Additional Mathematics 4037 / 13 — October/November 2025

DifficultyMedium-Easy

Worked solution

Approach

Factorise the quadratic to find its critical values, then take the region where the curve is on or above the $x$ -axis.

Working

x^2 - x - 6 = (x - 3)(x + 2)

Critical values: $x = 3$ and $x = -2$ .

Since the parabola opens upwards, $x^2 - x - 6 \geq 0$ outside the roots:

x \geq 3 \quad \text{or} \quad x \leq -2

Answer

x \geq 3 \quad \text{or} \quad x \leq -2

Final answer

x >= 3 or x <= -2

Detailed explanation

Walkthrough

We need the values of $x$ for which $x^2 - x - 6$ is zero or positive. First factorise: we need two numbers multiplying to $-6$ and adding to $-1$ , namely $-3$ and $+2$ , so $x^2 - x - 6 = (x - 3)(x + 2)$ . The curve crosses the $x$ -axis at $x = 3$ and $x = -2$ . Since the coefficient of $x^2$ is positive, the parabola opens upwards, so it is above the axis (or touching it) to the left of the smaller root and to the right of the larger root. Hence $x \geq 3$ or $x \leq -2$ .

Key Takeaways

Factorising a quadratic gives its roots, which are the critical values of any related inequality.
For an upwards-opening parabola, a $\geq 0$ inequality gives the two outer regions; a $\leq 0$ inequality gives the region between the roots.
The two parts of the answer must be joined by "or", not "and".

Common Mistakes

Writing the answer as $-2 \geq x \geq 3$ , which is impossible and is explicitly rejected by the mark scheme.
Joining the two inequalities with "and" instead of "or" — the mark scheme disallows this.
Sign errors in the factorisation (e.g. $(x+3)(x-2)$ ).
Reversing the inequality directions by forgetting the parabola opens upwards.

Things to Be Careful About

The final answer must state both inequalities correctly oriented: $x \geq 3$ and $x \leq -2$ . The mark scheme is strict: "Do not allow and between inequalities" and "Do not accept $-2 \geq x \geq 3$ ".
Include equality (the roots themselves) since the inequality is $\geq$ , not $>$ .

Techniques used

factorise the quadraticfind the critical valuesselect the correct outer region for a greater-than inequality

(b)

$|3x - 4| < x + 2$

DifficultyMedium

Worked solution

Approach

Split $|3x - 4| < x + 2$ into two cases: $3x - 4 \geq 0$ (so the modulus is $3x - 4$ ) and $3x - 4 < 0$ (so the modulus is $-(3x - 4)$ ). Solve each linear inequality and combine.

Working

Case 1: $3x - 4 \geq 0$ , so $|3x - 4| = 3x - 4$ :

3x - 4 < x + 2 \implies 2x < 6 \implies x < 3

This case requires $x \geq \frac{4}{3}$ , giving $\frac{4}{3} \leq x < 3$ .

Case 2: $3x - 4 < 0$ , so $|3x - 4| = -(3x - 4) = -3x + 4$ :

-3x + 4 < x + 2 \implies 2 < 4x \implies x > \frac{1}{2}

This case requires $x < \frac{4}{3}$ , giving $\frac{1}{2} < x < \frac{4}{3}$ .

Combining both cases:

\frac{1}{2} < x < 3

Answer

\frac{1}{2} < x < 3

Final answer

1/2 < x < 3

Detailed explanation

Walkthrough

The modulus $|3x - 4|$ equals $3x - 4$ when $3x - 4 \geq 0$ and $-(3x - 4)$ when $3x - 4 < 0$ , so we solve the inequality in each region.

Case 1 ( $x \geq \frac{4}{3}$ ): the inequality $3x - 4 < x + 2$ rearranges to $2x < 6$ , so $x < 3$ . Together with $x \geq \frac{4}{3}$ this contributes $\frac{4}{3} \leq x < 3$ .

Case 2 ( $x < \frac{4}{3}$ ): the inequality $-3x + 4 < x + 2$ rearranges to $2 < 4x$ , so $x > \frac{1}{2}$ . Together with $x < \frac{4}{3}$ this contributes $\frac{1}{2} < x < \frac{4}{3}$ .

The union of the two contributions is the single interval $\frac{1}{2} < x < 3$ .

Key Takeaways

A modulus inequality splits into two cases according to the sign of the expression inside the modulus.
Each case gives a linear inequality plus a domain restriction; the final answer is the union of the valid parts.
Equivalently, $|A| < B$ (with $B > 0$ ) means $-B < A < B$ , which leads to the same interval.

Common Mistakes

Solving only one case of the modulus and missing half the interval.
Writing the answer as two disconnected inequalities such as " $\frac{1}{2} < x$ , $x < 3$ " without "and" — the mark scheme explicitly rejects this.
Sign slips when removing the modulus in the negative case (forgetting that $|3x-4| = -3x + 4$ there).
Forgetting to check the domain restriction of each case, which here happens to merge cleanly but must be reasoned through.

Things to Be Careful About

The final answer must be written as a single connected statement $\frac{1}{2} < x < 3$ (or with "and"); the mark scheme says "Do not accept $\frac{1}{2} < x$ , $x < 3$ unless and is between the inequalities".
Strict inequalities throughout: the modulus inequality is $<$ , so the endpoints $\frac{1}{2}$ and $3$ are excluded.
An alternative accepted method is squaring both sides to get $9x^2 - 24x + 16 < x^2 + 4x + 4$ , i.e. $(2x-1)(x-3) < 0$ , giving the same critical values $\frac{1}{2}$ and $3$ — but squaring is only valid because both sides are non-negative on the solution set.

Techniques used

split the modulus into positive and negative casessolve each linear inequalitycombine the two cases into a single interval

Q23MMedium-EasyCalculus

Differentiate $x^2 \mathrm{e}^{3x}$ with respect to $x$ .

DifficultyMedium-Easy

Worked solution

Approach

Use the product rule on $x^2 \mathrm{e}^{3x}$ with $u = x^2$ and $v = \mathrm{e}^{3x}$ .

Working

\frac{\mathrm{d}u}{\mathrm{d}x} = 2x, \qquad \frac{\mathrm{d}v}{\mathrm{d}x} = 3\mathrm{e}^{3x}

(the derivative of $\mathrm{e}^{3x}$ uses the chain rule: multiply by the derivative of $3x$ ).

Product rule:

\frac{\mathrm{d}}{\mathrm{d}x}\left(x^2 \mathrm{e}^{3x}\right) = x^2 \cdot 3\mathrm{e}^{3x} + 2x \cdot \mathrm{e}^{3x}

= 3x^2\mathrm{e}^{3x} + 2x\mathrm{e}^{3x}

Answer

3x^2\mathrm{e}^{3x} + 2x\mathrm{e}^{3x}

Final answer

3x^2 e^(3x) + 2x e^(3x)

Detailed explanation

Walkthrough

The expression is a product of two functions of $x$ : the power $x^2$ and the exponential $\mathrm{e}^{3x}$ , so the product rule applies: $(uv)' = u'v + uv'$ .

First differentiate each factor. The derivative of $x^2$ is $2x$ . For $\mathrm{e}^{3x}$ , the chain rule gives the derivative as $\mathrm{e}^{3x}$ multiplied by the derivative of the exponent $3x$ , which is $3$ , so the derivative is $3\mathrm{e}^{3x}$ — this step earns the B1 in the mark scheme.

Substituting into the product rule:

u'v + uv' = (2x)(\mathrm{e}^{3x}) + (x^2)(3\mathrm{e}^{3x})

giving $2x\mathrm{e}^{3x} + 3x^2\mathrm{e}^{3x}$ , which may be left unsimplified or factored as $x\mathrm{e}^{3x}(2 + 3x)$ — later mis-simplification is ignored (isw).

Key Takeaways

Recognise when a function is a product and apply the product rule $u'v + uv'$ .
Differentiate $\mathrm{e}^{kx}$ by multiplying by $k$ (chain rule).
Either order of the two terms is acceptable; factoring is optional.

Common Mistakes

Writing only $2x\mathrm{e}^{3x}$ or only $3x^2\mathrm{e}^{3x}$ — both parts of the product rule are needed for full marks.
Forgetting the chain rule factor of $3$ when differentiating $\mathrm{e}^{3x}$ (giving just $\mathrm{e}^{3x}$ ), which loses the B1.
Using the quotient rule or adding derivatives instead of applying the product rule structure.
Sign slips are not an issue here, but mixing up $u'v$ and $uv'$ still yields the same sum here — the real trap is omitting a term entirely.

Things to Be Careful About

The mark scheme awards B1 for $3\mathrm{e}^{3x}$ , M1 for use of the product rule, and A1 FT for the correct combination with their $3\mathrm{e}^{3x}$ — so even with a wrong exponential derivative, correct product-rule structure can earn follow-through marks.
"isw" means subsequent incorrect simplification is ignored, so leaving the answer as $2x\mathrm{e}^{3x} + 3x^2\mathrm{e}^{3x}$ is perfectly fine.
Keep $\mathrm{e}$ upright as it denotes the constant, matching the printed notation.

