Additional Mathematics 4037 / 22 — October/November 2025

DifficultyMedium

Worked solution

Approach

Since $AC$ and $BD$ bisect each other at $O$ and have length $12$ cm, $OA = OB = OC = OD = 6$ cm — all four points lie on a circle of radius $6$ centred at $O$ . Chord $AD = BC = 4$ cm. Find angle $AOD$ (or $BOC$ ) from the isosceles triangle $AOD$ , then use angles on a straight line / round point $O$ to get the obtuse angle $AOB$ .

Working

In triangle $AOD$ : $OA = OD = 6$ , $AD = 4$ . By the cosine rule:

\cos(AOD) = \frac{6^2 + 6^2 - 4^2}{2 \times 6 \times 6} = \frac{36 + 36 - 16}{72} = \frac{7}{9}

AOD = \cos^{-1}\left(\frac{7}{9}\right) = 0.6796\ldots \text{ rad}

By symmetry $BOC = AOD = 0.6796\ldots$ rad, so

AOB = \pi - 2 \times 0.6796\ldots = 3.14159\ldots - 1.3593\ldots = 2.4618\ldots

(Equivalently, halving the isosceles triangle $AOB$ gives $\cos\left(\tfrac{AOB}{2}\right) = \frac{2}{6}$ , so $AOB = 2\cos^{-1}\left(\frac{1}{3}\right)$ .)

Answer

AOB = 2.46 \text{ rad} \quad (2.4618\ldots)

Final answer

2.46 rad

Detailed explanation

Walkthrough

The diagonals $AC$ and $BD$ are both $12$ cm long and bisect each other at $O$ , so each half is $6$ cm: $OA = OB = OC = OD = 6$ . This means $A$ , $B$ , $C$ , $D$ all sit on a circle of radius $6$ centred at $O$ , which matches the statement that arcs $AB$ and $DC$ belong to that circle.

To get angle $AOB$ , first find one of the other angles at $O$ . Triangle $AOD$ has sides $OA = OD = 6$ and $AD = 4$ , so the cosine rule gives $\cos(AOD) = \frac{7}{9}$ , i.e. $AOD \approx 0.680$ rad. The same holds for $BOC$ by symmetry. The four angles around $O$ add to $2\pi$ , and opposite pairs are equal, so $AOB = \pi - 2 \times 0.680 \approx 2.46$ rad.

Key Takeaways

Bisecting diagonals of equal length mean all four vertices lie on a circle centred at the intersection.
The cosine rule finds an angle of a triangle from three sides.
Angles around a point sum to $2\pi$ ; here symmetry halves the work.

Common Mistakes

Using degrees instead of radians — the question demands radians, and later parts need the radian value for arc length and sector area.
Finding the acute angle $BOC$ and forgetting to convert it into the obtuse angle $AOB$ by subtracting from $\pi$ .
Sign slips in the cosine rule numerator ( $36 + 36 - 16$ , not $+16$ ).
Rounding too early: keep $0.6796\ldots$ unrounded until the final subtraction.

Things to Be Careful About

The answer must be given to 3 significant figures: $2.46$ rad (accept $2.461$ – $2.462$ ).
The mark scheme awards M2 for a complete correct method such as $\pi - 2\sin^{-1}(\frac{1}{3})$ or $\cos^{-1}(-\frac{7}{9})$ ; a partial route (e.g. only finding $BOC$ ) earns M1.
Later parts follow through on your value of $2.46$ , but it must be less than $2\pi$ .

Techniques used

apply the cosine rule in a triangleuse the sine rule or right-angled trigonometry to find an anglecombine angles at a point to find the obtuse angleconvert the angle to radians

(b)

Use your answer to part (a) to find

5M

(i)

the perimeter of the shaded region

(ii)

the area of the shaded region.

DifficultyMedium

Worked solution

Approach

Split the region $ABCD$ into two sectors $AOB$ and $DOC$ (angle $2.4618\ldots$ rad each) plus two triangles $AOD$ and $BOC$ (angle $0.6796\ldots$ rad each, included between two radii of length $6$ ).

Working

Sector area: $\frac{1}{2}r^2\theta$ .

2 \times \frac{1}{2} \times 6^2 \times 2.4618\ldots = 36 \times 2.4618\ldots = 88.62\ldots

Triangle area: $\frac{1}{2}ab\sin C$ with $a = b = 6$ and $C = 0.6796\ldots$ :

2 \times \frac{1}{2} \times 6^2 \times \sin(0.6796\ldots) = 36 \times 0.6285\ldots = 22.63\ldots

Total area:

88.62\ldots + 22.63\ldots = 111.25\ldots

Answer

\text{Area} = 111 \text{ cm}^2

Final answer

111 cm^2

Detailed explanation

Walkthrough

The shaded region splits naturally along the dashed lines into four pieces around $O$ : two large sectors $AOB$ and $DOC$ with angle $2.4618\ldots$ rad, and two narrow triangles $AOD$ and $BOC$ with angle $0.6796\ldots$ rad squeezed between radii of length $6$ .

