4037/21 May/June 2025 — Additional Mathematics Solved Past Paper

Q1Medium-EasyQuadratic functionsEquations, inequalities and graphs

A curve has equation $y = x^2 + 2x - 3$ .

(a)

Use the method of completing the square to find the coordinates of the stationary point on the curve.

DifficultyMedium-Easy

Worked solution

Approach

Complete the square on $y = x^2 + 2x - 3$ ; the squared bracket is zero at the stationary point, so the vertex is read directly from the completed-square form.

Working

Take half of the coefficient of $x$ and square it:

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

So

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

The minimum occurs when $(x+1)^2 = 0$ , i.e. when $x = -1$ , giving $y = -4$ .

Answer

(-1,\ -4)

Final answer

(-1, -4)

Detailed explanation

Walkthrough

Completing the square rewrites a quadratic as a squared term plus a constant. For $y = x^2 + 2x - 3$ , we take half the coefficient of $x$ , which is $\frac{2}{2} = 1$ , and write $(x+1)^2$ . Expanding $(x+1)^2$ gives $x^2 + 2x + 1$ , which is one more than what we want, so we subtract 1: $x^2 + 2x = (x+1)^2 - 1$ . Substituting back, $y = (x+1)^2 - 1 - 3 = (x+1)^2 - 4$ .

Since $(x+1)^2 \geq 0$ for all real $x$ , the smallest possible value of $y$ is $-4$ , reached exactly when $(x+1)^2 = 0$ , that is $x = -1$ . So the stationary point is $(-1, -4)$ .

Key Takeaways

Completing the square converts $ax^2 + bx + c$ into vertex form, from which the turning point can be read without differentiating.
The vertex of $y = (x+p)^2 + q$ is at $(-p, q)$ — note the sign flip in the $x$ -coordinate.

Common Mistakes

Writing the $x$ -coordinate as $+1$ instead of $-1$ : the vertex of $(x+1)^2$ sits where $x + 1 = 0$ , so $x = -1$ .
Forgetting to subtract the squared half-coefficient ( $1^2$ ) after forming $(x+1)^2$ .
Dropping the constant $-3$ when combining terms.

Things to Be Careful About

The mark scheme awards B2 for the fully correct completed-square form $y = (x+1)^2 - 4$ (or equivalent), and only B1 if you stop at $y = (x+1)^2 + k$ with an unspecified constant — so finish the constant.
The answer must come from completing the square as instructed; the coordinates are cao (correct answer only).

Techniques used

complete the square on a quadraticread off the coordinates of the stationary point from the completed-square form

(b)

On the axes, sketch the curve, stating the intercepts with the coordinate axes.

Fig. 1

2M Similar

DifficultyMedium-Easy

Worked solution

Approach

Find the intercepts by factorising and substituting $x = 0$ , then sketch the upward-opening parabola with its minimum at $(-1, -4)$ from part (a).

Working

Factorise:

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

So $y = 0$ when $x = -3$ or $x = 1$ : the curve crosses the $x$ -axis at $(-3, 0)$ and $(1, 0)$ .

When $x = 0$ : $y = -3$ , so the curve crosses the $y$ -axis at $(0, -3)$ .

From part (a), the minimum point is $(-1, -4)$ , which lies in the third quadrant.

Answer

An upward-opening parabola crossing the $x$ -axis at $x = -3$ and $x = 1$ , crossing the $y$ -axis at $y = -3$ , with its minimum at $(-1, -4)$ in the third quadrant.

Final answer

Upward parabola with x-intercepts -3 and 1, y-intercept -3, minimum (-1, -4) in the third quadrant

Detailed explanation

Walkthrough

To sketch the parabola we need its key features. The coefficient of $x^2$ is positive, so it opens upwards. Factorising $x^2 + 2x - 3$ gives $(x+3)(x-1)$ , so the roots are $x = -3$ and $x = 1$ — these are where the curve cuts the $x$ -axis. Setting $x = 0$ gives $y = -3$ , the $y$ -intercept. Part (a) already located the minimum at $(-1, -4)$ , which lies below the axis between the two roots, in the third quadrant. Plotting these four features and joining them with a smooth U-shaped curve completes the sketch.

Key Takeaways

A quadratic sketch needs: direction of opening, both $x$ -intercepts, the $y$ -intercept, and the turning point.
The roots of $y = ax^2+bx+c$ come straight from factorising; the $y$ -intercept is just the constant term $c$ .

Common Mistakes

Marking the $y$ -intercept as $+3$ instead of $-3$ (sign slip on the constant).
Drawing the parabola opening downwards despite the positive $x^2$ coefficient.
Omitting one of the three intercepts — the mark scheme requires all of $-3$ on the $y$ -axis and $-3$ and $1$ on the $x$ -axis.
Placing the minimum outside the third quadrant or drawing a curve confined to fewer than four quadrants.

Things to Be Careful About

The first B1 requires the correct shape spanning all four quadrants with the minimum in the third quadrant; the second B1 requires the three intercepts clearly marked.
This is a sketch, not a plot: a smooth freehand curve through correctly placed key points is expected, but every labelled point must be present.

Techniques used

factorise the quadratic to find x-interceptsevaluate y at x = 0 for the y-interceptsketch a parabola marking all intercepts

Q25MMediumQuadratic functions

In this question, $k$ is a constant.

It is given that $2x^2 + (3k - 2)x + k = 0$ has roots that are real and distinct.

Find the set of possible values of $k$ .

DifficultyMedium

Worked solution

Approach

For real and distinct roots, the discriminant $b^2 - 4ac > 0$ . Apply this to the given quadratic in $x$ , then solve the resulting quadratic inequality in $k$ .

Working

For $2x^2 + (3k-2)x + k = 0$ , we have $a = 2$ , $b = 3k - 2$ , $c = k$ . Real and distinct roots require:

b^2 - 4ac > 0

(3k - 2)^2 - 4(2)(k) > 0

9k^2 - 12k + 4 - 8k > 0

9k^2 - 20k + 4 > 0

Factorise:

(9k - 2)(k - 2) > 0

Critical values: $k = \frac{2}{9}$ and $k = 2$ . Since the quadratic opens upwards, it is positive outside the roots:

k < \frac{2}{9} \quad \text{or} \quad k > 2

Answer

k < \frac{2}{9} \quad \text{or} \quad k > 2

Final answer

k < 2/9 or k > 2

Detailed explanation

Walkthrough

The condition for a quadratic $ax^2 + bx + c = 0$ to have two real, distinct roots is $b^2 - 4ac > 0$ . Here $a = 2$ , $b = 3k - 2$ and $c = k$ , so we form $(3k-2)^2 - 4(2)(k) > 0$ .

Expanding $(3k-2)^2 = 9k^2 - 12k + 4$ and subtracting $8k$ gives the three-term quadratic $9k^2 - 20k + 4 > 0$ . This factorises as $(9k-2)(k-2) > 0$ .

The critical values are where each factor is zero: $k = \frac{2}{9}$ and $k = 2$ . Since the parabola $y = 9k^2 - 20k + 4$ opens upwards, it is positive outside the interval between the roots, giving $k < \frac{2}{9}$ or $k > 2$ .

Key Takeaways

Real and distinct roots ⟺ $b^2 - 4ac > 0$ ; equal roots ⟺ $= 0$ ; no real roots ⟺ $< 0$ .
A quadratic inequality with an upward-opening parabola is satisfied outside the roots.
Always identify $a$ , $b$ , $c$ carefully when $b$ itself contains the unknown constant.

Common Mistakes

Using $b^2 - 4ac = 0$ (equal roots) instead of $> 0$ (distinct roots).
Sign errors expanding $(3k-2)^2$ , e.g. writing $9k^2 + 4$ and dropping the middle term.
Giving $\frac{2}{9} < k < 2$ — the inequality direction is reversed because the parabola opens upwards.
Forgetting to divide by $9$ correctly when solving $9k - 2 = 0$ .
The mark scheme marks the final answer, so the correct inequality signs are essential; a correct discriminant with wrong inequalities loses the last mark.

Things to Be Careful About

The discriminant step earns the first method mark even with any inequality sign or equals, but the final answer must have the correct strict inequalities.
Present the answer as a union of two regions: $k < \frac{2}{9}$ or $k > 2$ ; writing a single interval between the values is wrong.
Keep exact fractions: $\frac{2}{9}$ , not a decimal.

Techniques used

apply the discriminant condition for real distinct rootsexpand and simplify the discriminantfactorise the resulting quadraticfind critical values and solve the quadratic inequality

Approach

Keep all runners together, all swimmers together and all gymnasts together by treating each discipline as one block. Arrange the $3$ blocks, then arrange the members inside each block.