Techniques used

differentiate the exponential factor using the chain ruleapply the product rulefactorise the result

Q34MMediumCalculusTrigonometry

In this question you may use the values in the table below.

$\theta$ radians	$\sin\theta$	$\cos\theta$	$\tan\theta$
$\frac{\pi}{6}$	$\frac{1}{2}$	$\frac{\sqrt{3}}{2}$	$\frac{\sqrt{3}}{3}$
$\frac{\pi}{3}$	$\frac{\sqrt{3}}{2}$	$\frac{1}{2}$	$\sqrt{3}$

Variables $x$ and $y$ are related by the equation $y = \sin 5x$ where $0 \leq x \leq \frac{\pi}{10}$ .

Use calculus to find the approximate change in $x$ when $y$ increases from $\frac{\sqrt{3}}{2}$ by the small amount $0.01$ .

DifficultyMedium

Worked solution

Approach

First find the value of $x$ at which $y = \frac{\sqrt{3}}{2}$ , then differentiate $y = \sin 5x$ to get $\frac{\mathrm{d}y}{\mathrm{d}x}$ there, and finally use $\frac{\delta y}{\delta x} \approx \frac{\mathrm{d}y}{\mathrm{d}x}$ to estimate $\delta x$ .

Working

Solve for $x$ when $y = \frac{\sqrt{3}}{2}$ :

\sin 5x = \frac{\sqrt{3}}{2} \quad\Rightarrow\quad 5x = \frac{\pi}{3}

x = \frac{\pi}{15}

Differentiate using the chain rule:

\frac{\mathrm{d}y}{\mathrm{d}x} = 5\cos 5x

At $x = \frac{\pi}{15}$ , so that $5x = \frac{\pi}{3}$ , and using $\cos\frac{\pi}{3} = \frac{1}{2}$ from the table:

\frac{\mathrm{d}y}{\mathrm{d}x} = 5 \times \frac{1}{2} = \frac{5}{2}

Use the small-change approximation $\dfrac{\delta y}{\delta x} \approx \dfrac{\mathrm{d}y}{\mathrm{d}x}$ with $\delta y = 0.01$ :

\frac{0.01}{\delta x} = \frac{5}{2}

\delta x = \frac{0.01 \times 2}{5} = \frac{0.02}{5} = 0.004

Answer

\delta x = 0.004

Final answer

0.004

Detailed explanation

Walkthrough

The question asks how much $x$ changes when $y$ increases by a small amount, which is a small-increment (rates of change) problem: since $\frac{\mathrm{d}y}{\mathrm{d}x}$ is the rate of change of $y$ with respect to $x$ , we have $\frac{\delta y}{\delta x} \approx \frac{\mathrm{d}y}{\mathrm{d}x}$ .

Step 1 — locate the starting point. We are told $y$ increases from $\frac{\sqrt{3}}{2}$ , so we need the $x$ where $\sin 5x = \frac{\sqrt{3}}{2}$ . From the table, $\sin\theta = \frac{\sqrt{3}}{2}$ when $\theta = \frac{\pi}{3}$ , so $5x = \frac{\pi}{3}$ and $x = \frac{\pi}{15}$ . This lies inside the allowed range $0 \leq x \leq \frac{\pi}{10}$ , confirming it is the right solution.

Step 2 — find the rate of change. Differentiating $y = \sin 5x$ by the chain rule gives $\frac{\mathrm{d}y}{\mathrm{d}x} = 5\cos 5x$ . At our point, $5x = \frac{\pi}{3}$ and the table gives $\cos\frac{\pi}{3} = \frac{1}{2}$ , so $\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{5}{2}$ .

Step 3 — apply the approximation. Writing $\frac{\delta y}{\delta x} \approx \frac{5}{2}$ with $\delta y = 0.01$ gives $\delta x = \frac{0.01}{5/2} = 0.004$ , i.e. $\frac{1}{250}$ .

Key Takeaways

The small-change formula $\delta y \approx \frac{\mathrm{d}y}{\mathrm{d}x}\,\delta x$ links a tiny change in one variable to the resulting tiny change in the other; rearranged here it gives $\delta x \approx \delta y \div \frac{\mathrm{d}y}{\mathrm{d}x}$ .
The derivative must be evaluated at the point where the change starts, not left in general form.
The chain rule is needed whenever the argument of a trig function is a multiple of $x$ : differentiate the outer function, then multiply by the derivative of the inner function.

Common Mistakes

Forgetting the factor of 5 in the chain rule and writing $\frac{\mathrm{d}y}{\mathrm{d}x} = \cos 5x$ instead of $5\cos 5x$ .
Solving $\sin x = \frac{\sqrt{3}}{2}$ instead of $\sin 5x = \frac{\sqrt{3}}{2}$ , giving $x = \frac{\pi}{3}$ rather than $\frac{\pi}{15}$ .
Evaluating $\cos 5x$ at the wrong angle, or mixing up $\sin\frac{\pi}{3}$ and $\cos\frac{\pi}{3}$ when reading the table.
Inverting the approximation, e.g. writing $\delta x = \frac{5}{2} \times 0.01 = 0.025$ instead of dividing by the gradient.
Giving a rounded or calculator-dependent decimal when an exact fraction ( $\frac{1}{250}$ ) is available on this non-calculator paper.

Things to Be Careful About

The mark scheme awards B1 for $x = \frac{\pi}{15}$ (or equivalently $5x = \frac{\pi}{3}$ ), B1 for the correct differentiation, DM1 (dependent on the differentiation) for setting up $\frac{0.01}{\delta x} = \frac{5}{2}$ , and A1 for $\delta x = 0.004$ or $\frac{1}{250}$ — every step must be visible.
The final answer must be exact: $0.004$ or $\frac{1}{250}$ ; do not round further or express it via a calculator-only route.
Check the solution sits within the stated domain $0 \leq x \leq \frac{\pi}{10}$ ; $\frac{\pi}{15}$ does, but a candidate solving in degrees would leave the intended interval entirely.
The dependent mark means the setup $\frac{\delta y}{\delta x} = \frac{0.01}{\delta x}$ only scores if the derivative was found correctly.

Techniques used

solve a trigonometric equation to locate the given y-valuedifferentiate sin of a multiple angle using the chain ruleuse the small-increment approximation dy/dx ≈ δy/δxevaluate the approximation exactly as a fraction

Q4MediumCalculus

Fig. 1

The velocity–time graph represents the motion of a particle moving in a straight line.
The acceleration during the first $T$ seconds of the motion is $2\,\text{ms}^{-2}$ .
The total distance travelled is $27\,\text{m}$ .

(a)

Calculate $T$ .

DifficultyMedium

Worked solution

Approach

The particle accelerates uniformly at $2\,\text{ms}^{-2}$ from rest for $T$ seconds, so its velocity at time $T$ is $2T$ . The whole velocity–time graph is a trapezium with parallel sides of lengths $8$ and $4 - T$ , and height $2T$ ; since distance travelled equals the area under a velocity–time graph, set this area equal to $27$ .

Working

Velocity at time $T$ :

v = u + at = 0 + 2T = 2T

Area of the trapezium under the graph:

\frac{1}{2}\left(8 + (4 - T)\right) \times 2T = 27

Simplify:

(12 - T)T = 27

12T - T^2 = 27

T^2 - 12T + 27 = 0

Factorise:

(T - 3)(T - 9) = 0

so $T = 3$ or $T = 9$ . Since the graph shows $T < 4$ , only $T = 3$ is admissible.

Answer

T = 3

Final answer

3

Detailed explanation

Walkthrough

The first stage of motion is uniform acceleration at $2\,\text{ms}^{-2}$ starting from rest, so after $T$ seconds the velocity is $v = u + at = 2T$ . This is the height of the flat top of the graph. The whole graph is a trapezium: the parallel sides are the bottom edge (length $8$ , the full time) and the top edge (length $4 - T$ , the time spent at constant velocity), and the perpendicular height is the maximum velocity $2T$ . Distance is the area under a velocity–time graph, so we set the trapezium area equal to $27$ : $\frac{1}{2}(8 + 4 - T)(2T) = 27$ . Simplifying gives $T^2 - 12T + 27 = 0$ , which factorises as $(T-3)(T-9) = 0$ . Both roots are positive, but the diagram requires $T < 4$ (the constant-velocity section must have positive length), so $T = 9$ is rejected and $T = 3$ .

Key Takeaways

On a velocity–time graph, distance travelled equals the area between the graph and the time axis.
Uniform acceleration means the velocity rises linearly: $v = u + at$ .
A trapezoidal velocity–time profile has area $\frac{1}{2}(a + b)h$ where the parallel sides are the total time and the constant-velocity duration.
Quadratic equations arising from physical contexts often yield an extra root that must be checked against the conditions of the problem.

