Σ TMUA Practice
Home›Mock papers›Paper 1 Set 05

Paper 1 · Set 05

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

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
1Differentiation · spec MM6.3 · Paper 1 style · ★☆☆

The curve \(y = x^3 + px^2 + qx\) has stationary points at \(x = 1\) and \(x = 3\). Find \(q\).

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

\(y' = 3x^2 + 2px + q = 3(x - 1)(x - 3) = 3x^2 - 12x + 9\). So \(q = 9\) (and \(p = -6\)).

Full method — how and why
Known stationary points mean known roots of \(f'\). Compare the factorised derivative with its expanded form.
  1. Differentiate: \(y' = 3x^2 + 2px + q\).
    A cubic's derivative is a quadratic.
  2. The stationary points are its roots, so \(y' = 3(x-1)(x-3)\).
    The leading coefficient must be \(3\) to match \(3x^2\).
  3. Expand: \(3(x^2 - 4x + 3) = 3x^2 - 12x + 9\).
    Multiply out to compare coefficients.
  4. Compare constants: \(q = 9\) (and comparing \(x\) terms gives \(2p = -12\), so \(p = -6\)).
    The question asks only for \(q\).
Where it goes wrong
  • Forgetting the leading \(3\) and getting \(q = 3\).
  • Answering \(-6\), which is \(p\).
Takeaway Stationary points are roots of \(f'\): build \(f'\) in factorised form with the correct leading coefficient.
Answer C · Differentiation (MM6.3) · question DIF-10
2Quadratic functions · spec MM1.3 · Paper 1 style · ★☆☆

The expression \(x^2 + 6x + c\) can be written as \((x + p)^2 + q\). Given that \(q = -c\), find the value of \(c\).

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

\((x+3)^2 + (c - 9)\), so \(q = c - 9 = -c\), giving \(2c = 9\) and \(c = \frac92\).

Full method — how and why
Completing the square symbolically, then imposing a relationship between the two constants. It is an equation in \(c\), not in \(x\).
  1. Complete the square: \(x^2 + 6x + c = (x + 3)^2 - 9 + c\).
    Half of \(6\) is \(3\); subtracting \(9\) compensates for the \(+9\) the square introduces.
  2. Identify \(p = 3\) and \(q = c - 9\).
    Compare with the required form \((x+p)^2 + q\) term by term.
  3. Impose \(q = -c\): \(c - 9 = -c\).
    The condition given in the question is a link between \(q\) and \(c\), so substitute your expression for \(q\).
  4. Solve: \(2c = 9\), so \(c = \frac92\).
    A simple linear equation - just collect the \(c\) terms.
Where it goes wrong
  • Writing \(q = c + 9\) (sign error) and getting \(c = -\frac92\).
  • Solving \(c - 9 = c\) and concluding there is no solution.
Takeaway Complete the square with the constant left as a letter; the extra condition then becomes an ordinary equation in that letter.
Answer D · Quadratic functions (MM1.3) · question QUA-19
3Ratio, proportion and percentages · spec M3.5 · Paper 1 style · ★☆☆

A mixture contains sand and cement in the ratio \(5 : 2\). How much cement must be added to \(35\) kg of the mixture to make the ratio \(5 : 3\)?

  1. A\(2.5\) kg
  2. B\(15\) kg
  3. C\(3\) kg
  4. D\(7\) kg
  5. E\(5\) kg
  6. F\(10\) kg
Worked solution

\(35\) kg contains \(25\) kg sand and \(10\) kg cement. For \(5 : 3\) with \(25\) kg sand, cement must be \(15\) kg: add \(5\) kg.

Full method — how and why
A changing ratio where one quantity stays fixed. Anchor on the unchanged amount.
  1. Split the \(35\) kg in the ratio \(5:2\): seven parts of \(5\) kg, so \(25\) kg sand and \(10\) kg cement.
    \(35 \div 7 = 5\) kg per part.
  2. Only cement is added, so the sand stays at \(25\) kg.
    Identify what does not change - that is the anchor for the new ratio.
  3. In the new ratio \(5 : 3\), the \(5\) parts correspond to \(25\) kg, so one part is \(5\) kg and the cement must be \(3 \times 5 = 15\) kg.
    Scale the new ratio to the fixed quantity.
  4. Cement to add: \(15 - 10 = 5\) kg.
    The question asks for the amount added, not the final amount.
Where it goes wrong
  • Answering \(15\) kg (the new total cement) - an offered option.
  • Rescaling the sand as well, as though both quantities changed.
Takeaway In 'add to change the ratio' problems, anchor on the quantity that stays the same.
Answer E · Ratio, proportion and percentages (M3.5) · question RAT-07
4Number, units and estimation · spec M2.8 · Paper 1 style · ★☆☆

Evaluate \(\dfrac{(3 \times 10^{4}) \times (8 \times 10^{-2})}{6 \times 10^{-3}}\), giving your answer in standard form.

