Additional Mathematics 4037 / 12 — May/June 2026

DifficultyMedium-Easy

Worked solution

Approach

The graph of $y = 4\cos x - 2$ is the ordinary cosine wave stretched vertically by a factor of $4$ and then translated $2$ units down. So it oscillates between $-6$ and $2$ about the line $y = -2$ , and over $-360° \leq x \leq 360°$ it completes two full cycles.

Working

Maximum where $\cos x = 1$ : $y = 4(1) - 2 = 2$ , at $x = -360°, 0°, 360°$ .
Minimum where $\cos x = -1$ : $y = 4(-1) - 2 = -6$ , at $x = -180°, 180°$ .
Midline crossings where $\cos x = 0$ : $y = -2$ , at $x = -270°, -90°, 90°, 270°$ .
Axis intercepts where $\cos x = \frac{1}{2}$ : $y = 0$ , at $x = \pm 300°, \pm 60°$ .

Draw a smooth continuous cosine-shaped curve through these points, showing two complete cycles across the interval.

Answer

A cosine wave oscillating between a maximum of $2$ (at $x = -360°, 0°, 360°$ ) and a minimum of $-6$ (at $x = \pm 180°$ ), crossing the midline $y = -2$ at $x = -270°, -90°, 90°, 270°$ and crossing the $x$ -axis at $\pm 60°$ and $\pm 300°$ , drawn as two full continuous cycles.

Final answer

Cosine curve with maximum 2 at x = -360, 0 and 360, minimum -6 at x = +/-180, passing through (-270, -2), (-90, -2), (90, -2), (270, -2), two full cycles over -360 to 360

Detailed explanation

Walkthrough

The base cosine curve $y = \cos x$ runs between $-1$ and $1$ . Multiplying by $4$ stretches it so it runs between $-4$ and $4$ . Subtracting $2$ slides the whole wave down by $2$ , giving a range from $-6$ to $2$ with midline $y = -2$ . The period is unchanged ( $360°$ ), so on an interval of width $720°$ we see exactly two full cycles. The key plotted points are the maxima at $x = -360°, 0°, 360°$ (value $2$ ), the minima at $x = \pm 180°$ (value $-6$ ), and the four midline crossings at $x = -270°, -90°, 90°, 270°$ (value $-2$ ). Joining these smoothly gives the required sketch.

Key Takeaways

For $y = a\cos x + c$ : amplitude is $|a|$ , midline is $y = c$ , range is $c - |a|$ to $c + |a|$ .
Vertical stretch and translation do not change the period.
A sketch must show correct shape, correct amplitude and the correct number of cycles for the interval.

Common Mistakes

Drawing only one cycle instead of two across $-360°$ to $360°$ .
Plotting the wave about $y = 0$ instead of the shifted midline $y = -2$ (i.e. forgetting the $-2$ translation).
Using amplitude $2$ or drawing the wave between $-2$ and $6$ by misreading the transformation order.
Breaking the curve into separate pieces rather than one continuous smooth line — the mark scheme requires continuity and correct curvature at the end points.
Marking the midline crossings at wrong angles; they occur every $180°$ starting at $\pm 90°$ .

Things to Be Careful About

The mark scheme awards B2 only if the shape, both cycles, and the amplitude (or the four points $(-270,-2)$ , $(-90,-2)$ , $(90,-2)$ , $(270,-2)$ ) are all correct; otherwise a maximum of B1.
The curve must be continuous with correct curvature at the end points — no flat or kinked joins.
Values must be exact: maxima $2$ , minima $-6$ , midline value $-2$ .

Techniques used

identify amplitude and midline of a transformed cosine functionmark turning points and axis interceptssketch two full cycles over the given interval

(b)

Write down the amplitude of $4\cos x - 2$ .

(c)

Write down the period of $4\cos x - 2$ .

(d)

Write down the set of values of the constant $k$ for which the equation $4\cos x - 2 = k$ has at least two solutions.

DifficultyMedium-Easy

Worked solution

Approach

Solutions of $4\cos x - 2 = k$ correspond to intersections of the sketched curve with the horizontal line $y = k$ . Over $-360° \leq x \leq 360°$ the curve has two cycles, so any horizontal line meeting the curve cuts it at least twice — provided $k$ lies within the range of the function.

Working

From part (a) the curve takes all values between its minimum $-6$ and its maximum $2$ . For any $k$ with

-6 \leq k \leq 2,

the line $y = k$ meets the curve, and since the curve completes two full cycles on the interval, it meets it at least twice. Outside this range the line misses the curve entirely.

Answer

-6 \leq k \leq 2

Final answer

-6 <= k <= 2

Detailed explanation

Walkthrough

Each solution of $4\cos x - 2 = k$ is an $x$ -value where the curve reaches height $k$ , i.e. a point where the horizontal line $y = k$ crosses the sketch. Because the sketch shows two full cycles, any height actually attained by the curve is attained at least twice (at least once per cycle). The heights attained run from the minimum $-6$ up to the maximum $2$ , so the equation has at least two solutions precisely when $-6 \leq k \leq 2$ . At the extreme values $k = -6$ and $k = 2$ the line touches the minima/maxima, which occur twice each (at $x = \pm 180°$ and $x = -360°, 0°, 360°$ respectively), so the endpoints are included.

Key Takeaways

The number of solutions of $\mathrm{f}(x) = k$ equals the number of intersections of $y = \mathrm{f}(x)$ with $y = k$ .
Over two full cycles, every value in the range is hit at least twice.

Common Mistakes

Writing $k \geq -6$ or $k \leq 2$ (which allows all real numbers) instead of and — the mark scheme explicitly rejects 'or'.
Excluding the endpoints $-6$ and $2$ , even though the tangent lines still give solutions.
Giving the range of $\cos x$ itself ( $-1 \leq k \leq 1$ ) without applying the stretch and shift.

Things to Be Careful About

Follow-through is allowed on their graph provided part (a) earned at least B1, but the printed answer is $-6 \leq k \leq 2$ .
Both inequalities must hold simultaneously ('and'), not alternatively ('or').

Techniques used

interpret horizontal line intersections with a periodic graphread the range of the transformed cosine function

Q2Medium-EasyStraight-line graphs

Points $A$ and $B$ have coordinates $A(-2, 3)$ and $B(8, -3)$ .

(a)

The perpendicular bisector of $AB$ passes through the point $C(0, c)$ .

Find the value of $c$ .

3M Similar

DifficultyMedium-Easy

Worked solution

Approach

Find the midpoint of $AB$ and the gradient of $AB$ ; the perpendicular bisector passes through that midpoint with the negative reciprocal gradient. Substitute $x = 0$ to get $c$ .

Working

Midpoint of $A(-2, 3)$ and $B(8, -3)$ :

\left(\frac{-2 + 8}{2}, \frac{3 + (-3)}{2}\right) = (3, 0)

Gradient of $AB$ :

m_{AB} = \frac{-3 - 3}{8 - (-2)} = \frac{-6}{10} = -\frac{3}{5}

Gradient of the perpendicular bisector:

m_{\perp} = -\frac{1}{-\frac{3}{5}} = \frac{5}{3}

Equation through $(3, 0)$ :

y - 0 = \frac{5}{3}(x - 3)

At $x = 0$ (the point $C(0, c)$ ):

c = \frac{5}{3}(0 - 3) = -5

Answer

c = -5

Final answer

c = -5

Detailed explanation

Walkthrough

The perpendicular bisector of $AB$ is the line through the midpoint of $AB$ at right angles to $AB$ . So we need two ingredients: the midpoint and the perpendicular gradient.

The midpoint averages the coordinates: $\left(\frac{-2+8}{2}, \frac{3-3}{2}\right) = (3, 0)$ .

The gradient of $AB$ is the change in $y$ over the change in $x$ : $\frac{-3-3}{8-(-2)} = -\frac{3}{5}$ . Perpendicular lines have gradients whose product is $-1$ , so the perpendicular gradient is the negative reciprocal: $\frac{5}{3}$ .

The bisector therefore has equation $y = \frac{5}{3}(x - 3)$ . Point $C$ lies on it with $x = 0$ , so $c = \frac{5}{3}(-3) = -5$ .

Key Takeaways

Midpoint formula: average the $x$ -coordinates and average the $y$ -coordinates.
Perpendicular gradients multiply to $-1$ : flip the fraction and change the sign.
A point on the $y$ -axis has $x = 0$ ; substituting $x = 0$ gives the intercept directly.

Common Mistakes

Using $\frac{3}{5}$ instead of $\frac{5}{3}$ (forgetting to flip as well as change sign).
Taking the gradient of $AB$ itself rather than its negative reciprocal.
Sign slips in $\frac{-3-3}{8-(-2)}$ — the double negative in the denominator must become $+2$ .
Using the midpoint of the wrong pair of coordinates.

Things to Be Careful About

The mark scheme allows the answer embedded in the full line equation $y = \frac{5}{3}x - 5$ or as the point $(0, -5)$ , but the working must be shown — the gradient and midpoint each earn a mark.
The final answer is exact; no rounding is involved.

