(a) Expand $\left(2 - \frac{1}{2}x\right)^6$ in ascending powers of $x$ up to and including the term in $x^3$. (b) Hence find the coefficient of $x^3$ in the expansion of $(3 - x + 2x^3)\left(2 - \frac{1}{2}x\right)^6$.

(a) 64 - 96x + 60x^2 - 20x^3 · (b) 8

(a) Solve the equation $\tan^{-1}(5x - 3) = -\frac{1}{4}\pi$. (b) Solve the equation $5\cos^2\theta = 4\sin\theta + 4$ for $\frac{1}{2}\pi \le \theta \le 2\pi$.

(a) x = 2/5 · (b) theta = 2.94, 4.71

A function $f$ is defined by $f(x) = px^2 + 4x + q$ for $x \in \mathbb{R}$, where $p$ and $q$ are constants. (a) It is given that $p = 2$ and $q = 10$. (i) Express $f(x)$ in the form $a(x + b)^2 + c$, where $a$, $b$ and $c$ are constants. (ii) State the range of $f$. (b) It is given instead that $q = -5$ and the roots of $f(x) = 0$ are $5m$ and $-9m$, where $m$ is a constant. Find the values of $p$ and $m$.

(a)(i) f(x) = 2(x+1)^2 + 8 · (a)(ii) f(x) >= 8 · (b) p = 9, m = 1/9

The equation of a curve is $y = \frac{8}{3x - 8} - \frac{6}{x - 1}$. (a) Find the coordinates of the point at which the tangent to the curve at the point $(3, 5)$ intersects the line $y = -8x$. (i) Find the $x$-coordinates of each of the stationary points of the curve. (ii) Find $\frac{d^2y}{dx^2}$ and hence determine the nature of each of the stationary points.

(a) (5, -40) · (b)(i) x = 2 and x = 6 · (b)(ii) x = 2 is a maximum; x = 6 is a minimum

9709/13 Oct/Nov 2025 — Mathematics Solved Past Paper

Q1Medium-EasySeries

(a)

Expand $\left(2 - \frac{1}{2}x\right)^6$ in ascending powers of $x$ up to and including the term in $x^3$ .

3M Similar

DifficultyMedium-Easy

Worked solution

Approach

Use the binomial theorem to expand $(2 - \frac{1}{2}x)^6$ up to the term in $x^3$ . Write the first four terms with $r = 0,1,2,3$ .

Working

\begin{aligned} \left(2 - \frac{1}{2}x\right)^6 &= \binom{6}{0}2^6 + \binom{6}{1}2^5\left(-\frac{x}{2}\right) + \binom{6}{2}2^4\left(-\frac{x}{2}\right)^2 + \binom{6}{3}2^3\left(-\frac{x}{2}\right)^3 + \cdots \\ &= 64 + 6 \cdot 32 \cdot \left(-\frac{x}{2}\right) + 15 \cdot 16 \cdot \frac{x^2}{4} + 20 \cdot 8 \cdot \left(-\frac{x^3}{8}\right) + \cdots \\ &= 64 - 96x + 60x^2 - 20x^3 + \cdots \end{aligned}

Answer

64 - 96x + 60x^2 - 20x^3

Final answer

64 - 96x + 60x^2 - 20x^3

Detailed explanation

Walkthrough

We need the terms up to $x^3$ in the binomial expansion of $(2 - \frac{1}{2}x)^6$ . The binomial theorem states that $(a+b)^n = \sum_{r=0}^{n} \binom{n}{r} a^{n-r} b^r$ . Here $a = 2$ , $b = -\frac{x}{2}$ , and $n = 6$ . We only need $r = 0,1,2,3$ because higher values of $r$ produce powers of $x$ above $x^3$ . For each term, substitute the values into the formula and simplify the coefficient.

The $r=0$ term is $2^6 = 64$ . The $r=1$ term is $6 \cdot 2^5 \cdot \left(-\frac{x}{2}\right) = -96x$ . The $r=2$ term is $15 \cdot 2^4 \cdot \frac{x^2}{4} = 60x^2$ . The $r=3$ term is $20 \cdot 2^3 \cdot \left(-\frac{x^3}{8}\right) = -20x^3$ . Combining them gives the expansion up to $x^3$ .

Key Takeaways

This question tests the binomial expansion formula and the ability to handle a negative fractional term. It also reinforces that the binomial coefficient $\binom{6}{r}$ must be multiplied by the correct powers of $2$ and $\left(-\frac{1}{2}\right)^r$ .

Common Mistakes

Forgetting that the second term is $-\frac{x}{2}$ , not $+\frac{x}{2}$ , so the signs of odd-power terms are negative.
Using incorrect binomial coefficients, e.g. writing $\binom{6}{2} = 6$ instead of $15$ .
Not simplifying the coefficients fully, e.g. leaving $6 \cdot 32 \cdot \frac{x}{2}$ instead of $-96x$ .
Stopping at too few terms or including terms beyond $x^3$ when the question only asks up to $x^3$ .

Things to Be Careful About

The mark scheme allows the unsimplified expansion to earn the method mark, but the final simplified terms must be correct. Watch the sign of each term: terms with odd powers of $x$ are negative because of the $\left(-\frac{x}{2}\right)$ factor. Also, be careful with the arithmetic when simplifying fractions such as $2^4 \cdot \frac{x^2}{4}$ .

Techniques used

apply the binomial theoremexpand to the required powersimplify binomial coefficients

(b)

Hence find the coefficient of $x^3$ in the expansion of $(3 - x + 2x^3)\left(2 - \frac{1}{2}x\right)^6$ .

2M Similar

DifficultyMedium-Easy

Worked solution

Approach

Use the expansion from part (a). To find the coefficient of $x^3$ in the product, multiply each term of $(3 - x + 2x^3)$ by the term from the expansion that makes the total power $x^3$ , then sum the contributions.

Working

The relevant products are:

3 \cdot (-20x^3) = -60x^3

(-x) \cdot 60x^2 = -60x^3

2x^3 \cdot 64 = 128x^3

Adding these gives:

-60x^3 - 60x^3 + 128x^3 = 8x^3

So the coefficient of $x^3$ is $8$ .

Answer

8

Final answer

8

Detailed explanation

Walkthrough

We already have $(2 - \frac{1}{2}x)^6 = 64 - 96x + 60x^2 - 20x^3 + \cdots$ . To find the coefficient of $x^3$ in $(3 - x + 2x^3)\left(2 - \frac{1}{2}x\right)^6$ , multiply each term of $(3 - x + 2x^3)$ by a term from the expansion whose powers of $x$ add to $3$ .

$3$ times the $x^3$ term from the expansion: $3 \cdot (-20x^3) = -60x^3$ .
$(-x)$ times the $x^2$ term: $(-x)(60x^2) = -60x^3$ .
$2x^3$ times the constant term: $2x^3 \cdot 64 = 128x^3$ .

Adding these gives $8x^3$ , so the coefficient is $8$ .

Key Takeaways

This is a coefficient-extraction problem. When multiplying two polynomials, the coefficient of $x^k$ is the sum of all products of terms whose exponents add to $k$ . It is important to identify all relevant pairings, not just the most obvious one.

Common Mistakes

Missing one of the three contributing products, especially the $2x^3 \cdot 64$ term.
Using the wrong term from the expansion, e.g. using $64x^3$ or confusing $60x^2$ with $60x^3$ .
Forgetting the negative sign in $(-x) \cdot 60x^2$ .
Adding only the coefficients without attaching the correct powers of $x$ .

Things to Be Careful About

The question says "hence", so you should use your expansion from part (a). The mark scheme expects you to select the correct products using your expansion. Be careful with signs: the $x^3$ term in the expansion is $-20x^3$ , and the multiplier $-x$ makes the second contribution negative. The final coefficient is $8$ , not $8x^3$ , although $8x^3$ is accepted as the term.

Techniques used

identify product terms yielding x^3multiply and combine like terms

Q2MediumTrigonometry

(a)

Solve the equation $\tan^{-1}(5x - 3) = -\frac{1}{4}\pi$ .

2M Similar

DifficultyEasy

Worked solution

Approach

Apply the inverse tangent by taking the tangent of both sides, then solve the resulting linear equation.

Working

Since $\tan^{-1}(5x - 3) = -\frac{1}{4}\pi$ , take tangent of both sides:

5x - 3 = \tan\left(-\frac{1}{4}\pi\right)

Using the exact value $\tan\left(\frac{1}{4}\pi\right) = 1$ and the odd property of tangent,

\tan\left(-\frac{1}{4}\pi\right) = -1

So:

5x - 3 = -1

5x = 2

x = \frac{2}{5}

Answer

x = \frac{2}{5} \quad (0.4)

Final answer

x = 2/5

Detailed explanation

Walkthrough

The equation says that the inverse tangent of $5x - 3$ equals $-\frac{1}{4}\pi$ . Taking the tangent of both sides undoes the inverse tangent, giving $5x - 3 = \tan\left(-\frac{1}{4}\pi\right)$ . Since $\tan\left(\frac{1}{4}\pi\right) = 1$ and tangent is an odd function, $\tan\left(-\frac{1}{4}\pi\right) = -1$ . Solving $5x - 3 = -1$ gives $x = \frac{2}{5}$ . Because $\tan^{-1}$ is a one-to-one function, this is the only solution.

Key Takeaways

The notation $\tan^{-1}$ means the inverse tangent, not the reciprocal of tangent.
Taking the tangent of both sides isolates the input of the inverse function.
Exact values for special angles, including negative angles, are needed.

Common Mistakes

Writing $\tan\left(-\frac{1}{4}\pi\right) = 1$ instead of $-1$ because the negative sign is missed.
Treating $\tan^{-1}$ as $\frac{1}{\tan}$ and trying to use reciprocals.
Not simplifying the equation after taking tangent of both sides.