  1. A\(4 \times 10^{4}\)
  2. B\(4 \times 10^{6}\)
  3. C\(4 \times 10^{-1}\)
  4. D\(2.4 \times 10^{5}\)
  5. E\(1.5 \times 10^{5}\)
  6. F\(4 \times 10^{5}\)
Worked solution

\(\frac{24 \times 10^{2}}{6 \times 10^{-3}} = 4 \times 10^{5}\).

Full method — how and why
Standard-form arithmetic: handle the numbers and the powers of ten separately, then renormalise.
  1. Multiply the numerator: \((3 \times 8) \times 10^{4 + (-2)} = 24 \times 10^{2}\).
    Multiply the leading numbers, add the indices.
  2. Divide by \(6 \times 10^{-3}\): \(\dfrac{24}{6} \times 10^{2 - (-3)} = 4 \times 10^{5}\).
    Divide the numbers, subtract the indices - and subtracting \(-3\) adds \(3\).
  3. Check the form: \(1 \leq 4 < 10\) ✓, so no renormalising is needed.
    Standard form requires the leading number to be between \(1\) and \(10\).
Where it goes wrong
  • Getting \(10^{-1}\) by adding instead of subtracting the \(-3\).
  • Leaving \(24 \times 10^{2}\), which is not in standard form.
Takeaway Numbers multiply/divide; indices add/subtract. Then check \(1 \leq a < 10\).
Answer F · Number, units and estimation (M2.8) · question NUM-05
5Graphs of functions · spec MM8.1 · Paper 1 style · ★☆☆

The graph of \(y = \log_2 x\) is reflected in the line \(y = x\). The resulting graph has equation

  1. A\(y = 2^{x}\)
  2. B\(y = \log_{\frac12} x\)
  3. C\(y = \log_2(-x)\)
  4. D\(y = 2^{-x}\)
  5. E\(y = x^2\)
  6. F\(y = -\log_2 x\)
Worked solution

Reflection in \(y = x\) gives the inverse function: \(x = \log_2 y\), i.e. \(y = 2^x\).

Full method — how and why
Reflection in \(y = x\) produces the inverse function - swap \(x\) and \(y\) and re-solve.
  1. Reflecting in \(y = x\) swaps the roles of \(x\) and \(y\): the new graph satisfies \(x = \log_2 y\).
    Every point \((a,b)\) maps to \((b,a)\).
  2. Rewrite in exponential form: \(y = 2^x\).
    \(\log_2 y = x\) means \(y = 2^x\) - logs and exponentials are inverse functions.
Where it goes wrong
  • Choosing \(y = \log_2(-x)\) (a \(y\)-axis reflection) or \(-\log_2 x\) (an \(x\)-axis reflection).
  • Thinking the inverse of a log is \(x^2\) because of the base \(2\).
Takeaway Reflection in \(y = x\) gives the inverse; \(y = a^x\) and \(y = \log_a x\) are mirror images in that line.
Answer A · Graphs of functions (MM8.1) · question GRA-09
6Straight lines · spec MM3.1 · Paper 1 style · ★☆☆

The points \(A(1, 2)\), \(B(5, 4)\) and \(C(k, 8)\) are collinear. Find \(k\).

  1. A\(12\)
  2. B\(11\)
  3. C\(10\)
  4. D\(9\)
  5. E\(7\)
  6. F\(13\)
Worked solution

Gradient \(AB = \frac{2}{4} = \frac12\). Gradient \(AC = \frac{6}{k-1} = \frac12 \Rightarrow k - 1 = 12 \Rightarrow k = 13\).

