Additional Mathematics 4037 / 21 — May/June 2024

DifficultyMedium-Easy

Worked solution

Approach

The graph of $y = |4x - 6|$ is V-shaped. The vertex occurs where $4x - 6 = 0$ , and the $y$ -intercept is found by putting $x = 0$ .

Working

Vertex: solve $4x - 6 = 0$ , giving $x = \frac{6}{4} = 1.5$ . So the vertex is at $(1.5,\ 0)$ — it lies on the $x$ -axis.

$y$ -intercept: at $x = 0$ ,

y = |4(0) - 6| = |-6| = 6

so the graph passes through $(0,\ 6)$ .

Answer

A V-shaped graph with its vertex on the $x$ -axis at $(1.5,\ 0)$ and left branch crossing the $y$ -axis at $(0,\ 6)$ ; both branches rise upwards away from the vertex.

Final answer

V-shaped graph with vertex (1.5, 0) on the x-axis and y-intercept (0, 6)

Detailed explanation

Walkthrough

The modulus function $y = |4x - 6|$ takes the linear expression $4x - 6$ and reflects any negative values to positive ones. This produces the characteristic V shape: for $x \geq 1.5$ the graph is the straight line $y = 4x - 6$ with gradient $+4$ , and for $x < 1.5$ it is the reflected line $y = -(4x - 6)$ with gradient $-4$ .

The turning point of the V sits exactly where the inside of the modulus changes sign, i.e. where $4x - 6 = 0$ , so $x = 1.5$ . Because the question's expression equals zero there, the vertex lies on the $x$ -axis at $(1.5, 0)$ — this is also the only $x$ -intercept.

For the $y$ -intercept we substitute $x = 0$ : $y = |-6| = 6$ , so the left branch crosses the $y$ -axis at $(0, 6)$ . Both branches extend upward into the first quadrant as drawn in the mark scheme figure.

Key Takeaways

A graph of $y = |ax + b|$ is always V-shaped with vertex where $ax + b = 0$ .
The vertex of this particular modulus lies on the $x$ -axis because the constant term makes the zero occur at a real value of $x$ with no vertical shift.
Intercepts are found by direct substitution: $x = 0$ gives the $y$ -intercept, and solving $y = 0$ gives the $x$ -intercept(s).

Common Mistakes

Drawing two separate lines instead of one continuous V that meets at the vertex.
Placing the vertex at $x = 6$ or $x = -1.5$ through sign slips when solving $4x - 6 = 0$ .
Forgetting to mark the intercepts — the mark scheme awards B2 only for the fully correct graph with intercepts marked; B1 is available for correct shape with the vertex on the $x$ -axis.
Reflecting the wrong branch: the reflection happens below $x = 1.5$ , not above.

Things to Be Careful About

The mark scheme requires the intercepts $(1.5, 0)$ and $(0, 6)$ to be clearly labelled for full marks (B2); without them you cap at B1.
The vertex must sit exactly on the $x$ -axis — a V floating above or below loses the accuracy mark.
Sketches should show correct shape and key coordinates; they need not be to scale, but the relative positions (vertex right of the origin, $y$ -intercept above the origin) must be right.

Techniques used

recognise the V-shape of a modulus graphfind the vertex where the expression inside the modulus is zerofind the y-intercept by substituting x = 0

(b)

Solve the equation $|4x - 6| = |2x|$ .

3M Similar

DifficultyMedium-Easy

Worked solution

Approach

An equation of the form $|A| = |B|$ splits into two cases: $A = B$ and $A = -B$ . Solve each linear equation.

Working

Case 1:

4x - 6 = 2x \implies 2x = 6 \implies x = 3

Case 2:

4x - 6 = -2x \implies 6x = 6 \implies x = 1

Check: at $x = 3$ , $|12 - 6| = 6 = |6|$ ✓; at $x = 1$ , $|4 - 6| = 2 = |2|$ ✓.

Answer

x = 1 \quad \text{or} \quad x = 3

Final answer

x = 1 or x = 3

Detailed explanation

Walkthrough

The equation $|4x - 6| = |2x|$ says two quantities have equal size. This happens in exactly two ways: the insides are equal ( $4x - 6 = 2x$ ), or the insides are opposites ( $4x - 6 = -2x$ ). Solving each case gives one linear equation apiece.

Case 1: $4x - 6 = 2x$ gives $2x = 6$ , so $x = 3$ .

Case 2: $4x - 6 = -2x$ gives $6x = 6$ , so $x = 1$ .

Both check out in the original equation, so both are valid solutions. (The mark scheme also accepts squaring both sides to get $12x^2 - 48x + 36 = 0$ , which factorises to give the same roots.)

Key Takeaways

$|A| = |B|$ means $A = B$ or $A = -B$ — always two cases.
Each case here reduces to a simple linear equation.
Substituting solutions back confirms neither case produced an extraneous root.

Common Mistakes

Only solving one case and giving a single answer — the equation has two solutions and A2 requires both (A1 for either one alone).
Sign slips when moving terms across, e.g. writing $4x + 2x = 6$ incorrectly in case 2.
Squaring route errors: expanding $(4x-6)^2 = 4x^2$ wrongly or losing a factor when simplifying to $12x^2 - 48x + 36 = 0$ .

Things to Be Careful About

Both solutions must be given; the scheme awards A2 for both correct and only A1 if just one is right.
"oe" applies: equivalent correct working (such as the squaring method shown in the alternative) earns the same marks.
Answers should be exact integers here — no rounding issues arise.

Techniques used

split a modulus equation into positive and negative casessolve each resulting linear equationcheck both solutions satisfy the original equation

Q2Medium-EasyQuadratic functionsFunctions

(a)

Write $3 + 4x - 2x^2$ in the form $a + b(x + c)^2$ , where $a$ , $b$ and $c$ are integers.

3M Similar

DifficultyMedium-Easy

Worked solution

Approach

Factor $-2$ out of the terms in $x$ , complete the square inside the bracket, then multiply back through and simplify.

Working

3 + 4x - 2x^2 = -2\left(x^2 - 2x\right) + 3

Complete the square on $x^2 - 2x$ :

x^2 - 2x = (x - 1)^2 - 1

Substitute back and expand:

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

This is of the form $a + b(x + c)^2$ with $a = 5$ , $b = -2$ and $c = -1$ .

Answer

3 + 4x - 2x^2 = 5 - 2(x - 1)^2

Final answer

5 - 2(x - 1)^2

Detailed explanation

Walkthrough

The expression has a negative coefficient of $x^2$ , so we first pull that factor out of the two terms containing $x$ : $4x - 2x^2 = -2(x^2 - 2x)$ , leaving the constant $3$ outside.

Inside the bracket we complete the square: half of $-2$ is $-1$ , so $x^2 - 2x = (x - 1)^2 - 1$ . The $-1$ corrects for the extra $+1$ created by squaring $(x-1)$ .

Multiplying the $-2$ back through gives $-2(x-1)^2 + 2$ , and adding the original constant $3$ gives $5 - 2(x-1)^2$ . This matches the required form $a + b(x+c)^2$ with integers $a = 5$ , $b = -2$ , $c = -1$ — note that $b(x+c)^2$ with $c = -1$ is exactly $-2(x-1)^2$ .

Key Takeaways

When completing the square with a non-unit leading coefficient, always factor it out of the $x$ -terms first, then multiply it back through afterwards.
The form $a + b(x+c)^2$ allows negative $b$ and $c$ : here $b = -2$ and $c = -1$ .
Checking by expanding the final expression confirms no sign slips.

Common Mistakes

Forgetting to carry the factor $-2$ into the correction term, giving $-2(x-1)^2 + 3$ instead of $5 - 2(x-1)^2$ .
Writing $+2(x+1)^2$ instead of $-2(x-1)^2$ — both signs wrong at once.
Giving $c = 1$ because the printed bracket reads $(x-1)$ ; in the form $b(x+c)^2$ we need $c = -1$ .
The mark scheme awards B2 for just $-2(x-1)^2$ and B1 for just $(x-1)^2$ or for stating $b = -2$ , $c = -1$ , so partial answers still earn credit but the full expression is needed for all 3 marks.

Things to Be Careful About

The answer must be the complete expression $5 - 2(x-1)^2$ ; the mark scheme marks the final expression only, so an incorrect constant loses the accuracy mark even if the squared term is right.
All of $a$ , $b$ , $c$ must be integers as demanded by the question — they are here ( $5$ , $-2$ , $-1$ ).
Verify by expanding: $5 - 2(x^2 - 2x + 1) = 5 - 2x^2 + 4x - 2 = 3 + 4x - 2x^2$ .

Techniques used

factor out the coefficient of x squaredcomplete the square on the bracketed quadraticexpand and collect to verify the constant

(b)

Hence write down the range of the function $\mathrm{f}(x) = 3 + 4x - 2x^2$ , where $x \in \mathbb{R}$ .

1M Similar

DifficultyEasy

Worked solution

Approach

Use the completed-square form from part (a): since $-2(x-1)^2 \leq 0$ for all real $x$ , the function never exceeds its constant term.

