(a) Differentiate $y = \sqrt{x + 1} + \cos(5x + 3)$ with respect to $x$. (b) Find $\int \sec^2 \frac{x}{3}\,\mathrm{d}x$.

(a) dy/dx = (1/2)(x + 1)^(-1/2) - 5 sin(5x + 3) · (b) 3 tan(x/3) + c

4037/22 May/June 2026 — Additional Mathematics Solved Past Paper

Q1Medium-EasyQuadratic functionsStraight-line graphs

The point $A$ is the minimum point on the curve $y = x^2 - 6x + 10$ .

(a)

Using the method of completing the square, find the coordinates of $A$ .

2M Similar

DifficultyMedium-Easy

Worked solution

Approach

Complete the square on $y = x^2 - 6x + 10$ ; the vertex form immediately gives the minimum point $A$ .

Working

y = x^2 - 6x + 10 = \left(x - 3\right)^2 - 9 + 10 = (x - 3)^2 + 1

Since $(x - 3)^2 \geq 0$ , the least value of $y$ is $1$ , occurring when $x = 3$ .

Answer

A = (3,\ 1)

Final answer

(3, 1)

Detailed explanation

Walkthrough

To complete the square, take half of the coefficient of $x$ (which is $-6$ ), giving $-3$ , and write $x^2 - 6x$ as $(x - 3)^2 - 9$ . Then add the constant $10$ to get $(x - 3)^2 + 1$ .

Because a squared quantity is never negative, $(x - 3)^2$ is smallest (zero) when $x = 3$ , so the smallest value of $y$ is $0 + 1 = 1$ . Hence the minimum point is $A(3, 1)$ .

Key Takeaways

Vertex form $y = (x - h)^2 + k$ reveals the minimum point $(h, k)$ directly when the coefficient of $x^2$ is positive.
Half the $x$ -coefficient gives the shift; the constant is adjusted to compensate for the square.

Common Mistakes

Writing $(x - 3)^2 + 10$ and forgetting to subtract $3^2 = 9$ .
Using $(x + 3)^2$ , which gives the wrong vertex $(-3, \ldots)$ .
Setting $y = 0$ and solving, which finds $x$ -intercepts, not the minimum — the mark scheme only allows this route if the completed-square form $(x-3)^2 + 1$ is stated.

Things to Be Careful About

The mark scheme requires the completing-the-square method specifically; the vertex must not be found by differentiation here.
The scheme condones $(x - 3)^2 - 3^2 + 10$ followed by the correct answer, and accepts $x = 3$ , $y = 1$ as the answer.
A minor slip in the completed square still earns SC1 if the correct coordinates $(3, 1)$ result.

Techniques used

complete the square on a quadraticread off the minimum point from vertex form

(b)

The point $B$ has coordinates $(6, 7)$ .
The curve cuts the $y$ -axis at the point $C$ .

Find the equation of the line through $B$ parallel to the line $AC$ .

2M Similar

DifficultyMedium-Easy

Worked solution

Approach

The curve cuts the $y$ -axis where $x = 0$ , giving $C(0, 10)$ . Find the gradient of $AC$ using $A(3, 1)$ from part (a); the line through $B$ parallel to $AC$ has the same gradient.

Working

C = (0,\ 10)

\text{gradient of } AC = \frac{10 - 1}{0 - 3} = \frac{9}{-3} = -3

Using the point $B(6, 7)$ with gradient $-3$ :

y - 7 = -3(x - 6)

y = -3x + 18 + 7 = -3x + 25

Answer

y = -3x + 25

Final answer

y = -3x + 25

Detailed explanation

Walkthrough

First identify $C$ : the curve $y = x^2 - 6x + 10$ meets the $y$ -axis where $x = 0$ , so $C = (0, 10)$ .

Carrying forward $A(3, 1)$ from part (a), the gradient of $AC$ is

\frac{10 - 1}{0 - 3} = \frac{9}{-3} = -3.

Parallel lines share the same gradient, so the required line through $B(6, 7)$ has gradient $-3$ . Using the point–slope form $y - y_1 = m(x - x_1)$ :

y - 7 = -3(x - 6) \quad\Rightarrow\quad y = -3x + 25.

Key Takeaways

The $y$ -intercept of a curve is found by substituting $x = 0$ .
Parallel lines have equal gradients; only the intercept changes.
Point–slope form builds a line equation from one point and a gradient.

Common Mistakes

Using the perpendicular gradient $+\frac{1}{3}$ instead of the parallel gradient — the mark scheme awards M0 for this.
Sign slips in $\frac{10 - 1}{0 - 3}$ ; the denominator is $0 - 3 = -3$ , not $3$ .
Leaving an incomplete equation such as $y = -3x + c$ with $c = 25$ stated separately — a complete equation is required.

Things to Be Careful About

The mark scheme allows follow-through on the candidate's $A$ from part (a), but the gradient calculation must be shown explicitly.
Equivalent forms such as $y = 25 - 3x$ or $y - 7 = -3(x - 6)$ are accepted; later mis-simplification is ignored (isw).
The answer must be a complete equation in a standard form.

Techniques used

find the gradient between two pointsuse the gradient of a parallel lineform the equation of a line from a point and gradient

Q23MMedium-EasyQuadratic functions

Show that there is no real value of $k$ for which the equation

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

has two equal roots.

DifficultyMedium-Easy

Worked solution

Approach

A quadratic has two equal roots when its discriminant is zero. Compute the discriminant of $(k - 2)x^2 + (2k + 3)x + 3 = 0$ and show it can never be $0$ .

Working

For real roots, the discriminant condition is

b^2 - 4ac = (2k + 3)^2 - 4(k - 2)(3)

Expanding:

(2k + 3)^2 - 12(k - 2) = 4k^2 + 12k + 9 - 12k + 24 = 4k^2 + 33

For two equal roots we would need

4k^2 + 33 = 0 \quad\Rightarrow\quad k^2 = -\frac{33}{4}

But $4k^2 \geq 0$ for all real $k$ , so $4k^2 + 33 > 0$ always: the equation $4k^2 + 33 = 0$ is not possible for any real value of $k$ .

Hence there is no real value of $k$ for which the equation has two equal roots.

Answer

The discriminant simplifies to $4k^2 + 33$ , which is strictly positive for every real $k$ , so it can never equal $0$ ; therefore no real value of $k$ gives two equal roots.

Final answer

No real value of k exists, since the discriminant 4k^2 + 33 > 0 for all real k and can never be 0

Detailed explanation

Walkthrough

Two equal roots means the quadratic touches the x-axis at exactly one point, which happens precisely when the discriminant $b^2 - 4ac$ equals zero. Here $a = k - 2$ , $b = 2k + 3$ and $c = 3$ , so we form $(2k+3)^2 - 4(k-2)(3)$ . Expanding gives $4k^2 + 12k + 9 - 12k + 24 = 4k^2 + 33$ . Setting this equal to zero would require $k^2 = -\frac{33}{4}$ , but a square of a real number is never negative — indeed $4k^2 + 33$ is at least $33$ . So the discriminant can never be zero, and no real $k$ produces two equal roots.

Key Takeaways

Equal roots ⟺ discriminant $b^2 - 4ac = 0$ .
The discriminant may itself depend on a parameter; analysing whether it can reach $0$ answers existence questions.
A quadratic expression like $4k^2 + 33$ that is a positive multiple of a square plus a positive constant is always strictly positive.

Common Mistakes

Sign slips expanding $(2k+3)^2$ or distributing $-12(k-2)$ — the scheme allows $(2k-3)^2$ as a miscopy but not an incorrect final expression in $k$ (no follow-through).
Forgetting to state or clearly imply that the discriminant must equal $0$ — the ' $= 0$ ' step must be seen or implied.
Concluding without justification: you must argue why $4k^2 + 33 = 0$ is impossible (e.g. $k^2 = -\frac{33}{4}$ has no real solution).
The argument must not be contradicted anywhere in the working.

Things to Be Careful About

This is a 'show that' question: derive forward from the discriminant to the conclusion, never start from the answer.
Keep the exact form $4k^2 + 33$ ; the scheme accepts equivalents such as $k^2 = -\frac{33}{4}$ as the final argument.
Using an inequality sign earlier (e.g. noting the discriminant is $> 0$ ) is fine provided the ' $= 0$ ' requirement is clearly implied in the final argument.

Techniques used

state the equal-roots discriminant conditionexpand the discriminant in terms of kargue that a positive expression cannot equal zero

Similar questions

Q3Medium-EasyCalculus

(a)

Differentiate $y = \sqrt{x + 1} + \cos(5x + 3)$ with respect to $x$ .

3M Similar

DifficultyMedium-Easy

Worked solution

Approach

Differentiate each term separately using the chain rule: rewrite the square root as a power, and remember that the derivative of $\cos(5x+3)$ picks up the factor $5$ from the inner derivative. Combine both derivatives in a single expression.

Working

Write the first term as a power:

\sqrt{x + 1} = (x + 1)^{\frac{1}{2}}

Differentiate using the power rule with the chain rule (inner derivative is $1$ ):

\frac{\mathrm{d}}{\mathrm{d}x}(x + 1)^{\frac{1}{2}} = \frac{1}{2}(x + 1)^{-\frac{1}{2}}

Differentiate the second term by the chain rule, multiplying by the derivative of the inner function $5x + 3$ :

\frac{\mathrm{d}}{\mathrm{d}x}\cos(5x + 3) = -5\sin(5x + 3)

Combine into one expression:

\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{1}{2}(x + 1)^{-\frac{1}{2}} - 5\sin(5x + 3)

Answer

\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{1}{2}(x + 1)^{-\frac{1}{2}} - 5\sin(5x + 3)

Final answer

dy/dx = (1/2)(x + 1)^(-1/2) - 5 sin(5x + 3)

Detailed explanation

Walkthrough

The question asks for the derivative of a sum of two terms, so we differentiate each term on its own and add the results.