Full method — how and why
Collinear points have equal gradients between any pair. Set two gradients equal and solve.
  1. Gradient of \(AB\): \(\dfrac{4 - 2}{5 - 1} = \dfrac{2}{4} = \dfrac12\).
    Any two of the three points determine the line, so compute the gradient from the pair with known coordinates.
  2. Gradient of \(AC\): \(\dfrac{8 - 2}{k - 1} = \dfrac{6}{k-1}\).
    Use the same starting point \(A\) for both gradients to avoid sign confusion.
  3. Set them equal: \(\dfrac{6}{k-1} = \dfrac12\).
    Collinear means the gradient between any pair is the same.
  4. Cross-multiply: \(k - 1 = 12\), so \(k = 13\).
    Check: from \(B(5,4)\) to \(C(13,8)\) the gradient is \(\frac48 = \frac12\) ✓.
Where it goes wrong
  • Forgetting the \(-1\) and answering \(12\).
  • Inverting the gradient and solving \(\frac{k-1}{6} = \frac12\), which gives \(k = 4\).
Takeaway Collinearity is an equal-gradient condition; always verify with the third pair of points.
Answer F · Straight lines (MM3.1) · question LIN-02
7Polynomials, factor and remainder theorems · spec MM1.6 · Paper 1 style · ★☆☆

The polynomial \(p(x) = x^3 + ax^2 + bx + 4\) has \((x-1)\) and \((x+2)\) as factors. Find the third linear factor.

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

Write \(p(x) = (x-1)(x+2)(x-r)\). The constant term is \((-1)(2)(-r) = 2r = 4\), so \(r = 2\): the third factor is \((x - 2)\).

Full method — how and why
A cubic with two known factors has a third. The constant term alone pins it down - no division needed.
  1. Write \(p(x) = (x - 1)(x + 2)(x - r)\), where \((x-r)\) is the unknown factor.
    The leading coefficient is \(1\), so the product of the three monic linear factors is the whole cubic.
  2. Compare constant terms: the constant of the product is \((-1)(2)(-r) = 2r\), and \(p(x)\) has constant term \(4\).
    The constant term of a product of brackets is the product of their constant terms - much faster than expanding everything.
  3. Solve: \(2r = 4\), so \(r = 2\) and the third factor is \((x - 2)\).
    Check: \((-1)(2)(-2) = 4\) ✓.
Where it goes wrong
  • Reading the third factor as \((x + 2)\) by forgetting the sign flip between the root \(r\) and the factor \((x - r)\).
  • Trying to find \(a\) and \(b\) first - unnecessary work.
Takeaway Compare constant terms to find a missing factor: the product of the constants in the brackets equals the constant term.
Answer B · Polynomials, factor and remainder theorems (MM1.6) · question POL-05
8Probability · spec M7.7 · Paper 1 style · ★☆☆

A bag contains \(4\) red and \(6\) blue counters. Two counters are taken at random without replacement. Find the probability that they are different colours.

  1. A\(\dfrac{24}{100}\)
  2. B\(\dfrac{7}{15}\)
  3. C\(\dfrac{2}{5}\)
  4. D\(\dfrac{8}{15}\)
  5. E\(\dfrac{4}{15}\)
  6. F\(\dfrac{12}{25}\)
Worked solution

\(P(RB) + P(BR) = \frac{4}{10}\cdot\frac{6}{9} + \frac{6}{10}\cdot\frac{4}{9} = \frac{48}{90} = \frac{8}{15}\).

Full method — how and why
'Different colours' happens in two orders. Either add the two ordered cases or use the complement.
  1. Red then blue: \(\dfrac{4}{10} \times \dfrac{6}{9}= \dfrac{24}{90}\).
    After removing a red, \(9\) counters remain and all \(6\) blues are still there - without replacement.
  2. Blue then red: \(\dfrac{6}{10} \times \dfrac{4}{9} = \dfrac{24}{90}\).
    The same product, but the orders are distinct outcomes so both count.
  3. Add: \(\dfrac{48}{90} = \dfrac{8}{15}\).
    Mutually exclusive cases add; then cancel by \(6\).
  4. Check by complement: \(P(\text{same}) = \frac{4}{10}\frac{3}{9} + \frac{6}{10}\frac{5}{9} = \frac{12+30}{90} = \frac{42}{90}\), and \(1 - \frac{42}{90} = \frac{48}{90}\) ✓.
    Two routes agreeing is strong confirmation.
Where it goes wrong
  • Counting only one order, giving \(\frac{4}{15}\) - an offered option.
  • Keeping the denominator at \(10\) for the second draw (\(\frac{12}{25}\)), which is the with replacement answer.
Takeaway Without replacement: reduce both the relevant colour count and the total; and count both orders unless the question fixes one.
Answer D · Probability (M7.7) · question PRO-01
9Simultaneous equations and inequalities · spec MM1.4 · Paper 1 style · ★☆☆

Solve \(xy = 6\) and \(x - y = 1\). What is the sum of all possible values of \(x + y\)?

