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1
The function \(f(x) = 3\sin(2x) - 1\) is defined for all real \(x\). Find the difference between its maximum and minimum values.
- A\(1\)
- B\(2\)
- C\(4\)
- D\(3\)
- E\(6\)
- F\(7\)
Worked solution
Range is \([-4, 2]\); the difference is \(6\).
Full method — how and why
Range of \(a\sin(bx) + c\). Only \(a\) and \(c\) matter - the \(b\) affects the period, not the values taken.
- \(\sin(2x)\) takes every value in \([-1, 1]\).Compressing horizontally changes how fast the wave oscillates, not how high it goes.
- Multiply by \(3\): the range becomes \([-3, 3]\).Amplitude \(|a| = 3\).
- Subtract \(1\): the range becomes \([-4, 2]\).The whole wave shifts down by \(1\).
- Difference between maximum and minimum: \(2 - (-4) = 6\).Equivalently \(2 \times\) amplitude \(= 2 \times 3 = 6\) - the vertical shift cancels out of a difference.
Where it goes wrong
- Answering \(3\) (the amplitude) instead of the full range \(2a\).
- Letting the \(-1\) affect the difference, giving \(7\) or \(5\).
Takeaway Max minus min for \(a\sin(\cdot) + c\) is always \(2|a|\); vertical shifts cancel.
Answer E · Trigonometric functions and equations (MM4.4) · question TRG-11
2
The trapezium rule with \(2\) strips of equal width is used to estimate \(\displaystyle\int_{0}^{2} x^2\,dx\). Which of the following is true?
- AThe estimate is \(\dfrac{8}{3}\) and it is exact.
- BThe estimate is \(3\) and it is an overestimate.
- CThe estimate is \(\dfrac{5}{2}\) and it is an overestimate.
- DThe estimate is \(\dfrac{5}{2}\) and it is an underestimate.
- EThe estimate is \(3\) and it is an underestimate.
Worked solution
\(h = 1\): \(\frac{h}{2}[y_0 + 2y_1 + y_2] = \frac12[0 + 2(1) + 4] = 3\). The true value is \(\frac83\); the curve bends upwards (convex), so the chords lie above it and the rule overestimates.
Full method — how and why
Apply the trapezium rule and then decide over- or under-estimate from the curve's shape.
- Two strips on \([0,2]\) means \(h = 1\) and ordinates at \(x = 0, 1, 2\): \(y = 0, 1, 4\).\(h = \frac{b-a}{n}\) with \(n\) the number of strips (not the number of ordinates).
- Apply the rule: \(\dfrac h2\left[y_0 + 2y_1 + y_2\right] = \dfrac12\left[0 + 2(1) + 4\right] = 3\).Only the interior ordinates are doubled.
- Compare with the exact value \(\int_0^2 x^2 dx = \frac83 \approx 2.67\).The estimate \(3\) is larger.
- Explain it: \(y = x^2\) bends upwards (convex), so each chord lies above the curve and the trapezia overestimate.Convex \(\Rightarrow\) overestimate; concave \(\Rightarrow\) underestimate. This is the reasoning the question wants.
Where it goes wrong
- Forgetting to double the interior ordinate, giving \(\frac52\) - offered twice as a distractor.
- Saying 'underestimate' because the trapezia look like they cut corners - for a convex curve they do the opposite.
Takeaway Trapezium rule: \(\frac h2[y_0 + 2(\text{middles}) + y_n]\); convex curves are overestimated.
Answer B · Integration (MM7.5) · question INT-06
3
The price of a coat is reduced by \(20\%\) in a sale, and then the sale price is reduced by a further \(25\%\). The final price is what percentage of the original price?
- A\(50\%\)
- B\(55\%\)
- C\(65\%\)
- D\(60\%\)
- E\(40\%\)
- F\(45\%\)
Worked solution
\(0.8 \times 0.75 = 0.6\), so \(60\%\).
Full method — how and why
Successive percentage changes multiply. Convert each to a multiplier and combine.
- A \(20\%\) reduction multiplies by \(0.8\).You keep \(80\%\) of the price: \(1 - 0.20 = 0.8\).
- A further \(25\%\) reduction multiplies by \(0.75\) - applied to the sale price, not the original.The second discount acts on whatever the price already is; that is why the changes multiply.