Working

From part (a),

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

Since $(x - 1)^2 \geq 0$ for all $x \in \mathbb{R}$ , we have $-2(x-1)^2 \leq 0$ , so

\mathrm{f}(x) \leq 5

with equality when $x = 1$ .

Answer

\mathrm{f}(x) \leq 5

Final answer

f(x) <= 5

Detailed explanation

Walkthrough

The point of part (a) was to write the quadratic in completed-square form, which makes the range visible immediately. In $\mathrm{f}(x) = 5 - 2(x-1)^2$ , the squared term $(x-1)^2$ is never negative, so multiplying by $-2$ makes it never positive: the largest value of $\mathrm{f}$ occurs when $(x-1)^2 = 0$ , i.e. at $x = 1$ , giving $\mathrm{f}(1) = 5$ . Every other input subtracts something positive, so the outputs are all at most $5$ . Hence the range is $\mathrm{f}(x) \leq 5$ .

Key Takeaways

Completed-square form $a + b(x+c)^2$ reveals the maximum (if $b < 0$ ) or minimum (if $b > 0$ ) directly as the value $a$ .
The range of a quadratic over all reals is a one-sided inequality bounded by the turning-point value.

Common Mistakes

Writing $\mathrm{f}(x) \geq 5$ — the parabola opens downwards because $b = -2$ , so $5$ is a maximum, not a minimum.
Omitting the equality and writing $\mathrm{f}(x) < 5$ ; the value $5$ is attained at $x = 1$ , so $\leq$ is required.
The mark scheme applies strict follow-through: the bound must be their value of $a$ from part (a), so an error in part (a) carries forward here.

Things to Be Careful About

State the range as $\mathrm{f}(x) \leq 5$ (or equivalently $y \leq 5$ ); the mark scheme's B1 is strict follow-through on the part (a) constant.
Since this says "Hence", the completed-square result must actually be used rather than re-deriving the vertex another way.

Techniques used

read off the maximum value from completed-square formstate the range using the direction of the squared term

Q33MMedium-EasyQuadratic functions

Use algebra to show that the equation $5x(x - 3) = 5x - 26$ has no real solutions.

DifficultyMedium-Easy

Worked solution

Approach

Rearrange the equation into the standard form $ax^2 + bx + c = 0$ , then show that the discriminant $b^2 - 4ac$ is negative, which means the quadratic has no real roots.

Working

Expand and collect all terms on one side:

5x(x - 3) = 5x - 26 \implies 5x^2 - 15x - 5x + 26 = 0

5x^2 - 20x + 26 = 0

Here $a = 5$ , $b = -20$ , $c = 26$ . Compute the discriminant:

b^2 - 4ac = (-20)^2 - 4(5)(26) = 400 - 520 = -120

Since $-120 < 0$ , the discriminant is negative, so the quadratic equation has no real solutions.

Answer

The discriminant of $5x^2 - 20x + 26 = 0$ is $-120 < 0$ , so the equation has no real solutions.

Final answer

The discriminant is -120 < 0, so the equation has no real solutions.

Detailed explanation

Walkthrough

The equation $5x(x - 3) = 5x - 26$ is not in standard quadratic form, so the first step is to expand the left-hand side and bring every term to one side. Expanding gives $5x^2 - 15x$ , and subtracting $5x$ and adding $26$ gives $5x^2 - 20x + 26 = 0$ .

To decide whether a quadratic equation has real solutions, we examine the discriminant $b^2 - 4ac$ . With $a = 5$ , $b = -20$ and $c = 26$ , the discriminant is $(-20)^2 - 4 \times 5 \times 26 = 400 - 520 = -120$ . A negative discriminant means the square root in the quadratic formula would be the square root of a negative number, which is not real — so the equation has no real solutions. That completes the proof.

(An accepted alternative is to complete the square: $5x^2 - 20x + 26 = 5(x - 2)^2 + 6$ , whose minimum point $(2, 6)$ lies above the $x$ -axis, so the curve never meets the $x$ -axis and there are no real roots.)

Key Takeaways

To use the discriminant, the equation must first be written as $ax^2 + bx + c = 0$ with all terms collected on one side.
$b^2 - 4ac > 0$ gives two distinct real roots, $= 0$ gives one repeated root, and $< 0$ gives no real roots.
Completing the square is an equivalent way to prove no real roots: a minimum above the $x$ -axis means no $x$ -intercepts.

Common Mistakes

Sign slips when collecting terms: the equation must become $5x^2 - 20x + 26 = 0$ ; forgetting to move the $-26$ across (leaving $+26$ as $-26$ ) changes the discriminant.
Sign errors in the discriminant itself: $(-20)^2 = 400$ , not $-400$ .
Concluding without stating the meaning of the negative discriminant — the mark scheme requires the correct conclusion, not just the value $-120$ .
The mark scheme condones one slip in expansion or collection, but the follow-through only works if the discriminant for the candidate's equation is still negative.

Things to Be Careful About

This is a "show that" question, so every algebraic line must be visible: the rearranged equation, the substituted discriminant, and the conclusion.
The final answer is a statement, not a number — the value $-120$ alone does not answer the question; you must state that a negative discriminant means no real solutions.
Keep the exact value $-120$ ; no rounding or decimal is involved.

Techniques used

rearrange the equation into standard quadratic formcompute the discriminantargue that a negative discriminant means no real roots

Q4MediumCoordinate geometry of the circleSimultaneous equations[Legacy] Indices and surds

DO NOT USE A CALCULATOR IN THIS QUESTION.

(a)

Find the exact distance between the two points where the curve $9(x - 1)^2 + 4(y - 3)^2 = 36$ cuts the $y$ -axis.

DifficultyMedium-Easy

Worked solution

Approach

The curve cuts the $y$ -axis where $x = 0$ . Substitute $x = 0$ into the equation, solve for $y$ exactly, and take the difference of the two $y$ -values as the distance.

Working

Substituting $x = 0$ :

9(0-1)^2 + 4(y-3)^2 = 36

9 + 4(y-3)^2 = 36 \quad\Rightarrow\quad 4(y-3)^2 = 27

(y-3)^2 = \frac{27}{4} \quad\Rightarrow\quad y = 3 \pm \sqrt{\frac{27}{4}}

The distance between the two points is the difference of these $y$ -values:

\left(3 + \sqrt{\frac{27}{4}}\right) - \left(3 - \sqrt{\frac{27}{4}}\right) = 2\sqrt{\frac{27}{4}}

= 2 \times \frac{\sqrt{27}}{2} = \sqrt{27} = 3\sqrt{3}

Answer

3\sqrt{3}

Final answer

3√3

Detailed explanation

Walkthrough

Setting $x = 0$ finds where the curve meets the $y$ -axis, because every point on that axis has first coordinate zero. The substitution leaves a quadratic in $y$ : after moving the $9$ across, $(y-3)^2 = \frac{27}{4}$ , so taking square roots gives $y = 3 \pm \sqrt{\frac{27}{4}}$ . The two intersection points sit vertically above one another, so the distance between them is simply the larger $y$ minus the smaller $y$ — the two $3$ s cancel, leaving twice the square-root term. Finally, since $\sqrt{\frac{27}{4}} = \frac{\sqrt{27}}{2} = \frac{3\sqrt{3}}{2}$ , doubling gives $3\sqrt{3}$ .

Key Takeaways

To find where a curve crosses an axis, substitute the other coordinate as zero.
The distance between two points with the same $x$ -coordinate is just the difference of their $y$ -coordinates.
Surds should be simplified fully: $\sqrt{27} = 3\sqrt{3}$ .

Common Mistakes

Forgetting the $9(x-1)^2$ term still contributes when $x = 0$ ; writing $4(y-3)^2 = 36$ loses the M1.
Giving decimal answers instead of the exact surd form demanded by "DO NOT USE A CALCULATOR".
Leaving the answer as $\sqrt{27}$ without simplifying, or halving/doubling incorrectly when subtracting the roots.

Things to Be Careful About

The mark scheme requires the exact form ( $3\sqrt{3}$ or equivalent), and "nfww" means no marks if reached from wrong working.
Show each algebraic step: forming the quadratic, the exact roots, and the subtraction are separately credited.

Techniques used

substitute x = 0 to find y-axis intersectionssolve the resulting quadratic in ysubtract the two roots to find the distancesimplify the surd exactly

(b)

Find the coordinates of the points where the curve with equation $2x^2 + 83xy = x^3y - 20x$ intersects the curve with equation $y = \frac{1}{x}$ . Give each of your answers in the form $a + b\sqrt{c}$ , where $a$ and $b$ are rational and $c$ is the smallest integer possible.

6M Similar

DifficultyMedium

Worked solution

Approach

At any intersection point the same pair $(x, y)$ satisfies both equations, so substitute $y = \frac{1}{x}$ into the cubic equation. Multiplying through by $x$ removes the fractions and produces a quadratic in $x$ , whose exact roots give the $x$ -coordinates; then find $y = \frac{1}{x}$ and rationalise.

