The polynomial $\mathrm{p}$ is such that $\mathrm{p}(x) = 3x^3 - 7x^2 + ax + b$, where $a$ and $b$ are integers. It is given that $\mathrm{p}'(-1) = 21$ and that $x - 2$ is a factor of $\mathrm{p}(x)$. (a) Find the values of $a$ and $b$. (b) Hence write $\mathrm{p}(x)$ as a product of linear factors with integer coefficients. (c) Using your values of $a$ and $b$, solve the equation $3\mathrm{e}^{6y} - 7\mathrm{e}^{4y} + a\mathrm{e}^{2y} + b = 0$.

(a) a = -2, b = 8 · (b) (x - 2)(x + 1)(3x - 4) · (c) y = 0.5 ln 2 or y = 0.5 ln(4/3)

4037/12 May/June 2025 — Additional Mathematics Solved Past Paper

Q13MMedium-EasyEquations, inequalities and graphs

Fig. 1

The diagram shows the graph of $y = |\mathrm{f}(x)|$ , where $\mathrm{f}$ is a cubic polynomial.
Find expressions for the two possible functions $\mathrm{f}(x)$ .
Write each expression in fully factorised form.

DifficultyMedium-Easy

Worked solution

Approach

The graph of $y = |\mathrm{f}(x)|$ meets the $x$ -axis exactly where $\mathrm{f}(x) = 0$ , so the roots are read straight from the graph. The cubic is built from its factors, and the $y$ -intercept fixes the leading constant; the sign of the whole expression cannot be determined because the modulus makes $\pm\mathrm{f}(x)$ give the same graph.

Working

From the graph, $|\mathrm{f}(x)| = 0$ at $x = -2$ , $x = -0.5$ and $x = 3$ , so

\mathrm{f}(x) = k(x+2)\left(x+\tfrac{1}{2}\right)(x-3)

for some constant $k$ . Using the equivalent form with integer coefficients,

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

At $x = 0$ the graph shows $|\mathrm{f}(0)| = 2$ :

|A(2)(1)(-3)| = |-6A| = 2 \quad\Rightarrow\quad |A| = \frac{1}{3}

so $A = \pm\dfrac{1}{3}$ .

Answer

\mathrm{f}(x) = \pm\frac{1}{3}(x+2)(2x+1)(x-3)

Final answer

f(x) = +/- (1/3)(x + 2)(2x + 1)(x - 3)

Detailed explanation

Walkthrough

The graph shown is of $y = |\mathrm{f}(x)|$ . Wherever this graph touches or crosses the $x$ -axis, we have $|\mathrm{f}(x)| = 0$ , which happens exactly when $\mathrm{f}(x) = 0$ . So the three marked points on the axis, $x = -2$ , $x = -0.5$ and $x = 3$ , are the three roots of the cubic.

A cubic with roots $-2$ , $-\tfrac{1}{2}$ and $3$ has factors $(x+2)$ , $\left(x+\tfrac{1}{2}\right)$ and $(x-3)$ . Clearing the fraction by writing $\left(x+\tfrac{1}{2}\right) = \tfrac{1}{2}(2x+1)$ and absorbing the constant into a single unknown multiplier gives

\mathrm{f}(x) = A(x+2)(2x+1)(x-3).

To find $A$ , use the labelled point on the $y$ -axis: the graph passes through $(0, 2)$ , so $|\mathrm{f}(0)| = 2$ . Substituting $x = 0$ :

|A(2)(1)(-3)| = 6|A| = 2 \quad\Rightarrow\quad |A| = \frac{1}{3}.

Both $A = \tfrac{1}{3}$ and $A = -\tfrac{1}{3}$ are possible: taking the modulus destroys the overall sign, so $|\mathrm{f}|$ looks identical for either choice. Hence the two possible functions are

\mathrm{f}(x) = \frac{1}{3}(x+2)(2x+1)(x-3) \quad\text{or}\quad \mathrm{f}(x) = -\frac{1}{3}(x+2)(2x+1)(x-3).

Key Takeaways

The zeros of $y = |\mathrm{f}(x)|$ are exactly the zeros of $\mathrm{f}$ ; a modulus only reflects negative parts above the axis.
A polynomial can be reconstructed in factorised form directly from its roots.
A single known point (here the $y$ -intercept) determines the vertical scale factor.
Because the modulus erases an overall sign change, both $+\mathrm{f}$ and $-\mathrm{f}$ fit the same picture — hence the $\pm$ .

Common Mistakes

Forgetting the $\pm$ : the mark scheme explicitly awards a mark for it, since the modulus graph cannot distinguish the two signs.
Writing the root $-0.5$ as the factor $(x - 0.5)$ instead of $(x + 0.5)$ or $(2x+1)$ — signs inside factors must be reversed relative to the root.
Dropping the scale factor $\tfrac{1}{3}$ entirely, or computing it without the modulus (getting $A = -\tfrac{1}{3}$ only) and so missing one of the two functions.
Adding extra terms beyond the product of the three linear factors — the scheme requires "no extra terms".
Expanding the cubic when the question asks for fully factorised form.

Things to Be Careful About

The answer must be left fully factorised: $\pm\tfrac{1}{3}(x+2)(2x+1)(x-3)$ (equivalently $\pm 2(x+2)(x+0.5)(x-3)$ ). An expanded cubic loses the final mark.
The mark scheme says "isw": later incorrect simplification after a correct answer is ignored, but the printed exact form is what scores.
Check the $y$ -intercept calculation carefully: $(x+2)(2x+1)(x-3)$ at $x=0$ equals $-6$ , so $6|A| = 2$ gives $|A| = \tfrac{1}{3}$ , not $\tfrac{1}{6}$ or $\tfrac{2}{3}$ .

Techniques used

read the roots of the cubic from the modulus graphform the cubic from its factorsuse the y-intercept to find the scale factorallow both signs since the modulus removes the sign

Approach

Multiply both sides by $x^{\frac{1}{3}}$ to remove the fraction, giving a quadratic in $x^{\frac{1}{3}}$ . Factorise, solve for $x^{\frac{1}{3}}$ , then cube to find $x$ .

Working

x^{\frac{1}{3}} + 1 = \frac{6}{x^{\frac{1}{3}}}

Multiplying through by $x^{\frac{1}{3}}$ :

x^{\frac{2}{3}} + x^{\frac{1}{3}} = 6

x^{\frac{2}{3}} + x^{\frac{1}{3}} - 6 = 0

Factorising as a quadratic in $x^{\frac{1}{3}}$ :

\left(x^{\frac{1}{3}} + 3\right)\left(x^{\frac{1}{3}} - 2\right) = 0

So $x^{\frac{1}{3}} = -3$ or $x^{\frac{1}{3}} = 2$ .

Cubing each:

x = (-3)^3 = -27 \qquad \text{or} \qquad x = 2^3 = 8

Answer

x = -27 \quad \text{or} \quad x = 8

Final answer

x = -27 or x = 8

Detailed explanation

Walkthrough

The equation contains $x^{\frac{1}{3}}$ in two places, one of them in a denominator. Multiplying every term by $x^{\frac{1}{3}}$ clears the fraction and turns the equation into $x^{\frac{2}{3}} + x^{\frac{1}{3}} - 6 = 0$ . The key observation is that this is a quadratic in disguise: if we let $u = x^{\frac{1}{3}}$ , it reads $u^2 + u - 6 = 0$ , which factorises as $(u + 3)(u - 2) = 0$ . So $u = -3$ or $u = 2$ , meaning $x^{\frac{1}{3}} = -3$ or $x^{\frac{1}{3}} = 2$ . Cubing both sides of each gives $x = -27$ or $x = 8$ . Both must be kept — cubing is valid for negative numbers, so $(-3)^3 = -27$ is a genuine solution (check: $(-27)^{\frac{1}{3}} + 1 = -2$ and $\frac{6}{-3} = -2$ ). Discarding $x = -27$ loses an accuracy mark.

Key Takeaways

Equations with fractional powers can often be converted into quadratics by multiplying through by the lowest power present.
Cubing (unlike square-rooting) preserves sign, so negative roots are legitimate.
Always check both roots back in the original equation rather than rejecting one by habit.

Common Mistakes

Rejecting $x = -27$ because cube roots "can't be negative" — the mark scheme explicitly gives A0 if $x = -27$ is rejected.
Squaring instead of cubing when recovering $x$ , e.g. writing $x = 9$ from $x^{\frac{1}{3}} = -3$ .
Sign slips when factorising $u^2 + u - 6$ : it is $(u+3)(u-2)$ , not $(u-3)(u+2)$ .
Forgetting to multiply the constant term by $x^{\frac{1}{3}}$ when clearing the fraction.

Things to Be Careful About

Both answers are required; each carries its own A1 mark.
Keep exact values throughout — no decimals are involved here.
A quick check of each root in the original equation confirms validity and guards against accidental rejection of the negative root.

Techniques used

multiply through by a power of x to clear the fractionrecognise the hidden quadratic in x^(1/3)factorise the quadraticcube each root to recover x

Approach

Complete the square to identify the circle, then substitute $y = 2x - 3$ into the circle equation. If the resulting quadratic in $x$ has exactly one solution (discriminant zero), the line meets the circle once, so it is a tangent.