Working

Number of arrangements of the $3$ blocks:

3! = 6

Arrangements within the blocks:

5! \times 4! \times 3! = 120 \times 24 \times 6 = 17280

Total number of ways:

3! \times 5! \times 4! \times 3! = 6 \times 120 \times 24 \times 6 = 103\,680

Answer

103\,680

Final answer

103680

Detailed explanation

Walkthrough

The condition that all runners stand together, all swimmers stand together and all gymnasts stand together is handled by the 'block' method: glue each group into a single unit. There are then just $3$ units to line up, which can be done in $3! = 6$ orders. But the people inside each glued unit can still be shuffled among themselves: the $5$ runners in $5!$ ways, the $4$ swimmers in $4!$ ways, and the $3$ gymnasts in $3!$ ways. Since every choice of block order combines independently with every internal ordering, multiply all four factorials together.

3! \times 5! \times 4! \times 3! = 6 \times 120 \times 24 \times 6 = 103\,680

Key Takeaways

When specified items must stay together, treat them as one block, arrange the blocks, then arrange within each block.
The total is a product because the stages are independent choices.

Common Mistakes

Forgetting to permute the blocks themselves ( $k = 1$ instead of $k = 3$ ), giving $17\,280$ .
Adding the factorials instead of multiplying them.
Treating members within a group as indistinguishable and omitting the internal permutations.

Things to Be Careful About

The answer must be the exact integer $103\,680$ ; the mark scheme allows any multiplier $k \in \{1, 2, 3, 6\}$ times $5! \times 4! \times 3!$ , but only $k = 3$ satisfies the stated condition correctly here.
Show the product structure clearly so the method mark is visible.

Techniques used

treat each group of same-discipline members as a single blockarrange the blocks in a linearrange members within each block

(b)

Four of these members are selected for an event.

Find the number of ways that this can be done when at least one runner, at least one swimmer and at least one gymnast must be selected.

3M Similar

DifficultyMedium

Worked solution

Approach

Selecting $4$ members with at least one runner, one swimmer and one gymnast means the composition must be $(2, 1, 1)$ in some order. The possible splits are: 2 runners + 1 swimmer + 1 gymnast, or 1 runner + 2 swimmers + 1 gymnast, or 1 runner + 1 swimmer + 2 gymnasts. Count each with combinations and add.

Working

Case 1 — two runners, one swimmer, one gymnast:

{}^{5}\mathrm{C}_{2} \times {}^{4}\mathrm{C}_{1} \times {}^{3}\mathrm{C}_{1} = 10 \times 4 \times 3 = 120

Case 2 — one runner, two swimmers, one gymnast:

{}^{5}\mathrm{C}_{1} \times {}^{4}\mathrm{C}_{2} \times {}^{3}\mathrm{C}_{1} = 5 \times 6 \times 3 = 90

Case 3 — one runner, one swimmer, two gymnasts:

{}^{5}\mathrm{C}_{1} \times {}^{4}\mathrm{C}_{1} \times {}^{3}\mathrm{C}_{2} = 5 \times 4 \times 3 = 60

Total:

120 + 90 + 60 = 270

Answer

270

Final answer

270

Detailed explanation

Walkthrough

Since we need at least one member from each of the three disciplines and only $4$ people are chosen, the extra person beyond the compulsory one-per-group must come from one of the disciplines. So there are exactly three possible compositions: an extra runner, an extra swimmer, or an extra gymnast. Each composition is counted using combinations (order does not matter when selecting for an event), and since the cases are mutually exclusive, the counts are added.

{}^{5}\mathrm{C}_{2} \times {}^{4}\mathrm{C}_{1} \times {}^{3}\mathrm{C}_{1} + {}^{5}\mathrm{C}_{1} \times {}^{4}\mathrm{C}_{2} \times {}^{3}\mathrm{C}_{1} + {}^{5}\mathrm{C}_{1} \times {}^{4}\mathrm{C}_{1} \times {}^{3}\mathrm{C}_{2} = 120 + 90 + 60 = 270

Key Takeaways

'At least one of each' with a fixed total forces a small set of compositions — enumerate them as disjoint cases.
Use ${}^{n}\mathrm{C}_{r}$ for selections where order does not matter, and add counts of mutually exclusive cases.

Common Mistakes

Using permutations instead of combinations (multiplying by extra factors), which gives answers like $540$ — the mark scheme awards SC1 for $540$ .
Missing one of the three cases, e.g. forgetting the two-gymnast case, giving $264$ — also SC1.
Double-counting by treating the 'extra' person as special rather than simply choosing 2 from that group.

Things to Be Careful About

The three cases are exhaustive and disjoint: check they sum consistently before adding.
The final answer must be exactly $270$ ; wrong-working answers such as $540$ or $264$ score no marks beyond the single SC mark.

Techniques used

split the selection into cases by compositioncount each composition with combinationsadd the disjoint cases

Q47MMediumStraight-line graphs

The coordinates of points $A$ , $B$ , $C$ and $D$ are as follows.

A(-4, 3) \quad B(6, -9) \quad C(15, 10) \quad D(14, -1)

The line $L$ has equation $y = 11x - 75$ .
The perpendicular bisector of the line $AB$ meets $L$ at the point $E$ .

Find the area of triangle $CDE$ .

DifficultyMedium

Worked solution

Approach

Find the perpendicular bisector of $AB$ (its gradient is the negative reciprocal of $m_{AB}$ and it passes through the midpoint of $AB$ ), intersect it with $L$ to get $E$ , then compute the area of triangle $CDE$ from the coordinates.

Working

Gradient of $AB$ :

m_{AB} = \frac{-9 - 3}{6 - (-4)} = \frac{-12}{10} = -\frac{6}{5}

Perpendicular gradient:

m_{\perp} = -\frac{1}{-\frac{6}{5}} = \frac{5}{6}

Midpoint of $AB$ :

\left(\frac{-4 + 6}{2}, \frac{3 - 9}{2}\right) = (1, -3)

Equation of the perpendicular bisector:

y - (-3) = \frac{5}{6}(x - 1)

y = \frac{5}{6}x - \frac{5}{6} - 3 = \frac{5}{6}x - \frac{23}{6}

Intersect with $L$ : $y = 11x - 75$ .

11x - 75 = \frac{5}{6}x - \frac{23}{6}

66x - 450 = 5x - 23 \implies 61x = 427 \implies x = 7

y = 11(7) - 75 = 2 \implies E(7, 2)

Area of triangle $CDE$ with $C(15, 10)$ , $D(14, -1)$ , $E(7, 2)$ :

\text{Area} = \frac{1}{2}\left| 15(-1 - 2) + 14(2 - 10) + 7(10 - (-1)) \right|

= \frac{1}{2}\left| -45 - 112 + 77 \right| = \frac{1}{2} \times 80 = 40

Answer

\text{Area of triangle } CDE = 40

Final answer

40

Detailed explanation

Walkthrough

The question builds the point $E$ in stages. First we need the perpendicular bisector of $AB$ , which is the line through the midpoint of $AB$ with gradient perpendicular to $AB$ .

The gradient of $AB$ uses difference in $y$ over difference in $x$ : $\frac{-9-3}{6-(-4)} = -\frac{6}{5}$ . Perpendicular lines have gradients whose product is $-1$ , so the perpendicular gradient is the negative reciprocal, $\frac{5}{6}$ .

The midpoint averages the coordinates: $\left(\frac{-4+6}{2}, \frac{3-9}{2}\right) = (1, -3)$ .

Using $y - y_1 = m(x - x_1)$ with $m = \frac{5}{6}$ at $(1, -3)$ gives the bisector $y = \frac{5}{6}x - \frac{23}{6}$ .

$E$ is where this meets $L$ : $y = 11x - 75$ . Equating the two expressions for $y$ and clearing the fraction by multiplying by 6 gives $66x - 450 = 5x - 23$ , so $61x = 427$ and $x = 7$ , giving $y = 2$ , i.e. $E(7, 2)$ .

Finally the area of the triangle with vertices $C(15, 10)$ , $D(14, -1)$ , $E(7, 2)$ is computed with the coordinate (shoelace) formula, giving $\frac{1}{2}|{-45 - 112 + 77}| = 40$ .

Key Takeaways

Perpendicular gradients multiply to $-1$ : flip the fraction and change the sign.
A perpendicular bisector needs both the midpoint and the perpendicular gradient.
Intersections of two lines are found by equating their $y$ expressions.
The area of a triangle from three coordinates is $\frac{1}{2}|x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|$ .

Common Mistakes

Sign slips in $6 - (-4)$ , which must give $10$ , not $2$ .
Forgetting the negative in the negative reciprocal, giving $-\frac{5}{6}$ instead of $\frac{5}{6}$ .
Using the midpoint of the wrong pair of points, or using a point like $A$ or $B$ instead of the midpoint (the mark scheme explicitly disallows this).
Arithmetic errors when clearing the fractions in $11x - 75 = \frac{5}{6}x - \frac{23}{6}$ ; multiplying everything by 6 avoids this.
Dropping the modulus (absolute value) in the area formula, or halving incorrectly.
The final area mark is dependent on all previous marks, so an early error in $E$ loses the area mark too.

