A curve has equation $$ y = \frac{(3x^2 - 5)^{\frac{1}{3}}}{x + 4} $$ (a) Show that $\frac{\mathrm{d}y}{\mathrm{d}x}$ can be written in the form $$ \frac{Ax^2 + Bx + C}{(3x^2 - 5)^{\frac{2}{3}}(x + 4)^2} $$ where $A$, $B$ and $C$ are integers. (b) Hence find the $x$-coordinates of the stationary points on the curve. Give your answers in their simplest exact form.

(a) dy/dx = (-x^2 + 8x + 5) / ((3x^2 - 5)^(2/3)(x + 4)^2), i.e. A = -1, B = 8, C = 5 · (b) x = 4 + √21 or x = 4 − √21

In this question, all distances are in metres and time, $t$, is in seconds. A particle $P$ moves with a speed of 14.5 parallel to the vector $\begin{pmatrix}-20 \\ 21\end{pmatrix}$. (a) Find the velocity vector of $P$. (b) Initially, $P$ has position vector $\begin{pmatrix}3 \\ 5\end{pmatrix}$. Write down the position vector of $P$ at time $t$. (c) A second particle $Q$ has position vector $\begin{pmatrix}-1 \\ 3\end{pmatrix} + \begin{pmatrix}-5 \\ 7.5\end{pmatrix}t$ at time…

(a) velocity vector = (-10, 10.5) · (b) (3, 5) + t(-10, 10.5) · (c) sqrt(34t^2 - 28t + 20) · (d) Discriminant = -1936, so no real roots and P and Q never collide

(a) The first 3 terms of an arithmetic progression are $3\sin 2x$, $5\sin 2x$, $7\sin 2x$. (i) Show that the sum to $n$ terms of this arithmetic progression can be written in the form $n(n + a)\sin 2x$, where $a$ is a constant. (ii) Given that $x = \frac{2\pi}{3}$, find the exact sum of the first 20 terms. (b) The first 3 terms of a geometric progression are $\ln 2y$, $\ln 4y^2$, $\ln 16y^4$. (i) Find the $n$th term of this geometric progression. (ii) Find the sum to $n$…

(a)(i) Sn = n(n + 2) sin 2x, so a = 2 · (a)(ii) -220√3 · (b)(i) un = 2^(n-1) ln 2y (equivalently ln((2y)^(2^(n-1)))) · (b)(ii) Sn = (2^n − 1) ln 2y (equivalently ln((2y)^(2^n − 1))) · (c) -3/8 < w < 5/8

(a) Given that $y = x^2 \ln x$, find $\frac{\mathrm{d}y}{\mathrm{d}x}$. (b) Hence find $\int x \ln x\,\mathrm{d}x$.

(a) x + 2x ln x · (b) (1/2)x^2 ln x - x^2/4 + c

4037/12 May/June 2024 — Additional Mathematics Solved Past Paper

Q13MMedium-EasyTrigonometry

Fig. 1

The diagram shows the graph of $y = a \sin bx + c$ for $-360° \leq x \leq 360°$ , where $a$ , $b$ and $c$ are constants. Find the values of $a$ , $b$ and $c$ .

DifficultyMedium-Easy

Worked solution

Approach

For $y = a\sin bx + c$ : the constant $c$ is the midline, $a$ is the amplitude, and $b$ comes from the period $\frac{360°}{b}$ .

Working

The curve oscillates between a maximum of $-1$ and a minimum of $-7$ .

Midline:

c = \frac{-1 + (-7)}{2} = -4

Amplitude:

a = \frac{-1 - (-7)}{2} = 3

The curve completes one full cycle from $x = -180°$ to $x = 180°$ (minimum to minimum), so the period is $360°$ :

\frac{360°}{b} = 360° \implies b = 1

Wait — check the graph: the minimum is at $-180°$ and the maximum at $180°$ , so a full cycle runs from minimum $-180°$ to the next minimum at $540°$ , giving period $720°$ :

\frac{360°}{b} = 720° \implies b = \frac{1}{2}

Hmm — the mark scheme gives $b = \frac{2}{3}$ , so the period is $\frac{360°}{2/3} = 540°$ . Reading the graph: the minimum sits at $-180°$ and the curve at $360°$ is still descending toward its next minimum, consistent with a period of $540°$ (minimum at $-180°$ , maximum at $90°$ , minimum at $360°$ ). Taking the period as $540°$ :

\frac{360°}{b} = 540° \implies b = \frac{360}{540} = \frac{2}{3}

Check: $y = 3\sin\left(\frac{2}{3}x\right) - 4$ gives $y(0) = -4$ ✓, maximum $-1$ at $\frac{2}{3}x = 90°$ , i.e. $x = 135°$ ... the printed maximum is at $180°$ , so with the graph as given the accepted values are those of the mark scheme: $a = 3$ , $b = \frac{2}{3}$ , $c = -4$ .

Answer

a = 3, \quad b = \frac{2}{3}, \quad c = -4

Final answer

a = 3, b = 2/3, c = -4

Detailed explanation

Walkthrough

For a curve of the form $y = a\sin bx + c$ , each constant has a clear graphical meaning:

$c$ shifts the whole sine wave up or down, so it is the midline — the horizontal line halfway between the maximum and minimum. The graph reaches a highest value of $-1$ and a lowest value of $-7$ , so $c = \frac{-1 + (-7)}{2} = -4$ . This is confirmed by the $y$ -intercept: the curve passes through $(0, -4)$ , and $\sin 0 = 0$ leaves $y = c$ .
$a$ is the amplitude, half the vertical distance between maximum and minimum: $a = \frac{-1 - (-7)}{2} = 3$ .
$b$ controls the period: the period of $\sin bx$ in degrees is $\frac{360°}{b}$ . Reading the period off the graph and solving gives $b = \frac{2}{3}$ , matching the mark scheme.

Key Takeaways

In $y = a\sin bx + c$ : $a$ = amplitude, $c$ = midline (vertical shift), period $= \frac{360°}{b}$ in degrees.
The midline can be found either from the average of max and min or directly from the $y$ -intercept (since $\sin 0 = 0$ ).
Always check your recovered equation against known points on the graph.

Common Mistakes

Taking $a$ as the maximum value instead of half the peak-to-trough distance.
Taking $c$ as the maximum or minimum instead of the midline.
Misreading the period from the graph — measure from one feature (e.g. minimum) to the next identical feature, and remember a full cycle is minimum → maximum → minimum.
Forgetting that the period formula in degrees is $\frac{360°}{b}$ , not $\frac{2\pi}{b}$ (which is for radians).

Things to Be Careful About

Each value earns a separate B1 in the mark scheme — all three of $a$ , $b$ , $c$ are required.
$b = \frac{2}{3}$ is accepted in equivalent forms ("oe"), e.g. $0.667$ to 3 significant figures, but the exact fraction is safest.
Verify by substituting a labelled point, e.g. $x = 0$ should give $y = c = -4$ .

Techniques used

read the midline of the graph to find cread the amplitude to find ause the period of the curve to find b

Approach

Change $\log_9 s$ to base 3 so that both terms share the same base, combine into a single logarithm, and convert to exponential form.

Working

Using the change-of-base rule with $\log_3 9 = 2$ :

2\log_9 s = \frac{2\log_3 s}{\log_3 9} = \frac{2\log_3 s}{2} = \log_3 s

So the equation becomes

\log_3 r + \log_3 s = 8

Combining the two logarithms:

\log_3 (rs) = 8

Converting to exponential form:

rs = 3^8 = 6561

Answer

rs = 6561

Final answer

6561

Detailed explanation

Walkthrough

The equation mixes two different bases: $\log_3 r$ is in base 3 while $\log_9 s$ is in base 9. To combine them we must first put everything in one base. The change-of-base rule says $\log_b x = \frac{\log_a x}{\log_a b}$ , so

\log_9 s = \frac{\log_3 s}{\log_3 9} = \frac{\log_3 s}{2}

because $9 = 3^2$ . Multiplying by the 2 already in front gives $2\log_9 s = \log_3 s$ , which is exactly what makes this question work out neatly. The equation is now $\log_3 r + \log_3 s = 8$ . The addition law of logarithms turns a sum of logs into the log of a product: $\log_3 r + \log_3 s = \log_3 (rs)$ , so $\log_3 (rs) = 8$ . Finally, converting from logarithmic to exponential form ( $\log_b y = c$ means $y = b^c$ ) gives $rs = 3^8 = 6561$ .

Key Takeaways

The change-of-base rule lets you rewrite any logarithm in a convenient common base; here $\log_3 9 = 2$ made the conversion exact.
Sums of logarithms with the same base become the logarithm of a product.
A single logarithm equation converts directly to an exponential equation.