Each sector contributes $\frac{1}{2} \times 6^2 \times 2.4618\ldots = 44.31$ cm², so $88.62$ cm² for both. Each triangle contributes $\frac{1}{2} \times 6 \times 6 \times \sin(0.6796\ldots) = 11.31$ cm², so $22.63$ cm² for both. Adding gives $111.25\ldots \approx 111$ cm².

An equivalent route is two segments plus the rectangle between them, or the whole circle minus two segments cut off by $AD$ and $BC$ — all give the same total.

Key Takeaways

Sector area $A = \frac{1}{2}r^2\theta$ (radians).
Triangle area $\frac{1}{2}ab\sin C$ handles non-right-angled triangles between two radii.
Composite regions should be decomposed into standard shapes whose formulas you know.

Common Mistakes

Using $\frac{1}{2}r\theta$ instead of $\frac{1}{2}r^2\theta$ for sector area.
Using the wrong angle for the triangles (must be $BOC = 0.680$ rad, not $AOB$ ).
Forgetting there are two of each shape.
Subtracting when addition is needed, or mixing up segment and sector formulas.

Things to Be Careful About

The mark scheme requires a correct plan (M1 for the sector term, dependent M1 for the correct decomposition) before the final answer.
Follow-through on your part (a) angle is allowed if it is less than $2\pi$ .
Final answer to 3 significant figures: $111$ cm² (accept $111.2$ – $111.3$ ).

Techniques used

compute sector areas using half r squared thetadecompose a composite region into sectors and trianglesevaluate triangle area using half ab sin C

Q36MMediumSeries

Find the exact value of the term independent of $x$ in the expansion of $\left(2 + \frac{3}{x^2}\right)^{10} (1 - 4x^2)^2$ .

DifficultyMedium

Worked solution

Approach

Expand each factor far enough to capture all powers of $x$ that can cancel: the first factor gives powers $x^0$ , $x^{-2}$ , $x^{-4}$ , and the second gives $x^0$ , $x^2$ , $x^4$ . The term independent of $x$ comes from matching $x^0$ with $x^0$ , $x^{-2}$ with $x^2$ , and $x^{-4}$ with $x^4$ .

Working

Using the binomial theorem on the first three terms of $\left(2 + \dfrac{3}{x^2}\right)^{10}$ :

2^{10} + \binom{10}{1} 2^9 \left(\frac{3}{x^2}\right) + \binom{10}{2} 2^8 \left(\frac{3}{x^2}\right)^2

= 1024 + \frac{15360}{x^2} + \frac{103680}{x^4}

Expanding the second factor:

(1 - 4x^2)^2 = 1 - 8x^2 + 16x^4

The constant term is the sum of the products of matching powers:

1024 \times 1 + 15360 \times (-8) + 103680 \times 16

= 1024 - 122880 + 1658880 = 1\,537\,024

Answer

1\,537\,024

Final answer

1537024

Detailed explanation

Walkthrough

The expression is a product of two powers, so the term independent of $x$ is not simply the product of the two constant terms — negative powers from the first factor can be cancelled by positive powers from the second.

First, expand $\left(2 + \dfrac{3}{x^2}\right)^{10}$ using the binomial theorem. Only the first three terms are needed, because the second factor $(1-4x^2)^2$ contains powers up to $x^4$ , so only $x^0$ , $x^{-2}$ and $x^{-4}$ from the first factor can pair with something:

2^{10} + \binom{10}{1}2^9\left(\frac{3}{x^2}\right) + \binom{10}{2}2^8\left(\frac{3}{x^2}\right)^2 = 1024 + \frac{15360}{x^2} + \frac{103680}{x^4}

Next, expand the square directly:

(1-4x^2)^2 = 1 - 8x^2 + 16x^4

Now match powers of $x$ whose exponents add to zero:

$1024 \times 1$ (constant with constant),
$\dfrac{15360}{x^2} \times (-8x^2)$ , giving $15360 \times (-8) = -122880$ ,
$\dfrac{103680}{x^4} \times 16x^4$ , giving $103680 \times 16 = 1658880$ .

Adding these:

1024 - 122880 + 1658880 = 1\,537\,024

Key Takeaways

The binomial theorem gives coefficients $\binom{n}{r}$ for each term of $(a+b)^n$ .
When a binomial contains $\dfrac{1}{x^2}$ , its expansion has negative powers of $x$ .
In a product of two expansions, the term independent of $x$ comes from every pair of terms whose powers of $x$ sum to zero — not just the two constant terms.
$(1 - 4x^2)^2$ is best expanded directly rather than with the binomial theorem.