Things to Be Careful About

The mark scheme awards the perpendicular-gradient mark only if the gradient of $AB$ was attempted as difference in $y$ over difference in $x$ .
The midpoint must be the actual midpoint of $AB$ — using $A$ , $B$ , $C$ or $D$ scores no mark.
The area mark is 'dependent on all previous marks', so every earlier step must be correct for the final 40 to score.
Keep exact fractions throughout; the answer 40 is exact and 'cao' applies.

Techniques used

find the gradient of a line through two pointsuse the negative reciprocal for a perpendicular gradientfind the midpoint of a segmentform the equation of the perpendicular bisectorsolve simultaneous linear equations to find an intersection pointcompute the area of a triangle from coordinates

Approach

Differentiate $y = x^2 \tan \frac{x}{2}$ using the product rule together with the chain rule for $\tan \frac{x}{2}$ , evaluate the derivative at $x = \frac{\pi}{3}$ , and multiply by the small increment $h$ .

Working

Using the chain rule:

\frac{\mathrm{d}}{\mathrm{d}x}\left(\tan \frac{x}{2}\right) = \frac{1}{2}\sec^2 \frac{x}{2}

By the product rule:

\frac{\mathrm{d}y}{\mathrm{d}x} = 2x\tan \frac{x}{2} + x^2 \cdot \frac{1}{2}\sec^2 \frac{x}{2} = 2x\tan \frac{x}{2} + \frac{x^2}{2}\sec^2 \frac{x}{2}

At $x = \frac{\pi}{3}$ , so that $\frac{x}{2} = \frac{\pi}{6}$ with $\tan \frac{\pi}{6} = \frac{1}{\sqrt{3}}$ and $\sec \frac{\pi}{6} = \frac{2}{\sqrt{3}}$ :

\left.\frac{\mathrm{d}y}{\mathrm{d}x}\right|_{x=\frac{\pi}{3}} = 2 \cdot \frac{\pi}{3} \cdot \frac{1}{\sqrt{3}} + \frac{1}{2} \cdot \frac{\pi^2}{9} \cdot \frac{4}{3}

= \frac{2\pi}{3\sqrt{3}} + \frac{2\pi^2}{27} \approx 1.2092 + 0.7310 = 1.94

For small $h$ , the small-change relationship gives

\delta y \approx \left.\frac{\mathrm{d}y}{\mathrm{d}x}\right|_{x=\frac{\pi}{3}} \times h

Answer

\delta y \approx 1.94h

Final answer

1.94h

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: since $\frac{\mathrm{d}y}{\mathrm{d}x}$ measures the rate of change of $y$ with respect to $x$ , a small change $h$ in $x$ produces an approximate change $\delta y \approx \frac{\mathrm{d}y}{\mathrm{d}x} \times h$ , evaluated at the starting point $x = \frac{\pi}{3}$ .

First we need the derivative. The function is a product of two factors, $x^2$ and $\tan \frac{x}{2}$ , so the product rule applies: differentiate each factor in turn while keeping the other fixed. Differentiating $x^2$ gives $2x$ . Differentiating $\tan \frac{x}{2}$ needs the chain rule: the derivative of $\tan u$ is $\sec^2 u$ , and with $u = \frac{x}{2}$ we must multiply by $\frac{\mathrm{d}u}{\mathrm{d}x} = \frac{1}{2}$ , giving $\frac{1}{2}\sec^2 \frac{x}{2}$ . Putting these together:

\frac{\mathrm{d}y}{\mathrm{d}x} = 2x\tan \frac{x}{2} + \frac{x^2}{2}\sec^2 \frac{x}{2}

Next, substitute $x = \frac{\pi}{3}$ . Then $\frac{x}{2} = \frac{\pi}{6}$ , a special angle where $\tan \frac{\pi}{6} = \frac{1}{\sqrt{3}}$ and $\cos \frac{\pi}{6} = \frac{\sqrt{3}}{2}$ , so $\sec^2 \frac{\pi}{6} = \frac{4}{3}$ . Substituting:

\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{2\pi}{3\sqrt{3}} + \frac{2\pi^2}{27} \approx 1.94

Finally, multiplying this rate by the small increment $h$ gives the approximate change in $y$ : $\delta y \approx 1.94h$ .

Key Takeaways

The product rule $\frac{\mathrm{d}}{\mathrm{d}x}(uv) = u'v + uv'$ handles products of different function types such as a polynomial times a trigonometric function.
Chain rule inside trig functions: always multiply by the derivative of the inner argument, here the factor $\frac{1}{2}$ from $\frac{x}{2}$ .
The small-change approximation $\delta y \approx \frac{\mathrm{d}y}{\mathrm{d}x}\,\delta x$ estimates increments without recomputing the whole function.
Special angles ( $\frac{\pi}{6}$ ) allow exact evaluation before rounding to 3 significant figures.

Common Mistakes

Forgetting the factor $\frac{1}{2}$ when differentiating $\tan \frac{x}{2}$ — writing just $\sec^2 \frac{x}{2}$ loses the B1 mark.
Misapplying the product rule, e.g. multiplying the derivatives instead of cross-multiplying ( $2x \times \frac{1}{2}\sec^2 \frac{x}{2}$ alone).
Evaluating at the wrong point, or substituting $x = \frac{\pi}{6}$ instead of $x = \frac{\pi}{3}$ into the derivative.
Forgetting to multiply by $h$ at the end, leaving just the number $1.94$ as the answer.
Rounding intermediate values too early instead of carrying exact forms like $\frac{2\pi}{3\sqrt{3}} + \frac{2\pi^2}{27}$ through to the final 3-significant-figure coefficient.

Things to Be Careful About

The final answer must be expressed as a multiple of $h$ — the mark scheme's A1 is dependent on both the correct derivative (B2) and the correct small-change relationship (M1), so all three ingredients must appear.
Work in radians throughout: the argument $\frac{\pi}{3}$ is a radian measure, and the special-angle values used are radian-based.
Give the coefficient to 3 significant figures: $1.94h$ (awrt $1.94h$ ).
'oe' on the derivative means any equivalent correct form is accepted, but the structure of the product rule must be visible for the M1 FT mark.

Techniques used

differentiate tan of a half angle using the chain ruleapply the product ruleevaluate the derivative at the given pointuse the small-change approximation delta y = dy/dx times h

Approach

First, factorise the quadratic expression inside the bracket as a perfect square: $9x^2 + 12x + 4 = (3x + 2)^2$ .
Then apply the outer power of $5$ using index laws: $((3x + 2)^2)^5 = (2 + 3x)^{10}$ .
Finally, expand $(2 + 3x)^{10}$ in ascending powers of $x$ up to the term in $x^2$ .

Working

Factorising the quadratic base:

9x^2 + 12x + 4 = (3x + 2)^2

Raising to the power $5$ :

(9x^2 + 12x + 4)^5 = \left((3x + 2)^2\right)^5 = (3x + 2)^{10} = (2 + 3x)^{10}

Expanding $(2 + 3x)^{10}$ in ascending powers of $x$ :

(2 + 3x)^{10} = 2^{10} + \binom{10}{1} 2^9 (3x) + \binom{10}{2} 2^8 (3x)^2 + \ldots

Calculating each term:

2^{10} = 1024

\binom{10}{1} 2^9 (3x) = 10 \times 512 \times 3x = 15\,360x

\binom{10}{2} 2^8 (3x)^2 = 45 \times 256 \times 9x^2 = 103\,680x^2

Combining the first three terms:

1024 + 15\,360x + 103\,680x^2

Answer

1024 + 15\,360x + 103\,680x^2

Final answer

1024 + 15360x + 103680x^2

Detailed explanation

Walkthrough

Factorise the base: We are instructed to use an appropriate quadratic factorisation. Notice that $9x^2 + 12x + 4$ is a perfect square trinomial, $(3x + 2)^2$ .
Simplify using index laws: Raising $(3x + 2)^2$ to the power $5$ gives $((3x + 2)^2)^5 = (3x + 2)^{10} = (2 + 3x)^{10}$ .
Apply the binomial theorem: We write the binomial expansion in ascending powers of $x$ :

(a + b)^n = a^n + \binom{n}{1}a^{n-1}b + \binom{n}{2}a^{n-2}b^2 + \ldots

Here, $a = 2$ , $b = 3x$ , and $n = 10$ .
4. Evaluate each term:

Constant term: $2^{10} = 1024$
Linear term: $10 \times 2^9 \times (3x) = 10 \times 512 \times 3x = 15\,360x$
Quadratic term: $\frac{10 \times 9}{2} \times 2^8 \times (3x)^2 = 45 \times 256 \times 9x^2 = 103\,680x^2$

State the simplified sum: Add these three terms together to get $1024 + 15\,360x + 103\,680x^2$ .

