the question.
  2. The product of the roots is \(9 > 0\), so the roots have the same sign; their sum is \(k\), so both are positive exactly when \(k > 0\).
    Sum and product of roots decide the signs without solving.
  3. For exactly one value of \(x\), we need exactly one positive root, which (given both roots share a sign) means a repeated root: discriminant \(k^2 - 36 = 0\), so \(k = \pm6\).
    Two distinct positive roots would give two solutions for \(x\); two negative roots would give none.
  4. Check the signs: \(k = 6\) gives the repeated root \(y = 3 > 0\), so \(x = 1\) - exactly one solution. \(k = -6\) gives \(y = -3 < 0\), so no solutions at all. Hence \(k = 6\).
    The final sign check is what rules out \(-6\), which is offered.
Where it goes wrong
  • Answering \(\pm6\) from the discriminant alone.
  • Forgetting that \(y\) must be positive and counting \(y = -3\) as a solution.
Takeaway With \(y = a^x\), translate conditions on \(x\) into conditions on the positive roots in \(y\), using sum and product to check signs.
Answer E · Exponentials and logarithms (MM5.3) · question LOG-17
16Quadratic functions · spec MM1.3 · Paper 1 style · ★★☆

The equation \(x^2 - 2mx + m + 6 = 0\) has two distinct positive roots. Find the complete set of values of \(m\).

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

Distinct real roots: \(4m^2 - 4(m + 6) > 0 \Rightarrow m^2 - m - 6 > 0 \Rightarrow (m-3)(m+2) > 0\), so \(m < -2\) or \(m > 3\). Both roots positive needs sum \(2m > 0\) and product \(m + 6 > 0\), so \(m > 0\). Combined: \(m > 3\).

Full method — how and why
'Two distinct positive roots' is three conditions at once: discriminant positive, sum of roots positive, product of roots positive. All three must hold.
  1. Discriminant: \(4m^2 - 4(m + 6) > 0 \Rightarrow m^2 - m - 6 > 0 \Rightarrow (m-3)(m+2) > 0\), so \(m < -2\) or \(m > 3\).
    Distinctness and reality come from \(b^2 - 4ac > 0\); this alone allows negative \(m\).
  2. Sum of roots \(= 2m > 0\), so \(m > 0\).
    If both roots are positive their sum is positive. This is the condition that kills the \(m < -2\) branch.
  3. Product of roots \(= m + 6 > 0\), so \(m > -6\).
    If both roots are positive their product is positive; this also rules out one positive and one negative root.
  4. Intersect all three: \((m < -2 \text{ or } m > 3)\) and \(m > 0\) and \(m > -6\) gives \(m > 3\).
    Take the overlap on a number line - the answer must satisfy every condition simultaneously.
Where it goes wrong
  • Stopping after the discriminant and choosing \(m < -2\) or \(m > 3\) - which is offered.
  • Forgetting that sum and product must both be positive: either one alone is not enough.
Takeaway Both roots positive \(\Leftrightarrow\) discriminant \(\geq 0\), sum \(> 0\) and product \(> 0\); sign conditions on roots always come in this package.
Answer B · Quadratic functions (MM1.3) · question QUA-14
17Graphs of functions · spec MM8.7 · Paper 1 style · ★★☆

For how many values of \(k\) does the equation \(|x^2 - 4| = k\) have exactly three real solutions?

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

The graph of \(y = |x^2 - 4|\) is a W-shape with minima \(0\) at \(x = \pm2\) and a local maximum \(4\) at \(x = 0\). A horizontal line gives three solutions only when it passes through the local maximum: \(k = 4\). One value.

Full method — how and why
Sliding a horizontal line across a modulus graph. Three solutions happen only at the height of the local maximum.
  1. Sketch \(y = |x^2 - 4|\): the parabola \(x^2 - 4\) with the part between \(x = -2\) and \(x = 2\) flipped above the axis.
    Reflecting the negative part gives a W-shape with two zeros and a hump.
  2. Identify the key heights: zeros at \(x = \pm2\), and a local maximum of \(4\) at \(x = 0\).
    At \(x = 0\), \(|0 - 4| = 4\) - the top of the central hump.
  3. Count intersections with \(y = k\): for \(0 < k < 4\) there are four; for \(k > 4\) there are two; for \(k = 0\) there are two; for \(k = 4\) the line passes through the hump's peak and cuts the outer arms: three.
    At the peak, two of the four intersections merge into one.
  4. So exactly one value of \(k\) (namely \(4\)) gives three solutions.
    The question asks how many values of \(k\), not how many solutions.
Where it goes wrong
  • Answering \(4\) (the value of \(k\)) instead of \(1\) (the number of such \(k\)).
  • Forgetting \(k = 0\) gives two solutions, not three.
Takeaway Slide the horizontal line and note the heights where the count changes: those are the turning values.
Answer A · Graphs of functions (MM8.7) · question GRA-13
18Graphs of functions · spec MM8.1 · Paper 1 style · ★★☆

The graph of \(y = \sqrt{x}\) is transformed to \(y = \sqrt{4 - x}\). Which sequence of transformations achieves this?