Working

Complete the square in the circle equation:

x^2 - 10x + y^2 - 4y + 24 = 0

(x - 5)^2 - 25 + (y - 2)^2 - 4 + 24 = 0

(x - 5)^2 + (y - 2)^2 = 5

So the circle has centre $(5, 2)$ and radius $\sqrt{5}$ . Substitute $y = 2x - 3$ :

x^2 + (2x - 3)^2 - 10x - 4(2x - 3) + 24 = 0

x^2 + 4x^2 - 12x + 9 - 10x - 8x + 12 + 24 = 0

5x^2 - 30x + 45 = 0

The discriminant is

b^2 - 4ac = (-30)^2 - 4(5)(45) = 900 - 900 = 0

Since the discriminant is zero, the quadratic has exactly one (repeated) solution, so the line meets the circle at exactly one point and is a tangent.

Answer

The line $y = 2x - 3$ is a tangent to the circle, because substituting it gives $5x^2 - 30x + 45 = 0$ , whose discriminant is $0$ (one repeated solution).

Final answer

The discriminant of 5x^2 - 30x + 45 = 0 is zero, so the line meets the circle at exactly one point and is a tangent

Detailed explanation

Walkthrough

First we need to know the circle we are working with. Completing the square on $x^2 - 10x$ gives $(x-5)^2 - 25$ , and on $y^2 - 4y$ gives $(y-2)^2 - 4$ . Moving the constants to the right-hand side gives $(x-5)^2 + (y-2)^2 = 5$ , so the centre is $(5, 2)$ and the radius is $\sqrt{5}$ .

To test whether the line is a tangent, we find where it meets the circle. Substituting $y = 2x - 3$ into the circle equation replaces every $y$ , leaving a quadratic in $x$ alone. Expanding carefully: $(2x-3)^2 = 4x^2 - 12x + 9$ and $-4(2x-3) = -8x + 12$ . Collecting terms gives $5x^2 - 30x + 45 = 0$ .

A tangent touches the circle at exactly one point, which means this quadratic must have exactly one solution. The discriminant $b^2 - 4ac = 900 - 900 = 0$ confirms a repeated root, so the line is a tangent. (Equivalently, $5x^2 - 30x + 45 = 5(x-3)^2 = 0$ has the single solution $x = 3$ .)

Key Takeaways

Completing the square converts a circle equation into centre-radius form.
A line is a tangent to a circle exactly when substituting it produces a quadratic with discriminant zero (one repeated root).

Common Mistakes

Sign slips when completing the square or expanding $(2x-3)^2$ .
Stopping at $5x^2 - 30x + 45 = 0$ without stating the discriminant is zero or that there is one solution only — the mark scheme requires the conclusion.
Arithmetic errors in collecting the linear terms ( $-12x - 10x - 8x = -30x$ ).

Things to Be Careful About

The mark scheme awards the M1 for eliminating one variable and the A1 only with a stated conclusion (discriminant $= 0$ , or one solution only). Always finish the argument.
Alternative accepted routes exist (perpendicular radius through $(3,3)$ , or the distance condition $(-kp + q - n)^2 = r^2(1+k^2)$ ), but the substitution route shown is the standard one.

Techniques used

complete the square to find the circle centre and radiussubstitute the line equation into the circle equationshow the resulting quadratic has a zero discriminant (repeated root)

(b)

Given that this tangent touches the circle at the point $P$ , find the coordinates of $P$ .

2M Similar

(c)

Find the equation of the circle which has its centre at $P$ and passes through the origin.

3M Similar

Q4Medium-EasyCalculusTrigonometry

(a)

Find $\int_{0}^{\pi} \sin\theta\,\mathrm{d}\theta$ .

2M Similar

DifficultyEasy

Worked solution

Approach

Integrate $\sin\theta$ term by term to get $-\cos\theta$ , then substitute the limits $0$ and $\pi$ .

Working

\int_{0}^{\pi} \sin\theta\,\mathrm{d}\theta = \left[-\cos\theta\right]_{0}^{\pi}

= \left(-\cos\pi\right) - \left(-\cos 0\right) = -(-1) - (-1) = 1 + 1 = 2

Answer

2

Final answer

2

Detailed explanation

Walkthrough

The integral of $\sin\theta$ with respect to $\theta$ is $-\cos\theta$ , because differentiating $-\cos\theta$ gives $\sin\theta$ . Writing the antiderivative in square-bracket form with the limits attached earns the first mark. Then evaluate at the top limit $\theta = \pi$ , where $\cos\pi = -1$ , and at the bottom limit $\theta = 0$ , where $\cos 0 = 1$ , and subtract: $-(-1) - (-1) = 2$ .

Key Takeaways

$\int \sin\theta\,\mathrm{d}\theta = -\cos\theta + C$ ; the minus sign is essential.
Definite integrals are evaluated as (top limit) minus (bottom limit), and the constant of integration cancels so it is not needed.

Common Mistakes

Writing $+\cos\theta$ as the antiderivative instead of $-\cos\theta$ .
Substituting the limits in the wrong order, giving $-2$ instead of $2$ .
Misremembering $\cos\pi$ ; it equals $-1$ , not $1$ .

Things to Be Careful About

The answer is exact; no decimal is needed.
The angle is in radians here ( $0$ to $\pi$ ), which is consistent with the calculus result $\int \sin\theta\,\mathrm{d}\theta = -\cos\theta$ holding only in radians.

Techniques used

integrate sin theta to minus cos thetaevaluate the definite integral between the limits

(b)

Given that $0 < \alpha < \frac{\pi}{2}$ , show that $\frac{\sec\alpha}{\cot\alpha + \tan\alpha}$ can be written as $\sin\alpha$ .

3M Similar

DifficultyMedium-Easy

Worked solution

Approach

Rewrite $\sec\alpha$ , $\cot\alpha$ and $\tan\alpha$ in terms of $\sin\alpha$ and $\cos\alpha$ , combine the denominator into a single fraction, then use $\sin^2\alpha + \cos^2\alpha = 1$ to simplify.

Working

\frac{\sec\alpha}{\cot\alpha + \tan\alpha} = \frac{\dfrac{1}{\cos\alpha}}{\dfrac{\cos\alpha}{\sin\alpha} + \dfrac{\sin\alpha}{\cos\alpha}}

Combine the two terms of the denominator over the common denominator $\sin\alpha\cos\alpha$ :

\cot\alpha + \tan\alpha = \frac{\cos^2\alpha + \sin^2\alpha}{\sin\alpha\cos\alpha} = \frac{1}{\sin\alpha\cos\alpha}

using $\sin^2\alpha + \cos^2\alpha = 1$ . Therefore

\frac{\sec\alpha}{\cot\alpha + \tan\alpha} = \frac{1}{\dfrac{1}{\sin\alpha\cos\alpha}} = \sin\alpha\cos\alpha \times \frac{1}{\cos\alpha} = \sin\alpha

as required.

Answer

\frac{\sec\alpha}{\cot\alpha + \tan\alpha} = \sin\alpha

Final answer

Shown: sec(alpha)/(cot(alpha) + tan(alpha)) = sin(alpha)

Detailed explanation

Walkthrough

The expression mixes three different trigonometric ratios, so the first step is to convert everything into sines and cosines: $\sec\alpha = \dfrac{1}{\cos\alpha}$ , $\cot\alpha = \dfrac{\cos\alpha}{\sin\alpha}$ and $\tan\alpha = \dfrac{\sin\alpha}{\cos\alpha}$ . This earns the first mark.

The denominator is then a sum of two fractions with different denominators, so put them over the common denominator $\sin\alpha\cos\alpha$ : the numerator becomes $\cos^2\alpha + \sin^2\alpha$ , which equals $1$ by the Pythagorean identity. So the whole denominator collapses to the single term $\dfrac{1}{\sin\alpha\cos\alpha}$ — this earns the second mark.

Finally, dividing by a fraction means multiplying by its reciprocal:

\frac{1}{\frac{1}{\sin\alpha\cos\alpha}} = \sin\alpha\cos\alpha

and multiplying by $\sec\alpha = \dfrac{1}{\cos\alpha}$ gives $\dfrac{\sin\alpha\cos\alpha}{\cos\alpha} = \sin\alpha$ . Since $0 < \alpha < \dfrac{\pi}{2}$ , all the ratios are well defined and positive, so no sign issues arise. The target $\sin\alpha$ is reached, completing the proof.

Key Takeaways

For identity proofs, converting every ratio to sines and cosines is the standard first move.
$\sin^2\alpha + \cos^2\alpha = 1$ is the tool that collapses a two-term denominator into one term.
Dividing by a fraction is multiplying by its reciprocal: $\dfrac{1}{1/(ab)} = ab$ .
A proof must run forward from the given expression to the target, showing every algebraic line.