Key Takeaways

Always look for algebraic simplifications before expanding high powers of polynomials.
Perfect squares of the form $(ax + b)^2 = a^2x^2 + 2abx + b^2$ simplify trinomial powers into standard binomial expansions.
Remember that ascending powers means starting from the constant term ( $x^0$ ) and going up.

Common Mistakes

Expanding in descending powers of $x$ (e.g. starting with $x^{10}$ ), which does not answer the question's request for "ascending powers of $x$ ".
Forgetting to square the coefficient $3$ in $(3x)^2$ , writing $3x^2$ instead of $9x^2$ .
Leaving terms as an unsummed list rather than a polynomial expression.

Things to Be Careful About

Ensure powers of 2 and binomial coefficients are correctly evaluated and fully simplified without arithmetic slips.

Techniques used

recognise and factorise a perfect square quadraticapply the power of a power index lawexpand a binomial expression in ascending powers

(b)

Find the term independent of $x$ in the expansion of $\left(\frac{6}{x^2} + \frac{x^4}{2}\right)^{12}$ .

2M Similar

DifficultyMedium-Easy

Worked solution

Approach

Write down the general term in the expansion of $\left(\frac{6}{x^2} + \frac{x^4}{2}\right)^{12}$ . Determine the value of $r$ that makes the overall power of $x$ equal to zero (the term independent of $x$ ), then calculate the numerical value of that term.

Working

The general term $T_{r+1}$ in the expansion is:

T_{r+1} = \binom{12}{r} \left(6x^{-2}\right)^{12-r} \left(\frac{1}{2}x^4\right)^r

Separating coefficients and powers of $x$ :

T_{r+1} = \binom{12}{r} \times 6^{12-r} \times \left(\frac{1}{2}\right)^r \times \left(x^{-2}\right)^{12-r} \times \left(x^4\right)^r

x^{-2(12-r) + 4r} = x^{-24 + 2r + 4r} = x^{6r - 24}

For the term independent of $x$ , the power of $x$ must be $0$ :

6r - 24 = 0 \implies 6r = 24 \implies r = 4

Substitute $r = 4$ into the coefficient:

\text{Term} = \binom{12}{4} \times 6^{12-4} \times \left(\frac{1}{2}\right)^4

\binom{12}{4} = \frac{12 \times 11 \times 10 \times 9}{4 \times 3 \times 2 \times 1} = 495

6^8 = 1\,679\,616

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

\text{Term} = 495 \times 1\,679\,616 \times \frac{1}{16} = 495 \times 104\,976 = 51\,963\,120

Answer

51\,963\,120

Final answer

51963120

Detailed explanation

Walkthrough

General term formula: The general term of $(a + b)^n$ is $\binom{n}{r} a^{n-r} b^r$ . Here $n = 12$ , $a = 6x^{-2}$ , and $b = \frac{1}{2}x^4$ .
Find the power of $x$ : Multiplying the powers of $x$ gives $(x^{-2})^{12-r} \times (x^4)^r = x^{-24+2r} \times x^{4r} = x^{6r-24}$ .
Solve for $r$ : A term is "independent of $x$ " when the power of $x$ is $0$ . Setting $6r - 24 = 0$ yields $r = 4$ .
Evaluate the constant coefficient: Substitute $r = 4$ $r = 4$ into $\binom{12}{r} 6^{12-r} \left(\frac{1}{2}\right)^r$ $(r 12) 6^{12 - r} (\frac{1}{2})^{r}$ :
- $\binom{12}{4} = 495$
- $6^8 = 1\,679\,616$
- $\left(\frac{1}{2}\right)^4 = \frac{1}{16}$
- Total product: $495 \times 1\,679\,616 \div 16 = 51\,963\,120$ .

Key Takeaways

"Independent of $x$ " means the constant term ( $x^0$ ).
Finding the general term in index form and solving for the power of $x$ avoids expanding unnecessary terms.

Common Mistakes

Forgetting to apply the exponent to the constants in each term (e.g. forgetting $6^8$ or $\left(\frac{1}{2}\right)^4$ ).
Sign errors when working with negative powers, such as writing $(-2)(12-r) = -24 - 2r$ instead of $-24 + 2r$ .

Things to Be Careful About

Ensure the final answer is a single integer without any $x$ attached.

Techniques used

formulate the general term of a binomial expansionsolve for the index that gives a power of zero in xevaluate the constant coefficient

Q7MediumFunctions

(a)

The function $\mathrm{f}$ is defined by $\mathrm{f}(x) = 2\mathrm{e}^{-x} + 3$ for $x \in \mathbb{R}$ .

On the axes, sketch the graph of $y = \mathrm{f}(x)$ and hence, on the same axes, sketch the graph of $y = \mathrm{f}^{-1}(x)$ .
Show clearly

the positions of any points where your graphs meet the coordinate axes
the positions of any asymptotes.

Fig. 2

4M Similar

DifficultyMedium-Easy

Worked solution

Approach

The function $\mathrm{f}(x) = 2\mathrm{e}^{-x} + 3$ is an exponential decay curve: as $x$ increases, $2\mathrm{e}^{-x}$ decreases towards $0$ , so the curve falls towards the horizontal asymptote $y = 3$ . Its inverse is obtained by reflecting this graph in the line $y = x$ , which swaps every point $(a, b)$ for $(b, a)$ — so the $y$ -intercept becomes an $x$ -intercept and the horizontal asymptote becomes a vertical one.

Working

For $y = \mathrm{f}(x)$ :

At $x = 0$ : $\mathrm{f}(0) = 2\mathrm{e}^{0} + 3 = 5$ , so the curve crosses the $y$ -axis at $(0, 5)$ .
As $x \to \infty$ , $2\mathrm{e}^{-x} \to 0$ , so there is a horizontal asymptote at $y = 3$ .
The curve is decreasing, starting high on the left (in the second quadrant it rises steeply) and flattening towards $y = 3$ in the first quadrant.

For $y = \mathrm{f}^{-1}(x)$ , reflect in $y = x$ :

The point $(0, 5)$ reflects to $(5, 0)$ , so the inverse crosses the $x$ -axis at $(5, 0)$ .
The asymptote $y = 3$ reflects to a vertical asymptote at $x = 3$ .
The reflected curve rises steeply near $x = 3$ and then flattens out to the right.

Answer

A decreasing exponential curve through $(0, 5)$ approaching the horizontal asymptote $y = 3$ , together with its reflection in $y = x$ : a curve through $(5, 0)$ approaching the vertical asymptote $x = 3$ . The line $y = x$ itself need not be drawn.

Final answer

Decreasing exponential curve through (0, 5) with horizontal asymptote y = 3, and its reflection in y = x through (5, 0) with vertical asymptote x = 3

Detailed explanation

Walkthrough

The equation $\mathrm{f}(x) = 2\mathrm{e}^{-x} + 3$ tells us three things immediately. First, because $\mathrm{e}^{-x}$ decreases as $x$ increases, the whole curve is decreasing. Second, substituting $x = 0$ gives $\mathrm{f}(0) = 2(1) + 3 = 5$ , so the curve meets the $y$ -axis at $(0, 5)$ — the mark scheme awards a specific mark for showing this point. Third, since $\mathrm{e}^{-x} \to 0$ as $x \to \infty$ , the term $2\mathrm{e}^{-x}$ vanishes and the curve approaches the line $y = 3$ ; this horizontal asymptote earns another mark.

An inverse function's graph is always the mirror image of the original in the line $y = x$ , because swapping input and output swaps coordinates. So instead of computing $\mathrm{f}^{-1}$ explicitly, we simply reflect: the intercept $(0, 5)$ becomes $(5, 0)$ , and the horizontal asymptote $y = 3$ becomes the vertical asymptote $x = 3$ . The mark scheme allows the reflection of your curve, so even if your first sketch is slightly imperfect, a faithful reflection still scores that final mark.

Key Takeaways

An exponential of the form $a\mathrm{e}^{-x} + c$ has horizontal asymptote $y = c$ and $y$ -intercept $(0, a + c)$ .
The graph of $\mathrm{f}^{-1}$ is the reflection of the graph of $\mathrm{f}$ in the line $y = x$ ; intercepts and asymptotes swap axes under this reflection.
Sketches are marked on features: correct shape, labelled intercepts, labelled asymptotes.