Things to Be Careful About

Use the exact value $\tan\left(\frac{1}{4}\pi\right) = 1$ ; a calculator decimal is not needed.
The inverse tangent has a restricted range, but here the equation is linear, so no extra solutions arise.
The final answer may be written as $\frac{2}{5}$ or $0.4$ .

Techniques used

take tangent of both sides of an inverse-trig equationevaluate the tangent of a special negative anglesolve the resulting linear equation

(b)

Solve the equation $5\cos^2\theta = 4\sin\theta + 4$ for $\frac{1}{2}\pi \le \theta \le 2\pi$ .

4M Similar

DifficultyMedium

Worked solution

Approach

Use the Pythagorean identity $\cos^2\theta = 1 - \sin^2\theta$ to obtain a quadratic in $\sin\theta$ , factorise and solve it, then find all solutions in the interval $\frac{1}{2}\pi \le \theta \le 2\pi$ .

Working

Start from:

5\cos^2\theta = 4\sin\theta + 4

Replace $\cos^2\theta$ with $1 - \sin^2\theta$ :

5(1 - \sin^2\theta) = 4\sin\theta + 4

Expand and collect all terms on one side:

5 - 5\sin^2\theta = 4\sin\theta + 4

5\sin^2\theta + 4\sin\theta - 1 = 0

Factorise:

(5\sin\theta - 1)(\sin\theta + 1) = 0

So:

\sin\theta = \frac{1}{5} \quad \text{or} \quad \sin\theta = -1

For $\sin\theta = \frac{1}{5}$ , the principal value is $\alpha = \sin^{-1}\left(\frac{1}{5}\right) \approx 0.201$ radians. In the interval $\frac{1}{2}\pi \le \theta \le 2\pi$ , the solution is:

\theta = \pi - \alpha \approx 3.1416 - 0.2014 \approx 2.94

For $\sin\theta = -1$ , the solution in the interval is:

\theta = \frac{3\pi}{2} \approx 4.71

Answer

\theta = 2.94,\ 4.71 \quad (\text{radians})

Final answer

theta = 2.94, 4.71

Detailed explanation

Walkthrough

The equation mixes $\cos^2\theta$ and $\sin\theta$ , so use the Pythagorean identity $\cos^2\theta = 1 - \sin^2\theta$ to write everything in terms of $\sin\theta$ . Substituting gives $5(1 - \sin^2\theta) = 4\sin\theta + 4$ . Expanding and rearranging gives the three-term quadratic $5\sin^2\theta + 4\sin\theta - 1 = 0$ , which factorises as $(5\sin\theta - 1)(\sin\theta + 1) = 0$ . Hence $\sin\theta = \frac{1}{5}$ or $\sin\theta = -1$ .

For $\sin\theta = \frac{1}{5}$ , the calculator principal value is about $0.201$ radians, which is outside the interval $\frac{1}{2}\pi \le \theta \le 2\pi$ . Since sine is positive in the second quadrant, the required solution is $\theta = \pi - 0.201 \approx 2.94$ . For $\sin\theta = -1$ , the solution in the interval is $\theta = \frac{3\pi}{2} \approx 4.71$ .

Key Takeaways

The identity $\cos^2\theta = 1 - \sin^2\theta$ converts a mixed trigonometric equation into a quadratic.
A quadratic in $\sin\theta$ can be factorised and solved like an ordinary quadratic.
After solving for $\sin\theta$ , use the symmetry of the sine graph and the given interval to find every valid angle.

Common Mistakes

Forgetting to replace $\cos^2\theta$ with $1 - \sin^2\theta$ , or making a sign error when rearranging to $5\sin^2\theta + 4\sin\theta - 1 = 0$ .
Factorising incorrectly; the mark scheme requires the three-term quadratic to be formed.
Giving only the principal value $0.201$ for $\sin\theta = \frac{1}{5}$ , which is outside the interval.
Including solutions outside $\frac{1}{2}\pi \le \theta \le 2\pi$ or omitting $\frac{3\pi}{2}$ .

Things to Be Careful About

The interval is in radians: $\frac{1}{2}\pi \approx 1.5708$ and $2\pi \approx 6.2832$ .
For $\sin\theta = \frac{1}{5}$ , the other standard solution $2\pi - \alpha$ gives a negative sine, so it is not valid here.
$\theta = \frac{3\pi}{2}$ is exact and is accepted in place of $4.71$ ; $2.94$ is an approximate value (AWRT).
Ignore answers outside the given range; repetition of $\frac{3\pi}{2}$ is condoned but no other values within the range may be added.

Techniques used

apply the Pythagorean identity to express the equation in one trigonometric functionrearrange and factorise a quadratic in sin thetafind all solutions of a trigonometric equation on a given interval

Q3EasyDifferentiation

The equation of a curve is $y = f(x)$ , where $f(x) = \frac{1}{2}x^{\frac{2}{3}}(x - 2)^2$ . The following points lie on the curve. Non-exact values of the $y$ -coordinates are given correct to 6 decimal places.

A(8, 72), B(8.001, k), C(8.01, 72.300388), D(8.1, 75.038882)

(a)

Find the value of $k$ . Give your answer correct to 6 decimal places.

1M Similar

DifficultyEasy

Worked solution

Approach

The point $B(8.001, k)$ lies on the curve $y = f(x)$ , so its $y$ -coordinate is found by substituting $x = 8.001$ into $f$ . Round the result to 6 decimal places.

Working

Substituting $x = 8.001$ :

\begin{aligned} f(8.001) &= \frac{1}{2}(8.001)^{\frac{2}{3}}(8.001 - 2)^2 \\ &= \frac{1}{2}(8.001)^{\frac{2}{3}}(6.001)^2 \end{aligned}

Evaluate the parts:

(6.001)^2 = 36.012001

(8.001)^{\frac{2}{3}} = 8^{\frac{2}{3}}\left(\frac{8.001}{8}\right)^{\frac{2}{3}} = 4 \times (1.000125)^{\frac{2}{3}} \approx 4.0003333

So:

\begin{aligned} f(8.001) &= \frac{1}{2} \times 4.0003333 \times 36.012001 \\ &= 2.0001667 \times 36.012001 \\ &\approx 72.030004 \end{aligned}

Answer

k = 72.030004

Final answer

72.030004

Detailed explanation

Walkthrough

The point $B(8.001, k)$ is on the curve $y = f(x)$ . A point on a curve satisfies the equation of the curve, so the $y$ -coordinate equals $f(8.001)$ — we substitute $x = 8.001$ into the given expression.

The substitution gives $\frac{1}{2}(8.001)^{\frac{2}{3}}(6.001)^2$ . The bracket $(6.001)^2$ is computed directly as $36.012001$ .

For $(8.001)^{\frac{2}{3}}$ we use that $8^{\frac{2}{3}} = (2^3)^{\frac{2}{3}} = 2^2 = 4$ , and write $8.001 = 8 \times 1.000125$ . Then $(8.001)^{\frac{2}{3}} = 4(1.000125)^{\frac{2}{3}}$ . Using a calculator (or a binomial estimate) gives $4.0003333$ . Multiplying by $\frac{1}{2}$ and then by $36.012001$ yields $72.030004$ .

Key Takeaways

Substituting an $x$ -value into $f(x)$ gives the $y$ -coordinate of the corresponding point on the curve.
Knowing $8^{\frac{2}{3}} = 4$ helps evaluate the function near $x = 8$ .
Precision matters: keep full accuracy until the final rounding.

Common Mistakes

Rounding the intermediate value $(8.001)^{\frac{2}{3}}$ too early, which distorts the 6-decimal-place answer.
Forgetting the factor $\frac{1}{2}$ in the expression.
Mishandling the index $\frac{2}{3}$ (forgetting it applies only to $x$ , not to $(x-2)^2$ ).

Things to Be Careful About

The mark scheme requires the exact answer $72.030004$ (CAO) — approximations are not accepted.
Carry full calculator precision through the multiplication and round only at the end.

Techniques used

substitute a value into a functioncompute a rational-power expression accurately

(b)

The table below shows the gradients of the chords $AB$ and $AC$ , given correct to 4 decimal places.

Chord	$AB$	$AC$	$AD$
Gradient of chord	30.0039	30.0388

Find the gradient of the chord $AD$ . Give your answer correct to 4 decimal places.

1M Similar

DifficultyEasy

Worked solution

Approach

The gradient of the chord $AD$ is the difference quotient $\frac{y_D - y_A}{x_D - x_A}$ . Use $A(8, 72)$ and $D(8.1, 75.038882)$ and round to 4 decimal places.

Working

\text{Gradient of } AD = \frac{75.038882 - 72}{8.1 - 8}

= \frac{3.038882}{0.1} = 30.38882

Rounding to 4 decimal places:

\text{Gradient of } AD = 30.3888

Answer

30.3888

Final answer

30.3888

Detailed explanation

Walkthrough

The gradient of the chord joining two points $(x_1, y_1)$ and $(x_2, y_2)$ is given by the difference quotient $\frac{y_2 - y_1}{x_2 - x_1}$ — the change in $y$ divided by the change in $x$ . Here $A = (8, 72)$ and $D = (8.1, 75.038882)$ .

Apply the formula: the numerator is $75.038882 - 72 = 3.038882$ and the denominator is $8.1 - 8 = 0.1$ . Dividing gives $30.38882$ . Since the table rounds to 4 decimal places, we write $30.3888$ .

Key Takeaways

The gradient of a chord is the average rate of change between the two points.
This difference quotient is exactly the quantity that tends to the derivative $f'(8)$ as the second point approaches $A$ .