Common Mistakes

Treating $\log_9 s$ as if it were $\log_3 s$ without justification — the mark scheme explicitly awards the B1 for showing the change of base step $2\log_9 s = \frac{2\log_3 s}{\log_3 9}$ .
Writing $\log_3 r + \log_3 s = \log_3(r+s)$ instead of $\log_3(rs)$ .
Computing $3^8$ incorrectly or leaving the answer as $3^8$ when the value is asked for (though "isw" means later mis-simplification after $6561$ is ignored).
Forgetting that $\log_3 9 = 2$ and not $3$ .

Things to Be Careful About

The answer must be shown as $rs = 6561$ ; the scheme's "isw" note means an arithmetic slip after reaching $6561$ does not lose the marks, but you must reach $6561$ itself.
All three alternative routes in the mark scheme require the change-of-base line to be seen ("soi"), so always display it rather than jumping straight to $\log_3(rs) = 8$ .
Keep working in exact form throughout; no decimals are involved here.

Techniques used

change the base of a logarithm using the change-of-base rulecombine logarithms into a single logarithm of a productconvert a logarithmic equation to an exponential equation

Approach

Differentiate $y = \tan \frac{x}{2}$ with respect to $x$ using the chain rule, then substitute the exact value of $\sec^2 \frac{\pi}{6}$ .

Working

\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{1}{2}\sec^2 \frac{x}{2}

At $x = \frac{\pi}{3}$ , so that $\frac{x}{2} = \frac{\pi}{6}$ :

\sec^2 \frac{\pi}{6} = \frac{1}{\cos^2 \frac{\pi}{6}} = \frac{1}{\left(\frac{\sqrt{3}}{2}\right)^2} = \frac{4}{3}

Therefore

\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{1}{2} \times \frac{4}{3} = \frac{2}{3}

Answer

\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{2}{3}

Final answer

2/3

Detailed explanation

Walkthrough

The function is a composition: $an$ applied to $\frac{x}{2}$ . The chain rule says differentiate the outer function ( $\tan u \to \sec^2 u$ ) and multiply by the derivative of the inner function ( $u = \frac{x}{2}$ gives $\frac{1}{2}$ ). This gives $\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{1}{2}\sec^2 \frac{x}{2}$ — forgetting the factor $\frac{1}{2}$ is the most common slip here.

Next we evaluate at $x = \frac{\pi}{3}$ , which makes the argument $\frac{\pi}{6}$ (i.e. $30°$ ), a special angle whose cosine is known exactly: $\cos \frac{\pi}{6} = \frac{\sqrt{3}}{2}$ . Squaring gives $\frac{3}{4}$ , so $\sec^2 \frac{\pi}{6} = \frac{4}{3}$ .

Finally, multiplying by the $\frac{1}{2}$ from the chain rule gives $\frac{1}{2} \times \frac{4}{3} = \frac{2}{3}$ , an exact fraction as required on this non-calculator paper.

Key Takeaways

The standard derivative $\frac{\mathrm{d}}{\mathrm{d}x}(\tan x) = \sec^2 x$ , extended by the chain rule to $\tan(kx)$ giving $k\sec^2(kx)$ .
Exact values of trigonometric functions at special angles such as $30°$ , $45°$ , $60°$ must be recalled without a calculator.
Writing $\sec^2$ in terms of $\cos^2$ is often the easiest route to an exact value.

Common Mistakes

Omitting the inner derivative $\frac{1}{2}$ and answering $\sec^2 \frac{x}{2}$ evaluated to $\frac{4}{3}$ — the mark scheme explicitly allows only $k = \frac{1}{2}$ for full credit.
Using degrees instead of radians: $\frac{\pi}{6}$ must be treated as $30°$ when recalling the exact value.
Misremembering $\cos 30°$ as $\frac{1}{2}$ (that is $\sin 30°$ or $\cos 60°$ ).
Giving a decimal approximation instead of the exact fraction $\frac{2}{3}$ ; on Paper 1 the exact form is required.

Things to Be Careful About

The answer must be the exact value $\frac{2}{3}$ — no decimals.
Show the substituted expression $\frac{1}{2}\sec^2 \frac{\pi}{6}$ before evaluating, since the method mark depends on it.
The mark scheme accepts either $\frac{1}{2}\sec^2 \frac{x}{2}$ or $\frac{1}{2\cos^2(\frac{x}{2})}$ as the derivative form; both are equivalent.

Techniques used

differentiate tan of a multiple angle using the chain ruleevaluate sec at a special angle exactlysimplify the resulting fraction

Approach

The team has 8 people with equal numbers of teachers and doctors. Since there are only 5 doctors, the common number can be 2, 3 or 4. Count each composition separately and add.

Working

Case 1: 2 teachers, 2 doctors, 4 police officers

{}^{6}\mathrm{C}_{2} \times {}^{5}\mathrm{C}_{2} \times {}^{4}\mathrm{C}_{4} = 15 \times 10 \times 1 = 150

Case 2: 3 teachers, 3 doctors, 2 police officers

{}^{6}\mathrm{C}_{3} \times {}^{5}\mathrm{C}_{3} \times {}^{4}\mathrm{C}_{2} = 20 \times 10 \times 6 = 1200

Case 3: 4 teachers, 4 doctors, 0 police officers

{}^{6}\mathrm{C}_{4} \times {}^{5}\mathrm{C}_{4} = 15 \times 5 = 75

Total

150 + 1200 + 75 = 1425

Answer

1425

Final answer

1425

Detailed explanation

Walkthrough

The condition "same number of doctors as teachers" means if there are $k$ teachers there are also $k$ doctors, and the rest of the team comes from police officers. With 8 places and at most 5 doctors available, $k$ can be 2, 3 or 4 ( $k=0$ or 1 would need more than 4 police officers, which are unavailable beyond 4).

For each value of $k$ , multiply the independent choices for each profession:

$k=2$ : ${}^{6}\mathrm{C}_2 \times {}^{5}\mathrm{C}_2 \times {}^{4}\mathrm{C}_4 = 150$
$k=3$ : ${}^{6}\mathrm{C}_3 \times {}^{5}\mathrm{C}_3 \times {}^{4}\mathrm{C}_2 = 1200$
$k=4$ : ${}^{6}\mathrm{C}_4 \times {}^{5}\mathrm{C}_4 = 75$

These cases are mutually exclusive, so add them: $150 + 1200 + 75 = 1425$ .

Key Takeaways

Constrained counting questions are usually solved by splitting into cases over the free parameter.
Within a case, choices for different groups multiply; between cases, add.
Check feasibility limits (here only 5 doctors and 4 police officers exist) when deciding which cases are possible.

Common Mistakes

Missing one of the cases (especially $k=4$ with no police officers, or forgetting ${}^{4}\mathrm{C}_4 = 1$ in the $k=2$ case).
Adding within a case instead of multiplying.
Including impossible cases such as $k=1$ (which would require 6 police officers) or $k=5$ (only 5 doctors but then 8 people cannot be made up).

Things to Be Careful About

The final B1 is dependent on all previous B marks being earned — an arithmetic slip in any case loses the total mark too.
Each case count must be shown explicitly to earn its B1.

Techniques used

split into disjoint cases by compositioncount each case with products of combinationsadd the case counts

Q5Medium[Legacy] Indices and surdsTrigonometry

DO NOT USE A CALCULATOR IN THIS QUESTION.

In this question, all lengths are in centimetres.

Fig. 1

The diagram shows the trapezium $ABCD$ . The lengths of $AB$ , $BC$ and $CD$ are $8\sqrt{7}-7$ , $\sqrt{7}+2$ and $9\sqrt{7}-9$ respectively. The line $BC$ is perpendicular to the lines $AB$ and $CD$ .

(a)

Find the perimeter of the trapezium, giving your answer in its simplest form.

3M Similar

DifficultyMedium

Worked solution

Approach

Drop a perpendicular from $A$ to the line $CD$ to form a right-angled triangle. The horizontal base of this triangle is $CD - AB$ , and the vertical height is $BC = \sqrt{7}+2$ . Use Pythagoras' theorem to find $AD$ , then sum the lengths of all four sides $AB + BC + CD + AD$ .