Working

Substituting $y = \frac{1}{x}$ into $2x^2 + 83xy = x^3y - 20x$ :

2x^2 + 83x \cdot \frac{1}{x} = x^3 \cdot \frac{1}{x} - 20x

2x^2 + 83 = x^2 - 20x

Rearranging:

x^2 + 20x + 83 = 0

By the quadratic formula:

x = \frac{-20 \pm \sqrt{20^2 - 4(1)(83)}}{2} = \frac{-20 \pm \sqrt{400 - 332}}{2} = \frac{-20 \pm \sqrt{68}}{2}

x = -10 \pm \sqrt{17}

Now $y = \frac{1}{x}$ . Rationalising with $x = -10 + \sqrt{17}$ (and similarly for the other sign):

y = \frac{1}{-10 \pm \sqrt{17}} \times \frac{-10 \mp \sqrt{17}}{-10 \mp \sqrt{17}} = \frac{-10 \mp \sqrt{17}}{100 - 17} = \frac{-10 \mp \sqrt{17}}{83}

So the two points are:

\left(-10 + \sqrt{17},\; -\frac{10}{83} - \frac{\sqrt{17}}{83}\right) \quad\text{and}\quad \left(-10 - \sqrt{17},\; -\frac{10}{83} + \frac{\sqrt{17}}{83}\right)

Answer

x = -10 \pm \sqrt{17}, \qquad y = -\frac{10}{83} \mp \frac{\sqrt{17}}{83}

Final answer

Points: (-10 + √17, -(10+√17)/83) and (-10 - √17, (√17 - 10)/83)

Detailed explanation

Walkthrough

Two curves intersect where their equations hold simultaneously. Since the second curve fixes $y = \frac{1}{x}$ , replacing $y$ everywhere in the first equation eliminates $y$ at once. The terms $83xy$ and $x^3y$ become plain numbers because $x \cdot \frac{1}{x} = 1$ , giving $2x^2 + 83 = x^2 - 20x$ , which rearranges to the three-term quadratic $x^2 + 20x + 83 = 0$ . The discriminant $400 - 332 = 68$ is not a perfect square, so the roots stay as surds: $x = -10 \pm \sqrt{17}$ (since $\sqrt{68} = 2\sqrt{17}$ ). Each $x$ -value then determines its partner via $y = \frac{1}{x}$ ; multiplying top and bottom by the conjugate $-10 \mp \sqrt{17}$ makes the denominator $100 - 17 = 83$ , giving $y = \frac{-10 \mp \sqrt{17}}{83}$ . Note the sign pairing: the point with $x = -10 + \sqrt{17}$ has $y = -\frac{10}{83} - \frac{\sqrt{17}}{83}$ , and vice versa.

Key Takeaways

Substitution is the standard way to solve simultaneous equations when one equation already gives one variable explicitly.
A quadratic formula result like $\frac{-20 \pm \sqrt{68}}{2}$ must be simplified by cancelling the common factor before extracting squares from the surd.
Rationalising with the conjugate converts $\frac{1}{a \pm \sqrt{b}}$ into the required $p + q\sqrt{c}$ form.

Common Mistakes

Cancelling $x$ from terms like $83xy$ incorrectly, or dividing the whole equation by $x$ and losing solutions — here multiplying out is safe because $x = 0$ is not on the curve $y = \frac{1}{x}$ .
Sign slips in the discriminant ( $20^2 - 4 \times 83$ must be $68$ , not negative).
Leaving $x = \frac{-20 \pm \sqrt{68}}{2}$ unsimplified instead of $-10 \pm \sqrt{17}$ .
Mismatching the signs between $x$ and $y$ : the $\pm$ in $y$ runs opposite to the $\pm$ in $x$ .
Rationalising wrongly, e.g. using denominator $100 + 17$ instead of $100 - 17$ .

Things to Be Careful About

The final accuracy mark depends on all earlier marks, so every step must be correct — check the conjugate multiplication carefully.
The question demands the form $a + b\sqrt{c}$ with smallest possible $c$ : here $c = 17$ cannot be reduced further.
Keep everything exact throughout; decimals score nothing under the no-calculator instruction.

Techniques used

eliminate y by substituting y = 1/xform and solve a three-term quadraticapply the quadratic formularationalise the denominator of a surd fraction

Q5Medium-EasyPermutations and combinations

There are 3 women, 2 men and 4 children in a choir.

(a)

The choir stands in a single straight line.

4M

(i)

Find the number of possible arrangements if the first person and last person are both women.

DifficultyMedium-Easy

Worked solution

Approach

There are $3 + 2 + 4 = 9$ people. The first and last positions must both be women, so choose an ordered pair of women for the ends, then arrange the remaining $7$ people in the middle.

Working

Choose the woman for the first position and then the woman for the last position:

{}^{3}\mathrm{P}_{1} \times {}^{2}\mathrm{P}_{1} = 3 \times 2

Arrange the remaining $7$ people in the middle positions:

7! = 5040

Multiply:

3 \times 7! \times 2 = 6 \times 5040 = 30\,240

Answer

30\,240

Final answer

30240

Detailed explanation

Walkthrough

The choir has 9 distinct people in total. The condition fixes only the two end positions: both must be women. For the first position there are 3 choices of woman; once she is placed, 2 women remain for the last position. The remaining 7 people (1 woman, 2 men, 4 children) can stand in any order in the 7 middle positions, giving $7!$ arrangements. Since these choices are independent, multiply them together: $3 \times 2 \times 5040 = 30\,240$ .

Key Takeaways

In line-up problems, deal with restricted positions first, then permute everyone else.
Ordered placement at specific spots uses permutations ( ${}^{n}\mathrm{P}_{r}$ ), not combinations.
Independent stages of a count are multiplied.

Common Mistakes

Using ${}^{3}\mathrm{C}_{2}$ instead of ordered placements — the two end slots are distinguishable (first vs last), so order matters.
Permuting all 9 people instead of the remaining 7 after fixing the ends.
Forgetting that after placing one woman at one end, only 2 remain for the other end.

Things to Be Careful About

The answer is exact, so no rounding issues; but the method mark requires seeing $3 \times 7! \times 2$ or equivalent, so show the structure of the count rather than just the final number.

Techniques used

fix the first and last positions with a choice of womenarrange the remaining people in the middle positionsmultiply independent choices

(ii)

Find the number of possible arrangements if all the children stand next to each other.

(b)

Four of the choir are selected to sing in a group.

4M

(i)

Find the number of different selections if no man is chosen.

(ii)

Find the number of different selections if at least 2 women are chosen.

Q65MMediumCalculus

Variables $x$ and $y$ are such that $y = \cos x \sin^2 x$ . Use differentiation to find the approximate change in $y$ as $x$ increases from 3 to $3 + h$ , where $h$ is small.

DifficultyMedium

Worked solution

Approach

Differentiate $y = \cos x \sin^2 x$ using the product rule together with the chain rule for $\sin^2 x$ , then use the small-increment approximation $\delta y \approx \frac{\mathrm{d}y}{\mathrm{d}x}\,h$ evaluated at $x = 3$ .

Working

First differentiate $\sin^2 x$ by the chain rule:

\frac{\mathrm{d}}{\mathrm{d}x}\left(\sin^2 x\right) = 2\sin x \cos x

Now apply the product rule to $y = \cos x \cdot \sin^2 x$ :

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

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

For small $h$ , the approximate change in $y$ is:

\delta y \approx \frac{\mathrm{d}y}{\mathrm{d}x}\,h

Evaluating the derivative at $x = 3$ :

\frac{\mathrm{d}y}{\mathrm{d}x}\bigg|_{x=3} = 2\sin 3 \cos^2 3 - \sin^3 3 = 0.273808\ldots

Therefore:

\delta y \approx 0.274h

Answer

\delta y \approx 0.274h

Final answer

0.274h

Detailed explanation

Walkthrough

The question asks how much $y$ changes when $x$ increases by a tiny amount $h$ . The tool for this is the derivative as a rate of change: if $h$ is small, then $\delta y \approx \frac{\mathrm{d}y}{\mathrm{d}x}\,h$ . So everything hinges on finding $\frac{\mathrm{d}y}{\mathrm{d}x}$ correctly.

The function is a product of two factors of $x$ : $\cos x$ and $\sin^2 x$ . Before we can use the product rule we need the derivative of the second factor. Since $\sin^2 x$ is a function of a function ( $u = \sin x$ , squared), the chain rule gives $\frac{\mathrm{d}}{\mathrm{d}x}(\sin^2 x) = 2\sin x \cos x$ — differentiate the outer square to get $2\sin x$ , then multiply by the derivative of the inner function, $\cos x$ .

Applying the product rule, 'first times derivative of second plus second times derivative of first':

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

which simplifies to $2\sin x \cos^2 x - \sin^3 x$ .

Finally substitute $x = 3$ (radians, since no degree symbol appears): $2\sin 3\cos^2 3 - \sin^3 3 = 0.273808\ldots$ , so the approximate change in $y$ is $0.274h$ .