Common Mistakes

Using the wrong order: $(y_A - y_D)/(x_A - x_D)$ would still give the same value, but mixing coordinates wrongly (e.g. putting $y_A - y_D$ over $x_D - x_A$ ) changes the sign.
Dividing by the wrong difference in $x$ (some might use $8.1 - 8 = 0.1$ incorrectly as $0.01$ ).
Rounding incorrectly: $30.38882$ to 4 dp is $30.3888$ .

Things to Be Careful About

The final answer must be correct to 4 decimal places: $30.3888$ (CAO, not approximate).
Keep the full value $30.38882$ before rounding.

Techniques used

apply the difference quotient formula for gradientdivide the difference of coordinates

(c)

State what the values in the table suggest about the value of $f'(8)$ .

1M Similar

DifficultyEasy

Worked solution

Approach

The gradient of a chord between $A$ and a nearby point on the curve approximates the gradient of the curve at $A$ , i.e. $f'(8)$ . As the second point moves closer to $A$ , the chord gradients tend to a limiting value, which is $f'(8)$ .

Working

From the table (and part (b)):

\text{Gradient of } AB = 30.0039, \quad \text{Gradient of } AC = 30.0388, \quad \text{Gradient of } AD = 30.3888

As $x$ approaches 8 from above (the second point moves from $D$ to $C$ to $B$ ), the gradients approach 30. These gradients approximate the gradient of the tangent at $A$ , so $f'(8) \approx 30$ .

Answer

f'(8) = 30 \text{ (approximately)}

Final answer

30 (approximately)

Detailed explanation

Walkthrough

Each chord joins $A$ to a point on the curve with $x$ -coordinate slightly greater than 8. The gradient of such a chord is the average rate of change of $y$ with respect to $x$ over that interval, i.e. $\frac{f(x) - f(8)}{x - 8}$ .

As the second point moves closer to $A$ (from $D$ at $x = 8.1$ to $C$ at $8.01$ to $B$ at $8.001$ ), the chord gradients are $30.3888$ , $30.0388$ , $30.0039$ . They are clearly approaching 30. The limit of these difference quotients as $x \to 8$ is exactly the derivative $f'(8)$ — the gradient of the tangent at $A$ . Hence the table suggests $f'(8) = 30$ .

Key Takeaways

The derivative at a point is the limit of the gradient of chords (the difference quotient).
Numerical chord gradients give an excellent way to estimate the gradient of a tangent.
All the values here come from the right-hand side ( $x > 8$ ), giving a right-hand estimate approaching 30.

Common Mistakes

Quoting one of the chord gradients (e.g. 30.0039) as $f'(8)$ instead of the limiting value 30.
Thinking the derivative is exactly one of the table values rather than the value they approach.

Things to Be Careful About

The mark scheme accepts only 30 (or 30.0) with words like 'approximately' or 'around' — it must clearly be the limiting value.
This is an estimate from the right; the exact derivative is not computed here.

Techniques used

interpret chord gradients as approximations to the tangent gradienttake the limiting value as the second point approaches the fixed point

Q4MediumSeries

The first, second and third terms of a progression are 20, $k$ and $k - 5$ respectively.

(a)

Given that the progression is arithmetic, find the 30th term.

2M Similar

DifficultyMedium-Easy

Worked solution

Approach

For an arithmetic progression, consecutive terms differ by a constant $d$ . Find $d$ from the second and third terms, then use $u_n = a + (n-1)d$ .

Working

The common difference is

d = (k-5) - k = -5

The first term is $a = 20$ . For the 30th term, $n = 30$ :

u_{30} = 20 + (30-1)(-5)

u_{30} = 20 + 29(-5) = 20 - 145 = -125

Answer

u_{30} = -125

Final answer

-125

Detailed explanation

Walkthrough

We are told the progression is arithmetic, so the difference between consecutive terms is constant. The first two terms are $20$ and $k$ , and the second and third terms are $k$ and $k-5$ . The common difference is therefore $d = (k-5) - k = -5$ . The $k$ cancels, leaving $d = -5$ . The first term is $a = 20$ . The 30th term is found by adding the common difference 29 times to the first term, which is exactly what the formula $u_n = a + (n-1)d$ does: $u_{30} = 20 + 29(-5) = -125$ .

Key Takeaways

In an arithmetic progression, the common difference is the difference between any two consecutive terms.
The nth term formula $u_n = a + (n-1)d$ gives the term at position $n$ .
Here the $k$ cancels when finding $d$ , so the value of $k$ is not needed.

Common Mistakes

Using $n = 30$ as the multiplier instead of $n - 1 = 29$ .
Forgetting that $d$ is negative, which changes the sign of the final answer.
Confusing the first term $a$ with the second term $k$ .

Things to Be Careful About

The common difference is $-5$ , not $5$ .
The 30th term requires 29 steps from the first term, not 30.
The question only asks for the 30th term, so no sum formula is needed.

Techniques used

determine the common difference between consecutive termsapply the nth term formula for an arithmetic progression

(b)

Given instead that the progression is geometric, find the sum to infinity.

4M Similar

DifficultyMedium

Worked solution

Approach

For a geometric progression, the ratio of consecutive terms is constant. Write the common ratio using the first two terms and using the last two terms, equate them, solve for $k$ , then use the sum-to-infinity formula.

Working

The common ratio is
$r = \frac{k}{20} = \frac{k-5}{k}$

Cross-multiply:

$k^2 = 20(k-5)$

$k^2 = 20k - 100$

$k^2 - 20k + 100 = 0$

$(k-10)^2 = 0$

So $k = 10$ . Hence

$r = \frac{k}{20} = \frac{10}{20} = \frac{1}{2}$

Since $|r| = \frac{1}{2} < 1$ , the sum to infinity exists:

$S_\infty = \frac{a}{1-r} = \frac{20}{1-\frac{1}{2}}$

$S_\infty = \frac{20}{\frac{1}{2}} = 40$

Answer

$S_\infty = 40$

Final answer

40

Detailed explanation

Walkthrough

In a geometric progression, the ratio of any term to the previous term is constant. Using the first two terms gives $r = \frac{k}{20}$ . Using the second and third terms gives $r = \frac{k-5}{k}$ . Since both expressions represent the same ratio, set them equal: $\frac{k}{20} = \frac{k-5}{k}$ . Cross-multiplying gives $k^2 = 20(k-5)$ , which simplifies to $k^2 - 20k + 100 = 0$ . This is a perfect square, $(k-10)^2 = 0$ , so $k = 10$ . Substituting back gives $r = \frac{10}{20} = \frac{1}{2}$ . Because $|r| < 1$ , the geometric series converges, so the sum to infinity is $\frac{a}{1-r} = \frac{20}{1-\frac{1}{2}} = 40$ .

Key Takeaways

A geometric progression has a constant ratio between consecutive terms.
Equating two expressions for the common ratio turns the problem into a quadratic equation.
The sum to infinity of a geometric progression exists only when $|r| < 1$ .

Common Mistakes

Forgetting to equate the two ratio expressions and only using $r = \frac{k}{20}$ .
Making a sign error when cross-multiplying or rearranging the quadratic.
Using the finite sum formula $\frac{a(1-r^n)}{1-r}$ instead of the sum-to-infinity formula.
Forgetting to check that $|r| < 1$ before applying the sum-to-infinity formula.

Things to Be Careful About

The quadratic $k^2 - 20k + 100 = 0$ has a repeated root, so $k = 10$ is the only solution.
With $r = \frac{1}{2}$ , the series converges, so the infinite sum is finite.
The first term is $a = 20$ , not $k$ or $k-5$ .

Techniques used

find the common ratio in two waysequate the two ratio expressions and solve a quadraticapply the sum to infinity formulaverify the convergence condition

Q5MediumCircular Measure

Fig. 5.1

The diagram shows part of a circle with centre $O$ and radius $4\text{ cm}$ . The chord $PQ$ is of length $4\sqrt{3}\text{ cm}$ and angle $POQ = \theta$ radians. The point $X$ lies on the circle.

(a)

Find the exact value of $\theta$ .

2M Similar

DifficultyMedium-Easy

Worked solution

Approach

Triangle $OPQ$ is isosceles with $OP = OQ = 4$ and $PQ = 4\sqrt{3}$ . Drop the perpendicular from $O$ to the midpoint $M$ of $PQ$ . This perpendicular bisects both the chord $PQ$ and the angle $\theta$ . In the right triangle $OMP$ , use the sine ratio to find $\theta/2$ .

Working

The perpendicular from $O$ meets $PQ$ at its midpoint $M$ , so:

MP = \frac{1}{2} \times PQ = \frac{1}{2} \times 4\sqrt{3} = 2\sqrt{3}

In the right-angled triangle $OMP$ , the angle at $O$ is $\theta/2$ , so:

\sin\left(\frac{\theta}{2}\right) = \frac{MP}{OP} = \frac{2\sqrt{3}}{4} = \frac{\sqrt{3}}{2}

Since $0 < \theta < 2\pi$ , we have $0 < \theta/2 < \pi$ , so $\theta/2 = \pi/3$ .

Answer

\theta = \frac{2\pi}{3}

Final answer

θ = 2π/3

Detailed explanation

Walkthrough

The triangle $OPQ$ has two equal sides (the radii) of length $4$ , and a base (the chord) of length $4\sqrt{3}$ . This is an isosceles triangle.

In an isosceles triangle, the perpendicular from the apex to the base has two useful properties: it bisects the base, and it bisects the angle at the apex. Here, the apex is $O$ and the base is $PQ$ . So the perpendicular from $O$ to $PQ$ hits $PQ$ at the midpoint $M$ , and the angle at $O$ in the right triangle $OMP$ is exactly $\theta/2$ .