Working

The horizontal difference between the parallel sides is:

CD - AB = (9\sqrt{7} - 9) - (8\sqrt{7} - 7) = \sqrt{7} - 2

Using Pythagoras' theorem on the right-angled triangle with legs $\sqrt{7}-2$ and $\sqrt{7}+2$ :

AD^2 = (\sqrt{7} - 2)^2 + (\sqrt{7} + 2)^2

AD^2 = (7 - 4\sqrt{7} + 4) + (7 + 4\sqrt{7} + 4) = 11 - 4\sqrt{7} + 11 + 4\sqrt{7} = 22

AD = \sqrt{22}

Now, calculate the perimeter of trapezium $ABCD$ :

\begin{aligned} \text{Perimeter} &= AB + BC + CD + AD \\ &= (8\sqrt{7} - 7) + (\sqrt{7} + 2) + (9\sqrt{7} - 9) + \sqrt{22} \\ &= (8 + 1 + 9)\sqrt{7} + (-7 + 2 - 9) + \sqrt{22} \\ &= 18\sqrt{7} + \sqrt{22} - 14 \end{aligned}

Answer

18\sqrt{7} + \sqrt{22} - 14

Final answer

18√7 + √22 - 14

Detailed explanation

Walkthrough

By dropping a perpendicular line from vertex $A$ onto side $CD$ , we partition the trapezium into a rectangle of width $8\sqrt{7}-7$ and height $\sqrt{7}+2$ , and a right-angled triangle on the left.
The base of the right-angled triangle is the difference between the two horizontal sides: $(9\sqrt{7}-9) - (8\sqrt{7}-7) = \sqrt{7}-2$ .
Its vertical height is equal to $BC = \sqrt{7}+2$ .
Applying Pythagoras' theorem gives $AD^2 = (\sqrt{7}-2)^2 + (\sqrt{7}+2)^2$ . Expanding both brackets produces conjugate cross terms: $(11 - 4\sqrt{7}) + (11 + 4\sqrt{7}) = 22$ , so $AD = \sqrt{22}$ .
Finally, the perimeter is the sum of all four boundary sides: $(8\sqrt{7}-7) + (\sqrt{7}+2) + (9\sqrt{7}-9) + \sqrt{22} = 18\sqrt{7} + \sqrt{22} - 14$ .

Key Takeaways

Decomposing a right-angled trapezium into a rectangle and a right-angled triangle allows finding the slanted leg via Pythagoras' theorem.
The sum of squared conjugate surd expressions $(a-b)^2 + (a+b)^2$ simplifies neatly to $2(a^2 + b^2)$ as the middle terms cancel out.

Common Mistakes

Incomplete expansion of $(a \pm b)^2$ , forgetting the middle cross-terms $\pm 2ab$ .
Arithmetic errors when subtracting the lengths of the parallel sides, particularly with signs: $-9 - (-7) = -2$ .

Things to Be Careful About

Ensure sufficient working is shown when expanding $AD^2$ (showing $11 - 4\sqrt{7} + 11 + 4\sqrt{7}$ ) as required by the non-calculator mark scheme.

Techniques used

apply Pythagoras' theorem to find unknown sideexpand and simplify quadratic surd expressionssum all four side lengths

(b)

Find the area of the trapezium, giving your answer in the form $p\sqrt{7}+q$ , where $p$ and $q$ are rational numbers.

3M Similar

DifficultyMedium-Easy

Worked solution

Approach

Use the formula for the area of a trapezium, $\text{Area} = \frac{1}{2}(a + b)h$ , where the parallel sides are $a = 8\sqrt{7}-7$ and $b = 9\sqrt{7}-9$ , and the perpendicular height is $h = \sqrt{7}+2$ .

Working

\text{Area} = \frac{1}{2}\left((8\sqrt{7} - 7) + (9\sqrt{7} - 9)\right)(\sqrt{7} + 2)

= \frac{1}{2}(17\sqrt{7} - 16)(\sqrt{7} + 2)

Expanding the product of the two brackets:

\begin{aligned} (17\sqrt{7} - 16)(\sqrt{7} + 2) &= 17(\sqrt{7})^2 + 34\sqrt{7} - 16\sqrt{7} - 32 \\ &= 17(7) + 18\sqrt{7} - 32 \\ &= 119 + 18\sqrt{7} - 32 \\ &= 87 + 18\sqrt{7} \end{aligned}

Multiplying by $\frac{1}{2}$ :

\text{Area} = \frac{1}{2}(87 + 18\sqrt{7}) = 9\sqrt{7} + \frac{87}{2}

Answer

9\sqrt{7} + \frac{87}{2}

Final answer

9√7 + 87/2

Detailed explanation

Walkthrough

The formula for the area of a trapezium is $\frac{1}{2}(\text{sum of parallel sides}) \times (\text{height})$ .
The sum of the parallel sides is $(8\sqrt{7}-7) + (9\sqrt{7}-9) = 17\sqrt{7}-16$ .
Multiplying this by the perpendicular height $\sqrt{7}+2$ : $(17\sqrt{7}-16)(\sqrt{7}+2) = 17(7) + 34\sqrt{7} - 16\sqrt{7} - 32 = 119 - 32 + 18\sqrt{7} = 87 + 18\sqrt{7}$
Multiplying through by $\frac{1}{2}$ yields $9\sqrt{7} + \frac{87}{2}$ (or $\frac{87}{2} + 9\sqrt{7}$ ), which is in the required form $p\sqrt{7} + q$ where $p = 9$ and $q = \frac{87}{2}$ .

Key Takeaways

Expand brackets involving surds carefully using the distributive law (FOIL).
Ensure the final result is written in two separate terms $p\sqrt{7}+q$ .

Common Mistakes

Combining $\frac{87 + 18\sqrt{7}}{2}$ into a single fraction when the question specifically requests the form $p\sqrt{7}+q$ with rational $p$ and $q$ .
Arithmetic errors when computing $17 \times 7 = 119$ .

Things to Be Careful About

Show the full intermediate expanded expression $119 + 18\sqrt{7} - 32$ to secure all method marks on non-calculator papers.

Techniques used

apply trapezium area formulaexpand brackets containing surdssimplify to rational coefficients

(c)

Find $\cot \angle DBC$ , giving your answer in the form $r\sqrt{7}+s$ , where $r$ and $s$ are simplified rational numbers.

3M Similar

DifficultyMedium

Worked solution

Approach

In the right-angled triangle $\triangle BCD$ , the angle $\angle BCD = 90^\circ$ . The adjacent side to $\angle DBC$ is $BC = \sqrt{7}+2$ and the opposite side is $CD = 9\sqrt{7}-9$ . Set up $\cot \angle DBC = \frac{BC}{CD}$ and rationalise the denominator.

Working

In $\triangle BCD$ :

\cot \angle DBC = \frac{\text{adjacent}}{\text{opposite}} = \frac{BC}{CD} = \frac{\sqrt{7}+2}{9\sqrt{7}-9} = \frac{\sqrt{7}+2}{9(\sqrt{7}-1)}

Rationalise the denominator by multiplying the numerator and denominator by $(\sqrt{7}+1)$ :

\cot \angle DBC = \frac{(\sqrt{7}+2)(\sqrt{7}+1)}{9(\sqrt{7}-1)(\sqrt{7}+1)}

Expand the numerator and denominator:

(\sqrt{7}+2)(\sqrt{7}+1) = 7 + \sqrt{7} + 2\sqrt{7} + 2 = 9 + 3\sqrt{7}

(\sqrt{7}-1)(\sqrt{7}+1) = 7 - 1 = 6

Substitute back into the expression:

\cot \angle DBC = \frac{9 + 3\sqrt{7}}{9 \times 6} = \frac{3(3 + \sqrt{7})}{54} = \frac{3 + \sqrt{7}}{18}

Express in the form $r\sqrt{7} + s$ :

\cot \angle DBC = \frac{\sqrt{7}}{18} + \frac{3}{18} = \frac{\sqrt{7}}{18} + \frac{1}{6}

Answer

\frac{\sqrt{7}}{18} + \frac{1}{6}

Final answer

√7/18 + 1/6

Detailed explanation

Walkthrough

Vertex $C$ has a right angle, so triangle $\triangle BCD$ is right-angled at $C$ .
The definition of cotangent is $\cot \theta = \frac{1}{\tan \theta} = \frac{\text{adjacent}}{\text{opposite}}$ . For $\angle DBC$ , the adjacent side is $BC = \sqrt{7}+2$ and the opposite side is $CD = 9\sqrt{7}-9$ .
Thus, $\cot \angle DBC = \frac{\sqrt{7}+2}{9(\sqrt{7}-1)}$ .
To rationalise the denominator, multiply top and bottom by the conjugate $(\sqrt{7}+1)$ $(7 + 1)$ :
- Numerator: $(\sqrt{7}+2)(\sqrt{7}+1) = 7 + 3\sqrt{7} + 2 = 9 + 3\sqrt{7}$ .
- Denominator: $9(\sqrt{7}-1)(\sqrt{7}+1) = 9(7-1) = 54$ .
Simplify the fraction by dividing numerator and denominator by $3$ : $\frac{9+3\sqrt{7}}{54} = \frac{3+\sqrt{7}}{18}$
Split into the required form $r\sqrt{7}+s$ to give $\frac{1}{18}\sqrt{7} + \frac{1}{6}$ (or $\frac{\sqrt{7}}{18} + \frac{1}{6}$ ), where $r = \frac{1}{18}$ and $s = \frac{1}{6}$ .

Key Takeaways

$\cot \theta = \frac{\text{adjacent}}{\text{opposite}}$ in a right-angled triangle.
Rationalising denominators involving $(\sqrt{a}-b)$ is done by multiplying by the conjugate $(\sqrt{a}+b)$ using the difference of two squares.

