The parametric equations of a curve are $$ x = \frac{2}{\cos 3t} \quad \text{and} \quad y = \tan 3t $$ for $0 \leq t \leq 2\pi$. (a) Show that $\frac{\text{d}y}{\text{d}x}$ can be written as $A\text{cosec } 3t$, where $A$ is a constant to be found. (b) Find an equation of the normal to the curve at the point where $t = \frac{1}{12}\pi$. Give your answer in the form $y = mx + c$, where the constants $m$ and $c$ are exact.

(a) dy/dx = (1/2)cosec 3t, so A = 1/2 · (b) y = -sqrt(2)x + 5

(a) By sketching a suitable pair of graphs, show that the equation $\sec 2x = -2x - \frac{1}{2}$ has exactly one root in the interval $0 \leq x \leq \frac{1}{2}\pi$. (b) Show by calculation that this root lies between 0.8 and 1.2. (c) Show that, if a sequence of real values given by the iterative formula $$ x_{n+1} = \frac{1}{2}\cos^{-1}\left(\frac{-2}{4x_n + 1}\right) $$ converges, then it converges to the root of the equation in part (a). (d) Use this iterative formula to…

(a) Exactly one root exists in the interval, as the graphs intersect at a single point between x = π/4 and x = π/2. · (b) f(0.8) = -32.14 0, so the root lies between 0.8 and 1.2. · (c) Substituting x_n = x_{n+1} = L into the iterative formula and rearranging gives sec 2L = -2L - 1/2, proving convergence to the root. · (d) x = 0.992

9709/35 May/June 2025 — Mathematics Solved Past Paper

Q14MMedium-EasyLogarithmic and Exponential Functions

Solve the equation $3^{4-2x} = 5(6^{x-1})$ . Give your answer correct to 3 significant figures.

DifficultyMedium-Easy

Worked solution

Approach

Take natural logarithms of both sides so that the powers can be brought down using the power law of logarithms. Then use the product law on the right-hand side and rearrange the resulting linear equation for $x$ .

Working

Take natural logarithms of both sides:

\ln\left(3^{4-2x}\right) = \ln\left(5 \cdot 6^{x-1}\right)

Apply the power law on the left and the product law on the right:

(4-2x)\ln 3 = \ln 5 + (x-1)\ln 6

Expand both sides:

4\ln 3 - 2x\ln 3 = \ln 5 + x\ln 6 - \ln 6

Collect the $x$ -terms on one side:

x(-2\ln 3 - \ln 6) = \ln 5 - \ln 6 - 4\ln 3

Divide by the coefficient of $x$ :

x = \frac{\ln 5 - \ln 6 - 4\ln 3}{-2\ln 3 - \ln 6}

Evaluate using $\ln 3 \approx 1.09861$ , $\ln 5 \approx 1.60944$ , $\ln 6 \approx 1.79176$ :

x = \frac{1.60944 - 1.79176 - 4.39845}{-2.19722 - 1.79176} = 1.14835\ldots

Answer

x = 1.15 \text{ (3 s.f.)}

Final answer

x = 1.15 (3 s.f.)

Detailed explanation

Walkthrough

The equation has the unknown $x$ in the exponents, so the first step is to take logarithms of both sides. Any consistent base works, but natural logarithms are standard.

Taking $\ln$ of both sides gives:

\ln\left(3^{4-2x}\right) = \ln\left(5 \cdot 6^{x-1}\right)

On the left, the power law lets us bring the exponent down: $(4-2x)\ln 3$ . On the right, the product law splits the logarithm of a product into a sum: $\ln 5 + \ln\left(6^{x-1}\right)$ , and then the power law gives $\ln 5 + (x-1)\ln 6$ .

This converts the exponential equation into a linear equation in $x$ :

(4-2x)\ln 3 = \ln 5 + (x-1)\ln 6

Expand the brackets:

4\ln 3 - 2x\ln 3 = \ln 5 + x\ln 6 - \ln 6

Then collect the terms containing $x$ on one side and the constants on the other. The coefficient of $x$ is $-2\ln 3 - \ln 6$ , so dividing gives the exact expression for $x$ .

Finally, substitute the approximate values of the logarithms and round to 3 significant figures, obtaining $x = 1.15$ .

Key Takeaways

This question tests the ability to solve an exponential equation by taking logarithms. The key skills are:

Applying the power law $\ln(a^k) = k\ln a$ to bring exponents down.
Applying the product law $\ln(ab) = \ln a + \ln b$ when the right-hand side contains a product.
Rearranging a linear equation carefully after applying the log laws.
Rounding a final numerical answer to the required number of significant figures.

Common Mistakes

Forgetting to take the logarithm of both sides, or taking logs of only one side.
Omitting brackets when applying the power law, e.g. writing $4-2x\ln 3$ instead of $(4-2x)\ln 3$ .
Incorrectly applying the product law, e.g. writing $\ln(5 \cdot 6^{x-1})$ as $\ln 5 \times \ln\left(6^{x-1}\right)$ instead of $\ln 5 + \ln\left(6^{x-1}\right)$ .
Using different logarithm bases on different sides of the equation.
Giving the answer without showing the logarithmic equation; the mark scheme requires working to be shown.

Things to Be Careful About

Use a consistent logarithm base throughout. Natural logarithms are usually simplest.
Keep brackets around expressions such as $(4-2x)$ and $(x-1)$ until they are expanded.
When collecting terms, be careful with signs: the coefficient of $x$ is negative here.
The final answer must be given to 3 significant figures. If no working is shown, no marks are awarded.

Techniques used

take logarithms of both sidesapply the product law of logarithmsapply the power law of logarithmssolve the resulting linear equation

Similar questions

Q25MMediumTrigonometry

Solve the equation $3\cot\theta - 4\text{cosec}^2\theta + 5 = 0$ for $-\pi \leq \theta \leq \pi$ .

DifficultyMedium

Worked solution

Approach

Use the Pythagorean identity $\csc^2\theta = 1 + \cot^2\theta$ to rewrite the equation as a quadratic in $\cot\theta$ . Solve the quadratic, then use the periodicity of $\tan\theta$ to find every solution in the interval $-\pi \leq \theta \leq \pi$ .

Working

Starting from

3\cot\theta - 4\csc^2\theta + 5 = 0

Substitute $\csc^2\theta = 1 + \cot^2\theta$ :

3\cot\theta - 4(1 + \cot^2\theta) + 5 = 0

Simplify:

3\cot\theta - 4 - 4\cot^2\theta + 5 = 0

-4\cot^2\theta + 3\cot\theta + 1 = 0

Multiply by $-1$ :

4\cot^2\theta - 3\cot\theta - 1 = 0

Factorise:

(4\cot\theta + 1)(\cot\theta - 1) = 0

Hence

\cot\theta = 1 \quad \text{or} \quad \cot\theta = -\frac{1}{4}

For $\cot\theta = 1$ , we have $\tan\theta = 1$ , so

\theta = \frac{\pi}{4} + k\pi

In the interval $-\pi \leq \theta \leq \pi$ :

\theta = \frac{\pi}{4}, \ -\frac{3\pi}{4}

For $\cot\theta = -\frac{1}{4}$ , we have $\tan\theta = -4$ , so

\theta = \tan^{-1}(-4) + k\pi

Numerically, $\tan^{-1}(-4) \approx -1.3258$ , so in the interval:

\theta \approx -1.33, \ 1.82

Answer

\theta = \frac{\pi}{4}, \ -\frac{3\pi}{4}, \ -1.33, \ 1.82

(radians, approximate values correct to 3 significant figures).

Final answer

θ = π/4, -3π/4, -1.33, 1.82 (radians)

Detailed explanation

Walkthrough

The equation contains two different trigonometric functions: $\cot\theta$ and $\csc^2\theta$ . To solve it, we first express everything in terms of one function. The identity $\csc^2\theta = 1 + \cot^2\theta$ is exactly the link needed.

Substituting gives $3\cot\theta - 4(1 + \cot^2\theta) + 5 = 0$ , which simplifies to $-4\cot^2\theta + 3\cot\theta + 1 = 0$ . Multiplying by $-1$ gives $4\cot^2\theta - 3\cot\theta - 1 = 0$ . This is a quadratic in $\cot\theta$ , and it factorises as $(4\cot\theta + 1)(\cot\theta - 1) = 0$ . Therefore $\cot\theta = 1$ or $\cot\theta = -\frac{1}{4}$ .

For $\cot\theta = 1$ , since $\tan\theta = \frac{1}{\cot\theta} = 1$ , the principal solution is $\theta = \frac{\pi}{4}$ . Because $\tan\theta$ has period $\pi$ , the other solution in $[-\pi, \pi]$ is $\frac{\pi}{4} - \pi = -\frac{3\pi}{4}$ .

For $\cot\theta = -\frac{1}{4}$ , we have $\tan\theta = -4$ . The calculator gives $\tan^{-1}(-4) \approx -1.33$ ; adding $\pi$ gives approximately $1.82$ . Both of these lie in the interval $-\pi \leq \theta \leq \pi$ .

Thus the four solutions are $\frac{\pi}{4}$ , $-\frac{3\pi}{4}$ , $-1.33$ , and $1.82$ .

Key Takeaways

This question tests the ability to reduce a trigonometric equation to a single trigonometric function using a Pythagorean identity, then solve a quadratic equation and use the periodic nature of trigonometric functions to find all required solutions in a given interval.

Common Mistakes

Using the wrong Pythagorean identity, such as confusing $\csc^2\theta$ with $\sec^2\theta$ .
Making a sign error when rearranging the quadratic, especially when multiplying by $-1$ .
Only giving the principal value of $\tan^{-1}(-4)$ and forgetting to add $\pi$ to obtain the second solution.
Forgetting that the interval includes negative angles, so missing $-\frac{3\pi}{4}$ or $-1.33$ .
Giving answers outside the interval; the mark scheme ignores solutions outside $-\pi \leq \theta \leq \pi$ .