Common Mistakes

Taking only $2^{10} \times 1 = 1024$ as the answer, forgetting the cross terms $x^{-2} \cdot x^2$ and $x^{-4} \cdot x^4$ .
Sign errors: the middle term of $(1-4x^2)^2$ is $-8x^2$ , so the second contribution is negative ( $-122880$ ); dropping the minus sign gives a wrong total.
Arithmetic slips in $\binom{10}{2} = 45$ , $2^9 = 512$ , $2^8 = 256$ , or in $45 \times 256 \times 9 = 103680$ .
Expanding too few terms of the first factor — three terms are needed because the second factor reaches $x^4$ .
The mark scheme awards the accuracy mark only for the correct value reached from correct working (nfww), so a lucky correct total from wrong coefficients scores A0.

Things to Be Careful About

The answer must be exact: $1\,537\,024$ , an integer, with no rounding.
The mark scheme gives B3 for both expansions (B2 for the three-term binomial expansion, B1 for $1-8x^2+16x^4$ ), M2 for combining the three matching products, and A1 for the final value — so every expansion term and every product must be visible.
Check each power pairing carefully: $x^{-2}$ pairs with $x^2$ , and $x^{-4}$ pairs with $x^4$ ; pairing the wrong powers gives a term that still contains $x$ .
Keep the negative sign on the $-8x^2$ term throughout the multiplication.

Techniques used

expand a binomial power using the binomial theoremexpand a squared binomialidentify terms whose powers of x cancelcombine matching powers of x to find the constant term

Approach

Since plotting $\mathrm{e}^y$ against $x^3$ gives a straight line, write $\mathrm{e}^y = mx^3 + c$ where $m$ is the gradient and $c$ is the intercept. Find $m$ and $c$ from the two given points, then take natural logs.

Working

The line has equation

\mathrm{e}^y = mx^3 + c

Gradient using the points $(1, 13.5)$ and $(7.5, 0.5)$ :

m = \frac{13.5 - 0.5}{1 - 7.5} = \frac{13}{-6.5} = -2

Intercept, substituting $(1, 13.5)$ :

13.5 = -2(1) + c \quad\Rightarrow\quad c = 15.5

So

\mathrm{e}^y = 15.5 - 2x^3

Taking natural logarithms of both sides:

y = \ln\left(15.5 - 2x^3\right)

Answer

y = \ln\left(15.5 - 2x^3\right)

Final answer

y = ln(15.5 - 2x^3)

Detailed explanation

Walkthrough

The statement that plotting $\mathrm{e}^y$ against $x^3$ gives a straight line means there are constants $m$ and $c$ with $\mathrm{e}^y = m(x^3) + c$ . The coordinates of the plotted points give values of $x^3$ (horizontal axis) and $\mathrm{e}^y$ (vertical axis), so we can find the gradient in the usual way:

m = \frac{13.5 - 0.5}{1 - 7.5} = \frac{13}{-6.5} = -2

Then substitute one point to get the intercept: at $x^3 = 1$ , $\mathrm{e}^y = 13.5$ , so $13.5 = -2(1) + c$ , giving $c = 15.5$ . Hence $\mathrm{e}^y = 15.5 - 2x^3$ . To express $y$ itself we invert the exponential by taking natural logarithms: $y = \ln(15.5 - 2x^3)$ .

Key Takeaways

A straight-line plot of transformed variables corresponds to an equation of the form (vertical variable) = $m$ (horizontal variable) + $c$ .
Gradient and intercept come from the plotted points exactly as for any straight line.
Taking $\ln$ undoes $\mathrm{e}$ : if $\mathrm{e}^y = f(x)$ then $y = \ln(f(x))$ .

Common Mistakes

Using the raw $x$ instead of $x^3$ as the horizontal coordinate — the plotted points are $(x^3, \mathrm{e}^y)$ pairs.
Sign errors in the gradient calculation; here the gradient is negative because the second point has a smaller vertical value.
Forgetting to take logarithms at the end, leaving the answer as $\mathrm{e}^y = 15.5 - 2x^3$ .
Using log base 10 instead of the natural logarithm.

Things to Be Careful About

The mark scheme requires the final answer in the form $y = \ln(15.5 - 2x^3)$ ; intermediate values $m = -2$ and $c = 15.5$ must be shown for the method marks.
Keep exact values ( $-2$ , $15.5$ ) throughout rather than rounding.
The logarithm is only defined where $15.5 - 2x^3 > 0$ , which part (b) addresses.