Common Mistakes

Writing $\cot\alpha = \dfrac{\sin\alpha}{\cos\alpha}$ (confusing it with $\tan\alpha$ ).
Skipping the line where the denominator is combined into one fraction — the mark scheme requires sufficient correct detail, so jumping straight to the answer loses marks.
Forgetting $\sin^2\alpha + \cos^2\alpha = 1$ and leaving the numerator as $\cos^2\alpha + \sin^2\alpha$ .
Dropping the $\cos\alpha$ when cancelling at the end, or leaving the answer as $\sin\alpha\cos\alpha\sec\alpha$ without simplifying.
Working backwards from $\sin\alpha$ to the given expression, which is not a valid proof.

Things to Be Careful About

"Show that" (AG) parts require every intermediate line to be written out; the mark scheme states "Must show sufficient correct detail".
The condition $0 < \alpha < \dfrac{\pi}{2}$ guarantees $\sin\alpha$ , $\cos\alpha$ and $\tan\alpha$ are all non-zero, so no division by zero occurs and all ratios are positive.
Keep the notation consistent: the angle is $\alpha$ throughout, not $x$ or $\theta$ .

Techniques used

express sec, cot and tan in terms of sine and cosinecombine the fractions in the denominator over a common denominatorapply the identity sin squared plus cos squared equals onesimplify the compound fraction to sin alpha

Q5MediumFactors of polynomialsCalculusLogarithmic and exponential functions

The polynomial $\mathrm{p}$ is such that $\mathrm{p}(x) = 3x^3 - 7x^2 + ax + b$ , where $a$ and $b$ are integers.

It is given that $\mathrm{p}'(-1) = 21$ and that $x - 2$ is a factor of $\mathrm{p}(x)$ .

(a)

Find the values of $a$ and $b$ .

4M Similar

DifficultyMedium-Easy

Worked solution

Approach

Differentiate $\mathrm{p}(x)$ and use $\mathrm{p}'(-1) = 21$ to find $a$ ; then use the factor theorem, $\mathrm{p}(2) = 0$ , to find $b$ .

Working

\mathrm{p}'(x) = 9x^2 - 14x + a

Substituting $x = -1$ :

\mathrm{p}'(-1) = 9 + 14 + a = 21 \implies a = -2

Since $x - 2$ is a factor, $\mathrm{p}(2) = 0$ :

\mathrm{p}(2) = 24 - 28 + 2(-2) + b = 0 \implies b = 8

Answer

a = -2, \quad b = 8

Final answer

a = -2, b = 8

Detailed explanation

Walkthrough

The polynomial is $\mathrm{p}(x) = 3x^3 - 7x^2 + ax + b$ . Differentiating each term gives $\mathrm{p}'(x) = 9x^2 - 14x + a$ . The condition $\mathrm{p}'(-1) = 21$ means we substitute $x = -1$ : the first two terms become $9$ and $+14$ , so $9 + 14 + a = 21$ , giving $a = -2$ .

For $b$ , the statement that $x - 2$ is a factor invokes the factor theorem: a linear factor $x - c$ exists exactly when $\mathrm{p}(c) = 0$ . So $\mathrm{p}(2) = 24 - 28 + 2a + b = 0$ . With $a = -2$ this reads $24 - 28 - 4 + b = 0$ , i.e. $-8 + b = 0$ , so $b = 8$ .

Key Takeaways

Differentiating a cubic lowers each power by one and multiplies by the old index.
The factor theorem converts 'is a factor' into an equation by substituting the root.
Two independent conditions determine two unknown constants.

Common Mistakes

Sign slips when substituting $x = -1$ into $-14x$ (it becomes $+14$ , not $-14$ ).
Forgetting to substitute the found value of $a$ before solving for $b$ .
Misapplying the factor theorem by substituting $x = -2$ instead of $x = 2$ .

Things to Be Careful About

Each mark scheme line requires the substituted expression to be simplified ( $9 + 14 + a = 21$ , $24 - 28 + 2a + b = 0$ ), so show these lines explicitly.
This is a non-calculator paper: all arithmetic here is small integer work.

Techniques used

differentiate a cubic term by termevaluate the derivative at a given pointapply the factor theoremsolve simultaneous linear equations for the unknown coefficients

(b)

Hence write $\mathrm{p}(x)$ as a product of linear factors with integer coefficients.

3M Similar

DifficultyMedium-Easy

Worked solution

Approach

With $a = -2$ and $b = 8$ , divide $\mathrm{p}(x)$ by the known factor $(x - 2)$ , then factorise the quadratic quotient.

Working

\mathrm{p}(x) = 3x^3 - 7x^2 - 2x + 8

Dividing by $(x - 2)$ (e.g. by long division or comparing coefficients):

3x^3 - 7x^2 - 2x + 8 = (x - 2)(3x^2 - x - 4)

Factorising the quadratic:

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

since $(x+1)(3x-4) = 3x^2 - 4x + 3x - 4 = 3x^2 - x - 4$ .

Answer

\mathrm{p}(x) = (x - 2)(x + 1)(3x - 4)

Final answer

(x - 2)(x + 1)(3x - 4)

Detailed explanation

Walkthrough

Substituting $a = -2$ and $b = 8$ gives $\mathrm{p}(x) = 3x^3 - 7x^2 - 2x + 8$ . Since part (a) established that $x - 2$ is a factor, dividing the cubic by $(x - 2)$ leaves a quadratic. Long division (or matching coefficients in $(x-2)(3x^2 + px + q)$ ) gives the quotient $3x^2 - x - 4$ .

To factorise $3x^2 - x - 4$ , look for two numbers multiplying to $3 \times (-4) = -12$ and adding to $-1$ : these are $-4$ and $3$ . Splitting the middle term gives $3x^2 - 4x + 3x - 4 = x(3x - 4) + 1(3x - 4) = (x + 1)(3x - 4)$ .

So the full factorisation is $(x - 2)(x + 1)(3x - 4)$ .

Key Takeaways

A known linear factor reduces a cubic to a quadratic by division.
Factorising a non-monic quadratic uses the product-sum method on $ac$ .

Common Mistakes

Arithmetic slips in the division, giving a wrong quadratic quotient.
Failing to check that the factors multiply back to the original cubic.
Leaving the answer as $(x-2)(3x^2 - x - 4)$ instead of fully factorising into linear factors as asked.

Things to Be Careful About

The question says 'product of linear factors', so the quadratic must also be broken down.
The mark scheme requires both accuracy marks to come from the correct $a$ and $b$ — carry forward the values from part (a).

Techniques used

divide the cubic by the known linear factorfactorise the resulting quadraticwrite the cubic as a product of three linear factors

(c)

Using your values of $a$ and $b$ , solve the equation $3\mathrm{e}^{6y} - 7\mathrm{e}^{4y} + a\mathrm{e}^{2y} + b = 0$ .

3M Similar

DifficultyMedium

Worked solution

Approach

Let $u = \mathrm{e}^{2y}$ . Then the equation becomes the cubic from part (b) in $u$ , whose roots are $u = 2$ , $u = -1$ , $u = \frac{4}{3}$ . Since $\mathrm{e}^{2y} > 0$ , discard $u = -1$ , then take logs.

Working

With $u = \mathrm{e}^{2y}$ and $a = -2$ , $b = 8$ :

3u^3 - 7u^2 - 2u + 8 = 0

Using part (b):

(u - 2)(u + 1)(3u - 4) = 0 \implies u = 2, \; u = -1, \; u = \frac{4}{3}

Since $\mathrm{e}^{2y} > 0$ , reject $u = -1$ . Converting back:

\mathrm{e}^{2y} = 2 \implies 2y = \ln 2 \implies y = \frac{1}{2}\ln 2

\mathrm{e}^{2y} = \frac{4}{3} \implies 2y = \ln\frac{4}{3} \implies y = \frac{1}{2}\ln\frac{4}{3}

Answer

y = \frac{1}{2}\ln 2 \quad \text{or} \quad y = \frac{1}{2}\ln\frac{4}{3}

Final answer

y = 0.5 ln 2 or y = 0.5 ln(4/3)

Detailed explanation

Walkthrough

The equation $3\mathrm{e}^{6y} - 7\mathrm{e}^{4y} + a\mathrm{e}^{2y} + b = 0$ looks like the cubic from parts (a) and (b) if we write $u = \mathrm{e}^{2y}$ : then $\mathrm{e}^{4y} = u^2$ and $\mathrm{e}^{6y} = u^3$ . Substituting $a = -2$ and $b = 8$ gives exactly $3u^3 - 7u^2 - 2u + 8 = 0$ , which part (b) already factored as $(u - 2)(u + 1)(3u - 4) = 0$ .

The roots are $u = 2$ , $u = -1$ and $u = \frac{4}{3}$ . But $u = \mathrm{e}^{2y}$ is always positive, so $u = -1$ must be rejected — it corresponds to no real $y$ .

For each valid root, undo the substitution with natural logarithms: $\mathrm{e}^{2y} = 2$ gives $2y = \ln 2$ , so $y = \frac{1}{2}\ln 2$ ; similarly $\mathrm{e}^{2y} = \frac{4}{3}$ gives $y = \frac{1}{2}\ln\frac{4}{3}$ .

Key Takeaways

Equations in $\mathrm{e}^{ky}$ can be disguised polynomials under the substitution $u = \mathrm{e}^{2y}$ .
Exponentials are strictly positive, so negative roots of the substituted equation are extraneous.
Solving $\mathrm{e}^{2y} = k$ means taking $\ln$ of both sides and dividing by 2.