Common Mistakes

Drawing an increasing exponential ( $\mathrm{e}^{+x}$ shape) instead of the decay shape demanded by $\mathrm{e}^{-x}$ — the B1 for shape is lost.
Omitting or mislabelling the intercept $(0, 5)$ , e.g. writing $3$ or $2$ instead of $5$ .
Forgetting to state the asymptote $y = 3$ , or drawing the curve crossing it.
Reflecting incorrectly, e.g. reflecting in the $x$ -axis or $y$ -axis instead of the line $y = x$ ; the scheme gives a maximum of B3 if the pair of graphs is not fully correct.
Not labelling which curve is which.

Things to Be Careful About

Both curves must be on the same axes, each clearly identified.
The asymptote must be shown as a dashed line or clearly indicated, and the curve must visibly tend towards it without crossing.
The reflection mark is follow-through on your own curve, but the intercept and asymptote marks require the correct shape to have been attempted first.
The line $y = x$ does not have to be drawn, though drawing it lightly can help you reflect accurately.

Techniques used

identify the shape of an exponential decay curve from its equationfind the y-intercept by substituting x = 0identify the horizontal asymptote of freflect a graph in the line y = x to obtain the inverse graph

(b)

The function $\mathrm{g}$ is defined by $\mathrm{g}(x) = 2 - \frac{3}{\mathrm{e}^x + 2}$ for $x \geq 0$ .

Given that $\mathrm{g}^{-1}$ exists, find an expression for $\mathrm{g}^{-1}(x)$ and state its domain.

4M Similar

DifficultyMedium

Worked solution

Approach

Write $y = \mathrm{g}(x)$ , rearrange to make $\mathrm{e}^x$ the subject, take natural logarithms to get $x$ in terms of $y$ , then swap variables. The domain of $\mathrm{g}^{-1}$ is the range of $\mathrm{g}$ over $x \geq 0$ .

Working

Let $y = 2 - \dfrac{3}{\mathrm{e}^x + 2}$ . Rearranging:

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

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

\mathrm{e}^x = \frac{3}{2 - y} - 2 = \frac{3 - 2(2 - y)}{2 - y} = \frac{2y - 1}{2 - y}

Taking natural logarithms and swapping variables:

\mathrm{g}^{-1}(x) = \ln\left(\frac{3}{2 - x} - 2\right)

Domain: $\mathrm{g}$ is increasing, with $\mathrm{g}(0) = 2 - \dfrac{3}{1 + 2} = 1$ , and as $x \to \infty$ , $\dfrac{3}{\mathrm{e}^x + 2} \to 0$ so $\mathrm{g}(x) \to 2$ (never reaching it). Hence the range of $\mathrm{g}$ is $1 \leq \mathrm{g}(x) < 2$ , so:

\text{domain of } \mathrm{g}^{-1}: \quad 1 \leq x < 2

Answer

\mathrm{g}^{-1}(x) = \ln\left(\frac{3}{2 - x} - 2\right), \quad \text{with domain } 1 \leq x < 2

Final answer

g^-1(x) = ln(3/(2-x) - 2), domain 1 <= x < 2

Detailed explanation

Walkthrough

To find an inverse we start from $y = \mathrm{g}(x)$ and solve for $x$ . The obstacle is that $x$ sits inside an exponential, so we isolate $\mathrm{e}^x$ step by step: move the fraction across, invert both sides, subtract the $2$ , and combine into a single fraction to get $\mathrm{e}^x = \frac{2y-1}{2-y}$ . Taking natural logarithms of both sides then frees $x$ , giving $x = \ln\left(\frac{3}{2-y} - 2\right)$ ; swapping the variable names yields $\mathrm{g}^{-1}(x)$ .

The domain of $\mathrm{g}^{-1}$ equals the range of $\mathrm{g}$ . Since $\mathrm{g}$ is defined only for $x \geq 0$ , evaluate at the endpoint: $\mathrm{g}(0) = 2 - \frac{3}{3} = 1$ . As $x$ grows, $\mathrm{e}^x$ grows, so $\frac{3}{\mathrm{e}^x+2}$ shrinks towards $0$ and $\mathrm{g}(x)$ climbs towards $2$ without ever reaching it. So the range is $1 \leq \mathrm{g}(x) < 2$ , giving the domain $1 \leq x < 2$ .

Key Takeaways

To invert a function containing $\mathrm{e}^x$ , isolate the exponential and take $\ln$ of both sides.
Domain of the inverse = range of the original function; check endpoints of the restricted domain carefully, noting whether each end is included.
A fraction like $\frac{3}{2-x} - 2$ can be combined over a common denominator as an equivalent form.

Common Mistakes

Sign slips when rearranging: e.g. writing $\mathrm{e}^x = \frac{3}{2+y} - 2$ or losing the minus sign on the $2$ .
Forgetting to swap variables after solving for $x$ in terms of $y$ (the scheme accepts either route provided the swap happens).
Giving the domain as $x > 1$ or $x \leq 2$ — the endpoint $x = 1$ is included (since $\mathrm{g}(0) = 1$ ) and $x = 2$ is excluded (asymptote); the scheme awards B2 only for $1 \leq x < 2$ or equivalent, B1 for a partially correct bound such as $x \geq 1$ alone or $x < 2$ alone.
Stating the domain of $\mathrm{g}$ ( $x \geq 0$ ) instead of the domain of $\mathrm{g}^{-1}$ .
Simplifying $\ln\left(\frac{3}{2-x}-2\right)$ incorrectly afterwards — later mis-simplification is ignored (isw), but the printed form must be reached correctly first.

Things to Be Careful About

The full answer needs all three elements: the rearranged exponential form (M1), the correct logarithmic expression (A1), and the complete domain $1 \leq x < 2$ (B2). An incomplete domain caps at B1.
Equivalent forms of the answer are accepted (oe): $\ln\left(\frac{2x-1}{2-x}\right)$ or $\ln\left(\frac{1-2x}{x-2}\right)$ are fine.
Check the direction of each inequality: $\mathrm{g}(0) = 1$ exactly, so $1$ is included; $2$ is approached but never attained, so it is excluded.

Techniques used

rearrange an equation to make the exponential the subjecttake natural logarithms to isolate the inverse expressionderive the domain of the inverse from the range of g

Q8Medium-HardTrigonometry

(a)

Solve the equation $(2 - 3\cot x)\cos x = 0$ for $0 < x \leq \frac{\pi}{2}$ .

3M Similar

DifficultyMedium-Easy

Worked solution

Approach

The product $(2 - 3\cot x)\cos x$ is zero when either factor is zero. Solve each within $0 < x \leq \frac{\pi}{2}$ .

Working

Either factor may be zero:

\cos x = 0 \quad \text{or} \quad 2 - 3\cot x = 0

From $\cos x = 0$ , with $0 < x \leq \frac{\pi}{2}$ :

x = \frac{\pi}{2}

From $2 - 3\cot x = 0$ :

\cot x = \frac{2}{3} \quad \Rightarrow \quad \tan x = \frac{3}{2}

x = \tan^{-1}\left(\frac{3}{2}\right) = 0.98279\ldots \approx 0.983

Both values lie in the given interval and there are no others.

Answer

x = \frac{\pi}{2} \quad \text{or} \quad x = 0.983

Final answer

x = pi/2 or x = 0.983

Detailed explanation

Walkthrough

The equation is a product of two factors set equal to zero, so at least one factor must vanish — this is the same logic as solving $(x-1)(x+2)=0$ . Setting $\cos x = 0$ gives $x = \frac{\pi}{2}$ as the only solution in the interval (note that $x = 0$ is excluded because the interval is strict at the lower end). Setting the second factor to zero gives $\cot x = \frac{2}{3}$ ; since $\cot x = \frac{1}{\tan x}$ , this means $\tan x = \frac{3}{2}$ , and taking the inverse tangent gives $x \approx 0.983$ radians. Both solutions lie inside $0 < x \leq \frac{\pi}{2}$ , so both are reported.

Key Takeaways

A product equals zero exactly when one of its factors equals zero.
$\cot x = \frac{1}{\tan x}$ converts a cotangent equation into a tangent equation.
Radian-mode working is essential throughout this question.

Common Mistakes

Dividing through by $\cos x$ instead of treating the product as zero — this loses the solution $x = \frac{\pi}{2}$ entirely.
Including $x = 0$ from $\cos x = 0$ : the interval is $0 < x$ , so it is excluded.
Working in degrees instead of radians.
Inverting $\cot x = \frac{2}{3}$ incorrectly (writing $\tan x = \frac{2}{3}$ ).
The mark scheme penalises extra solutions in range (A2 only with none extra).

Things to Be Careful About

Both solutions are required; giving only one scores A1 at most.
Give $x = \frac{\pi}{2}$ exactly and $x = 0.983$ to 3 significant figures.
Check every candidate solution lies in $0 < x \leq \frac{\pi}{2}$ before reporting it.