Common Mistakes

Using the wrong parallel sides for the trapezium — the top side runs from $t = T$ to $t = 4$ , so its length is $4 - T$ , not $4$ or $8 - T$ .
Forgetting that distance is the area under the graph, not the final velocity or some other quantity.
Sign slips when rearranging $(12 - T)T = 27$ into the standard quadratic form; the scheme requires a correct 3-term quadratic before solving.
Accepting both roots $T = 3$ and $T = 9$ without checking against the figure, which demands $T < 4$ .
Confusing distance with displacement — here they coincide because velocity never goes negative, but the reasoning should be explicit.

Things to Be Careful About

The mark scheme awards B1 for stating the velocity at time $T$ is $2T$ (seen or implied), M1 for setting up the trapezium (or three separate areas) equal to 27 using their $2T$ , and a dependent M1 for correctly solving their quadratic — so every step must be visible.
The root $T = 9$ scores nothing: it contradicts the graph, where the deceleration starts at $t = 4 > T$ .
Keep exact working throughout; the answer is the integer $3$ , so no rounding issues arise.

Techniques used

read the maximum velocity from the constant acceleration phaseexpress total distance as the area of a trapezium on a velocity-time graphform and solve a quadratic equation by factorisingreject the inadmissible root

(b)

Calculate the acceleration during the last 4 seconds of the motion.

2M Similar

DifficultyMedium-Easy

Worked solution

Approach

Acceleration is the gradient of the velocity–time graph. On the last section (from $t = 4$ to $t = 8$ ) the velocity falls from $2T$ to $0$ .

Working

Using the result $T = 3$ from part (a), the velocity at $t = 4$ is $2T = 6$ . The gradient over the last 4 seconds is

a = \frac{0 - 6}{8 - 4} = \frac{-6}{4} = -1.5

Answer

a = -1.5\,\text{ms}^{-2}

Final answer

-1.5 ms^-2

Detailed explanation

Walkthrough

On any straight section of a velocity–time graph, the acceleration is the gradient. The final section runs from $t = 4$ to $t = 8$ , during which the velocity drops from its maximum value $2T$ down to $0$ . Carrying forward $T = 3$ from part (a), the maximum velocity is $6\,\text{ms}^{-1}$ . The gradient is therefore $\frac{0 - 6}{8 - 4} = -1.5$ , giving a deceleration of magnitude $1.5\,\text{ms}^{-2}$ but a negative acceleration, which must be reported as negative.

Key Takeaways

Acceleration on a velocity–time graph is the gradient of the line.
A decreasing velocity corresponds to a negative acceleration (deceleration).
'Hence'-style parts expect you to reuse the earlier answer rather than recompute.

Common Mistakes

Giving the answer as $+1.5$ — the mark scheme explicitly states the answer "must be negative"; the velocity is falling, so the acceleration is $-1.5\,\text{ms}^{-2}$ .
Using the wrong time interval, e.g. dividing by $8$ instead of $8 - 4 = 4$ .
Forgetting to substitute their own value of $T$ into $2T$ before computing the gradient.

Things to Be Careful About

The mark scheme's M1 is for an attempt at the gradient using their value of $T$ , so follow-through applies if your $T$ was wrong — but the sign requirement still holds.
State the unit: the acceleration is $-1.5\,\text{ms}^{-2}$ .

Techniques used

compute acceleration as the gradient of a velocity-time graph

Q58MMediumCalculusFactors of polynomials

The normal to the curve $y = x^3 + \frac{3}{2}x^2 - 2x + 1$ at the point where $x = 0$ cuts the curve again at two other points.

Find the $x$ -coordinates of these two points.

DifficultyMedium

Worked solution

Approach

Find the gradient of the curve at $x = 0$ by differentiation, take the negative reciprocal to get the normal's gradient, write the normal's equation through $(0, 1)$ , then set it equal to the curve and solve the resulting cubic.

Working

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

At $x = 0$ :

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

The gradient of the normal is the negative reciprocal:

-1 \div (-2) = \frac{1}{2}

When $x = 0$ , $y = 1$ , so the normal passes through $(0, 1)$ :

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

Intersect with the curve:

1 + \frac{1}{2}x = x^3 + \frac{3}{2}x^2 - 2x + 1

\frac{1}{2}x = x^3 + \frac{3}{2}x^2 - 2x

Multiply through by $2$ and factorise:

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

0 = 2x^3 + 3x^2 - 5x = x(2x^2 + 3x - 5)

0 = x(x - 1)(2x + 5)

The root $x = 0$ is the point of tangency, so the two other intersections are:

x = 1 \quad \text{and} \quad x = -\frac{5}{2}

Answer

x = 1 \quad \text{and} \quad x = -\frac{5}{2}

Final answer

x = 1 and x = -5/2

Detailed explanation

Walkthrough

The curve is $y = x^3 + \frac{3}{2}x^2 - 2x + 1$ . Differentiating gives $\frac{\mathrm{d}y}{\mathrm{d}x} = 3x^2 + 3x - 2$ . At $x = 0$ this equals $-2$ , which is the gradient of the tangent. The normal is perpendicular to the tangent, so its gradient is the negative reciprocal: $-1 \div (-2) = \frac{1}{2}$ .

At $x = 0$ the curve gives $y = 1$ , so the normal passes through $(0, 1)$ and its equation is $y - 1 = \frac{1}{2}x$ , i.e. $y = 1 + \frac{1}{2}x$ .

The normal cuts the curve again where the two equations are simultaneously true: set $1 + \frac{1}{2}x = x^3 + \frac{3}{2}x^2 - 2x + 1$ . Simplifying and multiplying by 2 gives $2x^3 + 3x^2 - 5x = 0$ , which factorises as $x(2x^2 + 3x - 5) = 0 = x(x - 1)(2x + 5)$ . The root $x = 0$ is the original point, so the two other intersection points have $x$ -coordinates $1$ and $-\frac{5}{2}$ .

Key Takeaways

The gradient of a normal is the negative reciprocal of the tangent's gradient, which itself comes from the derivative.
Intersections of a curve and a line are found by equating the two expressions and solving.
A cubic formed this way always contains the known intersection as a root; factorising isolates the remaining roots.

Common Mistakes

Forgetting the negative reciprocal and using $-2$ (or $2$ ) as the normal's gradient — the scheme marks the reciprocal step explicitly.
Dropping the root $x = 0$ from the factorisation but then reporting it as one of the answers, or failing to discard it since it is the given point.
Sign errors when rearranging $1 + \frac{1}{2}x = x^3 + \frac{3}{2}x^2 - 2x + 1$ ; the scheme requires the factorisation $x(2x^2 + 3x - 5) = 0$ from correct working (nfww).
Arithmetic slips in differentiating $\frac{3}{2}x^2$ — the scheme allows only one coefficient error at the M1 stage.

Things to Be Careful About

Answers must come from correct working (nfww): a wrong normal equation that happens to give the right roots scores nothing.
Show the substituted derivative value $-2$ explicitly, as it is a separate accuracy mark.
The quadratic $2x^2 + 3x - 5$ factorises as $(x - 1)(2x + 5)$ ; check by expanding before concluding.
Both roots are required — the question asks for two points, and omitting $x = -\frac{5}{2}$ loses the final mark.

Techniques used

differentiate the cubic to find the tangent gradientuse the negative reciprocal to get the normal gradientform the equation of the normal through the given pointsolve the curve and normal simultaneouslyfactorise the resulting cubic

Q6MediumTrigonometry

(a)

Fig. 1

The diagram shows an equilateral triangle $ABC$ with side $a$ .
$M$ is the midpoint of $AC$ and angle $AMB = 90°$ .

Use the diagram to find $\sec 30°$ .

DifficultyMedium-Easy

Worked solution

Approach

In the equilateral triangle $ABC$ of side $a$ , the midpoint $M$ of $AC$ gives $AM = \frac{a}{2}$ . The right-angled triangle $AMB$ has hypotenuse $AB = a$ , so Pythagoras' theorem gives the height $BM$ , from which $\cos 30° = \frac{AM}{AB}$ and hence $\sec 30°$ .

Working

Since $M$ is the midpoint of $AC$ :

AM = \frac{a}{2}

Applying Pythagoras' theorem in triangle $AMB$ :

BM^2 = AB^2 - AM^2 = a^2 - \frac{a^2}{4}

BM = \sqrt{\frac{3a^2}{4}} = \frac{\sqrt{3}\,a}{2}

In the right-angled triangle $AMB$ , the angle at $A$ is $30°$ (each angle of an equilateral triangle is $60°$ , and the altitude bisects it), so:

\cos 30° = \frac{AM}{AB} = \frac{\frac{a}{2}}{a} = \frac{1}{2}

Therefore:

\sec 30° = \frac{1}{\cos 30°} = \frac{1}{\frac{1}{2}}

Alternatively, using the sides of triangle $AMB$ directly:

\sec 30° = \frac{AB}{AM} = \frac{a}{\frac{a}{2}}

Answer

\sec 30° = \frac{2}{\sqrt{3}} = \frac{2\sqrt{3}}{3}

Final answer

sec 30° = 2/√3 = 2√3/3

Detailed explanation

Walkthrough

The question asks us to derive the exact value of $\sec 30°$ from the geometry of an equilateral triangle rather than quote it. The key observation is that dropping a perpendicular from a vertex of an equilateral triangle to the opposite side creates two congruent right-angled triangles with angles $30°$ , $60°$ and $90°$ .

