Additional Mathematics 4037 / 12 — October/November 2024

DifficultyMedium-Easy

Worked solution

Approach

The period of $y = a\cos bx + c$ is $\frac{2\pi}{b}$ , so equate this to $\frac{2\pi}{3}$ to find $b$ . Then substitute each given point into the equation to get two equations in $a$ and $c$ .

Working

\frac{2\pi}{b} = \frac{2\pi}{3} \quad \Rightarrow \quad b = 3

Substitute $\left(-\frac{\pi}{6}, -2\right)$ into $y = a\cos 3x + c$ :

-2 = a\cos\left(-\frac{\pi}{2}\right) + c = a(0) + c

c = -2

Substitute $\left(\frac{\pi}{9}, \frac{1}{2}\right)$ :

\frac{1}{2} = a\cos\left(\frac{\pi}{3}\right) + c = \frac{a}{2} - 2

\frac{a}{2} = \frac{5}{2} \quad \Rightarrow \quad a = 5

Answer

a = 5, \quad b = 3, \quad c = -2

Final answer

a = 5, b = 3, c = -2

Detailed explanation

Walkthrough

The curve $y = a\cos bx + c$ is a cosine wave with amplitude $|a|$ , midline $y = c$ , and period $\frac{2\pi}{b}$ (the full cosine cycle $0$ to $2\pi$ is compressed by the factor $b$ ). Since the period is given as $\frac{2\pi}{3}$ , we set $\frac{2\pi}{b} = \frac{2\pi}{3}$ and read off $b = 3$ .

Now the equation is $y = a\cos 3x + c$ , with two unknowns $a$ and $c$ . Each given point gives one equation. The point $\left(-\frac{\pi}{6}, -2\right)$ is chosen first because $3 \times \left(-\frac{\pi}{6}\right) = -\frac{\pi}{2}$ , where the cosine is exactly $0$ — so that equation contains only $c$ , giving $c = -2$ immediately.

The second point $\left(\frac{\pi}{9}, \frac{1}{2}\right)$ gives $3x = \frac{\pi}{3}$ , where $\cos\frac{\pi}{3} = \frac{1}{2}$ , so $\frac{1}{2} = \frac{a}{2} - 2$ , which rearranges to $a = 5$ .

Key Takeaways

For $y = a\cos bx + c$ : amplitude $|a|$ , vertical shift $c$ , period $\frac{2\pi}{b}$ .
Choosing which point to substitute first can make the algebra trivial — pick the point whose angle gives a special value of cosine ( $0$ , $\pm 1$ , $\pm\frac{1}{2}$ ).
Two unknowns need two equations; each coordinate pair supplies one.

Common Mistakes

Using period $= \frac{b}{2\pi}$ instead of $\frac{2\pi}{b}$ , giving $b = \frac{1}{3}$ or similar.
Substituting degrees instead of radians: here all angles are in radians, so $\cos\left(-\frac{\pi}{2}\right) = 0$ and $\cos\frac{\pi}{3} = \frac{1}{2}$ must be used as exact radian values.
Sign slips: $\cos(-\theta) = \cos\theta$ , but here the argument $-\frac{\pi}{2}$ makes the cosine zero anyway.
Mixing up which constant is $c$ (the vertical shift) and which is $a$ (the multiplier of the cosine).

Things to Be Careful About

All angles are in radians throughout — special values such as $\cos\frac{\pi}{3} = \frac{1}{2}$ should be known exactly.
The mark scheme requires both points to be used (or at least one, with $b$ correct); using only one point cannot determine both $a$ and $c$ .
Answers are exact integers here — no rounding involved.

Techniques used

read the period from the coefficient of xsubstitute given coordinates into the cosine equationsolve the resulting linear equations for a and c

(b)

Find the least value of $y$ on the curve for $0 \leq x \leq \frac{\pi}{2}$ , and state the value of $x$ at which this occurs.

DifficultyMedium-Easy

Worked solution

Approach

With $a = 5 > 0$ , the minimum of $y = 5\cos 3x - 2$ occurs when $\cos 3x = -1$ . Solve $3x = \pi$ for $x$ in $0 \leq x \leq \frac{\pi}{2}$ , then evaluate $y$ .

Working

Minimum when

\cos 3x = -1

Within $0 \leq x \leq \frac{\pi}{2}$ , i.e. $0 \leq 3x \leq \frac{3\pi}{2}$ , this gives

3x = \pi \quad \Rightarrow \quad x = \frac{\pi}{3}

Then

y = 5\cos\pi - 2 = 5(-1) - 2 = -7

Answer

x = \frac{\pi}{3}, \quad y = -7

Final answer

Least value y = -7 at x = pi/3

Detailed explanation

Walkthrough

Since $a = 5$ is positive, the smallest possible value of $5\cos 3x$ occurs when $\cos 3x$ takes its minimum value of $-1$ . So we need $3x = \pi$ (within the allowed range of $3x$ , namely $0$ to $\frac{3\pi}{2}$ ), giving $x = \frac{\pi}{3}$ , which lies inside $0 \leq x \leq \frac{\pi}{2}$ .

Substituting back: $y = 5\cos\pi - 2 = -5 - 2 = -7$ . This matches the structure of the curve: the minimum equals $c - a = -2 - 5 = -7$ .

An alternative route (accepted by the mark scheme) is calculus: differentiate to get $\frac{\mathrm{d}y}{\mathrm{d}x} = -15\sin 3x$ , set it to zero giving $x = \frac{\pi}{3}$ , then evaluate $y = -7$ .