Key Takeaways

For a small change $h$ in $x$ , the change in $y$ is estimated by $\delta y \approx \frac{\mathrm{d}y}{\mathrm{d}x}\,h$ evaluated at the given point.
Products of trig functions need the product rule; powers of trig functions need the chain rule.
Trigonometric arguments without a degree symbol are in radians.

Common Mistakes

Differentiating $\sin^2 x$ incorrectly, e.g. writing $2\cos x$ or forgetting the inner $\cos x$ from the chain rule.
Sign slip on the derivative of $\cos x$ : it is $-\sin x$ , so the second product-rule term is negative.
Forgetting to multiply by $h$ , or leaving the answer as just the value of the derivative at $x=3$ .
Working in degrees instead of radians when evaluating at $x = 3$ .
Rounding too early: the mark scheme requires the coefficient correct or rounded/truncated to 4 or more significant figures before the final answer, and the A1 depends on the derivative being correct.

Things to Be Careful About

The final accuracy mark is dependent ('dep') on a fully correct derivative being seen — an error in differentiation loses the last mark even if the approximation step is right.
Give the coefficient to at least 4 significant figures in working ( $0.273808\ldots$ ) and state the answer as $0.274h$ ; later mis-simplification is ignored (isw).
Keep the answer in terms of $h$ — the question asks for the change as $x$ increases from $3$ to $3+h$ , not a numerical value.

Techniques used

differentiate sin^2 x by the chain ruleapply the product rule to cos x times sin^2 xuse the small-increment approximation delta y = (dy/dx) hevaluate the derivative numerically at the given point

Q74MMediumCalculusQuadratic functions

It is given that $y = mx^2 + \frac{x}{2} + n$ , where $m$ and $n$ are non-zero constants. It is also given that

3\left(\frac{\mathrm{d}^2 y}{\mathrm{d}x^2}\right) = \left(\frac{\mathrm{d}y}{\mathrm{d}x}\right)^2 - y

for all values of $x$ . Find the values of $m$ and $n$ .

DifficultyMedium

Worked solution

Approach

Differentiate $y = mx^2 + \frac{x}{2} + n$ twice, substitute both derivatives and $y$ itself into the given identity, then compare coefficients of $x^2$ , $x$ and the constant term.

Working

\frac{\mathrm{d}y}{\mathrm{d}x} = 2mx + \frac{1}{2}

\frac{\mathrm{d}^2 y}{\mathrm{d}x^2} = 2m

Substituting into $3\left(\frac{\mathrm{d}^2 y}{\mathrm{d}x^2}\right) = \left(\frac{\mathrm{d}y}{\mathrm{d}x}\right)^2 - y$ :

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

6m = 4m^2x^2 + 2mx + \frac{1}{4} - mx^2 - \frac{x}{2} - n

This holds for all values of $x$ , so equate coefficients:

Coefficient of $x^2$ : $\;0 = 4m^2 - m = m(4m - 1)$ , so $m = \frac{1}{4}$ (since $m \neq 0$ ).

Coefficient of $x$ : $\;0 = 2m - \frac{1}{2} \Rightarrow m = \frac{1}{4}$ — consistent.

Constant term:

6m = \frac{1}{4} - n \Rightarrow \frac{3}{2} = \frac{1}{4} - n \Rightarrow n = \frac{1}{4} - \frac{3}{2} = -\frac{5}{4}

Answer

m = \frac{1}{4}, \quad n = -\frac{5}{4}

Final answer

m = 1/4, n = -5/4

Detailed explanation

Walkthrough

The equation given must be true for every value of $x$ , not just one value. That is the key idea: after substituting the derivatives, we get an identity in $x$ , which means the coefficients of each power of $x$ on both sides must match.

First differentiate: the derivative of $mx^2$ is $2mx$ , the derivative of $\frac{x}{2}$ is $\frac{1}{2}$ , and the derivative of the constant $n$ is zero. Differentiating again gives $2m$ .

Substituting these into $3\left(\frac{\mathrm{d}^2 y}{\mathrm{d}x^2}\right) = \left(\frac{\mathrm{d}y}{\mathrm{d}x}\right)^2 - y$ gives a quadratic expression in $x$ that must equal the constant $6m$ for all $x$ . Expanding $\left(2mx + \frac{1}{2}\right)^2$ gives $4m^2x^2 + 2mx + \frac{1}{4}$ .

Since the left side has no $x^2$ or $x$ terms, their coefficients on the right must vanish: $4m^2 - m = 0$ gives $m = \frac{1}{4}$ (rejecting $m = 0$ because the question says $m$ is non-zero), and $2m - \frac{1}{2} = 0$ confirms it. Finally, matching constant terms, $6m = \frac{1}{4} - n$ , so with $m = \frac{1}{4}$ we get $n = -\frac{5}{4}$ .

Key Takeaways

An equation that holds "for all values of $x$ " is an identity: equate coefficients of like powers of $x$ .
Second derivatives of polynomials are found by repeated term-by-term differentiation.
The condition that $m$ is non-zero lets you discard the root $m = 0$ from $m(4m-1) = 0$ .

Common Mistakes

Forgetting to square the whole of $\left(2mx + \frac{1}{2}\right)$ — writing only $(2mx)^2$ loses the cross term $2mx$ and the constant $\frac{1}{4}$ .
Keeping the root $m = 0$ , which the question explicitly excludes ( $m$ non-zero); the mark scheme requires no other values.
Only using one coefficient comparison and guessing $n$ instead of solving the constant-term equation.
Sign slips when moving terms across when solving $6m = \frac{1}{4} - n$ .

Things to Be Careful About

The mark scheme awards B1 for each correct derivative, M1 for the correct substitution, and then 2 marks for $m = \frac{1}{4}$ and $n = -\frac{5}{4}$ with no other values — so check consistency between the $x^2$ and $x$ coefficient equations rather than reporting extra roots.
Answers must be exact fractions here; decimal forms are not needed since the arithmetic is simple by hand.
The identity must hold for all $x$ — this is what justifies equating coefficients, so state that reasoning.

Techniques used

differentiate a polynomial twicesubstitute derivatives into a differential equationequate coefficients of powers of xsolve for the constants

Q8MediumSeries

(a)

In an arithmetic progression, the sum of the first 30 terms is $-1065$ .
The sum of the next 20 terms is $-2210$ .
Find the first term and the common difference.

5M Similar

DifficultyMedium

Worked solution

Approach

Use the formula $S_n = \frac{1}{2}n\{2a + (n-1)d\}$ twice. The first condition gives $S_{30} = -1065$ directly. The sum of the next 20 terms is $S_{50} - S_{30} = -2210$ , so $S_{50} = -3275$ . Solve the two linear equations for $a$ and $d$ .

Working

S_{30} = \frac{30}{2}\{2a + 29d\} = -1065 \quad\Rightarrow\quad 15(2a + 29d) = -1065

30a + 435d = -1065

The sum of the next 20 terms:

S_{50} - S_{30} = -2210 \quad\Rightarrow\quad S_{50} = -2210 + (-1065) = -3275

S_{50} = \frac{50}{2}\{2a + 49d\} = -3275 \quad\Rightarrow\quad 25(2a + 49d) = -3275

50a + 1225d = -3275

Multiply the first equation by $\frac{5}{3}$ to match the coefficient of $a$ : equivalently, subtract. From $150a + 2175d = -5325$ and $150a + 3675d = -9825$ :

1500d = -4500 \quad\Rightarrow\quad d = -3

Substitute into $30a + 435(-3) = -1065$ :

30a - 1305 = -1065 \quad\Rightarrow\quad 30a = 240 \quad\Rightarrow\quad a = 8

Answer

a = 8, \quad d = -3

Final answer

First term a = 8, common difference d = -3

Detailed explanation

Walkthrough

The sum of an arithmetic progression with first term $a$ and common difference $d$ is $S_n = \frac{1}{2}n\{2a + (n-1)d\}$ . The first sentence of the question translates straight into this formula with $n = 30$ .

The second condition needs care: "the sum of the next 20 terms" means terms 31 to 50, not the first 20 terms. The cleanest way to express that is as a difference of two totals: $S_{50} - S_{30}$ . Since $S_{30} = -1065$ is already known, we get $S_{50} = -2210 - 1065 = -3275$ , which feeds into the same formula with $n = 50$ .

We now have two linear equations in $a$ and $d$ . Eliminating $a$ (multiply the first equation through so both have the same coefficient of $a$ , then subtract) leaves a single equation in $d$ , giving $d = -3$ . Substituting back recovers $a = 8$ .

Key Takeaways

The AP sum formula $S_n = \frac{1}{2}n\{2a + (n-1)d\}$ converts any "sum of the first $n$ terms" statement into a linear equation in $a$ and $d$ .
A phrase like "the sum of the next $k$ terms" must be translated as $S_{n+k} - S_n$ , not as $S_k$ .
Two conditions on an AP give two simultaneous linear equations — solve by elimination.