First term: $y = \sqrt{x+1}$ . A square root is a power, so $(x+1)^{\frac{1}{2}}$ . The power rule gives $\frac{1}{2}(x+1)^{-\frac{1}{2}}$ ; because the inner function $x+1$ has derivative $1$ , no extra factor appears. This can also be written as $\frac{1}{2\sqrt{x+1}}$ — both forms are accepted ('oe').

Second term: $y = \cos(5x+3)$ . By the chain rule we differentiate the outer cosine to get $-\sin(5x+3)$ , then multiply by the derivative of the inside, which is $5$ . So this term contributes $-5\sin(5x+3)$ .

Adding them gives the final single expression. The mark scheme insists the two derivatives appear together as one expression for full marks.

Key Takeaways

Rewrite roots as fractional powers before differentiating.
The chain rule: differentiate the outer function, then multiply by the derivative of the inner function ( $5$ here).
A derivative of a sum is the sum of the derivatives, presented as one expression.

Common Mistakes

Forgetting the factor $5$ from the chain rule and writing just $-\sin(5x+3)$ (earns only partial credit).
Writing $-5\sin 5x + 3$ instead of $-5\sin(5x+3)$ — misplacing the bracket loses the mark.
Adding $+c$ : a constant of integration does not belong in a differentiation answer and destroys a mark here.
Leaving the two derivatives as separate statements rather than one combined expression caps the score at 2 marks.

Things to Be Careful About

The answer must be a single expression combining both terms ('maximum of 2 marks if not in a single expression').
'isw' applies: later incorrect simplification is ignored once the correct form is seen.
Either index form $\frac{1}{2}(x+1)^{-\frac{1}{2}}$ or surd form $\frac{1}{2\sqrt{x+1}}$ is acceptable ('oe').

Techniques used

differentiate a root function using the power rule with a negative fractional indexapply the chain rule to differentiate cos of a linear function of x

(b)

Find $\int \sec^2 \frac{x}{3}\,\mathrm{d}x$ .

3M Similar

DifficultyMedium-Easy

Worked solution

Approach

Use the standard result $\int \sec^2 u \,\mathrm{d}u = \tan u$ , adjusting for the inner function $\frac{x}{3}$ whose derivative is $\frac{1}{3}$ , so divide by $\frac{1}{3}$ (i.e. multiply by $3$ ). Include the constant of integration.

Working

With $u = \frac{x}{3}$ , we have $\frac{\mathrm{d}u}{\mathrm{d}x} = \frac{1}{3}$ , so:

\int \sec^2 \frac{x}{3}\,\mathrm{d}x = \frac{1}{\frac{1}{3}} \tan \frac{x}{3} + c

= 3\tan \frac{x}{3} + c

Check by differentiating: $\frac{\mathrm{d}}{\mathrm{d}x}\left(3\tan \frac{x}{3}\right) = 3 \cdot \sec^2 \frac{x}{3} \cdot \frac{1}{3} = \sec^2 \frac{x}{3}$ , as required.

Answer

3\tan \frac{x}{3} + c

Final answer

3 tan(x/3) + c

Detailed explanation

Walkthrough

This uses the standard derivative pair $\frac{\mathrm{d}}{\mathrm{d}x}\tan u = \sec^2 u \cdot \frac{\mathrm{d}u}{\mathrm{d}x}$ , so reversing it, $\int \sec^2 u\,\mathrm{d}u = \tan u + c$ .

Here the argument is $\frac{x}{3}$ rather than plain $x$ . When integrating a function of $ax$ , we divide by the coefficient $a$ : since $a = \frac{1}{3}$ , dividing by it means multiplying by $3$ . Hence the answer is $3\tan\frac{x}{3} + c$ .

Because this is an indefinite integral, the constant of integration $c$ must be included — the mark scheme awards an accuracy mark specifically for it.

A quick check: differentiating $3\tan\frac{x}{3}$ gives $3 \times \sec^2\frac{x}{3} \times \frac{1}{3} = \sec^2\frac{x}{3}$ , confirming the answer.

Key Takeaways

$\int \sec^2(ax)\,\mathrm{d}x = \frac{1}{a}\tan(ax) + c$ .
Always include $+c$ for an indefinite integral.
Verify integrals by differentiating the answer.

Common Mistakes

Omitting the $+c$ , losing the accuracy mark.
Writing $\tan\frac{x}{3} + c$ without the factor $3$ (ignoring the chain rule in reverse).
Writing $\frac{1}{3}\tan\frac{x}{3} + c$ — dividing instead of multiplying by the coefficient.

Things to Be Careful About

'isw' applies: later slips are ignored once $3\tan\frac{x}{3} + c$ is seen.
The scheme accepts $\frac{1}{\frac{1}{3}}\tan\frac{x}{3} + c$ as equivalent to $3\tan\frac{x}{3} + c$ .
An indefinite integral always needs the constant of integration.

Techniques used

recognise sec squared as the standard integral of tandivide by the coefficient of x from the chain rule when integrating

Q4MediumEquations, inequalities and graphs

(a)

On the axes, sketch the graph of $y = |x^2 - 5|$ .
State the intercepts with the coordinate axes.

Fig. 1

3M Similar

DifficultyMedium-Easy

Worked solution

Approach

Start from the parabola $y = x^2 - 5$ , which crosses the $x$ -axis at $x = \pm\sqrt{5}$ and has its minimum $(0, -5)$ . Taking the modulus reflects every point with negative $y$ -value above the axis, so the middle section becomes an inverted arch between the two roots.

Working

The underlying parabola $y = x^2 - 5$ has:

$y$ -intercept: $x = 0 \Rightarrow y = -5$ ;
$x$ -intercepts: $x^2 = 5 \Rightarrow x = \pm\sqrt{5}$ .

Applying the modulus, $y = |x^2 - 5|$ :

the points at $x = \pm\sqrt{5}$ stay on the axis — these become cusps;
the minimum $(0, -5)$ is reflected to $(0, 5)$ , giving a local maximum of $5$ on the $y$ -axis;
outside $[-\sqrt{5}, \sqrt{5}]$ the parabola is already positive, so the outer branches curve upwards unchanged.

Intercepts with the axes:

y\text{-intercept: } (0, 5), \qquad x\text{-intercepts: } (-\sqrt{5}, 0) \text{ and } (\sqrt{5}, 0)

Answer

A W-shaped smooth curve with cusps at $(-\sqrt{5}, 0)$ and $(\sqrt{5}, 0)$ , a local maximum $5$ on the $y$ -axis, and upward-curving outer branches. Intercepts: $(0, 5)$ , $(-\sqrt{5}, 0)$ , $(\sqrt{5}, 0)$ .

Final answer

W-shaped curve with cusps at x = +/-sqrt(5), local maximum 5 at x = 0; intercepts (0, 5), (-sqrt(5), 0) and (sqrt(5), 0)

Detailed explanation

Walkthrough

The modulus function $|u|$ equals $u$ when $u \geq 0$ and $-u$ when $u < 0$ . So to sketch $y = |x^2 - 5|$ we first sketch $y = x^2 - 5$ and then reflect any part lying below the $x$ -axis into the region above it.

The parabola $y = x^2 - 5$ crosses the $x$ -axis where $x^2 - 5 = 0$ , i.e. at $x = \pm\sqrt{5}$ , and its vertex is at $(0, -5)$ . Between $x = -\sqrt{5}$ and $x = \sqrt{5}$ the parabola lies below the axis, so that whole middle arch flips up: the vertex becomes a maximum at $(0, 5)$ , and the crossing points at $\pm\sqrt{5}$ become sharp corners (cusps) because the reflected curve meets the unreflected outer branches at an angle there. Outside those points the parabola was already positive, so nothing changes and the branches keep their upward curvature.

Finally, state the intercepts explicitly: the curve meets the $y$ -axis at $(0, 5)$ and the $x$ -axis at $(-\sqrt{5}, 0)$ and $(\sqrt{5}, 0)$ .

Key Takeaways

The graph of $y = |g(x)|$ is obtained by reflecting the parts of $y = g(x)$ below the $x$ -axis upwards.
Zeros of the inner function become cusps of the modulus graph.
Intercepts should be stated exactly ( $\pm\sqrt{5}$ ) or to at least 3 significant figures if decimalised.

Common Mistakes

Sketching $y = x^2 - 5$ without reflecting the middle section, or reflecting only the vertex instead of the whole arch.
Drawing ruled straight lines for the outer branches — the mark scheme requires a smooth curve with correct curvature and awards no shape marks for clearly ruled sections.
Giving intercepts as $\pm 5$ by confusing $x^2 = 5$ with $x = \pm 5$ .
Omitting the intercepts statement entirely — it carries its own B1.
Drawing a sharp V at the origin rather than a smooth inverted arch between the cusps.

Things to Be Careful About

The mark scheme gives B2 for the shape (cusps equidistant from $O$ , maximum on the $y$ -axis, correct curvature inside and out) and B1 for the intercepts, so both must be shown; a maximum of 2 marks is available if not fully correct.
The intercepts must be clearly identified as intercepts, not just listed as plotting points.
Decimals are accepted only to 3 or more significant figures (e.g. $2.24$ or $2.236$ ); exact surds $\pm\sqrt{5}$ are safest.
At each end of the curve a little flicking is condoned but the branches must not approach a turning point.

Techniques used

reflect the negative part of the parabola in the x-axisidentify cusps where the quadratic crosses the x-axisstate the coordinate-axis intercepts

(b)

It is given that $\mathrm{f}(x) = ax + b$ where $a$ and $b$ are non-zero integers.
The solutions of the equation $|\mathrm{f}(x)| = 4$ are $x = 2.5$ and $x = -1.5$ .

Find the possible expressions for $\mathrm{f}(x)$ .