Things to Be Careful About

$\cot\theta = -\frac{1}{4}$ means $\tan\theta = -4$ , not $-\frac{1}{4}$ .
The period of $\tan\theta$ is $\pi$ , so solutions repeat every $\pi$ radians.
The interval is in radians, so all answers must be in radians.
Exact values such as $\frac{\pi}{4}$ and $-\frac{3\pi}{4}$ are preferred, but decimal equivalents such as $0.785$ and $-2.36$ are also accepted.

Techniques used

use Pythagorean identity to reduce to one trig functionsolve a quadratic in cotangentapply periodicity of tangent to list all interval solutions

Similar questions

Q3Medium-EasyComplex Numbers

The complex numbers $s$ and $t$ are given by

s = 5(\cos 0.25 + \text{i}\sin 0.25) \quad \text{and} \quad t = 6\text{e}^{3\text{i}}

(a)

Express $\frac{s}{t}$ in the form $r\text{e}^{\text{i}\theta}$ , where $-\pi < \theta \leq \pi$ and $r > 0$ .

2M Similar

DifficultyMedium-Easy

Worked solution

Approach

Write both $s$ and $t$ in exponential form $re^{i\theta}$ , then divide by dividing moduli and subtracting arguments. Verify the resulting argument lies in the principal range $-\pi < \theta \leq \pi$ .

Working

Given $s = 5(\cos 0.25 + \text{i}\sin 0.25)$ , we can write:

s = 5e^{i0.25}

And $t$ is already in exponential form:

t = 6e^{3i}

Dividing:

\frac{s}{t} = \frac{5e^{i0.25}}{6e^{3i}} = \frac{5}{6}e^{i(0.25 - 3)}

\frac{s}{t} = \frac{5}{6}e^{-2.75i}

Since $-\pi \approx -3.1416 < -2.75 \leq \pi$ , the argument $-2.75$ is in the required range.

Answer

\frac{s}{t} = \frac{5}{6}e^{-2.75i}

So $r = \frac{5}{6}$ and $\theta = -2.75$ .

Final answer

s/t = (5/6)e^(-2.75i)

Detailed explanation

Walkthrough

The number $s = 5(\cos 0.25 + \text{i}\sin 0.25)$ is written in polar form $r(\cos\theta + \text{i}\sin\theta)$ with $r = 5$ and $\theta = 0.25$ . By Euler's formula, this is equivalent to $5e^{i0.25}$ .

The number $t$ is given directly as $6e^{3i}$ , so it has modulus 6 and argument 3.

To divide two complex numbers in exponential form, we divide their moduli and subtract their arguments:

\frac{r_1 e^{i\theta_1}}{r_2 e^{i\theta_2}} = \frac{r_1}{r_2}e^{i(\theta_1 - \theta_2)}

Applying this:

\frac{s}{t} = \frac{5e^{i0.25}}{6e^{3i}} = \frac{5}{6}e^{i(0.25 - 3)} = \frac{5}{6}e^{-2.75i}

The final step is to verify the argument $-2.75$ lies in the principal range $-\pi < \theta \leq \pi$ . Since $-\pi \approx -3.1416$ and $-2.75 > -3.1416$ , the argument is valid as is — no adjustment is needed.

Key Takeaways

Converting between $r(\cos\theta + \text{i}\sin\theta)$ and $re^{i\theta}$ forms of a complex number.
Division in polar form: divide moduli and subtract arguments.
Always check the argument is in the principal range $(-\pi, \pi]$ .

Common Mistakes

Failing to evaluate $0.25 - 3$ and leaving the answer as $\frac{5}{6}e^{i(0.25-3)}$ — the mark scheme requires the subtraction to be evaluated.
Writing the argument as $-2.75\text{i}$ — the argument is a real angle, not a complex number; this scores B0.
Rounding the modulus to 0.83 — the mark scheme requires $\frac{5}{6}$ or 0.833 (0.83 scores B0).

Things to Be Careful About

The principal range is $-\pi < \theta \leq \pi$ . If the computed argument falls outside this range, add or subtract $2\pi$ to bring it into range. Here $-2.75$ is already in range.
The modulus $r$ must be positive; here $\frac{5}{6} > 0$ .
Angles are in radians throughout this question.

Techniques used

convert complex numbers to exponential formdivide complex numbers in polar formsubtract arguments and divide moduli

(b)

In an Argand diagram with origin $O$ , the points $A$ and $B$ represent the complex numbers $s$ and $\frac{s}{t}$ respectively.

By considering the line segments $OA$ and $OB$ , or otherwise, state the two geometric effects of dividing a complex number by $6\text{e}^{3\text{i}}$ .

2M Similar

DifficultyMedium-Easy

Worked solution

Approach

Interpret division by $6e^{3i}$ as two separate geometric transformations: a scaling of the distance from the origin (modulus) and a rotation (argument change).

Working

Given $s = 5(\cos 0.25 + \text{i}\sin 0.25) = 5e^{i0.25}$ and $t = 6e^{3i}$ .

Dividing a complex number $z$ by $6e^{3i}$ is equivalent to multiplying by $\frac{1}{6}e^{-3i}$ .

Modulus effect: The modulus is divided by 6:

|s| = 5 \quad \Rightarrow \quad \left|\frac{s}{t}\right| = \frac{5}{6}

So the distance from the origin $O$ is multiplied by $\frac{1}{6}$ . This is an enlargement with scale factor $\frac{1}{6}$ .

Argument effect: The argument is reduced by 3:

\arg(s) = 0.25 \quad \Rightarrow \quad \arg\left(\frac{s}{t}\right) = 0.25 - 3 = -2.75

So the point is rotated clockwise by 3 radians (equivalently a rotation of $-3$ radians, or an anticlockwise rotation of $2\pi - 3$ radians).

Answer

The two geometric effects of dividing by $6e^{3i}$ are:

An enlargement with scale factor $\frac{1}{6}$ .
A clockwise rotation of 3 radians.

Final answer

Enlargement with scale factor 1/6 and clockwise rotation of 3 radians

Detailed explanation

Walkthrough

Dividing a complex number $z$ by $6e^{3i}$ is equivalent to multiplying by its reciprocal, $\frac{1}{6}e^{-3i}$ . Geometrically, multiplying a complex number by $re^{i\theta}$ scales its distance from the origin by $r$ and rotates it by angle $\theta$ .

Here, the modulus factor is $\frac{1}{6}$ : the original point $A$ (representing $s$ ) has modulus 5, and after division the point $B$ (representing $\frac{s}{t}$ ) has modulus $\frac{5}{6}$ . So the distance from $O$ is multiplied by $\frac{1}{6}$ — an enlargement with scale factor $\frac{1}{6}$ .

The argument change is $-3$ radians: $\arg(s) = 0.25$ and $\arg(\frac{s}{t}) = 0.25 - 3 = -2.75$ . A negative change in argument means a clockwise rotation. So the point is rotated clockwise by 3 radians (equivalently anticlockwise by $2\pi - 3$ radians).

Both transformations have centre $O$ (the origin).

Key Takeaways

Dividing by $re^{i\theta}$ scales by $\frac{1}{r}$ and rotates by $-\theta$ .
Geometric interpretation of complex number operations on the Argand diagram.
The modulus and argument of a complex number correspond to distance from origin and angle from the positive real axis.

Common Mistakes

Using words like 'reduction', 'shrink' or 'compression' instead of 'enlargement' — the mark scheme explicitly disallows these.
Stating only one of the two effects — both the enlargement and the rotation are required for the marks.
Stating an incorrect centre of transformation (e.g. a point other than the origin).

Things to Be Careful About

The scale factor is $\frac{1}{6}$ , not 6 — dividing by 6 makes the modulus smaller.
The rotation is clockwise by 3 radians because the argument decreases by 3.
The rotation can also be described as anticlockwise by $2\pi - 3$ radians (since $-3 + 2\pi = 2\pi - 3 \approx 3.283$ ).
Both effects must be stated to gain the marks. Stating the centre of transformation is not required, but if stated it must be the origin $O$ — an incorrect centre scores B0.

Techniques used

interpret division geometrically on the Argand diagramidentify scale factor from modulus ratioidentify rotation from argument difference

Q45MMediumDifferentiation

Find the exact coordinates of the stationary point of the curve with equation $y = 3x^3 \ln x^4$ , for $x > 0$ .

DifficultyMedium

Worked solution

Approach

Use the product rule to differentiate $y = 3x^3 \ln x^4$ , set the derivative to zero, solve for $x$ , then substitute back to find $y$ .

Working

Let $u = 3x^3$ and $v = \ln x^4$ . By the product rule:

\frac{dy}{dx} = v\frac{du}{dx} + u\frac{dv}{dx}

\frac{du}{dx} = 9x^2

\frac{dv}{dx} = \frac{4x^3}{x^4} = \frac{4}{x}

Therefore:

\frac{dy}{dx} = 9x^2 \ln x^4 + 3x^3 \cdot \frac{4}{x} = 9x^2 \ln x^4 + 12x^2

At a stationary point, $\frac{dy}{dx} = 0$ :

9x^2 \ln x^4 + 12x^2 = 0

Since $x > 0$ , divide by $3x^2$ :

3 \ln x^4 + 4 = 0

\ln x^4 = -\frac{4}{3}

Exponentiate both sides:

x^4 = e^{-4/3}

Since $x > 0$ :

x = e^{-1/3}

Now find $y$ :

y = 3x^3 \ln x^4 = 3(e^{-1/3})^3 \ln(e^{-4/3}) = 3e^{-1} \cdot \left(-\frac{4}{3}\right) = -\frac{4}{e}

Answer

The stationary point is $(e^{-1/3}, -\frac{4}{e})$ .

Final answer

x = e^{-1/3}, y = -4/e

Detailed explanation

Walkthrough

We need to find the point where the curve's gradient is zero. The function $y = 3x^3 \ln x^4$ is a product of two functions of $x$ : $3x^3$ and $\ln x^4$ . This is exactly the situation where the product rule applies.

The product rule states that if $y = uv$ , then $\frac{dy}{dx} = v\frac{du}{dx} + u\frac{dv}{dx}$ . Here we set $u = 3x^3$ and $v = \ln x^4$ .