Techniques used

solve a product of factors equal to zero by setting each factor to zeroconvert cot x to tan xuse inverse tangent in radians

(b)

Solve the equation $2\operatorname{cosec}(2\theta + 1) - 12\sin(2\theta + 1) = 5$ , where $\theta$ is in radians and $-1 \leq \theta \leq 2$ .

6M Similar

DifficultyMedium-Hard

Worked solution

Approach

Write everything in terms of $\sin(2\theta + 1)$ , solve the resulting quadratic, then find all angles $u = 2\theta + 1$ in the range corresponding to $-1 \leq \theta \leq 2$ , i.e. $-1 \leq u \leq 5$ , and finally convert back to $\theta$ .

Working

Since $\operatorname{cosec} u = \dfrac{1}{\sin u}$ with $u = 2\theta + 1$ :

\frac{2}{\sin(2\theta + 1)} - 12\sin(2\theta + 1) = 5

Multiply through by $\sin(2\theta + 1)$ :

2 - 12\sin^2(2\theta + 1) = 5\sin(2\theta + 1)

12\sin^2(2\theta + 1) + 5\sin(2\theta + 1) - 2 = 0

Factorise:

(4\sin(2\theta + 1) - 1)(3\sin(2\theta + 1) + 2) = 0

So $\sin(2\theta + 1) = \frac{1}{4}$ or $\sin(2\theta + 1) = -\frac{2}{3}$ .

The interval $-1 \leq \theta \leq 2$ corresponds to $-1 \leq 2\theta + 1 \leq 5$ .

For $\sin u = \frac{1}{4}$ :

u = \sin^{-1}\left(\frac{1}{4}\right) = 0.25268\ldots \quad \text{or} \quad u = \pi - 0.25268\ldots = 2.88891\ldots

For $\sin u = -\frac{2}{3}$ :

u = -\sin^{-1}\left(\frac{2}{3}\right) = -0.72973\ldots \quad \text{or} \quad u = \pi + 0.72973\ldots = 3.87132\ldots

All four values lie in $[-1, 5]$ . Converting back via $\theta = \dfrac{u - 1}{2}$ :

\theta = \frac{0.25268 - 1}{2} = -0.37366\ldots \approx -0.374

\theta = \frac{2.88891 - 1}{2} = 0.94446\ldots \approx 0.944

\theta = \frac{-0.72973 - 1}{2} = -0.86486\ldots \approx -0.865

\theta = \frac{3.87132 - 1}{2} = 1.43566\ldots \approx 1.44

Answer

\theta = -0.865, \quad -0.374, \quad 0.944, \quad 1.44

Final answer

theta = -0.865, -0.374, 0.944, 1.44

Detailed explanation

Walkthrough

The equation mixes $\operatorname{cosec}(2\theta+1)$ and $\sin(2\theta+1)$ . Using $\operatorname{cosec} u = \frac{1}{\sin u}$ puts everything into one trigonometric function. Multiplying through by $\sin(2\theta+1)$ turns the equation into a quadratic in $\sin(2\theta+1)$ , which factorises neatly as $(4s-1)(3s+2)=0$ , giving $\sin(2\theta+1) = \frac{1}{4}$ or $-\frac{2}{3}$ .

Next we need every angle $u = 2\theta+1$ with $-1 \leq u \leq 5$ (obtained by multiplying the $\theta$ bounds by 2 and adding 1). For $\sin u = \frac{1}{4}$ the principal value is $0.2527$ rad, and the sine graph symmetry gives the second solution $\pi - 0.2527 = 2.889$ rad. For $\sin u = -\frac{2}{3}$ the negative solution is $-0.7297$ rad, and the other solution in range is $\pi + 0.7297 = 3.871$ rad. All four lie between $-1$ and $5$ .

Finally each angle is converted back with $\theta = \frac{u-1}{2}$ , giving $-0.374$ , $0.944$ , $-0.865$ and $1.44$ respectively.

Key Takeaways

Rewrite reciprocal functions ( $\operatorname{cosec}$ , $\sec$ , $\cot$ ) in terms of the basic functions before solving.
Multiplying by $\sin u$ converts a cosec/sin mixture into a quadratic in $\sin u$ .
When the argument is a linear function like $2\theta + 1$ , transform the interval bounds first, solve for the argument, then invert the transformation.
Sine symmetry: if $�$ is a solution of $\sin u = k$ , so are $\pi - u$ and $u \pm 2\pi$ .

Common Mistakes

Solving for $\theta$ directly with $\sin^{-1}$ on $2\theta+1$ without dividing by 2 afterwards.
Missing solutions: taking only the principal value of each inverse sine instead of using the symmetry of the sine graph over the full transformed interval $[-1, 5]$ .
Including out-of-range angles such as $u = 0.2527 - 2\pi$ or $u = 2.889 + 2\pi$ , which fall outside $[-1, 5]$ .
Sign errors when rearranging $2 - 12\sin^2 u = 5\sin u$ into the standard quadratic form.
Working in degrees rather than radians.
The mark scheme awards A2 only when there are no extra solutions in range; A1 allows two correct solutions ignoring extras.

Things to Be Careful About

Four distinct answers are required; report them all to 3 significant figures: $-0.865$ , $-0.374$ , $0.944$ , $1.44$ .
Check each final $\theta$ lies in $-1 \leq \theta \leq 2$ — all four do here.
Keep calculator in radian mode throughout.
The quadratic step depends on correctly writing the equation in terms of $\sin(2\theta+1)$ alone first (M1), and the factorisation depends on that quadratic being correct.

Techniques used

rewrite cosec in terms of sinform and factorise a quadratic in sinfind all angles in an extended interval using the sine graph symmetryback-substitute through the linear argument transformation

Q9MediumCircular measureCalculus

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

Fig. 3

The diagram shows a sector $AOB$ of a circle, centre $O$ , radius 15.
Angle $AOB = \frac{6\pi}{5}$ .
The sector is made into a cone with points $A$ and $B$ touching, as shown.

(a)

Find the curved surface area of the cone.

2M Similar

(b)

The top of the cone is a horizontal circle.

Find the circumference of the circular top.

2M Similar

(c)

Hence find the radius of the circular top and the perpendicular height of the cone.

2M Similar

(d)

Water is poured into the cone.
When the depth of the water in the cone is $h$ , the radius of the circular top of the water is $r$ .

6M

(i)

Find an expression for $r$ in terms of $h$ .

1M Similar

(ii)

The water is poured into the cone at a constant rate of $27\text{ cm}^3$ per second.

Find the rate at which the depth of the water is rising when the depth of the water is 4.

5M Similar

DifficultyMedium

Worked solution

Approach

Write the water volume $V = \frac{1}{3}\pi r^2 h$ using $r = \frac{3}{4}h$ so it depends on $h$ alone, differentiate to get $\frac{\mathrm{d}V}{\mathrm{d}h}$ , then use the chain rule $\frac{\mathrm{d}h}{\mathrm{d}t} = \frac{\mathrm{d}V}{\mathrm{d}t} \div \frac{\mathrm{d}V}{\mathrm{d}h}$ with $\frac{\mathrm{d}V}{\mathrm{d}t} = 27$ and $h = 4$ .

Working

V = \frac{1}{3}\pi\left(\frac{3}{4}h\right)^2 h = \frac{1}{3}\pi \times \frac{9h^2}{16} \times h = \frac{3\pi h^3}{16}

Differentiate with respect to $h$ :

\frac{\mathrm{d}V}{\mathrm{d}h} = \frac{9\pi h^2}{16}

By the chain rule,

\frac{\mathrm{d}h}{\mathrm{d}t} = \frac{\mathrm{d}V}{\mathrm{d}t} \times \frac{\mathrm{d}h}{\mathrm{d}V} = \frac{\mathrm{d}V}{\mathrm{d}t} \div \frac{\mathrm{d}V}{\mathrm{d}h}

At $h = 4$ with $\frac{\mathrm{d}V}{\mathrm{d}t} = 27$ :

\frac{\mathrm{d}h}{\mathrm{d}t} = 27 \times \frac{16}{9\pi \times 16} = \frac{27}{9\pi} = \frac{3}{\pi} \approx 0.955

Answer

\frac{\mathrm{d}h}{\mathrm{d}t} \approx 0.955\ \text{cm s}^{-1}

Final answer

0.955 cm per second

Detailed explanation

Walkthrough

The water in the cone forms a smaller cone with radius $r = \frac{3}{4}h$ (from part (i)) and depth $h$ . Its volume is

V = \frac{1}{3}\pi r^2 h = \frac{1}{3}\pi\left(\frac{3}{4}h\right)^2 h = \frac{3\pi h^3}{16}

Writing $V$ in terms of $h$ only is the crucial move: it lets us differentiate $V$ with respect to $h$ directly:

\frac{\mathrm{d}V}{\mathrm{d}h} = \frac{9\pi h^2}{16}

The water is poured at $\frac{\mathrm{d}V}{\mathrm{d}t} = 27\,\text{cm}^3\text{s}^{-1}$ , and we want $\frac{\mathrm{d}h}{\mathrm{d}t}$ . The chain rule links them:

\frac{\mathrm{d}h}{\mathrm{d}t} = \frac{\mathrm{d}V}{\mathrm{d}t} \times \frac{\mathrm{d}h}{\mathrm{d}V} = \frac{\mathrm{d}V}{\mathrm{d}t} \div \frac{\mathrm{d}V}{\mathrm{d}h}

Substituting $\frac{\mathrm{d}V}{\mathrm{d}t} = 27$ and $h = 4$ (so $\frac{\mathrm{d}V}{\mathrm{d}h} = \frac{9\pi \times 16}{16} = 9\pi$ ):

\frac{\mathrm{d}h}{\mathrm{d}t} = 27 \times \frac{16}{9\pi \times 16} = \frac{3}{\pi} \approx 0.955\,\text{cm s}^{-1}

Key Takeaways

In related-rates problems, express the quantity in terms of a single variable before differentiating.
The chain rule connects rates: $\frac{\mathrm{d}h}{\mathrm{d}t} = \frac{\mathrm{d}V}{\mathrm{d}t} \div \frac{\mathrm{d}V}{\mathrm{d}h}$ .

Common Mistakes

Forgetting to substitute $r = \frac{3}{4}h$ , leaving $V$ in terms of two variables.
Dropping the $\frac{1}{3}$ or mis-squaring $\left(\frac{3}{4}h\right)^2$ .
Dividing the wrong way round (computing $\frac{\mathrm{d}V}{\mathrm{d}h} \div \frac{\mathrm{d}V}{\mathrm{d}t}$ ).
Substituting $h = 4$ before differentiating.

Things to Be Careful About

The mark scheme allows follow-through on the candidate's own $r = kh$ : B1 for $V = \frac{1}{3}\pi(kh)^2 h$ , B1 for $\frac{\mathrm{d}V}{\mathrm{d}h} = k^2\pi h^2$ , B1 for the correct chain rule statement, M1 for the numerical substitution, A1 for $0.955$ (or $0.9549$ ...).
The answer is a rate, so units are $\text{cm s}^{-1}$ ; 3 significant figures gives $0.955$ .

Techniques used

write the cone volume in terms of hdifferentiate V with respect to happly the chain rule to link dV/dt and dh/dtsubstitute the given depth

Q10Medium-HardCalculus

A particle $P$ moves in a straight line.
$t$ seconds after passing a fixed point, $O$ , the acceleration of $P$ , $a\,\text{ms}^{-2}$ , is given by

\begin{aligned} a &= t^2 - 2 && \text{for } 0 \leq t \leq 4 \\ a &= 19 - 5\mathrm{e}^{8-2t} && \text{for } 4 \leq t \leq 10. \end{aligned}

When $t = 3$ , the velocity of $P$ is $-\frac{1}{3}\,\text{ms}^{-1}$ and its displacement from $O$ is $-\frac{1}{4}\,\text{m}$ .

(a)

6M

(i)

Find the velocity of $P$ when $t = 4$ .

3M Similar

DifficultyMedium-Easy

Worked solution

Approach

For $0 \leq t \leq 4$ , acceleration is $a = t^2 - 2$ . Integrate $a$ with respect to $t$ to find the velocity expression $v(t)$ , then use the condition $v(3) = -\frac{1}{3}$ to determine the constant of integration. Finally, substitute $t = 4$ .

Working

v = \int (t^2 - 2)\,\mathrm{d}t = \frac{t^3}{3} - 2t + c

Using $v = -\frac{1}{3}$ when $t = 3$ :

-\frac{1}{3} = \frac{3^3}{3} - 2(3) + c

-\frac{1}{3} = 9 - 6 + c = 3 + c

c = -\frac{1}{3} - 3 = -\frac{10}{3}

Thus, the velocity function for $0 \leq t \leq 4$ is:

v = \frac{t^3}{3} - 2t - \frac{10}{3}

When $t = 4$ :

v = \frac{4^3}{3} - 2(4) - \frac{10}{3} = \frac{64}{3} - 8 - \frac{10}{3} = \frac{54}{3} - 8 = 18 - 8 = 10\,\text{ms}^{-1}

Answer

10\,\text{ms}^{-1}

Final answer

10 ms^-1

Detailed explanation

Walkthrough

To find velocity $v$ from acceleration $a$ , we integrate acceleration with respect to time $t$ .

Integrating $t^2 - 2$ yields $\frac{t^3}{3} - 2t + c$ .
Substitute $t = 3$ and $v = -\frac{1}{3}$ to find the arbitrary constant $c = -\frac{10}{3}$ .
Evaluate $v(4)$ by substituting $t = 4$ into the specific velocity function, giving $v = 10\,\text{ms}^{-1}$ .

Key Takeaways

Acceleration is the time derivative of velocity: $a = \frac{\mathrm{d}v}{\mathrm{d}t}$ , so $v = \int a\,\mathrm{d}t$ .
A boundary/initial condition allows us to solve for the constant of integration.

Common Mistakes

Forgetting the constant of integration $+ c$ .
Arithmetic slips when evaluating fractions like $\frac{64}{3} - \frac{10}{3}$ .

Things to Be Careful About

Ensure you use the correct acceleration model for the given domain $0 \leq t \leq 4$ .

Techniques used

integrate polynomial acceleration to find velocitydetermine constant of integration using initial conditionsevaluate velocity at given time

(ii)

Find the displacement of $P$ from $O$ when $t = 4$ .

3M Similar

DifficultyMedium-Easy

Worked solution

Approach

Integrate the velocity function $v(t) = \frac{t^3}{3} - 2t - \frac{10}{3}$ with respect to $t$ to find displacement $s(t)$ . Use the given boundary condition $s(3) = -\frac{1}{4}$ to find the constant of integration, then evaluate $s(4)$ .

Working

s = \int \left(\frac{t^3}{3} - 2t - \frac{10}{3}\right)\,\mathrm{d}t = \frac{t^4}{12} - t^2 - \frac{10}{3}t + d

Using $s = -\frac{1}{4}$ when $t = 3$ :

-\frac{1}{4} = \frac{3^4}{12} - 3^2 - \frac{10}{3}(3) + d

-\frac{1}{4} = \frac{81}{12} - 9 - 10 + d = \frac{27}{4} - 19 + d

d = -\frac{1}{4} - \frac{27}{4} + 19 = -\frac{28}{4} + 19 = -7 + 19 = 12

Thus, the displacement function for $0 \leq t \leq 4$ is:

s = \frac{t^4}{12} - t^2 - \frac{10}{3}t + 12

When $t = 4$ :

\begin{aligned} s &= \frac{4^4}{12} - 4^2 - \frac{10}{3}(4) + 12 \\ &= \frac{256}{12} - 16 - \frac{40}{3} + 12 \\ &= \frac{64}{3} - \frac{40}{3} - 4 \\ &= \frac{24}{3} - 4 \\ &= 8 - 4 = 4\,\text{m} \end{aligned}

Answer

4\,\text{m}

Final answer

4 m

Detailed explanation

Walkthrough

Integrate $v = \frac{t^3}{3} - 2t - \frac{10}{3}$ with respect to $t$ to get $s = \frac{t^4}{12} - t^2 - \frac{10}{3}t + d$ .
Substitute the given condition $t = 3, s = -\frac{1}{4}$ to find $d = 12$ .
Substitute $t = 4$ into the displacement formula to get the displacement at $t = 4$ , which simplifies directly to $4\,\text{m}$ .

Key Takeaways

Velocity is the derivative of displacement: $v = \frac{\mathrm{d}s}{\mathrm{d}t}$ , so $s = \int v\,\mathrm{d}t$ .
Keep fractions exact to avoid premature rounding.

Common Mistakes

Forgetting the constant of integration $d$ .
Applying the boundary condition incorrectly (e.g. assuming $s(0) = 0$ ).

Things to Be Careful About

Carry through the velocity equation from part (a)(i) accurately.

Techniques used

integrate velocity to find displacementdetermine constant of integration using given displacement conditionevaluate displacement at given time

(b)

Find the displacement of $P$ from $O$ when $t = 10$ .

5M Similar

DifficultyMedium-Hard

Worked solution

Approach

For $4 \leq t \leq 10$ , $a = 19 - 5\mathrm{e}^{8-2t}$ . Integrate $a$ to find $v(t)$ , using the continuity of velocity at $t = 4$ ( $v(4) = 10$ ) to find the integration constant. Then integrate $v(t)$ to find $s(t)$ , using the continuity of displacement at $t = 4$ ( $s(4) = 4$ ) to determine the new integration constant. Finally, evaluate $s(10)$ .