Techniques used

find the midpoint of a line segmentfind the gradient of a line from two pointsuse the negative reciprocal for a perpendicular gradientform the equation of a line and evaluate at a given x

(b)

The point $D$ has coordinates $(2, -1)$ .

The lengths of $AD$ and $BD$ are such that $BD = \frac{\sqrt{n}}{2} AD$ where $n$ is an integer.

Find the value of $n$ .

DifficultyMedium-Easy

Worked solution

Approach

Compute $AD$ and $BD$ with the distance formula, then substitute into $BD = \frac{\sqrt{n}}{2}AD$ and solve for $n$ .

Working

AD = \sqrt{(2 - (-2))^2 + (-1 - 3)^2} = \sqrt{16 + 16} = \sqrt{32}

BD = \sqrt{(2 - 8)^2 + (-1 - (-3))^2} = \sqrt{36 + 4} = \sqrt{40}

Substituting into $BD = \frac{\sqrt{n}}{2}AD$ :

\sqrt{40} = \frac{\sqrt{n}}{2}\sqrt{32}

\sqrt{n} = \frac{2\sqrt{40}}{\sqrt{32}} = 2\sqrt{\frac{40}{32}} = 2\sqrt{\frac{5}{4}} = \sqrt{5}

n = 5

Answer

n = 5

Final answer

n = 5

Detailed explanation

Walkthrough

The distance between two points is $\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$ .

For $A(-2,3)$ to $D(2,-1)$ : the $x$ -difference is $2 - (-2) = 4$ and the $y$ -difference is $-1 - 3 = -4$ , so $AD = \sqrt{4^2 + (-4)^2} = \sqrt{32}$ .

For $B(8,-3)$ to $D(2,-1)$ : the $x$ -difference is $2 - 8 = -6$ and the $y$ -difference is $-1 - (-3) = 2$ , so $BD = \sqrt{(-6)^2 + 2^2} = \sqrt{40}$ .

The relation $BD = \frac{\sqrt{n}}{2}AD$ gives $\sqrt{40} = \frac{\sqrt{n}}{2}\sqrt{32}$ . Rearranging: $\sqrt{n} = \frac{2\sqrt{40}}{\sqrt{32}}$ . Since $\frac{40}{32} = \frac{5}{4}$ , we get $\sqrt{n} = 2\sqrt{\frac{5}{4}} = \sqrt{5}$ , so $n = 5$ .

Key Takeaways

The distance formula is the 2-D version of Pythagoras; square the differences, add, then take the root.
Ratios of square roots can be handled by dividing under a single root: $\frac{\sqrt{40}}{\sqrt{32}} = \sqrt{\frac{40}{32}}$ .
Simplify fractions under the root before squaring to keep the arithmetic clean.

Common Mistakes

Sign slips in the differences, e.g. writing $-1 - (-3) = -4$ instead of $2$ — squaring hides some errors but not all.
Forgetting to square the differences before adding.
Mishandling the factor of 2: $\sqrt{n} = \frac{2\sqrt{40}}{\sqrt{32}}$ , not $\frac{\sqrt{40}}{2\sqrt{32}}$ .
Giving $\sqrt{5}$ as the answer instead of $n = 5$ .

Things to Be Careful About

The mark scheme accepts $AD^2 = 32$ and $BD^2 = 40$ as equivalents, so squared lengths are fine — but both lengths must be found for their marks.
The relation may be used in squared form: $40 = \frac{n}{4} \times 32$ gives $n = 5$ directly, which is a clean alternative.
$n$ must be an integer, as stated; the exact value $5$ is required, not a decimal.

Techniques used

apply the distance formula between two pointssquare both sides of a surd equationsolve for an integer under a square root

Q3MediumFactors of polynomials

The polynomial $\mathrm{p}$ is given by $\mathrm{p}(x) = 6x^3 - x^2 - 5x + 2$ .

(a)

Find the remainder when $\mathrm{p}(x)$ is divided by $x + 2$ .

(b)

Write $\mathrm{p}(x)$ as a product of linear factors.

DifficultyMedium

Worked solution

Approach

Find one linear factor by testing small integer values, divide the cubic by it to get a quadratic, then factorise the quadratic.

Working

Test $x = -1$ :

\mathrm{p}(-1) = 6(-1)^3 - (-1)^2 - 5(-1) + 2 = -6 - 1 + 5 + 2 = 0

So $(x + 1)$ is a factor. Dividing:

6x^3 - x^2 - 5x + 2 = (x + 1)(6x^2 - 7x + 2)

Factorising the quadratic by splitting the middle term ( $-12x + ...$ ): we need two numbers multiplying to $6 \times 2 = 12$ and adding to $-7$ : these are $-3$ and $-4$ .

6x^2 - 7x + 2 = 6x^2 - 3x - 4x + 2 = 3x(2x - 1) - 2(2x - 1) = (3x - 2)(2x - 1)

Therefore

\mathrm{p}(x) = (x + 1)(3x - 2)(2x - 1)

Answer

\mathrm{p}(x) = (x + 1)(3x - 2)(2x - 1)

Final answer

(x + 1)(3x - 2)(2x - 1)

Detailed explanation

Walkthrough

To write a cubic as a product of linear factors, first find one root by trial: substituting small integers, $\mathrm{p}(-1) = 0$ , so $(x + 1)$ is a factor. Dividing the cubic by $(x + 1)$ (by long division or synthetic division) gives the quadratic $6x^2 - 7x + 2$ . This quadratic does not factorise over integers directly as a product of simple brackets, so split the middle term: find two numbers multiplying to $ac = 12$ and summing to $-7$ , namely $-3$ and $-4$ , giving $(3x - 2)(2x - 1)$ . The full factorisation is $(x + 1)(3x - 2)(2x - 1)$ .

Key Takeaways

Trial values $\pm 1, \pm 2$ often reveal an integer root quickly.
Polynomial (or synthetic) division reduces a cubic to a quadratic once one factor is known.
Splitting the middle term handles quadratics whose leading coefficient exceeds 1.

Common Mistakes

Arithmetic errors in the division step producing a wrong quadratic coefficient.
Stopping at $(x + 1)(6x^2 - 7x + 2)$ without completing the factorisation into linear factors.
Sign errors when splitting the middle term.

Things to Be Careful About

The mark scheme requires the final product of three LINEAR factors for full credit; leaving a quadratic factor scores only part marks.
Factors spotted purely by inspection must still be written as a product to earn all marks.

Techniques used

find a linear factor by trial and error or the factor theoremdivide the cubic by the linear factor to obtain a quadraticfactorise the quadratic

(c)

Hence solve the equation $6u^6 - u^4 - 5u^2 + 2 = 0$ .

DifficultyMedium-Easy

Worked solution

Approach

Substitute $u^2 = x$ so the equation becomes $\mathrm{p}(u^2) = 0$ ; use the factors from part (b), then take square roots.

Working

Using part (b):

6u^6 - u^4 - 5u^2 + 2 = (u^2 + 1)(3u^2 - 2)(2u^2 - 1) = 0

From $3u^2 - 2 = 0$ :

u^2 = \frac{2}{3} \quad\Rightarrow\quad u = \pm\sqrt{\frac{2}{3}}

From $2u^2 - 1 = 0$ :

u^2 = \frac{1}{2} \quad\Rightarrow\quad u = \pm\sqrt{\frac{1}{2}}

(The factor $u^2 + 1 = 0$ has no real solutions.)

Answer

u = \pm\sqrt{\frac{2}{3}}, \quad u = \pm\sqrt{\frac{1}{2}}

Final answer

u = +/-sqrt(2/3), u = +/-sqrt(1/2)

Detailed explanation

Walkthrough

The equation in $u$ is exactly $\mathrm{p}(u^2) = 0$ , since replacing $x$ by $u^2$ turns $6x^3 - x^2 - 5x + 2$ into $6u^6 - u^4 - 5u^2 + 2$ . So we can reuse the factorisation from part (b): $(u^2 + 1)(3u^2 - 2)(2u^2 - 1) = 0$ . Setting each factor to zero: $u^2 + 1 = 0$ gives no real solutions; $3u^2 - 2 = 0$ gives $u = \pm\sqrt{2/3}$ ; $2u^2 - 1 = 0$ gives $u = \pm\sqrt{1/2}$ .

Key Takeaways

Recognising a disguised polynomial: substituting $x = u^2$ links the new equation back to the earlier factorisation.
Solving $u^2 = k$ yields TWO solutions, $\pm\sqrt{k}$ .

Common Mistakes

Giving only the positive square roots and forgetting the negative ones.
Solving from scratch instead of using the 'Hence' link to part (b).
Including spurious solutions from $u^2 + 1 = 0$ (these are condoned but should not be presented as real answers).

Things to Be Careful About

All four solutions are required; the mark scheme awards B1FT for any two correct solutions built on the part (b) factors and full credit for all four.
Later mis-simplification is ignored (isw), but the surd forms shown are the expected exact answers.