The half-chord $MP$ is $4\sqrt{3}/2 = 2\sqrt{3}$ , and the hypotenuse is the radius $OP = 4$ . The sine of the half-angle is the ratio of the opposite side to the hypotenuse, giving $\sin(\theta/2) = 2\sqrt{3}/4 = \sqrt{3}/2$ .

We recognise the exact value $\sin(\pi/3) = \sqrt{3}/2$ . Since $0 < \theta/2 < \pi$ , the only solution is $\theta/2 = \pi/3$ , hence $\theta = 2\pi/3$ .

Key Takeaways

A perpendicular from the vertex of an isosceles triangle to the base creates two congruent right triangles.
The relationship $\sin(\theta/2) = (\text{half-chord})/\text{radius}$ is a key result in circle geometry.
The exact value $\sin(60°) = \sqrt{3}/2$ is required.

Common Mistakes

Writing $\theta/2 = \pi/3$ as the final answer without doubling to $\theta = 2\pi/3$ .
Using $4\sqrt{3}$ instead of $2\sqrt{3}$ for the half-chord.

Things to Be Careful About

Ensure the answer is in radians, not degrees.
The half-chord is half the chord length, not the full chord.

Techniques used

bisect isosceles triangle with perpendicular from Oapply sin of half-angle from right triangleuse exact value sin(π/3) = √3/2

(b)

Find the exact area of the segment $PXQ$ .

3M Similar

DifficultyMedium

Worked solution

Approach

Since $X$ lies on the major arc, the segment $PXQ$ is the major segment. The area of the major segment equals the area of the major sector plus the area of the triangle $OPQ$ :

\text{Area of major segment} = \text{Area of major sector} + \text{Area of } \triangle OPQ

Working

The major sector has angle $2\pi - \theta = 2\pi - \frac{2\pi}{3} = \frac{4\pi}{3}$ and radius $r = 4$ .

Area of major sector:

\frac{1}{2} \times r^2 \times (2\pi - \theta) = \frac{1}{2} \times 4^2 \times \frac{4\pi}{3} = \frac{32\pi}{3}

Area of triangle $OPQ$ using $\frac{1}{2}ab\sin C$ with the two radii as the two sides and the included angle $\theta$ :

\frac{1}{2} \times 4 \times 4 \times \sin\left(\frac{2\pi}{3}\right) = 8 \times \frac{\sqrt{3}}{2} = 4\sqrt{3}

Area of segment $PXQ$ :

\frac{32\pi}{3} + 4\sqrt{3}

Answer

\frac{32\pi}{3} + 4\sqrt{3} \text{ cm}^2

Final answer

(32π/3 + 4√3) cm²

Detailed explanation

Walkthrough

The point $X$ lies on the major arc, so the segment $PXQ$ (bounded by the chord $PQ$ and the arc through $X$ ) is the major segment — the larger of the two segments cut off by the chord $PQ$ .

A useful identity is:

\text{Major segment} = \text{Major sector} + \triangle OPQ

This works because the major sector (the pie slice with the reflex angle) does not include the triangle $OPQ$ , but together the major sector and the triangle exactly fill the major segment.

Major sector area: The reflex angle is $2\pi - \theta = 2\pi - 2\pi/3 = 4\pi/3$ . Using the formula $\frac{1}{2}r^2 \theta$ with the reflex angle:

\frac{1}{2} \times 4^2 \times \frac{4\pi}{3} = 8 \times \frac{4\pi}{3} = \frac{32\pi}{3}

Triangle area: Using $\frac{1}{2}ab\sin C$ with $a = b = 4$ (the two radii) and the included angle $\theta = 2\pi/3$ :

\frac{1}{2} \times 4 \times 4 \times \sin\left(\frac{2\pi}{3}\right) = 8 \times \frac{\sqrt{3}}{2} = 4\sqrt{3}

Sum: $\frac{32\pi}{3} + 4\sqrt{3}$ .

An equivalent approach is:

\text{Major segment} = \pi r^2 - \text{Minor segment} = 16\pi - \left(\frac{16\pi}{3} - 4\sqrt{3}\right) = 16\pi - \frac{16\pi}{3} + 4\sqrt{3} = \frac{32\pi}{3} + 4\sqrt{3}

Both methods give the same final answer.

Key Takeaways

A segment is the region between a chord and an arc.
The major segment can be computed as the major sector plus the central triangle, or as the whole circle minus the minor segment.
Sector area: $\frac{1}{2}r^2\theta$ (with $\theta$ in radians).
Triangle area: $\frac{1}{2}ab\sin C$ .

Common Mistakes

Confusing the major and minor segments. The point $X$ on the major arc tells us the segment includes the major arc, so it is the major segment.
Subtracting the triangle from the major sector instead of adding it (this would give a different region, not the major segment).
Computing the sector area with the smaller angle $\theta$ instead of the reflex angle $2\pi - \theta$ .
Getting the sign of $\sin(2\pi/3)$ wrong (it is positive $\sqrt{3}/2$ , not negative).

Things to Be Careful About

The exact value $\sin(2\pi/3) = \sqrt{3}/2$ is positive (not negative).
The sector area formula requires the angle in radians.
The final answer should be in exact form, not decimal approximation.

Techniques used

identify that PXQ is the major segmentapply major sector area formula with reflex angle 2π - θapply triangle area formula (1/2)ab sin C with the included anglecombine sector area and triangle area to get the major segment

Q6MediumFunctions

Fig. 6.1

The diagram shows the graphs of $y = x^3$ and $y = f(x)$ . The graph of $y = x^3$ is transformed to the graph of $y = f(x)$ by a sequence of transformations.

(a)

Describe fully a suitable sequence of transformations. Make clear the order in which the transformations are applied.

5M Similar

DifficultyMedium

Worked solution

Approach

Compare key points on the graph of $y = x^3$ with corresponding points on the graph of $y = f(x)$ . The point of inflection on $y = x^3$ is at $(0, 0)$ . On $y = f(x)$ , the point of inflection is at $(4, -4)$ . We also check another point: on $y = x^3$ , $(1, 1)$ is on the curve. On $y = f(x)$ , $(5, -2)$ is on the curve.

Working

Method 1: Stretch then Translation

Apply a stretch parallel to the $y$ -axis with scale factor $2$ to $y = x^3$ :
$y = 2x^3$
The point $(1, 1)$ maps to $(1, 2)$ . The point of inflection remains at $(0, 0)$ .
Apply a translation by the vector $\binom{4}{-4}$ (i.e., $+4$ in the $x$ -direction and $-4$ in the $y$ -direction):
$y = 2(x - 4)^3 - 4$
The point $(1, 2)$ maps to $(1+4, 2-4) = (5, -2)$ , which matches the graph. The point of inflection maps to $(0+4, 0-4) = (4, -4)$ , which also matches.

Sequence:

Stretch parallel to the $y$ -axis (or $x$ -axis invariant) with scale factor $2$ .
Translation by vector $\binom{4}{-4}$ (or $+4$ in the $x$ -direction and $-4$ in the $y$ -direction).

Alternative Method: Translation then Stretch

Apply a translation by vector $\binom{4}{-2}$ to $y = x^3$ :
$y = (x - 4)^3 - 2$
The point $(1, 1)$ maps to $(1+4, 1-2) = (5, -1)$ .
Apply a stretch parallel to the $y$ -axis with scale factor $2$ (or $x$ -axis invariant):
$y = 2((x - 4)^3 - 2) = 2(x - 4)^3 - 4$
The point $(5, -1)$ maps to $(5, 2(-1)) = (5, -2)$ , which matches the graph.

Sequence:

Translation by vector $\binom{4}{-2}$ (or $+4$ in the $x$ -direction and $-2$ in the $y$ -direction).
Stretch parallel to the $y$ -axis (or $x$ -axis invariant) with scale factor $2$ .

Answer

A suitable sequence is:

Stretch parallel to the $y$ -axis with scale factor $2$ .
Translation by vector $\binom{4}{-4}$ .

(Alternatively: Translation by $\binom{4}{-2}$ followed by stretch parallel to the $y$ -axis with scale factor $2$ .)

Final answer

Stretch parallel to y-axis by scale factor 2, then translation by vector (4, -4).

Detailed explanation

Walkthrough

First, identify the key features of the original graph $y = x^3$ . The point of inflection is at $(0, 0)$ , and it passes through $(1, 1)$ and $(-1, -1)$ .

Next, identify the corresponding features on the transformed graph $y = f(x)$ . The point of inflection is at $(4, -4)$ . The curve passes through $(5, -2)$ and $(3, -6)$ .

To find the transformations, we can compare how the points move.

Method 1: Stretch then Translation
If we stretch $y = x^3$ parallel to the $y$ -axis by a scale factor of $2$ , the new equation is $y = 2x^3$ . The point $(1, 1)$ becomes $(1, 2)$ . The point of inflection is still at $(0, 0)$ .
Now we need to move $(1, 2)$ to $(5, -2)$ and $(0, 0)$ to $(4, -4)$ . This is a translation of $+4$ in the $x$ -direction and $-4$ in the $y$ -direction, which is the vector $\binom{4}{-4}$ .
Applying this translation to $y = 2x^3$ gives $y = 2(x - 4)^3 - 4$ .

Method 2: Translation then Stretch
If we translate $y = x^3$ first, we must translate the point $(1, 1)$ to a position that, when stretched vertically by $2$ , lands on $(5, -2)$ . The $y$ -coordinate before stretching must be $-2 / 2 = -1$ . The $x$ -coordinate is unaffected by a vertical stretch, so it must be $5$ . Thus, $(1, 1)$ must be translated to $(5, -1)$ .
This translation is $+4$ in the $x$ -direction and $-2$ in the $y$ -direction, which is the vector $\binom{4}{-2}$ .
Applying this translation gives $y = (x - 4)^3 - 2$ . Then applying the vertical stretch by $2$ gives $y = 2((x - 4)^3 - 2) = 2(x - 4)^3 - 4$ .