Key Takeaways

For $y = a\cos bx + c$ with $a > 0$ : maximum is $c + a$ , minimum is $c - a$ .
The minimum of $\cos(bx)$ occurs where $bx = \pi$ (plus multiples of $2\pi$ ); always check the candidate lies in the stated interval for $x$ .
A quick check on range: doubling the interval $0 \leq x \leq \frac{\pi}{2}$ gives $0 \leq 3x \leq \frac{3\pi}{2}$ , which contains exactly one minimum of cosine.

Common Mistakes

Solving $\cos 3x = -1$ as $3x = \frac{\pi}{2}$ (confusing with sine's maximum) or forgetting to divide by $3$ .
Giving $x = \pi$ instead of $x = \frac{\pi}{3}$ — the question asks for $x$ , not $3x$ .
Not checking that the solution lies in $0 \leq x \leq \frac{\pi}{2}$ .
Computing the minimum as $-(a + c)$ or $a - c$ instead of $c - a$ .
If using calculus, forgetting to confirm it is a minimum rather than a maximum (here $\sin 3x$ changes sign through $x = \frac{\pi}{3}$ , confirming a minimum).

Things to Be Careful About

The answer must be exact: $x = \frac{\pi}{3}$ and $y = -7$ , not decimal approximations.
Both parts of the answer are required: the least value of $y$ AND the value of $x$ at which it occurs — omitting either loses a mark.
The mark scheme allows follow-through on earlier values, but with the correct $a$ , $b$ , $c$ the answers are $x = \frac{\pi}{3}$ , $y = -7$ .

Techniques used

identify the minimum of a cosine curve from cos bx = -1solve for x within the restricted intervalevaluate y at the minimum

Q2Medium-EasyCalculusEquations, inequalities and graphsQuadratic functions

It is given that $y = \mathrm{f}(x)$ , where $\mathrm{f}(x) = (2x - 5)(x - 1)^2$ .

(a)

Find the coordinates of the stationary points on the curve $y = \mathrm{f}(x)$ .

DifficultyMedium-Easy

Worked solution

Approach

Differentiate $\mathrm{f}(x) = (2x - 5)(x - 1)^2$ by the product rule, set the derivative to zero, solve for $x$ , then substitute each value into $\mathrm{f}(x)$ .

Working

Using the product rule with $u = 2x - 5$ and $v = (x-1)^2$ :

\mathrm{f}'(x) = 2(x-1)^2 + (2x-5) \cdot 2(x-1)

= 2(x-1)\left[(x-1) + (2x-5)\right] = 2(x-1)(3x - 6) = 6(x-1)(x-2)

Setting $\mathrm{f}'(x) = 0$ :

6(x-1)(x-2) = 0 \quad \Rightarrow \quad x = 1 \text{ or } x = 2

Substituting back:

\mathrm{f}(1) = (2(1)-5)(1-1)^2 = (-3)(0) = 0

\mathrm{f}(2) = (2(2)-5)(2-1)^2 = (-1)(1) = -1

Answer

The stationary points are $(1, 0)$ and $(2, -1)$ .

Final answer

(1, 0) and (2, -1)

Detailed explanation

Walkthrough

The curve is a product of two factors, so the product rule is the natural tool: differentiate one factor while holding the other, then swap. Writing $\mathrm{f}'(x) = 2(x-1)^2 + (2x-5) \cdot 2(x-1)$ , both terms share a factor of $2(x-1)$ , so factoring it out gives a very simple quadratic: $6(x-1)(x-2)$ . Stationary points occur where the gradient is zero, so we set this equal to zero; since it is already factored, the roots $x = 1$ and $x = 2$ are immediate. Each root must then be substituted into the original function to obtain the full coordinates — note that $x = 1$ makes the squared factor zero, giving $y = 0$ , which is why the curve touches the axis there rather than crossing it.

Key Takeaways

The product rule: if $y = uv$ then $\frac{\mathrm{d}y}{\mathrm{d}x} = u'v + uv'$ .
Always factorise the derivative before solving — it makes finding the roots much easier.
A stationary point at a repeated root of the original cubic lies on the $x$ -axis (a touch point).
Coordinates need both values: substitute back into $\mathrm{f}(x)$ , not into $\mathrm{f}'(x)$ .

Common Mistakes

Forgetting the second term of the product rule or differentiating $(x-1)^2$ incorrectly (the chain rule gives $2(x-1)$ , not $2x - 1$ ).
Giving only the $x$ -values without substituting back for $y$ — the question asks for coordinates.
Sign slips when evaluating $\mathrm{f}(2) = (-1)(1)^2 = -1$ .
The mark scheme awards accuracy marks only from correct working, so an unsupported correct answer scores less.

Things to Be Careful About

Both stationary points are required — the scheme gives A2 for both pairs, A1 for any one correct pair from correct working.
Keep exact integer values throughout; no rounding is involved here.
These points feed directly into parts (b) and (c), so they must be right.

Techniques used

differentiate using the product ruleequate the derivative to zerosolve the resulting quadraticsubstitute back to find the y-coordinates

(b)

On the axes, sketch the graph of $y = \mathrm{f}(x)$ , stating the intercepts with the axes.

Fig. 1

DifficultyMedium-Easy

Worked solution

Approach

Use the factors to find the intercepts and the stationary points from part (a) to place the turning points, then sketch the cubic.

Working

Intercepts:

$x$ -axis: $\mathrm{f}(x) = 0$ when $2x - 5 = 0$ or $(x-1)^2 = 0$ , i.e. $x = 2.5$ (crossing) and $x = 1$ (touching).
$y$ -axis: $\mathrm{f}(0) = (-5)(-1)^2 = -5$ , so the curve passes through $(0, -5)$ .