Common Mistakes

Keeping the root $u = -1$ and trying to take its logarithm.
Forgetting to halve after taking $\ln$ , writing $y = \ln 2$ instead of $y = \frac{1}{2}\ln 2$ .
Not using the factorisation from part (b) and attempting to solve the cubic from scratch.
Giving extra solutions or decimal approximations when exact log form is required.

Things to Be Careful About

The mark scheme awards B2 only for both solutions with no extras — rejecting $u = -1$ is essential.
Answers may be given in any equivalent form ('oe'), but exact logarithmic form is expected on this non-calculator paper.
This is a 'Hence'-style part: the factorisation from part (b) must actually be used.

Techniques used

substitute a new variable for the exponential termuse the factorisation from part (b)reject invalid roots where the exponential substitution is non-positivetake logarithms to solve for y

Q6MediumStraight-line graphsLogarithmic and exponential functions

When $\ln y$ is plotted against $x^3$ , a straight line passing through the points $(2, 5)$ and $(-8, 25)$ is obtained.

(a)

Find $y$ in terms of $x$ .

4M Similar

DifficultyMedium

Worked solution

Approach

Since $\ln y$ is plotted against $x^3$ , the relationship is a straight line of the form $Y = mX + c$ , where $Y = \ln y$ and $X = x^3$ .
Find the gradient $m$ and vertical intercept $c$ using the given points $(2, 5)$ and $(-8, 25)$ , then convert the resulting equation for $\ln y$ into an expression for $y$ in terms of $x$ .

Working

The straight line has the form:

\ln y = m(x^3) + c

Calculate the gradient $m$ from the two points $(X_1, Y_1) = (2, 5)$ and $(X_2, Y_2) = (-8, 25)$ :

m = \frac{25 - 5}{-8 - 2} = \frac{20}{-10} = -2

Substitute $(2, 5)$ into the linear equation to find $c$ :

5 = -2(2) + c

5 = -4 + c \implies c = 9

Thus, the linear relation is:

\ln y = -2x^3 + 9

Exponentiate both sides with base $\mathrm{e}$ to make $y$ the subject:

y = \mathrm{e}^{9 - 2x^3}

Answer

y = \mathrm{e}^{9 - 2x^3}

Final answer

y = e^(9 - 2x^3)

Detailed explanation

Walkthrough

When variables are plotted as a straight line, the vertical axis variable $Y$ and horizontal axis variable $X$ are related by the standard equation $Y = mX + c$ . Here, $Y = \ln y$ and $X = x^3$ , so the relationship is:

\ln y = m(x^3) + c

The line passes through $(X, Y) = (2, 5)$ and $(-8, 25)$ . Find the gradient $m$ :

m = \frac{25 - 5}{-8 - 2} = -2

Using the point $(2, 5)$ , find the intercept $c$ :

5 = -2(2) + c \implies c = 9

So $\ln y = 9 - 2x^3$ .

To express $y$ in terms of $x$ , take the natural exponential of both sides:

y = \mathrm{e}^{9 - 2x^3}

Alternatively, this can be written as $y = \mathrm{e}^9 \mathrm{e}^{-2x^3}$ .

Key Takeaways

For linear law problems, identify the uppercase variables $Y$ and $X$ corresponding to the plotted axes.
The equation is always $Y = mX + c$ , where the given coordinates are values of $(X, Y)$ , not $(x, y)$ .
Invert the log function using the base: $\ln y = A \iff y = \mathrm{e}^A$ .

Common Mistakes

Confusing the coordinates: treating $(2, 5)$ as $(x, y)$ rather than $(x^3, \ln y)$ .
Sign errors when finding the gradient: $\frac{25 - 5}{-8 - 2} = \frac{20}{-10} = -2$ .
Incorrectly converting to exponential form, e.g. writing $y = 9 - 2x^3$ instead of $y = \mathrm{e}^{9 - 2x^3}$ .

Things to Be Careful About

Ensure the final answer has $y$ explicitly in terms of $x$ , meaning $y = \ldots$ with no $\ln$ remaining on the left-hand side.

Techniques used

identify transformed variables on linear axescalculate straight-line gradientfind vertical interceptconvert logarithmic form to exponential form

(b)

Find the value of $x$ when $y = \mathrm{e}^{25}$ .

2M Similar

Q7MediumSeries

A geometric progression has a 4th term of $\frac{8k^6}{27}$ and a 6th term of $\frac{32k^{10}}{243}$ , where $k$ is a constant.

The common ratio of this geometric progression is positive.

(a)

Find the common ratio in terms of $k$ and the value of the first term of this geometric progression.

4M Similar

DifficultyMedium

Worked solution

Approach

Write the 4th and 6th terms as $ar^3$ and $ar^5$ . Dividing eliminates $a$ and gives $r^2$ ; the positive-ratio condition then fixes the sign of $r$ . Substitute back to find $a$ .

Working

The two terms give

ar^3 = \frac{8k^6}{27}, \qquad ar^5 = \frac{32k^{10}}{243}

Dividing the second by the first:

r^2 = \frac{ar^5}{ar^3} = \frac{32k^{10}}{243} \times \frac{27}{8k^6} = \frac{4k^4}{9}

Since the common ratio is positive:

r = \frac{2k^2}{3}

Substituting into $ar^3 = \dfrac{8k^6}{27}$ :

a\left(\frac{2k^2}{3}\right)^3 = a \times \frac{8k^6}{27} = \frac{8k^6}{27}

a = 1

Answer

r = \frac{2k^2}{3}, \qquad a = 1

Final answer

r = 2k^2/3 and a = 1

Detailed explanation

Walkthrough

A geometric progression with first term $a$ and common ratio $r$ has $n$ th term $ar^{n-1}$ , so the 4th term is $ar^3$ and the 6th term is $ar^5$ . Writing both given values in this form is the first mark. Dividing $ar^5$ by $ar^3$ cancels $a$ and leaves $r^2$ , which is the standard trick for finding $r$ from two non-consecutive terms. Simplifying the fraction gives $r^2 = \frac{4k^4}{9}$ , so $r = \pm\frac{2k^2}{3}$ ; the question states the common ratio is positive, so we take $r = \frac{2k^2}{3}$ . Finally, substituting this $r$ into $ar^3$ gives $a \times \frac{8k^6}{27} = \frac{8k^6}{27}$ , so $a = 1$ .

Key Takeaways

The $n$ th term of a GP is $ar^{n-1}$ ; terms two apart differ by a factor of $r^2$ .
Dividing two term equations eliminates $a$ cleanly.
A sign condition on $r$ decides which square root to take.

Common Mistakes

Using $ar^4$ and $ar^6$ for the 4th and 6th terms (off-by-one in the index).
Forgetting the positive-ratio condition and giving $r = \pm\frac{2k^2}{3}$ .
Index slips when simplifying $\frac{k^{10}}{k^6}$ or $\frac{32 \times 27}{8 \times 243}$ .

Things to Be Careful About

The mark scheme allows the 5th term $\frac{16k^8}{81}$ as an implied route, but the division method shown is the standard one.
$r$ must be expressed in terms of $k^2$ (an unsimplified numeric constant times $k^2$ earns the method mark); $a$ must be a non-zero constant.
Leave $r$ as the exact fraction $\frac{2k^2}{3}$ — it is needed in part (b).

Techniques used

write the nth term of a geometric progression in terms of a and rdivide consecutive terms to eliminate a and obtain r squaredtake the positive square root using the sign conditionsubstitute r back to find the first term

(b)

Given that this geometric progression has a sum to infinity of 3, find the possible values of $k$ .

3M Similar

DifficultyMedium-Easy

Worked solution

Approach

Use $S_\infty = \dfrac{a}{1-r}$ with $a = 1$ and $r = \dfrac{2k^2}{3}$ from part (a), set it equal to 3, and solve for $k$ .

Working

Using the sum to infinity formula with the values from part (a):

\frac{1}{1 - \frac{2k^2}{3}} = 3

1 = 3\left(1 - \frac{2k^2}{3}\right) = 3 - 2k^2

2k^2 = 2 \quad \Rightarrow \quad k^2 = 1

k = \pm 1

Answer

k = \pm 1

Final answer

k = +/-1

Detailed explanation

Walkthrough

This part is a 'hence' question: it uses $a = 1$ and $r = \frac{2k^2}{3}$ found in part (a). The sum to infinity of a GP is $\frac{a}{1-r}$ , valid because $|r| = \frac{2k^2}{3}$ is less than 1 for the values found. Setting this equal to 3 and multiplying both sides by the denominator gives $1 = 3 - 2k^2$ , so $k^2 = 1$ and $k = \pm 1$ . Both signs are valid since $r$ depends on $k^2$ , so $r = \frac{2}{3}$ either way, which is positive as required.

Key Takeaways

$S_\infty = \frac{a}{1-r}$ requires $|r| < 1$ .
Squaring in the ratio means both signs of $k$ give the same progression.