Common Mistakes

Treating "the next 20 terms" as $S_{20}$ instead of $S_{50} - S_{30}$ — this produces wrong equations entirely.
Arithmetic slips when expanding $\frac{50}{2}\{2a + 49d\}$ ; note it is $49d$ , not $50d$ .
Sign errors when moving $-1065$ across the equation to find $S_{50}$ .
The mark scheme awards A1 for each of $a$ and $d$ , so both values must be correct; a single slip costs one mark even if the method was sound.

Things to Be Careful About

The M1 for solving depends on having formed at least one equation using a sum formula, so always show the substituted formula before simplifying.
Check your answer: with $a = 8$ , $d = -3$ , $S_{30} = 15(16 - 87) = 15(-71) = -1065$ ✓ and $S_{50} = 25(16 - 147) = 25(-131) = -3275$ ✓.
Keep exact integer working throughout; no rounding issues arise here.

Techniques used

apply the arithmetic series sum formulainterpret the sum of the next 20 terms as S50 minus S30solve simultaneous linear equations

(b)

A geometric progression is such that the first term is 4 and the sum of the first three terms is 7.
Find the two possible values of the common ratio and find the sum to infinity for the convergent progression.

5M Similar

DifficultyMedium

Worked solution

Approach

Write the sum of the first three terms $4 + 4r + 4r^2 = 7$ , rearrange into a quadratic, factorise to get the two possible ratios, then use $|r| < 1$ to identify the convergent one and apply $S_\infty = \frac{a}{1-r}$ .

Working

4 + 4r + 4r^2 = 7

4r^2 + 4r - 3 = 0

Factorise:

(2r - 1)(2r + 3) = 0

r = \frac{1}{2} \quad \text{or} \quad r = -\frac{3}{2}

A geometric progression converges only if $|r| < 1$ , so only $r = \frac{1}{2}$ qualifies. Then

S_\infty = \frac{a}{1 - r} = \frac{4}{1 - \tfrac{1}{2}} = \frac{4}{\tfrac{1}{2}} = 8

Answer

The two possible common ratios are $r = 0.5$ and $r = -1.5$ ; the sum to infinity of the convergent progression is $8$ .

Final answer

r = 0.5 or r = -1.5; sum to infinity = 8

Detailed explanation

Walkthrough

In a geometric progression with first term $a = 4$ and common ratio $r$ , the first three terms are $4$ , $4r$ , $4r^2$ . Their sum being $7$ gives the quadratic $4 + 4r + 4r^2 = 7$ , i.e. $4r^2 + 4r - 3 = 0$ .

This factorises neatly as $(2r - 1)(2r + 3) = 0$ , giving $r = \frac{1}{2}$ or $r = -\frac{3}{2}$ . Both are valid ratios for the finite sum condition — that is why the question asks for two possible values.

However, a sum to infinity exists only when $|r| < 1$ . Since $|-\tfrac{3}{2}| = 1.5 > 1$ , that progression diverges; only $r = \frac{1}{2}$ is convergent. Applying $S_\infty = \frac{a}{1-r}$ with $a = 4$ gives $S_\infty = \frac{4}{1/2} = 8$ .

Key Takeaways

The sum of the first three GP terms leads directly to a quadratic in $r$ .
A quadratic can yield two legitimate common ratios; both should be reported when asked.
The sum to infinity $S_\infty = \frac{a}{1-r}$ is valid only for $|r| < 1$ — always check which root satisfies this.

Common Mistakes

Computing $S_\infty$ for $r = -1.5$ as well — the mark scheme awards the final A1 for $8$ only, nfww (no marks from wrong working).
Sign errors forming the quadratic: $4r^2 + 4r - 3 = 0$ , not $+3$ .
Factorising incorrectly or misreading the roots of $(2r-1)(2r+3)$ .
Forgetting to state both values of $r$ and giving only the convergent one.

Things to Be Careful About

The final mark requires the correct value $8$ reached from correct working — an answer of $8$ obtained via $r = -1.5$ scores nothing.
State the convergence condition explicitly ( $|r| < 1$ ) to justify discarding $r = -1.5$ .
Exact fractions are preferred here: $r = \frac{1}{2}$ and $-\frac{3}{2}$ are equivalent to $0.5$ and $-1.5$ .

Techniques used

form an equation from the sum of the first three termsfactorise a quadraticselect the convergent ratio using the modulus conditionapply the sum to infinity formula

Q9MediumFunctionsQuadratic functions

The functions $\mathrm{f}$ and $\mathrm{g}$ are defined by

\begin{aligned} \mathrm{f}(x) &= \frac{3x^2}{4x - 1} \quad \text{for } x < 0 \\ \mathrm{g}(x) &= \frac{1}{x^2} \quad \text{for } x < 0. \end{aligned}

(a)

Explain why the function $\mathrm{fg}$ does not exist.

1M Similar

DifficultyEasy

Worked solution

Approach

For $\mathrm{fg}$ to exist, every output of $\mathrm{g}$ must be an acceptable input to $\mathrm{f}$ . Compare the range of $\mathrm{g}$ with the domain of $\mathrm{f}$ .

Working

For $x < 0$ , $\mathrm{g}(x) = \dfrac{1}{x^2}$ is always positive, so the range of $\mathrm{g}$ is $\mathrm{g}(x) > 0$ .

But $\mathrm{f}$ is defined only for $x < 0$ , so it cannot accept any output of $\mathrm{g}$ .

Answer

The range of $\mathrm{g}$ is $\mathrm{g}(x) > 0$ , which lies entirely outside the domain $x < 0$ of $\mathrm{f}$ , so $\mathrm{fg}$ does not exist.

Final answer

The range of g is g > 0, which is outside the domain x < 0 of f, so fg does not exist

Detailed explanation

Walkthrough

A composite function $\mathrm{fg}$ means 'apply $\mathrm{g}$ first, then apply $\mathrm{f}$ '. For this to make sense, each value produced by $\mathrm{g}$ must lie within the set of inputs that $\mathrm{f}$ accepts.

Since $\mathrm{g}(x) = \dfrac{1}{x^2}$ squares its input before taking the reciprocal, every output is strictly positive — the range of $\mathrm{g}$ is all values greater than $0$ . However, $\mathrm{f}$ is only defined for negative inputs ( $x < 0$ ). No positive number can be fed into $\mathrm{f}$ , so the composition fails at every point and $\mathrm{fg}$ does not exist.

Key Takeaways

A composite function exists only when the range of the inner function fits inside the domain of the outer function.
Squaring followed by reciprocating always gives positive outputs.

Common Mistakes

Checking the domains instead of comparing the range of $\mathrm{g}$ with the domain of $\mathrm{f}$ .
Stating ' $\mathrm{fg}$ does not exist' without giving the valid reason (the scheme requires the range argument).

Things to Be Careful About

The order matters: $\mathrm{fg}$ applies $\mathrm{g}$ first. The question's part (b) confirms $\mathrm{gf}$ does exist, so do not mix up the two orders.
A bare assertion earns nothing here — the explanation 'range of $\mathrm{g}$ is $> 0$ ' (or equivalent) is what carries the B1.

Techniques used

find the range of a functioncompare the range of the inner function with the domain of the outer function

(b)

Given that the function $\mathrm{gf}$ does exist, find and simplify an expression for $\mathrm{gf}(x)$ .

DifficultyMedium-Easy

Worked solution

Approach

$\mathrm{gf}$ means apply $\mathrm{f}$ first, then $\mathrm{g}$ . Substitute $\mathrm{f}(x)$ into $\dfrac{1}{x^2}$ and simplify.

Working

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

Note this is valid: $\mathrm{f}(x) < 0$ for $x<0$ , so the input to $\mathrm{g}$ is acceptable.

Simplify:

\mathrm{gf}(x) = \frac{1}{3 \cdot \dfrac{x^4}{(4x-1)^2}} = \frac{(4x - 1)^2}{9x^4}

Answer

\mathrm{gf}(x) = \frac{(4x - 1)^2}{9x^4}

Final answer

(4x - 1)^2 / (9x^4)

Detailed explanation

Walkthrough

The composite $\mathrm{gf}(x)$ is found by feeding $\mathrm{f}(x)$ into $\mathrm{g}$ : wherever $\mathrm{g}$ has an $x$ , write $\mathrm{f}(x)$ instead. Since $\mathrm{g}(x) = \dfrac{1}{x^2}$ , we get $\dfrac{1}{\left(\mathrm{f}(x)\right)^2}$ .

Substituting $\mathrm{f}(x) = \dfrac{3x^2}{4x-1}$ :

\mathrm{gf}(x) = \frac{1}{3\left(\dfrac{x^2}{4x-1}\right)^2}

Squaring the fraction gives $\dfrac{x^4}{(4x-1)^2}$ , so dividing by it multiplies by its reciprocal:

\mathrm{gf}(x) = \frac{(4x-1)^2}{9x^4}

This composite does exist because $\mathrm{f}(x) < 0$ when $x < 0$ (positive numerator over negative denominator), and negative numbers are valid inputs to $\mathrm{g}$ .