  1. Atranslate \(4\) units in the positive \(x\)-direction, then reflect in the \(y\)-axis
  2. Btranslate \(4\) units in the negative \(x\)-direction, then reflect in the \(y\)-axis
  3. Creflect in the \(x\)-axis, then translate \(4\) units in the positive \(y\)-direction
  4. Dstretch parallel to the \(x\)-axis with factor \(4\)
  5. Ereflect in the \(y\)-axis, then translate \(4\) units in the positive \(x\)-direction
Worked solution

\(\sqrt{4 - x} = \sqrt{-(x - 4)}\): reflect \(y = \sqrt x\) in the \(y\)-axis to get \(\sqrt{-x}\), then translate \(4\) to the right. (Translating first then reflecting gives \(\sqrt{-x - 4}\).)

Full method — how and why
Order matters when combining a reflection with a translation. Rewrite the target to reveal the correct sequence.
  1. Rewrite the target: \(\sqrt{4 - x} = \sqrt{-(x - 4)}\).
    Factoring out the \(-1\) exposes both operations: a negation of \(x\) and a shift.
  2. Reading it as \(f(-(x-4))\) where \(f(x) = \sqrt x\): first reflect in the \(y\)-axis (giving \(\sqrt{-x}\)), then translate \(4\) right (replacing \(x\) by \(x - 4\)).
    For nested transformations, the operation applied directly to \(x\) (here the shift) is done last.
  3. Check with a point: \(y = \sqrt{4-x}\) passes through \((4, 0)\); \(\sqrt{-x}\) passes through \((0,0)\), which moves to \((4,0)\) after shifting right ✓.
    A point check settles order questions decisively.
  4. Verify the other order fails: translating first gives \(\sqrt{x-4}\), then reflecting gives \(\sqrt{-x-4}\) - not the target.
    Different sequences give genuinely different graphs.
Where it goes wrong
  • Choosing 'translate then reflect', which gives \(\sqrt{-x-4}\).
  • Reflecting in the \(x\)-axis instead of the \(y\)-axis.
Takeaway Rewrite as \(f(\pm(x - h))\) and test a point; with mixed reflections and shifts the order is not interchangeable.
Answer E · Graphs of functions (MM8.1) · question GRA-14
19Ratio, proportion and percentages · spec M3.9 · Paper 1 style · ★★☆

\(p\) is directly proportional to \(\sqrt{q}\), and \(q\) is inversely proportional to \(r\). Which of the following is true?

  1. A\(p\) is directly proportional to \(\sqrt{r}\)
  2. B\(p\) is inversely proportional to \(r\)
  3. C\(p\) is inversely proportional to \(r^2\)
  4. D\(p\) is inversely proportional to \(\sqrt{r}\)
  5. E\(p\) is directly proportional to \(r\)
Worked solution

\(p = a\sqrt q\) and \(q = \frac{b}{r}\), so \(p = a\sqrt{\frac br} = \frac{a\sqrt b}{\sqrt r}\).

Full method — how and why
Chaining two proportionality statements. Write each with a constant and substitute.
  1. Write them algebraically: \(p = a\sqrt q\) and \(q = \dfrac{b}{r}\) for constants \(a, b > 0\).
    'Directly proportional' means a constant multiple; 'inversely' means a constant over.
  2. Substitute: \(p = a\sqrt{\dfrac br} = \dfrac{a\sqrt b}{\sqrt r}\).
    The square root distributes over the quotient.
  3. Since \(a\sqrt b\) is a constant, \(p = \dfrac{\text{constant}}{\sqrt r}\): \(p\) is inversely proportional to \(\sqrt r\).
    Collect all the constants into one; only the \(r\)-dependence matters.
Where it goes wrong
  • Forgetting the square root and concluding \(p \propto \frac1r\) - an offered option.
  • Losing the 'inversely' and choosing \(p \propto \sqrt r\).
Takeaway Chain proportionalities by substituting; the accumulated constants never affect the final relationship.
Answer D · Ratio, proportion and percentages (M3.9) · question RAT-09
20Straight lines · spec MM3.1 · Paper 1 style · ★★☆

For which value of \(k\) are the lines \(kx + 3y = 2\) and \(2x - (k-1)y = 5\) perpendicular?

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

Gradients \(-\frac k3\) and \(\frac{2}{k-1}\). Perpendicular when the product is \(-1\): \(\frac{-2k}{3(k-1)} = -1 \Rightarrow 2k = 3k - 3 \Rightarrow k = 3\).

Full method — how and why
A perpendicularity condition with the parameter in both equations. Write both gradients, multiply, set to \(-1\).
  1. First line \(kx + 3y = 2\) has gradient \(-\dfrac k3\).
    Using \(-\frac ab\) avoids rearranging.
  2. Second line \(2x - (k-1)y = 5\) has gradient \(-\dfrac{2}{-(k-1)} = \dfrac{2}{k-1}\).
    Here \(b = -(k-1)\), so the two minus signs cancel.
  3. Set the product to \(-1\): \(-\dfrac{k}{3} \times \dfrac{2}{k-1} = -1\), i.e. \(\dfrac{-2k}{3(k-1)} = -1\).
    Perpendicular lines have gradients multiplying to \(-1\).
  4. Solve: \(-2k = -3(k-1) \Rightarrow -2k = -3k + 3 \Rightarrow k = 3\).
    Check: gradients \(-1\) and \(1\), product \(-1\) ✓.
Where it goes wrong
  • Mishandling the sign of \(b\) in the second line and getting \(k = -3\) - which is offered.
  • Setting the product to \(+1\) or the gradients equal (that would be the parallel condition).
Takeaway Read gradients as \(-\frac ab\) straight from \(ax + by = c\), watching the sign of \(b\) carefully.
Answer A · Straight lines (MM3.1) · question LIN-08
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