First, since $M$ is the midpoint of $AC$ , we have $AM = \frac{a}{2}$ — this is the side adjacent to the $30°$ angle at $A$ . The hypotenuse is the full side $AB = a$ .

Pythagoras' theorem then gives the opposite side (the height):

BM^2 = a^2 - \left(\frac{a}{2}\right)^2 = a^2 - \frac{a^2}{4} = \frac{3a^2}{4}, \quad \text{so } BM = \frac{\sqrt{3}\,a}{2}

Now $\sec 30° = \frac{1}{\cos 30°}$ , and $\cos 30° = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{a/2}{a} = \frac{1}{2}$ . Taking the reciprocal gives $\sec 30° = 2$ . Equivalently, using all three sides at once, $\sec 30° = \frac{\text{hypotenuse}}{\text{adjacent}} = \frac{a}{a/2} = \frac{2}{\sqrt{3}}$ , which rationalises to $\frac{2\sqrt{3}}{3}$ .

Key Takeaways

The altitude of an equilateral triangle bisects the base and the vertex angle, producing the standard $30°$ - $60°$ - $90°$ triangle with side ratios $1 : \sqrt{3} : 2$ .
Exact trigonometric values can be derived geometrically rather than memorised.
$\sec x = \frac{1}{\cos x}$ , so $\sec 30° = \frac{\text{hypotenuse}}{\text{adjacent}}$ in the relevant right-angled triangle.

Common Mistakes

Using the whole base $AC = a$ instead of the half-base $AM = \frac{a}{2}$ in Pythagoras' theorem.
Confusing which angle is $30°$ : the angle at $A$ in triangle $AMB$ is $30°$ because the altitude bisects the $60°$ angle; taking $60°$ instead gives $\sec 60° = 2$ .
Leaving the answer as $\frac{2}{\sqrt{3}}$ when the scheme also accepts it — both forms are credited, but rationalising to $\frac{2\sqrt{3}}{3}$ is preferred presentation.
Arithmetic slips such as writing $a^2 - \frac{a^2}{4} = \frac{a^2}{2}$ instead of $\frac{3a^2}{4}$ .

Things to Be Careful About

The mark scheme awards B1 for seeing $BM^2 = a^2 - \frac{a^2}{4}$ (or equivalently $\cos^2 30° = 1 - \left(\frac{a/2}{a}\right)^2$ ) and B2 for reaching $BM = \frac{\sqrt{3}a}{2}$ or the equivalent identity route — so each stage must be visible in your working.
The answer must be exact: $\frac{2}{\sqrt{3}}$ or $\frac{2\sqrt{3}}{3}$ , not a decimal approximation.
This is a non-calculator paper, so all arithmetic must be shown by hand with surds kept in exact form.

Techniques used

use symmetry of the equilateral triangle to find the half-baseapply Pythagoras' theorem to find the altitude BMform cos 30° from the right-angled triangle and take its reciprocal

(b)

Show that

\frac{1}{\sec x - 1} + \frac{1}{\sec x + 1}

can be written as $2\operatorname{cosec} x \cot x$ .

DifficultyMedium

Worked solution

Approach

Add the two fractions over their common denominator $(\sec x - 1)(\sec x + 1)$ , simplify the denominator as a difference of two squares, use $\sec^2 x - 1 = \tan^2 x$ , and finally convert to sines and cosines to reach $2\operatorname{cosec} x \cot x$ .

Working

Adding the two fractions over the common denominator $(\sec x - 1)(\sec x + 1)$ :

\frac{1}{\sec x - 1} + \frac{1}{\sec x + 1} = \frac{(\sec x + 1) + (\sec x - 1)}{(\sec x - 1)(\sec x + 1)} = \frac{2\sec x}{(\sec x - 1)(\sec x + 1)}

Expanding the denominator as a difference of two squares:

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

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

= \frac{2\sec x}{\tan^2 x}

Splitting one factor of $\tan x$ off and converting to sines and cosines:

= \frac{2\sec x}{\tan x} \times \frac{1}{\tan x} = \frac{2\sec x}{\tan x} \times \cot x

and since $\frac{\sec x}{\tan x} = \frac{1/\cos x}{\sin x / \cos x} = \frac{1}{\sin x} = \operatorname{cosec} x$ :

= 2\operatorname{cosec} x \cot x

which is the required result.

Answer

\frac{1}{\sec x - 1} + \frac{1}{\sec x + 1} = 2\operatorname{cosec} x \cot x

Final answer

Shown: the expression equals 2 cosec x cot x

Detailed explanation

Walkthrough

This is a "show that" proof, so we must start from the given expression and derive forward to $2\operatorname{cosec} x \cot x$ without ever assuming the result.

Step 1 — combine the fractions. The two denominators $\sec x - 1$ and $\sec x + 1$ have product $(\sec x - 1)(\sec x + 1)$ as a natural common denominator. Adding gives numerator $(\sec x + 1) + (\sec x - 1) = 2\sec x$ — notice how the constants cancel, leaving a clean single fraction.

Step 2 — simplify the denominator. The product $(\sec x - 1)(\sec x + 1)$ is a difference of two squares: $\sec^2 x - 1$ . This is exactly where the Pythagorean identity $\sec^2 x = 1 + \tan^2 x$ becomes useful, because rearranged it says $\sec^2 x - 1 = \tan^2 x$ . The expression is now $\frac{2\sec x}{\tan^2 x}$ .

Step 3 — convert to the target form. The target contains $\operatorname{cosec} x$ and $\cot x$ . Splitting the denominator as $\tan x \times \tan x$ lets us pair $\frac{\sec x}{\tan x} = \frac{1/\cos x}{\sin x/\cos x} = \frac{1}{\sin x} = \operatorname{cosec} x$ , while the remaining $\frac{1}{\tan x} = \cot x$ . Hence the expression equals $2\operatorname{cosec} x \, \cot x$ , as required.

(The mark scheme also accepts an alternative route through $\frac{1}{\cos x}$ throughout, ending at $\frac{2\cos x}{\sin^2 x} = 2\operatorname{cosec} x \cot x$ ; either route earns the same four marks.)

Key Takeaways

To prove an identity, work on one side only, usually the more complicated one, and transform it into the other.
Difference of two squares, $(u-1)(u+1) = u^2 - 1$ , appears constantly with secant and tangent identities.
The rearranged Pythagorean identity $\sec^2 x - 1 = \tan^2 x$ is often the bridge between a sec-based expression and a tan-based one.
Converting everything to sines and cosines ( $\sec x = \frac{1}{\cos x}$ , $\tan x = \frac{\sin x}{\cos x}$ , $\operatorname{cosec} x = \frac{1}{\sin x}$ , $\cot x = \frac{\cos x}{\sin x}$ ) is a reliable way to finish a proof.

Common Mistakes

Adding the numerators but forgetting the common denominator, e.g. writing just $2\sec x$ over nothing.
Expanding $(\sec x - 1)(\sec x + 1)$ incorrectly, e.g. as $\sec^2 x + 1$ instead of $\sec^2 x - 1$ .
Using the identity the wrong way round: you need $\sec^2 x - 1 = \tan^2 x$ , not $\sec^2 x + \tan^2 x = 1$ .
Stopping at $\frac{2\sec x}{\tan^2 x}$ — the final B1 requires at least one further correct step reaching the printed answer, since the part is AG (answer given).
Sign slips when combining $(\sec x + 1) + (\sec x - 1)$ ; the $+1$ and $-1$ must cancel.

Things to Be Careful About

Every mark in this scheme is dependent on the previous one, so a single early error loses all subsequent marks — check each line before moving on.
Because the answer is given (AG), you must show sufficient correct detail: no steps may be jumped or asserted without working.
Keep the notation consistent with the paper: write $\operatorname{cosec} x$ (not csc) and keep function names upright.
The identity holds only where the original expression is defined, i.e. $\sec x \neq \pm 1$ and $\sin x \neq 0$ ; this need not be stated for the marks but explains why the manipulation is valid.

Techniques used

combine two fractions over a common denominatorexpand and simplify the denominator using the difference of two squaresapply the identity sec²x = 1 + tan²xconvert sec and tan into cosec and cot to reach the given answer

Q7MediumVectors in two dimensionsSimultaneous equations

The point $O$ is the origin.
Two points $P$ and $Q$ are such that $\overrightarrow{PQ}$ is in the same direction as $-\mathbf{i} + 5\mathbf{j}$ .

(a)

The point $R$ is such that $\overrightarrow{OR}$ is in the same direction as $\overrightarrow{PQ}$ and the magnitude of $\overrightarrow{OR}$ is $3\sqrt{26}$ .

Find $\overrightarrow{OR}$ .

DifficultyMedium-Easy

Worked solution

Approach

Since $\overrightarrow{OR}$ is in the same direction as $\overrightarrow{PQ}$ , it is a positive scalar multiple of $-\mathbf{i} + 5\mathbf{j}$ . We find the magnitude of $-\mathbf{i} + 5\mathbf{j}$ and scale the vector so that its magnitude becomes $3\sqrt{26}$ .