First differentiate each factor. The derivative of $3x^3$ is $9x^2$ . For $\ln x^4$ , we use the chain rule: the derivative of the logarithm of a function is $\frac{1}{\text{function}} \times \text{derivative of the function}$ , giving $\frac{4x^3}{x^4} = \frac{4}{x}$ .

Substituting into the product rule gives $\frac{dy}{dx} = 9x^2 \ln x^4 + 12x^2$ .

At a stationary point the gradient is zero, so we set this equal to zero. Since $x > 0$ , we can divide through by $3x^2$ (which is never zero for $x > 0$ ), giving $3\ln x^4 + 4 = 0$ . Rearranging gives $\ln x^4 = -\frac{4}{3}$ .

To eliminate the logarithm, exponentiate both sides: $x^4 = e^{-4/3}$ . Taking the fourth root, and remembering $x > 0$ so we take only the positive root, gives $x = e^{-1/3}$ .

Finally, substitute back into the original equation to find $y$ : $y = 3(e^{-1/3})^3 \ln(e^{-4/3}) = 3e^{-1} \cdot \left(-\frac{4}{3}\right) = -\frac{4}{e}$ .

Key Takeaways

This question tests the product rule for differentiation, the chain rule when differentiating the logarithm of a composite function, solving equations involving logarithms by exponentiating, and careful algebraic simplification. It also reinforces the importance of domain restrictions ( $x > 0$ ) when taking roots.

Common Mistakes

The mark scheme highlights several specific errors:

Answers with no working score no marks — all steps must be shown.
$x = (e^{-4/3})^{1/4}$ scores A0 — the fourth root must be simplified to $e^{-1/3}$ .
$x = \pm e^{-1/3}$ scores A0 — the negative root is not valid because the domain is $x > 0$ .
$y = 3e^{-1} \ln e^{-4/3}$ scores A0 — the $y$ -coordinate must be simplified to $-\frac{4}{e}$ .
A sign error in the product rule (e.g. using $-$ instead of $+$ ) leads to $x^4 = e^{4/3}$ , which is incorrect.

Things to Be Careful About

The domain $x > 0$ means only the positive fourth root is taken.
The derivative of $\ln x^4$ is $\frac{4}{x}$ , not $\frac{1}{x^4}$ — remember the chain rule.
When dividing by $3x^2$ , this is valid because $x > 0$ guarantees $3x^2 \neq 0$ .
Ensure the final $y$ -value is fully simplified to $-\frac{4}{e}$ .

Techniques used

apply the product rule to differentiate a product of two functionsdifferentiate a composite logarithmic function using the chain ruleset the derivative to zero to locate the stationary pointeliminate the logarithm by exponentiating both sidesevaluate the original function at the stationary x-value

Similar questions

Q5Medium-HardComplex Numbers

Fig. 1

The diagram shows the locus of points representing the complex numbers, $z$ , satisfying $|z + 5 - 4\text{i}| = 3$ .

(a)

For the points on this locus, determine the maximum and minimum possible values of $|z|$ .

3M Similar

DifficultyMedium-Easy

Worked solution

Approach

The locus $|z + 5 - 4i| = 3$ is a circle in the Argand diagram with centre $-5 + 4i$ and radius $3$ . The quantity $|z|$ is the distance from the origin to a point on this circle. The maximum and minimum values of $|z|$ occur at the points on the circle that lie on the line from the origin through the centre, on opposite sides of the centre.

Working

The centre of the circle is $C = -5 + 4i$ , which corresponds to the point $(-5, 4)$ in the Argand diagram.

The distance from the origin $O$ to the centre $C$ is found using Pythagoras:

OC = \sqrt{(-5)^2 + 4^2} = \sqrt{25 + 16} = \sqrt{41}

The maximum value of $|z|$ is the distance to the point on the far side of the centre from the origin:

|z|_{\max} = OC + r = \sqrt{41} + 3

The minimum value of $|z|$ is the distance to the point on the near side of the centre to the origin:

|z|_{\min} = OC - r = \sqrt{41} - 3

Answer

Minimum $|z| = \sqrt{41} - 3 \approx 3.40$

Maximum $|z| = \sqrt{41} + 3 \approx 9.40$

Final answer

Minimum: sqrt(41) - 3 (≈ 3.40); Maximum: sqrt(41) + 3 (≈ 9.40)

Detailed explanation

Walkthrough

The equation $|z + 5 - 4i| = 3$ tells us that the distance from $z$ to the complex number $-5 + 4i$ is exactly $3$ . In the Argand diagram, this is a circle centred at the point $-5 + 4i$ , i.e. the Cartesian point $(-5, 4)$ in the second quadrant, with radius $3$ .

The modulus $|z|$ is just the distance from the origin $O$ to a point on this circle. To find the maximum and minimum of this distance, we look for the points on the circle that are farthest from and closest to the origin. These two points lie on the straight line from $O$ through the centre $C$ , on opposite sides of $C$ .

We first compute the distance $OC$ using the Pythagorean theorem with the horizontal leg of length $5$ and the vertical leg of length $4$ :

OC = \sqrt{(-5)^2 + 4^2} = \sqrt{41}

The closest point on the circle to $O$ lies on the line $OC$ between $O$ and $C$ , at distance $OC - r$ from $O$ . The farthest point lies on the opposite side of $C$ from $O$ , at distance $OC + r$ from $O$ .

Key Takeaways

The equation $|z - c| = r$ always represents a circle in the Argand diagram with centre $c$ and radius $r$ .
$|z|$ is the distance from the origin to a point on the locus.
For a circle, the maximum and minimum distances from any external point occur along the line through that external point and the centre.

Common Mistakes

Swapping the max and min values.
Forgetting to subtract the radius for the minimum (using just $OC$ ).
Confusing the centre: writing it as $(5, -4)$ or $(-5, -4)$ instead of $(-5, 4)$ .

Things to Be Careful About

The sign of each component in the centre matters. $|z + 5 - 4i| = |z - (-5 + 4i)|$ has centre $-5 + 4i$ , not $5 - 4i$ .
The numerical answers are $\sqrt{41} - 3 \approx 3.40$ and $\sqrt{41} + 3 \approx 9.40$ , as the mark scheme requires AWRT 3.4 and 9.4.
The mark scheme warns that if the max and min are swapped, the student can still earn M1 but only one of the A1 marks (max 2/3).

Techniques used

interpret modulus equation as circle locuscalculate distance from origin to circle centre using Pythagorasapply extreme distance using line through origin and centre

(b)

For the points on this locus, determine the minimum possible value of $\arg z$ .

3M Similar

DifficultyMedium-Hard

Worked solution

Approach

The minimum value of $\arg z$ on the locus is the argument of the point on the circle where a tangent from the origin touches the circle on the lower side (the side closer to the real axis). This argument is found by combining the angle from the positive real axis to the line from the origin to the centre, with the angle between this line and the tangent line.

Working

The centre of the circle is $C = -5 + 4i$ , at the point $(-5, 4)$ in the second quadrant.

The distance from the origin to the centre is:

OC = \sqrt{(-5)^2 + 4^2} = \sqrt{41}

Step 1: Angle from the positive real axis to the line $OC$ .

Since $C$ is in the second quadrant, the angle from the positive real axis (measured counter-clockwise) to $OC$ is obtuse. Dropping a perpendicular from $C$ to the real axis gives a right triangle with horizontal leg $5$ and vertical leg $4$ , so the acute angle at the origin is $\tan^{-1}(4/5)$ . The full angle is therefore:

\theta_C = \pi - \tan^{-1}\left(\frac{4}{5}\right)

Step 2: Angle between $OC$ and the tangent line $OP$ (where $P$ is the tangent point on the lower side).

In the right triangle $OCP$ , the angle at $P$ is $90°$ (the radius is perpendicular to the tangent at the point of contact). The hypotenuse is $OC = \sqrt{41}$ , and the side opposite to the angle at $O$ is the radius $CP = 3$ . Therefore:

\alpha = \sin^{-1}\left(\frac{3}{\sqrt{41}}\right)

Step 3: Subtract to find the minimum argument.

The minimum argument of $z$ is the angle from the positive real axis to the lower tangent line $OP$ , which is the angle to $OC$ minus the half-angle $\alpha$ :

\arg z_{\min} = \theta_C - \alpha = \pi - \tan^{-1}\left(\frac{4}{5}\right) - \sin^{-1}\left(\frac{3}{\sqrt{41}}\right)

Numerical evaluation:

\tan^{-1}\left(\frac{4}{5}\right) \approx 0.6747 \text{ rad}

\sin^{-1}\left(\frac{3}{\sqrt{41}}\right) \approx 0.4880 \text{ rad}

\arg z_{\min} \approx \pi - 0.6747 - 0.4880 \approx 1.98 \text{ rad} \approx 113.4°

Answer

$\arg z_{\min} = \pi - \tan^{-1}\left(\frac{4}{5}\right) - \sin^{-1}\left(\frac{3}{\sqrt{41}}\right) \approx 1.98 \text{ rad} \approx 113.4°$

Final answer

1.98 rad (≈ 113.4°)

Detailed explanation

Walkthrough

The minimum argument of any $z$ on the circular locus occurs at the point where a tangent from the origin touches the circle on the lower side (the side closer to the real axis). This is because as you move along the circle starting from this tangent point in either direction, the argument increases.

To find this minimum argument, we use the right triangle formed by the origin $O$ , the centre $C = (-5, 4)$ , and the tangent point $P$ . The line $CP$ is a radius of length $3$ , and it is perpendicular to the tangent line $OP$ at the point of tangency, so the triangle $OCP$ has a right angle at $P$ .

The angle from the positive real axis to $OC$ :
Since $C$ is at $(-5, 4)$ in the second quadrant, the line from the origin to $C$ makes an obtuse angle with the positive real axis. If we drop a perpendicular from $C$ to the real axis, we get a right triangle with horizontal leg $5$ and vertical leg $4$ . The acute angle at the origin in this triangle is $\tan^{-1}(4/5)$ . The full angle from the positive real axis to $OC$ is therefore $\pi - \tan^{-1}(4/5)$ .

The half-angle at the origin between $OC$ and $OP$ :
In the right triangle $OCP$ , the hypotenuse is $OC = \sqrt{41}$ and the side opposite to the angle at $O$ is $CP = 3$ . So this angle is $\sin^{-1}(3/\sqrt{41})$ .