Key Takeaways

In $\mathrm{gf}$ , the function written second ( $\mathrm{f}$ ) acts first.
Dividing by a fraction means multiplying by its reciprocal.

Common Mistakes

Computing $\mathrm{fg}$ instead of $\mathrm{gf}$ (order reversed).
Writing $\dfrac{1}{3}\left(\dfrac{x^2}{4x-1}\right)^2$ instead of $\dfrac{1}{3\left(\cdots\right)^2}$ — the whole expression including the 3 is squared's denominator.
Leaving the answer unsimplified; the scheme allows 'isw' after a correct simplified form but the simplified form itself carries the A1.

Things to Be Careful About

The M1 is for the correct substitution structure; the A1 requires the fully simplified single fraction $\dfrac{(4x-1)^2}{9x^4}$ or an equivalent.

Techniques used

form the composite gf by substituting f into gsimplify using laws of indices

(c)

Show that $\mathrm{f}^{-1}(x)$ can be written as $\frac{px - \sqrt{x(qx + r)}}{3}$ where $p$ , $q$ and $r$ are integers.

DifficultyMedium

Worked solution

Approach

To find $\mathrm{f}^{-1}$ , write $y = \dfrac{3x^2}{4x-1}$ , rearrange into a quadratic in $x$ , solve for $x$ using the quadratic formula, and choose the root consistent with $x < 0$ .

Working

Let $y = \dfrac{3x^2}{4x - 1}$ . Then

y(4x - 1) = 3x^2 \quad\Longrightarrow\quad 3x^2 - 4xy + y = 0

This is a quadratic in $x$ . Applying the quadratic formula:

x = \frac{-(-4y) \pm \sqrt{(-4y)^2 - 4(3)(y)}}{2(3)} = \frac{4y \pm \sqrt{16y^2 - 12y}}{6}

Factorising under the root and simplifying:

x = \frac{4y \pm \sqrt{4y(4y - 3)}}{6} = \frac{4y \pm 2\sqrt{y(4y-3)}}{6} = \frac{2y \pm \sqrt{y(4y-3)}}{3}

Since $x < 0$ throughout the domain of $\mathrm{f}$ , we must take the negative square root: with $2y$ already negative (as $y = \mathrm{f}(x) < 0$ ), adding $+\sqrt{y(4y-3)}$ could give a non-negative result, whereas subtracting guarantees $x < 0$ .

Swapping letters to express the inverse as a function of $x$ :

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

This has the required form $\dfrac{px - \sqrt{x(qx + r)}}{3}$ with integers $p = 2$ , $q = 4$ , $r = -3$ .

Answer

\mathrm{f}^{-1}(x) = \frac{2x - \sqrt{x(4x - 3)}}{3}, \quad p = 2,\ q = 4,\ r = -3

Final answer

f^-1(x) = (2x - sqrt(x(4x - 3))) / 3, i.e. p = 2, q = 4, r = -3

Detailed explanation

Walkthrough

An inverse function swaps the roles of input and output. Starting from $y = \dfrac{3x^2}{4x-1}$ , multiply both sides by $(4x-1)$ and collect everything on one side:

3x^2 - 4xy + y = 0

This is a quadratic equation in $x$ (with $y$ acting as a constant), so the quadratic formula solves it:

x = \frac{4y \pm \sqrt{16y^2 - 12y}}{6}

Extracting the factor $4$ from under the square root and cancelling the common factor $2$ gives $\dfrac{2y \pm \sqrt{y(4y-3)}}{3}$ .

Now the crucial step: the original function only accepts $x < 0$ , so the inverse must also produce negative outputs. Since $y = \mathrm{f}(x)$ is negative on this domain (positive numerator, negative denominator), the term $2y$ is negative. Taking the plus sign risks producing a zero or positive result; taking the minus sign keeps the whole numerator negative. Hence the negative square root is chosen.

Finally, relabelling $y$ as $x$ gives the inverse in standard form, matching $\dfrac{px - \sqrt{x(qx+r)}}{3}$ with $p=2$ , $q=4$ , $r=-3$ .

Key Takeaways

To invert a function algebraically, set $y = \mathrm{f}(x)$ , rearrange to make $x$ the subject, then swap the letters.
When a quadratic formula produces two branches, the domain restriction of the original function tells you which branch belongs to the inverse.
A quadratic in $x$ with coefficients involving $y$ is still solved by the standard formula.

Common Mistakes

Taking the wrong (plus) square root branch — the scheme explicitly awards a B1 for justifying the negative root.
Sign errors in the discriminant: $(-4y)^2 - 4(3)(y) = 16y^2 - 12y$ , not $16y^2 + 12y$ .
Failing to simplify $\dfrac{4y \pm \sqrt{4y(4y-3)}}{6}$ down to $\dfrac{2y \pm \sqrt{y(4y-3)}}{3}$ , so the printed target form is never reached.
Forgetting to swap $x$ and $y$ at the end, leaving the answer in terms of $y$ .

Things to Be Careful About

This is a 'show that' (AG) part: derive forward to exactly $\dfrac{px - \sqrt{x(qx+r)}}{3}$ and confirm the match — never start from the printed answer.
The justification of the negative root must reference the domain $x < 0$ (equivalently the range of $\mathrm{f}$ being negative); simply picking a root without reason loses the B1.
The final A1 requires exactly $\dfrac{2x - \sqrt{x(4x-3)}}{3}$ or an equivalent exact form — no decimals.

Techniques used

rearrange the equation of the curve into a quadratic in xapply the quadratic formulaselect the root consistent with the restricted domain

Q10MediumTrigonometry

(a)

Show that $(\tan x + \sec x)^2$ can be written as $\frac{1 + \sin x}{1 - \sin x}$ .

DifficultyMedium

Worked solution

Approach

Express $\tan x$ and $\sec x$ in terms of $\sin x$ and $\cos x$ , combine the terms under a single denominator, apply the identity $\cos^2 x = 1 - \sin^2 x$ , and factorise the denominator as a difference of two squares to cancel the common factor $(1 + \sin x)$ .

Working

\begin{aligned} (\tan x + \sec x)^2 &= \left(\frac{\sin x}{\cos x} + \frac{1}{\cos x}\right)^2 \\ &= \left(\frac{1 + \sin x}{\cos x}\right)^2 \\ &= \frac{(1 + \sin x)^2}{\cos^2 x} \\ &= \frac{(1 + \sin x)^2}{1 - \sin^2 x} \\ &= \frac{(1 + \sin x)(1 + \sin x)}{(1 - \sin x)(1 + \sin x)} \\ &= \frac{1 + \sin x}{1 - \sin x} \end{aligned}

Answer

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

Final answer

(1 + sin x)/(1 - sin x)

Detailed explanation

Walkthrough

Replace $\tan x = \frac{\sin x}{\cos x}$ and $\sec x = \frac{1}{\cos x}$ inside the bracket.
Since both terms share the denominator $\cos x$ , combine them to get $\frac{1 + \sin x}{\cos x}$ .
Squaring numerator and denominator gives $\frac{(1 + \sin x)^2}{\cos^2 x}$ .
Use the fundamental Pythagorean identity $\cos^2 x = 1 - \sin^2 x$ in the denominator to rewrite the expression entirely in terms of $\sin x$ .
Recognise that $1 - \sin^2 x$ is a difference of two squares, factorising into $(1 - \sin x)(1 + \sin x)$ .
Cancel the common factor $(1 + \sin x)$ from numerator and denominator, which leaves the target expression $\frac{1 + \sin x}{1 - \sin x}$ .

Key Takeaways

Expressing compound trigonometric expressions in terms of $\sin$ and $\cos$ is often the most direct route to algebraic simplification.
Always look to apply $\sin^2 x + \cos^2 x = 1$ when relating squared trigonometric functions.
Difference of two squares ( $a^2 - b^2 = (a-b)(a+b)$ ) frequently allows simplification of trigonometric fractions.

Common Mistakes

Incomplete factorisation step: omitting the visible cancellation step $\frac{(1+\sin x)(1+\sin x)}{(1-\sin x)(1+\sin x)}$ can lose the final accuracy mark on "Show that" questions requiring full justification.
Squaring errors: mis-expanding $(\sin x + 1)^2$ as $\sin^2 x + 1$ .

Things to Be Careful About

Every intermediate step must be shown cleanly forward from the left-hand side to the right-hand side; do not assume the result or work backwards without showing the proper forward derivation.

Techniques used

express tan and sec in terms of sin and coscombine fractions over a common denominatoruse the identity cos^2 x = 1 - sin^2 xfactorise the difference of two squares and cancel

(b)

Hence solve the equation $(\tan 3\theta + \sec 3\theta)^2 = 6$ for $0° \leq \theta \leq 180°$ .