4M Similar

DifficultyMedium

Worked solution

Approach

Since $|\mathrm{f}(x)| = 4$ , at each given $x$ the value $\mathrm{f}(x) = ax + b$ is either $4$ or $-4$ . Pairing one choice at $x = 2.5$ with the opposite choice at $x = -1.5$ gives a non-zero gradient; solving the resulting simultaneous equations gives $a$ and $b$ , and swapping all signs gives the second expression.

Working

Take $\mathrm{f}(2.5) = 4$ and $\mathrm{f}(-1.5) = -4$ :

2.5a + b = 4, \qquad -1.5a + b = -4

Subtracting eliminates $b$ :

4a = 8 \quad\Rightarrow\quad a = 2

Then $2.5(2) + b = 4$ gives $b = -1$ .

Negating everything gives the other solution: $a = -2$ , $b = 1$ .

Check: $|2(2.5) - 1| = 4$ ✓, $|2(-1.5) - 1| = |-4| = 4$ ✓.

Answer

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

Final answer

f(x) = 2x - 1 or f(x) = -2x + 1

Detailed explanation

Walkthrough

The equation $|\mathrm{f}(x)| = 4$ means $\mathrm{f}(x) = 4$ or $\mathrm{f}(x) = -4$ . We know the two solutions are $x = 2.5$ and $x = -1.5$ , so at these two $x$ -values the line $y = ax + b$ takes values $\pm 4$ .

If both points gave the same sign (e.g. both equal to $4$ ), subtracting the equations would give $4a = 0$ , so $a = 0$ — but $a$ must be a non-zero integer. So the signs must be opposite. Choose $\mathrm{f}(2.5) = 4$ and $\mathrm{f}(-1.5) = -4$ :

2.5a + b = 4, \qquad -1.5a + b = -4

Subtracting removes $b$ : $4a = 8$ , so $a = 2$ , and substituting back gives $b = -1$ . Hence $\mathrm{f}(x) = 2x - 1$ .

The mirror-image assignment ( $\mathrm{f}(2.5) = -4$ , $\mathrm{f}(-1.5) = 4$ ) simply negates the whole line, giving $a = -2$ , $b = 1$ , i.e. $\mathrm{f}(x) = -2x + 1$ . Both check out: $|2(2.5)-1| = 4$ and $|-2(2.5)+1| = 4$ .

An equivalent route notes that $y = |ax+b|$ has its vertex where $ax + b = 0$ , i.e. midway between the solutions at $x = 0.5$ , and finds the gradient from $(0.5, 0)$ to $(2.5, 4)$ , which is $2$ .

Key Takeaways

$|u| = k$ splits into $u = k$ or $u = -k$ ; applying this at each known solution generates simultaneous equations.
With a modulus equation whose solutions are known, opposite signs at the two solutions are forced when the coefficient of $x$ is non-zero.
Negating $a$ and $b$ together always produces the second valid expression, since $|-(ax+b)| = |ax+b|$ .

Common Mistakes

Assigning the same value ( $4$ and $4$ , or $-4$ and $-4$ ) at both $x$ -values, which forces $a = 0$ and contradicts the condition that $a$ is non-zero.
Mixing up which $x$ goes with which equation, e.g. writing $-1.5a + b = 4$ alongside $2.5a + b = 4$ inconsistently.
Finding only $\mathrm{f}(x) = 2x - 1$ and forgetting the negated alternative $-2x + 1$ — the final A1 requires both and no others.
Arithmetic slips with the decimals: $2.5 - (-1.5) = 4$ , not $1$ or $3$ .
Giving $a = \pm 2$ without matching $b$ correctly (the pairings must be $a = 2, b = -1$ and $a = -2, b = 1$ ).

Things to Be Careful About

Both expressions must be given, and no others — the mark scheme's final A1 is for " $2x - 1$ and $-2x + 1$ and no others".
The pairings matter: $a = 2$ goes with $b = -1$ , and $a = -2$ with $b = 1$ ; mismatched pairs score no accuracy mark.
Correct expressions with little or no working still earn full credit, but showing the simultaneous-equation step secures the M1 safely.
Answers may be embedded, e.g. $|\mathrm{f}(x)| = |2x - 1|$ , and are accepted.

Techniques used

interpret the modulus equation as two linear equationssolve simultaneous linear equations for a and bform both possible expressions for f

Q5MediumPermutations and combinations

(a)

In this question, 6-digit numbers do not start with 0.

Find how many 6-digit numbers have 6 different digits and are divisible by 2.

3M Similar

DifficultyMedium

Worked solution

Approach

A 6-digit number with 6 different digits divisible by 2 must end in an even digit. Split into two disjoint cases: last digit $0$ , and last digit one of $2, 4, 6, 8$ . In each case count the arrangements of the remaining five digits, remembering the first digit cannot be $0$ .

Working

Case 1: last digit is $0$ .
The first digit can be any of the remaining $9$ digits, then $8, 7, 6, 5$ for the next positions:

9 \times 8 \times 7 \times 6 \times 5 = 15120

Case 2: last digit is $2$ , $4$ , $6$ or $8$ ( $4$ choices).
Now $0$ is available for the first position, so the first digit has $8$ choices (any of the remaining $9$ digits except the one used last), then $8, 7, 6, 5$ :

4 \times 8 \times 8 \times 7 \times 6 \times 5 = 4 \times 13440 = 53760

Total:

15120 + 53760 = 68880

Answer

68880

Final answer

68880

Detailed explanation

Walkthrough

The number must be even, so its last digit is even. The two sub-cases behave differently because of the rule that the first digit cannot be $0$ .

If the last digit is $0$ , the restriction on the first digit is already satisfied by the other digits: the first position has $9$ choices (digits $1$ to $9$ ), then $8$ , $7$ , $6$ , $5$ for the remaining positions, giving $9 \times 8 \times 7 \times 6 \times 5 = 15120$ .

If the last digit is $2$ , $4$ , $6$ or $8$ , there are $4$ choices for the last digit. Once it is fixed, $9$ digits remain, but one of them is $0$ , so the first position has only $8$ valid choices; then $8$ , $7$ , $6$ , $5$ follow. This gives $4 \times 8 \times 8 \times 7 \times 6 \times 5 = 53760$ .

The two cases cannot overlap, so the total is $15120 + 53760 = 68880$ .

Key Takeaways

Counting problems with a leading-zero restriction are best handled by conditioning on the digit that creates the restriction.
Splitting into disjoint cases and adding is the standard technique; the cases here are separated by whether the final digit is $0$ .
When a digit is used in a fixed position, the remaining positions are filled by permutations of the remaining digits.

Common Mistakes

Treating the first digit as having $9$ choices in the second case, forgetting that $0$ is now available and one even digit has been used (this gives the wrong $9 \times 8 \times 7 \times 6 \times 5 \times 4$ ).
Forgetting the leading-zero restriction entirely and counting $10 \times 9 \times 8 \times 7 \times 6 \times 5$ arrangements.
Counting only the case where the last digit is $2, 4, 6, 8$ and missing the last digit $0$ case.
Double-counting by not checking the cases are disjoint.
Arithmetic slips in $4 \times 13440$ or the final addition; the answer $68880$ is cao.

Things to Be Careful About

The mark scheme awards M2 for a complete correct case analysis (such as $41 \times 8 \times 7 \times 6 \times 5$ , i.e. $15120 + 53760$ combined) and M1 for any one correct individual calculation such as $53760$ or $15120$ — so both cases must appear for full credit.
Several alternative routes are accepted (complement methods such as all arrangements minus those starting with $0$ or ending oddly), but whichever route is used, every case must be handled consistently.
The final answer must be exactly $68880$ ; wrong working scores nothing for the answer mark.

Techniques used

split the count into disjoint cases by the final digitcount arrangements of distinct digits with the first digit non-zeroadd the case counts

(b)

A committee of 10 people is to be selected from 9 students and 7 teachers.

Find the number of different committees that can be selected if the committee must have at least 4 students and at least 4 teachers.

3M Similar

DifficultyMedium

Worked solution

Approach

A committee of $10$ from $9$ students and $7$ teachers needs at least $4$ of each. Since there are only $7$ teachers, the possible compositions are $4$ students with $6$ teachers, $5$ students with $5$ teachers, and $6$ students with $4$ teachers. Count each with combinations and add.

Working

{}^{9}\mathrm{C}_{4} \times {}^{7}\mathrm{C}_{6} + {}^{9}\mathrm{C}_{5} \times {}^{7}\mathrm{C}_{5} + {}^{9}\mathrm{C}_{6} \times {}^{7}\mathrm{C}_{4}

= (126 \times 7) + (126 \times 21) + (84 \times 35)

= 882 + 2646 + 2940 = 6468

Answer

6468

Final answer

6468

Detailed explanation

Walkthrough

The committee has $10$ people drawn from $9$ students and $7$ teachers, with at least $4$ students and at least $4$ teachers. If $s$ students are chosen, then $10 - s$ teachers are chosen. The constraints give $s \geq 4$ and $10 - s \geq 4$ , so $4 \leq s \leq 6$ . Also $10 - s \leq 7$ is automatic for these values. So the three cases are $(s, t) = (4, 6), (5, 5), (6, 4)$ .

Each case is counted with combinations since a committee is an unordered selection: ${}^{9}\mathrm{C}_{4} \times {}^{7}\mathrm{C}_{6} = 126 \times 7 = 882$ , ${}^{9}\mathrm{C}_{5} \times {}^{7}\mathrm{C}_{5} = 126 \times 21 = 2646$ , and ${}^{9}\mathrm{C}_{6} \times {}^{7}\mathrm{C}_{4} = 84 \times 35 = 2940$ .

The cases are disjoint, so the total is $882 + 2646 + 2940 = 6468$ .