Techniques used

recognise the equation as p(u squared)solve each linear factor equal to zerotake square roots including both signs

Q4Medium-EasySeries

An arithmetic progression has first term $u_1 = 400$ and common difference $d = -2$ .

(a)

Find the 81st term and the 100th term.

(b)

Find the sum of all the terms from the 81st to the 100th, $u_{81} + u_{82} + \ldots + u_{100}$ .

3M Similar

DifficultyMedium-Easy

Worked solution

Approach

The terms from $u_{81}$ to $u_{100}$ form an arithmetic series of $20$ terms whose first term is $u_{81} = 240$ and last term is $u_{100} = 202$ . Use $S_n = \frac{n}{2}(\text{first} + \text{last})$ .

Working

Number of terms: $100 - 81 + 1 = 20$ .

u_{81} + u_{82} + \ldots + u_{100} = \frac{20}{2}\{240 + 202\}

= 10 \times 442 = 4420

(An equivalent route is $S_{100} - S_{80}$ , since the required sum is the total of the first 100 terms minus the total of the first 80.)

Answer

4420

Final answer

4420

Detailed explanation

Walkthrough

The sum asked for runs from the 81st term to the 100th term inclusive. Counting inclusively gives $100 - 81 + 1 = 20$ terms — forgetting the $+1$ would give only 19. The first term of this block is $u_{81} = 240$ and the last is $u_{100} = 202$ , both found in part (a).

The sum formula using first and last terms is

S_n = \frac{n}{2}(\text{first term} + \text{last term})

so

S = \frac{20}{2}(240 + 202) = 10 \times 442 = 4420

An equally valid method is to compute $S_{100}$ and $S_{80}$ separately and subtract: $S_{100} = 30100$ and $S_{80} = 25680$ , giving $30100 - 25680 = 4420$ again.

Key Takeaways

The number of terms from position $m$ to position $n$ inclusive is $n - m + 1$ .
The formula $S_n = \frac{n}{2}(a + l)$ is efficient when both end terms are known.
A partial run of terms can also be found as a difference of two full sums, $S_n - S_{m-1}$ .

Common Mistakes

Taking the number of terms as $19$ or $21$ instead of $20$ (the mark scheme allows follow-through on 19 or 21 but the correct count is 20).
Using $S_{100} - S_{81}$ instead of $S_{100} - S_{80}$ when subtracting sums, which drops the 81st term.
Adding instead of averaging inside the bracket, i.e. omitting the division by 2.

Things to Be Careful About

The answer must be exactly $4420$ ; the mark scheme awards the final A1 for this value reached from correct working. Listing and adding all twenty individual terms also earns full credit (SC B3), but is far slower.

Techniques used

apply the sum formula using first and last termscount the number of terms between two positionssubtract two partial sums as an alternative route

Q54MMediumCalculusLogarithmic and exponential functions

Find, in exact form, the equation of the tangent to the curve $y = \ln\left((2x - 4)^5\right)$ at the point where $x = 3$ .

DifficultyMedium

Worked solution

Approach

Differentiate $y = \ln\left((2x-4)^5\right)$ using the chain rule, evaluate the gradient at $x = 3$ , find the exact $y$ -coordinate there, then use $y - y_1 = m(x - x_1)$ .

Working

Using the chain rule:

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

Cancelling $(2x-4)^4$ :

\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{10}{2x-4} = \frac{5}{x-2}

At $x = 3$ :

\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{5}{3-2} = 5

The $y$ -coordinate when $x = 3$ :

y = \ln\left((6-4)^5\right) = \ln(2^5) = \ln 32 = 5\ln 2

Tangent through $\left(3, \ln 32\right)$ with gradient $5$ :

y - \ln 32 = 5(x - 3)

Equivalently:

y = 5x - 15 + \ln 32

Answer

y - \ln 32 = 5(x - 3) \quad \text{oe} \quad (y = 5x - 15 + \ln 32)

Final answer

y - ln 32 = 5(x - 3), i.e. y = 5x - 15 + ln 32

Detailed explanation

Walkthrough

The curve is $y = \ln\left((2x-4)^5\right)$ , a composition of functions: an inner linear function $2x-4$ , raised to the fifth power, inside a logarithm. To differentiate it we apply the chain rule from the outside in. Writing $u = (2x-4)^5$ , we have $y = \ln u$ , so $\frac{\mathrm{d}y}{\mathrm{d}u} = \frac{1}{u}$ , while $\frac{\mathrm{d}u}{\mathrm{d}x} = 5(2x-4)^4 \times 2$ . Multiplying gives $\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{5(2x-4)^4 \times 2}{(2x-4)^5}$ .

This simplifies neatly: cancelling $(2x-4)^4$ leaves $\frac{10}{2x-4}$ , which further reduces to $\frac{5}{x-2}$ . (Alternatively, using the law $\ln(a^k) = k\ln a$ first gives $y = 5\ln(2x-4)$ , so $\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{5 \times 2}{2x-4} = \frac{10}{2x-4}$ — the same result.)

A tangent needs both a gradient and a point. The gradient at $x = 3$ is $\frac{5}{3-2} = 5$ . The point itself comes from substituting $x = 3$ into the original curve: $y = \ln(2^5) = \ln 32$ , which is also $5\ln 2$ — kept in exact form since this is a non-calculator paper.

Finally, the equation of a straight line through $(x_1, y_1)$ with gradient $m$ is $y - y_1 = m(x - x_1)$ , giving $y - \ln 32 = 5(x-3)$ , or rearranged, $y = 5x - 15 + \ln 32$ .

Key Takeaways

The chain rule handles any nesting depth: differentiate the outer function, multiply by the derivative of the inner function.
Logarithm laws ( $\ln a^k = k\ln a$ ) can simplify differentiation before you start — often the fastest route.
A tangent requires two ingredients: the gradient (from the derivative at the point) and the coordinates of the point (from the original curve, not the derivative).
Keep answers in exact form when asked: $\ln 32$ stays as $\ln 32$ (or $5\ln 2$ ), never a decimal.

Common Mistakes

Forgetting the inner factor of $2$ from differentiating $2x - 4$ , giving $\frac{5}{2x-4}$ instead of $\frac{10}{2x-4}$ — this loses the B1 for the derivative.
Reading the point's $y$ -value off the derivative rather than the original curve — the mark scheme awards a separate B1 for $y = \ln 32$ found from the curve.
Giving a decimal approximation for $\ln 32$ ; the question demands exact form, and the A1 is for the exact printed form.
Sign slips when expanding $5(x-3)$ if rearranging to $y = mx + c$ : it is $5x - 15 + \ln 32$ , not $5x + 15 + \ln 32$ .
Substituting $x = 3$ into $\frac{5}{x-2}$ incorrectly (e.g. writing $\frac{5}{1} = 5$ correctly but then misusing it as the intercept).

Things to Be Careful About

This is a non-calculator paper: all values must be exact — $\ln 32$ (or equivalently $5\ln 2$ ) must appear in the answer; a decimal like $3.47$ scores nothing for the final A1.
The mark scheme accepts equivalent forms ('oe'): $y - \ln 32 = 5(x-3)$ , $y = 5x - 15 + \ln 32$ , or $y - 5\ln 2 = 5(x-3)$ are all fine.
Show the unsimplified derivative before cancelling — the B1 is awarded for a correct application of the chain rule, so the intermediate step should be visible.
'nfww' applies: the final answer must come from correct working; a correct-looking line from a wrong derivative does not score.

Techniques used

differentiate a composite logarithmic function using the chain rulesimplify the derivative using laws of logarithms or cancellationevaluate the gradient at the given pointfind the exact y-coordinate of the point on the curveform the tangent equation from a point and a gradient

Q6MediumCircular measure

In this question lengths are in centimetres.

Fig. 2

The diagram shows a circle with centre $O$ and radius $r$ .
Points $A$ , $B$ and $C$ are on the circumference of the circle.
Angle $AOB = \alpha$ radians.

(a)

The perimeter of the shaded sector is equal to the length of the major arc, $ACB$ .

Find the exact value of $\alpha$ .

DifficultyMedium-Easy

Worked solution

Approach

The perimeter of the shaded sector consists of two radii plus its arc. The major arc $ACB$ subtends the reflex angle $2\pi - \alpha$ at the centre. Equate the two lengths and solve for $\alpha$ .

Working

Perimeter of shaded sector:

2r + r\alpha

Length of major arc $ACB$ (angle $2\pi - \alpha$ radians):

r(2\pi - \alpha)

Equating:

2r + r\alpha = r(2\pi - \alpha)

Divide by $r$ :

2 + \alpha = 2\pi - \alpha

2\alpha = 2\pi - 2

\alpha = \pi - 1

Answer

\alpha = \pi - 1

Final answer

alpha = pi - 1

Detailed explanation

Walkthrough

The sector's boundary is made of two straight radii (total $2r$ ) and the minor arc. Since the angle at the centre is $\alpha$ radians, the minor arc length is $r\alpha$ , so the perimeter is $2r + r\alpha$ . The rest of the circle's circumference is the major arc $ACB$ , which subtends the remaining angle $2\pi - \alpha$ radians, so its length is $r(2\pi - \alpha)$ . Setting these equal and cancelling the common factor $r$ gives a simple linear equation: $2 + \alpha = 2\pi - \alpha$ , so $2\alpha = 2\pi - 2$ and $\alpha = \pi - 1$ .