  1. A\(10\)
  2. B\(5\)
  3. C\(-5\)
  4. D\(7\)
  5. E\(1\)
  6. F\(0\)
Worked solution

\(x = y + 1\): \(y^2 + y - 6 = 0 \Rightarrow (y+3)(y-2) = 0\). Solutions \((3, 2)\) with \(x + y = 5\) and \((-2, -3)\) with \(x + y = -5\). The sum is \(0\).

Full method — how and why
A product and a difference. Substituting the linear relation gives a quadratic with two solution pairs, and the question wants a total over both.
  1. From \(x - y = 1\), write \(x = y + 1\).
    Substituting the linear equation keeps things to one variable.
  2. Substitute into \(xy = 6\): \((y + 1)y = 6 \Rightarrow y^2 + y - 6 = 0\).
    Expand before rearranging into standard form.
  3. Factorise: \((y + 3)(y - 2) = 0\), so \(y = -3\) or \(y = 2\).
    Both are valid - nothing in the question restricts the sign.
  4. Find the pairs: \(y = 2 \Rightarrow x = 3\), giving \(x + y = 5\); \(y = -3 \Rightarrow x = -2\), giving \(x + y = -5\). Sum of these: \(0\).
    The question asks for the sum of all possible values of \(x + y\), so compute \(x+y\) for each pair and add.
Where it goes wrong
  • Finding only the positive pair and answering \(5\).
  • Adding all four numbers \(3 + 2 + (-2) + (-3)\) - which happens to be \(0\) too, but for the wrong reason; on another question that luck runs out.
Takeaway When a system has several solution pairs, compute the requested quantity for each pair and then combine.
Answer F · Simultaneous equations and inequalities (MM1.4) · question SIM-06
10Polynomials, factor and remainder theorems · spec MM1.6 · Paper 1 style · ★☆☆

Find the coefficient of \(x^2\) in the expansion of \((x + 1)(x - 2)(2x + 3)\).

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

\((x+1)(x-2) = x^2 - x - 2\); multiply by \((2x + 3)\): the \(x^2\) term is \(3x^2 - 2x^2 = x^2\), coefficient \(1\).

Full method — how and why
Only one coefficient is wanted, so do not expand everything: collect just the ways of making \(x^2\).
  1. Multiply the first two brackets: \((x+1)(x-2) = x^2 - x - 2\).
    Two brackets at a time keeps the bookkeeping small.
  2. To get \(x^2\) from \((x^2 - x - 2)(2x + 3)\), pair \(x^2\) with \(3\), and \(-x\) with \(2x\).
    An \(x^2\) term arises from (degree 2)\(\times\)(degree 0) and (degree 1)\(\times\)(degree 1) - list the pairings rather than expanding.
  3. Add the contributions: \(3x^2 + (-x)(2x) = 3x^2 - 2x^2 = x^2\), so the coefficient is \(1\).
    Only these two products can contribute.
Where it goes wrong
  • Missing the \((-x)(2x)\) contribution and answering \(3\).
  • Expanding everything under time pressure and losing a sign.
Takeaway For a single coefficient, list the pairs of terms whose degrees add to the one you want.
Answer A · Polynomials, factor and remainder theorems (MM1.6) · question POL-11
11Triangles, radians and sectors · spec MM4.1 · Paper 1 style · ★☆☆

Two sides of a triangle are \(3\) and \(5\) and the included angle is \(\theta\), where \(\cos\theta = -\dfrac{1}{5}\). Find the length of the third side.

  1. A\(\sqrt{34}\)
  2. B\(6\)
  3. C\(\sqrt{40}\)
  4. D\(7\)
  5. E\(\sqrt{28}\)
  6. F\(\sqrt{31}\)
Worked solution

\(c^2 = 9 + 25 - 2\cdot3\cdot5\cdot\left(-\frac15\right) = 34 + 6 = 40\).