Key Takeaways

Translate 'at least' constraints into a range for one variable, then enumerate the feasible compositions.
Committees are unordered, so combinations (not permutations) are used, and the two groups are multiplied within a case.
An alternative valid method is the complement: ${}^{16}\mathrm{C}_{10}$ minus all committees violating a constraint.

Common Mistakes

Missing one of the three cases (commonly forgetting $(4, 6)$ because $6$ teachers feels large, or forgetting $(6, 4)$ ).
Including an invalid case such as $3$ students with $7$ teachers or $7$ students with $3$ teachers, which violate the 'at least 4' conditions.
Using permutations instead of combinations, or adding instead of multiplying within a case.
Using the complement method but omitting one of the excluded compositions (e.g. forgetting ${}^{9}\mathrm{C}_{9} \times {}^{7}\mathrm{C}_{1}$ ).
Arithmetic slips in the products or the final sum; the answer $6468$ is cao.

Things to Be Careful About

The mark scheme gives M2 for the full three-term sum (or the full complement expression) and M1 for any two correct products or values — so all three cases must be present for full credit.
If using the complement route, all four excluded compositions ${}^{9}\mathrm{C}_{3} \times {}^{7}\mathrm{C}_{7}$ , ${}^{9}\mathrm{C}_{7} \times {}^{7}\mathrm{C}_{3}$ , ${}^{9}\mathrm{C}_{8} \times {}^{7}\mathrm{C}_{2}$ and ${}^{9}\mathrm{C}_{9} \times {}^{7}\mathrm{C}_{1}$ must be subtracted, giving $8008 - 1540 = 6468$ .
The final answer must be exactly $6468$ ; wrong working scores nothing for the answer mark.

Techniques used

split the committee into cases by the number of studentscount each case with combinationsadd the disjoint case counts

Q68MMediumSeries

In this question, $a$ , $b$ and $c$ are integers.

When the expansion of $(a + x)^4(1 + ax)^5$ is written in ascending powers of $x$ , the first three terms are $16 + bx + cx^2$ .

Find all the possible values of $a$ , $b$ and $c$ .

DifficultyMedium

Worked solution

Approach

Expand $(a + x)^4$ and $(1 + ax)^5$ up to the term in $x^2$ , multiply, and match the coefficients of $x^0$ , $x$ and $x^2$ with $16$ , $b$ and $c$ .

Working

(a + x)^4 = a^4 + 4a^3x + 6a^2x^2

(1 + ax)^5 = 1 + 5ax + 10a^2x^2

Matching the constant term:

a^4 = 16 \quad\Rightarrow\quad a = \pm 2

Coefficient of $x$ :

b = 4a^3 + 5a \cdot a^4 = 5a^5 + 4a^3

For $a = 2$ :

b = 5(2)^5 + 4(2)^3 = 160 + 32 = 192

For $a = -2$ :

b = 5(-2)^5 + 4(-2)^3 = -160 - 32 = -192

Coefficient of $x^2$ :

c = 10a^2 \cdot a^4 + 20a^3 \cdot a + 6a^2 = 10a^6 + 20a^4 + 6a^2

For $a = 2$ :

c = 10(64) + 20(16) + 6(4) = 640 + 320 + 24 = 984

For $a = -2$ :

c = 10(64) + 20(16) + 6(4) = 984

Answer

a = 2,\; b = 192,\; c = 984 \qquad \text{or} \qquad a = -2,\; b = -192,\; c = 984

Final answer

a = 2, b = 192, c = 984 or a = -2, b = -192, c = 984

Detailed explanation

Walkthrough

The product $(a + x)^4(1 + ax)^5$ must begin $16 + bx + cx^2$ , so we expand each factor far enough to see the constant, $x$ and $x^2$ contributions.

Using the binomial theorem, $(a + x)^4 = a^4 + 4a^3x + 6a^2x^2$ and $(1 + ax)^5 = 1 + 5ax + 10a^2x^2$ . The constant term of the product is $a^4 \times 1$ , so $a^4 = 16$ , giving the two integer solutions $a = 2$ and $a = -2$ .

The $x$ term collects two contributions: $4a^3x$ from the first factor times $1$ , and $a^4$ times $5ax$ . So $b = 5a^5 + 4a^3$ . Substituting $a = 2$ gives $b = 160 + 32 = 192$ ; substituting $a = -2$ gives $b = -160 - 32 = -192$ . Note the sign of $b$ follows the sign of $a$ because $5a^5 + 4a^3$ is an odd function of $a$ .

The $x^2$ term collects three contributions: $6a^2x^2$ from the first factor, $4a^3x \times 5ax = 20a^4x^2$ , and $a^4 \times 10a^2x^2 = 10a^6x^2$ . So $c = 10a^6 + 20a^4 + 6a^2$ , which involves only even powers of $a$ , so both $a = 2$ and $a = -2$ give the same value: $c = 640 + 320 + 24 = 984$ .

Key Takeaways

The binomial theorem gives each coefficient as a binomial coefficient times powers of the terms: $(1 + ax)^5$ starts $1 + 5ax + 10a^2x^2$ .
When multiplying two expansions, the coefficient of $x^k$ in the product is the sum over all ways of splitting $k$ between the two factors.
Matching coefficients lets an unknown parameter be found from a known term (here $a$ from the constant), and the remaining coefficients then follow.
Even powers of $a$ are unchanged when $a$ changes sign; odd powers change sign — this explains why $c$ is the same for both solutions but $b$ is not.

Common Mistakes

Forgetting one of the two solutions $a = \pm 2$ — the mark scheme allows a maximum of 6 of the 8 marks if only $a = 2$ is used.
Missing one of the three contributions to the $x^2$ coefficient (especially the cross term $20a^4$ ).
Writing $(1 + ax)^5$ with the wrong coefficients, e.g. using $5a^2x^2$ instead of $10a^2x^2$ .
Sign slips when substituting $a = -2$ into $5a^5 + 4a^3$ — both terms are negative.
Leaving binomial coefficients unevaluated; the mark scheme requires them evaluated for the method marks.
Leaving $b$ or $c$ containing $x$ — the values must be free of $x$ .

Things to Be Careful About

The question asks for ALL possible values, so both $a = 2$ and $a = -2$ cases must be fully worked with their own $b$ values.
The mark scheme awards method marks for the correct coefficient expressions ( $b = 5a^5 + 4a^3$ , $c = 10a^6 + 20a^4 + 6a^2$ ) with follow-through on the candidate's integer $a$ , but the substitution must be explicitly shown.
Terms may be listed rather than summed, and extra terms beyond $x^2$ are ignored.
All answers here are exact integers — no rounding involved.

Techniques used

expand each binomial factor to the first three termsequate the constant term to find acollect coefficients of like powers of xsolve for both integer values of a and compute b and c for each

Similar questions

Q7MediumVectors in two dimensions

In this question, $\mathbf{i}$ is a unit vector due east and $\mathbf{j}$ is a unit vector due north.
Distances are measured in kilometres and time is measured in hours.

Point $P$ has position vector $2\mathbf{i} + 5\mathbf{j}$ relative to an origin $O$ .
At 1300, boat $A$ sails from point $P$ with velocity vector $8\mathbf{i} + 8\mathbf{j}$ .

(a)

Find the position vector of $A$ at the end of 2 hours of sailing.

1M Similar

(b)

Write down the position vector of $A$ at the end of $t$ hours of sailing.

1M Similar

(c)

Point $Q$ has position vector $20\mathbf{i} + 8\mathbf{j}$ relative to the origin $O$ .
At 1500, boat $B$ sails from point $Q$ on a bearing of $300^\circ$ with a constant speed of $5\sqrt{3}$ .

Find the position vector of $B$ at 1700.

3M Similar

DifficultyMedium

Worked solution

Approach

A bearing of $300^\circ$ is measured $300^\circ$ clockwise from north. Resolve the speed $5\sqrt{3}$ into east ( $x$ ) and north ( $y$ ) components: east component $= 5\sqrt{3}\sin 300^\circ$ , north component $= 5\sqrt{3}\cos 300^\circ$ . Then use position = start + velocity × time over the 2 hours from 1500 to 1700.

Working

x = 5\sqrt{3}\sin 300^\circ = 5\sqrt{3}\left(-\frac{\sqrt{3}}{2}\right) = -\frac{15}{2} = -7.5

y = 5\sqrt{3}\cos 300^\circ = 5\sqrt{3}\left(\frac{1}{2}\right) = \frac{5\sqrt{3}}{2}

So the velocity of $B$ is $-7.5\mathbf{i} + \frac{5\sqrt{3}}{2}\mathbf{j}$ .

From 1500 to 1700 is 2 hours, so

\overrightarrow{OB} = 20\mathbf{i} + 8\mathbf{j} + 2\left(-7.5\mathbf{i} + \frac{5\sqrt{3}}{2}\mathbf{j}\right)

= (20 - 15)\mathbf{i} + \left(8 + 5\sqrt{3}\right)\mathbf{j} = 5\mathbf{i} + (8 + 5\sqrt{3})\mathbf{j}

Answer

5\mathbf{i} + (8 + 5\sqrt{3})\mathbf{j}

Final answer

5i + (8 + 5√3)j

Detailed explanation

Walkthrough

The key step is converting a bearing into vector components. Bearings are measured clockwise from north, so the north component uses cosine and the east component uses sine of the bearing angle. At $300^\circ$ the direction points north-west, so the east component must be negative: $5\sqrt{3}\sin 300^\circ = -7.5$ , and the north component positive: $5\sqrt{3}\cos 300^\circ = \frac{5\sqrt{3}}{2} \approx 4.33$ . Boat $B$ sails for 2 hours (1500 to 1700), so its displacement is twice this velocity vector. Adding to its start $Q = 20\mathbf{i} + 8\mathbf{j}$ gives $5\mathbf{i} + (8 + 5\sqrt{3})\mathbf{j}$ .