From part (a), the stationary points are a local maximum at $(1, 0)$ — on the $x$ -axis — and a local minimum at $(2, -1)$ in the fourth quadrant. Since the leading coefficient is positive, the curve rises to the right and falls to the left.

Answer

A positive cubic passing through $(0, -5)$ , rising to touch the $x$ -axis at $(1, 0)$ (local maximum), dipping to the local minimum $(2, -1)$ below the $x$ -axis, then crossing the $x$ -axis at $(2.5, 0)$ and rising steeply.

Final answer

Positive cubic through (0, -5), touching the x-axis at (1, 0), local minimum (2, -1) in the fourth quadrant, crossing the x-axis at (2.5, 0)

Detailed explanation

Walkthrough

The factors of $\mathrm{f}(x)$ tell us where the curve meets the $x$ -axis: the single factor $2x - 5$ gives a simple root at $x = 2.5$ where the curve crosses, while the repeated factor $(x-1)^2$ gives a double root at $x = 1$ where the curve touches. Setting $x = 0$ gives the $y$ -intercept $-5$ . From part (a) we know the turning points sit exactly at $(1, 0)$ and $(2, -1)$ , so the maximum lies on the axis and the minimum is just below it in the fourth quadrant. Because the coefficient of $x^3$ is positive, the overall shape rises to the right. Joining these features smoothly completes the sketch.

Key Takeaways

A repeated linear factor produces a touch point (no sign change); a simple factor produces a crossing.
The $y$ -intercept comes from evaluating $\mathrm{f}(0)$ .
Stationary points found earlier pin down the shape between the roots.
Positive leading coefficient means the cubic falls to the left and rises to the right.

Common Mistakes

Drawing the curve crossing at $x = 1$ instead of touching it.
Omitting the $y$ -intercept $(0, -5)$ or adding spurious extra intercepts — the third B1 requires all intercepts and no extras.
Placing the minimum above the axis or mislabelling which root is the touch point.
Getting the overall cubic orientation wrong (falling to the right).

Things to Be Careful About

The mark scheme pays B1 for correct cubic shape, B1 for correct position (touching once in the fourth quadrant region and intersecting the positive $x$ -axis once), and B1 for all intercepts with no extras — all three features must appear.
Label the intercepts clearly: $(0, -5)$ , $(1, 0)$ and $(2.5, 0)$ .

Techniques used

identify intercepts of a cubic from its factorsuse stationary points to position the turning pointssketch a positive cubic with a repeated root

(c)

Hence find the values of $k$ for which $\mathrm{f}(x) = k$ has exactly one solution.

DifficultyMedium-Easy

Worked solution

Approach

$\mathrm{f}(x) = k$ has exactly one solution when the horizontal line $y = k$ cuts the sketch from part (b) exactly once. Read off the relevant ranges of $k$ .

Working

From the sketch: the local maximum has $y = 0$ and the local minimum has $y = -1$ . A horizontal line meets the cubic exactly once when it lies entirely above the local maximum or entirely below the local minimum:

Above the maximum: $k > 0$ — the line cuts only the far-right branch.
Below the minimum: $k < -1$ — the line cuts only the far-left branch.

(Between $-1$ and $0$ the line cuts three times; at $k = 0$ or $k = -1$ it cuts twice counting the touch.)

Answer

k < -1 \quad \text{or} \quad k > 0

Final answer

k < -1 or k > 0

Detailed explanation

Walkthrough

Solving $\mathrm{f}(x) = k$ geometrically means finding where the horizontal line $y = k$ intersects the sketched curve. Looking at the sketch from part (b): the curve's highest turning point is at height $0$ and its lowest at height $-1$ . If the line sits above $y = 0$ , it can only meet the steeply rising right-hand branch — one solution. If it sits below $y = -1$ , it only meets the falling left-hand branch — again one solution. Between these heights the line weaves through the wiggle and meets the curve three times (or twice at the boundary heights, because of the touch at the maximum). Hence the answer is $k < -1$ or $k > 0$ .

Key Takeaways

The number of real solutions of $\mathrm{f}(x) = k$ equals the number of intersections of $y = k$ with the graph of $\mathrm{f}$ .
Turning-point heights are the critical boundary values for such questions.
Strict inequalities apply here because at $k = 0$ and $k = -1$ there are two solutions, not one.

Common Mistakes

Including the boundary values ( $k \leq -1$ or $k \geq 0$ ) — at those heights the line meets the curve twice, not once.
Reversing the inequalities or reading the minimum height as $+1$ instead of $-1$ .
Not using the sketch ('Hence') and instead attempting algebra, which is far harder for a cubic.

Things to Be Careful About

Both B1 marks are strict: $k < -1$ and $k > 0$ separately, each worth one mark.
This part says 'Hence', so the argument must be based on the graph from part (b) — the turning points $(1, 0)$ and $(2, -1)$ supply the critical values.

Techniques used

interpret horizontal line intersections with a sketched cubicread off ranges of k giving exactly one intersection

Q3MediumCircular measureTrigonometry

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

Fig. 2

The diagram shows a circle with centre $O$ and radius 12, and a circle with centre $C$ and radius 5.
The circles intersect at the points $A$ and $B$ , such that $OA$ and $OB$ are tangents to the circle with centre $C$ .

(a)

Show that the obtuse angle $ACB$ is 2.35 radians, correct to 2 decimal places.