Full method — how and why
Cosine rule with a negative cosine. The only difficulty is handling the double negative.
  1. Write the rule: \(c^2 = 3^2 + 5^2 - 2(3)(5)\cos\theta\).
    Two sides and the included angle again.
  2. Substitute \(\cos\theta = -\frac15\): \(c^2 = 9 + 25 - 30\left(-\frac15\right) = 34 + 6\).
    Subtracting a negative adds: \(-30 \times (-\frac15) = +6\).
  3. So \(c^2 = 40\) and \(c = \sqrt{40}\).
    Leave it in surd form; \(\sqrt{40} = 2\sqrt{10}\) if you prefer.
Where it goes wrong
  • Getting \(34 - 6 = 28\) by mishandling the sign, which is offered as \(\sqrt{28}\).
  • Forgetting the angle is obtuse, so the third side must be longer than \(\sqrt{34}\) - a useful sanity check.
Takeaway A negative cosine makes the third side longer than Pythagoras would give; check your answer against \(\sqrt{a^2+b^2}\).
Answer C · Triangles, radians and sectors (MM4.1) · question TRI-09
12Quadratic functions · spec MM1.3 · Paper 1 style · ★☆☆

By completing the square, find the minimum value of \(2x^2 - 12x + 23\).

  1. A\(-13\)
  2. B\(5\)
  3. C\(23\)
  4. D\(7\)
  5. E\(-5\)
  6. F\(3\)
Worked solution

\(2x^2 - 12x + 23 = 2(x-3)^2 - 18 + 23 = 2(x-3)^2 + 5\), so the minimum is \(5\) (at \(x = 3\)).

Full method — how and why
Minimum value of a quadratic: complete the square. The leading coefficient \(2\) must be factored out before completing the square.
  1. Factor the \(2\) out of the \(x\) terms: \(2(x^2 - 6x) + 23\).
    Completing the square only works on \(x^2 + (\text{something})x\); the coefficient of \(x^2\) must be \(1\) inside the bracket.
  2. Complete the square inside: \(x^2 - 6x = (x-3)^2 - 9\), so the expression is \(2\left[(x-3)^2 - 9\right] + 23\).
    Halve the coefficient of \(x\) (\(-6 \to -3\)) and subtract the square of that half.
  3. Multiply out the \(2\): \(2(x-3)^2 - 18 + 23 = 2(x-3)^2 + 5\).
    The \(-9\) is inside the bracket, so it is multiplied by \(2\) as well - this is where marks are lost.
  4. \(2(x-3)^2 \geq 0\), so the least value is \(5\), at \(x = 3\).
    A square is never negative, and it is zero exactly at the vertex.
Where it goes wrong
  • Forgetting to multiply the \(-9\) by \(2\), giving \(2(x-3)^2 + 14\) and the answer \(14\).
  • Quoting the \(x\)-coordinate \(3\) instead of the minimum value \(5\) - both are offered.
Takeaway Take the leading coefficient outside first, complete the square, then multiply back in; the constant left outside is the minimum.
Answer B · Quadratic functions (MM1.3) · question QUA-02
13Graphs of functions · spec MM8.6 · Paper 1 style · ★☆☆

How many times does the graph of \(y = x^3 - x^2 - 6x\) cross or touch the \(x\)-axis?

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

\(x(x^2 - x - 6) = x(x - 3)(x + 2)\): three distinct real roots, so three crossings.

Full method — how and why
Axis crossings are roots. Factorise fully and count distinct roots.
  1. Factor out \(x\): \(x^3 - x^2 - 6x = x(x^2 - x - 6)\).
    There is no constant term, so \(x\) is a factor - always check this first.
  2. Factorise the quadratic: \(x^2 - x - 6 = (x-3)(x+2)\).
    Two numbers multiplying to \(-6\), adding to \(-1\).
  3. Roots: \(x = 0, 3, -2\) - three distinct values, each a simple root, so the graph crosses three times.
    Distinct simple roots mean genuine crossings; a repeated root would be a touch.
Where it goes wrong
  • Missing the root \(x = 0\) and answering \(2\).
  • Counting the \(y\)-intercept as an extra crossing of the \(x\)-axis.
Takeaway Factor out \(x\) when there is no constant term; count distinct roots for the number of axis meetings.
Answer A · Graphs of functions (MM8.6) · question GRA-01
14Integration · spec MM7.2 · Paper 1 style · ★☆☆

Find the value of \(\displaystyle\int_{1}^{8} x^{-\frac{2}{3}}\,dx\).

  1. A\(\dfrac{21}{2}\)
  2. B\(9\)
  3. C\(3\)
  4. D\(6\)
  5. E\(2\)
  6. F\(\dfrac{3}{2}\)
Worked solution

\(\left[3x^{1/3}\right]_1^8 = 3(2 - 1) = 3\).