- Combine: \(0.8 \times 0.75 = 0.6\).Multiplying the multipliers gives the overall multiplier.
- So the final price is \(60\%\) of the original.Equivalently a \(40\%\) reduction overall - not \(45\%\).
Where it goes wrong
- Adding the percentages to get \(45\%\) off, hence \(55\%\) - an offered option.
- Applying the second reduction to the original price.
Takeaway Percentage changes compose by multiplying multipliers, never by adding percentages.
Answer D · Ratio, proportion and percentages (M3.8) · question RAT-01
4
A chord of a circle of radius \(13\) is at perpendicular distance \(5\) from the centre. Find the length of the chord.
- A\(26\)
- B\(\sqrt{144}\)
- C\(18\)
- D\(12\)
- E\(10\)
- F\(24\)
Worked solution
The perpendicular from the centre bisects the chord: half-length \(= \sqrt{13^2 - 5^2} = 12\), so the chord is \(24\).
Full method — how and why
The perpendicular from the centre bisects the chord, creating a right-angled triangle with the radius as hypotenuse.
- Draw the radius to one end of the chord and the perpendicular from the centre: they form a right-angled triangle.The perpendicular distance, half the chord and the radius are the three sides.
- Apply Pythagoras: half-chord \(= \sqrt{13^2 - 5^2} = \sqrt{169 - 25} = \sqrt{144} = 12\).The radius \(13\) is the hypotenuse because it is the longest side.
- Double it: the chord is \(24\).The perpendicular bisects the chord, so the triangle gives only half of it.
Where it goes wrong
- Answering \(12\) by forgetting to double.
- Adding instead of subtracting inside the square root.
Takeaway Perpendicular from centre to chord: it bisects the chord, so use Pythagoras and then double.
Answer F · Circles and circle theorems (MM3.3) · question CIR-08
5
Let \(f(x) = 2x + 1\) and \(g(x) = x^2 - 3\). Find \(f(g(2)) - g(f(2))\).
- A\(-3\)
- B\(-17\)
- C\(-19\)
- D\(3\)
- E\(25\)
- F\(19\)
Worked solution
\(g(2) = 1\), \(f(1) = 3\). \(f(2) = 5\), \(g(5) = 22\). So \(3 - 22 = -19\).
Full method — how and why
Composite functions evaluated at a number. Work strictly from the inside out, and keep the two compositions separate.
- \(g(2) = 2^2 - 3 = 1\), then \(f(1) = 2(1) + 1 = 3\). So \(f(g(2)) = 3\).In \(f(g(2))\) the inner function acts first: feed \(2\) into \(g\), then its output into \(f\).
- \(f(2) = 2(2) + 1 = 5\), then \(g(5) = 25 - 3 = 22\). So \(g(f(2)) = 22\).The other order gives a completely different value - composition is not commutative.
- Subtract in the order asked: \(3 - 22 = -19\).The question is \(f(g(2)) - g(f(2))\), so the small one comes first.
Where it goes wrong
- Computing \(g(f(2)) - f(g(2))\) and answering \(19\).
- Applying the functions in the wrong order inside one composition.
Takeaway \(f(g(x))\) means \(g\) first; check the order twice, because the wrong order is always an offered option.
Answer C · Functions and transformations (MM8.2) · question FUN-03
6
A car travels at \(90\) km/h. What is this speed in metres per second?
- A\(30\) m/s
- B\(54\) m/s
- C\(15\) m/s
- D\(25\) m/s
- E\(324\) m/s
- F\(1.5\) m/s
Worked solution
\(90\,000\) m in \(3600\) s: \(90000/3600 = 25\) m/s.
Full method — how and why
A compound-unit conversion: convert distance and time separately.
- \(90\) km \(= 90 \times 1000 = 90\,000\) m.Convert the distance unit first.
- \(1\) hour \(= 60 \times 60 = 3600\) s.Convert the time unit.
- Divide: \(\dfrac{90\,000}{3600} = 25\) m/s.Equivalently, divide km/h by \(3.6\) - worth memorising.
Where it goes wrong
- Multiplying by \(3.6\) instead of dividing, giving \(324\) - an offered option.
- Converting only one of the two units.
Takeaway km/h \(\div 3.6 =\) m/s; convert numerator and denominator units separately.