Key Takeaways

Arc length is $s = r\theta$ only when $\theta$ is in radians.
The two arcs of a circle correspond to angles $\theta$ and $2\pi - \theta$ .
Perimeter of a sector includes both radii, not just the arc.

Common Mistakes

Using degrees instead of radians — the mark scheme insists angles must be in radians.
Forgetting to include the two radii ( $2r$ ) in the sector's perimeter.
Writing the major arc length as $r(\pi - \alpha)$ or another wrong reflex angle.
Dividing by $r$ incorrectly or losing a factor of 2 when solving.

Things to Be Careful About

The answer must be exact: $\pi - 1$ , not a decimal approximation.
The equation must be formed from arc lengths (not areas) for the method mark.
Consistent use of $\theta$ instead of $\alpha$ is condoned, but the working must be in radians throughout.

Techniques used

express the arc length of a sector as r times the angle in radiansexpress the major arc length using the reflex angleform and solve a linear equation in alpha

(b)

The area of the shaded sector is $18(\pi^2 - 1)$ .

Find the exact value of $r$ .
Simplify your answer.

DifficultyMedium

Worked solution

Approach

Use the sector area formula $A = \frac{1}{2}r^2\alpha$ with $\alpha = \pi - 1$ from part (a), set it equal to $18(\pi^2 - 1)$ , and solve for $r$ , simplifying with the difference of two squares.

Working

Using the area of the shaded sector:

\frac{1}{2}r^2(\pi - 1) = 18(\pi^2 - 1)

Solve for $r^2$ :

r^2 = \frac{36(\pi^2 - 1)}{\pi - 1}

Factorise using $\pi^2 - 1 = (\pi - 1)(\pi + 1)$ :

r^2 = \frac{36(\pi - 1)(\pi + 1)}{\pi - 1} = 36(\pi + 1)

Take the positive square root:

r = \sqrt{36(\pi + 1)} = 6\sqrt{\pi + 1}

Answer

r = 6\sqrt{\pi + 1}

Final answer

r = 6√(π + 1)

Detailed explanation

Walkthrough

The area of a sector with radius $r$ and angle $\alpha$ radians is $\frac{1}{2}r^2\alpha$ . From part (a), $\alpha = \pi - 1$ , so we write $\frac{1}{2}r^2(\pi - 1) = 18(\pi^2 - 1)$ and multiply both sides by 2 to get $r^2(\pi - 1) = 36(\pi^2 - 1)$ . Dividing by $(\pi - 1)$ gives $r^2 = \frac{36(\pi^2 - 1)}{\pi - 1}$ . The key simplification is recognising $\pi^2 - 1$ as a difference of two squares: $(\pi - 1)(\pi + 1)$ . Cancelling $(\pi - 1)$ leaves $r^2 = 36(\pi + 1)$ , so $r = 6\sqrt{\pi + 1}$ , taking the positive root since $r$ is a length.

Key Takeaways

Sector area is $A = \frac{1}{2}r^2\theta$ with $\theta$ in radians.
Difference of two squares, $a^2 - b^2 = (a-b)(a+b)$ , often simplifies fractions involving $\pi$ .
Lengths require the positive square root.

Common Mistakes

Using $\frac{1}{2}r^2\theta$ with $\theta$ in degrees.
Forgetting the factor $\frac{1}{2}$ in the sector area formula.
Leaving the answer as $\sqrt{36(\pi+1)}$ without taking out the factor 6 — the question says "Simplify your answer".
Giving a negative root alongside or instead of the positive one.
Not carrying forward the value of $\alpha$ from part (a).

Things to Be Careful About

The final answer must be the fully simplified exact form $6\sqrt{\pi + 1}$ ; unsimplified forms score the method mark but not the final accuracy mark.
This is a calculator-permitted paper, but the answer is still required exactly, not as a decimal.
The method mark for taking the square root depends on having correctly reached $r^2$ first.

Techniques used

apply the sector area formula half r squared thetasubstitute the value of alpha found earliersimplify using difference of two squarestake the square root and simplify the surd

Q73MMedium-EasyTrigonometry

Show that $\operatorname{cosec} x - \sin x$ can be written as $\frac{\cos x}{\tan x}$ .

DifficultyMedium-Easy

Worked solution

Approach

Rewrite $\operatorname{cosec} x$ as $\frac{1}{\sin x}$ , combine over a common denominator, use $\cos^2 x = 1 - \sin^2 x$ , and finish by writing $\frac{\cos x}{\sin x} = \cot x$ .

Working

\operatorname{cosec} x - \sin x = \frac{1}{\sin x} - \sin x

Combine over the common denominator $\sin x$ :

= \frac{1 - \sin^2 x}{\sin x}

Using the identity $\sin^2 x + \cos^2 x = 1$ , so that $1 - \sin^2 x = \cos^2 x$ :

= \frac{\cos^2 x}{\sin x}

Split and convert to cotangent:

= \frac{\cos x}{\sin x} \times \cos x = \cot x \times \cos x = \frac{\cos x}{\tan x}

This is exactly the required form, so $\operatorname{cosec} x - \sin x = \frac{\cos x}{\tan x}$ as required.

Answer

\operatorname{cosec} x - \sin x = \frac{\cos x}{\tan x}

Final answer

Shown: cosec x - sin x = cos x / tan x

Detailed explanation

Walkthrough

The proof starts from the left-hand side. The first move is to express everything in terms of sine and cosine only: since $\operatorname{cosec} x$ is defined as $\frac{1}{\sin x}$ , the expression becomes $\frac{1}{\sin x} - \sin x$ . Putting both terms over the common denominator $\sin x$ gives $\frac{1 - \sin^2 x}{\sin x}$ . The Pythagorean identity $\sin^2 x + \cos^2 x = 1$ rearranges to $1 - \sin^2 x = \cos^2 x$ , so the numerator becomes $\cos^2 x$ , giving $\frac{\cos^2 x}{\sin x}$ . The mark scheme insists on one more step after this: split the fraction as $\frac{\cos x}{\sin x} \times \cos x$ , recognise $\frac{\cos x}{\sin x}$ as $\cot x$ , and then use $\cot x = \frac{1}{\tan x}$ so that $\cot x \times \cos x = \frac{\cos x}{\tan x}$ — exactly the printed target, reached forward from the left-hand side.

Key Takeaways

Convert every trig function back to sines and cosines as the standard opening move of any identity proof.
The Pythagorean identity $\sin^2 x + \cos^2 x = 1$ is the bridge between expressions like $1 - \sin^2 x$ and $\cos^2 x$ .
For "show that" questions you must derive forward to the exact printed form; an extra simplifying step (here, converting $\frac{\cos x}{\sin x}$ to $\cot x$ ) is often explicitly required rather than stopping at an intermediate fraction.

Common Mistakes

Stopping at $\frac{\cos^2 x}{\sin x}$ : the mark scheme requires an additional justified step before the given answer, so this scores the final A1 only if converted further.
Omitting the argument $x$ consistently — the scheme withholds the last A mark for consistent omission of $x$ .
Cancelling $\cos^2 x / \sin x$ incorrectly or claiming it equals $\tan x$ instead of handling the split correctly.
Starting from the right-hand side is allowed, but working backwards from the target without justifying each line loses marks.

Things to Be Careful About

This is an AG (Answer Given) part: every algebraic line between the start and $\frac{\cos x}{\tan x}$ must be shown — insufficient detail forfeits the accuracy marks.
Keep the variable $x$ on every trig function throughout; dropping it costs the final A mark.
The final step must be fully justified: show $\frac{\cos x}{\sin x} = \cot x$ and $\cot x = \frac{1}{\tan x}$ explicitly.

Techniques used

rewrite cosecant in terms of sinecombine into a single fractionapply the Pythagorean identity sin^2 x + cos^2 x = 1convert cos x / sin x to cot x

Q85MMediumCalculus

Variables $x$ and $y$ are related by the equation

y = \frac{5x + 1}{x - 2}

Using calculus, find the approximate change in $x$ when $y$ increases from 4 by the small amount $h$ .

DifficultyMedium

Worked solution

Approach

First find the value of $x$ at which $y = 4$ . Then differentiate $y$ with respect to $x$ using the quotient rule, evaluate the derivative at that point, and use the small-increment relation $\delta x = \dfrac{h}{\frac{\mathrm{d}y}{\mathrm{d}x}}$ .