Working

Let $\mathbf{v} = -\mathbf{i} + 5\mathbf{j}$ .

The magnitude of $\mathbf{v}$ is:

|\mathbf{v}| = \sqrt{(-1)^2 + 5^2} = \sqrt{1 + 25} = \sqrt{26}

Since $\overrightarrow{OR}$ is in the same direction as $\mathbf{v}$ , we can write:

\overrightarrow{OR} = k(-\mathbf{i} + 5\mathbf{j}) \quad (k > 0)

The magnitude of $\overrightarrow{OR}$ is given as $3\sqrt{26}$ :

|\overrightarrow{OR}| = k\sqrt{26} = 3\sqrt{26} \implies k = 3

Multiplying $\mathbf{v}$ by $k = 3$ :

\overrightarrow{OR} = 3(-\mathbf{i} + 5\mathbf{j}) = -3\mathbf{i} + 15\mathbf{j}

Answer

-3\mathbf{i} + 15\mathbf{j}

Final answer

-3i + 15j

Detailed explanation

Walkthrough

Vectors in the same direction are positive scalar multiples of one another. We write $\overrightarrow{OR} = k(-\mathbf{i} + 5\mathbf{j})$ with $k > 0$ .
We calculate the magnitude of the direction vector $-\mathbf{i} + 5\mathbf{j}$ using the formula $\sqrt{x^2 + y^2}$ , giving $\sqrt{(-1)^2 + 5^2} = \sqrt{26}$ .
Since the required magnitude is $|\overrightarrow{OR}| = 3\sqrt{26}$ , the scalar multiplier $k$ must be $\frac{3\sqrt{26}}{\sqrt{26}} = 3$ .
Multiplying each component by $3$ gives $\overrightarrow{OR} = -3\mathbf{i} + 15\mathbf{j}$ or $\begin{pmatrix}-3 \\ 15\end{pmatrix}$ .

Key Takeaways

A vector $\mathbf{u}$ in the same direction as $\mathbf{v}$ is given by $\mathbf{u} = k\mathbf{v}$ where $k = \frac{|\mathbf{u}|}{|\mathbf{v}|} > 0$ .

Common Mistakes

Forgetting to square the negative sign properly: $(-1)^2 = 1$ , not $-1$ .
Reversing the direction by using a negative scalar.

Things to Be Careful About

Ensure the answer is left in standard vector form (in terms of $\mathbf{i}$ and $\mathbf{j}$ or column vector form).

Techniques used

express a parallel vector as a scalar multiplecalculate vector magnitudescale by magnitude ratio

(b)

$\overrightarrow{OP}$ is in the same direction as $2\mathbf{i} - 3\mathbf{j}$ and $\overrightarrow{OQ} = 10\mathbf{i} + 6\mathbf{j}$ .

Find $\overrightarrow{OP}$ .

DifficultyMedium

Worked solution

Approach

Use the vector triangle relation $\overrightarrow{OP} + \overrightarrow{PQ} = \overrightarrow{OQ}$ . Express $\overrightarrow{OP}$ and $\overrightarrow{PQ}$ as scalar multiples of their direction vectors, equate components to form simultaneous linear equations, solve for the scalars, and evaluate $\overrightarrow{OP}$ .

Working

Let $\overrightarrow{OP} = a(2\mathbf{i} - 3\mathbf{j})$ and $\overrightarrow{PQ} = b(-\mathbf{i} + 5\mathbf{j})$ for constants $a, b > 0$ .

Using the vector addition law:

\overrightarrow{OP} + \overrightarrow{PQ} = \overrightarrow{OQ}

\begin{aligned} a(2\mathbf{i} - 3\mathbf{j}) + b(-\mathbf{i} + 5\mathbf{j}) &= 10\mathbf{i} + 6\mathbf{j} \\ (2a - b)\mathbf{i} + (-3a + 5b)\mathbf{j} &= 10\mathbf{i} + 6\mathbf{j} \end{aligned}

Equating the $\mathbf{i}$ and $\mathbf{j}$ components gives the system of linear equations:

\begin{aligned} (1) \quad 2a - b &= 10 \\ (2) \quad -3a + 5b &= 6 \end{aligned}

From equation (1), express $b$ in terms of $a$ :

b = 2a - 10

Substitute into equation (2):

\begin{aligned} -3a + 5(2a - 10) &= 6 \\ -3a + 10a - 50 &= 6 \\ 7a &= 56 \\ a &= 8 \end{aligned}

Then:

b = 2(8) - 10 = 6

Now, find $\overrightarrow{OP}$ :

\overrightarrow{OP} = 8(2\mathbf{i} - 3\mathbf{j}) = 16\mathbf{i} - 24\mathbf{j}

Answer

16\mathbf{i} - 24\mathbf{j}

Final answer

16i - 24j

Detailed explanation

Walkthrough

From the vector triangle, the position vector $\overrightarrow{OQ}$ can be reached by going from $O$ to $P$ and then $P$ to $Q$ , so $\overrightarrow{OQ} = \overrightarrow{OP} + \overrightarrow{PQ}$ .
We are given the direction vectors for $\overrightarrow{OP}$ and $\overrightarrow{PQ}$ . We represent them as $\overrightarrow{OP} = a(2\mathbf{i} - 3\mathbf{j})$ and $\overrightarrow{PQ} = b(-\mathbf{i} + 5\mathbf{j})$ .
Equating the sum of these vectors to $\overrightarrow{OQ} = 10\mathbf{i} + 6\mathbf{j}$ allows us to separate the $\mathbf{i}$ -components and $\mathbf{j}$ -components into two independent linear equations: $2a - b = 10$ and $-3a + 5b = 6$ .
Solving this system gives $a = 8$ and $b = 6$ .
Finally, we substitute $a = 8$ back into the expression for $\overrightarrow{OP}$ to get $8(2\mathbf{i} - 3\mathbf{j}) = 16\mathbf{i} - 24\mathbf{j}$ .

Key Takeaways

Any position vector can be found via vector paths: $\overrightarrow{OQ} = \overrightarrow{OP} + \overrightarrow{PQ}$ .
Two vectors are equal if and only if their corresponding $\mathbf{i}$ and $\mathbf{j}$ components are equal.

Common Mistakes

Confusing $\overrightarrow{PQ}$ with $\overrightarrow{OP} - \overrightarrow{OQ}$ or setting $\overrightarrow{OP} + \overrightarrow{OQ} = \overrightarrow{PQ}$ .
Finding only the scalar value $a = 8$ without substituting it back to give the final vector $\overrightarrow{OP}$ .
Sign errors when solving the simultaneous linear equations.

Things to Be Careful About

Ensure the final answer is stated as a vector ( $16\mathbf{i} - 24\mathbf{j}$ or $\begin{pmatrix}16 \\ -24\end{pmatrix}$ ), not just the scalar $a=8$ .

Techniques used

express vector relationship using triangle lawequate vector components to set up simultaneous equationssolve a pair of linear simultaneous equations

Q87MMediumCalculusFactors of polynomials

Fig. 1

The diagram shows part of each of the curves $y = 12 - x^2$ and $y = x^4 - 4x^2 + 8$ .

Find the area of the shaded region enclosed by the two curves.

DifficultyMedium

Worked solution

Approach

The area between two curves is found by integrating (upper curve $-$ lower curve) between the intersection points. First find the intersections by equating the two curves, then integrate the difference from $x = -2$ to $x = 2$ .

Working

Solve simultaneously:

12 - x^2 = x^4 - 4x^2 + 8

x^4 - 3x^2 - 4 = 0

Factorise as a quadratic in $x^2$ :

(x^2 - 4)(x^2 + 1) = 0

Since $x^2 + 1 = 0$ gives no real solutions, $x^2 = 4$ , so

x = 2 \quad \text{and} \quad x = -2

Between these points the parabola $y = 12 - x^2$ lies above the quartic, so the shaded area is

\int_{-2}^{2} \left[(12 - x^2) - (x^4 - 4x^2 + 8)\right]\,\mathrm{d}x = \int_{-2}^{2} \left(-x^4 + 3x^2 + 4\right)\,\mathrm{d}x

Integrate term by term:

\left[-\frac{x^5}{5} + x^3 + 4x\right]_{-2}^{2}

Substitute the limits:

\left(-\frac{32}{5} + 8 + 8\right) - \left(\frac{32}{5} - 8 - 8\right) = \frac{48}{5} + \frac{48}{5} = \frac{96}{5}

Answer

\frac{96}{5} = 19.2

Final answer

96/5 = 19.2

Detailed explanation

Walkthrough

To find where the two curves meet, set their equations equal: $12 - x^2 = x^4 - 4x^2 + 8$ . Rearranging everything to one side gives $x^4 - 3x^2 - 4 = 0$ . This is a quartic, but it contains only even powers of $x$ , so treating $x^2$ as the unknown turns it into a quadratic: $(x^2)^2 - 3x^2 - 4 = 0$ , which factorises as $(x^2 - 4)(x^2 + 1) = 0$ . The factor $x^2 + 1$ cannot be zero for real $x$ , so $x^2 = 4$ , giving $x = 2$ and $x = -2$ .