Answer D · Number, units and estimation (M1.2) · question NUM-07
7
Given that \(\log_a x = 3\) and \(\log_a y = 5\), find \(\log_a \left(\dfrac{x^2 \sqrt{y}}{a}\right)\).
- A\(8\)
- B\(\dfrac{15}{2}\)
- C\(\dfrac{13}{2}\)
- D\(\dfrac{17}{2}\)
- E\(10\)
- F\(\dfrac{11}{2}\)
Worked solution
\(2\log_a x + \frac12\log_a y - \log_a a = 6 + \frac52 - 1 = \frac{15}{2}\).
Full method — how and why
Break the expression apart with the log laws, then substitute the given values.
- Split the quotient and product: \(\log_a\dfrac{x^2\sqrt y}{a} = \log_a x^2 + \log_a \sqrt y - \log_a a\).Products become sums, quotients become differences.
- Bring the powers down: \(2\log_a x + \frac12\log_a y - \log_a a\).\(\sqrt y = y^{1/2}\), so its log is \(\frac12\log_a y\).
- Substitute \(\log_a x = 3\), \(\log_a y = 5\), \(\log_a a = 1\): \(6 + \frac52 - 1\).\(\log_a a = 1\) always - a fact worth having instantly available.
- Add: \(\frac{12 + 5 - 2}{2} = \frac{15}{2}\).Put everything over \(2\).
Where it goes wrong
- Forgetting \(\log_a a = 1\) and answering \(\frac{17}{2}\).
- Using \(\log_a\sqrt y = \sqrt{\log_a y}\).
Takeaway Expand fully with the three laws before substituting; \(\log_a a = 1\) and \(\log_a 1 = 0\).
Answer B · Exponentials and logarithms (MM5.2) · question LOG-04
8
If \(x = 2^{p}\), \(y = 2^{q}\) and \(\dfrac{x^{3}}{y^{2}} = 2^{5}\) with \(xy = 2^{10}\), find \(p\).
- A\(3\)
- B\(7\)
- C\(8\)
- D\(4\)
- E\(6\)
- F\(5\)
Worked solution
\(3p - 2q = 5\) and \(p + q = 10\). From the second, \(q = 10 - p\); then \(3p - 20 + 2p = 5\), so \(5p = 25\) and \(p = 5\).
Full method — how and why
Two equations in disguise. Writing both conditions in base \(2\) turns them into simultaneous linear equations in \(p\) and \(q\).
- \(\dfrac{x^3}{y^2} = \dfrac{2^{3p}}{2^{2q}} = 2^{3p - 2q}\), so \(3p - 2q = 5\).Powers of powers multiply indices, division subtracts them; equating indices is valid because base \(2\) is one-to-one.
- \(xy = 2^{p}\cdot 2^{q} = 2^{p+q}\), so \(p + q = 10\).Multiplication adds indices.
- Solve: from the second, \(q = 10 - p\); substituting, \(3p - 2(10 - p) = 5 \Rightarrow 5p = 25 \Rightarrow p = 5\).Substitution is quickest with only two unknowns; note the bracket when substituting a negative.
Where it goes wrong
- Dropping the bracket to get \(3p - 20 - 2p = 5\) and hence \(p = 25\).
- Equating \(x^3/y^2 = 2^5\) to \(3p/2q = 5\) - indices subtract, they do not divide.
Takeaway Converting everything to one base turns exponential conditions into linear equations in the exponents.
Answer F · Indices and surds (MM1.1) · question IND-21
9
Simplify \(\dfrac{1 - \cos^2\theta}{\cos\theta \tan\theta}\) for values of \(\theta\) where the expression is defined.
- A\(1\)
- B\(\sin\theta\)
- C\(\tan\theta\)
- D\(\sin^2\theta\)
- E\(\cos\theta\)
- F\(\dfrac{\sin\theta}{\cos\theta}\)
Worked solution
\(\cos\theta\tan\theta = \sin\theta\), and \(1 - \cos^2\theta = \sin^2\theta\). So the expression is \(\sin\theta\).
Full method — how and why
Simplification using the two permitted identities. Work on the denominator and numerator separately.
- Denominator: \(\cos\theta\tan\theta = \cos\theta \times \dfrac{\sin\theta}{\cos\theta} = \sin\theta\).\(\tan\theta = \frac{\sin\theta}{\cos\theta}\), and the \(\cos\theta\) cancels (legitimate where the expression is defined).