Working

When $y = 4$ :

\frac{5x + 1}{x - 2} = 4 \implies 5x + 1 = 4(x - 2) \implies 5x + 1 = 4x - 8 \implies x = -9

Differentiate using the quotient rule with $u = 5x + 1$ and $v = x - 2$ :

\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{(x - 2)(5) - (5x + 1)(1)}{(x - 2)^2} = \frac{5x - 10 - 5x - 1}{(x - 2)^2} = \frac{-11}{(x - 2)^2}

At $x = -9$ :

\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{-11}{(-9 - 2)^2} = \frac{-11}{121} = -\frac{1}{11}

Since $y$ increases by the small amount $h$ , the corresponding change in $x$ is

\delta x = \frac{h}{\frac{\mathrm{d}y}{\mathrm{d}x}} = \frac{h}{-\frac{1}{11}} = -11h

Answer

\delta x = -11h

Final answer

-11h

Detailed explanation

Walkthrough

The question gives the change in $y$ (an increase of $h$ from 4) and asks for the corresponding change in $x$ , so we need the rate of change of $x$ with respect to $y$ at the point where $y = 4$ .

First, we must locate the point. Setting $y = 4$ in $y = \frac{5x+1}{x-2}$ and cross-multiplying gives $5x + 1 = 4x - 8$ , so $x = -9$ . This step is essential: the derivative must be evaluated at the right point.

Next we differentiate $y = \frac{5x+1}{x-2}$ with the quotient rule, taking $u = 5x+1$ and $v = x-2$ :

\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{v\frac{\mathrm{d}u}{\mathrm{d}x} - u\frac{\mathrm{d}v}{\mathrm{d}x}}{v^2} = \frac{5(x-2) - (5x+1)}{(x-2)^2} = \frac{-11}{(x-2)^2}

At $x = -9$ this equals $\frac{-11}{121} = -\frac{1}{11}$ .

Finally, the small-change approximation: since $\delta y \approx \frac{\mathrm{d}y}{\mathrm{d}x}\,\delta x$ , rearranging gives $\delta x = \frac{\delta y}{\frac{\mathrm{d}y}{\mathrm{d}x}} = \frac{h}{-1/11} = -11h$ . The negative sign is meaningful: as $y$ increases, $x$ decreases.

(Equivalently, one may invert the relation to $x = \frac{1+2y}{y-5}$ and differentiate $\frac{\mathrm{d}x}{\mathrm{d}y}$ directly, then multiply by $h$ at $y = 4$ — the mark scheme accepts this route too.)

Key Takeaways

The small-increment approximation $\delta y \approx \frac{\mathrm{d}y}{\mathrm{d}x}\,\delta x$ works in both directions; when the change in the dependent variable is given, rearrange to $\delta x = \frac{\delta y}{\frac{\mathrm{d}y}{\mathrm{d}x}}$ .
The quotient rule requires the correct structure $\frac{v u' - u v'}{v^2}$ ; the denominator must be $v^2$ .
Before evaluating a derivative, always find the exact point of evaluation — here it had to be recovered from the given $y$ -value.

Common Mistakes

Evaluating the derivative at $x = 4$ instead of first solving for the $x$ -value where $y = 4$ (the mark scheme explicitly disallows this).
Getting the quotient rule numerator backwards in a way that changes the sign, or omitting the square in the denominator $(x-2)^2$ .
Forgetting the negative sign: $\delta x = -11h$ , not $11h$ — as $y$ increases, $x$ must decrease here.
Confusing the direction of the approximation and writing $\delta y = \frac{\mathrm{d}y}{\mathrm{d}x}\,\delta x$ but then dividing incorrectly.
Sign slips when simplifying $5x - 10 - 5x - 1$ .

Things to Be Careful About

The answer must be given in terms of $h$ : $\delta x = -11h$ , an exact symbolic answer, not a number.
The mark scheme awards the M1 for $\delta x = \frac{h}{\text{their } \frac{\mathrm{d}y}{\mathrm{d}x}}$ only when evaluated at the correct point ( $x = -9$ or a correct follow-through value), and it depends on the derivative being attempted.
The quotient rule mark requires the correct structure with the correct $u$ and $v$ ; minor errors in $\frac{\mathrm{d}u}{\mathrm{d}x}$ or $\frac{\mathrm{d}v}{\mathrm{d}x}$ may still earn it, but the denominator must be $v^2$ .
Keep the negative sign through to the end; the final answer is negative because $y$ increases as $x$ decreases near this point.

Techniques used

differentiate a quotient using the quotient rulesolve for the x-value corresponding to a given y-valueuse the small-increment relation delta x = h / (dy/dx)

Q9MediumLogarithmic and exponential functions

(a)

Write $\frac{1}{\log_x \mathrm{e}} - \ln(x - 1)$ as a single logarithm to base $\mathrm{e}$ .

DifficultyMedium-Easy

Worked solution

Approach

Use the change-of-base result $\frac{1}{\log_x \mathrm{e}} = \ln x$ , then combine the two natural logarithms with the quotient law.

Working

By the change of base rule,

\frac{1}{\log_x \mathrm{e}} = \frac{1}{\frac{\ln \mathrm{e}}{\ln x}} = \frac{1}{\frac{1}{\ln x}} = \ln x

So the expression becomes

\ln x - \ln(x - 1)

and by the quotient law of logarithms,

\ln x - \ln(x-1) = \ln\left(\frac{x}{x-1}\right)

Answer

\ln\left(\frac{x}{x-1}\right)

Final answer

ln(x / (x - 1))

Detailed explanation

Walkthrough

The key fact is that $\log_x \mathrm{e}$ means 'the power to which $x$ must be raised to give $\mathrm{e}$ '. By the change-of-base formula, $\log_x \mathrm{e} = \frac{\ln \mathrm{e}}{\ln x} = \frac{1}{\ln x}$ , since $\ln \mathrm{e} = 1$ . Its reciprocal is therefore just $\ln x$ .

Once the first term has become $\ln x$ , the whole expression is $\ln x - \ln(x-1)$ , and subtracting logarithms corresponds to dividing their arguments: $\ln A - \ln B = \ln\frac{A}{B}$ . This gives the single logarithm $\ln\frac{x}{x-1}$ .

Key Takeaways

The reciprocal identity $\log_a b = \frac{1}{\log_b a}$ , and in particular $\frac{1}{\log_x \mathrm{e}} = \ln x$ .
The quotient law: $\ln A - \ln B = \ln\frac{A}{B}$ .

Common Mistakes

Writing $\frac{1}{\log_x \mathrm{e}} = \log_{\mathrm{e}} x$ incorrectly as $\log_x \mathrm{e}$ itself — the reciprocal relation must be used.
Trying to combine $\ln x - \ln(x-1)$ into $\frac{\ln x}{\ln(x-1)}$ : the quotient law applies to the arguments, not to the logarithms themselves.
Forgetting that the answer must be a single logarithm to base $\mathrm{e}$ (i.e. a $\ln$ ), not base $x$ or base 10.

Things to Be Careful About

The mark scheme awards B1 for seeing $\frac{1}{\log_x \mathrm{e}} = \ln x$ (or equivalent), so this step must be visible even if it feels obvious.
The final answer must be exactly one logarithm: $\ln\frac{x}{x-1}$ ; leaving $\ln x - \ln(x-1)$ does not answer the question.

Techniques used

use the change of base rule to rewrite 1 over log_x e as ln xcombine two logarithms into one using the quotient law

(b)

Solve the equation $2\log_5 2x = 1 + \log_5(9x - 10)$ .

5M Similar

DifficultyMedium

Worked solution

Approach

Rewrite both sides as single logarithms to base 5: use the power law on the left and write $1 = \log_5 5$ on the right. Then equate the arguments and solve the resulting quadratic, discarding any solution that makes a logarithm undefined.

Working

Using the power law $k\log_5 A = \log_5 A^k$ :

2\log_5 2x = \log_5 (2x)^2 = \log_5 (4x^2)

Also write the constant as a logarithm:

1 = \log_5 5

So the equation becomes

\log_5 (4x^2) = \log_5 5 + \log_5 (9x - 10) = \log_5 \left(5(9x - 10)\right)

Equating the arguments:

4x^2 = 5(9x - 10)

4x^2 - 45x + 50 = 0

Factorising:

(4x - 5)(x - 10) = 0

x = \frac{5}{4} \quad \text{or} \quad x = 10

Check validity: for $x = \frac{5}{4}$ , $2x = \frac{5}{2} > 0$ and $9x - 10 = \frac{45}{4} - 10 = \frac{5}{4} > 0$ ; for $x = 10$ , $2x = 20 > 0$ and $9x - 10 = 80 > 0$ . Both are valid.

Answer

x = \frac{5}{4}, \quad x = 10

Final answer

x = 5/4, x = 10

Detailed explanation

Walkthrough

The equation mixes a multiple of a logarithm with a constant plus another logarithm. The strategy is always the same: get everything into the form 'single log = single log', then drop the logs.

First, the left side $2\log_5 2x$ becomes $\log_5 (2x)^2 = \log_5 4x^2$ by the power law. Second, the constant $1$ on the right is written as $\log_5 5$ , so the right side is $\log_5 5 + \log_5(9x-10) = \log_5\left(5(9x-10)\right)$ by the product law.