The area between two curves is the integral of (upper $-$ lower). From the figure, the parabola $y = 12 - x^2$ sits above the quartic between the intersection points, so the integrand is $(12 - x^2) - (x^4 - 4x^2 + 8) = -x^4 + 3x^2 + 4$ . Integrating term by term gives $-\frac{x^5}{5} + x^3 + 4x$ . Evaluating at $x = 2$ gives $-\frac{32}{5} + 8 + 8 = \frac{48}{5}$ , and at $x = -2$ it gives $\frac{32}{5} - 8 - 8 = -\frac{48}{5}$ . Subtracting, the area is $\frac{48}{5} - \left(-\frac{48}{5}\right) = \frac{96}{5} = 19.2$ square units.

Key Takeaways

The area between two curves is $\int_{a}^{b} (y_{\text{upper}} - y_{\text{lower}})\,\mathrm{d}x$ between their intersection points.
Quartics containing only even powers can be solved by treating $x^2$ as the unknown and factorising as a quadratic.
The integrand must be the correct way round: subtracting the curves the wrong way gives a negative answer, and its absolute value is not automatically the area unless you justify which curve is on top.

Common Mistakes

Subtracting the curves in the wrong order and reporting a negative answer — the mark scheme requires the correct difference $-x^4 + 3x^2 + 4$ (or its negative with the limits reversed).
Keeping both roots $x = \pm 2$ and also inventing roots from $x^2 + 1 = 0$ , or forgetting the negative root $x = -2$ .
Sign slips when integrating: the antiderivative is $-\frac{x^5}{5} + x^3 + 4x$ , and errors in the signs of $x^3$ or $4x$ are common.
Forgetting to substitute both limits, or substituting only $x = 2$ and forgetting the curve is symmetric so the answer must be doubled if integrating from $0$ to $2$ .
Arithmetic slips in $\left(-\frac{32}{5} + 8 + 8\right) - \left(\frac{32}{5} - 8 - 8\right)$ — note the whole second bracket is subtracted, so both its signs flip.

Things to Be Careful About

Show the factorisation $(x^2 - 4)(x^2 + 1) = 0$ explicitly — the mark scheme awards a method mark for solving the quadratic in $x^2$ .
The integration mark requires at least one term correctly integrated; the evaluation mark requires the limits $-2$ and $2$ (or $0$ to $2$ with doubling) substituted into your integral.
The exact answer is $\frac{96}{5}$ ; the decimal $19.2$ is also accepted. Since a calculator is allowed on this paper, either exact or decimal form is fine, but show the substituted expression before the final value so the method marks are visible.

Techniques used

solve the curves simultaneously by substituting into a quadratic in x squaredfactorise using the difference of two squaressubtract the lower curve from the upper curve to form the integrandintegrate term by termevaluate the definite integral using symmetric limits

Q9MediumLogarithmic and exponential functionsQuadratic functions

Solve the following equations.

(a)

$\log_5(5x - 2) - \log_{25} x = \frac{1}{2}$

5M Similar

DifficultyMedium

Worked solution

Approach

Change $\log_{25}x$ to base $5$ , then use the laws of logarithms to combine everything into a single logarithm equal to $1$ , and solve the resulting quadratic.

Working

Using the change of base formula:

\log_{25}x = \frac{\log_5 x}{\log_5 25} = \frac{\log_5 x}{2}

So the equation becomes

\log_5(5x-2) - \frac{1}{2}\log_5 x = \frac{1}{2}

Multiplying through by $2$ :

2\log_5(5x-2) - \log_5 x = 1

Since $2\log_5(5x-2) = \log_5(5x-2)^2$ and $1 = \log_5 5$ :

\log_5\frac{(5x-2)^2}{x} = 1

Removing the logarithm:

\frac{(5x-2)^2}{x} = 5

25x^2 - 20x + 4 = 5x

25x^2 - 25x + 4 = 0

Factorising:

(5x-1)(5x-4) = 0

so $x = \frac{1}{5}$ or $x = \frac{4}{5}$ .

Checking validity: if $x = \frac{1}{5}$ then $5x - 2 = -1 < 0$ , so $\log_5(5x-2)$ is not defined. This root must be rejected.

Answer

x = \frac{4}{5}

Final answer

x = 4/5 only

Detailed explanation

Walkthrough

The equation mixes two different bases ( $5$ and $25$ ), so the first job is to put every logarithm on the same base. Since $25 = 5^2$ , the change-of-base formula gives $\log_{25}x = \frac{\log_5 x}{2}$ , i.e. $\log_{25}x = \frac{1}{2}\log_5 x$ . Substituting this turns the whole equation into base- $5$ logarithms.

Next we tidy up with the laws of logarithms. Multiplying the equation by $2$ clears the halves, giving $2\log_5(5x-2) - \log_5 x = 1$ . The power law rewrites $2\log_5(5x-2)$ as $\log_5(5x-2)^2$ , and the right-hand side $1$ is just $\log_5 5$ . The subtraction law then combines the two left-hand logarithms into one: $\log_5\frac{(5x-2)^2}{x} = 1$ .

Now that a single logarithm equals $\log_5 5$ , the arguments must be equal: $\frac{(5x-2)^2}{x} = 5$ . Expanding $(5x-2)^2 = 25x^2 - 20x + 4$ and multiplying by $x$ gives $25x^2 - 25x + 4 = 0$ , which factorises as $(5x-1)(5x-4) = 0$ .

Both roots must be checked against the original equation, because logarithms are only defined for positive arguments. With $x = \frac{1}{5}$ , the quantity $5x - 2 = -1$ is negative, so that root is invalid. Only $x = \frac{4}{5}$ survives (it gives $5x-2 = 2 > 0$ ).

Key Takeaways

Change of base: $\log_{a^k} x = \frac{1}{k}\log_a x$ , so $\log_{25}x = \frac{1}{2}\log_5 x$ .
Power, product and quotient laws let you collapse a multi-logarithm equation into one single logarithm.
Once you have $\log_b A = \log_b B$ , equate the arguments.
Always check solutions against the domains of the original logarithms; extraneous roots from squaring or combining logs must be rejected.

Common Mistakes

Writing $\log_{25}x = 2\log_5 x$ instead of $\frac{1}{2}\log_5 x$ — the coefficient divides, it does not multiply.
Forgetting that $2\log_5(5x-2)$ becomes $\log_5(5x-2)^2$ , not $\log_5(10x-4)$ .
Giving both roots $x = \frac{1}{5}$ and $x = \frac{4}{5}$ without checking them — the mark scheme awards the final mark for $x = \frac{4}{5}$ only; keeping the invalid root loses the A1.
Sign slips when expanding $(5x-2)^2$ or when moving terms to form $25x^2 - 25x + 4 = 0$ .
Attempting to divide by $x$ early instead of multiplying through by $x$ .

Things to Be Careful About

The final answer must be $x = \frac{4}{5}$ alone — the mark scheme explicitly says " $x = \frac{4}{5}$ only", so an unchecked second root scores nothing on the last mark.
Every mark-scheme step (change of base, combining logs, forming the quadratic, solving) must be visible in your working; the M marks depend on correct use of the log rules.
Keep exact fractions throughout — no decimals are needed here.

Techniques used

change the base of a logarithm using change-of-base formulacombine logarithms into a single logarithm using quotient and power lawsconvert a logarithmic equation into an equation without logarithmssolve the resulting quadratic and reject the inadmissible root

(b)

$\mathrm{e}^{3y - 7} + \frac{4}{\mathrm{e}^3} = \frac{5}{\mathrm{e}^{3y - 1}}$

6M Similar

DifficultyMedium

Worked solution

Approach

Multiply the equation by a suitable power of $\mathrm{e}$ so that all terms become positive powers, then recognise the result as a quadratic in $\mathrm{e}^{3y-4}$ , factorise, and take natural logarithms.

Working

Start with

\mathrm{e}^{3y-7} + \frac{4}{\mathrm{e}^3} = \frac{5}{\mathrm{e}^{3y-1}}

Multiply every term by $\mathrm{e}^{3y-1}$ :

\mathrm{e}^{(3y-7)+(3y-1)} + 4\mathrm{e}^{(3y-1)-3} = 5

\mathrm{e}^{6y-8} + 4\mathrm{e}^{3y-4} - 5 = 0

Let $u = \mathrm{e}^{3y-4}$ , so $u^2 = \mathrm{e}^{6y-8}$ . Then

u^2 + 4u - 5 = 0

Factorising:

(u - 1)(u + 5) = 0

So $\mathrm{e}^{3y-4} = 1$ or $\mathrm{e}^{3y-4} = -5$ . The second has no solution since $\mathrm{e}^{3y-4} > 0$ always.