- Numerator: \(1 - \cos^2\theta = \sin^2\theta\).Rearranged Pythagorean identity.
- Divide: \(\dfrac{\sin^2\theta}{\sin\theta} = \sin\theta\).One power of \(\sin\theta\) cancels.
Where it goes wrong
- Cancelling to \(\sin^2\theta\) by forgetting to simplify the denominator.
- Writing \(1 - \cos^2\theta = \cos^2\theta - 1\) (sign flip), giving \(-\sin\theta\).
Takeaway Convert \(\tan\) into \(\frac{\sin}{\cos}\) and use \(1 - \cos^2 = \sin^2\); then cancel.
Answer B · Trigonometric functions and equations (MM4.5) · question TRG-09
10
Which of the following could be the equation of a cubic curve that touches the \(x\)-axis at \(x = 2\) and crosses it at \(x = -1\), and has a negative \(y\)-intercept?
- A\(y = -(x - 2)^2(x + 1)\)
- B\(y = (x - 2)(x + 1)^2\)
- C\(y = (x + 2)^2(x - 1)\)
- D\(y = -(x + 2)^2(x - 1)\)
- E\(y = (x - 2)^2(x + 1)\)
Worked solution
Touching at \(2\) needs the factor \((x - 2)^2\) and crossing at \(-1\) the factor \((x + 1)\). \(y = (x-2)^2(x+1)\) has \(y\)-intercept \(4 > 0\), so the sign is reversed: \(y = -(x-2)^2(x+1)\) with intercept \(-4\).
Full method — how and why
Build the cubic from its root behaviour, then use the \(y\)-intercept to fix the sign.
- 'Touches at \(x = 2\)' means a repeated factor \((x-2)^2\); 'crosses at \(x = -1\)' means a simple factor \((x+1)\).Even-power factors touch the axis; odd-power factors cross it.
- So the cubic is \(y = a(x-2)^2(x+1)\) for some constant \(a \neq 0\).Degree \(2 + 1 = 3\) ✓.
- Compute the \(y\)-intercept: at \(x = 0\), \(y = a(4)(1) = 4a\).Substitute \(x = 0\) into the factorised form.
- A negative intercept needs \(a < 0\), so \(y = -(x-2)^2(x+1)\) works (intercept \(-4\)).The option with \(a = +1\) has intercept \(+4\) and is the trap.
Where it goes wrong
- Choosing \((x-2)^2(x+1)\) and ignoring the sign of the intercept.
- Swapping which root is repeated, e.g. \((x-2)(x+1)^2\).
Takeaway Repeated factor \(\Rightarrow\) touch; simple factor \(\Rightarrow\) cross; the \(y\)-intercept fixes the leading sign.
Answer A · Graphs of functions (MM8.1) · question GRA-03
11
The curve \(y = x^2 - 2ax + a^2 + a - 6\) has its vertex on the \(x\)-axis. Find the value of \(a\).
- A\(0\)
- B\(6\)
- C\(-6\)
- D\(3\)
- E\(2\)
- F\(-2\)
Worked solution
Complete the square: \(y = (x-a)^2 + a - 6\). The vertex is \((a, a-6)\), which lies on the \(x\)-axis when \(a = 6\).
Full method — how and why
'Vertex on the \(x\)-axis' means the minimum value is zero, i.e. the quadratic has a repeated root. Completing the square is quicker than the discriminant here.
- Complete the square in \(x\), treating \(a\) as a constant: \(x^2 - 2ax + a^2 = (x - a)^2\), so \(y = (x-a)^2 + (a - 6)\).The first three terms are already a perfect square - that is why the question is built with \(a^2\) in it.
- Read off the vertex: \((a,\; a - 6)\).In \((x - h)^2 + k\) the vertex is \((h, k)\).
- On the \(x\)-axis means the \(y\)-coordinate is zero: \(a - 6 = 0\), so \(a = 6\).'On the \(x\)-axis' is a statement about \(y\), not \(x\) - the commonest misread in this question.
Where it goes wrong
- Setting the \(x\)-coordinate to zero and answering \(a = 0\).
- Using the discriminant but mis-expanding \(4a^2 - 4(a^2 + a - 6)\), which should collapse to \(-4(a - 6)\).
Takeaway Vertex on the \(x\)-axis \(\Leftrightarrow\) repeated root \(\Leftrightarrow\) discriminant zero - three descriptions of the same situation.