Key Takeaways

For a bearing $\theta$ : velocity $= (v\sin\theta)\mathbf{i} + (v\cos\theta)\mathbf{j}$ . Signs of sine and cosine in the fourth quadrant ( $270^\circ$ to $360^\circ$ ) give east negative, north positive.

Common Mistakes

Swapping sine and cosine (using $v\cos\theta$ for east), which the mark scheme penalises with only an SC1 for reversed components; forgetting that $\sin 300^\circ$ is negative; using the wrong elapsed time (the boat sails from 1500, not 1300).

Things to Be Careful About

The mark scheme condones the decimal $4.33$ for the north component and the final answer $\begin{pmatrix}5 \\ 16.7\end{pmatrix}$ , but the exact form $8 + 5\sqrt{3}$ is preferred since part (d) builds on it. The two velocity components must be clearly identified as $x$ and $y$ (or written as a vector) to earn the B2.

Techniques used

resolve a speed on a bearing into east and north components using sine and cosinemultiply the velocity vector by the elapsed timeadd the displacement to the initial position vector

(d)

Find the distance between the two boats at 1700.

3M Similar

DifficultyMedium-Easy

Worked solution

Approach

At 1700, boat $A$ has been sailing for 4 hours (from 1300), so its position is $2\mathbf{i} + 5\mathbf{j} + 4(8\mathbf{i} + 8\mathbf{j}) = 34\mathbf{i} + 37\mathbf{j}$ . Subtract the position vectors and take the magnitude.

Working

\overrightarrow{OA} = 2\mathbf{i} + 5\mathbf{j} + 4(8\mathbf{i} + 8\mathbf{j}) = 34\mathbf{i} + 37\mathbf{j}

\overrightarrow{AB} = \overrightarrow{OB} - \overrightarrow{OA} = \left(5\mathbf{i} + (8 + 5\sqrt{3})\mathbf{j}\right) - \left(34\mathbf{i} + 37\mathbf{j}\right)

= -29\mathbf{i} + (5\sqrt{3} - 29)\mathbf{j}

d^2 = (-29)^2 + (29 - 5\sqrt{3})^2 = 841 + \left(841 - 290\sqrt{3} + 75\right) = 1757 - 290\sqrt{3}

d = \sqrt{1757 - 290\sqrt{3}} = 35.421\ldots

Answer

35.4

Final answer

35.4 km

Detailed explanation

Walkthrough

First find where each boat is at 1700. Boat $A$ left at 1300, so it has sailed 4 hours: $2\mathbf{i} + 5\mathbf{j} + 4(8\mathbf{i} + 8\mathbf{j}) = 34\mathbf{i} + 37\mathbf{j}$ . Boat $B$ 's position came from part (c): $5\mathbf{i} + (8 + 5\sqrt{3})\mathbf{j}$ . The distance between two points with position vectors $\mathbf{a}$ and $\mathbf{b}$ is the magnitude of their difference: $|\overrightarrow{OB} - \overrightarrow{OA}|$ . Subtracting component-wise gives $-29\mathbf{i} + (5\sqrt{3} - 29)\mathbf{j}$ , and Pythagoras on the components gives $d^2 = 29^2 + (29 - 5\sqrt{3})^2 = 1757 - 290\sqrt{3}$ , so $d = 35.421\ldots \approx 35.4$ .

Key Takeaways

Distance between two points = magnitude of the difference of their position vectors; $|a\mathbf{i} + b\mathbf{j}| = \sqrt{a^2 + b^2}$ .

Common Mistakes

Using $A$ 's 2-hour position $18\mathbf{i} + 21\mathbf{j}$ from part (a) instead of the 4-hour position at 1700 (this still earns M1 FT if stated, but gives a wrong answer); adding the vectors instead of subtracting; sign slips when squaring $(29 - 5\sqrt{3})^2$ .

Things to Be Careful About

The answer must be given to 3 significant figures: $35.4$ . The mark scheme requires either the difference of vectors plus the squared magnitude, or the difference of components in a distance calculation, to be visible for the method marks. Follow-through is allowed on stated position vectors of the form $a\mathbf{i} + b\mathbf{j}$ with non-zero components.

Techniques used

subtract position vectors to find the displacement between two pointscompute the magnitude of a vector using Pythagoras

Q8MediumFunctionsLogarithmic and exponential functionsTrigonometry

The function $\mathrm{f}$ is defined, for $x \geq -1$ , by $\mathrm{f}(x) = 1 + 2^x$ .

(a)

Find the range of $\mathrm{f}$ .

1M Similar

(b)

7M

(i)

Write down the domain of $\mathrm{f}^{-1}$ .

1M Similar

(ii)

Find an expression for $\mathrm{f}^{-1}(x)$ .

2M Similar

DifficultyMedium-Easy

Worked solution

Approach

Let $y = \mathrm{f}(x)$ , swap the variables, then take $\log_2$ of both sides to make $y$ the subject.

Working

y = 1 + 2^x

Swap the variables:

x = 1 + 2^y

Make $2^y$ the subject:

2^y = x - 1

Take $\log_2$ of both sides:

y = \log_2(x - 1)

Answer

\mathrm{f}^{-1}(x) = \log_2(x - 1)

Final answer

f^{-1}(x) = log_2(x - 1)

Detailed explanation

Walkthrough

To find an inverse function, start by writing $y = 1 + 2^x$ and interchanging the roles of $x$ and $y$ , giving $x = 1 + 2^y$ . The goal is now to make $y$ the subject. Subtracting $1$ gives $2^y = x - 1$ . Since $y$ sits in the exponent, undo the exponential by taking logarithms base $2$ of both sides: $\log_2(2^y) = y$ , so $y = \log_2(x - 1)$ . This is the expression for $\mathrm{f}^{-1}(x)$ .

Key Takeaways

Inverse: swap variables, then make $y$ the subject.
$\log_2(2^y) = y$ is the tool for extracting a variable from an exponent.
Equivalent forms such as $\frac{\lg(x-1)}{\lg 2}$ are accepted.

Common Mistakes

Writing $y = \log_2 x - 1$ instead of $\log_2(x - 1)$ — a bracketing/sign error that still earns the method mark but loses the accuracy mark.
Solving $x = 1 + 2^y$ as $y = 2^{x-1}$ — no logarithms used, so no marks.
Taking $\lg$ of each term separately, e.g. $\lg x = \lg 1 + \lg 2^y$ — an invalid application of the logarithm laws.

Things to Be Careful About

The mark is awarded for the final answer only (nfww), and the answer must not be left as $x = \ldots$ ; it must be $y = \ldots$ , $\mathrm{f}^{-1}(x) = \ldots$ or equivalent.

Techniques used

swap the variablesrearrange an exponential equation to make the subject lineartake logarithms base 2 to solve for the subject

(iii)

On the axes, sketch the graph of $y = \mathrm{f}(x)$ and hence the graph of $y = \mathrm{f}^{-1}(x)$ .
State any intercepts with the coordinate axes.

Fig. 2

4M Similar

DifficultyMedium

Worked solution

Approach

Sketch $y = 1 + 2^x$ starting at its endpoint $(-1, 1.5)$ with the correct increasing exponential shape, mark its $y$ -intercept $(0, 2)$ , then reflect it in the line $y = x$ to obtain $y = \mathrm{f}^{-1}(x)$ .

Working

For $y = \mathrm{f}(x) = 1 + 2^x$ on $x \geq -1$ :

Endpoint: $\mathrm{f}(-1) = 1.5$ , so the curve starts at $(-1, 1.5)$ and nothing is drawn to the left of it.
$y$ -intercept: $\mathrm{f}(0) = 1 + 2^0 = 2$ , so the curve passes through $(0, 2)$ .
Shape: smooth, increasing, steepening exponential curve through quadrants 1 and 2, tending to the horizontal asymptote $y = 1$ (approached but never reached) on the left.

The inverse is the reflection of this curve in the line $y = x$ :

It starts at $(1.5, -1)$ (the reflection of $(-1, 1.5)$ ).
It passes through the $x$ -intercept $(2, 0)$ (the reflection of $(0, 2)$ ).
Shape: smooth increasing curve flattening out as $x$ grows, tending to a vertical asymptote $x = 1$ .

Answer

A smooth increasing exponential curve for $\mathrm{f}$ from $(-1, 1.5)$ through $(0, 2)$ curving up into quadrant 1, and its mirror image for $\mathrm{f}^{-1}$ from $(1.5, -1)$ through $(2, 0)$ curving rightwards into quadrant 1, the two being reflections of each other in the line $y = x$ .

Final answer

Exponential curve f from (-1, 1.5) through y-intercept (0, 2) curving upward, with its inverse as its reflection in y = x from (1.5, -1) through x-intercept (2, 0)

Detailed explanation

Walkthrough

The graph of $\mathrm{f}(x) = 1 + 2^x$ is the standard exponential growth curve lifted by $1$ . On the restricted domain $x \geq -1$ it begins at the endpoint $(-1, 1.5)$ — the mark scheme insists nothing may be drawn to the left of this point — and rises smoothly through the $y$ -intercept $(0, 2)$ , since $2^0 = 1$ gives $1 + 1 = 2$ . The curve steepens as it moves right and, on its left, flattens towards the horizontal asymptote $y = 1$ , which it approaches but never touches.

The graph of the inverse is obtained by reflecting this curve in the line $y = x$ : every point $(a, b)$ on $\mathrm{f}$ becomes $(b, a)$ on $\mathrm{f}^{-1}$ . So the endpoint $(-1, 1.5)$ reflects to $(1.5, -1)$ , and the intercept $(0, 2)$ reflects to the $x$ -intercept $(2, 0)$ . The reflected curve is smooth and increasing, flattening out to the right. The line $y = x$ itself does not need to be drawn, though it is helpful as a construction guide.