Now both sides are single logarithms to the same base, so their arguments must be equal: $(2x)^2 = 5(9x-10)$ . Expanding gives $4x^2 = 45x - 50$ , i.e. $4x^2 - 45x + 50 = 0$ , which factorises as $(4x-5)(x-10) = 0$ , giving $x = \frac{5}{4}$ or $x = 10$ .

Finally each candidate must be checked against the domains of the original logarithms: we need $2x > 0$ and $9x - 10 > 0$ . Both candidates satisfy these, so both are accepted.

Key Takeaways

Power law: $k\log_b A = \log_b A^k$ ; product law: $\log_b A + \log_b B = \log_b(AB)$ .
Any constant $c$ can be written as $\log_b b^c$ to match bases.
If $\log_b P = \log_b Q$ then $P = Q$ — but only after both sides are genuine single logarithms.
Logarithm arguments must be positive, so solutions must be checked against the original equation's domain.

Common Mistakes

The mark scheme explicitly flags M0 XP for writing $\log_5(2x)^2 = \log_5 5 \times \log_5(9x-10)$ — treating addition of logarithms as multiplication of logarithms. Addition inside logs means multiplying the arguments.
Rejecting one of the two valid solutions: the mark scheme awards A0 if either solution is rejected. Both $x = \frac{5}{4}$ and $x = 10$ satisfy the domain conditions here.
Forgetting to square the whole of $2x$ : $\log_5(2x)^2 = \log_5 4x^2$ , not $\log_5 2x^2$ .
Sign slips forming $4x^2 - 45x + 50 = 0$ from $4x^2 = 45x - 50$ .
Accepting a solution without checking $2x > 0$ and $9x - 10 > 0$ — in general such checks can remove extraneous roots.

Things to Be Careful About

The mark scheme requires 'no other solutions' and rejects answers where either root is discarded (A0 if either solution is rejected).
The quadratic must be solved correctly from the candidate's own three-term quadratic (follow-through allowed via factorising, the formula, or completing the square), but the final values must be exact: $x = \frac{5}{4}$ and $x = 10$ .
Missing brackets around $2x$ before squaring changes the equation entirely; the scheme allows recovery of missing brackets but the correct form $(2x)^2$ earns the marks.

Techniques used

apply the power law of logarithmswrite a constant as a logarithm of the same baseequate arguments when the logarithms are equalsolve the resulting quadratic and check for extraneous solutions

Q10MediumSeries

It is given that a geometric progression has the following terms.

\begin{aligned} \text{2nd term} & \quad 11x - 1 \\ \text{3rd term} & \quad x + 5 \\ \text{4th term} & \quad x - 1 \end{aligned}

(a)

Given that $x > 0$ , find the common ratio and first term of the progression.

6M Similar

DifficultyMedium

Worked solution

Approach

In a geometric progression each term is obtained by multiplying by the common ratio $r$ , so the ratio of consecutive terms is constant:

\frac{\text{3rd term}}{\text{2nd term}} = \frac{\text{4th term}}{\text{3rd term}}

Substitute the given expressions in $x$ , solve for $x$ using $x > 0$ , then read off $r$ and find the first term $a$ .

Working

\frac{x + 5}{11x - 1} = \frac{x - 1}{x + 5}

Cross-multiplying and expanding both sides:

(x + 5)^2 = (11x - 1)(x - 1)

x^2 + 10x + 25 = 11x^2 - 12x + 1

10x^2 - 22x - 24 = 0

Dividing by 2:

5x^2 - 11x - 12 = 0

Factorising:

(5x + 4)(x - 3) = 0

so $x = -\frac{4}{5}$ or $x = 3$ . Since $x > 0$ :

x = 3

Then the terms are: 2nd term $= 11(3) - 1 = 32$ , 3rd term $= 3 + 5 = 8$ , 4th term $= 3 - 1 = 2$ . Hence

r = \frac{8}{32} = \frac{1}{4}

The first term satisfies $ar = 32$ , so

a = \frac{32}{\frac{1}{4}} = 128

Answer

r = \frac{1}{4}, \qquad a = 128

Final answer

common ratio = 1/4, first term = 128

Detailed explanation

Walkthrough

A geometric progression has a constant ratio between consecutive terms, so the ratio of the 3rd term to the 2nd must equal the ratio of the 4th term to the 3rd. Writing this single equation in $x$ gives $(x+5)/(11x-1) = (x-1)/(x+5)$ . Cross-multiplying produces a quadratic: expanding $(x+5)^2 = x^2 + 10x + 25$ and $(11x-1)(x-1) = 11x^2 - 12x + 1$ , then collecting everything onto one side gives $10x^2 - 22x - 24 = 0$ , which halves to $5x^2 - 11x - 12 = 0$ . This factorises as $(5x+4)(x-3) = 0$ , giving roots $x = -4/5$ and $x = 3$ ; the condition $x > 0$ selects $x = 3$ . Substituting back gives the actual terms 32, 8, 2, so the common ratio is $r = 8/32 = 1/4$ . Since the 2nd term is $ar = 32$ , the first term is $a = 32 \div \frac{1}{4} = 128$ .

Key Takeaways

The defining property of a GP — equal ratios of consecutive terms — turns unknown-term problems into equations.
Conditions such as $x > 0$ are there to discard extraneous roots; always check which root survives.
Once the terms are known numerically, $r$ follows directly and $a$ follows from any known term via $u_n = ar^{n-1}$ .

Common Mistakes

Expanding $(11x-1)(x-1)$ incorrectly (sign slips give $11x^2 + 12x + 1$ or similar); the mark scheme allows one expansion error but not more.
Keeping both roots and reporting $r$ or $a$ from $x = -4/5$ , ignoring the given $x > 0$ .
Dividing the equation by $x$ at some stage and losing a root.
Finding $r$ correctly but computing $a$ as $32 \times \frac{1}{4}$ instead of $32 \div \frac{1}{4}$ .
The mark scheme notes $x = 3$ scores only 'nfww' — no marks if reached from wrong working.

Things to Be Careful About

Show every algebraic line: the scheme awards M1 for the correct set-up, A1 for the simplified quadratic, A1 for the correct root, and A1 for both $r$ and $a$ together — both values are needed for that final mark.
The second solution method (writing terms as $a$ , $ar$ , $ar^2$ ) is also accepted, but follow one route fully rather than mixing them.
Answers should be exact fractions here ( $r = \frac{1}{4}$ ), not decimals.

Techniques used

equate successive ratios of consecutive terms of a geometric progressionexpand and simplify to form a quadratic equationfactorise the quadratic and select the root satisfying the given conditionrecover the common ratio and first term from the terms

(b)

The sum to infinity of this progression is $\frac{k}{3}$ where $k$ is an integer.

Find the value of $k$ .

DifficultyMedium-Easy

Worked solution

Approach

Use the sum-to-infinity formula $S_\infty = \dfrac{a}{1 - r}$ with the values $a = 128$ and $r = \frac{1}{4}$ carried forward from part (a), then compare with $\frac{k}{3}$ .

Working

Since $|r| = \frac{1}{4} < 1$ , the sum to infinity exists:

S_\infty = \frac{a}{1 - r} = \frac{128}{1 - \frac{1}{4}} = \frac{128}{\frac{3}{4}} = 128 \times \frac{4}{3} = \frac{512}{3}

Comparing with $S_\infty = \frac{k}{3}$ :

k = 512

Answer

k = 512

Final answer

512

Detailed explanation

Walkthrough

This part is a direct application of the sum-to-infinity formula, which applies because the common ratio satisfies $|r| < 1$ . Carrying forward $a = 128$ and $r = \frac{1}{4}$ from part (a), we compute $S_\infty = \frac{128}{1 - \frac{1}{4}} = \frac{128}{\frac{3}{4}} = \frac{512}{3}$ . The question states this equals $\frac{k}{3}$ with $k$ an integer, so comparing denominators gives $k = 512$ .

Key Takeaways

$S_\infty = \frac{a}{1-r}$ is valid only when $|r| < 1$ — always check this before applying it.
Multi-part GP questions thread results forward: part (b) depends entirely on the $a$ and $r$ found in part (a).

Common Mistakes

Using $1 + r$ instead of $1 - r$ in the denominator, giving $\frac{128}{1.25} = 102.4$ instead of $\frac{512}{3}$ .
Forgetting to divide by $\frac{3}{4}$ correctly — dividing by a fraction means multiplying by its reciprocal.
Reporting $\frac{512}{3}$ as the answer when the question asks specifically for the integer $k$ .
Using values of $a$ or $r$ inconsistent with part (a); the mark scheme allows follow-through only on numeric values with $|r| < 1$ .

Things to Be Careful About

The M1 depends on having earned at least one method mark in part (a), so part (a)'s working must be shown properly.
Give the exact value $k = 512$ ; no rounding is involved here.