Answer B · Quadratic functions (MM1.3) · question QUA-11
12
In triangle \(ABC\), \(D\) lies on \(AB\) and \(E\) lies on \(AC\) with \(DE\) parallel to \(BC\). \(AD = 4\), \(DB = 6\) and \(DE = 5\). Find \(BC\).
- A\(12.5\)
- B\(8\)
- C\(15\)
- D\(10\)
- E\(7.5\)
- F\(12\)
Worked solution
Triangles \(ADE\) and \(ABC\) are similar with ratio \(AD : AB = 4 : 10\). \(BC = 5 \times \frac{10}{4} = 12.5\).
Full method — how and why
A line parallel to one side creates a similar triangle. Use the full side, not just the part.
- \(DE \parallel BC\) makes triangles \(ADE\) and \(ABC\) similar (equal corresponding angles).Parallel lines give equal corresponding angles, which is the AAA similarity condition.
- Compute the ratio: \(AB = AD + DB = 4 + 6 = 10\), so \(AD : AB = 4 : 10 = 2 : 5\).The ratio must compare corresponding whole sides - \(AD\) to \(AB\), not \(AD\) to \(DB\).
- Scale up \(DE\): \(BC = 5 \times \dfrac{10}{4} = 12.5\).Multiply by the reciprocal of the ratio to go from small to large.
Where it goes wrong
- Using the ratio \(4 : 6\) and answering \(7.5\) - an offered option.
- Dividing instead of multiplying, giving \(2\).
Takeaway In similar triangles compare corresponding full sides: \(\frac{AD}{AB}\), not \(\frac{AD}{DB}\).
Answer A · Angles, polygons, congruence and vectors (M5.5) · question GEO-14
13
Find the set of values of \(x\) satisfying both \(3x - 7 < 2\) and \(x^2 > 4\).
- A\(x < -2\) or \(2 < x < 3\)
- B\(2 < x < 3\)
- C\(-2 < x < 3\)
- D\(x < 3\)
- E\(x > 2\)
- F\(x < -2\)
Worked solution
\(3x < 9 \Rightarrow x < 3\). \(x^2 > 4 \Rightarrow x < -2\) or \(x > 2\). Combining: \(x < -2\) or \(2 < x < 3\).
Full method — how and why
Two inequalities joined by 'and'. Solve separately, then take the overlap - the second one produces two intervals, which is where the care is needed.
- Solve the linear one: \(3x - 7 < 2 \Rightarrow 3x < 9 \Rightarrow x < 3\).Dividing by a positive number leaves the inequality direction unchanged.
- Solve the quadratic one: \(x^2 > 4 \Rightarrow |x| > 2 \Rightarrow x < -2\) or \(x > 2\).\(x^2 > 4\) is satisfied by large positive and large negative \(x\) - taking only the positive branch is the classic error.
- Intersect the sets on a number line: from \(x < 3\), keep the parts of \(x < -2\) (all of it) and of \(x > 2\) (only up to \(3\)).'Both' means intersection; drawing both solution sets on a single line prevents mistakes.
- Answer: \(x < -2\) or \(2 < x < 3\).The answer is itself two intervals - a perfectly legitimate solution set.
Where it goes wrong
- Dropping the branch \(x < -2\) and answering \(2 < x < 3\).
- Solving \(x^2 > 4\) as \(x > \pm 2\), which is not a meaningful statement.
Takeaway \(x^2 > k\) gives two intervals; intersect solution sets on a number line rather than in your head.
Answer A · Simultaneous equations and inequalities (MM1.5) · question SIM-03
14
A sequence satisfies \(x_{n+1} = \dfrac{1}{1 - x_n}\) with \(x_1 = 2\). Find \(x_{100}\).
- A\(1\)
- B\(2\)
- C\(\dfrac{1}{2}\)
- D\(-\dfrac12\)
- E\(0\)
- F\(-1\)
Worked solution
\(x_2 = \frac{1}{1-2} = -1\), \(x_3 = \frac{1}{1-(-1)} = \frac12\), \(x_4 = \frac{1}{1 - 1/2} = 2 = x_1\): the sequence has period \(3\). Since \(100 = 3\times33 + 1\), \(x_{100} = x_1 = 2\).