Techniques used

recognise the linear form of the transformed variablescalculate the gradient from two plotted pointssubstitute a point to find the intercepttake natural logarithms to recover y in terms of x

(b)

Find the values of $x$ for which your equation is valid.

DifficultyMedium-Easy

Worked solution

Approach

The logarithm is only defined when its argument is positive, so require $15.5 - 2x^3 > 0$ and solve for $x$ .

Working

15.5 - 2x^3 > 0

x^3 < \frac{15.5}{2} = 7.75

x < \sqrt[3]{7.75}

Numerically, $\sqrt[3]{7.75} \approx 1.98$ .

Answer

x < \sqrt[3]{7.75} \approx 1.98

Final answer

x < cube root of 7.75 (approximately 1.98)

Detailed explanation

Walkthrough

An equation like $y = \ln(15.5 - 2x^3)$ only makes sense where the inside of the logarithm is positive. So we solve $15.5 - 2x^3 > 0$ : rearranging gives $x^3 < 7.75$ , and taking cube roots (which preserves the direction of the inequality since cubing is increasing) gives $x < \sqrt[3]{7.75} \approx 1.98$ . This is the set of $x$ -values over which the straight-line model was valid.

Key Takeaways

The domain of $\ln(f(x))$ is where $f(x) > 0$ .
Cube-root inequalities keep their direction because the cube function is strictly increasing.

Common Mistakes

Solving $15.5 - 2x^3 > 0$ incorrectly, e.g. writing $x^3 > 7.75$ after dividing by $-2$ without flipping the inequality sign correctly.
Giving only the decimal without showing the inequality step, or rounding before stating the condition.
Forgetting that the answer must follow from correct working — the mark scheme states nfww (no follow-through from wrong working).

Things to Be Careful About

The mark scheme awards M1 for setting up $15.5 - 2x^3 > 0$ (or equivalent) and the accuracy marks for $x < \sqrt[3]{7.75}$ ; the exact form $\sqrt[3]{7.75}$ should be shown alongside the decimal approximation.
Follow-through is allowed from part (a)'s equation in the form $a + bx^3 > 0$ , but the working must be correct.

Techniques used

impose positivity of the argument of a logarithmsolve a cubic inequality

Q5MediumPermutations and combinations

A 6-character password is to be formed from the following characters.

Letters	b	f	g	k	m
Numbers	3	5	7	9
Symbols	*	!	@

Each character can be used at most once in any 6-character password.

(a)

Find the number of 6-character passwords that can be formed if there are no further restrictions.

(b)

Find the number of 6-character passwords that can be formed if the password starts and ends with a symbol.

(c)

Find the number of 6-character passwords that can be formed if the password:

starts with either a symbol and then a number, or a number and then a symbol and
ends with 2 letters.

DifficultyMedium

Worked solution

Approach

The first two positions are either (symbol, number) or (number, symbol) — 2 cases. The last two positions must be letters. The middle two positions take any remaining characters.

Working

First two positions: $2 \times 3 \times 4$ (2 orders, then 3 symbols and 4 numbers).
Last two positions: $5 \times 4$ letters.
Middle two positions: $8 \times 7$ from the remaining 8 characters.

(2 \times 3 \times 4) \times (8 \times 7) \times (5 \times 4)

= 24 \times 56 \times 20 = 26\,880

Answer

26\,880

Final answer

26880

Detailed explanation

Walkthrough

The start condition has two cases: symbol-then-number or number-then-symbol. Each case gives $3 \times 4 = 12$ choices, so together $2 \times 3 \times 4 = 24$ . The end condition fixes the last two positions as letters: $5 \times 4 = 20$ ways. Four characters are now used, leaving 8, so the middle two positions give $8 \times 7 = 56$ ways. Multiplying: $24 \times 56 \times 20 = 26\,880$ .

Key Takeaways

"Either ... or ..." at a position pair introduces a factor of 2 for the two orders.
Restricted groups (both ends here) are counted independently and multiplied.
After fixing positions, the pool for free positions shrinks by the number used.

Common Mistakes

Forgetting the factor of 2 for the two possible orders at the start (this gives $13\,440$ , which earns B1 only).
Using $5 \times 5$ for the last two letters (allowing repetition of a letter).
Not reducing the middle pool from 12 to 8 after using 4 characters.

Things to Be Careful About

A final answer of $13\,440$ (missing the order factor) scores B1 only; the full $26\,880$ requires the M1 structure $2 \times 3 \times 4 \times 8 \times 7 \times 5 \times 4$ (oe).

Techniques used

split into cases with the multiplication rulefill restricted positions at both endsmultiply disjoint cases

Q6MediumVectors in two dimensions

In this question lengths are in centimetres and time, $t$ , is in seconds.