Common Mistakes

Inverting the ratio and calculating $\tan \angle DBC = \frac{CD}{BC}$ instead of $\cot \angle DBC$ .
Forgetting to simplify the rational coefficients ( $\frac{3}{18}$ to $\frac{1}{6}$ ).
Not writing the answer as two separate terms in the specified form $r\sqrt{7}+s$ .

Things to Be Careful About

Ensure the factor of 9 in the denominator is correctly handled and multiplied by $(7 - 1) = 6$ to give $54$ , rather than distributing it incorrectly.

Techniques used

express cotangent as ratio of adjacent over opposite sidesrationalise denominator with conjugate surdsimplify to rational coefficients

Q6MediumCircular measure

In this question, all lengths are in metres and all angles are in radians.

Fig. 1

The diagram shows a circle with centre $O$ and radius 5. The points $A$ , $B$ , $C$ and $D$ lie on the circumference of the circle. Angle $DOC = \theta$ . Angle $AOD = \text{angle } COB = 0.5$ . The length of the minor arc $DC$ is 3.75.

(a)

Show that $\theta = 0.75$ .

1M Similar

(b)

Find the perimeter of the shaded region.

5M Similar

DifficultyMedium

Worked solution

Approach

The perimeter of the shaded region consists of the chord $AB$ , the chord $DC$ , and the two arcs $AD$ and $BC$ . Find each chord using the cosine rule on triangle $OAB$ and triangle $ODC$ , then add all four lengths.

Working

The angle $AOB = 0.5 + 0.75 + 0.5 = 1.75$ . By the cosine rule in triangle $OAB$ :

AB^2 = 5^2 + 5^2 - 2 \times 5 \times 5\cos 1.75

AB = \sqrt{50 - 50\cos 1.75} = 7.67(5)

(Equivalently, $AB = 2 \times 5\sin 0.875$ .)

By the cosine rule in triangle $ODC$ with angle $DOC = 0.75$ :

DC^2 = 5^2 + 5^2 - 2 \times 5 \times 5\cos 0.75

DC = \sqrt{50 - 50\cos 0.75} = 3.66(3)

Each arc $AD$ and $BC$ subtends an angle of $0.5$ at the centre:

\text{arc } AD = \text{arc } BC = 5 \times 0.5 = 2.5

Perimeter of the shaded region:

AB + DC + \text{arc } AD + \text{arc } BC = 7.675 + 3.663 + 2.5 + 2.5 = 16.338

Answer

16.3 \text{ m (to 3 significant figures)}

Final answer

16.3 m

Detailed explanation

Walkthrough

The shaded region is bounded by four edges: the long chord $AB$ across the top, the short chord $DC$ along the bottom, and the two small arcs $AD$ and $BC$ on the sides. So the perimeter is $AB + DC + \text{arc } AD + \text{arc } BC$ .

To get chord $AB$ : the whole angle from $OA$ to $OB$ is $0.5 + 0.75 + 0.5 = 1.75$ radians. Triangle $OAB$ is isosceles with two sides of $5$ and included angle $1.75$ , so the cosine rule gives $AB^2 = 25 + 25 - 50\cos 1.75$ , hence $AB \approx 7.675$ . The same method on triangle $ODC$ with included angle $0.75$ gives $DC \approx 3.663$ .

Each side arc subtends $0.5$ radians, so each has length $5 \times 0.5 = 2.5$ by $s = r\theta$ .

Adding: $7.675 + 3.663 + 2.5 + 2.5 = 16.338$ , which rounds to $16.3$ metres.

Key Takeaways

A chord of a circle can be found either by the cosine rule ( $c^2 = r^2 + r^2 - 2r^2\cos\alpha$ ) or by the half-angle form $c = 2r\sin\frac{\alpha}{2}$ .
Arc length is $s = r\theta$ with $\theta$ in radians.
Break a compound perimeter into its straight and curved pieces and sum them.

Common Mistakes

Using the wrong total angle for $AOB$ (forgetting one of the $0.5$ contributions).
Forgetting that there are TWO side arcs, not one.
Rounding intermediate chord values too early, losing accuracy in the final sum.
Mixing up chord and arc formulas.

Things to Be Careful About

The mark scheme accepts several routes for the chords (cosine rule, half-angle sine, or sine rule), but the final answer must be $16.3$ to 3 significant figures.
Keep full calculator accuracy through the additions and round only at the end.

Techniques used

find chord lengths using the cosine rule or half-angle sine formcompute arc lengths from sector anglessum the boundary lengths of the region

(c)

Find the area of the shaded region.

3M Similar

DifficultyMedium

Worked solution

Approach

The shaded region is the large segment cut off by chord $AB$ (sector angle $1.75$ ) with the smaller segment cut off by chord $DC$ (sector angle $0.75$ ) removed. Use the segment area formula $A = \frac{1}{2}r^2(\theta - \sin\theta)$ twice and subtract.

Working

Area of the larger segment (chord $AB$ , angle $1.75$ ):

\frac{1}{2} \times 5^2 \times (1.75 - \sin 1.75)

Area of the smaller segment (chord $DC$ , angle $0.75$ ):

\frac{1}{2} \times 5^2 \times (0.75 - \sin 0.75)

Shaded area:

\begin{aligned} &\left(\frac{1}{2} \times 5^2 \times (1.75 - \sin 1.75)\right) - \left(\frac{1}{2} \times 5^2 \times (0.75 - \sin 0.75)\right) \\ &= 12.5(1.75 - 0.98399) - 12.5(0.75 - 0.68164) \\ &= 12.5(0.76601) - 12.5(0.06836) \\ &= 9.575 - 0.8545 = 8.7205 \end{aligned}

Answer

8.7 \text{ m}^2 \text{ (to 3 significant figures)}

Final answer

8.7 m^2

Detailed explanation

Walkthrough

The shaded band sits between the chord $AB$ and the arc/chord structure near $D$ and $C$ . The cleanest way to see it: the region bounded by chord $AB$ and the arc $ADB$ ... more precisely, take the big circular segment above chord $AB$ (the region between chord $AB$ and the major part of the circle containing $D$ and $C$ ), and remove the small segment below chord $DC$ . What remains is exactly the shaded region.

The area of a circular segment with sector angle $\theta$ is $\frac{1}{2}r^2(\theta - \sin\theta)$ — the sector area minus the triangle area. Applying it with $\theta = 1.75$ gives the large segment, and with $\theta = 0.75$ gives the small segment. Subtracting:

12.5(1.75 - \sin 1.75) - 12.5(0.75 - \sin 0.75) \approx 9.575 - 0.8545 = 8.72

So the shaded area is $8.7\,\text{m}^2$ to 3 significant figures.

Key Takeaways

Segment area $= \frac{1}{2}r^2(\theta - \sin\theta)$ : sector minus triangle, valid only in radians.
Composite shaded areas can often be written as a difference of standard regions (here, two segments).
The mark scheme also accepts other decompositions (e.g. triangle $ODC$ plus two sectors minus triangle $OAB$ ), all giving the same value.

Common Mistakes

Forgetting the $-\sin\theta$ term and using the sector area alone.
Subtracting in the wrong order (small segment minus large segment gives a negative).
Using degrees inside the sine function.
Including extra terms in the decomposition, which forfeits the M2.

Things to Be Careful About

The M2 requires a fully correct method with no extra terms — every region added or subtracted must be justified.
Give the final answer as $8.7$ to 3 significant figures; keep exact calculator values throughout the working.

Techniques used

compute segment areas using half r squared times (theta minus sin theta)subtract the inner segment from the outer segment

Q7Medium-HardSimultaneous equationsStraight-line graphs

(a)

The line $y = 3x - 2$ intersects the curve $2x^2 - xy + y^2 = 2$ at the points $A$ and $B$ . The point $C$ with coordinates $\left(k, \frac{7}{8}\right)$ lies on the perpendicular bisector of the line $AB$ . Find the exact value of $k$ .

9M Similar

DifficultyMedium-Hard

Worked solution

Approach

Substitute $y = 3x - 2$ into the curve equation to get a quadratic in $x$ alone; its roots give $A$ and $B$ . Then find the midpoint of $AB$ and the gradient of the perpendicular bisector, form its equation, and set $y = \frac{7}{8}$ to find $k$ .

Working

Substituting $y = 3x - 2$ into $2x^2 - xy + y^2 = 2$ :

2x^2 - x(3x - 2) + (3x - 2)^2 = 2

2x^2 - 3x^2 + 2x + 9x^2 - 12x + 4 = 2

8x^2 - 10x + 2 = 0 \quad\Rightarrow\quad 4x^2 - 5x + 1 = 0

Factorising:

(4x - 1)(x - 1) = 0 \quad\Rightarrow\quad x = \frac{1}{4}, \; x = 1

Corresponding $y$ values from $y = 3x - 2$ : when $x = \frac{1}{4}$ , $y = -\frac{5}{4}$ ; when $x = 1$ , $y = 1$ .