Key Takeaways

A restricted exponential sketch must show the endpoint, the correct increasing shape, and the asymptote approached.
The inverse graph is the reflection of the original in $y = x$ ; intercepts swap coordinates.
Intercepts of $\mathrm{f}^{-1}$ can be read directly: $\mathrm{f}^{-1}(2) = 0$ because $\mathrm{f}(0) = 2$ .

Common Mistakes

Drawing the exponential to the left of $x = -1$ — the domain restriction forbids it, and the mark scheme caps the part at 3 marks if the sketch is not fully correct.
Omitting the $y$ -intercept $(0, 2)$ , which carries its own mark.
Reflecting incorrectly, e.g. reflecting in the $x$ -axis or $y$ -axis instead of the line $y = x$ .
Marking conflicting intercepts when using follow-through values.

Things to Be Careful About

Curves must be smooth, not joined by straight segments.
The intercept mark for $\mathrm{f}$ requires a correct smooth curve to have been attempted.
The fourth mark is follow-through on the candidate's own curve: the reflection must match their $\mathrm{f}$ , and there must be no contradictory points marked.

Techniques used

sketch an exponential curve with its endpoint and asymptotemark the y-intercept of the exponentialreflect a curve in the line y = x to obtain its inverse

(c)

The function $\mathrm{g}$ is defined, for $x \geq -1$ , by $\mathrm{g}(x) = x(x + k)$ where $k$ is a constant.

4M

(i)

Explain why $\mathrm{gf}$ exists.

1M Similar

DifficultyMedium-Easy

Worked solution

Approach

$\mathrm{gf}$ exists provided every output of $\mathrm{f}$ is an acceptable input of $\mathrm{g}$ , i.e. the range of $\mathrm{f}$ lies inside the domain of $\mathrm{g}$ .

Working

From part (a), the range of $\mathrm{f}$ is $\mathrm{f}(x) \geq 1.5$ . The domain of $\mathrm{g}$ is $x \geq -1$ . Since every value of $1.5$ or more is also at least $-1$ :

\{\mathrm{f}(x) : x \geq -1\} \subseteq \{x : x \geq -1\}

so every output of $\mathrm{f}$ lies in the domain of $\mathrm{g}$ .

Answer

$\mathrm{gf}$ exists because the range of $\mathrm{f}$ is a subset of the domain of $\mathrm{g}$ (every value $\mathrm{f}(x) \geq 1.5$ satisfies $x \geq -1$ ).

Final answer

The range of f is a subset of the domain of g, so gf exists

Detailed explanation

Walkthrough

The composite $\mathrm{gf}$ means 'apply $\mathrm{f}$ first, then apply $\mathrm{g}$ to the result'. For this to make sense, every number that $\mathrm{f}$ can produce must be a number that $\mathrm{g}$ is willing to accept. Part (a) showed that $\mathrm{f}$ only produces values of $1.5$ or more, and $\mathrm{g}$ accepts any input of $-1$ or more. Since $1.5 \geq -1$ , all outputs of $\mathrm{f}$ fall inside the domain of $\mathrm{g}$ , so $\mathrm{gf}$ exists.

Key Takeaways

$\mathrm{gf}$ exists if and only if the range of $\mathrm{f}$ is contained in the domain of $\mathrm{g}$ .
Earlier parts of a question often supply exactly the fact needed here (the range from part (a)).

Common Mistakes

Saying ' $\mathrm{gf}$ exists because both functions exist' — the condition is about range versus domain, not mere existence.
Comparing the domains of $\mathrm{f}$ and $\mathrm{g}$ instead of the range of $\mathrm{f}$ with the domain of $\mathrm{g}$ .

Things to Be Careful About

The mark depends on a correct range from part (a) (or $\mathrm{f} \geq 1.5$ stated here); a wrong range in part (a) undermines this explanation.

Techniques used

compare the range of f with the domain of g to justify existence of the composite

(ii)

Find an expression for $\mathrm{gf}(x)$ .

1M Similar

(iii)

It is given that $\mathrm{gf}(3) = 126$ .

Find the value of $k$ .

2M Similar

Q9MediumCalculusLogarithmic and exponential functions

A particle $P$ moves in a straight line such that $t$ seconds after passing through a fixed point $O$ , its displacement from $O$ , $s$ metres, is given by

s = \mathrm{e}^{4t} - 10\mathrm{e}^{2t} + 9

(a)

The velocity of $P$ is $v\,\text{ms}^{-1}$ .

Find the value of $t$ for which $v$ has a stationary value.

5M Similar

DifficultyMedium

Worked solution

Approach

Velocity is the derivative of displacement, so differentiate $s$ with respect to $t$ . A stationary value of $v$ occurs when $\frac{\mathrm{d}v}{\mathrm{d}t} = 0$ , so differentiate again, set to zero, and solve.

Working

v = \frac{\mathrm{d}s}{\mathrm{d}t} = 4\mathrm{e}^{4t} - 20\mathrm{e}^{2t}

Differentiate again:

\frac{\mathrm{d}v}{\mathrm{d}t} = 16\mathrm{e}^{4t} - 40\mathrm{e}^{2t}

For a stationary value of $v$ :

16\mathrm{e}^{4t} - 40\mathrm{e}^{2t} = 0

Factorise (or divide by $8\mathrm{e}^{2t}$ ):

8\mathrm{e}^{2t}\left(2\mathrm{e}^{2t} - 5\right) = 0

Since $\mathrm{e}^{2t} \neq 0$ :

2\mathrm{e}^{2t} = 5 \quad \Rightarrow \quad \mathrm{e}^{2t} = \frac{5}{2}

Taking natural logarithms:

2t = \ln\frac{5}{2} \quad \Rightarrow \quad t = \frac{1}{2}\ln\frac{5}{2}

Answer

t = \frac{1}{2}\ln\frac{5}{2} \approx 0.458

Final answer

t = 0.458 s

Detailed explanation

Walkthrough

The displacement is given as a function of time, and velocity is defined as the rate of change of displacement. Differentiating each exponential term using $\frac{\mathrm{d}}{\mathrm{d}t}\left(\mathrm{e}^{kt}\right) = k\mathrm{e}^{kt}$ gives $v = 4\mathrm{e}^{4t} - 20\mathrm{e}^{2t}$ .

A stationary value of $v$ means the graph of $v$ against $t$ has zero gradient at that instant — it does NOT mean $v = 0$ . So we must differentiate $v$ again to get $\frac{\mathrm{d}v}{\mathrm{d}t}$ (which is also the acceleration), giving $16\mathrm{e}^{4t} - 40\mathrm{e}^{2t}$ , and set this equal to zero.

The equation $16\mathrm{e}^{4t} - 40\mathrm{e}^{2t} = 0$ looks like a quadratic in disguise: writing $u = \mathrm{e}^{2t}$ gives $16u^2 - 40u = 0$ , i.e. $8u(2u - 5) = 0$ . The factor $\mathrm{e}^{2t}$ can never be zero, so we need $\mathrm{e}^{2t} = \frac{5}{2}$ .

Taking natural logarithms of both sides undoes the exponential: $2t = \ln\frac{5}{2}$ , so $t = \frac{1}{2}\ln\frac{5}{2} \approx 0.458$ seconds.

Key Takeaways

Velocity is $\frac{\mathrm{d}s}{\mathrm{d}t}$ ; acceleration is $\frac{\mathrm{d}v}{\mathrm{d}t} = \frac{\mathrm{d}^2 s}{\mathrm{d}t^2}$ .
A "stationary value" of any quantity means its derivative with respect to the independent variable equals zero — here $\frac{\mathrm{d}v}{\mathrm{d}t} = 0$ , not $v = 0$ .
Equations like $a\mathrm{e}^{4t} + b\mathrm{e}^{2t} + c = 0$ are quadratics in $\mathrm{e}^{2t}$ ; solve for $\mathrm{e}^{2t}$ then take logs.
$\mathrm{e}^{2t} > 0$ always, so discard any non-positive root for $\mathrm{e}^{2t}$ .

Common Mistakes

Setting $v = 0$ instead of $\frac{\mathrm{d}v}{\mathrm{d}t} = 0$ — the mark scheme explicitly states that using $v$ rather than $\frac{\mathrm{d}v}{\mathrm{d}t}$ scores M0.
Forgetting the chain-rule factor when differentiating: $\frac{\mathrm{d}}{\mathrm{d}t}\left(\mathrm{e}^{4t}\right) = 4\mathrm{e}^{4t}$ , not $\mathrm{e}^{4t}$ .
Dividing through by $\mathrm{e}^{2t}$ without noting it cannot be zero — here it is safe, but the reasoning should be stated.
Sign slips when rearranging $16\mathrm{e}^{4t} = 40\mathrm{e}^{2t}$ .
Leaving the answer as $\ln\frac{5}{2}$ instead of halving it.

Things to Be Careful About

The final answer may be given in exact form $\frac{1}{2}\ln\frac{5}{2}$ or as a decimal $0.458$ (3 significant figures); later mis-simplification is ignored (isw).
Follow-through applies on the second differentiation provided at least one term is correct, but the method mark for setting up $\frac{\mathrm{d}v}{\mathrm{d}t} = 0$ depends on having differentiated correctly first.
Keep the notation consistent: $\frac{\mathrm{d}s}{\mathrm{d}t}$ or even $\frac{\mathrm{d}y}{\mathrm{d}x}$ is condoned for the first B1, but the second derivative must be with respect to $t$ .

Techniques used

differentiate an exponential expression to find velocitydifferentiate again to find dv/dtset the derivative equal to zero for a stationary valuesolve an equation in e^(2t) using logarithms

(b)

Find the acceleration of $P$ when $t = \ln 2$ .

2M Similar

DifficultyMedium-Easy

Worked solution

Approach

Acceleration is $\frac{\mathrm{d}v}{\mathrm{d}t}$ , found in part (a). Substitute $t = \ln 2$ and use $\mathrm{e}^{k\ln 2} = 2^k$ .