Full method — how and why
A negative fractional index. Add one to it, and choose limits that make the cube root easy.
  1. Add one to the index: \(-\frac23 + 1 = \frac13\), and divide by \(\frac13\) (i.e. multiply by \(3\)): \(\displaystyle\int x^{-2/3}dx = 3x^{1/3}\).
    Dividing by a fraction multiplies by its reciprocal.
  2. Evaluate at the limits: \(3\left(8^{1/3}\right) - 3\left(1^{1/3}\right) = 3(2) - 3(1)\).
    \(8^{1/3} = 2\) - the limits were chosen to be perfect cubes.
  3. Subtract: \(6 - 3 = 3\).
    Simple arithmetic.
Where it goes wrong
  • Forgetting the factor \(3\) and answering \(1\) or \(2\).
  • Adding one to the index the wrong way and using \(x^{-5/3}\).
Takeaway \(\int x^n dx = \frac{x^{n+1}}{n+1}\) works for negative and fractional \(n\) (except \(n = -1\)); dividing by \(\frac13\) multiplies by \(3\).
Answer C · Integration (MM7.2) · question INT-12
15Functions and transformations · spec MM8.2 · Paper 1 style · ★☆☆

The graph of \(y = \sin x\) is transformed to \(y = 2\sin(3x) + 1\). What is the maximum value of the new function, and what is its period (in degrees)?

  1. Amaximum \(1\), period \(120^\circ\)
  2. Bmaximum \(3\), period \(360^\circ\)
  3. Cmaximum \(2\), period \(120^\circ\)
  4. Dmaximum \(3\), period \(120^\circ\)
  5. Emaximum \(3\), period \(1080^\circ\)
  6. Fmaximum \(2\), period \(180^\circ\)
Worked solution

\(\sin(3x)\) has period \(360^\circ/3 = 120^\circ\). \(2\sin(3x)\) ranges from \(-2\) to \(2\), so \(2\sin(3x) + 1\) has maximum \(3\).

Full method — how and why
For \(y = a\sin(bx) + c\), the amplitude, period and shift can be read straight off - as long as you know which letter does what.
  1. The \(3\) inside compresses horizontally: period \(= \dfrac{360^\circ}{3} = 120^\circ\).
    \(\sin(bx)\) completes a cycle when \(bx\) increases by \(360^\circ\), i.e. when \(x\) increases by \(360^\circ/b\).
  2. The \(2\) outside is the amplitude: \(2\sin(3x)\) ranges from \(-2\) to \(2\).
    Multiplying the function stretches the range by that factor, symmetrically about zero.
  3. The \(+1\) raises everything: the range becomes \(-1\) to \(3\), so the maximum is \(3\).
    A vertical translation moves the whole wave, including its maximum.
Where it goes wrong
  • Multiplying the period by \(3\) to get \(1080^\circ\) instead of dividing.
  • Giving the amplitude \(2\) as the maximum and ignoring the \(+1\).
Takeaway In \(a\sin(bx) + c\): amplitude \(|a|\), period \(360^\circ/b\), midline \(y = c\), maximum \(c + |a|\).
Answer D · Functions and transformations (MM8.2) · question FUN-09
16Pythagoras, mensuration and 3D · spec M5.16 · Paper 1 style · ★★☆

A cylinder has radius \(r\) and height \(h\). Its volume is \(54\pi\) and \(h = 2r\). Find its total surface area.

  1. A\(27\pi\)
  2. B\(45\pi\)
  3. C\(108\pi\)
  4. D\(54\pi\)
  5. E\(72\pi\)
  6. F\(36\pi\)
Worked solution

\(\pi r^2 \cdot 2r = 2\pi r^3 = 54\pi \Rightarrow r = 3\), \(h = 6\). Surface area \(= 2\pi r^2 + 2\pi rh = 18\pi + 36\pi = 54\pi\).

Full method — how and why
Use the volume condition to find \(r\), then substitute into the surface-area formula. Both formulas must be recalled.
  1. Write the volume with \(h = 2r\): \(V = \pi r^2 h = \pi r^2 (2r) = 2\pi r^3\).
    Substituting the relationship immediately reduces to one unknown.
  2. Solve \(2\pi r^3 = 54\pi\): \(r^3 = 27\), so \(r = 3\) and \(h = 6\).
    The \(\pi\) cancels, which is a good sign.
  3. Total surface area \(= 2\pi r^2\) (two circular ends) \(+\ 2\pi r h\) (curved surface).
    'Total' means include both ends; 'curved surface area' would be just \(2\pi rh\).
  4. Substitute: \(2\pi(9) + 2\pi(3)(6) = 18\pi + 36\pi = 54\pi\).
    Coincidentally equal to the volume's numerical value - a deliberate touch.
Where it goes wrong
  • Giving only the curved surface \(36\pi\) - an offered option.
  • Forgetting that there are two circular ends.
Takeaway Cylinder: \(V = \pi r^2h\), curved surface \(2\pi rh\), total surface \(2\pi r^2 + 2\pi rh\).
Answer D · Pythagoras, mensuration and 3D (M5.16) · question MEN-02
17Trigonometric functions and equations · spec MM4.5 · Paper 1 style · ★★☆