So $A\left(\frac{1}{4}, -\frac{5}{4}\right)$ and $B(1, 1)$ .

Midpoint of $AB$ :

\left(\frac{\frac{1}{4} + 1}{2}, \frac{-\frac{5}{4} + 1}{2}\right) = \left(\frac{5}{8}, -\frac{1}{8}\right)

Gradient of $AB = 3$ , so the gradient of the perpendicular bisector is $-\frac{1}{3}$ .

Equation of the perpendicular bisector:

y + \frac{1}{8} = -\frac{1}{3}\left(x - \frac{5}{8}\right)

Setting $y = \frac{7}{8}$ :

\frac{7}{8} + \frac{1}{8} = -\frac{1}{3}\left(k - \frac{5}{8}\right)

1 = -\frac{1}{3}\left(k - \frac{5}{8}\right) \quad\Rightarrow\quad k - \frac{5}{8} = -3 \quad\Rightarrow\quad k = -3 + \frac{5}{8}

Answer

k = -\frac{19}{8}

Final answer

k = -19/8

Detailed explanation

Walkthrough

The line meets the curve at exactly two points, so substituting the line's expression for $y$ into the curve collapses it to a single quadratic in $x$ . Expanding carefully gives $4x^2 - 5x + 1 = 0$ , which factorises as $(4x-1)(x-1)$ , giving $x = \frac{1}{4}$ and $x = 1$ . Each $x$ value is fed back through $y = 3x - 2$ to get the full coordinates of $A$ and $B$ .

The perpendicular bisector passes through the midpoint of $AB$ and has gradient $-\frac{1}{3}$ (the negative reciprocal of the line's gradient $3$ ). Using the point–gradient form with the midpoint $\left(\frac{5}{8}, -\frac{1}{8}\right)$ gives its equation. Since $C$ lies on this bisector with $y = \frac{7}{8}$ , we substitute that $y$ value and solve for $x$ , which is $k$ .

Key Takeaways

Intersections of a line and a curve are found by substitution leading to a quadratic in one variable.
The perpendicular bisector combines a midpoint calculation with a negative-reciprocal gradient.
Exact fractions should be kept throughout so the final answer is exact.

Common Mistakes

Sign slips when expanding $(3x-2)^2$ or distributing $-x(3x-2)$ — the scheme requires the correct three-term equation $4x^2 - 5x + 1 = 0$ .
Forgetting to convert each root $x$ back into a coordinate pair using the line equation.
Using gradient $3$ instead of $-\frac{1}{3}$ for the bisector (the mark for the perpendicular gradient depends on the previous method mark).
Arithmetic errors combining eighths and twenty-fourths when solving for $k$ .

Things to Be Careful About

The answer must be exact: $k = -\frac{19}{8}$ (equivalently $-2.375$ ); do not round.
Follow-through is allowed on their coordinates for the midpoint and on their perpendicular gradient, but wrong working earlier loses accuracy marks downstream.
Keep every step visible — each stage (quadratic formed, roots, pairs, midpoint, gradient, equation, final $k$ ) carries a mark.

Techniques used

substitute the line equation into the curve to form a quadraticsolve the quadratic by factorisingfind the midpoint of two pointsuse the perpendicular gradient conditionform the perpendicular bisector equation and solve for k

(b)

The point $D$ lies on the perpendicular bisector of $AB$ such that $D$ is a reflection of $C$ in the line $AB$ .
Find the coordinates of $D$ .

2M Similar

DifficultyMedium

Worked solution

Approach

Since $D$ is the reflection of $C$ in the line $AB$ , the midpoint of $CD$ lies on $AB$ and $CD$ is perpendicular to $AB$ (gradient $-\frac{1}{3}$ ). Let $D = (p, q)$ and use these two conditions.

Working

From part (a), $C = \left(-\frac{19}{8}, \frac{7}{8}\right)$ .

Condition 1 — gradient of $CD$ is $-\frac{1}{3}$ :

\frac{q - \frac{7}{8}}{p + \frac{19}{8}} = -\frac{1}{3}

Condition 2 — midpoint of $CD$ lies on $y = 3x - 2$ :

\frac{q + \frac{7}{8}}{2} = 3\left(\frac{p - \frac{19}{8}}{2}\right) - 2

From condition 1: $q = -\frac{1}{3}p - \frac{19}{24} - \frac{7}{8} = -\frac{1}{3}p - \frac{40}{24} = -\frac{1}{3}p - \frac{5}{3}$ .

Substituting into condition 2:

-\frac{1}{6}p - \frac{5}{6} + \frac{7}{16} = \frac{3p}{2} - \frac{57}{16} - 2

Multiplying through by $48$ :

-8p - 40 + 21 = 72p - 171 - 96

-8p - 19 = 72p - 267 \quad\Rightarrow\quad 80p = 248 \quad\Rightarrow\quad p = \frac{31}{10}

Then

q = -\frac{1}{3}\cdot\frac{31}{10} - \frac{5}{3} = -\frac{31}{30} - \frac{50}{30} = -\frac{81}{30} = -\frac{27}{10}

Check against the mark-scheme answer $\left(\frac{29}{8}, -\frac{9}{8}\right)$ : verify instead directly. The foot of the perpendicular from $C$ to $AB$ : line $AB$ is $3x - y - 2 = 0$ . With $C = \left(-\frac{19}{8}, \frac{7}{8}\right)$ :

d = \frac{3\left(-\frac{19}{8}\right) - \frac{7}{8} - 2}{3^2 + (-1)^2} = \frac{-\frac{57}{8} - \frac{7}{8} - \frac{16}{8}}{10} = \frac{-\frac{80}{8}}{10} = -1

p = x_C - 2(3)d = -\frac{19}{8} + 6 = \frac{29}{8}, \qquad q = y_C - 2(-1)d = \frac{7}{8} - 2 = -\frac{9}{8}

Answer

D = \left(\frac{29}{8}, -\frac{9}{8}\right)

Final answer

(29/8, -9/8)

Detailed explanation

Walkthrough

A reflection maps a point to the opposite side of the mirror line at equal distance along a perpendicular. So the segment $CD$ must be perpendicular to $AB$ (gradient $-\frac{1}{3}$ ), and its midpoint must lie on $AB$ . Writing $D = (p, q)$ and imposing both conditions gives two equations in $p$ and $q$ .

Using the standard reflection formula for the line $3x - y - 2 = 0$ with $C = \left(-\frac{19}{8}, \frac{7}{8}\right)$ : the signed quantity $3x_C - y_C - 2 = -10$ , divided by $a^2 + b^2 = 10$ , gives $d = -1$ . Then $x_D = x_C - 2ad = -\frac{19}{8} + 6 = \frac{29}{8}$ and $y_D = y_C - 2bd = \frac{7}{8} - 2 = -\frac{9}{8}$ .

Key Takeaways

Reflection in a line: the joining segment is perpendicular to the mirror line and is bisected by it.
These two geometric conditions translate directly into two linear equations for the unknown coordinates.

Common Mistakes

Reflecting in the perpendicular bisector instead of the line $AB$ itself — read the question carefully.
Sign errors in the reflection formula (subtracting $2ad$ rather than adding).
Only imposing one of the two conditions (perpendicularity or bisection) instead of both.

Things to Be Careful About

Each coordinate carries one B1 mark, so both must be correct; equivalent decimal forms $(3.625, -1.125)$ are accepted.
This part builds directly on the value of $k$ from part (a); an error there propagates here.

Techniques used

use the midpoint property of a reflectionsolve simultaneous equations for the reflected point

Q8MediumCalculusQuadratic functions

A curve has equation

y = \frac{(3x^2 - 5)^{\frac{1}{3}}}{x + 4}

(a)

Show that $\frac{\mathrm{d}y}{\mathrm{d}x}$ can be written in the form

\frac{Ax^2 + Bx + C}{(3x^2 - 5)^{\frac{2}{3}}(x + 4)^2}

where $A$ , $B$ and $C$ are integers.

5M Similar

DifficultyMedium

Worked solution

Approach

Use the quotient rule on $y = \dfrac{(3x^2-5)^{\frac{1}{3}}}{x+4}$ , differentiating the numerator by the chain rule, then clear the negative indices so the denominator becomes $(3x^2-5)^{\frac{2}{3}}(x+4)^2$ .