DifficultyMedium-Easy

Worked solution

Approach

Since $OA$ is a tangent to the circle with centre $C$ at point $A$ , the angle $O\hat{A}C = 90^\circ = \frac{\pi}{2}\text{ rad}$ . In the right-angled triangle $OAC$ , $OA = 12$ and $CA = 5$ . By symmetry, $A\hat{C}B = 2 \times A\hat{C}O$ .

Working

In right-angled triangle $OAC$ :

\tan(A\hat{C}O) = \frac{OA}{AC} = \frac{12}{5}

A\hat{C}O = \tan^{-1}\left(\frac{12}{5}\right) \approx 1.176005\text{ rad}

By symmetry about line $OC$ :

A\hat{C}B = 2 \times A\hat{C}O = 2 \times 1.176005\ldots = 2.35201\ldots\text{ rad}

A\hat{C}B = 2.35\text{ rad}\quad (\text{correct to } 2\text{ decimal places})

Answer

2.35\text{ rad}

Final answer

2.35 rad

Detailed explanation

Walkthrough

Tangent-radius property: $OA$ is tangent to the circle centre $C$ at $A$ , so the radius $CA$ is perpendicular to $OA$ , making $\triangle OAC$ a right-angled triangle at $A$ .
Right-angled trigonometry: With opposite side $OA = 12$ and adjacent side $AC = 5$ , we have $\tan(A\hat{C}O) = \frac{12}{5}$ .
Calculate the angle in radians: $A\hat{C}O = \tan^{-1}(2.4) \approx 1.176005\text{ rad}$ .
Double the angle: Since the figure is symmetric across $OC$ , the total angle $A\hat{C}B = 2 \times A\hat{C}O = 2.352\ldots \approx 2.35\text{ rad}$ to $2\text{ decimal places}$ .

Key Takeaways

The angle between a tangent and radius at the point of contact is $90^\circ$ .
Ensure your calculator is in radian mode when solving geometric problems specified in radians.

Common Mistakes

Calculating the angle in degrees rather than radians.
Showing insufficient intermediate precision before rounding to $2.35$ (the mark scheme requires showing at least $2 \times 1.176\ldots$ ).

Things to Be Careful About

For "Show that" questions, always provide the unrounded value (e.g. $1.176\ldots$ or $2.352\ldots$ ) before stating the rounded target to gain full credit.

Techniques used

use tangent-radius perpendicularityapply right-angled trigonometry in radiansdouble the angle for symmetry

(b)

Find the perimeter of the shaded region.

(c)

Find the area of the shaded region.

DifficultyMedium

Worked solution

Approach

The area of the shaded region is found by taking the area of the quadrilateral (kite) $OACB$ and subtracting the area of the sector $CAB$ of the circle with centre $C$ .

Working

The area of quadrilateral $OACB$ is the sum of the areas of the two congruent right-angled triangles $\triangle OAC$ and $\triangle OBC$ :

\text{Area of } OACB = 2 \times \left(\frac{1}{2} \times OA \times AC\right) = 2 \times \left(\frac{1}{2} \times 12 \times 5\right) = 60\text{ cm}^2

The area of sector $CAB$ with radius $r = 5$ and angle $\theta = 2.35201\ldots\text{ rad}$ is:

\text{Area of sector} = \frac{1}{2} r^2 \theta = \frac{1}{2} \times 5^2 \times 2.35201\ldots = 29.400\text{ cm}^2

Using $\theta = 2.35$ :

\text{Area of sector} = \frac{1}{2} \times 25 \times 2.35 = 29.375\text{ cm}^2

Subtract the sector area from the quadrilateral area:

\text{Area of shaded region} = 60 - 29.375 = 30.625\text{ cm}^2 \approx 30.6\text{ cm}^2

Answer

30.6\text{ cm}^2

Final answer

30.6 cm^2

Detailed explanation

Walkthrough

Break the enclosing polygon $OACB$ into two congruent right-angled triangles, $\triangle OAC$ and $\triangle OBC$ .
Each triangle has base $12$ and height $5$ , so its area is $\frac{1}{2} \times 12 \times 5 = 30\text{ cm}^2$ . The total area of kite $OACB$ is $2 \times 30 = 60\text{ cm}^2$ .
Calculate the area of sector $CAB$ using the formula $\text{Area} = \frac{1}{2}r^2\theta = \frac{1}{2} \times 5^2 \times 2.35 = 29.375\text{ cm}^2$ .
Subtract the sector area from the kite area: $60 - 29.375 = 30.625\text{ cm}^2$ , which rounds to $30.6\text{ cm}^2$ (3 s.f.).

Key Takeaways

Composite areas are frequently found by subtracting a circular sector from a polygon composed of right-angled triangles.
Sector area formula in radians: $A = \frac{1}{2}r^2\theta$ .

Common Mistakes

Subtracting from the area of sector $OAB$ of the large circle instead of the kite $OACB$ .
Leaving answers as fractions when decimals are expected.

Things to Be Careful About

Ensure correct substitution of $r = 5$ into the sector formula, not $r = 12$ .
Round the final answer appropriately to 3 significant figures ( $30.6$ ).

Techniques used

calculate area of quadrilateral from right-angled trianglescalculate sector area using (1/2) r^2 thetasubtract sector area from quadrilateral area

Q4MediumFunctionsLogarithmic and exponential functions

The function $\mathrm{f}$ is such that $\mathrm{f}(x) = 4\ln(3x - 2)$ , for $x > a$ , where $a$ is as small as possible.

(a)

10M

(i)

Write down the value of $a$ .

(ii)

Write down the range of $\mathrm{f}$ .