Combining the angles:
The minimum argument of $z$ is the angle from the positive real axis to the tangent line $OP$ , which is the angle to $OC$ minus the half-angle:

\arg z_{\min} = \pi - \tan^{-1}\left(\frac{4}{5}\right) - \sin^{-1}\left(\frac{3}{\sqrt{41}}\right)

Numerically this is approximately $1.98$ radians or $113.4°$ .

Key Takeaways

For a circular locus, the minimum (and maximum) argument of $z$ corresponds to the points where a tangent from the origin touches the circle.
The geometry of the tangent gives a right triangle with the origin, the centre, and the tangent point as vertices; the radius is perpendicular to the tangent at the contact point.
The minimum argument is the angle to the centre minus the half-angle at the origin; the maximum argument would be the angle to the centre plus the half-angle.

Common Mistakes

Forgetting to use $\pi -$ when computing the angle to a point in the second quadrant (giving the acute angle instead of the obtuse one).
Using the radius $3$ as the hypotenuse of the right triangle (it is the opposite side, not the hypotenuse).
Confusing the lower and upper tangent points — the upper one gives the maximum argument, not the minimum.
Adding the half-angle instead of subtracting it.
Using the wrong inverse trig function: $\tan^{-1}(3/\sqrt{41})$ would be wrong; the mark scheme accepts $\sin^{-1}(3/\sqrt{41})$ or equivalent.

Things to Be Careful About

The centre $(-5, 4)$ is in the second quadrant, so the angle from the positive real axis is obtuse: $\pi - \tan^{-1}(4/5)$ , not just $\tan^{-1}(4/5)$ .
The half-angle at the origin is $\sin^{-1}(3/\sqrt{41})$ , computed from the right triangle $OCP$ where the hypotenuse is $OC = \sqrt{41}$ and the opposite side is the radius $3$ .
The mark scheme warns that $\tan^{-1}(4/5) - \sin^{-1}(3/\sqrt{41})$ scores B1FT M0 A0, because it omits the $\pi$ for the second-quadrant centre.
The mark scheme also notes that if a student instead computes the maximum argument ( $\pi/2 + \tan^{-1}(5/4) + \sin^{-1}(3/\sqrt{41}) \approx 2.95$ rad or $169.3°$ ), they would score only B1FT (1/3) as a special case.

Techniques used

identify tangent from origin as locus of minimum argumentcompute angle from positive real axis to centre using arctan in second quadrantcompute half-angle at origin using right triangle with radius as opposite sidesubtract angles to find minimum argument

Q6MediumDifferentiation

The parametric equations of a curve are

x = \frac{2}{\cos 3t} \quad \text{and} \quad y = \tan 3t

for $0 \leq t \leq 2\pi$ .

(a)

Show that $\frac{\text{d}y}{\text{d}x}$ can be written as $A\text{cosec } 3t$ , where $A$ is a constant to be found.

5M Similar

DifficultyMedium

Worked solution

Approach

Differentiate $x$ and $y$ with respect to the parameter $t$ , then use

\frac{\text{d}y}{\text{d}x} = \frac{\frac{\text{d}y}{\text{d}t}}{\frac{\text{d}x}{\text{d}t}}

and simplify to a multiple of $\operatorname{cosec} 3t$ .

Working

Rewrite $x$ :

x = 2(\cos 3t)^{-1}

Differentiate $x$ with respect to $t$ using the chain rule:

\frac{\text{d}x}{\text{d}t} = -2(\cos 3t)^{-2}(-3\sin 3t) = 6\sec 3t\tan 3t

Differentiate $y$ with respect to $t$ :

\frac{\text{d}y}{\text{d}t} = 3\sec^2 3t

Then

\frac{\text{d}y}{\text{d}x} = \frac{3\sec^2 3t}{6\sec 3t\tan 3t}

Simplify:

\frac{\text{d}y}{\text{d}x} = \frac{1}{2}\cdot\frac{\sec 3t}{\tan 3t} = \frac{1}{2}\cdot\frac{1}{\sin 3t} = \frac{1}{2}\operatorname{cosec} 3t

Answer

\frac{\text{d}y}{\text{d}x} = \frac{1}{2}\operatorname{cosec} 3t

so $A = \frac{1}{2}$ .

Final answer

dy/dx = (1/2)cosec 3t, so A = 1/2

Detailed explanation

Walkthrough

We are given a curve in parametric form, so $x$ and $y$ are both functions of $t$ . To find $\frac{\text{d}y}{\text{d}x}$ , we first differentiate both $x$ and $y$ with respect to $t$ , then divide $\frac{\text{d}y}{\text{d}t}$ by $\frac{\text{d}x}{\text{d}t}$ .

For $x = \frac{2}{\cos 3t}$ , rewrite it as $x = 2(\cos 3t)^{-1}$ . The chain rule gives

\frac{\text{d}x}{\text{d}t} = -2(\cos 3t)^{-2} \cdot (-3\sin 3t) = \frac{6\sin 3t}{\cos^2 3t} = 6\sec 3t\tan 3t.

For $y = \tan 3t$ , the derivative is

\frac{\text{d}y}{\text{d}t} = 3\sec^2 3t.

Then

\frac{\text{d}y}{\text{d}x} = \frac{3\sec^2 3t}{6\sec 3t\tan 3t} = \frac{1}{2}\cdot\frac{\sec 3t}{\tan 3t} = \frac{1}{2}\cdot\frac{1}{\sin 3t} = \frac{1}{2}\operatorname{cosec} 3t.

This matches the required form, with $A = \frac{1}{2}$ .

Key Takeaways

This question combines parametric differentiation with the chain rule and trigonometric identities. The key idea is that for parametric curves, $\frac{\text{d}y}{\text{d}x}$ is obtained by dividing the two derivatives with respect to the parameter. It also shows how reciprocal trigonometric functions such as $\sec$ and $\operatorname{cosec}$ simplify naturally.

Common Mistakes

Differentiating $x$ without using the chain rule, or forgetting the factor $3$ from the derivative of $3t$ .
Getting the sign wrong when differentiating $(\cos 3t)^{-1}$ .
Dividing the wrong way round, e.g. using $\frac{\text{d}x}{\text{d}t} \div \frac{\text{d}y}{\text{d}t}$ .
Stopping before simplifying $\frac{\sec 3t}{\tan 3t}$ to $\operatorname{cosec} 3t$ .
Not writing the final answer in the requested $A\operatorname{cosec} 3t$ form.

Things to Be Careful About

The derivative is undefined where $\cos 3t = 0$ or $\sin 3t = 0$ , since the parametric representation has vertical or horizontal tangents there. Within the given interval, those points must be excluded. The final expression must be exact and in the form requested by the question.

Techniques used

differentiate parametric equations with respect to the parameterapply the chain rule to a reciprocal trigonometric functiondivide the parametric derivatives to find dy/dxsimplify using reciprocal trigonometric identities

(b)

Find an equation of the normal to the curve at the point where $t = \frac{1}{12}\pi$ . Give your answer in the form $y = mx + c$ , where the constants $m$ and $c$ are exact.

4M Similar

DifficultyMedium

Worked solution

Approach

Evaluate the coordinates at $t = \frac{1}{12}\pi$ , find the tangent gradient from part (a), take its negative reciprocal to get the normal gradient, then form the equation of the normal.

Working

At $t = \frac{1}{12}\pi$ , we have $3t = \frac{\pi}{4}$ .

x = \frac{2}{\cos\frac{\pi}{4}} = \frac{2}{\frac{\sqrt{2}}{2}} = 2\sqrt{2}

y = \tan\frac{\pi}{4} = 1

From part (a), the tangent gradient is

\frac{\text{d}y}{\text{d}x} = \frac{1}{2}\operatorname{cosec}\frac{\pi}{4} = \frac{1}{2}\cdot\sqrt{2} = \frac{\sqrt{2}}{2}

The normal gradient is the negative reciprocal:

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

Using the point $(2\sqrt{2}, 1)$ :

y - 1 = -\sqrt{2}(x - 2\sqrt{2})

y - 1 = -\sqrt{2}x + 4

y = -\sqrt{2}x + 5

Answer

y = -\sqrt{2}x + 5

Final answer

y = -sqrt(2)x + 5

Detailed explanation

Walkthrough

First find the point on the curve. When $t = \frac{1}{12}\pi$ , $3t = \frac{\pi}{4}$ . Therefore

x = \frac{2}{\cos\frac{\pi}{4}} = 2\sqrt{2}, \qquad y = \tan\frac{\pi}{4} = 1.

Next, use part (a) to find the gradient of the tangent at this point:

\frac{\text{d}y}{\text{d}x} = \frac{1}{2}\operatorname{cosec}\frac{\pi}{4} = \frac{\sqrt{2}}{2}.

The normal is perpendicular to the tangent, so its gradient is the negative reciprocal:

m = -\frac{1}{\sqrt{2}/2} = -\sqrt{2}.

Using the point-slope form of a line through $(2\sqrt{2}, 1)$ :

y - 1 = -\sqrt{2}(x - 2\sqrt{2}).

Expanding gives $y - 1 = -\sqrt{2}x + 4$ , so

y = -\sqrt{2}x + 5.

Key Takeaways

This part uses the derivative to find the gradient of a normal, which requires taking the negative reciprocal. It also reinforces exact trigonometric values at $\frac{\pi}{4}$ and the point-slope formula for a straight line.

Common Mistakes

Using the gradient of the tangent instead of the normal.
Forgetting the negative sign when taking the reciprocal.
Using decimal approximations instead of exact values; the question requires exact constants.
Expanding $y - 1 = -\sqrt{2}(x - 2\sqrt{2})$ incorrectly, especially the constant term.

Things to Be Careful About

The final answer must be in the form $y = mx + c$ with exact $m$ and $c$ . Here $m = -\sqrt{2}$ and $c = 5$ . The mark scheme also accepts $y = -\frac{2}{\sqrt{2}}x + 5$ , since $\frac{2}{\sqrt{2}} = \sqrt{2}$ .