Common Mistakes

Forgetting to use the part (a) values and inventing new ones.
Giving only $k = 1$ and missing $k = -1$ .
Sign slips when multiplying out $3(1 - \frac{2k^2}{3})$ .

Things to Be Careful About

The mark scheme awards each value of $k$ only 'from correct work' — the equation $\frac{1}{1-\frac{2k^2}{3}} = 3$ must be seen.
Both values are required; a single value scores only one of the two answer marks.
Check consistency: with $k = \pm 1$ , $r = \frac{2}{3} > 0$ , satisfying the condition in the stem.

Techniques used

apply the sum to infinity formulasolve a linear equation in k squaredtake both square roots to obtain two values

Q8MediumCalculusLogarithmic and exponential functions

It is given that $y = \frac{\ln(3x^2 + 16)}{x + 2}$ .

(a)

Find $\frac{\mathrm{d}y}{\mathrm{d}x}$ when $x = 0$ .

Give your answer in the form $\ln p$ , where $p$ is a constant.

5M Similar

DifficultyMedium

Worked solution

Approach

Differentiate $y = \dfrac{\ln(3x^2 + 16)}{x + 2}$ using the quotient rule, then substitute $x = 0$ and simplify to a single logarithm.

Working

By the quotient rule,

\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{(x + 2) \cdot \frac{6x}{3x^2 + 16} - \ln(3x^2 + 16)}{(x + 2)^2}

since $\dfrac{\mathrm{d}}{\mathrm{d}x}\ln(3x^2 + 16) = \dfrac{6x}{3x^2 + 16}$ .

Substituting $x = 0$ :

\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{(0 + 2) \cdot \frac{0}{16} - \ln 16}{(0 + 2)^2} = \frac{0 - \ln 16}{4} = -\frac{\ln 16}{4}

Writing as a single logarithm:

-\frac{\ln 16}{4} = \ln\left(16^{-\frac{1}{4}}\right) = \ln\left(\frac{1}{16^{\frac{1}{4}}}\right) = \ln\frac{1}{2}

Answer

\frac{\mathrm{d}y}{\mathrm{d}x} = \ln\frac{1}{2}

Final answer

ln(1/2)

Detailed explanation

Walkthrough

The function is a quotient: a logarithm on top divided by a linear expression. The quotient rule gives the derivative as (bottom times derivative of top minus top times derivative of bottom) over bottom squared.

First find the derivative of the top using the chain rule: differentiating $\ln(3x^2 + 16)$ multiplies by the derivative of the inside bracket, giving $\frac{6x}{3x^2 + 16}$ .

Assembling the quotient rule:

\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{(x + 2)\frac{6x}{3x^2 + 16} - \ln(3x^2 + 16)}{(x + 2)^2}

Now substitute $x = 0$ . The first term vanishes because it contains a factor $6x = 0$ , leaving $\frac{-\ln 16}{4}$ . Finally use the law of logs in reverse: $-\frac{1}{4}\ln 16 = \ln(16^{-1/4}) = \ln\frac{1}{2}$ , since $16^{1/4} = 2$ .

Key Takeaways

The chain rule turns $\frac{\mathrm{d}}{\mathrm{d}x}\ln u$ into $\frac{u'}{u}$ .
The quotient rule structure must be memorised exactly; sign errors are common.
A coefficient multiplying a logarithm can be pulled inside as a power: $k\ln a = \ln(a^k)$ .

Common Mistakes

Writing the quotient rule with the terms reversed or adding instead of subtracting — the scheme requires all other terms correct for the A1.
Forgetting the chain-rule factor $\frac{6x}{3x^2 + 16}$ (the B1 step).
Leaving the answer as $-\frac{\ln 16}{4}$ instead of converting to the required form $\ln p$ .
Sign slips when moving the negative index inside the logarithm: $-\frac{1}{4}\ln 16 = \ln\frac{1}{2}$ , not $\ln(-2)$ or $\ln 2$ .

Things to Be Careful About

The final answer must be a single logarithm $\ln p$ ; intermediate forms score method marks only if all earlier marks were earned (the scheme's dependency note).
Substitution at $x = 0$ must be shown explicitly — the M1 depends on substituting and reaching one log term from correct work.

Techniques used

differentiate ln of a function of x using the chain ruleapply the quotient rulesubstitute the given value of xcombine a multiple of a logarithm into a single logarithm

(b)

Given that $x$ increases from $0$ to $h$ , where $h$ is small, write down the approximate change in $y$ .

1M Similar

Q9MediumFunctionsLogarithmic and exponential functions

It is given that $\mathrm{f}(x) = 2\ln(3x - 4)$ , for $x > a$ , and that $\mathrm{f}^{-1}$ exists.

(a)

Find the least possible value of $a$ .

1M Similar

(b)

For your value of $a$ , find the range of $\mathrm{f}$ .

1M Similar

(c)

For your value of $a$ , find an expression for $\mathrm{f}^{-1}(x)$ .

2M Similar

DifficultyMedium-Easy

Worked solution

Approach

Write $y = 2\ln(3x - 4)$ , interchange $x$ and $y$ , then solve for $y$ by exponentiating.

Working

y = 2\ln(3x - 4)

Interchanging $x$ and $y$ :

x = 2\ln(3y - 4)

Dividing by $2$ and exponentiating:

\frac{x}{2} = \ln(3y - 4) \implies 3y - 4 = \mathrm{e}^{\frac{x}{2}}

Solving for $y$ :

y = \frac{4 + \mathrm{e}^{\frac{x}{2}}}{3}

Answer

\mathrm{f}^{-1}(x) = \frac{4 + \mathrm{e}^{\frac{x}{2}}}{3}

Final answer

f^-1(x) = (4 + e^(x/2))/3

Detailed explanation

Walkthrough

To find an inverse, start from $y = 2\ln(3x - 4)$ and swap the roles of $x$ and $y$ , giving $x = 2\ln(3y - 4)$ . The logarithm is undone by exponentiating: dividing by $2$ gives $\frac{x}{2} = \ln(3y - 4)$ , so $3y - 4 = \mathrm{e}^{\frac{x}{2}}$ . Finally, adding $4$ and dividing by $3$ isolates $y$ , giving $\mathrm{f}^{-1}(x) = \frac{4 + \mathrm{e}^{\frac{x}{2}}}{3}$ . Note the domain of $\mathrm{f}^{-1}$ is $\mathbb{R}$ , matching the range of $\mathrm{f}$ found in part (b).

Key Takeaways

The inverse is found by interchanging $x$ and $y$ and solving; $\mathrm{e}^{\ln u} = u$ is the tool that undoes the logarithm. The domain of $\mathrm{f}^{-1}$ equals the range of $\mathrm{f}$ .

Common Mistakes

Forgetting to divide by $2$ before exponentiating; writing $\mathrm{e}^{x}$ instead of $\mathrm{e}^{\frac{x}{2}}$ ; sign errors when moving the $4$ ; and omitting the $\mathrm{f}^{-1}(x)$ notation, which the mark scheme requires.

Things to Be Careful About

The mark scheme allows one sign error for the method mark but the final answer must use correct $\mathrm{f}^{-1}(x)$ notation. The constant $\mathrm{e}$ is written upright and the exponent is $\frac{x}{2}$ , not $2x$ .

Techniques used

swap x and y in the function equationexponentiate to undo the natural logarithmrearrange for the inverse function

(d)

It is given that the equation $\mathrm{f}(x) = \mathrm{f}^{-1}(x)$ has two roots.

For your value of $a$ , sketch the graphs of $y = \mathrm{f}(x)$ and $y = \mathrm{f}^{-1}(x)$ on the axes.

Label each graph.
State the intercepts of each graph with the axes.
State the equations of any asymptotes.

Fig. 2

4M Similar

DifficultyMedium

Worked solution

Approach

$y = \mathrm{f}(x) = 2\ln(3x - 4)$ is a logarithmic curve with vertical asymptote $x = \frac{4}{3}$ , defined for $x > \frac{4}{3}$ . Its inverse $y = \mathrm{f}^{-1}(x) = \frac{4 + \mathrm{e}^{\frac{x}{2}}}{3}$ is the reflection of it in the line $y = x$ , with horizontal asymptote $y = \frac{4}{3}$ . Find the intercepts, then sketch both curves showing their two intersections on $y = x$ .

Working

Intercept of $\mathrm{f}$ : set $\mathrm{f}(x) = 0$ :

2\ln(3x - 4) = 0 \implies 3x - 4 = 1 \implies x = \frac{5}{3}

So $y = \mathrm{f}(x)$ crosses the $x$ -axis at $\left(\frac{5}{3}, 0\right)$ ; it has no $y$ -intercept since $\frac{5}{3} > 0$ lies to the right of the asymptote and the curve is not defined at $x = 0$ .

Intercept of $\mathrm{f}^{-1}$ : set $x = 0$ :

\mathrm{f}^{-1}(0) = \frac{4 + \mathrm{e}^{0}}{3} = \frac{4 + 1}{3} = \frac{5}{3}

So $y = \mathrm{f}^{-1}(x)$ crosses the $y$ -axis at $\left(0, \frac{5}{3}\right)$ ; it has no $x$ -intercept since $\mathrm{f}^{-1}(x) > \frac{4}{3}$ for all $x$ .