Taking natural logarithms of the first:

3y - 4 = \ln 1 = 0

y = \frac{4}{3}

Answer

y = \frac{4}{3}

Final answer

y = 4/3

Detailed explanation

Walkthrough

The equation contains $\mathrm{e}^{3y-7}$ , $\mathrm{e}^{-3}$ and $\mathrm{e}^{-(3y-1)}$ — three different-looking exponentials, but they are related. Multiplying every term by $\mathrm{e}^{3y-1}$ clears the fraction on the right and, using the index law $\mathrm{e}^a \times \mathrm{e}^b = \mathrm{e}^{a+b}$ , produces $\mathrm{e}^{6y-8} + 4\mathrm{e}^{3y-4} = 5$ .

The key observation is that $\mathrm{e}^{6y-8} = \left(\mathrm{e}^{3y-4}\right)^2$ , because $2(3y-4) = 6y-8$ . So if we write $u = \mathrm{e}^{3y-4}$ , the equation becomes the simple quadratic $u^2 + 4u - 5 = 0$ , which factorises as $(u-1)(u+5) = 0$ .

This gives $\mathrm{e}^{3y-4} = 1$ or $\mathrm{e}^{3y-4} = -5$ . An exponential can never be negative, so the second possibility is discarded immediately. For the first, taking natural logarithms gives $3y - 4 = \ln 1 = 0$ , hence $y = \frac{4}{3}$ .

Key Takeaways

Index laws turn products of exponentials into sums of indices — choose the multiplier that makes all exponents work out.
Equations like $\mathrm{e}^{2A} + k\,\mathrm{e}^{A} + c = 0$ are quadratics in disguise: substitute $u = \mathrm{e}^{A}$ .
$\mathrm{e}^{\text{anything}} > 0$ , so any factorisation yielding a negative exponential value contributes no solution.
$\ln 1 = 0$ finishes the problem without any calculator work.

Common Mistakes

Getting the index arithmetic wrong when multiplying by $\mathrm{e}^{3y-1}$ — e.g. writing $\mathrm{e}^{6y-6}$ instead of $\mathrm{e}^{6y-8}$ ; check $(3y-7)+(3y-1) = 6y-8$ carefully.
Not recognising the hidden quadratic and trying to take logs of individual terms, which does not work because $\ln(a+b) \neq \ln a + \ln b$ .
Keeping the root $\mathrm{e}^{3y-4} = -5$ and attempting to take its logarithm — exponentials are strictly positive, so this yields no real solution.
Sign errors when rearranging into the three-term form $\mathrm{e}^{6y-8} + 4\mathrm{e}^{3y-4} - 5 = 0$ before factorising.
Arithmetic slips in the final step, e.g. answering $y = \frac{3}{4}$ instead of $y = \frac{4}{3}$ .

Things to Be Careful About

The mark scheme requires the equation arranged as a three-term quadratic (B1 dependent), so show the line $\mathrm{e}^{6y-8} + 4\mathrm{e}^{3y-4} - 5 = 0$ explicitly.
The answer is exact: $y = \frac{4}{3}$ , no decimals needed.
Show the rejection of $\mathrm{e}^{3y-4} = -5$ (or at least silently drop it correctly) — taking $\ln$ of a negative number is a serious error.
All working here is doable by hand since $\ln 1 = 0$ ; no calculator evaluation is required.

Techniques used

multiply through by a power of e to clear denominatorsrecognise a hidden quadratic in e^(3y-4)factorise the quadratic in the exponential expressiontake natural logarithms to solve for y

Q10MediumSeries

(a)

An arithmetic progression has first term $a$ and common difference $d$ .

Given that $S_{20} = 3 \times S_{10}$ , find $a$ in terms of $d$ .

DifficultyMedium-Easy

Worked solution

Approach

Use the formula $S_n = \frac{n}{2}\{2a + (n-1)d\}$ for both $S_{20}$ and $S_{10}$ , set $S_{20} = 3S_{10}$ , and solve for $a$ .

Working

S_{20} = \frac{20}{2}(2a + 19d) = 10(2a + 19d)

S_{10} = \frac{10}{2}(2a + 9d) = 5(2a + 9d)

Given $S_{20} = 3 \times S_{10}$ :

10(2a + 19d) = 3 \times 5(2a + 9d)

20a + 190d = 30a + 135d

190d - 135d = 30a - 20a

55d = 10a

a = \frac{11}{2}d

Answer

a = \frac{11}{2}d

Final answer

a = (11/2)d

Detailed explanation

Walkthrough

The sum of the first $n$ terms of an arithmetic progression is $S_n = \frac{n}{2}\{2a + (n-1)d\}$ . We write this out twice: once with $n = 20$ , giving $\frac{20}{2}(2a + 19d)$ , and once with $n = 10$ , giving $\frac{10}{2}(2a + 9d)$ . The condition $S_{20} = 3S_{10}$ turns these into a single linear equation in $a$ and $d$ . Expanding both sides gives $20a + 190d = 30a + 135d$ . Collecting the $a$ terms on one side and the $d$ terms on the other gives $55d = 10a$ , so dividing by 10 yields $a = \frac{11}{2}d$ .

Key Takeaways

The arithmetic series formula $S_n = \frac{n}{2}\{2a + (n-1)d\}$ must be applied carefully — the $(n-1)$ multiplier on $d$ changes with each value of $n$ .
A condition linking two partial sums usually reduces to a simple linear relation between $a$ and $d$ .

Common Mistakes

Writing $S_{20}$ as $\frac{20}{2}(2a + 20d)$ instead of $\frac{20}{2}(2a + 19d)$ — the last term is the 20th term, but the bracket needs $n - 1 = 19$ times $d$ .
Expanding brackets incorrectly; the mark scheme awards a specific mark for correct removal of brackets ( $20a + 190d = 30a + 135d$ ).
Sign slips when collecting terms across the equals sign.

Things to Be Careful About

The answer is an expression, not a number: leave it as $a = \frac{11}{2}d$ (or any equivalent form such as $a = 5.5d$ ).
Show the substituted formulas before simplifying so the method mark is clearly earned.

Techniques used

apply the arithmetic series sum formula to both S20 and S10expand and simplify the resulting equationrearrange to express a in terms of d

(b)

A geometric progression, A, has common ratio $r$ , where $|r| < 1$ .
The terms of this progression are $a_1$ , $a_2$ , $a_3$ $\ldots$ .

Another geometric progression, B, has terms $b_1$ , $b_2$ , $b_3$ $\ldots$ , where

b_1 = a_2, \quad b_2 = a_4, \quad b_3 = a_6 \; \ldots .

The sum to infinity of A is $S_A$ and the sum to infinity of B is $S_B$ .

Find $\frac{S_B}{S_A}$ in terms of $r$ .

Give your answer in its simplest form.

5M Similar

DifficultyMedium

Worked solution

Approach

Progression A has first term $a_1$ and ratio $r$ , so its sum to infinity is $\frac{a_1}{1-r}$ . Progression B takes every second term of A, so its ratio is $r^2$ and its first term is $a_2 = a_1 r$ . Write both sums to infinity and divide.

Working

Sum to infinity of A:

S_A = \frac{a_1}{1 - r}

Since $b_1 = a_2$ , $b_2 = a_4$ , $b_3 = a_6$ , each term of B is obtained from the previous by multiplying by $r^2$ :

r_b = r^2

First term of B: $b_1 = a_2 = a_1 r$ . Sum to infinity of B:

S_B = \frac{a_1 r}{1 - r^2}

Forming the ratio:

\frac{S_B}{S_A} = \frac{\dfrac{a_1 r}{1 - r^2}}{\dfrac{a_1}{1 - r}} = \frac{a_1 r}{1 - r^2} \times \frac{1 - r}{a_1}

Cancelling $a_1$ and factorising $1 - r^2 = (1-r)(1+r)$ :

\frac{S_B}{S_A} = \frac{r(1 - r)}{(1-r)(1+r)} = \frac{r}{1 + r}

Answer

\frac{S_B}{S_A} = \frac{r}{1 + r}

Final answer

r / (1 + r)

Detailed explanation

Walkthrough

Progression A is geometric with first term $a_1$ and common ratio $r$ , where $|r| < 1$ , so its sum to infinity exists and equals $S_A = \frac{a_1}{1-r}$ .

Progression B is built from the even-positioned terms of A: $b_1 = a_2$ , $b_2 = a_4$ , $b_3 = a_6$ , and so on. Since consecutive terms of A differ by a factor of $r$ , skipping every other term means consecutive terms of B differ by a factor of $r^2$ : for example $\frac{a_4}{a_2} = \frac{a_1 r^3}{a_1 r} = r^2$ . So B is itself geometric with first term $b_1 = a_2 = a_1 r$ and ratio $r^2$ (and $|r^2| < 1$ since $|r| < 1$ , so its sum to infinity also converges).

Thus $S_B = \frac{a_1 r}{1 - r^2}$ . Dividing gives

\frac{S_B}{S_A} = \frac{a_1 r}{1 - r^2} \times \frac{1 - r}{a_1},

the factors $a_1$ cancel, and factorising $1 - r^2 = (1-r)(1+r)$ lets $(1-r)$ cancel too, leaving $\frac{r}{1+r}$ .