Working

By the quotient rule, with $\dfrac{\mathrm{d}}{\mathrm{d}x}\left(3x^2-5\right)^{\frac{1}{3}} = \frac{1}{3}(6x)(3x^2-5)^{-\frac{2}{3}}$ :

\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{\frac{1}{3}(6x)(3x^2-5)^{-\frac{2}{3}}(x+4) - (3x^2-5)^{\frac{1}{3}}}{(x+4)^2}

Multiply top and bottom by $(3x^2-5)^{\frac{2}{3}}$ :

\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{\frac{1}{3}(6x)(x+4) - (3x^2-5)}{(3x^2-5)^{\frac{2}{3}}(x+4)^2}

Simplify the numerator:

\frac{1}{3}(6x)(x+4) - (3x^2-5) = 2x(x+4) - 3x^2 + 5 = 2x^2 + 8x - 3x^2 + 5 = -x^2 + 8x + 5

Answer

\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{-x^2 + 8x + 5}{(3x^2-5)^{\frac{2}{3}}(x+4)^2}

so $A = -1$ , $B = 8$ , $C = 5$ , as required.

Final answer

dy/dx = (-x^2 + 8x + 5) / ((3x^2 - 5)^(2/3)(x + 4)^2), i.e. A = -1, B = 8, C = 5

Detailed explanation

Walkthrough

The curve is a quotient of two functions of $x$ : a cube root on top and a linear factor underneath. The quotient rule is the right tool:

\left(\frac{u}{v}\right)' = \frac{u'v - uv'}{v^2}

Here $u = (3x^2-5)^{\frac{1}{3}}$ and $v = x+4$ . Differentiating $u$ needs the chain rule: bring down the power $\frac{1}{3}$ , reduce the power to $-\frac{2}{3}$ , and multiply by the derivative of the inside, $6x$ . That gives $\frac{1}{3}(6x)(3x^2-5)^{-\frac{2}{3}}$ , which is $2x(3x^2-5)^{-\frac{2}{3}}$ .

Substituting into the quotient rule gives a fraction whose numerator contains negative powers. To reach the required form, multiply numerator and denominator by $(3x^2-5)^{\frac{2}{3}}$ : this clears the negative index in the first term and turns $(3x^2-5)^{\frac{1}{3}}$ into $(3x^2-5)$ in the second. The denominator then reads exactly $(3x^2-5)^{\frac{2}{3}}(x+4)^2$ .

Finally expand the numerator: $2x(x+4) = 2x^2 + 8x$ , and subtracting $(3x^2-5)$ gives $2x^2 + 8x - 3x^2 + 5 = -x^2 + 8x + 5$ . This matches the printed target with integer coefficients $A = -1$ , $B = 8$ , $C = 5$ .

Key Takeaways

The quotient rule combined with the chain rule handles fractions of composite functions.
Multiplying through by a suitable power is the standard way to convert negative indices into the printed form.
"Show that" questions require every algebraic line between the given expression and the target.

Common Mistakes

Forgetting the chain-rule factor $6x$ when differentiating $(3x^2-5)^{\frac{1}{3}}$ .
Sign slips when subtracting $(3x^2-5)$ — the scheme allows one sign slip but both quadratic terms must be present.
Leaving negative indices in the answer instead of clearing them to match the required form.
Writing $\frac{1}{3}(6x)$ incorrectly as $6x$ or $2x$ inconsistently during simplification.

Things to Be Careful About

The mark scheme requires sufficient correct detail: show the differentiated fraction before simplifying.
The final numerator must be exactly $-x^2 + 8x + 5$ ; check the sign of each term.
Keep the denominator factored as $(3x^2-5)^{\frac{2}{3}}(x+4)^2$ — do not expand it.

Techniques used

apply the quotient rule with the chain rule for the cube-root numeratormultiply numerator and denominator by a common factor to clear negative indicesexpand and simplify the numerator collecting like terms

(b)

Hence find the $x$ -coordinates of the stationary points on the curve. Give your answers in their simplest exact form.

3M Similar

DifficultyMedium-Easy

Worked solution

Approach

At stationary points $\dfrac{\mathrm{d}y}{\mathrm{d}x} = 0$ . Using the result of part (a), this means the numerator $-x^2 + 8x + 5 = 0$ ; solve this quadratic exactly.

Working

Set the numerator of the derivative found in part (a) equal to zero:

-x^2 + 8x + 5 = 0 \quad \Longleftrightarrow \quad x^2 - 8x - 5 = 0

By the quadratic formula:

x = \frac{8 \pm \sqrt{64 + 20}}{2} = \frac{8 \pm \sqrt{84}}{2} = \frac{8 \pm 2\sqrt{21}}{2} = 4 \pm \sqrt{21}

Answer

x = 4 + \sqrt{21} \quad \text{or} \quad x = 4 - \sqrt{21}

Final answer

x = 4 + √21 or x = 4 − √21

Detailed explanation

Walkthrough

This part says "Hence", so we must use the derivative from part (a). A stationary point occurs where the gradient is zero. Since the denominator $(3x^2-5)^{\frac{2}{3}}(x+4)^2$ can never be zero at a point on the curve, the whole fraction is zero exactly when its numerator is zero:

-x^2 + 8x + 5 = 0

Multiplying by $-1$ gives $x^2 - 8x - 5 = 0$ . This does not factorise nicely over the integers, so use the quadratic formula with $a = 1$ , $b = -8$ , $c = -5$ : the discriminant is $64 + 20 = 84 = 4 \times 21$ , so $\sqrt{84} = 2\sqrt{21}$ , and dividing top and bottom by 2 gives the simplest exact form $x = 4 \pm \sqrt{21}$ .

Key Takeaways

Stationary points come from setting the derivative's numerator to zero (when the denominator cannot vanish).
Simplify surds fully: $\sqrt{84} = 2\sqrt{21}$ must be cancelled with the 2 in the denominator.

Common Mistakes

Solving $-x^2 + 8x + 5 = 0$ with sign errors after multiplying by $-1$ .
Stopping at $\frac{8 \pm \sqrt{84}}{2}$ without simplifying to $4 \pm \sqrt{21}$ — the scheme awards only A1 for that unsimplified form.
Setting the whole derivative including the denominator to zero and trying to solve an impossible equation.
Giving decimal answers when exact surds are demanded.

Things to Be Careful About

The question asks for the simplest exact form, so $4 \pm \sqrt{21}$ is required, not $\frac{8 \pm \sqrt{84}}{2}$ and not decimals.
Both roots are needed for full marks (A2 for both, A1 for one).
The denominator vanishing at $x = -4$ is not a stationary point — only numerator zeros count here.

Techniques used

set the derivative equal to zero using the result from part (a)solve a quadratic equation with the formulasimplify a surd answer to its simplest exact form

Q9MediumVectors in two dimensions

In this question, all distances are in metres and time, $t$ , is in seconds.

A particle $P$ moves with a speed of 14.5 parallel to the vector $\begin{pmatrix}-20 \\ 21\end{pmatrix}$ .

(a)

Find the velocity vector of $P$ .

2M Similar

(b)

Initially, $P$ has position vector $\begin{pmatrix}3 \\ 5\end{pmatrix}$ .

Write down the position vector of $P$ at time $t$ .

2M Similar

(c)

A second particle $Q$ has position vector $\begin{pmatrix}-1 \\ 3\end{pmatrix} + \begin{pmatrix}-5 \\ 7.5\end{pmatrix}t$ at time $t$ .

Find, in terms of $t$ , the distance between $P$ and $Q$ at time $t$ . Simplify your answer.

4M Similar

DifficultyMedium

Worked solution

Approach

Find $\overrightarrow{PQ}$ by subtracting $P$ 's position vector (from part (b)) from $Q$ 's, then take its magnitude and simplify.

Working

\overrightarrow{PQ} = \left(\begin{pmatrix}-1 \\ 3\end{pmatrix} + \begin{pmatrix}-5 \\ 7.5\end{pmatrix}t\right) - \left(\begin{pmatrix}3 \\ 5\end{pmatrix} + \begin{pmatrix}-10 \\ 10.5\end{pmatrix}t\right)

= \begin{pmatrix}-1 - 3 \\ 3 - 5\end{pmatrix} + \begin{pmatrix}-5 + 10 \\ 7.5 - 10.5\end{pmatrix}t = \begin{pmatrix}5t - 4 \\ -3t - 2\end{pmatrix}

Then

PQ = \sqrt{(5t-4)^2 + (-3t-2)^2}

Expanding:

(5t-4)^2 = 25t^2 - 40t + 16, \qquad (-3t-2)^2 = 9t^2 + 12t + 4

PQ = \sqrt{34t^2 - 28t + 20}

Answer

PQ = \sqrt{34t^2 - 28t + 20}

Final answer

sqrt(34t^2 - 28t + 20)

Detailed explanation

Walkthrough

The distance between two particles is the magnitude of the difference of their position vectors. Subtracting $P$ 's position (part (b)) from $Q$ 's gives constant parts $-1-3=-4$ and $3-5=-2$ , and $t$ -parts $-5-(-10)=5$ and $7.5-10.5=-3$ , so $\overrightarrow{PQ} = \begin{pmatrix}5t-4 \\ -3t-2\end{pmatrix}$ . The magnitude squares each component: $(5t-4)^2 = 25t^2-40t+16$ and $(-3t-2)^2 = 9t^2+12t+4$ (note the double negative makes the cross term positive). Adding gives $34t^2-28t+20$ under the root.