(a)

A particle $P$ is moving in a straight line with a speed of $26$ in the direction of the vector $\begin{pmatrix}5 \\ -12\end{pmatrix}$ .

Find the velocity vector of $P$ .

DifficultyMedium-Easy

Worked solution

Approach

The velocity vector is the speed multiplied by the unit vector in the given direction, so first find the magnitude of $\begin{pmatrix}5 \\ -12\end{pmatrix}$ .

Working

\left|\begin{pmatrix}5 \\ -12\end{pmatrix}\right| = \sqrt{5^2 + (-12)^2} = \sqrt{25 + 144} = \sqrt{169} = 13

The unit vector is $\frac{1}{13}\begin{pmatrix}5 \\ -12\end{pmatrix}$ , so the velocity is

26 \times \frac{1}{13}\begin{pmatrix}5 \\ -12\end{pmatrix} = 2\begin{pmatrix}5 \\ -12\end{pmatrix} = \begin{pmatrix}10 \\ -24\end{pmatrix}

Answer

\begin{pmatrix}10 \\ -24\end{pmatrix}

Final answer

(10, -24) cm/s

Detailed explanation

Walkthrough

A velocity vector combines a speed (a scalar) with a direction. The direction is given by the vector $\begin{pmatrix}5 \\ -12\end{pmatrix}$ , but this vector has length 13, not 1. So we first compute its magnitude using Pythagoras: $\sqrt{5^2 + (-12)^2} = \sqrt{169} = 13$ . Dividing by 13 gives the unit direction vector, and multiplying by the speed 26 gives the velocity: $26 \times \frac{1}{13} = 2$ , so the velocity is $2\begin{pmatrix}5 \\ -12\end{pmatrix} = \begin{pmatrix}10 \\ -24\end{pmatrix}$ .

Key Takeaways

Velocity vector = speed × unit direction vector.
Magnitude of $\begin{pmatrix}a \\ b\end{pmatrix}$ is $\sqrt{a^2 + b^2}$ ; the triple $(5, 12, 13)$ is worth recognising instantly.

Common Mistakes

Multiplying the direction vector by the speed directly without first dividing by its magnitude (this gives a speed of $26 \times 13$ ).
Writing the answer as $\begin{pmatrix}10\mathbf{i} \\ -24\mathbf{j}\end{pmatrix}$ — the mark scheme warns this mixed form does not earn full marks unless the correct column-vector form is recovered.

Things to Be Careful About

Keep the sign of the $y$ -component negative: the direction is $(5, -12)$ , not $(5, 12)$ .
The answer is a vector, so it must be presented in column-vector (or clearly equivalent) form.

Techniques used

find the magnitude of a direction vectorscale the unit direction vector by the speed

(b)

When $t = 0$ , $P$ passes through a point $A$ which has position vector $\begin{pmatrix}3 \\ 6\end{pmatrix}$ .

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

(c)

At the same time that $P$ passes through $A$ , a particle $Q$ passes through a point $B$ .

The position vector of $Q$ at time $t$ is given by $\begin{pmatrix}8t - 5 \\ 2 - 25t\end{pmatrix}$ .

The distance between $P$ and $Q$ at time $t$ is $d$ .

Show that $d^2 = mt^2 + nt + r$ , where $m$ , $n$ and $r$ are integers to be found.

DifficultyMedium

Worked solution

Approach

Find the vector from $Q$ to $P$ by subtracting position vectors, then square and add its components to get $d^2$ .

Working

The position vector of $P$ is $\begin{pmatrix}3 + 10t \\ 6 - 24t\end{pmatrix}$ from part (b). The displacement from $Q$ to $P$ is

\begin{pmatrix}3 + 10t \\ 6 - 24t\end{pmatrix} - \begin{pmatrix}8t - 5 \\ 2 - 25t\end{pmatrix} = \begin{pmatrix}3 + 10t - 8t + 5 \\ 6 - 24t - 2 + 25t\end{pmatrix} = \begin{pmatrix}8 + 2t \\ 4 + t\end{pmatrix}

Then

d^2 = (8 + 2t)^2 + (4 + t)^2

= \left(64 + 32t + 4t^2\right) + \left(16 + 8t + t^2\right)

= 5t^2 + 40t + 80

Answer

d^2 = 5t^2 + 40t + 80

Final answer

d^2 = 5t^2 + 40t + 80

Detailed explanation

Walkthrough

To find the distance between two moving points, first find the vector joining them: subtract one position vector from the other. Using $P$ 's position from part (b) and $Q$ 's given position $\begin{pmatrix}8t - 5 \\ 2 - 25t\end{pmatrix}$ , the difference is $\begin{pmatrix}8 + 2t \\ 4 + t\end{pmatrix}$ (subtracting in the other order just changes both signs and gives the same $d^2$ ). The squared distance is the sum of the squares of the components: $d^2 = (8+2t)^2 + (4+t)^2$ . Expanding each bracket carefully — $(8+2t)^2 = 64 + 32t + 4t^2$ and $(4+t)^2 = 16 + 8t + t^2$ — and collecting like terms gives $d^2 = 5t^2 + 40t + 80$ , with integer coefficients as required.