Working

v = \int \left(19 - 5\mathrm{e}^{8-2t}\right)\,\mathrm{d}t = 19t - 5\left(\frac{\mathrm{e}^{8-2t}}{-2}\right) + c = 19t + \frac{5}{2}\mathrm{e}^{8-2t} + c

Since $v = 10$ when $t = 4$ :

10 = 19(4) + \frac{5}{2}\mathrm{e}^{8-2(4)} + c

10 = 76 + \frac{5}{2}\mathrm{e}^{0} + c = 76 + \frac{5}{2} + c = \frac{157}{2} + c

c = 10 - \frac{157}{2} = -\frac{137}{2}

Thus, for $4 \leq t \leq 10$ :

v = 19t + \frac{5}{2}\mathrm{e}^{8-2t} - \frac{137}{2}

Now integrate $v$ to obtain $s$ :

\begin{aligned} s &= \int \left(19t + \frac{5}{2}\mathrm{e}^{8-2t} - \frac{137}{2}\right)\,\mathrm{d}t \\ &= \frac{19t^2}{2} + \frac{5}{2}\left(\frac{\mathrm{e}^{8-2t}}{-2}\right) - \frac{137}{2}t + d \\ &= \frac{19t^2}{2} - \frac{5}{4}\mathrm{e}^{8-2t} - \frac{137}{2}t + d \end{aligned}

Since $s = 4$ when $t = 4$ :

4 = \frac{19(4^2)}{2} - \frac{5}{4}\mathrm{e}^{8-2(4)} - \frac{137}{2}(4) + d

4 = 152 - \frac{5}{4} - 274 + d = -122 - \frac{5}{4} + d = -\frac{493}{4} + d

d = 4 + \frac{493}{4} = \frac{509}{4}

Thus, the displacement function for $4 \leq t \leq 10$ is:

s = \frac{19t^2}{2} - \frac{5}{4}\mathrm{e}^{8-2t} - \frac{137}{2}t + \frac{509}{4}

When $t = 10$ :

\begin{aligned} s &= \frac{19(10^2)}{2} - \frac{5}{4}\mathrm{e}^{8-2(10)} - \frac{137}{2}(10) + \frac{509}{4} \\ &= \frac{1900}{2} - \frac{5}{4}\mathrm{e}^{-12} - 685 + \frac{509}{4} \\ &= 950 - 685 + \frac{509}{4} - \frac{5}{4}\mathrm{e}^{-12} \\ &= 265 + 127.25 - \frac{5}{4}\mathrm{e}^{-12} \\ &= 392.25 - \frac{5}{4}\mathrm{e}^{-12} \\ &\approx 392\,\text{m} \quad (\text{or } 392.25\,\text{m}) \end{aligned}

Answer

392\,\text{m}

Final answer

392 m

Detailed explanation

Walkthrough

In the interval $4 \leq t \leq 10$ , we integrate the new acceleration function $a = 19 - 5\mathrm{e}^{8-2t}$ . Integrating an exponential term $\mathrm{e}^{kx}$ divides by the constant $k$ , here $-2$ , turning $-5\mathrm{e}^{8-2t}$ into $+\frac{5}{2}\mathrm{e}^{8-2t}$ .
The particle's velocity is continuous at $t = 4$ , so we use $v(4) = 10$ (from part (a)(i)) to find $c = -\frac{137}{2}$ .
Integrate the velocity function again to find displacement $s(t)$ . Integrating $+\frac{5}{2}\mathrm{e}^{8-2t}$ gives $-\frac{5}{4}\mathrm{e}^{8-2t}$ .
Use the displacement at $t = 4$ found in part (a)(ii) ( $s(4) = 4$ ) to find $d = \frac{509}{4}$ .
Substitute $t = 10$ into the expression. Since $\mathrm{e}^{-12} \approx 6.14 \times 10^{-6}$ is extremely small, $s(10) \approx 392.25\,\text{m}$ , which rounds to $392\,\text{m}$ to 3 significant figures.

Key Takeaways

When functions are defined piecewise in kinematics, velocity and displacement are continuous across the boundaries, allowing the values at the end of one interval to act as initial conditions for the next.
$\int \mathrm{e}^{ax+b}\,\mathrm{d}x = \frac{1}{a}\mathrm{e}^{ax+b} + C$ .

Common Mistakes

Sign errors when integrating $\mathrm{e}^{8-2t}$ (forgetting to divide by $-2$ ).
Forgetting that $\mathrm{e}^0 = 1$ when substituting $t = 4$ .
Using $s(4) = 0$ or $v(4) = 0$ instead of the values found in part (a).

Things to Be Careful About

Ensure you do not drop the exponential term when writing the exact formula, even though its value at $t = 10$ is negligible numerically.

Techniques used

integrate exponential and linear accelerationmatch boundary conditions at section boundaryintegrate velocity to find displacement across piecewise domainevaluate displacement at specific time

Q11MediumSeries

A geometric progression has first term $a$ and common ratio $r$ , where $r > 0$ .
The sum of the 2nd and 3rd terms of the progression is 168.
The sum of the 4th and 5th terms of the progression is 94.5.

(a)

Find the 6th term of the progression.

7M Similar

DifficultyMedium

Worked solution

Approach

Write the 2nd and 3rd terms as $ar$ and $ar^2$ , and the 4th and 5th terms as $ar^3$ and $ar^4$ . The two given sums give two equations, and dividing them eliminates $a$ to give $r^2$ .

Working

ar + ar^2 = 168

ar^3 + ar^4 = 94.5

Divide the second equation by the first:

\frac{ar^3(1 + r)}{ar(1 + r)} = \frac{94.5}{168}

r^2 = \frac{94.5}{168} = 0.5625

Since $r > 0$ :

r = \frac{3}{4}

Substitute back into the first equation:

\frac{3}{4}a + \frac{9}{16}a = 168

\frac{21}{16}a = 168 \quad\Rightarrow\quad a = 168 \times \frac{16}{21} = 128

The 6th term is:

u_6 = ar^5 = 128 \times \left(\frac{3}{4}\right)^5 = 128 \times \frac{243}{1024} = \frac{243}{8}

Answer

u_6 = \frac{243}{8}

Final answer

243/8

Detailed explanation

Walkthrough

The terms of a geometric progression are $a, ar, ar^2, ar^3, ar^4, \ldots$ , so the 2nd and 3rd terms are $ar$ and $ar^2$ , and the 4th and 5th are $ar^3$ and $ar^4$ . Translating the two given sums into equations gives $ar + ar^2 = 168$ and $ar^3 + ar^4 = 94.5$ .

Each equation contains two unknowns, $a$ and $r$ , so we eliminate one. Both equations factor nicely: $ar(1 + r) = 168$ and $ar^3(1 + r) = 94.5$ . Dividing the second by the first cancels both $a$ and $(1 + r)$ , leaving $r^2 = \frac{94.5}{168} = 0.5625$ . Since the question states $r > 0$ , we take only the positive root $r = \frac{3}{4}$ .

Substituting $r = \frac{3}{4}$ back into the first equation gives $\frac{3}{4}a + \frac{9}{16}a = \frac{21}{16}a = 168$ , so $a = 168 \times \frac{16}{21} = 128$ .

Finally the 6th term is $ar^5 = 128 \times \left(\frac{3}{4}\right)^5 = \frac{243}{8}$ .

Key Takeaways

The $n$ th term of a GP is $ar^{n-1}$ ; sums of specific terms translate directly into equations in $a$ and $r$ .
Dividing two GP equations eliminates $a$ cleanly because both share the factor $a(1+r)$ in this structure.
The condition $r > 0$ tells you to discard the negative square root.

Common Mistakes

Using $ar^n$ instead of $ar^{n-1}$ for the $n$ th term, giving wrong powers of $r$ .
Taking $r = -0.75$ as well; the condition $r > 0$ forbids it, and the mark scheme says $r = \frac{3}{4}$ only.
Arithmetic slips in $\frac{94.5}{168}$ — it is exactly $0.5625 = \frac{9}{16}$ .
Giving a decimal for $u_6$ instead of the exact $\frac{243}{8}$ ; the scheme accepts equivalent exact forms (isw after a correct exact answer).

Things to Be Careful About

The answer $r = \frac{3}{4}$ is marked 'only, nfww' — it must come from correct working, not guessed and fitted.
Keep $a = 128$ exact; it feeds part (b).
$u_6 = \frac{243}{8} = 30.375$ ; the exact fraction is the preferred printed form.

Techniques used

form equations from the sums of pairs of termsdivide the equations to eliminate the first termsolve for the common ratiosubstitute back to find the first termevaluate the sixth term

(b)

Find the sum to infinity of the progression.

1M Similar

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