Techniques used

apply the sum-to-infinity formula for a geometric progressionsubstitute the first term and common ratio found earlier

Q11MediumVectors in two dimensions

The position vectors of points $A$ and $B$ relative to an origin $O$ are

\overrightarrow{OA} = \begin{pmatrix} 4 \\ -2 \end{pmatrix} \quad \text{and} \quad \overrightarrow{OB} = \begin{pmatrix} -4 \\ 6 \end{pmatrix}.

The point $C$ lies on $AB$ and is such that $AC : CB = 3 : 1$ .

The point $D$ lies on $OA$ such that $\overrightarrow{OD} = \lambda \overrightarrow{OA}$ and $\overrightarrow{CD} = \begin{pmatrix} 5 \\ -5.5 \end{pmatrix}$ .

(a)

Find $\overrightarrow{OD}$ .

5M Similar

DifficultyMedium

Worked solution

Approach

First find $\overrightarrow{AB}$ from the two position vectors. Since $AC : CB = 3 : 1$ , point $C$ is $\frac{3}{4}$ of the way along $AB$ , so $\overrightarrow{AC} = \frac{3}{4}\overrightarrow{AB}$ . Then $\overrightarrow{OC} = \overrightarrow{OA} + \overrightarrow{AC}$ , and finally $\overrightarrow{OD} = \overrightarrow{OC} + \overrightarrow{CD}$ .

Working

\overrightarrow{AB} = \overrightarrow{OB} - \overrightarrow{OA} = \begin{pmatrix}-4 \\ 6\end{pmatrix} - \begin{pmatrix}4 \\ -2\end{pmatrix} = \begin{pmatrix}-8 \\ 8\end{pmatrix}

Since $AC : CB = 3 : 1$ :

\overrightarrow{AC} = \frac{3}{4}\overrightarrow{AB} = \frac{3}{4}\begin{pmatrix}-8 \\ 8\end{pmatrix} = \begin{pmatrix}-6 \\ 6\end{pmatrix}

Then:

\overrightarrow{OC} = \overrightarrow{OA} + \overrightarrow{AC} = \begin{pmatrix}4 \\ -2\end{pmatrix} + \begin{pmatrix}-6 \\ 6\end{pmatrix} = \begin{pmatrix}-2 \\ 4\end{pmatrix}

Since $\overrightarrow{CD} = \overrightarrow{OD} - \overrightarrow{OC}$ :

\overrightarrow{OD} = \overrightarrow{OC} + \overrightarrow{CD} = \begin{pmatrix}-2 \\ 4\end{pmatrix} + \begin{pmatrix}5 \\ -5.5\end{pmatrix} = \begin{pmatrix}3 \\ -1.5\end{pmatrix}

Answer

\overrightarrow{OD} = \begin{pmatrix}3 \\ -1.5\end{pmatrix}

Final answer

(3, -1.5)

Detailed explanation

Walkthrough

The key idea in vector geometry questions like this is that every displacement can be written as the difference of two position vectors. We start by computing $\overrightarrow{AB}$ , which tells us how to travel from $A$ to $B$ . The ratio $AC : CB = 3 : 1$ means $C$ divides the segment into four equal parts, taking three of them from $A$ , so $\overrightarrow{AC} = \frac{3}{4}\overrightarrow{AB}$ . Adding this to $\overrightarrow{OA}$ gives the position vector of $C$ . Finally, we are told $\overrightarrow{CD} = \begin{pmatrix}5 \\ -5.5\end{pmatrix}$ ; rearranging $\overrightarrow{CD} = \overrightarrow{OD} - \overrightarrow{OC}$ gives $\overrightarrow{OD} = \overrightarrow{OC} + \overrightarrow{CD}$ , which evaluates directly.

Key Takeaways

A displacement vector between two points is found by subtracting their position vectors component by component.
A ratio $m : n$ on a line segment means the fraction $\frac{m}{m+n}$ of the whole segment from the first point.
Vector equations such as $\overrightarrow{OC} = \overrightarrow{OA} + \overrightarrow{AC}$ let you hop between known and unknown points.

Common Mistakes

Subtracting in the wrong order when finding $\overrightarrow{AB}$ (it must be $B - A$ , giving $(-8, 8)$ , not $(8, -8)$ ).
Using $\frac{3}{1}$ or $\frac{3}{3}$ instead of $\frac{3}{4}$ as the fraction of $\overrightarrow{AB}$ for $\overrightarrow{AC}$ .
Writing $\overrightarrow{OD} = \overrightarrow{CD} - \overrightarrow{OC}$ instead of adding — check the direction of each arrow carefully.
Giving coordinates instead of a column vector: the mark scheme states the final answer must be a vector.

Things to Be Careful About

The final answer must be presented as a column vector $\begin{pmatrix}3 \\ -1.5\end{pmatrix}$ — the scheme marks the final answer only if it is a vector.
Keep components exact; $-1.5$ is exact here so no rounding issues arise.
The scheme allows equivalent routes (e.g. via $\overrightarrow{AD} = \overrightarrow{AC} + \overrightarrow{CD}$ ), but every route must show the intermediate vectors.

Techniques used

find a displacement vector by subtracting position vectorsuse a given ratio to locate a point dividing a line segmentcombine position vectors using vector addition

(b)

Hence find the value of $\lambda$ .

DifficultyEasy

Worked solution

Approach

Using the result from part (a), compare $\overrightarrow{OD} = \lambda\overrightarrow{OA}$ component by component.

Working

From part (a):

\overrightarrow{OD} = \begin{pmatrix}3 \\ -1.5\end{pmatrix}, \qquad \overrightarrow{OA} = \begin{pmatrix}4 \\ -2\end{pmatrix}

Using the first component:

4\lambda = 3 \quad \Rightarrow \quad \lambda = \frac{3}{4}

Check with the second component: $-2 \times \frac{3}{4} = -1.5$ ✓

Answer

\lambda = \frac{3}{4}

Final answer

0.75

Detailed explanation

Walkthrough

Part (a) gave us $\overrightarrow{OD} = \begin{pmatrix}3 \\ -1.5\end{pmatrix}$ , and the question defines $D$ by $\overrightarrow{OD} = \lambda\overrightarrow{OA}$ . Two vectors are equal only when all matching components are equal, so equating the top components gives $4\lambda = 3$ , hence $\lambda = \frac{3}{4}$ . The bottom components confirm this: $-2 \times \frac{3}{4} = -1.5$ , exactly as required.

Key Takeaways

Equality of vectors means equality of every component, giving one equation per component.
A scalar multiple $\lambda\mathbf{a}$ scales both components of $\mathbf{a}$ by the same factor, so either component can be used to find $\lambda$ .

Common Mistakes

Dividing the wrong way round ( $\lambda = \frac{4}{3}$ instead of $\frac{3}{4}$ ).
Not checking the second component, which would catch an arithmetic slip in part (a).

Things to Be Careful About

The mark scheme requires full marks in part (a) for this B1 — an error in $\overrightarrow{OD}$ costs this mark too, since $\lambda$ depends on it.
Both $0.75$ and $\frac{3}{4}$ are accepted forms.

Techniques used

compare corresponding components of parallel vectorssolve for a scalar multiple

Q1211MMedium-HardCalculusLogarithmic and exponential functions

Fig. 3

The diagram shows part of the curve $y = \frac{10}{4x + 1}$ and the line $L$ .

$L$ passes through the points $(0, 6)$ and $(1.5, 0)$ .
The curve and $L$ intersect at the points $P$ and $Q$ .

Find the exact area of the shaded region.

DifficultyMedium-Hard

Worked solution

Approach

The shaded region lies between the line $L$ (above) and the curve (below), between their intersection points $P$ and $Q$ . Find the equation of $L$ , solve for the intersection $x$ -coordinates, then compute (area under line) − (area under curve).

Working

Equation of $L$ . Through $(0,6)$ and $(1.5,0)$ :

\text{gradient} = \frac{0 - 6}{1.5 - 0} = -4 \quad \Rightarrow \quad y = -4x + 6

Intersection points. Set curve equal to line:

\frac{10}{4x+1} = -4x + 6

10 = (-4x + 6)(4x + 1) = -16x^2 + 20x + 6

16x^2 - 20x + 4 = 0 \quad \Rightarrow \quad 4x^2 - 5x + 1 = 0

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

Corresponding $y$ -values: at $x = \frac{1}{4}$ , $y = -4\left(\frac{1}{4}\right) + 6 = 5$ ; at $x = 1$ , $y = 2$ . So $P = \left(\frac{1}{4}, 5\right)$ , $Q = (1, 2)$ .

Area under the line (trapezium with parallel sides $5$ and $2$ , width $1 - \frac{1}{4} = \frac{3}{4}$ ):

\text{Trapezium} = \frac{1}{2}(5 + 2) \times \frac{3}{4} = \frac{21}{8}

(Equivalently, $\displaystyle\int_{\frac14}^{1}(-4x+6)\,\mathrm{d}x = \left[-2x^2 + 6x\right]_{\frac14}^{1} = 4 - \frac{11}{8} = \frac{21}{8}$ .)