(iii)

Find $\mathrm{f}^{-1}(x)$ , stating its domain and range.

DifficultyMedium-Easy

Worked solution

Approach

Write $y = 4\ln(3x - 2)$ , make $x$ the subject by exponentiating, then swap $x$ and $y$ . The domain of $\mathrm{f}^{-1}$ equals the range of $\mathrm{f}$ , and the range of $\mathrm{f}^{-1}$ equals the domain of $\mathrm{f}$ .

Working

y = 4\ln(3x - 2) \quad \Rightarrow \quad \frac{y}{4} = \ln(3x - 2)

Exponentiating both sides:

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

Swapping $x$ and $y$ :

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

Since the range of $\mathrm{f}$ is $\mathbb{R}$ , the domain of $\mathrm{f}^{-1}$ is $x \in \mathbb{R}$ . Since the domain of $\mathrm{f}$ is $x > \frac{2}{3}$ , the range of $\mathrm{f}^{-1}$ is $\mathrm{f}^{-1}(x) > \frac{2}{3}$ .

Answer

\mathrm{f}^{-1}(x) = \frac{1}{3}\left(\mathrm{e}^{\frac{x}{4}} + 2\right), \quad x \in \mathbb{R}, \quad \mathrm{f}^{-1}(x) > \frac{2}{3}

Final answer

f^-1(x) = (1/3)(e^(x/4) + 2), domain x in R, range f^-1(x) > 2/3

Detailed explanation

Walkthrough

To find an inverse, set $y = \mathrm{f}(x)$ and solve for $x$ . Dividing by 4 gives $\frac{y}{4} = \ln(3x - 2)$ . Exponentiating removes the logarithm: $3x - 2 = \mathrm{e}^{\frac{y}{4}}$ , so $x = \frac{1}{3}\left(\mathrm{e}^{\frac{y}{4}} + 2\right)$ . Swapping $x$ and $y$ gives the inverse function. The domain of the inverse is the range of the original ( $\mathbb{R}$ ), and the range of the inverse is the domain of the original ( $>\frac{2}{3}$ ).

Key Takeaways

The standard inversion routine: set $y$ , rearrange, swap.
$\mathrm{e}^{\ln u} = u$ is the tool that undoes a logarithm.
Domain and range of a function swap to become range and domain of its inverse.

Common Mistakes

Forgetting to divide by 4 before exponentiating, giving $\mathrm{e}^{y}$ instead of $\mathrm{e}^{\frac{y}{4}}$ .
Adding 2 before dividing by 3, i.e. writing $\frac{\mathrm{e}^{\frac{x}{4}} + 2}{3}$ incorrectly as $\frac{\mathrm{e}^{\frac{x}{4}}}{3} + 2$ .
Stating the domain and range of $\mathrm{f}^{-1}$ the wrong way round.
Not using correct notation for the domain and range (the mark scheme requires proper notation for the B2).

Things to Be Careful About

The mark scheme awards M1 for reaching $3y - 2 = \mathrm{e}^{\frac{x}{4}}$ (or with $x$ / $y$ swapped) and A1 for the correct final form — show both stages.
The range must be written as $\mathrm{f}^{-1}(x) > \frac{2}{3}$ , not $y > \frac{2}{3}$ loosely attached to the wrong variable.

Techniques used

rearrange an equation to make the subject the inverse functionexponentiate to remove a logarithmswap variables to express the inversededuce domain and range of the inverse from the original function

(iv)

On the axes sketch the graphs of $y = \mathrm{f}(x)$ and $y = \mathrm{f}^{-1}(x)$ , stating the intercepts with the axes.

Fig. 3

DifficultyMedium

Worked solution

Approach

$y = \mathrm{f}(x) = 4\ln(3x - 2)$ is a logarithmic curve with vertical asymptote $x = \frac{2}{3}$ , crossing the $x$ -axis where $3x - 2 = 1$ , i.e. at $(1, 0)$ . Its inverse $y = \mathrm{f}^{-1}(x) = \frac{1}{3}\left(\mathrm{e}^{\frac{x}{4}} + 2\right)$ is the reflection in the line $y = x$ , with horizontal asymptote $y = \frac{2}{3}$ and $y$ -intercept $(0, 1)$ . The two curves meet on the line $y = x$ .

Working

For $y = \mathrm{f}(x)$ : the $x$ -intercept satisfies $4\ln(3x - 2) = 0$ , so $3x - 2 = 1$ and $x = 1$ , giving $(1, 0)$ . The curve rises slowly to the right of the asymptote $x = \frac{2}{3}$ , lying in the first quadrant (and dropping steeply in the fourth just right of the asymptote).

For $y = \mathrm{f}^{-1}(x)$ : the $y$ -intercept is $\mathrm{f}^{-1}(0) = \frac{1}{3}(1 + 2) = 1$ , giving $(0, 1)$ . The curve approaches the horizontal asymptote $y = \frac{2}{3}$ as $x \to -\infty$ and grows exponentially for large $x$ .

The curves intersect where $\mathrm{f}(x) = \mathrm{f}^{-1}(x)$ , which lies on $y = x$ ; solving $x = 4\ln(3x - 2)$ gives $x \approx 2.31$ , so they cross at approximately $(2.31, 2.31)$ in the first quadrant.

Answer

A logarithmic curve through $(1, 0)$ with vertical asymptote $x = \frac{2}{3}$ , and its reflection, an exponential curve through $(0, 1)$ with horizontal asymptote $y = \frac{2}{3}$ , the two curves intersecting once in the first quadrant on the line $y = x$ .