Key Takeaways

Taking every $k$ -th term of a geometric progression produces another geometric progression whose ratio is the original ratio raised to the power $k$ .
The sum to infinity $\frac{a}{1-r}$ only applies when $|r| < 1$ — here that condition guarantees both sums converge.
Simplifying ratios of fractions often hinges on factorising a difference of squares.

Common Mistakes

Using $r$ as the common ratio of B instead of $r^2$ .
Taking the first term of B to be $a_1$ rather than $a_2 = a_1 r$ .
Forgetting to cancel or mis-cancelling the $a_1$ factors when forming the ratio.
Failing to factorise $1 - r^2$ and so leaving the answer as $\frac{r(1-r)}{1-r^2}$ instead of the required simplest form $\frac{r}{1+r}$ .

Things to Be Careful About

The final answer is 'cao' (correct answer only) and must be in simplest form: $\frac{r}{1+r}$ , not $\frac{r(1-r)}{1-r^2}$ .
The method mark for forming the ratio depends on having already stated both $S_A = \frac{a_1}{1-r}$ and the ratio $r^2$ correctly, so write those steps explicitly.
Note $|r|<1$ ensures $1-r > 0$ and $1-r^2 > 0$ , so no sign issues arise.

Techniques used

write the sum to infinity of a geometric progressionidentify that progression B has common ratio r squaredexpress the first term of B using a1 and rform and simplify the ratio of two sums to infinity

Q11MediumCoordinate geometry of the circleQuadratic functions

The lines $x = 0$ , $x = 4$ , $y = 3$ and $y = -1$ are tangents to a circle.

(a)

Find the equation of the circle.

DifficultyMedium-Easy

Worked solution

Approach

The vertical tangents $x = 0$ and $x = 4$ are parallel and equidistant from the centre, so the centre lies halfway between them; similarly for $y = 3$ and $y = -1$ . The radius is half the distance between each pair.

Working

Centre: midway between $x = 0$ and $x = 4$ , and between $y = -1$ and $y = 3$ :

(2,\ 1)

Radius: half of $4$ (the gap between $x=0$ and $x=4$ , or between $y=-1$ and $y=3$ ):

r = 2

Equation:

(x - 2)^2 + (y - 1)^2 = 4

Answer

(x - 2)^2 + (y - 1)^2 = 4

Final answer

(x - 2)^2 + (y - 1)^2 = 4

Detailed explanation

Walkthrough

Four tangent lines to a circle come in two parallel pairs: the pair $x = 0$ and $x = 4$ touch the circle on opposite sides, so the centre must sit exactly halfway between them at $x = 2$ . Likewise $y = 3$ and $y = -1$ force the centre's $y$ -coordinate to be $1$ . The radius is the distance from the centre to any one tangent line: from $(2,1)$ to $x = 0$ is $2$ units (and this matches the distance to every other tangent). Substituting centre $(2,1)$ and radius $2$ into $(x-a)^2 + (y-b)^2 = r^2$ gives the equation.

Key Takeaways

Parallel tangent pairs straddle the circle symmetrically, so their midline passes through the centre.
The radius equals the perpendicular distance from the centre to any tangent line.
Standard form of a circle: $(x - a)^2 + (y - b)^2 = r^2$ with centre $(a,b)$ , radius $r$ .

Common Mistakes

Writing the right-hand side as $r$ instead of $r^2$ — the mark scheme explicitly rejects leaving it as $2^2$ unexpanded... actually it requires the value $4$ , not the unsquared form.
Taking the radius as the full gap ( $4$ ) instead of half the gap ( $2$ ).
Sign slips inside the brackets: $(x+2)$ instead of $(x-2)$ for centre $x = 2$ .

Things to Be Careful About

The answer must show the squared radius as $4$ ; an equivalent form (expanded) is accepted, but the unsquared " $2^2$ " alone is not.
Follow-through applies only to your own correct centre and radius.

Techniques used

locate the centre from the four tangent linesread off the radiuswrite the equation of the circle in standard form

(b)

The line $y = 2x + a$ , where $a$ is a constant, is also a tangent to the circle.

Show that $5x^2 + 4(a - 2)x + (a - 1)^2 = 0$ , and hence find the possible values of $a$ .

Give your answers in exact form.

DifficultyMedium

Worked solution

Approach

Substitute $y = 2x + a$ into the circle equation from part (a); a tangent touches at exactly one point, so the resulting quadratic in $x$ has discriminant zero. Solve that condition for $a$ exactly.

Working

Substituting $y = 2x + a$ into $(x-2)^2 + (y-1)^2 = 4$ :

(x - 2)^2 + (2x + a - 1)^2 = 4

Expanding:

x^2 - 4x + 4 + 4x^2 + 4(a-1)x + (a-1)^2 = 4

Collecting terms (the constants $4$ cancel):

5x^2 + \left(-4 + 4a - 4\right)x + (a-1)^2 = 0

5x^2 + 4(a-2)x + (a-1)^2 = 0

as required (AG).

For tangency there is exactly one solution, so the discriminant is zero:

\left[4(a-2)\right]^2 - 4(5)(a-1)^2 = 0

16(a-2)^2 - 20(a-1)^2 = 0

Dividing by $4$ and expanding:

4(a^2 - 4a + 4) - 5(a^2 - 2a + 1) = 0

4a^2 - 16a + 16 - 5a^2 + 10a - 5 = 0

-a^2 - 6a + 11 = 0 \quad\Longrightarrow\quad a^2 + 6a - 11 = 0

By the quadratic formula:

a = \frac{-6 \pm \sqrt{36 + 44}}{2} = \frac{-6 \pm \sqrt{80}}{2} = -3 \pm 2\sqrt{5}

Answer

a = -3 \pm 2\sqrt{5}

Final answer

a = -3 +/- 2√5

Detailed explanation

Walkthrough

First we use the result of part (a): the circle is $(x-2)^2 + (y-1)^2 = 4$ . Since the line $y = 2x + a$ meets the circle where both equations hold, we substitute the line's expression for $y$ directly into the circle equation. This removes $y$ and leaves a quadratic in $x$ whose coefficients involve the unknown constant $a$ .

Expanding carefully: $(x-2)^2 = x^2 - 4x + 4$ , and $(2x + a - 1)^2 = 4x^2 + 4(a-1)x + (a-1)^2$ . Adding these and setting equal to $4$ , the constant $4$ s cancel, giving $5x^2 + 4(a-2)x + (a-1)^2 = 0$ — exactly the printed target, confirming our algebra.

Now the key geometric fact: a tangent touches the circle at exactly one point, so this quadratic must have exactly one root. A quadratic has a repeated root precisely when its discriminant $b^2 - 4ac$ equals zero. Applying this with $b = 4(a-2)$ , $a_{coef} = 5$ , $c = (a-1)^2$ gives $16(a-2)^2 - 20(a-1)^2 = 0$ .

Dividing through by $4$ and expanding both squares turns this into $a^2 + 6a - 11 = 0$ . This does not factorise over the integers, so we use the quadratic formula: $a = \frac{-6 \pm \sqrt{36+44}}{2} = \frac{-6 \pm \sqrt{80}}{2}$ . Simplifying $\sqrt{80} = 4\sqrt{5}$ gives the exact answers $a = -3 \pm 2\sqrt{5}$ .

The two values correspond to the two parallel lines of gradient $2$ that touch the circle on opposite sides.

Key Takeaways

Substituting a line into a circle converts a geometry problem into a quadratic in one variable.
Tangency ⟺ the intersection quadratic has discriminant zero (one repeated root).
Exact-form answers should stay as simplified surds: $\sqrt{80} = 4\sqrt{5}$ .

Common Mistakes

Expanding $(2x + a - 1)^2$ incorrectly — the middle term is $2 \cdot 2x \cdot (a-1) = 4(a-1)x$ , not $2(a-1)x$ or $4ax$ .
Forgetting the "show that" requirement: every algebraic line from the substitution to the printed quadratic must appear.
Sign errors forming the discriminant, e.g. writing $-4 \times 5 \times (a-1)^2$ as $+20(a-1)^2$ .
Giving decimal answers instead of exact surds — the question demands exact form.
Stopping at $a = \frac{-6 \pm \sqrt{80}}{2}$ without simplifying (this is accepted as "oe", but $-3 \pm 2\sqrt{5}$ is cleanest).

Things to Be Careful About

This is a calculator-permitted paper, but the answer must still be given in exact form — no decimals.
The "AG" step means you must derive the quadratic forward from the substitution; starting from the target loses the method marks.
"isw" applies after a correct $a$ : later slips are ignored, but the discriminant equation itself must be correct.
Both values of $a$ are required — there are two parallel tangents of gradient $2$ .

Techniques used

substitute the line into the circle equationexpand and collect terms to reach the given quadraticset the discriminant equal to zero for tangencysolve the resulting quadratic exactly with surds

Q12MediumSeries

(a)

Write down the coefficient of $x^r$ in the binomial expansion of $(2 + x)^{59}$ .

1M Similar

(b)

For this expansion, find the value of $r$ for which the coefficient of $x^r$ is equal to the coefficient of $x^{r+1}$ .