Working

From part (a):

a = \frac{\mathrm{d}v}{\mathrm{d}t} = 16\mathrm{e}^{4t} - 40\mathrm{e}^{2t}

Substitute $t = \ln 2$ :

a = 16\mathrm{e}^{4\ln 2} - 40\mathrm{e}^{2\ln 2}

Using $\mathrm{e}^{4\ln 2} = 2^4 = 16$ and $\mathrm{e}^{2\ln 2} = 2^2 = 4$ :

a = 16(16) - 40(4) = 256 - 160 = 96

Answer

a = 96\,\text{ms}^{-2}

Final answer

96 ms^-2

Detailed explanation

Walkthrough

This part uses the result carried forward from part (a): the acceleration is $\frac{\mathrm{d}v}{\mathrm{d}t} = 16\mathrm{e}^{4t} - 40\mathrm{e}^{2t}$ . We simply substitute $t = \ln 2$ .

The key simplification is the logarithm–exponential inverse relationship: $\mathrm{e}^{4\ln 2} = \left(\mathrm{e}^{\ln 2}\right)^4 = 2^4 = 16$ , and similarly $\mathrm{e}^{2\ln 2} = 2^2 = 4$ . Then $16 \times 16 - 40 \times 4 = 256 - 160 = 96$ .

Key Takeaways

Acceleration is the derivative of velocity (the second derivative of displacement).
$\mathrm{e}^{k\ln a} = a^k$ — evaluating exponentials at logarithmic inputs converts them to simple powers.
Substitution questions often reuse earlier results; keep your part (a) expressions tidy.

Common Mistakes

Mis-evaluating $\mathrm{e}^{4\ln 2}$ as $(\mathrm{e}^{\ln 2}) \times 4 = 8$ instead of $2^4 = 16$ — the power applies to the whole exponential, not just the log.
Using $v$ instead of $\frac{\mathrm{d}v}{\mathrm{d}t}$ — the substitution must go into the acceleration expression.
Arithmetic slips: $16 \times 16 = 256$ and $40 \times 4 = 160$ , so the difference is $96$ , not $86$ or $116$ .

Things to Be Careful About

The mark scheme requires visible evidence of substituting $t = \ln 2$ into $\frac{\mathrm{d}v}{\mathrm{d}t}$ before the answer — write the substituted line out fully.
Follow-through is allowed only if your $\frac{\mathrm{d}v}{\mathrm{d}t}$ has the form $p\mathrm{e}^{4t} + q\mathrm{e}^{2t}$ .
Include the unit: acceleration is in $\text{ms}^{-2}$ since displacement is in metres and time in seconds.

Techniques used

use acceleration as the second derivative of displacementsubstitute t = ln 2 into the acceleration expressionevaluate exponentials using e^(k ln 2) = 2^k

Q106MMedium-HardTrigonometry

Solve the equation $\tan(3x + 1) = 5\sin(3x + 1)$ for $-\frac{\pi}{3} < x < \frac{\pi}{3}$ .

DifficultyMedium-Hard

Worked solution

Approach

Rewrite $\tan$ as $\frac{\sin}{\cos}$ and bring everything to one side, then factorise — do not cancel $\sin(3x+1)$ , since that would lose solutions. Each factor gives an equation for $3x+1$ , which is solved over $0 < 3x+1 < 2\pi$ , and finally divided by $3$ .

Working

\frac{\sin(3x+1)}{\cos(3x+1)} = 5\sin(3x+1)

\sin(3x+1) - 5\sin(3x+1)\cos(3x+1) = 0

\sin(3x+1)\left(1 - 5\cos(3x+1)\right) = 0

So either $\sin(3x+1) = 0$ or $\cos(3x+1) = \frac{1}{5}$ .

Since $-\frac{\pi}{3} < x < \frac{\pi}{3}$ , we have $0 < 3x+1 < 2\pi$ .

Case 1: $\sin(3x+1) = 0$

3x+1 = 0 \quad \text{or} \quad 3x+1 = \pi

x = -\frac{1}{3} \approx -0.333 \quad \text{or} \quad x = \frac{\pi - 1}{3} \approx 0.714

Case 2: $\cos(3x+1) = \frac{1}{5}$

3x+1 = \pm \arccos(0.2) = \pm 1.3694\ldots

Both lie in $(0, 2\pi)$ :

x = \frac{1.3694\ldots - 1}{3} = 0.1231\ldots \approx 0.123

x = \frac{-1.3694\ldots - 1}{3} = -0.7898\ldots \approx -0.790

Answer

x = -\frac{1}{3}, \;\; x = \frac{\pi-1}{3}, \;\; x = 0.123, \;\; x = -0.790

Final answer

x = -1/3, x = 0.714, x = 0.123, x = -0.790

Detailed explanation

Walkthrough

The equation mixes tangent and sine, so the first move is to express everything in sines and cosines using $\tan A = \frac{\sin A}{\cos A}$ . This gives $\frac{\sin(3x+1)}{\cos(3x+1)} = 5\sin(3x+1)$ .

Multiplying through by $\cos(3x+1)$ and collecting terms gives $\sin(3x+1)\left(1 - 5\cos(3x+1)\right) = 0$ . The crucial point is that we factorise rather than divide both sides by $\sin(3x+1)$ : dividing would silently discard the solutions where $\sin(3x+1) = 0$ .

Each factor is set to zero in turn. For $\sin(3x+1) = 0$ , the angles in one full turn are $3x+1 = 0$ and $3x+1 = \pi$ . Dividing by $3$ and subtracting $1$ : $x = -\frac{1}{3}$ and $x = \frac{\pi-1}{3} \approx 0.714$ .

For $\cos(3x+1) = \frac{1}{5}$ , taking the inverse cosine gives the principal value $1.3694\ldots$ , and because cosine is positive in both the first and fourth quadrants, the second solution is $-1.3694\ldots$ . Both converted values satisfy $0 < 3x+1 < 2\pi$ , so both are kept: $x = 0.123$ and $x = -0.790$ .

Key Takeaways

Never cancel a trigonometric factor from an equation — factorise so no solutions are lost.
When the argument is a linear function like $3x+1$ , first find its range from the interval for $x$ ( $0 < 3x+1 < 2\pi$ here), solve in that range, then convert back.
$\cos u = k > 0$ has two solutions per revolution, $u = \pm\arccos k$ .
Exact answers such as $-\frac{1}{3}$ and $\frac{\pi-1}{3}$ should be kept exact; calculator-derived answers go to 3 significant figures.

Common Mistakes

Cancelling $\sin(3x+1)$ from both sides at the start — this loses the pair of solutions $x = -\frac{1}{3}$ and $x = 0.714$ entirely (the scheme marks this B0 unless recovered).
Writing $\frac{\sin}{\cos}(3x+1)$ instead of $\frac{\sin(3x+1)}{\cos(3x+1)}$ — the scheme awards B0 for this incorrect form.
Finding only the principal value $3x+1 = 1.3694\ldots$ and missing the fourth-quadrant solution $-1.3694\ldots$ .
Working in degrees without converting to radians — no method marks until conversion.
Truncating instead of rounding: $-0.789$ is not accepted because it truncates $-0.7898\ldots$ ; the correct 3 sf value is $-0.790$ .
Including extra values outside $-\frac{\pi}{3} < x < \frac{\pi}{3}$ costs the final accuracy mark.

Things to Be Careful About

The mark scheme says "nfww" (not from wrong working): the equations $\sin(3x+1)=0$ and $\cos(3x+1)=\frac{1}{5}$ must follow from correct algebra.
All four values are required; any two correct earn A1, all four with no extras earn A2.
Premature rounding of intermediate values is tolerated ( $\pm 1.36$ , $\pm 1.37$ , or even $3.1$ for $\pi$ ), but the final answers must round correctly to 3 significant figures.
Angles here are in radians throughout — there is no degree symbol anywhere in the question.

Techniques used

rewrite tan as sin over cosfactorise instead of cancelling a common factorsolve each factor separatelyfind all angles of the form arccos(1/5) in the intervalconvert values of 3x+1 back to values of x

Similar questions

Q119MMedium-HardCoordinate geometry of the circle

Solutions to this question by accurate drawing will not be accepted.

A circle $C_1$ has centre $(3, 5)$ .
The point $P(1, 4)$ lies on circle $C_1$ .

A circle $C_2$ has centre $(12, 8)$ .
The line with equation $y = 19 - 3x$ is the common chord of circles $C_1$ and $C_2$ .

Find the radius of circle $C_2$ .

DifficultyMedium-Hard

Worked solution

Approach

First find the radius of $C_1$ using the point $P(1,4)$ on it, then write down the equation of $C_1$ . The common chord $y = 19 - 3x$ meets $C_1$ at the two intersection points of the circles, so substitute the line into the equation of $C_1$ to find those points. The radius of $C_2$ is then the distance from $(12, 8)$ to either intersection point.

Working

Radius of $C_1$ using $P(1, 4)$ :

r_1^2 = (1 - 3)^2 + (4 - 5)^2 = 4 + 1 = 5

Equation of $C_1$ :

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

Substitute $y = 19 - 3x$ :

(x - 3)^2 + (19 - 3x - 5)^2 = 5

(x - 3)^2 + (14 - 3x)^2 = 5

x^2 - 6x + 9 + 9x^2 - 84x + 196 = 5

10x^2 - 90x + 200 = 0

Divide by 10 and factorise:

x^2 - 9x + 20 = 0 \quad\Rightarrow\quad (x - 4)(x - 5) = 0

x = 4 \quad \text{or} \quad x = 5

Corresponding $y$ -values from $y = 19 - 3x$ :

x = 4: \; y = 19 - 12 = 7 \quad\Rightarrow\quad (4, 7)

x = 5: \; y = 19 - 15 = 4 \quad\Rightarrow\quad (5, 4)

Radius of $C_2$ is the distance from $(12, 8)$ to either point:

r_2 = \sqrt{(12 - 5)^2 + (8 - 4)^2} = \sqrt{49 + 16} = \sqrt{65}

(Check with $(4, 7)$ : $\sqrt{(12-4)^2 + (8-7)^2} = \sqrt{64 + 1} = \sqrt{65}$ .)