Key Takeaways

Transformations can be applied in different orders (stretch then translate, or translate then stretch), but the translation vector will be different depending on the order.
A stretch parallel to the $y$ -axis by scale factor $k$ transforms $y = f(x)$ to $y = k f(x)$ .
A translation by vector $\binom{a}{b}$ transforms $y = f(x)$ to $y = f(x - a) + b$ .
Always check your answer by applying the transformations to a known point on the original graph and verifying it lands on the transformed graph.

Common Mistakes

Incorrect order or vector: Applying a translation of $\binom{4}{-4}$ after a stretch is different from applying it before. If you translate by $\binom{4}{-4}$ first, you get $y = (x-4)^3 - 4$ , and then stretching gives $y = 2((x-4)^3 - 4) = 2(x-4)^3 - 8$ , which is incorrect.
Wrong stretch direction: A stretch parallel to the $x$ -axis would change the $x$ -coordinates, not the $y$ -coordinates. Here, the $y$ -values are scaled, so it must be parallel to the $y$ -axis.
Forgetting to specify the axis invariant: When describing a stretch, it is important to state that the $x$ -axis is invariant (or that it is parallel to the $y$ -axis) to be fully correct.

Things to Be Careful About

Order of transformations: The mark scheme accepts two main sequences. If you use a translation before a stretch, the $y$ -component of the translation must be half the final $y$ -displacement because the stretch will double it. Specifically, the $y$ -displacement is $-4$ , so the pre-stretch translation must be $-2$ .
Notation: Use vector notation $\binom{a}{b}$ or clearly state the $x$ and $y$ directions. Avoid ambiguous terms like 'right' and 'down' without specifying the axes, although some leniency may apply if the intention is clear.
Final equation check: Always derive the final equation $y = a(x-h)^3 + k$ and verify it matches the given points on the graph.

Techniques used

identify key points on original and transformed graphsdetermine stretch factor from vertical scalingdetermine translation vector from point displacementspecify sequence and order of transformations

(b)

You are given that $f(x) = a(x + b)^3 + c$ .

State the values of the constants $a$ , $b$ and $c$ .

3M Similar

DifficultyMedium-Easy

Worked solution

Approach

From part (a), we determined that the function is $f(x) = 2(x - 4)^3 - 4$ . We are given the form $f(x) = a(x + b)^3 + c$ . We can match the constants directly.

Working

The equation derived from the transformations is:

f(x) = 2(x - 4)^3 - 4

Comparing this with the given form:

f(x) = a(x + b)^3 + c

We can identify the constants by matching the corresponding parts:

The coefficient of the cubic term is $a$ , so $a = 2$ .
The term inside the cube is $(x - 4)$ , which matches $(x + b)$ , so $b = -4$ .
The constant term is $c$ , so $c = -4$ .

Answer

a = 2, \quad b = -4, \quad c = -4

Final answer

a = 2, b = -4, c = -4

Detailed explanation

Walkthrough

In part (a), we found the equation for $f(x)$ to be $f(x) = 2(x - 4)^3 - 4$ . The question asks us to state the values of $a$ , $b$ , and $c$ for the form $f(x) = a(x + b)^3 + c$ .

We simply equate the two expressions:

2(x - 4)^3 - 4 = a(x + b)^3 + c

By comparing the coefficients and constants:

$a$ is the multiplier outside the cube, which is $2$ .
$b$ is the value added to $x$ inside the cube. Since we have $(x - 4)$ , this is $(x + (-4))$ , so $b = -4$ .
$c$ is the constant added at the end, which is $-4$ .

Key Takeaways

The form $a(x + b)^3 + c$ represents a cubic function with a vertical stretch by $a$ , a horizontal translation by $-b$ , and a vertical translation by $c$ .
Be careful with the sign of $b$ . The term $(x + b)$ means a translation to the left by $b$ if $b > 0$ , or to the right by $|b|$ if $b < 0$ . Here, $(x - 4)$ means $b = -4$ .

Common Mistakes

Sign error for b: Writing $b = 4$ instead of $b = -4$ because they see $(x - 4)$ and assume $b = 4$ . Remember the form is $(x + b)$ , so $x - 4 = x + (-4)$ .
Confusing c with the y-coordinate of the inflection: While $c$ is indeed the y-coordinate of the point of inflection for this form (since the inflection is at $(-b, c)$ ), it's important to extract it directly from the equation.

Things to Be Careful About

Individual values vs. equation: The mark scheme states that individual values are considered the final answer. Ensure you clearly state $a = 2$ , $b = -4$ , and $c = -4$ rather than just writing the full equation, although the equation implies the values.
Form matching: Ensure the given form matches your derived equation exactly. If the given form was $a(x - b)^3 + c$ , then $b$ would be $4$ . Always check the sign in the given form.

Techniques used

compare given function form with derived equationextract constants by matching coefficients

Q74MMediumFunctions

The function $g$ is defined by $g(x) = \frac{2}{ax - 3} + \frac{1}{2}$ for $x > \frac{3}{a}$ , where $a$ is a positive constant.

Find $g^{-1}(x)$ and hence verify that if $a = 6$ then $g^{-1}(x) \equiv g(x)$ .

DifficultyMedium

Worked solution

Approach

Set $y = g(x)$ and rearrange to make $x$ the subject. Interchange $x$ and $y$ to obtain $g^{-1}(x)$ , then substitute $a = 6$ into both $g^{-1}(x)$ and $g(x)$ and simplify to confirm the two functions are identical.

Working

Let $y = g(x)$ :

y = \frac{2}{ax - 3} + \frac{1}{2}

Subtract $\frac{1}{2}$ and multiply through:

\left(y - \frac{1}{2}\right)(ax - 3) = 2

So

ax - 3 = \frac{2}{y - \frac{1}{2}}

Since $y - \frac{1}{2} = \frac{2y - 1}{2}$ ,

ax - 3 = \frac{4}{2y - 1}

Therefore

ax = \frac{4}{2y - 1} + 3

and

x = \frac{4}{a(2y - 1)} + \frac{3}{a}

Now interchange $x$ and $y$ , so

g^{-1}(x) = \frac{4}{a(2x - 1)} + \frac{3}{a}

or equivalently

g^{-1}(x) = \frac{6x + 1}{a(2x - 1)}

Verify for $a = 6$ :

g^{-1}(x) = \frac{4}{6(2x - 1)} + \frac{3}{6} = \frac{2}{3(2x - 1)} + \frac{1}{2} = \frac{2}{6x - 3} + \frac{1}{2}

which is exactly $g(x)$ when $a = 6$ . Hence $g^{-1}(x) \equiv g(x)$ when $a = 6$ .

Answer

g^{-1}(x) = \frac{4}{a(2x - 1)} + \frac{3}{a}

With $a = 6$ , $g^{-1}(x) \equiv g(x)$ .

Final answer

g^-1(x) = 4/[a(2x - 1)] + 3/a; when a = 6, g^-1(x) is identically equal to g(x).

Detailed explanation

Walkthrough

Start by writing the defining equation with $y$ in place of $g(x)$ :

y = \frac{2}{ax - 3} + \frac{1}{2}.

This is the required first step because an inverse is found by solving $y = g(x)$ for $x$ .

Subtract $\frac{1}{2}$ from both sides and multiply by $ax - 3$ . This moves the denominator out of the fraction:

\left(y - \frac{1}{2}\right)(ax - 3) = 2.

Next isolate $ax - 3$ by dividing by $y - \frac{1}{2}$ :

ax - 3 = \frac{2}{y - \frac{1}{2}}.

Notice that $y - \frac{1}{2}$ is not a simple denominator, so rewrite it as $\frac{2y - 1}{2}$ . Dividing by this fraction means multiplying by its reciprocal, giving:

ax - 3 = \frac{4}{2y - 1}.

Now make $ax$ the subject by adding 3:

ax = \frac{4}{2y - 1} + 3.

Then divide by $a$ to make $x$ the subject:

x = \frac{4}{a(2y - 1)} + \frac{3}{a}.

Finally, swap the roles of $x$ and $y$ . The resulting expression is the inverse:

g^{-1}(x) = \frac{4}{a(2x - 1)} + \frac{3}{a}.

This can also be written as a single fraction:

g^{-1}(x) = \frac{4}{a(2x - 1)} + \frac{3}{a} = \frac{4 + 3(2x - 1)}{a(2x - 1)} = \frac{6x + 1}{a(2x - 1)}.

For the verification, put $a = 6$ into $g^{-1}(x)$ :

g^{-1}(x) = \frac{4}{6(2x - 1)} + \frac{3}{6} = \frac{2}{6x - 3} + \frac{1}{2}.

But the right-hand side is exactly $g(x)$ with $a = 6$ :

g(x) = \frac{2}{6x - 3} + \frac{1}{2}.

Therefore $g^{-1}(x) \equiv g(x)$ when $a = 6$ .

Key Takeaways

The central skill is reversing a function by algebra: write $y = g(x)$ , rearrange to make $x$ the subject, then interchange $x$ and $y$ . You must also be comfortable simplifying compound fractions such as $\frac{2}{(2y-1)/2} = \frac{4}{2y-1}$ . Finally, verifying equality of functions requires substituting the given parameter value and simplifying until both sides look exactly the same.