Solve \(2\cos^2 x + 3\sin x - 3 = 0\) for \(0^\circ \leq x \leq 360^\circ\). How many solutions are there?

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

\(2(1 - \sin^2 x) + 3\sin x - 3 = 0 \Rightarrow 2\sin^2 x - 3\sin x + 1 = 0 \Rightarrow (2\sin x - 1)(\sin x - 1) = 0\). \(\sin x = \frac12\): \(30^\circ, 150^\circ\); \(\sin x = 1\): \(90^\circ\). Three solutions.

Full method — how and why
A mixed \(\sin\)/\(\cos\) equation. Use \(\cos^2 = 1 - \sin^2\) to make it a quadratic in \(\sin x\), then count solutions carefully.
  1. Replace \(\cos^2 x\) with \(1 - \sin^2 x\): \(2(1 - \sin^2 x) + 3\sin x - 3 = 0\).
    The identity \(\sin^2+\cos^2=1\) is the only way to get a single trig function - always convert the squared one.
  2. Tidy: \(-2\sin^2 x + 3\sin x - 1 = 0\), or \(2\sin^2 x - 3\sin x + 1 = 0\).
    Multiply through by \(-1\) so the leading coefficient is positive.
  3. Factorise: \((2\sin x - 1)(\sin x - 1) = 0\), giving \(\sin x = \frac12\) or \(\sin x = 1\).
    Treat \(\sin x\) as a single unknown.
  4. Count solutions in \(0^\circ \leq x \leq 360^\circ\): \(\sin x = \frac12\) gives \(30^\circ\) and \(150^\circ\); \(\sin x = 1\) gives only \(90^\circ\).
    A maximum value like \(\sin x = 1\) is reached just once per period, not twice - this is what makes the total odd.
  5. Total: \(3\).
    Count, do not assume 'two roots means four solutions'.
Where it goes wrong
  • Assuming each value of \(\sin x\) gives two solutions and answering \(4\).
  • Sign slips when substituting the identity, producing an unfactorisable quadratic.
Takeaway Convert to one trig function, factorise, then count solutions per value - endpoints \(\pm1\) give only one each.
Answer E · Trigonometric functions and equations (MM4.5) · question TRG-02
18Straight lines · spec MM3.1 · Paper 1 style · ★★☆

The lines \(y = mx + 4\) and \(y = 3x - 2\) intersect at a point with positive \(x\)- and \(y\)-coordinates. Find the complete set of values of \(m\).

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

\(mx + 4 = 3x - 2 \Rightarrow x = \frac{6}{3 - m}\), which is positive iff \(m < 3\). Then \(y = 3x - 2 > 0\) iff \(x > \frac23\), i.e. \(\frac{6}{3-m} > \frac23\); with \(3 - m > 0\) this is \(18 > 6 - 2m\), i.e. \(m > -6\). So \(-6 < m < 3\).

Full method — how and why
Two conditions on the intersection point. Find the intersection in terms of \(m\), then impose positivity on each coordinate.
  1. Find the intersection: \(mx + 4 = 3x - 2 \Rightarrow x(m - 3) = -6 \Rightarrow x = \dfrac{6}{3 - m}\).
    Keep the denominator as \(3 - m\) so that positivity is easy to read.
  2. \(x > 0\) requires \(3 - m > 0\), i.e. \(m < 3\).
    The numerator \(6\) is positive, so the sign of \(x\) is the sign of \(3 - m\).
  3. \(y > 0\): since \(y = 3x - 2\), this needs \(x > \frac23\), i.e. \(\dfrac{6}{3-m} > \dfrac23\).
    Use whichever line is simpler to express \(y\) in terms of \(x\).
  4. With \(3 - m > 0\) we may multiply safely: \(18 > 2(3 - m) \Rightarrow 18 > 6 - 2m \Rightarrow m > -6\).
    Multiplying by a positive quantity is safe - and we already know \(3 - m > 0\) from the first condition.
  5. Combine: \(-6 < m < 3\).
    Both conditions must hold at once.
Where it goes wrong
  • Checking only \(x > 0\) and answering \(m < 3\).
  • Multiplying by \(3 - m\) without noting its sign - legitimate here only because the first condition established it.
Takeaway Impose one condition at a time, and use the first to justify the algebra needed for the second.
Answer C · Straight lines (MM3.1) · question LIN-12
19Exponentials and logarithms · spec MM5.1 · Paper 1 style · ★★☆