Answer

r = \sqrt{65}

Final answer

√65

Detailed explanation

Walkthrough

The question gives the centre of $C_1$ and a point $P$ on it, so the first job is to find the radius of $C_1$ by the distance formula: $(1-3)^2 + (4-5)^2 = 5$ , giving the equation $(x-3)^2 + (y-5)^2 = 5$ .

The line $y = 19 - 3x$ is the common chord — the line through the two points where the circles cross. So the points where this line cuts $C_1$ are exactly the intersection points of the two circles. Substituting $y = 19 - 3x$ into the equation of $C_1$ gives $(x-3)^2 + (14-3x)^2 = 5$ , which expands and simplifies to $10x^2 - 90x + 200 = 0$ , i.e. $x^2 - 9x + 20 = 0$ . Factorising gives $(x-4)(x-5) = 0$ , so $x = 4$ or $x = 5$ , and the line gives $y = 7$ or $y = 4$ : the intersection points are $(4, 7)$ and $(5, 4)$ .

Finally, the radius of $C_2$ is the distance from its centre $(12, 8)$ to either intersection point: $\sqrt{(12-5)^2 + (8-4)^2} = \sqrt{65}$ (checking with $(4,7)$ gives the same value, as it must).

Key Takeaways

The equation of a circle with centre $(a, b)$ and radius $r$ is $(x-a)^2 + (y-b)^2 = r^2$ ; the radius can be found from any point on the circle.
A common chord of two circles is the line through their intersection points, so intersecting that line with one circle recovers the intersection points.
The radius of a circle is the distance from its centre to any point on it.

Common Mistakes

Sign slips when expanding $(14 - 3x)^2$ — the cross term is $-84x$ , not $+84x$ .
Forgetting to subtract the 5 (or moving it to the wrong side) when forming the quadratic.
Finding $x = 4, 5$ but forgetting to substitute back into $y = 19 - 3x$ to get the actual points.
Using the distance from $(12, 8)$ to the centre of $C_1$ instead of to an intersection point.
The mark scheme awards A0 for a final answer of $\pm\sqrt{65}$ — a radius is positive.
The final mark is nfww: $\sqrt{65}$ must come from correct working, not guessed.

Things to Be Careful About

Show the unsimplified substituted equation before simplifying — the scheme awards separate marks for the correct unsimplified equation and the correct 3-term quadratic.
The answer should be left as the exact surd $\sqrt{65}$ (the scheme condones $8.06[225\ldots]$ but the exact form is expected); do not write $\pm\sqrt{65}$ .
The final A1 is dependent on all previous marks, so every intermediate step must be correct.
An alternative accepted route subtracts the two expanded circle equations to recover the common chord line and solve for $r^2$ ; the method above follows the primary scheme route.

Techniques used

form the equation of a circle from its centre and a point on itsubstitute a line equation into a circle equation to find intersection pointssolve the resulting quadraticfind the radius of a circle as the distance between its centre and a point on it

Similar questions

Q124MMediumCalculus

A sphere has volume $V\,\text{cm}^3$ and radius $r\,\text{cm}$ .

Given that $r$ varies with time $t$ , find $V$ at the instant when

\frac{\mathrm{d}V}{\mathrm{d}t} = 9\frac{\mathrm{d}r}{\mathrm{d}t}

DifficultyMedium

Worked solution

Approach

The sphere's volume is $V = \frac{4}{3}\pi r^3$ . Since both $V$ and $r$ vary with time, link their rates of change with the chain rule, then use the given condition $\frac{\mathrm{d}V}{\mathrm{d}t} = 9\frac{\mathrm{d}r}{\mathrm{d}t}$ to find $r$ at that instant, and substitute back to get $V$ .

Working

By the chain rule:

\frac{\mathrm{d}V}{\mathrm{d}t} = \frac{\mathrm{d}V}{\mathrm{d}r} \times \frac{\mathrm{d}r}{\mathrm{d}t}

Differentiate the volume formula:

V = \frac{4}{3}\pi r^3 \quad \Rightarrow \quad \frac{\mathrm{d}V}{\mathrm{d}r} = 4\pi r^2

Given $\frac{\mathrm{d}V}{\mathrm{d}t} = 9\frac{\mathrm{d}r}{\mathrm{d}t}$ :

4\pi r^2 \times \frac{\mathrm{d}r}{\mathrm{d}t} = 9 \times \frac{\mathrm{d}r}{\mathrm{d}t}

Since $\frac{\mathrm{d}r}{\mathrm{d}t} \neq 0$ :

4\pi r^2 = 9 \quad \Rightarrow \quad r^2 = \frac{9}{4\pi} \quad \Rightarrow \quad r = \frac{3}{2\sqrt{\pi}}

(only the positive value of $r$ is valid). Substitute into the volume formula:

V = \frac{4}{3}\pi \left(\frac{3}{2\sqrt{\pi}}\right)^3 = \frac{4}{3}\pi \times \frac{27}{8\pi\sqrt{\pi}} = \frac{9}{2\sqrt{\pi}} \approx 2.54

Answer

V = \frac{9}{2\sqrt{\pi}} \approx 2.54\,\text{cm}^3

Final answer

V = 9/(2√π) ≈ 2.54 cm^3

Detailed explanation

Walkthrough

The question gives two quantities that both change with time: the volume $V$ and the radius $r$ of a sphere. We are told that at some instant $\frac{\mathrm{d}V}{\mathrm{d}t} = 9\frac{\mathrm{d}r}{\mathrm{d}t}$ , and asked for the volume at that instant.

The bridge between $\frac{\mathrm{d}V}{\mathrm{d}t}$ and $\frac{\mathrm{d}r}{\mathrm{d}t}$ is the chain rule: since $V$ depends on $r$ , and $r$ depends on $t$ , we have $\frac{\mathrm{d}V}{\mathrm{d}t} = \frac{\mathrm{d}V}{\mathrm{d}r} \times \frac{\mathrm{d}r}{\mathrm{d}t}$ . So the first job is to find $\frac{\mathrm{d}V}{\mathrm{d}r}$ from the sphere volume formula $V = \frac{4}{3}\pi r^3$ , which differentiates to $4\pi r^2$ .

Substituting into the chain rule gives $\frac{\mathrm{d}V}{\mathrm{d}t} = 4\pi r^2 \frac{\mathrm{d}r}{\mathrm{d}t}$ . Setting this equal to $9\frac{\mathrm{d}r}{\mathrm{d}t}$ and cancelling the common factor $\frac{\mathrm{d}r}{\mathrm{d}t}$ (legitimate since the radius is genuinely changing) gives $4\pi r^2 = 9$ , so $r = \frac{3}{2\sqrt{\pi}}$ , taking only the positive root because a radius must be positive.

Finally, substitute this radius back into $V = \frac{4}{3}\pi r^3$ :

V = \frac{4}{3}\pi \left(\frac{3}{2\sqrt{\pi}}\right)^3 = \frac{4}{3}\pi \times \frac{27}{8\pi\sqrt{\pi}} = \frac{9}{2\sqrt{\pi}} \approx 2.54\,\text{cm}^3

Key Takeaways

Related-rates problems are solved by the chain rule: $\frac{\mathrm{d}V}{\mathrm{d}t} = \frac{\mathrm{d}V}{\mathrm{d}r} \times \frac{\mathrm{d}r}{\mathrm{d}t}$ .
Differentiate the geometric formula first, then impose the given rate condition to pin down the variable's value at the instant in question.
Once $r$ is known, substitute into the original formula to get the required quantity.

Common Mistakes

Writing $\frac{\mathrm{d}V}{\mathrm{d}t} = 9\frac{\mathrm{d}r}{\mathrm{d}t}$ and treating it as if it directly gives $\frac{\mathrm{d}V}{\mathrm{d}r} = 9$ without the chain rule — the mark scheme explicitly says this given equation alone does not imply the first M1.
Forgetting the factor $4$ in $\frac{\mathrm{d}V}{\mathrm{d}r} = 4\pi r^2$ .
Taking the negative square root of $r^2 = \frac{9}{4\pi}$ — the mark scheme requires the positive value of $r$ only.
Leaving the answer as $\frac{4}{3}\pi\left(\frac{3}{2\sqrt{\pi}}\right)^3$ unevaluated — this only implies the first A1; the volume must be simplified to $\frac{9}{2\sqrt{\pi}}$ (or an equivalent form such as $2.54$ ).
Rounding $r = 0.8462843\ldots$ too early and losing accuracy in $V$ .

Things to Be Careful About

The chain rule statement must be shown (or clearly implied by correct substitution) to earn the first M1; the given equation $\frac{\mathrm{d}V}{\mathrm{d}t} = 9\frac{\mathrm{d}r}{\mathrm{d}t}$ on its own earns nothing.
The second M1 is dependent on the first — the equation $4\pi r^2 = 9$ must come from a correct chain-rule route.
The exact answer $\frac{9}{2\sqrt{\pi}}$ is accepted, as is $2.54$ (awrt $2.54$ ); the unsimplified substitution expression alone is not enough for full marks.
Only the positive value of $r$ may be stated or used.
"isw" applies to the final answer: later mis-simplification after reaching $\frac{9}{2\sqrt{\pi}}$ is ignored.

Techniques used

apply the chain rule to related rates of changedifferentiate the sphere volume formulaequate the rate expressions to find rsubstitute r back into the volume formula

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