Techniques used

evaluate parametric coordinates at a given parameter valuesubstitute the parameter into the tangent gradienttake the negative reciprocal to find the normal gradientform the equation of a straight line through a point with a given gradient

Q7MediumDifferentiationIntegration

The equation of a curve is $y = \tan^{-1}(4x)$ .

(a)

Find the exact values of $x$ when the gradient of the curve is $\frac{1}{4}$ .

3M Similar

DifficultyMedium-Easy

Worked solution

Approach

Differentiate $y = \tan^{-1}(4x)$ using the chain rule, then set the derivative equal to $\frac{1}{4}$ and solve for $x$ .

Working

y = \tan^{-1}(4x)

Using the derivative of $\tan^{-1} x$ , which is $\frac{1}{1+x^2}$ , and the chain rule:

\frac{dy}{dx} = \frac{4}{1+(4x)^2} = \frac{4}{1+16x^2}

Set the gradient equal to $\frac{1}{4}$ :

\frac{4}{1+16x^2} = \frac{1}{4}

Multiply both sides:

16 = 1 + 16x^2

16x^2 = 15

x^2 = \frac{15}{16}

Hence:

x = \pm \frac{\sqrt{15}}{4}

Answer

x = \pm \frac{\sqrt{15}}{4}

Final answer

x = ±√15/4

Detailed explanation

Walkthrough

The curve is $y = \tan^{-1}(4x)$ . To find where the gradient is $\frac{1}{4}$ , we first differentiate. The derivative of $\tan^{-1} u$ is $\frac{1}{1+u^2}\frac{du}{dx}$ , so with $u = 4x$ we get $\frac{dy}{dx} = \frac{4}{1+16x^2}$ . Then set this equal to $\frac{1}{4}$ and solve:

\frac{4}{1+16x^2} = \frac{1}{4}

This gives $16 = 1 + 16x^2$ , so $x^2 = \frac{15}{16}$ and $x = \pm \frac{\sqrt{15}}{4}$ . Both values are valid because the derivative formula is defined for all real $x$ .

Key Takeaways

This question tests differentiating an inverse trigonometric function with a linear composite, and solving a simple equation. The chain rule is essential when differentiating $\tan^{-1}(ax)$ .

Common Mistakes

Forgetting the chain rule and writing $\frac{1}{1+4x^2}$ instead of $\frac{4}{1+16x^2}$ .
Losing the negative solution when taking square roots.
Not simplifying $\sqrt{\frac{15}{16}}$ to $\frac{\sqrt{15}}{4}$ .

Things to Be Careful About

The derivative of $\tan^{-1}(4x)$ has $16x^2$ in the denominator, not $4x^2$ . When solving $x^2 = \frac{15}{16}$ , remember both positive and negative roots. The mark scheme accepts exact equivalents such as $\pm \frac{\sqrt{15}}{\sqrt{16}}$ .

Techniques used

differentiate inverse tangent using the chain ruleset derivative equal to a given valuesolve the resulting quadratic equation

(b)

Find the exact value of $\int_0^{0.25} y \, \text{d}x$ .

5M Similar

DifficultyMedium

Worked solution

Approach

Use integration by parts with $u = \tan^{-1}(4x)$ and $\text{d}v = \text{d}x$ . Then integrate the remaining term using the form $k f'(x)/f(x)$ , and evaluate between $0$ and $0.25$ .

Working

Let

I = \int_0^{0.25} \tan^{-1}(4x) \, \text{d}x

Use integration by parts, $\int u \, \text{d}v = uv - \int v \, \text{d}u$ , with $u = \tan^{-1}(4x)$ and $\text{d}v = \text{d}x$ :

\frac{du}{dx} = \frac{4}{1+16x^2}, \quad v = x

So

I = \left[ x \tan^{-1}(4x) \right]_0^{0.25} - \int_0^{0.25} \frac{4x}{1+16x^2} \, \text{d}x

Now integrate $\frac{4x}{1+16x^2}$ . Since the derivative of $1+16x^2$ is $32x$ , we have

\int \frac{4x}{1+16x^2} \, \text{d}x = \frac{1}{8} \ln(1+16x^2)

Therefore

I = \left[ x \tan^{-1}(4x) - \frac{1}{8} \ln(1+16x^2) \right]_0^{0.25}

Evaluate at $x = 0.25$ :

0.25 \tan^{-1}(1) - \frac{1}{8} \ln(1+1) = \frac{1}{4} \cdot \frac{\pi}{4} - \frac{1}{8} \ln 2 = \frac{\pi}{16} - \frac{1}{8} \ln 2

At $x = 0$ , both terms are $0$ , so the value is

\frac{\pi}{16} - \frac{1}{8} \ln 2

Answer

\frac{\pi}{16} - \frac{1}{8} \ln 2

Final answer

π/16 - (1/8)ln 2

Detailed explanation

Walkthrough

We need to integrate $y = \tan^{-1}(4x)$ from $0$ to $0.25$ . Since $\tan^{-1}(4x)$ is not a standard direct integral, integration by parts is appropriate: choose $u = \tan^{-1}(4x)$ and $\text{d}v = \text{d}x$ . Then $\text{d}u = \frac{4}{1+16x^2}\,\text{d}x$ and $v = x$ . This gives

\left[ x\tan^{-1}(4x) \right]_0^{0.25} - \int_0^{0.25} \frac{4x}{1+16x^2} \, \text{d}x

The remaining integral is of the form $k f'(x)/f(x)$ because the derivative of $1+16x^2$ is $32x$ , so

\int \frac{4x}{1+16x^2} \, \text{d}x = \frac{1}{8} \ln(1+16x^2)

Substituting the limits, at $x=0.25$ we get $\frac{\pi}{16} - \frac{1}{8}\ln 2$ , and at $x=0$ the expression is $0$ .

Key Takeaways

This question combines integration by parts with recognising a logarithmic integral of the form $k f'(x)/f(x)$ . It also requires careful substitution of exact limits involving $\tan^{-1}(1)=\pi/4$ .

Common Mistakes

Choosing the wrong $u$ and $\text{d}v$ , making the integral harder.
Forgetting the coefficient when integrating $\frac{4x}{1+16x^2}$ ; the correct antiderivative is $\frac{1}{8}\ln(1+16x^2)$ , not $\frac{1}{4}\ln(1+16x^2)$ .
Forgetting to subtract the lower-limit value, although here it is zero.
Writing $\tan^{-1}(1)$ as $\pi/2$ instead of $\pi/4$ .

Things to Be Careful About

At $x=0.25$ , $4x=1$ , so $\tan^{-1}(1)=\pi/4$ . Also $1+16x^2 = 1+1=2$ . The mark scheme allows equivalent forms such as $\frac{1}{16}\pi + \frac{1}{4}\ln\left(\frac{\sqrt{2}}{2}\right)$ or $\frac{1}{16}\pi + \frac{1}{8}\ln\frac{1}{2}$ , so any exact equivalent is acceptable. Since $\ln 1 = 0$ , the lower limit contributes nothing, but it must still be stated for full method marks.

Techniques used

apply integration by partsrecognise and integrate k f'(x)/f(x)substitute exact limits and simplify

Q8MediumTrigonometryNumerical Solution of Equations

(a)

By sketching a suitable pair of graphs, show that the equation $\sec 2x = -2x - \frac{1}{2}$ has exactly one root in the interval $0 \leq x \leq \frac{1}{2}\pi$ .

2M Similar

DifficultyMedium-Easy

Worked solution

Approach

Sketch $y = \sec 2x$ and $y = -2x - \frac{1}{2}$ on the same axes for $0 \leq x \leq \frac{1}{2}\pi$ . Identify the number of intersection points to show there is exactly one root.

Working

Graph of $y = \sec 2x$ :

Recall that $\sec 2x = \frac{1}{\cos 2x}$ . Key features for $0 \leq x \leq \frac{1}{2}\pi$ :

At $x = 0$ : $\sec 0 = 1$ , so the graph passes through $(0, 1)$ .
As $x \to \frac{\pi}{4}^{-}$ : $\cos 2x \to 0^{+}$ , so $\sec 2x \to +\infty$ . There is a vertical asymptote at $x = \frac{\pi}{4}$ .
As $x \to \frac{\pi}{4}^{+}$ : $\cos 2x \to 0^{-}$ , so $\sec 2x \to -\infty$ .
At $x = \frac{\pi}{2}$ : $\sec \pi = \frac{1}{\cos \pi} = \frac{1}{-1} = -1$ , so the graph passes through $\left(\frac{\pi}{2}, -1\right)$ .

The graph has two branches: one from $(0,1)$ rising to $+\infty$ as $x \to \frac{\pi}{4}^{-}$ , and another from $-\infty$ (as $x \to \frac{\pi}{4}^{+}$ ) rising to $-1$ at $x = \frac{\pi}{2}$ .

Graph of $y = -2x - \frac{1}{2}$ :

This is a straight line with gradient $-2$ and $y$ -intercept $-\frac{1}{2}$ .

At $x = 0$ : $y = -\frac{1}{2}$ , so the line passes through $(0, -0.5)$ .
At $x = \frac{\pi}{2}$ : $y = -\pi - \frac{1}{2} \approx -3.57$ .

The line decreases steadily from $(0, -0.5)$ to approximately $(\frac{\pi}{2}, -3.57)$ .

Intersection:

The first branch of $\sec 2x$ (where $y \geq 1$ ) lies entirely above the line $y = -2x - \frac{1}{2}$ (which is at most $-0.5$ in this region), so there is no intersection here.

The second branch of $\sec 2x$ (where $x > \frac{\pi}{4}$ ) rises from $-\infty$ to $-1$ , while the line decreases from approximately $-2.09$ at $x = \frac{\pi}{4}$ to $-3.57$ at $x = \frac{\pi}{2}$ . Since the secant curve goes from $-\infty$ to $-1$ and the line goes from about $-2.09$ to $-3.57$ , they must cross exactly once in the interval $\frac{\pi}{4} < x < \frac{\pi}{2}$ .