Full method — how and why
A recurrence with a huge index. Compute a few terms, spot the cycle, then use arithmetic on the index.
- Iterate: \(x_2 = \frac{1}{1-2} = -1\), \(x_3 = \frac{1}{1-(-1)} = \frac12\), \(x_4 = \frac{1}{1 - \frac12} = 2\).When an index like \(100\) appears, always compute the first few terms - the sequence is almost certainly periodic.
- \(x_4 = x_1\), so the sequence repeats with period \(3\): \(2, -1, \frac12, 2, -1, \frac12, \ldots\)Once a value repeats, the whole sequence from that point repeats identically.
- Reduce the index: \(100 = 3 \times 33 + 1\), so \(x_{100}\) is in the same position of the cycle as \(x_1\).Terms with indices differing by a multiple of the period are equal; compare remainders on division by \(3\).
- Therefore \(x_{100} = x_1 = 2\).Check the convention: the indices \(1, 4, 7, \ldots\) all leave remainder \(1\) on division by \(3\), and \(100\) does too.
Where it goes wrong
- Finding that \(100\) leaves remainder \(1\) but then reading off \(x_0\) or \(x_3\) - line the remainders up with the actual starting index.
- Giving up after two terms and guessing \(-1\) or \(\frac12\).
Takeaway Large-index recurrence questions are periodicity questions: find the cycle length, then work with the index modulo it.
Answer B · Sequences and series (MM2.1) · question SEQ-07
15
The point \(P(x, y)\) moves so that its distance from \((3, 0)\) is twice its distance from \((0, 0)\). The locus of \(P\) is a circle. Find its radius.
- A\(\sqrt{2}\)
- B\(3\)
- C\(2\)
- D\(4\)
- E\(\sqrt{3}\)
- F\(1\)
Worked solution
\((x-3)^2 + y^2 = 4(x^2 + y^2) \Rightarrow 3x^2 + 3y^2 + 6x - 9 = 0 \Rightarrow x^2 + 2x + y^2 = 3 \Rightarrow (x+1)^2 + y^2 = 4\). Radius \(2\).
Full method — how and why
A locus defined by a distance ratio. Write the condition algebraically, square to remove roots, and complete the square.
- Write the condition: \(\sqrt{(x-3)^2 + y^2} = 2\sqrt{x^2 + y^2}\).'Distance from \(A\) is twice the distance from \(B\)' - write both distances with the formula.
- Square both sides: \((x-3)^2 + y^2 = 4(x^2 + y^2)\).Squaring is safe because both sides are non-negative; and it removes both roots at once. Note the right-hand side gains a factor \(4\), not \(2\).
- Expand and collect: \(x^2 - 6x + 9 + y^2 = 4x^2 + 4y^2 \Rightarrow 3x^2 + 3y^2 + 6x - 9 = 0\).Bring everything to one side.
- Divide by \(3\) and complete the square: \(x^2 + 2x + y^2 = 3 \Rightarrow (x+1)^2 + y^2 = 4\), so \(r = 2\).Dividing by the common factor first makes the coefficients of \(x^2\) and \(y^2\) equal to \(1\) - necessary before completing the square.
Where it goes wrong
- Squaring the right side as \(2(x^2+y^2)\) instead of \(4(x^2+y^2)\).
- Forgetting to divide by \(3\) before completing the square, which corrupts the radius.
Takeaway Distance-ratio loci (Apollonius circles): square the condition, divide so \(x^2\) and \(y^2\) have coefficient \(1\), then complete the square.
Answer C · Circles and circle theorems (MM3.2) · question CIR-16
16
Simplify \(\dfrac{x^2 - 9}{x^2 + x - 12} \times \dfrac{x^2 - 16}{x^2 - 3x}\).
- A\(\dfrac{(x + 3)(x - 4)}{x(x - 3)}\)
- B\(\dfrac{x - 4}{x - 3}\)
- C\(\dfrac{x + 3}{x}\)
- D\(\dfrac{x - 4}{x}\)
- E\(\dfrac{(x+3)(x+4)}{x(x-3)}\)
- F\(\dfrac{(x - 3)(x - 4)}{x(x + 3)}\)
Worked solution
Factorise everything: \(\dfrac{(x-3)(x+3)}{(x+4)(x-3)} \times \dfrac{(x-4)(x+4)}{x(x-3)}\). The \((x+4)\) factors cancel and one \((x-3)\) cancels, leaving \(\dfrac{(x+3)(x-4)}{x(x-3)}\).