Key Takeaways

Distance between points with position vectors $\mathbf{p}$ and $\mathbf{q}$ is $|\mathbf{p} - \mathbf{q}|$ ; squaring avoids the square root.
$(a + bt)^2 = a^2 + 2abt + b^2t^2$ — expand, don't forget the middle term.

Common Mistakes

Sign errors when subtracting the components of $\begin{pmatrix}8t - 5 \\ 2 - 25t\end{pmatrix}$ (note the $-(-5) = +5$ and $-(-25t) = +25t$ ).
Omitting the middle term when squaring a binomial.
Using $d$ instead of $d^2$ and leaving a square root in the answer.

Things to Be Careful About

The question asks for $d^2$ , so no square root should be taken at the end.
The mark scheme accepts either order of subtraction, since $d^2$ is unchanged.

Techniques used

subtract position vectors to find the displacement between particlessquare and add components to obtain the squared distanceexpand and collect terms into a quadratic in t

(d)

Hence show that $P$ and $Q$ do not collide.

DifficultyMedium-Easy

Worked solution

Approach

A collision means $d = 0$ . Factorise $d^2$ from part (c) and find when it is zero, then check whether that time is allowed.

Working

Using $d^2 = 5t^2 + 40t + 80$ from part (c):

d^2 = 5\left(t^2 + 8t + 16\right) = 5(t + 4)^2

So $d = 0$ only when $t = -4$ . But time must satisfy $t \geq 0$ , so $d \neq 0$ for all permitted $t$ ; in fact $d > 0$ when $t \geq 0$ .

Answer

Since $d = 0$ only at $t = -4$ , which is not an allowed time ( $t \geq 0$ ), $P$ and $Q$ never meet and do not collide.

Final answer

d = 0 only at t = -4, but t >= 0, so P and Q do not collide

Detailed explanation

Walkthrough

A collision would mean the two particles are at the same point at the same time, i.e. the distance between them is zero. From part (c), $d^2 = 5t^2 + 40t + 80$ , which factorises as $5(t+4)^2$ — a perfect square. So $d^2 = 0$ only when $t = -4$ , giving $d = 0$ only at $t = -4$ . But time in this model starts at $t = 0$ , so negative times are not valid. Hence $d > 0$ for every allowed value of $t$ , and the particles never occupy the same point: no collision.

Key Takeaways

Collision ⟺ distance between particles is 0 at a valid time.
A quadratic in $t$ that is a perfect square, like $5(t+4)^2$ , touches zero at exactly one value of $t$ .

Common Mistakes

Solving $d^2 = 0$ but forgetting to state the condition $t \geq 0$ — the mark scheme requires the argument to include $d$ and the time condition.
Setting $t = 0$ and checking positions instead of examining when $d = 0$ .

Things to Be Careful About

This part depends on part (c) being correct — the mark is dependent on the earlier result.
The argument must mention $d$ explicitly (e.g. " $d = 0$ at $t = -4$ , but $t \geq 0$ "), not just quote the quadratic.

Techniques used

factorise the quadratic for d squaredargue that the zero of d occurs only at a negative time

Q7MediumCalculus

(a)

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

DifficultyMedium-Easy

Worked solution

Approach

$y = x\cos 2x$ is a product, so differentiate it with the product rule; the derivative of $\cos 2x$ needs the chain rule.

Working

\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{\mathrm{d}}{\mathrm{d}x}(x) \cdot \cos 2x + x \cdot \frac{\mathrm{d}}{\mathrm{d}x}(\cos 2x)

Using the chain rule, $\frac{\mathrm{d}}{\mathrm{d}x}(\cos 2x) = -2\sin 2x$ , so

\frac{\mathrm{d}y}{\mathrm{d}x} = \cos 2x + x(-2\sin 2x) = \cos 2x - 2x\sin 2x

Answer

\frac{\mathrm{d}y}{\mathrm{d}x} = \cos 2x - 2x\sin 2x

Final answer

dy/dx = cos 2x - 2x sin 2x

Detailed explanation

Walkthrough

The function is a product of two factors: $x$ and $\cos 2x$ . The product rule says the derivative of a product $uv$ is $u'v + uv'$ . Here $u = x$ gives $u' = 1$ , and $v = \cos 2x$ needs the chain rule: differentiating the outer cosine gives $-\sin 2x$ , then multiplying by the derivative of the inner function $2x$ , which is $2$ , giving $-2\sin 2x$ . Combining these gives $\frac{\mathrm{d}y}{\mathrm{d}x} = \cos 2x - 2x\sin 2x$ .