Common Mistakes

Not showing the initial step $y = g(x)$ and the rearrangement. Mark scheme requires method; an unsupported inverse is not accepted.
Losing $a$ when dividing by $a$ in the final step. Always divide the whole right-hand side by $a$ .
Forgetting to swap $x$ and $y$ after making $x$ the subject.
Treating $\frac{2}{y - \frac{1}{2}}$ as $\frac{2}{y} - \frac{2}{1/2}$ ; this is not valid. The whole denominator $y-\frac{1}{2}$ must be used.
Not simplifying the verification far enough to observe that the expression matches $g(x)$ .

Things to Be Careful About

The domain restriction $x > \frac{3}{a}$ ensures $ax - 3 > 0$ , so the fraction is defined for the original function. The inverse expression is equivalent in either form, but if you use the single-fraction form $(6x+1)/(a(2x-1))$ , take care that $2x - 1 \neq 0$ when substituting values. For the substitution $a=6$ , show the final simplification clearly to earn the dependent verification mark.

Techniques used

set y = g(x) and rearrange to isolate the denominatormake x the subject and swap variables to write the inversesimplify a compound fraction to an equivalent rational formsubstitute the special value and compare expressions to verify identity

Approach

The centre of a circle is the midpoint of a diameter, and the radius is the distance from this centre to any endpoint. So we use the standard form $(x-a)^2+(y-b)^2=r^2$ .

Working

The centre of the circle is the midpoint of $(6,1)$ and $(-2,7)$ :

\left(\frac{6+(-2)}{2},\frac{1+7}{2}\right)=(2,4)

Using the endpoint $(6,1)$ , the radius squared is:

(6-2)^2+(1-4)^2=4^2+(-3)^2=16+9=25

So the equation of the circle is:

(x-2)^2+(y-4)^2=25

Answer

(x-2)^2+(y-4)^2=25

Final answer

(x - 2)^2 + (y - 4)^2 = 25

Detailed explanation

Walkthrough

The endpoints of the diameter are $(6,1)$ and $(-2,7)$ . The centre of the circle is exactly halfway between them, so we find the midpoint by averaging the $x$ -coordinates and the $y$ -coordinates:

\left(\frac{6+(-2)}{2},\frac{1+7}{2}\right)=(2,4)

Once the centre is known, the radius is the distance from the centre to any point on the circle. Using $(6,1)$ :

(6-2)^2+(1-4)^2=4^2+(-3)^2=16+9=25

So $r^2=25$ . Since the circle has centre $(2,4)$ and radius $5$ , its equation is:

(x-2)^2+(y-4)^2=25

Key Takeaways

The centre of a circle is the midpoint of any diameter. The equation of a circle with centre $(a,b)$ and radius $r$ is $(x-a)^2+(y-b)^2=r^2$ . To find the radius, use the distance from the centre to a known point on the circle.

Common Mistakes

Forgetting to halve the coordinates when finding the midpoint, so using the diameter endpoints as the centre.
Using the full diameter length as the radius instead of the distance from the centre to a point.
Mixing up the signs in the standard equation: $(x-a)^2+(y-b)^2=r^2$ uses the coordinates of the centre.

Things to Be Careful About

The radius is squared in the equation, so it is enough to compute $r^2$ directly. The centre must be the midpoint of the diameter, not one of the endpoints. The standard form is accepted, but an expanded form such as $x^2+y^2-4x-8y-5=0$ is also correct.

Techniques used

find the midpoint of a diameterfind the radius squared using the distance from the centre to a pointwrite the circle equation in centre-radius form

(b)

There are two tangents to the circle which have gradient $-\frac{1}{2}$ .

Find the exact values of the $x$ -coordinates of the points at which these tangents touch the circle.

5M Similar

DifficultyMedium

Worked solution

Approach

The tangent at a point on a circle is perpendicular to the radius at that point. Since the tangent has gradient $-\frac{1}{2}$ , the radius has gradient $2$ . The points of tangency lie on this normal line through the centre, so we find where this line meets the circle.

Working

The gradient of the normal is:

-\frac{1}{-\frac{1}{2}}=2

The normal line through the centre $(2,4)$ is:

y-4=2(x-2)

which simplifies to $y=2x$ .

Substitute $y=2x$ into the circle equation:

(x-2)^2+(2x-4)^2=25

Since $2x-4=2(x-2)$ :

(x-2)^2+4(x-2)^2=5(x-2)^2=25

So:

(x-2)^2=5

Taking square roots:

x-2=\pm\sqrt{5}

Therefore:

x=2\pm\sqrt{5}

Answer

x=2+\sqrt{5}\quad\text{or}\quad x=2-\sqrt{5}

Final answer

x = 2 + sqrt(5) or x = 2 - sqrt(5)

Detailed explanation

Walkthrough

At the point where a tangent touches a circle, the tangent is perpendicular to the radius. The tangent has gradient $-\frac{1}{2}$ , so the radius has gradient $2$ , because the product of perpendicular gradients is $-1$ .

The radius through the centre $(2,4)$ therefore has equation:

y-4=2(x-2)

which simplifies to $y=2x$ . The points where the tangent touches the circle are exactly the points where this normal line meets the circle. Substituting $y=2x$ into the circle equation gives:

(x-2)^2+(2x-4)^2=25

Since $2x-4=2(x-2)$ , this becomes

(x-2)^2+4(x-2)^2=5(x-2)^2=25

so

(x-2)^2=5

Taking square roots gives $x-2=\pm\sqrt{5}$ , so the two $x$ -coordinates are $2+\sqrt{5}$ and $2-\sqrt{5}$ .

Key Takeaways

For a circle, the tangent at a point is perpendicular to the radius at that point. The product of the gradients of two perpendicular lines is $-1$ (unless one is vertical). To find where a tangent with a given gradient touches the circle, find the normal line through the centre and intersect it with the circle.

Common Mistakes

Using the tangent gradient $-\frac{1}{2}$ instead of the normal gradient $2$ when forming the line through the centre.
Forgetting to take the reciprocal when finding the perpendicular gradient.
Losing the $\pm$ when solving $(x-2)^2=5$ .
Giving decimal answers instead of exact surds.

Things to Be Careful About

The normal line must pass through the centre of the circle, not through the diameter endpoints. When substituting $y=2x$ , note that $2x-4=2(x-2)$ , which simplifies the algebra. The final answer must be exact: $2+\sqrt{5}$ and $2-\sqrt{5}$ .

Techniques used

use the perpendicular gradient relationshipform the equation of the normal line through the centresubstitute the line into the circle equationsolve the resulting quadratic for the x-coordinate

Q98MMediumIntegrationQuadratics

Fig. 9.1

The diagram shows part of the curve with equation $y = \frac{1}{2}x + \frac{4}{x}$ and the line $y = 4.5$ .

Find the exact volume of the solid formed when the shaded region is rotated through $360^\circ$ about the $x$ -axis.

DifficultyMedium

Worked solution

Approach

The shaded region is bounded above by the line $y = 4.5$ and below by the curve $y = \frac{1}{2}x + \frac{4}{x}$ . When rotated about the $x$ -axis, the solid formed is a washer shape at each cross-section. The volume is found by subtracting the volume generated by the curve from the volume generated by the line, between the $x$ -coordinates of their intersection points.

First, find the intersection points by equating the two equations. Then set up the volume integral using the formula $V = \pi \int_{a}^{b} (y_1^2 - y_2^2) \, dx$ , where $y_1 = 4.5$ is the outer radius and $y_2 = \frac{1}{2}x + \frac{4}{x}$ is the inner radius.

Working

Step 1: Find the limits of integration

Equate the curve and the line:

\frac{1}{2}x + \frac{4}{x} = 4.5

Multiply through by $2x$ to clear denominators:

x^2 + 8 = 9x

Rearrange into a quadratic equation:

x^2 - 9x + 8 = 0

Factorise:

(x - 1)(x - 8) = 0

So the limits are $x = 1$ and $x = 8$ .

Step 2: Set up the volume integral

Using the washer method, the volume $V$ is:

V = \pi \int_{1}^{8} \left[ (4.5)^2 - \left( \frac{1}{2}x + \frac{4}{x} \right)^2 \right] dx

Step 3: Expand the integrand

First, expand $\left( \frac{1}{2}x + \frac{4}{x} \right)^2$ :

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

Now substitute back into the integrand:

(4.5)^2 - \left( \frac{1}{4}x^2 + 4 + \frac{16}{x^2} \right) = 20.25 - \frac{1}{4}x^2 - 4 - \frac{16}{x^2}

Simplify:

= 16.25 - \frac{1}{4}x^2 - 16x^{-2} = \frac{65}{4} - \frac{1}{4}x^2 - 16x^{-2}

Step 4: Integrate

\int \left( \frac{65}{4} - \frac{1}{4}x^2 - 16x^{-2} \right) dx = \frac{65}{4}x - \frac{1}{12}x^3 + \frac{16}{x}

Step 5: Evaluate at the limits

At $x = 8$ :

\frac{65}{4}(8) - \frac{1}{12}(8)^3 + \frac{16}{8} = 130 - \frac{512}{12} + 2 = 132 - \frac{128}{3} = \frac{396 - 128}{3} = \frac{268}{3}

At $x = 1$ :

\frac{65}{4}(1) - \frac{1}{12}(1)^3 + \frac{16}{1} = \frac{65}{4} - \frac{1}{12} + 16 = \frac{195 - 1 + 192}{12} = \frac{386}{12} = \frac{193}{6}

Subtract the lower limit value from the upper limit value:

\frac{268}{3} - \frac{193}{6} = \frac{536}{6} - \frac{193}{6} = \frac{343}{6}

Step 6: Final volume

Multiply by $\pi$ :

V = \frac{343}{6}\pi

Answer

\frac{343}{6}\pi

Final answer

343pi/6

Detailed explanation

Walkthrough

The problem asks for the volume of a solid of revolution. The shaded region is bounded above by the horizontal line $y = 4.5$ and below by the curve $y = \frac{1}{2}x + \frac{4}{x}$ . When this region is rotated $360^\circ$ about the $x$ -axis, each vertical slice forms a washer (a disc with a hole). The volume of such a solid is found by integrating the area of the washer cross-sections along the $x$ -axis.