Full method — how and why
Multiplying algebraic fractions: factorise everything first, then cancel common factors across the product.
- Factorise each part: \(x^2-9 = (x-3)(x+3)\); \(x^2+x-12 = (x+4)(x-3)\); \(x^2-16 = (x-4)(x+4)\); \(x^2-3x = x(x-3)\).Nothing can be cancelled until everything is in factor form.
- Write the product: \(\dfrac{(x-3)(x+3)}{(x+4)(x-3)} \times \dfrac{(x-4)(x+4)}{x(x-3)}\).In a product of fractions any top factor can cancel with any bottom factor.
- Cancel: the \((x+4)\) pair, and one \((x-3)\) from the first numerator with the first denominator.There are two \((x-3)\)s on the bottom but only one on top, so one survives below.
- What remains: \(\dfrac{(x+3)(x-4)}{x(x-3)}\).Count the leftovers carefully - this is where marks are lost.
Where it goes wrong
- Cancelling both \((x-3)\) factors and losing one from the denominator.
- Cancelling terms rather than factors, e.g. striking out the \(x^2\)s.
Takeaway Factorise fully, then cancel factor-by-factor and count how many copies of each remain.
Answer A · GCSE algebra (M4.6) · question ALG-03
17
Given \(f(x) = x^{\frac{3}{2}} - 6x^{\frac{1}{2}}\) for \(x > 0\), find the value of \(x\) at which \(f\) has its minimum.
- A\(2\)
- B\(4\)
- C\(1\)
- D\(6\)
- E\(\sqrt{2}\)
- F\(3\)
Worked solution
\(f'(x) = \frac32 x^{1/2} - 3x^{-1/2} = \frac{3}{2\sqrt x}(x - 2) = 0\) at \(x = 2\); \(f'\) changes from negative to positive there, so it is a minimum.
Full method — how and why
Differentiating fractional powers. Factorising the derivative makes the sign change easy to see.
- Differentiate: \(f'(x) = \frac32 x^{1/2} - 3 \times \frac12 x^{-1/2} = \frac32 x^{1/2} - 3x^{-1/2}\).\(\frac{d}{dx}x^{3/2} = \frac32 x^{1/2}\) and \(\frac{d}{dx}(6x^{1/2}) = 3x^{-1/2}\).
- Factor out \(\dfrac{3}{2\sqrt x}\): \(f'(x) = \dfrac{3}{2\sqrt x}\left(x - 2\right)\).\(\frac32 x^{1/2} = \frac{3}{2\sqrt x}\cdot x\) and \(3x^{-1/2} = \frac{3}{2\sqrt x}\cdot 2\). Factorising isolates the sign-changing part.
- Set \(f' = 0\): since \(\frac{3}{2\sqrt x} > 0\) for \(x > 0\), the only root is \(x = 2\).A positive factor can never make the product zero.
- Check it is a minimum: \(f'\) is negative for \(x < 2\) and positive for \(x > 2\), so the curve falls then rises.The sign of \(f'\) is the sign of \((x-2)\) - immediate from the factorised form.
Where it goes wrong
- Making index errors: \(\frac{d}{dx}x^{1/2} = \frac12 x^{-1/2}\), not \(\frac12 x^{1/2}\).
- Squaring both sides carelessly when solving \(\frac32\sqrt x = \frac{3}{\sqrt x}\) and getting \(x = 4\).
Takeaway Factor a common power out of the derivative: the remaining bracket controls the sign.
Answer A · Differentiation (MM6.2) · question DIF-08
18
For how many integer values of \(k\) does \(x^2 + kx + 12 = 0\) have integer roots?
- A\(8\)
- B\(4\)
- C\(2\)
- D\(12\)
- E\(3\)
- F\(6\)
Worked solution
Integer roots \(r, s\) with \(rs = 12\): pairs \((1,12), (2,6), (3,4)\) and their negatives, giving \(k = -(r+s) = -13, -8, -7, 13, 8, 7\): six values.
Full method — how and why
Integer roots means an integer factorisation of the constant term. Count systematically and do not forget negative pairs.
- If the roots are integers \(r\) and \(s\) then \(rs = 12\) and \(r + s = -k\).For a monic quadratic, product of roots \(= c\) and sum \(= -b\).