Final answer

Log curve through (1, 0) with asymptote x = 2/3; inverse exponential curve through (0, 1) with asymptote y = 2/3; the curves intersect once in the first quadrant on y = x

Detailed explanation

Walkthrough

The graph of a logarithm $4\ln(3x - 2)$ has a vertical asymptote where the argument vanishes, $x = \frac{2}{3}$ , and crosses the $x$ -axis where the logarithm is zero, i.e. where $3x - 2 = 1$ , giving $(1, 0)$ . The inverse function's graph is the mirror image in the line $y = x$ , so its asymptote is the reflection of $x = \frac{2}{3}$ , namely the horizontal line $y = \frac{2}{3}$ , and its intercept is the reflection of $(1, 0)$ , namely $(0, 1)$ . Because a function and its inverse swap $x$ and $y$ , any point where they intersect must lie on $y = x$ ; here they cross once, at approximately $(2.31, 2.31)$ .

Key Takeaways

The graph of $\mathrm{f}^{-1}$ is the reflection of the graph of $\mathrm{f}$ in the line $y = x$ .
Asymptotes and intercepts reflect too: a vertical asymptote $x = c$ becomes a horizontal asymptote $y = c$ .
A function and its inverse can only intersect on the line $y = x$ .

Common Mistakes

Drawing the logarithmic curve to the left of its asymptote, or in the second/third quadrants — the mark scheme awards B1 only for the correct shape in the first and fourth quadrants.
Omitting the intercept $(1, 0)$ (B1 dependent on the shape being right) or the intercept $(0, 1)$ .
Not showing the two curves intersecting in the first quadrant — the final B1 requires at least one correct intersection point.
Reflecting in the wrong line, or sketching the inverse with a vertical instead of horizontal asymptote.

Things to Be Careful About

The intercept marks are dependent on the curve shape being correct first, so get the shape right before labelling points.
The asymptotes $x = \frac{2}{3}$ and $y = \frac{2}{3}$ should be shown (dashed) to make the shapes clear.
State the intercepts explicitly: $(1, 0)$ for $\mathrm{f}$ and $(0, 1)$ for $\mathrm{f}^{-1}$ .

Techniques used

sketch a logarithmic curve with its vertical asymptotesketch the inverse exponential curve reflected in y = xmark intercepts with the axesidentify the intersection of a function and its inverse on y = x

(b)

Given that $\mathrm{g}(x) = (2x + 1)^{\frac{1}{2}} + 4$ , for $x > 0$ , solve the equation $\mathrm{gg}(x) = 9$ .

DifficultyMedium-Easy

Worked solution

Approach

Form the composite $\mathrm{gg}(x) = \mathrm{g}\left(\mathrm{g}(x)\right)$ by substituting $\mathrm{g}(x)$ into itself, set it equal to 9, and solve by peeling off the outer layers.

Working

\mathrm{gg}(x) = \left(2\left(2(2x+1)^{\frac{1}{2}} + 4\right)^{\frac{1}{2}} + 1\right)^{\frac{1}{2}} + 4

Setting this equal to 9:

\left(2\left(2(2x+1)^{\frac{1}{2}} + 4\right)^{\frac{1}{2}} + 1\right)^{\frac{1}{2}} + 4 = 9

\left(2\left(2(2x+1)^{\frac{1}{2}} + 4\right)^{\frac{1}{2}} + 1\right)^{\frac{1}{2}} = 5

Squaring:

2\left(2(2x+1)^{\frac{1}{2}} + 4\right)^{\frac{1}{2}} + 1 = 25 \quad \Rightarrow \quad \left(2(2x+1)^{\frac{1}{2}} + 4\right)^{\frac{1}{2}} = 12

Squaring again:

2(2x+1)^{\frac{1}{2}} + 4 = 144 \quad \Rightarrow \quad (2x+1)^{\frac{1}{2}} = 70

Squaring a third time:

2x + 1 = 4900 \quad \Rightarrow \quad x = \frac{4900 - 1}{2} = 2449.5

Alternatively, using the inverse route: $\mathrm{g}(x) = 9$ gives $(2x+1)^{\frac{1}{2}} = 4$ , so $x = 7.5$ ; then $\mathrm{g}(x) = 7.5$ gives $(2x+1)^{\frac{1}{2}} = 1.75$ , so $2x + 1 = 3.0625$ and $x = 1.03125$ .

Answer

x = 2449.5

Final answer

x = 2449.5

Detailed explanation

Walkthrough

The composite $\mathrm{gg}(x)$ means applying $\mathrm{g}$ twice: substitute the whole expression $(2x+1)^{\frac{1}{2}} + 4$ in place of $x$ inside $\mathrm{g}$ . Setting the result equal to 9, we undo the function layer by layer: subtract 4, square, subtract 1, divide by 2, square, subtract 4, divide by 2, square. Each squaring removes one square root. This yields $(2x+1)^{\frac{1}{2}} = 70$ , so $2x + 1 = 4900$ and $x = 2449.5$ . The mark scheme's alternative route works backwards through $\mathrm{g}(x) = 9$ then $\mathrm{g}(x) = 12$ ... but the direct composite route shown is the primary one.

Key Takeaways

$\mathrm{gg}(x)$ means $\mathrm{g}$ applied to $\mathrm{g}(x)$ — substitute the entire output expression back into the function.
Radical equations are solved by isolating the root and squaring, one layer at a time.

Common Mistakes

Treating $\mathrm{gg}(x)$ as $\left(\mathrm{g}(x)\right)^{2}$ instead of the composition.
Making an arithmetic slip when expanding the nested composite — the mark scheme's B1 is for the correct composite expression.
Squaring without isolating the root first, introducing spurious terms.
Forgetting the domain restriction $x > 0$ (satisfied here since $2449.5 > 0$ ).