Step 1: Find the limits of integration.
The region is bounded between the two points where the line and the curve intersect. We find these by setting the equations equal: $\frac{1}{2}x + \frac{4}{x} = 4.5$ . Multiplying by $2x$ gives $x^2 + 8 = 9x$ , which rearranges to $x^2 - 9x + 8 = 0$ . Factoring yields $(x-1)(x-8) = 0$ , so the limits are $x = 1$ and $x = 8$ . These are the $x$ -coordinates of the left and right edges of the shaded region.

Step 2: Set up the volume integral.
The formula for the volume of a solid of revolution using the washer method is $V = \pi \int_{a}^{b} (R^2 - r^2) \, dx$ , where $R$ is the outer radius (distance from the axis of rotation to the outer boundary) and $r$ is the inner radius (distance to the inner boundary). Here, the outer boundary is the line $y = 4.5$ , so $R = 4.5$ . The inner boundary is the curve $y = \frac{1}{2}x + \frac{4}{x}$ , so $r = \frac{1}{2}x + \frac{4}{x}$ . Thus, $V = \pi \int_{1}^{8} \left[ 4.5^2 - \left( \frac{1}{2}x + \frac{4}{x} \right)^2 \right] dx$ .

Step 3: Expand and simplify the integrand.
We need to square the curve equation: $\left( \frac{1}{2}x + \frac{4}{x} \right)^2 = \frac{1}{4}x^2 + 4 + \frac{16}{x^2}$ . Subtracting this from $4.5^2 = 20.25 = \frac{81}{4}$ gives $\frac{81}{4} - 4 - \frac{1}{4}x^2 - 16x^{-2} = \frac{65}{4} - \frac{1}{4}x^2 - 16x^{-2}$ . This is the expression we will integrate.

Step 4: Integrate term by term.
Using the power rule for integration: $\int x^n \, dx = \frac{x^{n+1}}{n+1}$ for $n \neq -1$ , and $\int x^{-2} \, dx = -x^{-1}$ . We get $\frac{65}{4}x - \frac{1}{12}x^3 + \frac{16}{x}$ . Note that the $-16x^{-2}$ integrates to $-16 \cdot \frac{x^{-1}}{-1} = \frac{16}{x}$ .

Step 5: Evaluate at the limits.
Substitute $x = 8$ : $\frac{65}{4}(8) - \frac{512}{12} + \frac{16}{8} = 130 - \frac{128}{3} + 2 = 132 - \frac{128}{3} = \frac{396 - 128}{3} = \frac{268}{3}$ . Substitute $x = 1$ : $\frac{65}{4} - \frac{1}{12} + 16 = \frac{195 - 1 + 192}{12} = \frac{386}{12} = \frac{193}{6}$ . Subtract: $\frac{268}{3} - \frac{193}{6} = \frac{536 - 193}{6} = \frac{343}{6}$ . Finally, multiply by $\pi$ to get $\frac{343}{6}\pi$ .

Key Takeaways

The volume of a solid of revolution about the $x$ -axis between two curves $y = f(x)$ and $y = g(x)$ (where $f(x) \geq g(x) \geq 0$ ) is $V = \pi \int_{a}^{b} (f(x)^2 - g(x)^2) \, dx$ .
Finding the limits of integration often requires solving an equation formed by equating the two bounding curves.
Expanding squared binomial fractions like $\left( \frac{a}{x} + bx \right)^2$ carefully is essential to avoid algebraic errors.
Integrating negative powers of $x$ (like $x^{-2}$ ) requires the power rule: $\int x^{-2} \, dx = -x^{-1}$ , which flips the sign.

Common Mistakes

Forgetting to square the $y$ -values in the volume formula. Students sometimes integrate $(y_1 - y_2)$ instead of $(y_1^2 - y_2^2)$ .
Incorrectly expanding $\left( \frac{1}{2}x + \frac{4}{x} \right)^2$ . A common error is writing $\frac{1}{4}x^2 + \frac{16}{x^2}$ , missing the middle term $4$ .
Sign errors when integrating $-16x^{-2}$ . The integral is $+\frac{16}{x}$ , not $-\frac{16}{x}$ .
Arithmetic mistakes when evaluating fractions at the limits, especially finding common denominators for $\frac{268}{3} - \frac{193}{6}$ .
Omitting the $\pi$ factor in the final answer.

Things to Be Careful About

The limits of integration must be the $x$ -coordinates of the intersection points, not the $y$ -coordinates. Here, $x = 1$ and $x = 8$ are positive, which is consistent with the diagram showing the region in the first quadrant.
Ensure the outer radius is the larger $y$ -value. In the shaded region, the line $y = 4.5$ is above the curve, so $R = 4.5$ and $r = \frac{1}{2}x + \frac{4}{x}$ . Reversing these would give a negative volume.
The question asks for the exact volume, so the answer must be left in terms of $\pi$ as a fraction ( $\frac{343}{6}\pi$ ), not as a decimal approximation.
When multiplying through by $2x$ to solve the intersection equation, remember that $x > 0$ in this region, so no extraneous solutions are introduced by this operation.

Techniques used

equate curve and line to find limits of integrationexpand square of binomial expressionintegrate powers of x and x^-2apply washer method for volume of revolutionevaluate definite integral at limits

Approach

Since the roots are $5m$ and $-9m$ , write $f(x)$ as $p(x-5m)(x+9m)$ . Expand and compare coefficients with $px^2+4x-5$ to obtain equations in $p$ and $m$ , then solve.

Working

With $q=-5$ :

f(x)=px^2+4x-5

Because the roots are $5m$ and $-9m$ , we can write

f(x)=p(x-5m)(x+9m)

Expand:

f(x)=p(x^2+4mx-45m^2)

f(x)=px^2+4pmx-45pm^2

Compare coefficients with $px^2+4x-5$ .

Coefficient of $x$ :

4pm=4 \Rightarrow pm=1

Constant term:

-45pm^2=-5 \Rightarrow 45pm^2=5

Since $pm=1$ , we have $45pm^2=45(pm)m=45m$ :

45m=5

Then

p=\frac{1}{m}=9

Answer

p=9,\quad m=\frac{1}{9}

Final answer

p = 9, m = 1/9

Detailed explanation

Walkthrough

We know that if a quadratic has roots $5m$ and $-9m$ , it can be written as $p(x-5m)(x+9m)$ . Expanding gives $p(x^2+4mx-45m^2)=px^2+4pmx-45pm^2$ . This must equal $px^2+4x-5$ . Comparing the coefficient of $x$ gives $4pm=4$ , so $pm=1$ . Comparing the constant terms gives $-45pm^2=-5$ , or $45pm^2=5$ . Since $pm=1$ , we have $45pm^2=45(pm)m=45m$ , so $45m=5$ , hence $m=\frac{1}{9}$ . Then $p=\frac{1}{m}=9$ . The solution is $p=9$ , $m=\frac{1}{9}$ .

Key Takeaways

This question tests the relationship between roots and coefficients. If the roots are $r_1$ and $r_2$ , the quadratic can be written as $a(x-r_1)(x-r_2)$ , and comparing coefficients gives equations linking the roots and the parameters.

Common Mistakes

Forgetting the factor $p$ when expanding $p(x-5m)(x+9m)$ .
Sign error with the constant term: the constant is $-45pm^2$ , so equating it to $-5$ gives $45pm^2=5$ , not $-45pm^2=5$ .
Dividing by $m$ without noting $m\neq 0$ ; here $pm=1$ already ensures $m\neq 0$ .

Things to Be Careful About

Compare the coefficient of $x$ and the constant term separately. Keep track of $pm^2$ : it is not the same as $pm$ . Also, $m=0$ is impossible because $pm=1$ . The final values can be checked by substituting into $f(x)$ .

Techniques used

express a quadratic in factored form from its rootsexpand and compare coefficientssolve simultaneous equationssubstitute to find constants

Q11Medium-HardDifferentiationCoordinate Geometry

The equation of a curve is $y = \frac{8}{3x - 8} - \frac{6}{x - 1}$ .

(a)

Find the coordinates of the point at which the tangent to the curve at the point $(3, 5)$ intersects the line $y = -8x$ .

6M Similar

DifficultyMedium-Hard

Worked solution

Approach

Differentiate the curve to obtain the gradient function $\frac{dy}{dx}$ , evaluate it at $x = 3$ to find the gradient of the tangent, write the equation of the tangent at $(3, 5)$ , then substitute $y = -8x$ to find the intersection.

Working

Use the chain rule on each term of $y = \frac{8}{3x - 8} - \frac{6}{x - 1}$ :

\frac{dy}{dx} = 8(-1)(3x - 8)^{-2}(3) - 6(-1)(x - 1)^{-2}(1)

\frac{dy}{dx} = -\frac{24}{(3x - 8)^2} + \frac{6}{(x - 1)^2}

At $x = 3$ , the gradient is:

m = -\frac{24}{(3 \cdot 3 - 8)^2} + \frac{6}{(3 - 1)^2} = -\frac{24}{1^2} + \frac{6}{2^2} = -24 + \frac{3}{2} = -\frac{45}{2}

The tangent at $(3, 5)$ :

y - 5 = -\frac{45}{2}(x - 3)

y = -\frac{45}{2}x + \frac{135}{2} + 5 = -\frac{45}{2}x + \frac{145}{2}

Intersection with $y = -8x$ :

-8x = -\frac{45}{2}x + \frac{145}{2}

Multiply by 2:

-16x = -45x + 145

29x = 145

x = 5

Then $y = -8(5) = -40$ .