Area under the curve:

\int \frac{10}{4x+1}\,\mathrm{d}x = \frac{10}{4}\ln(4x+1)

Evaluating between the limits:

\left[\frac{10}{4}\ln(4x+1)\right]_{\frac{1}{4}}^{1} = \frac{5}{2}\ln 5 - \frac{5}{2}\ln 2

Shaded area:

\frac{21}{8} - \left(\frac{5}{2}\ln 5 - \frac{5}{2}\ln 2\right)

Answer

\frac{21}{8} - \frac{5}{2}\ln\frac{5}{2}

Final answer

21/8 − (5/2)ln(5/2)

Detailed explanation

Walkthrough

The shaded region is bounded above by the straight line $L$ and below by the curve $y = \frac{10}{4x+1}$ , so its area is the integral of (line − curve) between the two intersection points.

First we need the equation of $L$ . It passes through $(0, 6)$ and $(1.5, 0)$ , so its gradient is $\frac{0-6}{1.5-0} = -4$ , giving $y = -4x + 6$ .

Next, find where the line meets the curve by solving $\frac{10}{4x+1} = -4x + 6$ . Multiplying both sides by $(4x+1)$ gives a quadratic: $10 = -16x^2 + 20x + 6$ , which rearranges to $4x^2 - 5x + 1 = 0$ . This factorises as $(4x-1)(x-1) = 0$ , so $x = \frac{1}{4}$ or $x = 1$ . Substituting back into the line gives $y = 5$ and $y = 2$ : these are points $P$ and $Q$ .

The area under the line between these $x$ -values is a trapezium with parallel sides of lengths $5$ and $2$ and width $\frac{3}{4}$ , so its area is $\frac{1}{2}(5+2)\times\frac{3}{4} = \frac{21}{8}$ .

The area under the curve needs integration of a reciprocal function. Using the standard result $\int \frac{1}{ax+b}\,\mathrm{d}x = \frac{1}{a}\ln(ax+b)$ , we get $\int \frac{10}{4x+1}\,\mathrm{d}x = \frac{10}{4}\ln(4x+1) = \frac{5}{2}\ln(4x+1)$ . Evaluating from $\frac{1}{4}$ to $1$ gives $\frac{5}{2}\ln 5 - \frac{5}{2}\ln 2$ .

Subtracting, the shaded area is $\frac{21}{8} - \frac{5}{2}\ln\frac{5}{2}$ , using the log law $\ln 5 - \ln 2 = \ln\frac{5}{2}$ .

Key Takeaways

The area between a curve and a line equals the integral of (upper function − lower function) over the interval between their intersections.
Intersections are found by equating the two expressions and solving — here producing a quadratic that factorises nicely.
$\int \frac{k}{ax+b}\,\mathrm{d}x = \frac{k}{a}\ln|ax+b| + c$ ; when evaluating a definite integral of this form, the constant of integration cancels.
Log laws allow $\ln A - \ln B$ to be combined into $\ln\frac{A}{B}$ for a compact exact answer.
When one boundary is a straight line, its 'area' can be found by trapezium geometry instead of integration.

Common Mistakes

Subtracting the areas the wrong way round: the mark scheme explicitly awards A0 if the subtraction is done incorrectly — the line is above the curve on this interval, so it must be trapezium minus curve integral.
Forgetting to divide by $a = 4$ when integrating $\frac{10}{4x+1}$ ; writing just $10\ln(4x+1)$ loses the accuracy mark.
Sign errors when expanding $(-4x+6)(4x+1)$ or rearranging into standard quadratic form.
Discarding one root of the quadratic or misidentifying which point is $P$ and which is $Q$ — both roots are needed as limits.
Rounding the final answer to a decimal; the question demands the exact form involving $\ln$ .
Omitting the constant of integration in the indefinite integral (harmless here once limits are applied, but bad practice if limits are substituted late).

Things to Be Careful About

The final answer must be exact: $\frac{21}{8} - \frac{5}{2}\ln\frac{5}{2}$ (equivalently $\frac{21}{8} + \frac{5}{2}\ln\frac{2}{5}$ ). Do not evaluate numerically.
The substitution of limits into the $k\ln(4x+1)$ form depends on all earlier method marks being correct ('dep' marking), so keep every step visible.
Transcription errors in copying values forward are allowed ('allow transcription error'), but wrong-way-round subtraction is not.
Check the direction of subtraction by testing a point inside the interval: at $x = \frac{1}{2}$ , the line gives $y = 4$ while the curve gives $\frac{10}{3} \approx 3.33$ , confirming the line is above.

Techniques used

find the equation of the line from two pointssolve a quadratic formed by equating curve and line to locate intersection pointsintegrate a reciprocal linear function to give a logarithmsubtract the area under the curve from the trapezium area under the line

Q134MMediumPermutations and combinations

Show that, for all values of $n \geq 3$ , $^{n+2}\mathrm{C}_{3} - {}^{n}\mathrm{C}_{3} = n^2$ .

DifficultyMedium

Worked solution

Approach

Write each combination in factorial form, cancel the factorials down to a product of three factors, expand both quadratics, then take out the common factor $\frac{n}{6}$ and simplify to reach $n^2$ .

Working

{}^{n+2}\mathrm{C}_{3} - {}^{n}\mathrm{C}_{3} = \frac{(n+2)!}{(n-1)!\,3!} - \frac{n!}{(n-3)!\,3!}

Cancelling the factorials:

= \frac{(n+2)(n+1)n}{6} - \frac{n(n-1)(n-2)}{6}

Expanding both numerators:

(n+2)(n+1)n = n(n^2 + 3n + 2), \qquad n(n-1)(n-2) = n(n^2 - 3n + 2)

So

= \frac{n}{6}\left[(n^2 + 3n + 2) - (n^2 - 3n + 2)\right]

= \frac{n}{6}(6n)

= n^2

as required, for all values of $n \geq 3$ (so that all factorials are defined).

Answer

{}^{n+2}\mathrm{C}_{3} - {}^{n}\mathrm{C}_{3} = n^2

Final answer

Shown: (n+2)C3 - nC3 = n^2

Detailed explanation

Walkthrough

The identity must be proved by starting from the left-hand side. The first step is to convert each binomial coefficient into its factorial definition: ${}^{m}\mathrm{C}_{r} = \frac{m!}{(m-r)!\,r!}$ . This gives $\frac{(n+2)!}{(n-1)!\,3!}$ and $\frac{n!}{(n-3)!\,3!}$ — this conversion is what earns the method mark.

Next, cancel the factorials. Since $(n+2)! = (n+2)(n+1)(n)(n-1)!$ , dividing by $(n-1)!$ leaves just the product of three consecutive factors; similarly $n! = n(n-1)(n-2)(n-3)!$ . With $3! = 6$ this gives two fractions over 6.

Expanding the brackets: $(n+2)(n+1) = n^2 + 3n + 2$ and $(n-1)(n-2) = n^2 - 3n + 2$ , so each numerator is $n$ times one of these quadratics. Taking out the common factor $\frac{n}{6}$ leaves the difference of the two quadratics, which is $(n^2 + 3n + 2) - (n^2 - 3n + 2) = 6n$ . Finally $\frac{n}{6} \times 6n = n^2$ , matching the right-hand side exactly.

Key Takeaways

The factorial form of a binomial coefficient works even when the top entry is an expression like $n+2$ : ${}^{n+2}\mathrm{C}_{3} = \frac{(n+2)(n+1)n}{3!}$ .
Factorials telescope: $(n+2)!/(n-1)!$ is just the product of the three factors from $n$ to $n+2$ .
Proving an identity means working from one side to the other without ever assuming the result.

Common Mistakes

Writing ${}^{n}\mathrm{C}_{3}$ as $\frac{n!}{3!(n-3)!}$ incorrectly, e.g. using $(n-r)!$ with the wrong $r$ or forgetting the $3!$ denominator entirely.
Sign slips when subtracting: the whole quadratic $n^2 - 3n + 2$ must be negated, giving $-(n^2 - 3n + 2)$ , not $-n^2 - 3n + 2$ .
Expanding $(n-1)(n-2)$ wrongly (a common error is $n^2 - 3n - 2$ ).
Starting from $n^2$ and working backwards — the scheme requires working from the left-hand side to obtain the right-hand side.
Cancelling factorials incorrectly, e.g. claiming $(n+2)!/(n-1)! = (n+2)! - (n-1)!$ or leaving unmatched factors.

Things to Be Careful About

The condition $n \geq 3$ matters because $(n-3)!$ must be defined; mention it in the conclusion if asked "for all values of $n \geq 3$ ".
The mark scheme allows $3!$ to be left unexpanded instead of writing 6.
The final line must read exactly $n^2$ — the answer is given (AG), so every intermediate line of correct algebra must be shown; insufficient detail loses marks.
Keep the subtraction bracketed until the expansion is complete to avoid sign errors.

Techniques used

write binomial coefficients in factorial formcancel factorials to products of three consecutive factorsexpand the resulting quadraticsfactorise out n/6 and simplify