Things to Be Careful About

The final B1 is dependent on both earlier B marks, so the composite must be formed correctly before solving.
Check the answer lies in the stated domain $x > 0$ — it does.

Techniques used

form the composite function gg(x)solve the resulting equation by isolating the square rootsquare to eliminate the root

Q5MediumTrigonometryCalculus

(a)

Show that $\frac{1 + \cot^2 \theta}{\cot^2 \theta} = \sec^2 \theta$ .

DifficultyMedium-Easy

Worked solution

Approach

Rewrite the left-hand side using $1 + \cot^2 \theta = \operatorname{cosec}^2 \theta$ , then convert everything into sines and cosines and simplify.

Working

\frac{1 + \cot^2 \theta}{\cot^2 \theta} = \frac{\operatorname{cosec}^2 \theta}{\cot^2 \theta}

Using $\operatorname{cosec}^2 \theta = \frac{1}{\sin^2 \theta}$ and $\cot^2 \theta = \frac{\cos^2 \theta}{\sin^2 \theta}$ :

= \frac{1}{\sin^2 \theta} \times \frac{\sin^2 \theta}{\cos^2 \theta} = \frac{1}{\cos^2 \theta} = \sec^2 \theta

This is exactly the required result.

Answer

\frac{1 + \cot^2 \theta}{\cot^2 \theta} = \sec^2 \theta

Final answer

(1 + cot^2 theta)/cot^2 theta = sec^2 theta, shown by writing the numerator as cosec^2 theta and simplifying to 1/cos^2 theta

Detailed explanation

Walkthrough

The key identity here is the Pythagorean identity $1 + \cot^2 \theta = \operatorname{cosec}^2 \theta$ , which is just $\sin^2 \theta + \cos^2 \theta = 1$ divided through by $\sin^2 \theta$ . Replacing the numerator gives $\frac{\operatorname{cosec}^2 \theta}{\cot^2 \theta}$ . Then both functions are converted to their definitions: $\operatorname{cosec} \theta = \frac{1}{\sin \theta}$ and $\cot \theta = \frac{\cos \theta}{\sin \theta}$ . Dividing by a fraction means multiplying by its reciprocal, so the $\sin^2 \theta$ factors cancel, leaving $\frac{1}{\cos^2 \theta}$ , which is by definition $\sec^2 \theta$ .

Key Takeaways

The three Pythagorean identities ( $1 + \tan^2 = \sec^2$ , $1 + \cot^2 = \operatorname{cosec}^2$ , $\sin^2 + \cos^2 = 1$ ) are interchangeable via division by $\sin^2$ or $\cos^2$ .
Converting everything to sine and cosine almost always simplifies a trigonometric fraction.

Common Mistakes

Omitting the argument $\theta$ throughout — the mark scheme explicitly says the mark is not awarded if $\theta$ is consistently omitted.
Misquoting the identity as $1 + \cot^2 \theta = \sec^2 \theta$ or similar.
Cancelling incorrectly when dividing by the compound fraction $\frac{\cos^2 \theta}{\sin^2 \theta}$ .
Not showing sufficient intermediate detail — the scheme requires every algebraic line between the given expression and the target.

Things to Be Careful About

This is an "Answer Given" (AG) part: you must derive forward from the printed left-hand side to exactly $\sec^2 \theta$ ; starting from the answer loses the mark. Write $\theta$ on every line, keep each step visible, and finish at $\sec^2 \theta$ rather than stopping at $\frac{1}{\cos^2 \theta}$ without stating the equality.

Techniques used

use the Pythagorean identity to rewrite the numeratorexpress cotangent in terms of sine and cosinesimplify the compound fraction to secant squared

(b)

Write down the derivative of $\tan \theta$ with respect to $\theta$ .

(c)

Using part (a) and part (b), find the exact value of $\int_{0}^{\frac{\pi}{3}} \left(\frac{1 + \cot^2 \theta}{\cot^2 \theta} - \sin \theta\right) \mathrm{d}\theta$ .

DifficultyMedium

Worked solution

Approach

Use part (a) to replace $\frac{1 + \cot^2 \theta}{\cot^2 \theta}$ with $\sec^2 \theta$ , integrate using part (b), then substitute the exact limits $0$ and $\frac{\pi}{3}$ .

Working

Using part (a), the integrand becomes:

\int_{0}^{\frac{\pi}{3}} \left(\sec^2 \theta - \sin \theta\right)\mathrm{d}\theta

Integrating term by term, using part (b) for the first term:

\left[\tan \theta + \cos \theta\right]_{0}^{\frac{\pi}{3}}

Evaluating at the limits:

= \left(\tan \frac{\pi}{3} + \cos \frac{\pi}{3}\right) - \left(\tan 0 + \cos 0\right)

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

Answer

\sqrt{3} - \frac{1}{2}

Final answer

√3 − 1/2

Detailed explanation

Walkthrough

The question says "using part (a) and part (b)", so the first step is to carry those results forward: part (a) showed that $\frac{1 + \cot^2 \theta}{\cot^2 \theta} = \sec^2 \theta$ , so the integrand is simply $\sec^2 \theta - \sin \theta$ . Part (b) gave the antiderivative of $\sec^2 \theta$ as $\tan \theta$ , and the standard integral of $\sin \theta$ is $-\cos \theta$ , so subtracting $\sin \theta$ contributes $+\cos \theta$ . The definite integral is therefore $[\tan \theta + \cos \theta]$ evaluated between $0$ and $\frac{\pi}{3}$ . At the upper limit, the special values give $\tan \frac{\pi}{3} = \sqrt{3}$ and $\cos \frac{\pi}{3} = \frac{1}{2}$ ; at the lower limit, $\tan 0 = 0$ and $\cos 0 = 1$ . Subtracting gives $\sqrt{3} + \frac{1}{2} - 1 = \sqrt{3} - \frac{1}{2}$ .