DifficultyMedium

Worked solution

Approach

Use the identity established in part (a) by setting $x = 3\theta$ to replace $(\tan 3\theta + \sec 3\theta)^2$ with $\frac{1 + \sin 3\theta}{1 - \sin 3\theta} = 6$ . Solve this linear equation for $\sin 3\theta$ , find all possible values of $3\theta$ in the range $0° \leq 3\theta \leq 540°$ , and then divide by $3$ to obtain $\theta$ .

Working

Using the result from part (a) with $x = 3\theta$ :

\frac{1 + \sin 3\theta}{1 - \sin 3\theta} = 6

Multiply through by $(1 - \sin 3\theta)$ :

1 + \sin 3\theta = 6(1 - \sin 3\theta)

1 + \sin 3\theta = 6 - 6\sin 3\theta

7\sin 3\theta = 5

\sin 3\theta = \frac{5}{7}

For $0° \leq \theta \leq 180°$ , the interval for $3\theta$ is $0° \leq 3\theta \leq 540°$ .

Find the principal angle and all solutions in $[0°, 540°]$ :

\text{Principal angle} = \sin^{-1}\left(\frac{5}{7}\right) \approx 45.5846°

\begin{aligned} 3\theta_1 &= 45.5846° \\ 3\theta_2 &= 180° - 45.5846° = 134.4154° \\ 3\theta_3 &= 360° + 45.5846° = 405.5846° \\ 3\theta_4 &= 540° - 45.5846° = 494.4154° \end{aligned}

Dividing each by $3$ :

\begin{aligned} \theta_1 &= \frac{45.5846°}{3} = 15.2° \\ \theta_2 &= \frac{134.4154°}{3} = 44.8° \\ \theta_3 &= \frac{405.5846°}{3} = 135.2° \\ \theta_4 &= \frac{494.4154°}{3} = 164.8° \end{aligned}

Answer

\theta = 15.2°, \; 44.8°, \; 135.2°, \; 164.8°

Final answer

15.2°, 44.8°, 135.2°, 164.8°

Detailed explanation

Walkthrough

The word "Hence" directs us to apply the identity from part (a). Setting $x = 3\theta$ gives $\frac{1 + \sin 3\theta}{1 - \sin 3\theta} = 6$ .
Multiply both sides by $(1 - \sin 3\theta)$ and collect terms to find $\sin 3\theta = \frac{5}{7}$ .
Determine the required interval for the angle $3\theta$ : since $0° \leq \theta \leq 180°$ , multiplying by 3 gives $0° \leq 3\theta \leq 540°$ .
Compute the principal angle: $\sin^{-1}\left(\frac{5}{7}\right) \approx 45.58°$ .
Find all values of $3\theta$ in the range $[0°, 540°]$ : in the first revolution ( $0° \leq 3\theta < 360°$ ), the values are $45.58°$ and $180° - 45.58° = 134.42°$ . In the next half-revolution ( $360° \leq 3\theta \leq 540°$ ), add $360°$ to each to get $405.58°$ and $494.42°$ .
Divide each of these four values by $3$ to obtain the values of $\theta$ , giving $15.2°$ , $44.8°$ , $135.2°$ , and $164.8°$ to 1 decimal place.

Key Takeaways

When solving trigonometric equations with a multiple angle such as $3\theta$ , adjust the domain interval before finding all solution branches.
In degrees, final angles must be given to 1 decimal place unless otherwise stated.

Common Mistakes

Finding only the first two solutions in $[0°, 360°]$ for $3\theta$ , thereby missing the solutions $\theta = 135.2°$ and $164.8°$ .
Dividing by 3 before finding the other quadrant solutions in the full domain.
Premature rounding of intermediate values of $3\theta$ , which can introduce rounding errors into the 1 decimal place final answers.

Things to Be Careful About

Ensure all four solutions lie strictly within the given domain $0° \leq \theta \leq 180°$ and no extraneous solutions are included.
Remember that Cambridge Additional Mathematics requires angles in degrees to be given to 1 decimal place.

Techniques used

use the identity proven in part (a)rearrange a linear equation in sin(3theta)find all solutions within the transformed domaindivide by the frequency multiplier and round to 1 decimal place

Q117MMediumCalculusCircular measure

In this question all lengths are in centimetres.

Fig. 2

The diagram shows a rectangle $ABCD$ with $BC = x$ .
The area of the rectangle is $400\text{ cm}^2$ .

Two identical quarter-circles of radius $\frac{x}{2}$ , with centres $A$ and $C$ , are removed from the rectangle to make the shaded shape.

Given that $x$ can vary, find the value of $x$ that gives the minimum value of the perimeter of the shaded shape and hence find this minimum value.

DifficultyMedium

Worked solution

Approach

The rectangle has height $BC = x$ and area $400$ , so its length is $\frac{400}{x}$ . The shaded shape's perimeter consists of two quarter-circle arcs (radius $\frac{x}{2}$ ) plus the remaining straight portions of the sides. Form $P$ in terms of $x$ , differentiate, set to zero, solve, then substitute back.

Working

Each quarter-circle has radius $\frac{x}{2}$ , so each arc length is

\frac{1}{4}(2\pi \cdot \tfrac{x}{2}) = \frac{\pi x}{4}

Two such arcs contribute $2 \times \frac{\pi x}{4} = \frac{\pi x}{2}$ .

The two vertical sides each lose a length $\frac{x}{2}$ to the removed quarter-circles, leaving $\left(\frac{400}{x} - \frac{x}{2}\right)$ each, while the full side $BC = x$ remains intact. Hence

P = 2\left(\frac{\pi x}{4}\right) + x + 2\left(\frac{400}{x} - \frac{x}{2}\right)

P = \frac{\pi x}{2} + \frac{800}{x}

Differentiate:

\frac{\mathrm{d}P}{\mathrm{d}x} = \frac{\pi}{2} - \frac{800}{x^2}

Set equal to zero and solve:

\frac{\pi}{2} = \frac{800}{x^2} \quad\Rightarrow\quad x^2 = \frac{1600}{\pi} \quad\Rightarrow\quad x = \frac{40}{\sqrt{\pi}} \approx 22.6

Substitute back:

P = \frac{\pi}{2}\left(\frac{40}{\sqrt{\pi}}\right) + \frac{800}{\frac{40}{\sqrt{\pi}}} = 20\sqrt{\pi} + 20\sqrt{\pi} = 40\sqrt{\pi} \approx 70.9

Since $\frac{\mathrm{d}^2P}{\mathrm{d}x^2} = \frac{1600}{x^3} > 0$ for $x > 0$ , this is a minimum.

Answer

x = \frac{40}{\sqrt{\pi}} \approx 22.6, \qquad P_{\min} = 40\sqrt{\pi} \approx 70.9

Final answer

x = 22.6 cm, minimum perimeter = 70.9 cm

Detailed explanation

Walkthrough

The rectangle's area fixes its length: since $BC = x$ and the area is $400$ , the horizontal length must be $\frac{400}{x}$ . This is the key first step — everything else hangs on expressing both dimensions in terms of the single variable $x$ .

Next we build the perimeter of the shaded shape by walking around it. Two pieces are curved: the quarter-circle arcs centred at $A$ and $C$ . Each has radius $\frac{x}{2}$ , so each arc is a quarter of a full circumference: $\frac{1}{4} \times 2\pi \times \frac{x}{2} = \frac{\pi x}{4}$ , giving $\frac{\pi x}{2}$ for both together. Note this only works because the angle is measured in radians — arc length $= r\theta$ with $\theta = \frac{\pi}{2}$ .

The straight pieces: looking at the figure, the two vertical sides have had a length $\frac{x}{2}$ eaten away by the quarter-circles, so each contributes $\frac{400}{x} - \frac{x}{2}$ . One full side of length $x$ survives untouched. Adding everything:

P = \frac{\pi x}{2} + x + \frac{800}{x} - x = \frac{\pi x}{2} + \frac{800}{x}

(the $+x$ and $-x$ cancel). To minimise, differentiate: $\frac{\mathrm{d}P}{\mathrm{d}x} = \frac{\pi}{2} - \frac{800}{x^2}$ . Setting this to zero gives $x^2 = \frac{1600}{\pi}$ , so $x = \frac{40}{\sqrt{\pi}} \approx 22.6$ (taking the positive root, since a length cannot be negative).

Substituting back gives $P = 20\sqrt{\pi} + 20\sqrt{\pi} = 40\sqrt{\pi} \approx 70.9$ . The second derivative $\frac{1600}{x^3}$ is positive, confirming a minimum rather than a maximum.

Key Takeaways

When a quantity depends on one variable through a constraint (here the fixed area), use the constraint to eliminate the other dimension before differentiating.
Arc length requires radians: a quarter circle of radius $r$ has arc $\frac{\pi r}{2}$ .
Perimeter optimisation problems follow the pattern: form the function, differentiate, set to zero, solve, substitute back, and justify the nature of the stationary point.