Key Takeaways

The product rule: $(uv)' = u'v + uv'$ .
Chain rule for trig functions of a multiple angle: $\frac{\mathrm{d}}{\mathrm{d}x}\cos kx = -k\sin kx$ .

Common Mistakes

Forgetting the factor of $2$ from the chain rule and writing $-\sin 2x$ instead of $-2\sin 2x$ (the mark scheme explicitly awards M1 for this inner derivative).
Dropping one term of the product rule, e.g. writing only $-2x\sin 2x$ .
Sign errors on the cosine derivative.

Things to Be Careful About

The mark scheme notes "isw" — later incorrect simplification is ignored once the correct form appears — but "nfww": no marks if reached from wrong working.
Keep the answer in terms of both $\cos 2x$ and $\sin 2x$ ; part (b) depends on it.

Techniques used

apply the product ruledifferentiate cos of a multiple angle using the chain rule

(b)

Hence find $\int x\sin 2x\,\mathrm{d}x$ .

4M Similar

DifficultyMedium

Worked solution

Approach

"Hence" means use part (a). Since $\frac{\mathrm{d}}{\mathrm{d}x}(x\cos 2x) = \cos 2x - 2x\sin 2x$ , rearranging isolates $x\sin 2x$ , and integrating both sides gives the required integral.

Working

From part (a),

x\cos 2x = \int \left(\cos 2x - 2x\sin 2x\right)\,\mathrm{d}x = \int \cos 2x\,\mathrm{d}x - 2\int x\sin 2x\,\mathrm{d}x

Rearranging,

2\int x\sin 2x\,\mathrm{d}x = \int \cos 2x\,\mathrm{d}x - x\cos 2x

Now $\int \cos 2x\,\mathrm{d}x = \frac{1}{2}\sin 2x$ , so

2\int x\sin 2x\,\mathrm{d}x = \frac{1}{2}\sin 2x - x\cos 2x

Dividing by $2$ :

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

Answer

-\frac{1}{2}x\cos 2x + \frac{1}{4}\sin 2x + c

Final answer

-1/2 x cos 2x + 1/4 sin 2x + c

Detailed explanation

Walkthrough

The word "Hence" tells us to build on part (a). We know that differentiating $x\cos 2x$ produces exactly the expression $\cos 2x - 2x\sin 2x$ . Writing that statement in reverse — as an integral equation — lets us solve algebraically for the unknown integral $\int x\sin 2x\,\mathrm{d}x$ : integrate both sides of the derivative result, move the known integral $\int \cos 2x\,\mathrm{d}x$ across, and divide by $2$ . The remaining integral is a standard one, giving $\frac{1}{2}\sin 2x$ (the half comes from dividing by the inner coefficient $2$ ). Finally divide everything by $2$ and add the constant of integration $c$ , since this is an indefinite integral. This route is equivalent to integration by parts with $u = x$ , $\mathrm{d}v = \sin 2x\,\mathrm{d}x$ .

Key Takeaways

A "Hence" question expects you to reuse the previous result, not start afresh.
Reversing a product rule gives integration by parts: $\int u\,\mathrm{d}v = uv - \int v\,\mathrm{d}u$ .
Integrating $\cos kx$ gives $\frac{1}{k}\sin kx$ — always divide by the inner coefficient.
An indefinite integral must include $+c$ .

Common Mistakes

Omitting the constant of integration $c$ .
Getting the sign wrong when moving $x\cos 2x$ across the equation, leading to $+\frac{1}{2}x\cos 2x$ instead of $-\frac{1}{2}x\cos 2x$ .
Forgetting to halve when integrating $\cos 2x$ , writing $\sin 2x$ instead of $\frac{1}{2}\sin 2x$ .
Dividing incorrectly at the end, e.g. leaving $\frac{1}{2}\sin 2x$ without the final factor of $\frac{1}{4}$ .
Ignoring the "Hence" instruction and attempting integration by parts from scratch — allowed by the mark scheme's alternative routes but more error-prone.

Things to Be Careful About

The mark scheme awards M3 for reaching $-\frac{1}{2}x\cos 2x + \frac{1}{4}\sin 2x$ (or the equivalent rearranged forms shown), so every intermediate line matters; "nfww" applies — no marks from wrong working.
The final answer must be exact, with $c$ present; "isw" means later slips in simplification are forgiven but the printed form is the target.
Check your answer by differentiating it: you should recover $x\sin 2x$ .