- List the positive factor pairs of \(12\): \((1, 12), (2, 6), (3, 4)\).Work through divisors in order so none is missed.
- Each pair also has a negative version \((-1, -12)\), \((-2, -6)\), \((-3, -4)\), since two negatives also multiply to \(+12\).The product is positive, so the roots share a sign - both positive or both negative.
- Each of the six pairs gives a different \(k = -(r+s)\): \(-13, -8, -7, 13, 8, 7\). So six values.Check they are distinct before counting.
Where it goes wrong
- Listing only the positive pairs and answering \(3\).
- Counting \((1,12)\) and \((12,1)\) as different, which double-counts.
Takeaway Integer-root problems become divisor-counting problems; the sign of the product tells you whether the roots share a sign.
Answer F · Quadratic functions (MM1.3) · question QUA-20
19
The trapezium rule is used to estimate \(\displaystyle\int_{a}^{b} f(x)\,dx\). For which of the following functions is the estimate guaranteed to be an underestimate for every choice of \(a < b\)?
- A\(f(x) = -x^2\)
- B\(f(x) = x^4\)
- C\(f(x) = 2x + 1\)
- D\(f(x) = x^2\)
- E\(f(x) = x^3\)
Worked solution
The trapezium rule underestimates when the curve is concave (bends downwards) so that chords lie below the curve: \(y = -x^2\) everywhere. \(x^3\) changes concavity at \(0\).
Full method — how and why
A conceptual question about the trapezium rule: it is the curve's concavity, not its sign or steepness, that decides.
- Recall the rule: the trapezium rule replaces the curve by straight chords.Each strip's true area is compared with a trapezium whose top is the chord.
- A chord lies below a concave (downward-bending) curve, so the estimate is too small: an underestimate.Concave means \(f'' < 0\) throughout - the curve bulges above its chords.
- Check the options: \(-x^2\) has \(f'' = -2 < 0\) everywhere ✓. \(x^2\) and \(x^4\) are convex (overestimate); \(2x+1\) is straight (exact); \(x^3\) changes concavity at \(0\), so it depends on the interval.'Guaranteed for every choice of \(a < b\)' rules out \(x^3\), whose behaviour varies.
- So \(f(x) = -x^2\).The only universally concave option.
Where it goes wrong
- Choosing \(x^3\) without noticing that it is concave for \(x<0\) but convex for \(x>0\).
- Thinking a negative function must give an underestimate - sign is irrelevant; shape is everything.
Takeaway Trapezium rule: concave (\(f''<0\)) underestimates, convex (\(f''>0\)) overestimates, linear is exact.
Answer A · Integration (MM7.5) · question INT-14
20
The coefficient of \(x^2\) in \((1 + x)^n + (1 + x)^{n+1}\) is \(36\). Find \(n\).
- A\(4\)
- B\(8\)
- C\(6\)
- D\(5\)
- E\(9\)
- F\(7\)
Worked solution
\(\binom n2 + \binom{n+1}2 = \frac{n(n-1) + (n+1)n}{2} = n^2 = 36\), so \(n = 6\).
Full method — how and why
Two expansions added. Write both coefficients in factorial form and the sum simplifies remarkably.
- The \(x^2\) coefficients are \(\binom n2\) and \(\binom{n+1}{2}\).One from each expansion; the total coefficient of \(x^2\) is their sum.
- Write them out: \(\dfrac{n(n-1)}{2} + \dfrac{(n+1)n}{2} = \dfrac{n\left[(n-1) + (n+1)\right]}{2}\).Factor out \(\frac n2\) - the brackets then collapse.
- Simplify: \(\dfrac{n \times 2n}{2} = n^2\).A surprisingly clean result; recognising it makes the final step trivial.
- Solve \(n^2 = 36\), so \(n = 6\) (positive integer).\(n\) counts an index, so the negative root is rejected.
Where it goes wrong
- Using Pascal's rule to write the sum as \(\binom{n+1}{3}\) - that rule adds adjacent entries in the same row, not the same entry in different rows.
- Taking \(n = \pm 6\) and offering \(-6\).
Takeaway \(\binom n2 + \binom{n+1}{2} = n^2\): write coefficients in factorial form and look for common factors before solving.
Answer C · Binomial expansion (MM2.4) · question BIN-15
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