Asymptotes: $y = \mathrm{f}(x)$ has vertical asymptote $x = \frac{4}{3}$ ; $y = \mathrm{f}^{-1}(x)$ has horizontal asymptote $y = \frac{4}{3}$ .

The two curves are reflections of each other in $y = x$ , so their intersections lie on $y = x$ . The equation $\mathrm{f}(x) = \mathrm{f}^{-1}(x)$ has two roots, so the curves cross twice, both points on the line $y = x$ : one in the first quadrant (between $\frac{5}{3}$ and larger values) and one in the second quadrant (negative $x$ , positive $y$ ).

Answer

The sketch shows $y = \mathrm{f}(x)$ , a logarithmic curve in the first and fourth quadrants rising steeply, with vertical asymptote $x = \frac{4}{3}$ and $x$ -intercept $\left(\frac{5}{3}, 0\right)$ ; and $y = \mathrm{f}^{-1}(x)$ , an exponential-type curve in the first and second quadrants, with horizontal asymptote $y = \frac{4}{3}$ and $y$ -intercept $\left(0, \frac{5}{3}\right)$ . The curves intersect twice, both intersection points lying on the line $y = x$ .

Final answer

Logarithmic curve y = f(x) with vertical asymptote x = 4/3 and x-intercept (5/3, 0); reflected exponential curve y = f^-1(x) with horizontal asymptote y = 4/3 and y-intercept (0, 5/3); the two curves intersect twice, both points on the line y = x

Detailed explanation

Walkthrough

The function $\mathrm{f}(x) = 2\ln(3x - 4)$ is a standard logarithmic curve shifted right so that its vertical asymptote sits at $x = \frac{4}{3}$ , the domain boundary from part (a). It rises from $-\infty$ near the asymptote, crosses the $x$ -axis where $\ln(3x - 4) = 0$ , i.e. where $3x - 4 = 1$ , giving $x = \frac{5}{3}$ , and then grows without bound.

The inverse $\mathrm{f}^{-1}(x) = \frac{4 + \mathrm{e}^{\frac{x}{2}}}{3}$ is the mirror image of this curve in the line $y = x$ : the vertical asymptote $x = \frac{4}{3}$ becomes the horizontal asymptote $y = \frac{4}{3}$ , and the $x$ -intercept $\left(\frac{5}{3}, 0\right)$ becomes the $y$ -intercept $\left(0, \frac{5}{3}\right)$ . This is why the same number $\frac{5}{3}$ appears on both axes — the mark scheme awards a mark for marking $\frac{5}{3}$ on each axis with no other intercepts.

Because the curves are reflections in $y = x$ , any point where they meet must be fixed by the reflection, i.e. must lie on $y = x$ itself. Since the equation $\mathrm{f}(x) = \mathrm{f}^{-1}(x)$ has two roots, the sketch must show exactly two intersection points, both on $y = x$ : one in the first quadrant (where both coordinates are positive and greater than $\frac{4}{3}$ ) and one in the second quadrant (negative $x$ , positive $y$ ).

Key Takeaways

A function and its inverse are reflections of each other in the line $y = x$ ; consequently vertical and horizontal asymptotes swap, and intercepts swap coordinates. Intersections of a function with its inverse lie on $y = x$ .

Common Mistakes

Drawing only one intersection point; placing an intersection off the line $y = x$ ; marking a $y$ -intercept for $\mathrm{f}$ (it has none, since its domain starts at $\frac{4}{3}$ ); drawing the exponential curve crossing below its asymptote $y = \frac{4}{3}$ ; forgetting to label the curves; and marking extra intercepts, which loses the intercept mark.

Things to Be Careful About

The mark scheme is dependent: the $\mathrm{f}^{-1}$ mark depends on the $\mathrm{f}$ mark being correct, and the intercept mark depends on the first $\mathrm{B}$ . The value $\frac{5}{3}$ must appear on each axis (or be stated) and no other intercepts may be marked. Both asymptote equations, $x = \frac{4}{3}$ and $y = \frac{4}{3}$ , must be given, on the graph or at the side. The curves must show correct asymptotic behaviour: $\mathrm{f}$ plunging to $-\infty$ near $x = \frac{4}{3}$ , and $\mathrm{f}^{-1}$ flattening towards $y = \frac{4}{3}$ as $x \to -\infty$ .

Techniques used

sketch a logarithmic curve with its vertical asymptotesketch the reflected exponential curve with its horizontal asymptoteuse symmetry in the line y = x to locate intersectionsidentify intercepts of logarithmic and exponential graphs

Q10MediumCircular measure

Fig. 3

The diagram shows the shape $OABCDEF$ .
$AOF$ is a straight line.
$OAB$ and $OEF$ are sectors of a circle with centre $O$ and radius $r$ .
Angle $BOA = {}$ angle $EOF$ .

$OCD$ is a sector of a circle with centre $O$ and radius $\frac{4r}{3}$ .

Angle $COD$ is $\theta$ radians.
The point $B$ lies on the line $OC$ and the point $E$ lies on the line $OD$ .
The line $BE$ is parallel to the line $AOF$ .

(a)

Find, in terms of $r$ and $\theta$ , the area of the shaded region $BCDE$ .

3M Similar

DifficultyMedium-Easy

Worked solution

Approach

The shaded region $BCDE$ is the sector $OCD$ of radius $\frac{4r}{3}$ minus the triangle $OBE$ formed by the two radii $OB$ and $OE$ (each of length $r$ ) with included angle $\theta$ .

Working

Area of sector $OCD$ :

\frac{1}{2}\left(\frac{4r}{3}\right)^2 \theta

Since $B$ lies on $OC$ and $E$ lies on $OD$ , we have $OB = OE = r$ , so the area of triangle $OBE$ is:

\frac{1}{2}r^2 \sin\theta

Therefore the area of the shaded region is:

\frac{1}{2}\left(\frac{4r}{3}\right)^2 \theta - \frac{1}{2}r^2 \sin\theta

Answer

\text{Area of } BCDE = \frac{1}{2}\left(\frac{4r}{3}\right)^2 \theta - \frac{1}{2}r^2 \sin\theta

Final answer

(1/2)(4r/3)^2(theta) - (1/2)r^2 sin(theta)

Detailed explanation

Walkthrough

The shaded region sits between the large arc $CD$ and the straight segment $BE$ . The natural way to find its area is to take the whole sector $OCD$ — a sector of radius $\frac{4r}{3}$ and angle $\theta$ radians — and remove the unshaded triangle $OBE$ that lies underneath $BE$ inside that sector. Because $B$ lies on the line $OC$ and $E$ lies on the line $OD$ , the sides $OB$ and $OE$ are exactly the radius $r$ of the smaller sectors, and the angle between them at $O$ is $\theta$ . A triangle with two known sides and the included angle has area $\frac{1}{2}ab\sin C$ , giving $\frac{1}{2}r^2\sin\theta$ . Subtracting leaves the shaded area.

Key Takeaways

Sector area in radians: $A = \frac{1}{2}r^2\theta$ .
Triangle area with two sides and included angle: $A = \frac{1}{2}ab\sin C$ .
Composite areas are found by subtracting a simpler inner region from an outer one.

Common Mistakes

Using degrees instead of radians in the sector formula — the formula $\frac{1}{2}r^2\theta$ only works when $\theta$ is in radians.
Forgetting to square the radius $\frac{4r}{3}$ correctly; it gives $\frac{16r^2}{9}$ , not $\frac{4r^2}{3}$ .
Subtracting a small sector instead of the triangle — the region below $BE$ inside sector $OCD$ is a triangle, since $OB$ and $OE$ are straight segments.

Things to Be Careful About

The mark scheme allows the answer unsimplified, so leaving $\frac{1}{2}\left(\frac{4r}{3}\right)^2\theta - \frac{1}{2}r^2\sin\theta$ as it stands earns full marks.
Both B1 marks (sector area and triangle area) must be visible separately in your working.

Techniques used

compute the area of a sector using half r squared thetasubtract the area of triangle OBE from the larger sector

(b)

Fig. 4

The diagram shows the shape from part (a) with region $OABEF$ shaded.
Find, in terms of $r$ and $\theta$ , the perimeter of the shaded region.

5M Similar

DifficultyMedium

Worked solution

Approach

The perimeter of the shaded region $OABEF$ consists of: the two arcs $AB$ and $EF$ , the straight base $AF$ (length $2r$ ), and the straight top $BE$ . Each arc subtends angle $\pi - \theta$ at $O$ (since $BOA = EOF$ and together with $\theta$ they make the straight angle $\pi$ ).

Working

Each of the angles $BOA$ and $EOF$ equals $\frac{\pi - \theta}{2}$ , so each arc length is:

r \times \frac{\pi - \theta}{2}

For the chord $BE$ : triangle $OBE$ has $OB = OE = r$ and included angle $\theta$ . By the cosine rule:

BE^2 = r^2 + r^2 - 2r^2\cos\theta = 2r^2(1 - \cos\theta)

so

BE = \sqrt{2r^2(1 - \cos\theta)}

(equivalently $BE = 2r\sin\frac{\theta}{2}$ ).