Techniques used

recognise the integrand as related to the earlier derivativerearrange the differentiated result to isolate the integralintegrate cos of a multiple angleinclude the constant of integration

Q8MediumSeries

An arithmetic progression has first term $t$ and common difference $1.5$ .

The 4th, 8th and 20th terms of this arithmetic progression form the 1st, 2nd and 3rd terms of a geometric progression.

(a)

Find the value of $t$ .

5M Similar

DifficultyMedium

Worked solution

Approach

The 4th, 8th and 20th terms of the AP are $t + 3(1.5)$ , $t + 7(1.5)$ and $t + 19(1.5)$ . Since these form a GP, the square of the middle term equals the product of the outer terms (equivalently the two ratios are equal). Solve for $t$ .

Working

The three AP terms are:

t + 4.5, \qquad t + 10.5, \qquad t + 28.5

Since they form a GP, the common ratio is the same for both pairs:

\frac{t + 10.5}{t + 4.5} = \frac{t + 28.5}{t + 10.5}

Cross-multiplying and expanding both sides:

(t + 10.5)^2 = (t + 4.5)(t + 28.5)

t^2 + 21t + 110.25 = t^2 + 33t + 128.25

Cancelling $t^2$ from both sides:

21t - 33t = 128.25 - 110.25

-12t = 18 \quad\Rightarrow\quad t = -1.5

Answer

t = -1.5

Final answer

-1.5

Detailed explanation

Walkthrough

An arithmetic progression with first term $t$ and common difference $d$ has nth term $t + (n-1)d$ . With $d = 1.5$ , the 4th term uses $(4-1) = 3$ differences, giving $t + 4.5$ ; the 8th term uses $7$ differences, giving $t + 10.5$ ; the 20th term uses $19$ differences, giving $t + 28.5$ . These three numbers must behave like a geometric progression.

In a GP, each term divided by the previous one is the same constant $r$ , so we can write one equation by equating the two ratios:

\frac{t + 10.5}{t + 4.5} = \frac{t + 28.5}{t + 10.5}

This is exactly the condition that the middle term squared equals the product of the outer terms. Cross-multiplying removes the fractions, and expanding both sides gives quadratics in $t$ . The key observation is that the $t^2$ terms cancel — the same thing happens on either side because both sides have leading coefficient $1$ — leaving a simple linear equation:

21t + 110.25 = 33t + 128.25

Rearranging gives $-12t = 18$ , so $t = -1.5$ . A quick check: the three GP terms become $3$ , $9$ , $27$ , which indeed have common ratio $3$ .

Key Takeaways

The nth term of an AP is $a + (n-1)d$ ; count the number of differences carefully ( $n-1$ , not $n$ ).
Three consecutive GP terms satisfy "middle squared equals product of outers", which lets you turn a GP condition into a single equation.
When unknowns appear on both sides of a fraction equation, cross-multiplication clears the denominators cleanly.

Common Mistakes

Using $nd$ instead of $(n-1)d$ when writing the 4th, 8th and 20th terms — this shifts every term by $d$ and produces a wrong value of $t$ .
Expanding $(t+10.5)^2$ incorrectly, e.g. forgetting the middle term $2 \times t \times 10.5 = 21t$ .
Sign slips when rearranging after cancelling $t^2$ : it should be $-12t = 18$ , not $+12t = 18$ .
The mark scheme awards the final mark only if $-1.5$ is the sole answer reached from correct working (nfww), so arithmetic errors earlier lose everything at the end.

Things to Be Careful About

The final answer must be exact: $t = -1.5$ (or $-\frac{3}{2}$ ), with no other values offered.
The mark scheme requires the expansion step to be seen explicitly — do not jump straight to the answer.
Check your result by substituting back: with $t = -1.5$ the terms are $3, 9, 27$ , confirming a valid GP.

Techniques used

write the required AP terms using the nth term formulaequate ratios of consecutive GP termsclear algebraic fractions and expandsolve the resulting linear equation

(b)

Find the common ratio of the geometric progression.

Q9MediumFunctionsLogarithmic and exponential functionsQuadratic functions

It is given that $\mathrm{f}(x) = \ln(2x + 5)$ for $x > a$ , where $a$ is a constant.

(a)

Write down the least possible value of $a$ .

(b)

Using your value of $a$ , write down the range of $\mathrm{f}$ .

(c)

It is also given that $\mathrm{g}(x) = x^2 + 1$ for $x \in \mathbb{R}$ .

Using your value of $a$ , solve the equation $\mathrm{fg}(x) = 4$ .

Give your answers in exact form.