Answer

(5, -40)

Final answer

(5, -40)

Detailed explanation

Walkthrough

We need the tangent line to the curve at $(3,5)$ and its intersection with $y = -8x$ . Start by differentiating. Each term is a constant times a linear expression raised to the power $-1$ , so the chain rule applies: the outer derivative contributes the power $-1$ as a multiplier, and we must also multiply by the derivative of the inside — $3$ for $3x-8$ and $1$ for $x-1$ . This gives the gradient function $\frac{dy}{dx} = -\frac{24}{(3x-8)^2} + \frac{6}{(x-1)^2}$ .

Next, substitute $x = 3$ into $\frac{dy}{dx}$ , because a tangent's gradient equals the curve's gradient at the point of contact. This yields $m = -\frac{45}{2}$ .

A straight line through $(3,5)$ with gradient $m$ has equation $y - 5 = m(x-3)$ , which we rearrange to $y = -\frac{45}{2}x + \frac{145}{2}$ .

Finally, the intersection with $y = -8x$ is found by equating the two $y$ -expressions. Solving then gives $x = 5$ and hence $y = -40$ .

Key Takeaways

This problem ties together differentiation (chain rule on rational functions), gradient evaluation, and straight-line geometry. The key skill is recognising that once the gradient is found, everything else is a standard line equation and a linear solve.

Common Mistakes

Forgetting the factor $3$ (the derivative of $3x - 8$ ) when differentiating — the mark scheme awards a mark for each correctly differentiated element, so a missing factor costs a mark.
Substituting $x = 3$ into the original $y$ rather than into $\frac{dy}{dx}$ to obtain the gradient.
Sign slips when forming the tangent: from $y - 5 = -\frac{45}{2}(x-3)$ the constant simplifies to $\frac{145}{2}$ .
When intersecting, forgetting to replace $y$ by $-8x$ .

Things to Be Careful About

The chain-rule factors: $\frac{d}{dx}(3x-8)^{-1} = -3(3x-8)^{-2}$ and $\frac{d}{dx}(x-1)^{-1} = -(x-1)^{-2}$ .
Verify the point: at $x = 3$ , the curve value is $\frac{8}{1} - \frac{6}{2} = 8 - 3 = 5$ , matching the given point.
The final answer must be given as coordinates $(5, -40)$ ; the mark scheme accepts this form.

Techniques used

differentiate rational terms using the chain ruleevaluate the gradient at a given pointform the equation of the tangent linesolve the linear system for the intersection

(b)

7M

(i)

Find the $x$ -coordinates of each of the stationary points of the curve.

3M Similar

DifficultyMedium

Worked solution

Approach

Stationary points occur where the gradient function is zero. Set $\frac{dy}{dx} = 0$ , clear the fractions to obtain a quadratic, and solve for the $x$ -coordinates.

Working

The gradient function is:

\frac{dy}{dx} = -\frac{24}{(3x - 8)^2} + \frac{6}{(x - 1)^2}

Set it equal to zero:

-\frac{24}{(3x - 8)^2} + \frac{6}{(x - 1)^2} = 0

\frac{6}{(x - 1)^2} = \frac{24}{(3x - 8)^2}

Cross-multiply:

24(x - 1)^2 = 6(3x - 8)^2

Divide by 6:

4(x - 1)^2 = (3x - 8)^2

Taking square roots (keeping the $\pm$ ):

2(x - 1) = \pm(3x - 8)

With the plus sign:

2x - 2 = 3x - 8 \quad \Rightarrow \quad x = 6

With the minus sign:

2x - 2 = -3x + 8 \quad \Rightarrow \quad 5x = 10 \quad \Rightarrow \quad x = 2

(Equivalently, expanding gives $5x^2 - 40x + 60 = 0$ , i.e. $x^2 - 8x + 12 = 0$ , with roots $x = 2$ and $x = 6$ .)

Answer

x = 2 \quad \text{and} \quad x = 6

Final answer

x = 2 and x = 6

Detailed explanation

Walkthrough

Stationary points are places where the gradient is zero. Using the gradient function, set $\frac{dy}{dx} = 0$ . Move one fraction to the other side and cross-multiply to clear denominators, obtaining $24(x-1)^2 = 6(3x-8)^2$ , then divide by 6 to get $4(x-1)^2 = (3x-8)^2$ . Taking square roots introduces a $\pm$ , giving two linear equations. Solving $2(x-1) = 3x-8$ and $2(x-1) = -(3x-8)$ yields $x = 6$ and $x = 2$ . Expanding instead gives the quadratic $x^2 - 8x + 12 = 0$ , with the same roots.

Key Takeaways

A stationary point satisfies $\frac{dy}{dx} = 0$ .
When clearing fractions, multiply both sides by the denominators without losing the squared structure.
When taking square roots, the $\pm$ must be included to capture both stationary points.

Common Mistakes

Omitting the $\pm$ when taking square roots — the mark scheme only allows this method if the $\pm$ is present, otherwise solutions are lost.
Sign errors when expanding $4(x-1)^2 = (3x-8)^2$ : be careful with $9x^2 - 48x + 64$ .
Stopping after one root instead of giving both $x = 2$ and $x = 6$ .

Things to Be Careful About

Neither denominator is zero at the found points: $3x - 8 \ne 0$ and $x - 1 \ne 0$ for $x \in \{2, 6\}$ , so no extraneous roots.
When using the square-root method, keep the two cases separate and solve each.
The mark scheme awards the final A1 only for both values.

Techniques used

set the first derivative to zeroclear fractions to form a quadraticsolve the quadratic to find stationary points

(ii)

Find $\frac{d^2y}{dx^2}$ and hence determine the nature of each of the stationary points.

4M Similar

DifficultyMedium

Worked solution

Approach

Differentiate the gradient function to obtain $\frac{d^2y}{dx^2}$ , then evaluate it at each stationary point and use its sign to classify each point.

Working

Differentiate $\frac{dy}{dx} = -24(3x - 8)^{-2} + 6(x - 1)^{-2}$ :

\frac{d^2y}{dx^2} = -24(-2)(3x - 8)^{-3}(3) + 6(-2)(x - 1)^{-3}(1)

\frac{d^2y}{dx^2} = \frac{144}{(3x - 8)^3} - \frac{12}{(x - 1)^3}

At $x = 2$ :

\frac{d^2y}{dx^2}\bigg|_{x=2} = \frac{144}{(3(2) - 8)^3} - \frac{12}{(2 - 1)^3} = \frac{144}{(-2)^3} - \frac{12}{1} = -18 - 12 = -30

Since $-30 < 0$ , the stationary point at $x = 2$ is a maximum.

At $x = 6$ :

\frac{d^2y}{dx^2}\bigg|_{x=6} = \frac{144}{(3(6) - 8)^3} - \frac{12}{(6 - 1)^3} = \frac{144}{10^3} - \frac{12}{5^3} = \frac{144}{1000} - \frac{12}{125} = \frac{6}{125}

Since $\frac{6}{125} > 0$ , the stationary point at $x = 6$ is a minimum.

Answer

At $x = 2$ the stationary point is a maximum, and at $x = 6$ it is a minimum.

Final answer

x = 2 is a maximum; x = 6 is a minimum

Detailed explanation

Walkthrough

To classify each stationary point we look at the sign of the second derivative. First find $\frac{d^2y}{dx^2}$ by differentiating each term of $\frac{dy}{dx}$ using the chain rule again. The term $-24(3x-8)^{-2}$ differentiates to $-24 \cdot (-2) \cdot 3 \cdot (3x-8)^{-3} = 144(3x-8)^{-3}$ , and the term $6(x-1)^{-2}$ differentiates to $-12(x-1)^{-3}$ . So $\frac{d^2y}{dx^2} = \frac{144}{(3x-8)^3} - \frac{12}{(x-1)^3}$ .

At $x = 2$ , the denominators are $(-2)^3 = -8$ and $1$ , so $\frac{d^2y}{dx^2} = -18 - 12 = -30 < 0$ . A negative second derivative means the curve is concave down — a maximum.

At $x = 6$ , the denominators are $10^3 = 1000$ and $5^3 = 125$ , so $\frac{d^2y}{dx^2} = \frac{144}{1000} - \frac{12}{125} = \frac{6}{125} > 0$ . A positive second derivative means concave up — a minimum.

Key Takeaways

The second derivative test: $\frac{d^2y}{dx^2} < 0$ indicates a maximum and $\frac{d^2y}{dx^2} > 0$ a minimum.
Differentiating negative-power expressions requires multiplying by the exponent (which changes the sign) and by the derivative of the inner function.

Common Mistakes

Forgetting the factor $3$ when differentiating $(3x-8)^{-2}$ , which would give the wrong coefficient in $\frac{d^2y}{dx^2}$ .
Sign errors: the first term's exponent $-2$ multiplies $-24$ to give $+144$ ; the second term's $-2$ multiplies $6$ to give $-12$ .
Evaluating at the wrong $x$ or swapping the two stationary points.

Things to Be Careful About

At $x = 2$ , $3x - 8 = -2$ and $(-2)^3 = -8$ — handle the sign carefully.
At $x = 6$ , $\frac{144}{1000} - \frac{12}{125} = \frac{6}{125}$ , which is small but positive.
The mark scheme requires you to state the sign condition ( $< 0$ for max, $> 0$ for min) explicitly.

Techniques used

obtain the second derivative via the chain ruleevaluate the second derivative at each stationary pointclassify stationary points by the sign of the second derivative

Mathematics 9709/13 — October/November 2025

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