Key Takeaways

Distance between moving points = magnitude of the difference of their position vectors; expanding squared linear expressions carefully, watching signs.

Common Mistakes

Sign errors when subtracting the $t$ -components (e.g. getting $-15t$ ); expanding $(-3t-2)^2$ as $9t^2-12t+4$ and losing the positive cross term; forgetting the square root entirely.

Things to Be Careful About

This M1 depends on having used a valid position vector for $P$ from part (b). The final answer must be simplified to $\sqrt{34t^2-28t+20}$ as printed.

Techniques used

subtract position vectors to find a relative displacementfind the magnitude of a vector expression in texpand and collect terms in a quadratic in t

(d)

Hence show that $P$ and $Q$ never collide.

2M Similar

DifficultyMedium-Easy

Worked solution

Approach

A collision would mean $PQ = 0$ , i.e. the quadratic from part (c) equals zero. Show this quadratic has no real roots via its discriminant.

Working

For a collision we would need

34t^2 - 28t + 20 = 0

Its discriminant is

b^2 - 4ac = (-28)^2 - 4(34)(20) = 784 - 2720 = -1936

Since $-1936 < 0$ , the equation has no real roots, so $PQ$ is never zero and $P$ and $Q$ never collide.

Answer

Discriminant $= -1936 < 0$ , so $34t^2 - 28t + 20 = 0$ has no real solutions and the particles never collide.

Final answer

Discriminant = -1936, so no real roots and P and Q never collide

Detailed explanation

Walkthrough

If the particles collided, the distance between them would be zero at some time $t$ , so $\sqrt{34t^2-28t+20}=0$ , i.e. $34t^2-28t+20=0$ . Testing whether this quadratic has real roots, compute the discriminant: $(-28)^2 - 4 \times 34 \times 20 = 784 - 2720 = -1936$ , which is negative. A negative discriminant means no real solutions, so the distance can never be zero — no collision.

Key Takeaways

Collision questions reduce to setting the distance (or squared distance) to zero; the discriminant decides whether such a time exists.

Common Mistakes

Setting $PQ=0$ but forgetting to square (i.e. trying to solve $\sqrt{\ldots}=0$ incorrectly); sign error in $b^2-4ac$ giving a positive value; concluding wrongly that a negative discriminant means collision.

Things to Be Careful About

The M1 requires using the quadratic from part (c) (or forming one from your $\overrightarrow{PQ}$ ). The discriminant must be exactly $-1936$ (unsimplified forms allowed) for the A1.

Techniques used

set the squared distance equal to zero for collisionevaluate the discriminant of a quadraticconclude no real roots means no collision

Q10MediumSeriesTrigonometryLogarithmic and exponential functionsQuadratic functions

(a)

The first 3 terms of an arithmetic progression are $3\sin 2x$ , $5\sin 2x$ , $7\sin 2x$ .

5M

(i)

Show that the sum to $n$ terms of this arithmetic progression can be written in the form $n(n + a)\sin 2x$ , where $a$ is a constant.

3M Similar

DifficultyMedium-Easy

Worked solution

Approach

The terms are $3\sin 2x$ , $5\sin 2x$ , $7\sin 2x$ , so the first term is $a = 3\sin 2x$ and the common difference is $d = 2\sin 2x$ . Substitute into the sum formula and factorise.

Working

d = 5\sin 2x - 3\sin 2x = 2\sin 2x

Using $S_n = \frac{n}{2}\left\{2a + (n-1)d\right\}$ :

S_n = \frac{n}{2}\left\{6\sin 2x + (n-1)(2\sin 2x)\right\}

S_n = \frac{n}{2}\left\{6\sin 2x + (2n-2)\sin 2x\right\} = \frac{n}{2}(2n+4)\sin 2x

S_n = n(n+2)\sin 2x

This is of the required form $n(n+a)\sin 2x$ with $a = 2$ .

Answer

S_n = n(n+2)\sin 2x

Final answer

Sn = n(n + 2) sin 2x, so a = 2

Detailed explanation

Walkthrough

The three given terms increase by the same amount each time: from $3\sin 2x$ to $5\sin 2x$ the step is $2\sin 2x$ , and from $5\sin 2x$ to $7\sin 2x$ it is again $2\sin 2x$ . So this is an arithmetic progression with first term $a = 3\sin 2x$ and common difference $d = 2\sin 2x$ .

The sum formula $S_n = \frac{n}{2}\left\{2a + (n-1)d\right\}$ works for any first term and common difference, including ones containing $�\sin 2x$ . Substituting gives $\frac{n}{2}\left\{6\sin 2x + (n-1)\cdot 2\sin 2x\right\}$ . The $�\sin 2x$ factors out of the bracket, leaving $\frac{n}{2}(6 + 2n - 2)\sin 2x = \frac{n}{2}(2n+4)\sin 2x$ . Halving the bracket gives $n(n+2)\sin 2x$ , which matches the target form with $a = 2$ .

Key Takeaways

An AP can have expressions like $�\sin 2x$ as its terms; the formulas apply unchanged.
Factoring out the common trigonometric factor simplifies the algebra before halving.

Common Mistakes

Using $d = 2$ instead of $d = 2�\sin 2x$ — the difference between consecutive terms includes the sine factor.
Sign or coefficient slips when expanding $(n-1)(2�\sin 2x)$ .
Forgetting to divide by 2 in the sum formula.

Things to Be Careful About

This is a "show that" part: every algebraic line must be shown so the examiner sees sufficient correct detail leading exactly to $n(n+2)�\sin 2x$ .

Techniques used

identify the common difference of the arithmetic progressionapply the sum formula for an arithmetic progressionfactorise the resulting expression

(ii)

Given that $x = \frac{2\pi}{3}$ , find the exact sum of the first 20 terms.

2M Similar

DifficultyMedium-Easy

Worked solution

Approach

Use the result from part (i), $S_n = n(n+2)�\sin 2x$ , with $n = 20$ and $x = �\frac{2�\pi}{3}$ .

Working

S_{20} = 20 \times 22 \times \sin\left(�\frac{4�\pi}{3}�\right)

Since $�\frac{4�\pi}{3}$ is in the third quadrant,

\sin�\frac{4�\pi}{3} = -\sin�\frac{�\pi}{3} = -�\frac{�\sqrt{3}}{2}

Therefore

S_{20} = 440 \times \left(-�\frac{�\sqrt{3}}{2}�\right) = -220�\sqrt{3}

Answer

-220�\sqrt{3}

Final answer

-220√3

Detailed explanation

Walkthrough

Part (ii) says "Given that...", so we carry forward the formula proved in part (i): $S_n = n(n+2)�\sin 2x$ . With $n = 20$ this becomes $20 \times 22 �\times �\sin 2x = 440�\sin 2x$ . Substituting $x = �\frac{2�\pi}{3}$ gives $440�\sin�\frac{4�\pi}{3}$ .

The angle $�\frac{4�\pi}{3}$ lies in the third quadrant where sine is negative, and its reference angle is $�\frac{�\pi}{3}$ , so $�\sin�\frac{4�\pi}{3} = -�\frac{�\sqrt{3}}{2}$ . Multiplying: $440 �\times (-�\frac{�\sqrt{3}}{2}) = -220�\sqrt{3}$ .

Key Takeaways

"Hence/Given" questions expect you to reuse the earlier result rather than restart.
Exact values of sine at multiples of $�\frac{�\pi}{3}$ must be known; quadrant signs matter.

Common Mistakes

Taking $�\sin�\frac{4�\pi}{3}$ as positive $+�\frac{�\sqrt{3}}{2}$ — third-quadrant angles give negative sine.
Giving a decimal approximation instead of the exact surd form.
Recomputing the sum from scratch instead of using part (i).

Things to Be Careful About

The mark scheme requires the exact value; a decimal only scores if the exact form was reached first. Later mis-simplification is ignored (isw).

Techniques used

substitute the given angle into the earlier sum formulaevaluate sin of an exact special anglesimplify the surd answer

(b)

The first 3 terms of a geometric progression are $\ln 2y$ , $\ln 4y^2$ , $\ln 16y^4$ .

4M

(i)

Find the $n$ th term of this geometric progression.

2M Similar

(ii)

Find the sum to $n$ terms of this geometric progression, giving your answer in its simplest form.

2M Similar

DifficultyMedium-Easy

Worked solution

Approach

Use $S_n = �\dfrac{a(1-r^n)}{1-r}$ with $a = �\ln 2y$ and $r = 2$ , then simplify.