Common Mistakes

Using the diameter instead of the radius in the arc length formula, or forgetting the arcs are quarter-circles (a factor-of-4 error).
Misidentifying which straight edges survive: the mark scheme expects $x + 2\left(\frac{400}{x} - \frac{x}{2}\right)$ for the straight portions — taking all four full sides or omitting the cut-outs gives a wrong $P$ .
Forgetting to convert the area into the length $\frac{400}{x}$ before forming $P$ .
Keeping the negative root $x = -\frac{40}{\sqrt{\pi}}$ : a length must be positive.
Rounding $x$ too early (e.g. using $22.6$ in the substitution) instead of the exact $\frac{40}{\sqrt{\pi}}$ , which can shift the final answer off 3 significant figures.
Omitting units or failing to confirm the stationary point is a minimum when the question asks for the minimum value.

Things to Be Careful About

The final answers should be given to 3 significant figures: $x \approx 22.6$ and $P \approx 70.9$ ; the exact forms are $x = \frac{40}{\sqrt{\pi}}$ and $P = 40\sqrt{\pi}$ .
The mark scheme awards B2 for the correct perimeter expression (with B1 for seeing the rectangle length $\frac{400}{x}$ ), so the expression itself carries significant weight — show it clearly before differentiating.
Follow-through applies to the differentiation and solving steps provided your $P$ is of the form $\frac{a}{x} + bx$ , but the printed answers require correct working throughout (nfww).
All lengths here are in centimetres, so state the unit with the final answers.

Techniques used

express the rectangle's other side from its areafind arc lengths of quarter-circles in radiansform an expression for the perimeter of the shaded shapedifferentiate the perimeter functionset the derivative to zero and solvesubstitute back to find the minimum perimeter

Q12MediumVectors in two dimensions

Fig. 3

The diagram shows a triangle $OBC$ .
$OA : OB = 4 : 7$ and $OD : OC = 4 : 7$ .

\overrightarrow{OB} = \mathbf{b} \quad \text{and} \quad \overrightarrow{OC} = \mathbf{c}

The point $P$ is the point of intersection of $AC$ and $BD$ such that $\overrightarrow{AP} = \lambda\overrightarrow{AC}$ and $\overrightarrow{BP} = \mu\overrightarrow{BD}$ where $\lambda$ and $\mu$ are scalars.

(a)

Find two expressions for $\overrightarrow{OP}$ , each in terms of $\mathbf{b}$ , $\mathbf{c}$ and a scalar, and hence show that $P$ divides both $AC$ and $DB$ in the ratio $4 : 7$ .

7M Similar

DifficultyMedium

Worked solution

Approach

Since $OA : OB = 4 : 7$ , we have $\overrightarrow{OA} = \frac{4}{7}\mathbf{b}$ , and since $OD : OC = 4 : 7$ , we have $\overrightarrow{OD} = \frac{4}{7}\mathbf{c}$ . Write $\overrightarrow{OP}$ in two ways: once travelling from $O$ to $A$ then part of the way along $AC$ , and once from $O$ to $B$ then part of the way along $BD$ . Equate the two expressions and compare coefficients of the independent vectors $\mathbf{b}$ and $\mathbf{c}$ .

Working

First expression, via $A$ on the line $AC$ :

\overrightarrow{OP} = \overrightarrow{OA} + \lambda\overrightarrow{AC} = \frac{4}{7}\mathbf{b} + \lambda\left(\mathbf{c} - \frac{4}{7}\mathbf{b}\right)

Second expression, via $B$ on the line $BD$ :

\overrightarrow{OP} = \overrightarrow{OB} + \mu\overrightarrow{BD} = \mathbf{b} + \mu\left(\frac{4}{7}\mathbf{c} - \mathbf{b}\right)

Equating the two expressions and collecting the coefficients of $\mathbf{b}$ and $\mathbf{c}$ :

\frac{4}{7}(1-\lambda)\,\mathbf{b} + \lambda\,\mathbf{c} = (1-\mu)\,\mathbf{b} + \frac{4}{7}\mu\,\mathbf{c}

Comparing coefficients of the independent vectors $\mathbf{b}$ and $\mathbf{c}$ :

\frac{4}{7}(1-\lambda) = 1-\mu \qquad \text{and} \qquad \lambda = \frac{4}{7}\mu

Substituting $\lambda = \frac{4}{7}\mu$ into the first equation:

\frac{4}{7}\left(1 - \frac{4}{7}\mu\right) = 1 - \mu

\frac{4}{7} - \frac{16}{49}\mu = 1 - \mu \quad\Longrightarrow\quad \mu - \frac{16}{49}\mu = 1 - \frac{4}{7} = \frac{3}{7}

\frac{33}{49}\mu = \frac{3}{7} \quad\Longrightarrow\quad \mu = \frac{3}{7} \times \frac{49}{33} = \frac{7}{11}, \qquad \lambda = \frac{4}{7} \times \frac{7}{11} = \frac{4}{11}

Interpreting the parameters as ratios:

AP : PC = \lambda : (1-\lambda) = \frac{4}{11} : \frac{7}{11} = 4 : 7

BP : PD = \mu : (1-\mu) = \frac{7}{11} : \frac{4}{11} = 7 : 4, \quad \text{so} \quad DP : PB = 4 : 7

Answer

$P$ divides $AC$ in the ratio $AP : PC = 4 : 7$ and divides $DB$ in the ratio $DP : PB = 4 : 7$ , as required.

Final answer

P divides AC in the ratio AP : PC = 4 : 7 and divides DB in the ratio DP : PB = 4 : 7

Detailed explanation

Walkthrough

The key idea is that any point on a straight line can be reached by going to one end of the line and then travelling a fraction of the way to the other end. Since $A$ lies on $OB$ with $OA : OB = 4 : 7$ , the position vector of $A$ is $\frac{4}{7}$ of $\overrightarrow{OB}$ , i.e. $\frac{4}{7}\mathbf{b}$ . Similarly $D$ has position vector $\frac{4}{7}\mathbf{c}$ .

The point $P$ lies on both lines $AC$ and $BD$ . Going via $A$ : $\overrightarrow{OP} = \overrightarrow{OA} + \lambda\overrightarrow{AC}$ , where $\overrightarrow{AC} = \overrightarrow{OC} - \overrightarrow{OA} = \mathbf{c} - \frac{4}{7}\mathbf{b}$ . This gives the first expression. Going via $B$ : $\overrightarrow{OP} = \mathbf{b} + \mu\left(\frac{4}{7}\mathbf{c} - \mathbf{b}\right)$ , since $\overrightarrow{BD} = \overrightarrow{OD} - \overrightarrow{OB} = \frac{4}{7}\mathbf{c} - \mathbf{b}$ .

Both expressions describe the same point $P$ , so they are equal. Because $\mathbf{b}$ and $\mathbf{c}$ are independent vectors (not parallel), their coefficients must match separately — this is why we get two simultaneous equations rather than just one. Solving gives $\lambda = \frac{4}{11}$ and $\mu = \frac{7}{11}$ .

Finally, the parameter tells us how far along each line $P$ sits: $\lambda = \frac{4}{11}$ means $AP$ is $\frac{4}{11}$ of $AC$ , so $PC$ is the remaining $\frac{7}{11}$ , giving $AP : PC = 4 : 7$ . Likewise $BP = \frac{7}{11}$ of $BD$ and $PD = \frac{4}{11}$ of $BD$ , so reading from $D$ , $DP : PB = 4 : 7$ .

Key Takeaways

A point dividing a line internally can be written as one end plus a scalar multiple of the direction vector.
If two expressions for the same vector involve independent vectors, equating coefficients gives simultaneous equations.
A parameter $\lambda$ on segment $XY$ converts directly to the ratio $XP : PY = \lambda : (1-\lambda)$ .

Common Mistakes

Writing $\overrightarrow{AC}$ or $\overrightarrow{BD}$ back-to-front, giving sign errors in the direction vectors.
Equating the whole expressions but forgetting that $\mathbf{b}$ and $\mathbf{c}$ must be compared coefficient by coefficient.
Reporting $BP : BD = 7 : 11$ instead of converting it to the asked ratio $DP : PB = 4 : 7$ — the conclusion must state the ratio in which $P$ divides the line.
Arithmetic slips when clearing fractions; keep exact fractions throughout.

Things to Be Careful About

The final A marks require both the values of $\lambda$ and $\mu$ AND an explicit concluding statement about the ratios — stopping at $\lambda = \frac{4}{11}$ loses marks.
Note the asymmetry: on $AC$ the parameter is measured from $A$ , but on $DB$ the ratio is quoted from $D$ , so $\mu = \frac{7}{11}$ corresponds to $DP : PB = 4 : 7$ , not $7 : 4$ .
Keep all working in exact fractional form; no decimals are needed.

Techniques used

express position vectors along a line using a scalar parameterequate two expressions for the same vectorcompare coefficients of independent vectorssolve simultaneous equations for the scalarsconvert scalar parameters into division ratios

(b)

The point $Q$ is such that $\overrightarrow{OQ} = \frac{2}{7}\mathbf{b} + \frac{2}{7}\mathbf{c}$ .

Use a vector method to show that $O$ , $Q$ and $P$ are collinear. Justify your answer.