Key Takeaways

"Hence/using" questions expect earlier results to be reused, not re-derived.
$\int \sec^2 \theta\,\mathrm{d}\theta = \tan \theta$ and $\int \sin \theta\,\mathrm{d}\theta = -\cos \theta$ are core standard integrals.
Definite integrals of trig functions need the angle limits treated in radians, with special-angle values recalled exactly.

Common Mistakes

Integrating $\sin \theta$ as $+\cos \theta$ instead of $-\cos \theta$ , which flips the sign of the whole answer.
Substituting degree values such as $\tan 60°$ while the limits are written in radians — the working must stay consistent in radians.
Forgetting to evaluate at the lower limit $0$ , giving $\sqrt{3} + \frac{1}{2}$ instead of $\sqrt{3} - \frac{1}{2}$ .
Giving a decimal approximation instead of the exact surd form demanded by the non-calculator paper.
Re-deriving the identity of part (a) inside part (c) instead of quoting it.

Things to Be Careful About

The final B1 mark depends on all three previous marks being earned, and the answer must be exact: $\sqrt{3} - \frac{1}{2}$ (or an exact equivalent). Show the substituted expression before the evaluated value so the method is visible, and quote the special values $\tan \frac{\pi}{3} = \sqrt{3}$ , $\cos \frac{\pi}{3} = \frac{1}{2}$ explicitly.

Techniques used

rewrite the integrand using the results of parts (a) and (b)integrate secant squared and sine term by termevaluate the definite integral at exact radian limits

Q6MediumSeries

(a)

Find, in descending powers of $x$ , the first 3 terms in the expansion of $\left(x + \frac{2}{x^2}\right)^{10}$ . Simplify each term as far as possible.

DifficultyMedium-Easy

Worked solution

Approach

Use the binomial theorem on $\left(x + \dfrac{2}{x^2}\right)^{10}$ , taking terms with $r = 0, 1, 2$ and simplifying each power of $x$ .

Working

\left(x + \frac{2}{x^2}\right)^{10} = \sum_{r=0}^{10} \binom{10}{r} x^{10-r} \left(\frac{2}{x^2}\right)^{r}

First term ( $r = 0$ ):

\binom{10}{0} x^{10} = x^{10}

Second term ( $r = 1$ ):

\binom{10}{1} x^{9} \cdot \frac{2}{x^{2}} = 10 \times 2 \, x^{7} = 20x^{7}

Third term ( $r = 2$ ):

\binom{10}{2} x^{8} \cdot \frac{4}{x^{4}} = 45 \times 4 \, x^{4} = 180x^{4}

Answer

x^{10} + 20x^{7} + 180x^{4}

Final answer

x^10 + 20x^7 + 180x^4

Detailed explanation

Walkthrough

The binomial theorem gives each term of $(a+b)^n$ as $\binom{n}{r}a^{n-r}b^{r}$ . Here $a = x$ , $b = \dfrac{2}{x^2}$ and $n = 10$ . The first three terms come from $r = 0, 1, 2$ .

For $r = 0$ : the coefficient is $\binom{10}{0} = 1$ and we get $x^{10}$ .

For $r = 1$ : the coefficient is $\binom{10}{1} = 10$ , giving $10 \times x^9 \times \dfrac{2}{x^2} = 20x^{7}$ , since $x^9 \div x^2 = x^7$ by subtracting indices.

For $r = 2$ : the coefficient is $\binom{10}{2} = 45$ , and $\left(\dfrac{2}{x^2}\right)^2 = \dfrac{4}{x^4}$ , so the term is $45 \times x^8 \times \dfrac{4}{x^4} = 180x^{4}$ .

Key Takeaways

The general term of $(a+b)^n$ is $\binom{n}{r}a^{n-r}b^{r}$ ; the first three terms use $r = 0, 1, 2$ .
Powers of $x$ combine by subtracting indices: $x^{10-r} \times x^{-2r} = x^{10-3r}$ .
Binomial coefficients needed: $\binom{10}{0}=1$ , $\binom{10}{1}=10$ , $\binom{10}{2}=45$ .

Common Mistakes

Starting at $r = 1$ instead of $r = 0$ , losing the $x^{10}$ term (the mark scheme awards B1 per correct term).
Writing $\left(\dfrac{2}{x^2}\right)^2$ as $\dfrac{2}{x^4}$ instead of $\dfrac{4}{x^4}$ — the 2 must be squared too.
Adding indices instead of subtracting when dividing powers of $x$ .
Using $\binom{10}{2} = 90$ or other arithmetic slips in the coefficients.

Things to Be Careful About

Each term is marked separately (B1 each), so even if one term is wrong, correct terms still score.
Simplify fully: the scheme expects $20x^7$ and $180x^4$ , not unsimplified products like $10 \times 2x^7$ .
Keep the answer in descending powers of $x$ as asked.

Techniques used

apply the binomial theorem to a two-term expressioncompute binomial coefficientssimplify powers of x using index laws

(b)

Find the term independent of $x$ in the expansion of $\left(4x^2 + \frac{1}{2x^2}\right)^8$ .