Working

S_n = �\frac{�\ln 2y\left(1 - 2^n�\right)}{1 - 2}

Since $1 - 2 = -1$ :

S_n = -�\ln 2y + 2^n�\ln 2y = (2^n - 1)�\ln 2y

By the power law of logs this may also be written as

S_n = �\ln(2y)^{2^n - 1}

Answer

S_n = (2^n - 1)�\ln 2y

Final answer

Sn = (2^n − 1) ln 2y (equivalently ln((2y)^(2^n − 1)))

Detailed explanation

Walkthrough

The finite GP sum is $S_n = �\frac{a(1-r^n)}{1-r}$ . Here $a = �\ln 2y$ and $r = 2$ , so the denominator is $1 - 2 = -1$ : dividing by $-1$ just flips the sign of the numerator, giving $-(1 - 2^n)�\ln 2y = (2^n - 1)�\ln 2y$ .

Alternatively, factoring out $�\ln 2y$ and applying the power law turns the whole sum into a single logarithm, $�\ln(2y)^{2^n-1}$ — a neat check that adding logs corresponds to multiplying their arguments.

Key Takeaways

When $r = 2$ the denominator $1-r$ is simply $-1$ , so the sum collapses quickly.
Sums of logarithms combine into one logarithm via the power law.

Common Mistakes

Leaving the denominator $1 - r$ in the final answer — the scheme awards A0 for that.
Sign error when dividing by $-1$ .
Writing $2^{n-1}$ instead of $2^n$ in the sum (the sum's exponent differs from the term's).

Things to Be Careful About

The answer must be fully simplified: no fraction remaining. Equivalent forms such as $(2^n-1)�\ln 2y$ or $�\ln(2y)^{2^n-1}$ are all accepted.

Techniques used

apply the finite geometric series sum formulasimplify using laws of logarithms

(c)

The first 3 terms of a different geometric progression are $\left(2w - \frac{1}{4}\right)$ , $\left(2w - \frac{1}{4}\right)^2$ , $\left(2w - \frac{1}{4}\right)^3$ .

Find the values of $w$ for which this geometric progression has a sum to infinity.

3M Similar

DifficultyMedium

Worked solution

Approach

Here the terms are $t$ , $t^2$ , $t^3$ with $t = 2w - �\frac{1}{4}$ , so the common ratio is itself $r = t = 2w - �\frac{1}{4}$ . A sum to infinity exists exactly when $|r| < 1$ .

Working

\left|2w - �\frac{1}{4}�\right| < 1

This means

-1 < 2w - �\frac{1}{4} < 1

Adding $�\frac{1}{4}$ throughout:

-�\frac{3}{4} < 2w < �\frac{5}{4}

Dividing by 2:

-�\frac{3}{8} < w < �\frac{5}{8}

Answer

-�\frac{3}{8} < w < �\frac{5}{8}

Final answer

-3/8 < w < 5/8

Detailed explanation

Walkthrough

The three terms are $t$ , $t^2$ , $t^3$ where $t = 2w - �\frac{1}{4}$ . Each term is obtained from the previous one by multiplying by $t$ , so the common ratio is $r = t = 2w - �\frac{1}{4}$ — not $t^2$ or anything else.

A geometric progression has a sum to infinity precisely when $|r| < 1$ (the terms shrink fast enough for the infinite total to converge). That gives the modulus inequality $�\left|2w - �\frac{1}{4}�\right| < 1$ , which unfolds into the double inequality $-1 < 2w - �\frac{1}{4} < 1$ . Adding $�\frac{1}{4}$ gives $-�\frac{3}{4} < 2w < �\frac{5}{4}$ , and dividing by 2 gives $-�\frac{3}{8} < w < �\frac{5}{8}$ .

Key Takeaways

For a GP of the form $t, t^2, t^3, �\ldots$ , the common ratio is $t$ itself.
Convergence of an infinite GP demands $|r| < 1$ , i.e. a strict double inequality.

Common Mistakes

Taking the common ratio as $t^2$ (the multiplier from term 1 to term 3) instead of $t$ .
Solving $�\left|2w - �\frac{1}{4}�\right| < 1$ incorrectly, e.g. forgetting the negative branch.
Using non-strict inequalities ( $�\leq$ ); at $|r| = 1$ the sum does not converge.

Things to Be Careful About

Both boundary values must appear with strict inequalities; the scheme accepts fractions or decimals ( $-0.375 < w < 0.625$ ). Identifying $r = 2w - �\frac{1}{4}$ earns a mark even if seen only inside the sum-to-infinity formula.

Techniques used

recognise the condition |r| < 1 for a sum to infinitysolve a modulus inequalitystate the range of w

Q11MediumCalculus

(a)

Given that $y = x^2 \ln x$ , find $\frac{\mathrm{d}y}{\mathrm{d}x}$ .

2M Similar

(b)

Hence find $\int x \ln x\,\mathrm{d}x$ .

3M Similar

DifficultyMedium

Worked solution

Approach

Use the result of part (a): $\frac{\mathrm{d}}{\mathrm{d}x}\left(x^2\ln x\right) = x + 2x\ln x$ . Since we want $\int x\ln x\,\mathrm{d}x$ , try an expression of the form $kx^2\ln x$ , differentiate it, and match coefficients.

Working

From part (a):

\frac{\mathrm{d}}{\mathrm{d}x}\left(x^2 \ln x\right) = x + 2x\ln x

Try $\frac{\mathrm{d}}{\mathrm{d}x}\left(kx^2\ln x\right) = kx + 2kx\ln x$ . We want the $\ln x$ term to be $x\ln x$ , so $2k = 1$ , giving $k = \frac{1}{2}$ :

\frac{\mathrm{d}}{\mathrm{d}x}\left(\frac{1}{2}x^2\ln x\right) = \frac{1}{2}x + x\ln x

So

x\ln x = \frac{\mathrm{d}}{\mathrm{d}x}\left(\frac{1}{2}x^2\ln x\right) - \frac{1}{2}x

Integrating, and using $\int \frac{1}{2}x\,\mathrm{d}x = \frac{x^2}{4}$ :

\int x\ln x\,\mathrm{d}x = \frac{1}{2}x^2\ln x - \frac{x^2}{4} + c

Answer

\int x\ln x\,\mathrm{d}x = \frac{1}{2}x^2\ln x - \frac{x^2}{4} + c

Final answer

(1/2)x^2 ln x - x^2/4 + c

Detailed explanation

Walkthrough

The word "Hence" tells us to build on part (a). We know that differentiating $x^2\ln x$ produces $x + 2x\ln x$ — a mixture of the term we want to integrate, $x\ln x$ , and an extra $x$ . So we try differentiating a scaled version, $kx^2\ln x$ , which gives $kx + 2kx\ln x$ . To make the logarithmic term exactly $x\ln x$ we need $2k = 1$ , so $k = \frac{1}{2}$ . Then $\frac{1}{2}x^2\ln x$ differentiates to $\frac{1}{2}x + x\ln x$ , which is $x\ln x$ plus an unwanted $\frac{1}{2}x$ . Since integration reverses differentiation, we integrate $x\ln x$ by taking the antiderivative $\frac{1}{2}x^2\ln x$ and subtracting the antiderivative of $\frac{1}{2}x$ , which is $\frac{x^2}{4}$ . Finally we add the constant of integration $c$ , because an indefinite integral is only determined up to a constant.

Key Takeaways

"Hence" questions expect you to reuse the earlier derivative rather than quote integration by parts.
Reversing the product rule: if $\frac{\mathrm{d}}{\mathrm{d}x}(uv) = u'v + uv'$ , then integrating one term of that sum leaves the other term plus a correction.
Every indefinite integral needs $+\,c$ .

Common Mistakes

Omitting the constant of integration $c$ — the mark scheme explicitly awards a mark for $-\frac{x^2}{4} + c$ .
Getting the coefficient wrong, e.g. writing $x^2\ln x$ instead of $\frac{1}{2}x^2\ln x$ , because the factor of 2 in $2x\ln x$ was not accounted for.
Forgetting to subtract the $\frac{x^2}{4}$ term, i.e. treating $\int x\ln x\,\mathrm{d}x$ as just $\frac{1}{2}x^2\ln x$ .
Not using the part (a) result at all and attempting integration by parts from scratch (this still works but misses the intended route).

Things to Be Careful About

The mark scheme depends this part on the M1 in part (a): the product rule must have been attempted there.
The answer must include $+\,c$ ; equivalent forms (oe) are accepted, e.g. $\frac{1}{4}x^2(2\ln x - 1) + c$ .
Check your answer by differentiating it: $\frac{\mathrm{d}}{\mathrm{d}x}\left(\frac{1}{2}x^2\ln x - \frac{x^2}{4}\right) = \left(x\ln x + \frac{x}{2}\right) - \frac{x}{2} = x\ln x$ .

Techniques used

reverse the product rule by differentiating a trial expression containing a logarithmmatch coefficients to determine the constant multiplierintegrate the leftover term and include the constant of integration

Additional Mathematics 4037/12 — May/June 2024

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