The perimeter is then the sum of both arcs, the base $AF = OA + OF = 2r$ , and $BE$ :

\text{Perimeter} = 2 \times r\frac{\pi - \theta}{2} + 2r + \sqrt{2r^2(1 - \cos\theta)}

= r(\pi - \theta) + 2r + \sqrt{2r^2(1 - \cos\theta)}

Answer

\text{Perimeter} = r(\pi - \theta) + 2r + \sqrt{2r^2(1 - \cos\theta)}

Final answer

r(pi - theta) + 2r + sqrt(2r^2(1 - cos(theta)))

Detailed explanation

Walkthrough

To find the perimeter of the shaded lower region, walk around its boundary: arc $AB$ , then the straight line $AF$ , then arc $FE$ , then the straight line $EB$ back to the start.

First work out the angle of each small sector. Since $AOF$ is a straight line, the three angles at $O$ add to $\pi$ : the two equal sector angles plus $\theta$ . So each sector angle is $\frac{\pi - \theta}{2}$ , and each arc has length $s = r\theta_{\text{sector}} = r\frac{\pi - \theta}{2}$ .

Next find the length $BE$ . Triangle $OBE$ is isosceles with $OB = OE = r$ and vertex angle $\theta$ . The cosine rule gives $BE^2 = r^2 + r^2 - 2r^2\cos\theta = 2r^2(1 - \cos\theta)$ , so $BE = \sqrt{2r^2(1 - \cos\theta)}$ . (Using the half-angle identity this also equals $2r\sin\frac{\theta}{2}$ .)

Finally add up: two arcs give $r(\pi - \theta)$ , the base $AF$ is $OA + OF = 2r$ , and adding $BE$ completes the perimeter.

Key Takeaways

Arc length in radians: $s = r\theta$ .
Angles on a straight line sum to $\pi$ radians — used here to get each sector's angle.
Cosine rule finds a chord from two radii and the included angle; the half-angle form $2r\sin\frac{\theta}{2}$ is a useful shortcut for chords.
Perimeters of composite shapes require identifying every boundary piece exactly once.

Common Mistakes

Taking each sector angle to be $\pi - \theta$ instead of $\frac{\pi - \theta}{2}$ , doubling the arc contribution incorrectly.
Forgetting to include both arcs, or including the outer arc $CD$ which is not part of the shaded boundary.
Making an algebra slip rearranging the cosine rule to get $BE$ — the mark scheme awards A0 if this rearrangement is wrong.
Working in degrees: all angles here are in radians because they come from the straight-line angle $\pi$ .

Things to Be Careful About

The mark scheme allows unsimplified answers, so $2 \times r\frac{\pi - \theta}{2} + 2r + \sqrt{2r^2(1 - \cos\theta)}$ scores full marks.
The M1 for $BE$ requires a complete method (cosine rule, sine rule, or basic trigonometry) using $r$ and a correct angle — show the full working, not just the result.
The final M1 is for a correct plan using your own earlier values, so even if $BE$ was slightly wrong, assembling the perimeter correctly still earns the method mark.

Techniques used

find the reflex-sector angle by subtracting theta from picompute arc lengths using s equals r thetafind chord length BE using the cosine rule or half-angle sine rulesum the boundary lengths of the shaded region

Q11MediumVectors in two dimensions

Fig. 5

The diagram shows the triangle $OAB$ , where $\overrightarrow{OA} = \mathbf{a}$ and $\overrightarrow{OB} = \mathbf{b}$ .

The point $P$ lies on $OA$ such that $\overrightarrow{OP} = \frac{3}{4}\overrightarrow{OA}$ .

The point $Q$ lies on $AB$ such that $\overrightarrow{AQ} = \frac{1}{3}\overrightarrow{AB}$ .

The straight line through $P$ and $Q$ meets the straight line through $O$ and $B$ at the point $R$ .
It is given that $\overrightarrow{OR} = \lambda\mathbf{b}$ and $\overrightarrow{PR} = \mu\overrightarrow{PQ}$ , where $\lambda$ and $\mu$ are constants.

(a)

Find $\overrightarrow{OR}$ in terms of $\mathbf{a}$ , $\mathbf{b}$ and $\mu$ .

6M Similar

DifficultyMedium

Worked solution

Approach

First find $\overrightarrow{OQ}$ by going from $O$ to $A$ and then a third of the way along $\overrightarrow{AB}$ . Then find $\overrightarrow{PQ}$ , and finally write $\overrightarrow{OR}$ as $\overrightarrow{OP} + \mu\,\overrightarrow{PQ}$ .

Working

\overrightarrow{AB} = \mathbf{b} - \mathbf{a}, \qquad \overrightarrow{AQ} = \frac{1}{3}(\mathbf{b} - \mathbf{a})

\overrightarrow{OQ} = \mathbf{a} + \frac{1}{3}(\mathbf{b} - \mathbf{a}) = \frac{2}{3}\mathbf{a} + \frac{1}{3}\mathbf{b}

Since $\overrightarrow{OP} = \frac{3}{4}\mathbf{a}$ :

\overrightarrow{PQ} = \overrightarrow{OQ} - \overrightarrow{OP} = -\frac{3}{4}\mathbf{a} + \frac{2}{3}\mathbf{a} + \frac{1}{3}\mathbf{b}

\overrightarrow{PQ} = \frac{1}{3}\mathbf{b} - \frac{1}{12}\mathbf{a}

The route from $O$ to $R$ goes from $O$ to $P$ , then a multiple $\mu$ of $\overrightarrow{PQ}$ :

\overrightarrow{OR} = \overrightarrow{OP} + \mu\,\overrightarrow{PQ} = \frac{3}{4}\mathbf{a} + \mu\left(\frac{1}{3}\mathbf{b} - \frac{1}{12}\mathbf{a}\right)

Simplifying:

\overrightarrow{OR} = \left(\frac{3}{4} - \frac{\mu}{12}\right)\mathbf{a} + \frac{\mu}{3}\mathbf{b}

Answer

\overrightarrow{OR} = \left(\frac{3}{4} - \frac{\mu}{12}\right)\mathbf{a} + \frac{\mu}{3}\mathbf{b}

Final answer

OR = (3/4 - mu/12)a + (mu/3)b

Detailed explanation

Walkthrough

The key idea in vector geometry questions like this is that any route between two points can be built by hopping through known points. We want $\overrightarrow{OR}$ , but $R$ is not directly connected to anything we know — except that it lies on the line through $P$ and $Q$ . So we build the route $O \to P \to Q \to R$ : that is $\overrightarrow{OP} + \mu\,\overrightarrow{PQ}$ .

To get $\overrightarrow{PQ}$ we first need $\overrightarrow{OQ}$ . Since $Q$ lies on $AB$ with $\overrightarrow{AQ} = \frac{1}{3}\overrightarrow{AB}$ , we go from $O$ to $A$ ( $\mathbf{a}$ ) and then add one third of $\overrightarrow{AB} = \mathbf{b} - \mathbf{a}$ . Collecting terms gives $\overrightarrow{OQ} = \frac{2}{3}\mathbf{a} + \frac{1}{3}\mathbf{b}$ .

Then $\overrightarrow{PQ} = \overrightarrow{OQ} - \overrightarrow{OP}$ , using $\overrightarrow{OP} = \frac{3}{4}\mathbf{a}$ given in the stem. Subtracting gives $-\frac{3}{4}\mathbf{a} + \frac{2}{3}\mathbf{a} + \frac{1}{3}\mathbf{b} = \frac{1}{3}\mathbf{b} - \frac{1}{12}\mathbf{a}$ (note $-\frac{3}{4} + \frac{2}{3} = -\frac{9}{12} + \frac{8}{12} = -\frac{1}{12}$ ).

Finally, $\overrightarrow{OR} = \overrightarrow{OP} + \mu\,\overrightarrow{PQ} = \frac{3}{4}\mathbf{a} + \mu\left(\frac{1}{3}\mathbf{b} - \frac{1}{12}\mathbf{a}\right)$ , which simplifies to $\left(\frac{3}{4} - \frac{\mu}{12}\right)\mathbf{a} + \frac{\mu}{3}\mathbf{b}$ .

Key Takeaways

A point dividing a segment in ratio $k : (1-k)$ has position vector found by adding $k$ times the direction vector to one endpoint's position vector.
The vector from one point to another is 'destination minus start': $\overrightarrow{PQ} = \overrightarrow{OQ} - \overrightarrow{OP}$ .
Any path through known points gives the same resultant vector — choose the route that uses what you know.

Common Mistakes

Writing $\overrightarrow{AQ} = \frac{1}{3}\mathbf{b}$ instead of $\frac{1}{3}(\mathbf{b} - \mathbf{a})$ — the direction must be measured from $A$ , not from $O$ .
Sign errors when subtracting vectors: $\overrightarrow{PQ} = \overrightarrow{OQ} - \overrightarrow{OP}$ , not the other way round.
Arithmetic slips combining fractions such as $-\frac{3}{4} + \frac{2}{3}$ ; use a common denominator of 12.
Leaving the final answer unsimplified — the mark scheme requires the simplified form.