We can verify this by noting:

At $x = \frac{\pi}{4}^{+}$ : $\sec 2x \to -\infty < -2(\frac{\pi}{4}) - \frac{1}{2} \approx -2.09$ .
At $x = \frac{\pi}{2}$ : $\sec \pi = -1 > -2(\frac{\pi}{2}) - \frac{1}{2} \approx -3.57$ .

By the Intermediate Value Theorem, there is at least one intersection. Since $\sec 2x$ is strictly increasing on $(\frac{\pi}{4}, \frac{\pi}{2})$ and $-2x - \frac{1}{2}$ is strictly decreasing, there is exactly one point of intersection.

Answer

The equation $\sec 2x = -2x - \frac{1}{2}$ has exactly one root in $0 \leq x \leq \frac{1}{2}\pi$ , located between $x = \frac{\pi}{4}$ and $x = \frac{\pi}{2}$ .

Final answer

Exactly one root exists in the interval, as the graphs intersect at a single point between x = π/4 and x = π/2.

Detailed explanation

Walkthrough

First, we sketch $y = \sec 2x$ on the interval $[0, \frac{\pi}{2}]$ . The function $\sec 2x = \frac{1}{\cos 2x}$ has a vertical asymptote where $\cos 2x = 0$ , which occurs at $2x = \frac{\pi}{2}$ , i.e., $x = \frac{\pi}{4}$ . This divides the interval into two branches:

Branch 1 ( $0 \leq x < \frac{\pi}{4}$ ): $\cos 2x$ is positive and decreasing from 1 to 0, so $\sec 2x$ increases from 1 to $+\infty$ .
Branch 2 ( $\frac{\pi}{4} < x \leq \frac{\pi}{2}$ ): $\cos 2x$ is negative, going from $0^{-}$ to $-1$ , so $\sec 2x$ goes from $-\infty$ to $-1$ .

Next, we sketch $y = -2x - \frac{1}{2}$ , which is a straight line with negative gradient $-2$ passing through $(0, -0.5)$ .

On Branch 1, $\sec 2x \geq 1$ while the line is at most $-0.5$ , so no intersection is possible. On Branch 2, the secant curve rises from $-\infty$ to $-1$ while the line falls from about $-2.09$ to $-3.57$ . Since one is increasing and the other decreasing, and they cross in sign relative to each other, there is exactly one intersection point.

Key Takeaways

The secant function $\sec 2x$ has vertical asymptotes at $x = \frac{\pi}{4} + \frac{k\pi}{2}$ and alternates between branches where it is $\geq 1$ and $\leq -1$ .
To show an equation has exactly one root graphically, sketch both sides and argue that the curves intersect exactly once, using monotonicity to rule out multiple intersections.

Common Mistakes

Forgetting the vertical asymptote at $x = \frac{\pi}{4}$ and drawing a continuous curve.
Not showing the correct values at key points: $(0, 1)$ and $(\frac{\pi}{2}, -1)$ .
Simply stating "only one root" without justification from the sketch (the mark scheme requires a dot at the intersection or a dotted line to the x-axis).
Sketching regions outside $0 \leq x \leq \frac{\pi}{2}$ (these should be ignored).

Things to Be Careful About

The interval is $0 \leq x \leq \frac{1}{2}\pi$ , not $0 \leq x \leq 2\pi$ . Only sketch within this range.
The asymptote at $x = \frac{\pi}{4}$ must be clearly indicated; the curve does not cross it.
Justification requires more than just stating there is one intersection — show why by comparing values at the endpoints of the branch or by noting monotonicity.

Techniques used

sketch secant graph with asymptotessketch linear functionidentify intersection point

(b)

Show by calculation that this root lies between 0.8 and 1.2.

2M Similar

DifficultyMedium-Easy

Worked solution

Approach

Define $f(x) = \sec 2x + 2x + \frac{1}{2}$ . Calculate $f(0.8)$ and $f(1.2)$ . If there is a sign change between these values, then by the Intermediate Value Theorem, the root lies between 0.8 and 1.2.

Working

Calculate $f(0.8)$ :

\begin{aligned} 2(0.8) &= 1.6 \text{ radians} \\ \cos(1.6) &= -0.02920 \quad \text{(to 5 d.p.)} \\ \sec(1.6) &= \frac{1}{\cos(1.6)} = -34.241 \quad \text{(to 5 d.p.)} \\ f(0.8) &= \sec(1.6) + 2(0.8) + \frac{1}{2} \\ &= -34.241 + 1.6 + 0.5 \\ &= -32.141 \end{aligned}

So $f(0.8) < 0$ .

Calculate $f(1.2)$ :

\begin{aligned} 2(1.2) &= 2.4 \text{ radians} \\ \cos(2.4) &= -0.73739 \quad \text{(to 5 d.p.)} \\ \sec(2.4) &= \frac{1}{\cos(2.4)} = -1.3563 \quad \text{(to 5 d.p.)} \\ f(1.2) &= \sec(2.4) + 2(1.2) + \frac{1}{2} \\ &= -1.3563 + 2.4 + 0.5 \\ &= 1.5437 \end{aligned}

So $f(1.2) > 0$ .

Since $f(0.8) = -32.14 < 0$ and $f(1.2) = 1.544 > 0$ , there is a sign change in the interval $[0.8, 1.2]$ . Since $f(x)$ is continuous on this interval (the asymptote at $x = \frac{\pi}{4} \approx 0.785$ is less than 0.8), by the Intermediate Value Theorem, there is at least one root in $(0.8, 1.2)$ . From part (a), there is exactly one root in the entire interval $[0, \frac{\pi}{2}]$ , so this root must lie between 0.8 and 1.2.

Answer

The root lies between 0.8 and 1.2, as $f(0.8) < 0$ and $f(1.2) > 0$ .

Final answer

f(0.8) = -32.14 < 0 and f(1.2) = 1.544 > 0, so the root lies between 0.8 and 1.2.

Detailed explanation

Walkthrough

We want to show the root of $\sec 2x = -2x - \frac{1}{2}$ lies between 0.8 and 1.2. Rearrange this as $f(x) = \sec 2x + 2x + \frac{1}{2} = 0$ .

We evaluate $f(x)$ at the endpoints $x = 0.8$ and $x = 1.2$ . Remember that all trigonometric calculations must be in radians.

At $x = 0.8$ : $2x = 1.6$ rad. $\cos(1.6) \approx -0.02920$ , so $\sec(1.6) \approx -34.241$ . Then $f(0.8) = -34.241 + 1.6 + 0.5 = -32.141 < 0$ .

At $x = 1.2$ : $2x = 2.4$ rad. $\cos(2.4) \approx -0.73739$ , so $\sec(2.4) \approx -1.3563$ . Then $f(1.2) = -1.3563 + 2.4 + 0.5 = 1.5437 > 0$ .

Since $f$ changes sign between $x = 0.8$ and $x = 1.2$ , and $f$ is continuous on this interval (the asymptote at $x = \pi/4 \approx 0.785$ is outside this interval), there must be a root between 0.8 and 1.2.

Key Takeaways

The sign-change method (bisection principle) requires evaluating a continuous function at two points and checking for a sign change.
Always ensure your calculator is in radian mode when working with trigonometric functions in calculus/numerical methods.
The function must be continuous on the interval; check that no asymptotes lie within it.

Common Mistakes

Working in degrees instead of radians — this gives completely wrong values and earns no marks.
Not showing both evaluations clearly; the mark scheme requires seeing both $x = 0.8$ and $x = 1.2$ explicitly.
Forgetting to check continuity: if there were an asymptote between the two points, a sign change would not guarantee a root.
Rounding too early: answers must be correct to at least 2 significant figures for the A1 mark.

Things to Be Careful About

The asymptote at $x = \frac{\pi}{4} \approx 0.7854$ is just below 0.8, so the interval $[0.8, 1.2]$ does not contain the asymptote and $f(x)$ is continuous there.
Use at least 2 significant figures in the final values for the A1 mark.
The mark scheme also accepts alternative approaches: comparing $\sec 2x$ and $-2x - \frac{1}{2}$ directly at both endpoints, or using the rearranged form $g(x) = \frac{1}{2}\cos^{-1}\left(\frac{-2}{4x+1}\right) - x$ .

Techniques used

define function f(x) = sec 2x + 2x + 1/2evaluate f at endpoints 0.8 and 1.2check sign change to locate root

(c)

Show that, if a sequence of real values given by the iterative formula

x_{n+1} = \frac{1}{2}\cos^{-1}\left(\frac{-2}{4x_n + 1}\right)

converges, then it converges to the root of the equation in part (a).

2M Similar

DifficultyMedium

Worked solution

Approach

If the sequence converges, let the limit be $L$ . Then as $n \to \infty$ , both $x_n \to L$ and $x_{n+1} \to L$ . Substitute $x_n = x_{n+1} = L$ into the iterative formula and rearrange to show that $L$ satisfies $\sec 2L = -2L - \frac{1}{2}$ .

Working

Assume the sequence converges to a limit $L$ . Then:

\lim_{n \to \infty} x_n = L \quad \text{and} \quad \lim_{n \to \infty} x_{n+1} = L

Substitute $x_n = x_{n+1} = L$ into the iterative formula:

L = \frac{1}{2} \cos^{-1}\left(\frac{-2}{4L + 1}\right)

Multiply both sides by 2:

2L = \cos^{-1}\left(\frac{-2}{4L + 1}\right)

Take cosine of both sides:

\cos 2L = \frac{-2}{4L + 1}

Take the reciprocal of both sides:

\frac{1}{\cos 2L} = \frac{4L + 1}{-2}

Simplify the right-hand side:

\sec 2L = -\frac{4L + 1}{2} = -2L - \frac{1}{2}

Therefore:

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

This shows that $L$ is a root of the equation $\sec 2x = -2x - \frac{1}{2}$ . Since part (a) establishes that there is exactly one root in the interval $[0, \frac{\pi}{2}]$ , the sequence must converge to this unique root.

Answer

If the sequence converges to $L$ , then $L$ satisfies $\sec 2L = -2L - \frac{1}{2}$ , so it converges to the root of the equation.

Final answer

Substituting x_n = x_{n+1} = L into the iterative formula and rearranging gives sec 2L = -2L - 1/2, proving convergence to the root.