The graph of \(y = 2^{x}\) is translated \(3\) units in the positive \(x\)-direction. The resulting graph has equation

  1. A\(y = \dfrac{1}{8}\cdot 2^{x}\)
  2. B\(y = 2^{x/3}\)
  3. C\(y = 8\cdot 2^{x}\)
  4. D\(y = 2^{x} - 3\)
  5. E\(y = 2^{3x}\)
  6. F\(y = 2^{x} + 3\)
Worked solution

\(y = 2^{x - 3} = 2^{-3}\cdot 2^x = \frac18 \cdot 2^x\). (A translation of an exponential is also a stretch.)

Full method — how and why
A translation of an exponential graph is also a stretch. The algebra shows why.
  1. Translating \(3\) right replaces \(x\) by \(x - 3\): \(y = 2^{x-3}\).
    A shift to the right by \(a\) substitutes \(x - a\).
  2. Split the index: \(2^{x-3} = 2^x \times 2^{-3} = \dfrac{1}{8}\cdot 2^x\).
    \(a^{m-n} = \frac{a^m}{a^n}\) - the translation has become multiplication by a constant.
  3. So the new graph is the old one stretched vertically by factor \(\frac18\).
    This is special to exponentials: horizontal translation and vertical stretch are the same thing.
Where it goes wrong
  • Writing \(2^{x-3} = 2^x - 3\) and choosing that option.
  • Using \(2^{+3} = 8\) as the factor, which corresponds to a shift to the left.
Takeaway \(a^{x-k} = a^{-k}a^x\): for exponentials, translating horizontally is the same as stretching vertically.
Answer A · Exponentials and logarithms (MM5.1) · question LOG-06
20Circles and circle theorems · spec MM3.2 · Paper 1 style · ★★☆

The line \(y = x + c\) is a tangent to the circle \((x-1)^2 + (y-2)^2 = 8\). Find the possible values of \(c\).

  1. A\(c = -3\) or \(c = 5\)
  2. B\(c = 1 \pm 2\sqrt{2}\)
  3. C\(c = 1\) or \(c = 5\)
  4. D\(c = -1\) or \(c = 3\)
  5. E\(c = \pm 4\)
  6. F\(c = -5\) or \(c = 3\)
Worked solution

The distance from \((1, 2)\) to \(x - y + c = 0\) is \(\frac{|1 - 2 + c|}{\sqrt2} = \sqrt8 = 2\sqrt2\), so \(|c - 1| = 4\): \(c = 5\) or \(c = -3\).

Full method — how and why
Tangency means the perpendicular distance from the centre to the line equals the radius. Faster than substituting and using the discriminant.
  1. Identify centre \((1,2)\) and radius \(\sqrt8 = 2\sqrt2\).
    Read them straight from the centre-radius form.
  2. Write the line in the form \(x - y + c = 0\) and apply the distance formula: \(\dfrac{|1 - 2 + c|}{\sqrt{1^2 + (-1)^2}} = \dfrac{|c - 1|}{\sqrt2}\).
    Rearranging \(y = x + c\) to \(x - y + c = 0\) gives \(a = 1\), \(b = -1\).
  3. Set it equal to the radius: \(\dfrac{|c-1|}{\sqrt2} = 2\sqrt2\), so \(|c - 1| = 4\).
    Multiply both sides by \(\sqrt2\): \(2\sqrt2 \times \sqrt2 = 4\).
  4. Solve the modulus: \(c - 1 = \pm 4\), so \(c = 5\) or \(c = -3\).
    Two tangents with that gradient exist - one on each side of the circle.
Where it goes wrong
  • Dropping the modulus and finding only \(c = 5\).
  • Using \(r = 8\) instead of \(\sqrt8\).
Takeaway Tangency \(\Leftrightarrow\) distance from centre to line \(=\) radius; the modulus always gives two answers.
Answer A · Circles and circle theorems (MM3.2) · question CIR-05
After submitting, every question shows the correct answer and its worked solution; wrong answers are added to your needs work list and correct ones to secure.