Things to Be Careful About

Keep $�$ and $\mathbf{b}$ separate throughout; only collect like vectors at the end.
The mark scheme allows unsimplified intermediate lines, but the final $\overrightarrow{OR}$ must be simplified ('Must be simplified').
An equivalent alternative route via $\overrightarrow{PA} = \frac{1}{4}\mathbf{a}$ and $\overrightarrow{AQ}$ earns the same marks.

Techniques used

express a point on a line segment as a position vector plus a fraction of a direction vectorfind the vector between two points by subtractionbuild the route from O to R via P and a multiple of PQ

(b)

Hence find the values of $\lambda$ and $\mu$ .

3M Similar

DifficultyMedium-Easy

Worked solution

Approach

Since $R$ lies on the line $OB$ , we have $\overrightarrow{OR} = \lambda\mathbf{b}$ : no $\mathbf{a}$ component. Equate this with the expression from part (a) and match coefficients of $\mathbf{a}$ and $\mathbf{b}$ .

Working

From part (a):

\overrightarrow{OR} = \left(\frac{3}{4} - \frac{\mu}{12}\right)\mathbf{a} + \frac{\mu}{3}\mathbf{b}

and since $R$ lies on $OB$ , $\overrightarrow{OR} = \lambda\mathbf{b}$ .

Equating the coefficient of $\mathbf{b}$ :

\lambda = \frac{\mu}{3}

Equating the coefficient of $\mathbf{a}$ to zero:

\frac{3}{4} - \frac{\mu}{12} = 0 \quad\Rightarrow\quad \mu = 9

Then:

\lambda = \frac{9}{3} = 3

Answer

\mu = 9, \qquad \lambda = 3

Final answer

mu = 9, lambda = 3

Detailed explanation

Walkthrough

This part uses the result from part (a) — hence the word 'Hence'. Because $R$ lies on the straight line through $O$ and $B$ , its position vector must be a pure multiple of $\mathbf{b}$ : there can be no $\mathbf{a}$ component. Since $\mathbf{a}$ and $\mathbf{b}$ are independent vectors (they are not parallel), the only way two expressions in them can be equal is if their coefficients match separately.

Setting the $\mathbf{a}$ coefficient to zero gives $\frac{3}{4} - \frac{\mu}{12} = 0$ , so $\mu = 9$ . Substituting into the $\mathbf{b}$ coefficient gives $\lambda = \frac{\mu}{3} = 3$ .

Geometrically this says $R$ is three times as far from $O$ as $B$ is, i.e. $OB = BR$ — consistent with the diagram where $R$ sits beyond $B$ on the extension of $OB$ .

Key Takeaways

If two expressions in independent vectors are equal, their coefficients must be equal term by term.
A point on the line through $O$ and $B$ has position vector $\lambda\mathbf{b}$ with no $\mathbf{a}$ component.
'Hence' signals that the previous answer should be carried forward rather than re-derived.

Common Mistakes

Forgetting that the $\mathbf{a}$ coefficient must equal zero, not just be ignored.
Solving $\frac{3}{4} - \frac{\mu}{12} = 0$ incorrectly; multiplying through by 12 gives $9 - \mu = 0$ .
Not using the part (a) result and attempting a fresh derivation, which wastes time and invites errors.

Things to Be Careful About

Both values are required for the final mark: $\mu = 9$ AND $\lambda = 3$ .
Check consistency: $\lambda = 3 > 1$ confirms $R$ lies beyond $B$ , matching the figure.

Techniques used

equate coefficients of independent vectorssolve simultaneous linear equations for lambda and mu

Q126MMediumCalculus

A curve is such that its gradient at the point $(x, y)$ is given by $(5x - 2)^{\frac{1}{3}}$ .

The curve passes through the point $\left(2, \frac{32}{5}\right)$ .

Find the coordinates of the stationary point on the curve.

DifficultyMedium

Worked solution

Approach

At a stationary point $\frac{\mathrm{d}y}{\mathrm{d}x} = 0$ , so set $(5x-2)^{\frac{1}{3}} = 0$ to find the $x$ -coordinate. Then integrate the gradient function to recover the equation of the curve, using the given point $\left(2, \frac{32}{5}\right)$ to find the constant of integration, and finally substitute the stationary $x$ -value into that equation.

Working

At a stationary point:

(5x - 2)^{\frac{1}{3}} = 0 \quad\Rightarrow\quad 5x - 2 = 0 \quad\Rightarrow\quad x = \frac{2}{5}

Integrate the gradient function:

y = \int (5x - 2)^{\frac{1}{3}}\,\mathrm{d}x = \frac{(5x-2)^{\frac{4}{3}}}{\frac{4}{3} \times 5} + c = \frac{3}{20}(5x - 2)^{\frac{4}{3}} + c

Use the point $\left(2, \frac{32}{5}\right)$ :

\frac{32}{5} = \frac{3}{20}\left((5 \times 2) - 2\right)^{\frac{4}{3}} + c = \frac{3}{20}(8)^{\frac{4}{3}} + c

Since $8^{\frac{4}{3}} = 16$ :

\frac{32}{5} = \frac{3}{20} \times 16 + c = \frac{12}{5} + c \quad\Rightarrow\quad c = \frac{32}{5} - \frac{12}{5} = 4

So the curve is $y = \frac{3}{20}(5x-2)^{\frac{4}{3}} + 4$ . At $x = \frac{2}{5}$ :

y = \frac{3}{20}(0)^{\frac{4}{3}} + 4 = 4

Answer

The stationary point is $\left(\frac{2}{5},\ 4\right)$ .

Final answer

(2/5, 4)

Detailed explanation

Walkthrough

A stationary point is where the gradient is zero, so we begin by setting the given derivative $(5x-2)^{\frac{1}{3}}$ equal to zero. A power equals zero only when its base is zero, giving $5x - 2 = 0$ , so $x = \frac{2}{5}$ .

To find the $y$ -coordinate at this point we need the equation of the curve itself, so we integrate the gradient function. The integrand is a linear expression raised to a power, which calls for the reverse chain rule: raise the power by one (from $\frac{1}{3}$ to $\frac{4}{3}$ ), divide by the new power, and also divide by the inner derivative $5$ . This gives $y = \frac{3}{20}(5x-2)^{\frac{4}{3}} + c$ .

The constant $c$ is found using the fact that the curve passes through $\left(2, \frac{32}{5}\right)$ : substituting $x = 2$ gives $(8)^{\frac{4}{3}} = 16$ , so $\frac{32}{5} = \frac{48}{20} + c = \frac{12}{5} + c$ , hence $c = 4$ .

Finally, substitute $x = \frac{2}{5}$ into the full equation. Since $5 \times \frac{2}{5} - 2 = 0$ , the whole bracket term vanishes and $y = 4$ . The stationary point is therefore $\left(\frac{2}{5}, 4\right)$ .

Key Takeaways

A stationary point satisfies $\frac{\mathrm{d}y}{\mathrm{d}x} = 0$ ; find its $x$ -coordinate directly from the derivative before integrating.
Integrating $(ax+b)^n$ gives $\frac{(ax+b)^{n+1}}{a(n+1)} + c$ — remember to divide by the coefficient of $x$ as well as the new power.
A point on the curve fixes the constant of integration; always add $+c$ when recovering an equation from a derivative.
Substituting the stationary $x$ -value back into the completed equation gives the $y$ -coordinate.

Common Mistakes

Forgetting to divide by $5$ when integrating $(5x-2)^{\frac{1}{3}}$ , writing $\frac{3}{4}(5x-2)^{\frac{4}{3}}$ instead of $\frac{3}{20}(5x-2)^{\frac{4}{3}}$ — the mark scheme requires the correct $k$ in $k(5x-2)^{\frac{4}{3}}$ for M1.
Omitting the constant of integration $+c$ entirely, making it impossible to use the given point.
Mis-evaluating $8^{\frac{4}{3}}$ : it is $(\sqrt[3]{8})^4 = 2^4 = 16$ , not $8^{\frac{1}{3}} \times 4$ or similar.
Stopping after finding $x = \frac{2}{5}$ without finding the $y$ -coordinate — the question asks for coordinates.
Trying to find the stationary point from an integrated (second-derivative-type) route rather than setting the given first derivative to zero — the scheme explicitly requires the first derivative.

Things to Be Careful About

The final B1 mark depends on the earlier marks: the answer must be $\left(\frac{2}{5}, 4\right)$ with $c = 4$ correctly found.
The scheme allows an unsimplified fraction for $c$ , but the final coordinates should be given exactly — no decimals needed here since all values are exact.
Show the substitution of the point $\left(2, \frac{32}{5}\right)$ explicitly, as the dependent M1 for finding $c$ requires this attempt to be visible.
Keep the fractional-index arithmetic exact: $8^{\frac{4}{3}} = 16$ is best done by taking the cube root first.

Techniques used

set the first derivative equal to zero to locate the stationary point's x-coordinateintegrate a function of a linear expression using the reverse chain ruledetermine the constant of integration from a given point on the curveevaluate the curve equation at the stationary x-value

Additional Mathematics 4037/12 — May/June 2025

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