Detailed explanation

Walkthrough

The key idea is that if a sequence defined by $x_{n+1} = F(x_n)$ converges, then in the limit, $x_{n+1}$ and $x_n$ both approach the same value $L$ . This is called a fixed point of the function $F$ .

So we set $x_{n+1} = x_n = L$ in the formula:

$L = \frac{1}{2} \cos^{-1}\left(\frac{-2}{4L + 1}\right)$

Now we need to manipulate this to recover the original equation $\sec 2x = -2x - \frac{1}{2}$ .

Step 1: Multiply by 2 to isolate the inverse cosine:
$2L = \cos^{-1}\left(\frac{-2}{4L + 1}\right)$

Step 2: Apply cosine to both sides. Since $\cos(\cos^{-1}(y)) = y$ for $y \in [-1, 1]$ :
$\cos 2L = \frac{-2}{4L + 1}$

Step 3: Take reciprocals:
$\frac{1}{\cos 2L} = \frac{4L + 1}{-2}$

Step 4: Recognize $\frac{1}{\cos 2L} = \sec 2L$ and simplify the right side:
$\sec 2L = -\frac{4L + 1}{2} = -2L - \frac{1}{2}$

This is exactly the original equation with $L$ in place of $x$ . Therefore, the limit $L$ must be a root of the original equation.

Key Takeaways

If an iterative sequence $x_{n+1} = F(x_n)$ converges to $L$ , then $L$ is a fixed point: $L = F(L)$ .
To show convergence to a specific root, substitute the fixed-point condition and rearrange to recover the original equation.
The rearrangement must be algebraically correct at every step — slips lose the accuracy mark.

Common Mistakes

Forgetting to take cosine of both sides after isolating $\cos^{-1}$ .
Making algebraic errors when taking reciprocals or simplifying $\frac{4L+1}{-2}$ .
Not showing enough working — the mark scheme requires full correct working with no slips for the A1 mark.
Using inconsistent variables (mixing $x_n$ and $x_{n+1}$ without justification).

Things to Be Careful About

The rearrangement must be done carefully: $\frac{4L+1}{-2} = -2L - \frac{1}{2}$ , not $-2L + \frac{1}{2}$ .
The mark scheme accepts either direction of rearrangement: from the iterative formula to the original equation, or vice versa.
Ensure all steps are reversible and valid (e.g., the reciprocal step requires $\cos 2L \neq 0$ , which is satisfied since the root is not at an asymptote).

Techniques used

assume convergence to limit Lsubstitute x_n = x_{n+1} = Lrearrange to recover original equation

(d)

Use this iterative formula to calculate this root correct to 3 decimal places. Give the result of each iteration to 5 decimal places.

3M Similar

DifficultyMedium

Worked solution

Approach

Choose an initial value $x_0$ between 0.8 and 1.2 (from part b). Apply the iterative formula $x_{n+1} = \frac{1}{2} \cos^{-1}\left(\frac{-2}{4x_n + 1}\right)$ repeatedly, giving each result to 5 decimal places. Continue until the answer is stable to 3 decimal places.

Working

Choose $x_0 = 1.0$ (a value between 0.8 and 1.2).

Iteration 1:

\begin{aligned} x_1 &= \frac{1}{2} \cos^{-1}\left(\frac{-2}{4(1.0) + 1}\right) \\ &= \frac{1}{2} \cos^{-1}\left(\frac{-2}{5}\right) \\ &= \frac{1}{2} \cos^{-1}(-0.4) \\ &= \frac{1}{2}(1.98231) \\ &= 0.99116 \end{aligned}

Iteration 2:

\begin{aligned} x_2 &= \frac{1}{2} \cos^{-1}\left(\frac{-2}{4(0.99116) + 1}\right) \\ &= \frac{1}{2} \cos^{-1}\left(\frac{-2}{4.96464}\right) \\ &= \frac{1}{2} \cos^{-1}(-0.40286) \\ &= \frac{1}{2}(1.98542) \\ &= 0.99271 \end{aligned}

Iteration 3:

\begin{aligned} x_3 &= \frac{1}{2} \cos^{-1}\left(\frac{-2}{4(0.99271) + 1}\right) \\ &= \frac{1}{2} \cos^{-1}\left(\frac{-2}{4.97084}\right) \\ &= \frac{1}{2} \cos^{-1}(-0.40235) \\ &= \frac{1}{2}(1.98488) \\ &= 0.99244 \end{aligned}

Iteration 4:

\begin{aligned} x_4 &= \frac{1}{2} \cos^{-1}\left(\frac{-2}{4(0.99244) + 1}\right) \\ &= \frac{1}{2} \cos^{-1}\left(\frac{-2}{4.96976}\right) \\ &= \frac{1}{2} \cos^{-1}(-0.40244) \\ &= \frac{1}{2}(1.98497) \\ &= 0.99249 \end{aligned}

Iteration 5:

\begin{aligned} x_5 &= \frac{1}{2} \cos^{-1}\left(\frac{-2}{4(0.99249) + 1}\right) \\ &= \frac{1}{2} \cos^{-1}\left(\frac{-2}{4.96996}\right) \\ &= \frac{1}{2} \cos^{-1}(-0.40242) \\ &= \frac{1}{2}(1.98495) \\ &= 0.99248 \end{aligned}

Iteration 6:

\begin{aligned} x_6 &= \frac{1}{2} \cos^{-1}\left(\frac{-2}{4(0.99248) + 1}\right) \\ &= \frac{1}{2} \cos^{-1}\left(\frac{-2}{4.96992}\right) \\ &= \frac{1}{2} \cos^{-1}(-0.40243) \\ &= \frac{1}{2}(1.98496) \\ &= 0.99248 \end{aligned}

Since $x_5 = 0.99248$ and $x_6 = 0.99248$ , the sequence has converged to 5 decimal places. To 3 decimal places:

x = 0.992

We can verify this is correct to 3 d.p. because $x_5$ and $x_6$ agree to 5 d.p., and both round to $0.992$ to 3 d.p.

Answer

x = 0.992

Final answer

x = 0.992

Detailed explanation

Walkthrough

We use the iterative formula $x_{n+1} = \frac{1}{2} \cos^{-1}\left(\frac{-2}{4x_n + 1}\right)$ with an initial value $x_0 = 1.0$ (chosen between 0.8 and 1.2 from part b).

Each iteration requires:

Compute $4x_n + 1$ .
Compute $\frac{-2}{4x_n + 1}$ .
Find $\cos^{-1}$ of this value (in radians).
Divide by 2 to get $x_{n+1}$ .

We continue until two consecutive values agree to 5 decimal places, which guarantees the answer is correct to 3 decimal places.

The iterations give:

$x_0 = 1.00000$
$x_1 = 0.99116$
$x_2 = 0.99271$
$x_3 = 0.99244$
$x_4 = 0.99249$
$x_5 = 0.99248$
$x_6 = 0.99248$

Since $x_5 = x_6 = 0.99248$ to 5 d.p., the sequence has converged. Rounding to 3 d.p. gives $0.992$ .

Key Takeaways

Always choose an initial value within the interval where the root is known to lie.
Carry at least 5 decimal places during iterations to justify the answer to 3 d.p.
Continue until two consecutive values agree to the required accuracy (or more).
Always work in radians for inverse trigonometric functions in this context.

Common Mistakes

Starting with a value outside $[0.8, 1.2]$ — the mark scheme gives M0 if $x_0$ is not in this range.
Working in degrees instead of radians — this gives completely wrong iterations and earns no marks.
Stopping too early — need to show enough iterations to justify the answer. The mark scheme requires either convergence to 5 d.p. or a sign change in $(0.9915, 0.9925)$ .
Writing $x_n = 0.992$ instead of simply stating $x = 0.992$ — the mark scheme awards A0 for the latter phrasing.
Rounding intermediate results too aggressively — always keep at least 5 d.p. during iterations.

Things to Be Careful About

The iterative formula must be entered correctly into the calculator: $\frac{1}{2} \cos^{-1}\left(\frac{-2}{4x + 1}\right)$ .
Ensure the calculator is in radian mode throughout.
The mark scheme provides iteration sequences for several starting values (0.8, 0.9, 1.0, 1.1, 1.2). All should converge to the same value $0.99248\ldots$ .
To justify 3 d.p., you can either show convergence to 5 d.p. (as done above) or show that $f(0.9915)$ and $f(0.9925)$ have opposite signs.

Techniques used

choose initial value between 0.8 and 1.2apply iterative formula repeatedlycheck convergence to required accuracy

Q9MediumAlgebra

(a)

Express $\frac{12x^2 + 55x - 2}{(3x - 2)(x + 6)}$ in partial fractions.

5M Similar

DifficultyMedium

Worked solution

Approach

Since the numerator and denominator have the same degree, first divide to express the fraction as a constant plus a proper partial fraction. Then decompose the remainder into $\frac{B}{3x-2}+\frac{C}{x+6}$ and solve for $B$ and $C$ .

Working

Let

\frac{12x^2+55x-2}{(3x-2)(x+6)} = A + \frac{B}{3x-2} + \frac{C}{x+6}.

Divide the numerator by the denominator:

(3x-2)(x+6)=3x^2+16x-12

4(3x^2+16x-12)=12x^2+64x-48

Subtract:

12x^2+55x-2 - (12x^2+64x-48) = -9x+46

so $A=4$ and

\frac{12x^2+55x-2}{(3x-2)(x+6)} = 4 + \frac{-9x+46}{(3x-2)(x+6)}.

Now set

\frac{-9x+46}{(3x-2)(x+6)} = \frac{B}{3x-2}+\frac{C}{x+6}.

Multiply through by the denominator:

-9x+46 = B(x+6)+C(3x-2).

Compare coefficients:

B+3C=-9,\quad 6B-2C=46.

From the first equation, $B=-9-3C$ . Substitute:

6(-9-3C)-2C=46

-54-18C-2C=46 \implies -20C=100 \implies C=-5.

Then $B=-9-3(-5)=6$ .

Thus

\frac{12x^2+55x-2}{(3x-2)(x+6)} = 4 + \frac{6}{3x-2} - \frac{5}{x+6}.

Answer

4+\frac{6}{3x-2}-\frac{5}{x+6}

Final answer

4 + 6/(3x - 2) - 5/(x + 6)

Detailed explanation

Walkthrough

The numerator has degree 2 and the denominator has degree 2, so the fraction is improper. Before using the usual partial fraction method, we extract the whole-number part by division. Dividing gives quotient $4$ and remainder $-9x+46$ , so the fraction becomes $4+\frac{-9x+46}{(3x-2)(x+6)}$ .

Next, because the denominator has two distinct linear factors, write the proper part as $\frac{B}{3x-2}+\frac{C}{x+6}$ . Multiplying through by the denominator gives $-9x+46 = B(x+6)+C(3x-2)$ . Comparing coefficients of $x$ and the constant term produces two linear equations. Solving them gives $C=-5$ and $B=6$ . Thus the original expression is $4+\frac{6}{3x-2}-\frac{5}{x+6}$ .

Key Takeaways

This question tests partial fractions for an improper rational function. The key idea is that when the numerator degree is not smaller than the denominator degree, divide first. It also tests solving a small system of linear equations by comparing coefficients.

Common Mistakes

Forgetting the constant term $A$ when numerator and denominator have the same degree.
Using the form $\frac{B}{3x-2}+\frac{C}{x+6}$ without the leading $A$ ; this loses the mark for the correct form.
Making sign errors when substituting $B=-9-3C$ into the constant equation.
Confusing the remainder when dividing.

Things to Be Careful About

The denominator factors are distinct linear factors, so the partial fraction numerators are constants, not linear expressions. When comparing coefficients, check both the $x$ coefficient and the constant term. The mark scheme allows the alternative method of dividing first and then decomposing the remainder; either route should give the same final answer.

Techniques used

divide numerator by denominatorset up partial fraction decompositionsolve linear equations for constants

(b)

Hence obtain the expansion of $\frac{12x^2 + 55x - 2}{(3x - 2)(x + 6)}$ in ascending powers of $x$ , up to and including the term in $x^2$ .

4M Similar

DifficultyMedium

Worked solution

Approach

Use the partial fraction result from part (a). Rewrite each denominator as a constant times $(1 + \text{multiple of }x)$ , then apply the binomial expansion for negative powers up to $x^2$ and collect like terms.

Working

From part (a),

\frac{12x^2+55x-2}{(3x-2)(x+6)} = 4 + \frac{6}{3x-2} - \frac{5}{x+6}.

Expand the first partial fraction:

\frac{6}{3x-2} = \frac{6}{-2(1-\frac{3}{2}x)} = -3\left(1-\frac{3}{2}x\right)^{-1}.

Using $(1+u)^{-1}=1-u+u^2-\cdots$ with $u=-\frac{3}{2}x$ :

-3\left(1+\frac{3}{2}x+\left(\frac{3}{2}x\right)^2+\cdots\right) = -3 - \frac{9}{2}x - \frac{27}{4}x^2+\cdots.

Expand the second partial fraction:

-\frac{5}{x+6} = -\frac{5}{6(1+\frac{x}{6})} = -\frac{5}{6}\left(1+\frac{x}{6}\right)^{-1}.

-\frac{5}{6}\left(1-\frac{x}{6}+\frac{x^2}{36}-\cdots\right) = -\frac{5}{6}+\frac{5}{36}x-\frac{5}{216}x^2+\cdots.

Add the constant $4$ and collect terms.

Constant term:

4-3-\frac{5}{6}=\frac{1}{6}.

Coefficient of $x$ :

-\frac{9}{2}+\frac{5}{36}=-\frac{157}{36}.

Coefficient of $x^2$ :

-\frac{27}{4}-\frac{5}{216}=-\frac{1463}{216}.

Therefore

\frac{12x^2+55x-2}{(3x-2)(x+6)} = \frac{1}{6} - \frac{157}{36}x - \frac{1463}{216}x^2 + \cdots.

Answer

\frac{1}{6} - \frac{157}{36}x - \frac{1463}{216}x^2

Final answer

1/6 - 157/36 x - 1463/216 x^2

Detailed explanation

Walkthrough

We use the partial fraction result from part (a). Each term must be expanded separately. For $\frac{6}{3x-2}$ , factor $-2$ out of the denominator to get $-3(1-\frac{3}{2}x)^{-1}$ . For $-\frac{5}{x+6}$ , factor $6$ out to get $-\frac{5}{6}(1+\frac{x}{6})^{-1}$ . Then apply the binomial expansion for negative index: $(1+u)^{-1}=1-u+u^2-\cdots$ . We only need terms up to $x^2$ , so we stop after the $u^2$ term. After expanding both terms and adding the constant $4$ , collect the constant, $x$ and $x^2$ coefficients to obtain the final expansion.

Key Takeaways

This question tests the binomial expansion for rational and negative powers, which produces an infinite series. It also shows how partial fractions can break a complicated rational function into simpler pieces that are easy to expand. The expansion is valid only when $|x|$ is small enough for both geometric-type series to converge.

Common Mistakes

Forgetting to multiply the binomial expansion by the constant factor in front, such as $-3$ or $-\frac{5}{6}$ .
Using the wrong sign in $(1-\frac{3}{2}x)^{-1}$ ; the signs alternate.
Stopping at the $x$ term instead of including $x^2$ .
Making arithmetic errors when adding fractions with denominators $36$ and $216$ .

Things to Be Careful About

The expansion of $(1+u)^{-1}$ is $1-u+u^2-\cdots$ , so the $x$ term is negative and the $x^2$ term is positive before multiplying by the outer constant. Combine the constant terms carefully: $4-3-\frac{5}{6}=\frac{1}{6}$ . The mark scheme allows follow-through on $B$ and $C$ from part (a), but the final simplified coefficients must be correct. Since $(1-\frac{3}{2}x)^{-1}$ requires $|\frac{3}{2}x|<1$ and $(1+\frac{x}{6})^{-1}$ requires $|\frac{x}{6}|<1$ , the combined expansion is valid for $|x|<\frac{2}{3}$ .

Techniques used

rewrite denominators in binomial formexpand negative-power binomialscollect coefficients up to x^2

Q10MediumVectors

With respect to the origin $O$ , the points $A$ , $B$ and $C$ have position vectors given by

\vec{OA} = 2\mathbf{i} - \mathbf{j} - 6\mathbf{k}, \quad \vec{OB} = b\mathbf{i} - 2\mathbf{j} + 3\mathbf{k} \quad \text{and} \quad \vec{OC} = -4\mathbf{i} + 5\mathbf{j} - 2\mathbf{k}.

(a)

It is given that $|\vec{AB}| = |\vec{BC}|$ .

Find the value of $b$ .

3M Similar

DifficultyMedium-Easy

Worked solution

Approach

Form the displacement vectors $\vec{AB}$ and $\vec{BC}$ , equate their squared magnitudes to avoid square roots, and solve the resulting equation for $b$ .

Working

\vec{AB} = \vec{OB} - \vec{OA} = (b-2)\mathbf{i} + (-2+1)\mathbf{j} + (3+6)\mathbf{k} = (b-2)\mathbf{i} - \mathbf{j} + 9\mathbf{k}

\vec{BC} = \vec{OC} - \vec{OB} = (-4-b)\mathbf{i} + (5+2)\mathbf{j} + (-2-3)\mathbf{k} = -(b+4)\mathbf{i} + 7\mathbf{j} - 5\mathbf{k}

Equate the squared magnitudes:

|\vec{AB}|^2 = (b-2)^2 + 1 + 81 = (b-2)^2 + 82

|\vec{BC}|^2 = (b+4)^2 + 49 + 25 = (b+4)^2 + 74

Setting $|\vec{AB}|^2 = |\vec{BC}|^2$ :

(b-2)^2 + 82 = (b+4)^2 + 74

(b-2)^2 - (b+4)^2 = -8

Using the difference of squares $\big[(b-2)-(b+4)\big]\big[(b-2)+(b+4)\big] = -8$ :

(-6)(2b+2) = -8

-12b - 12 = -8

-12b = 4

b = -\frac{1}{3}

Answer

b = -\frac{1}{3}

Final answer

b = -1/3

Detailed explanation

Walkthrough

We are told that the lengths of $\vec{AB}$ and $\vec{BC}$ are equal. To make use of this, we first need the two displacement vectors themselves, which we obtain by subtracting position vectors. The position vector $\vec{OA} = 2\mathbf{i} - \mathbf{j} - 6\mathbf{k}$ gives the coordinates of $A$ , and similarly for $B$ and $C$ .

For $\vec{AB}$ , we subtract component-wise: the $i$ -component is $b-2$ , the $j$ -component is $-2-(-1) = -1$ , the $k$ -component is $3-(-6) = 9$ . For $\vec{BC}$ , the $i$ -component is $-4-b$ , the $j$ -component is $5-(-2) = 7$ , the $k$ -component is $-2-3 = -5$ .

Because both magnitudes are non-negative, the equation $|\vec{AB}| = |\vec{BC}|$ is equivalent to $|\vec{AB}|^2 = |\vec{BC}|^2$ . This removes the square roots and gives a polynomial equation in $b$ . Expanding the squares and simplifying leads to a linear equation once the quadratic terms cancel, which happens because the squared terms $(b-2)^2$ and $(b+4)^2$ both contain $b^2$ . Using the difference of squares makes the simplification quick: the factor $(2b+2)$ contains the only remaining $b$ . Solving gives $b = -1/3$ .

Key Takeaways

Displacement vectors are obtained by subtracting position vectors component-wise.
The equation $|\vec{u}| = |\vec{v}|$ is equivalent to $|\vec{u}|^2 = |\vec{v}|^2$ , avoiding square roots.
When the squared expressions both contain $b^2$ , those terms cancel and the resulting equation is linear.

Common Mistakes

Forgetting to square both sides, leaving an equation with two square roots.
Sign errors when forming $\vec{AB}$ (should be $\vec{OB} - \vec{OA}$ , not the other way round).
Errors when squaring terms like $(b+4)^2$ — remember the cross term is $2 \cdot b \cdot 4 = 8b$ .

Things to Be Careful About

The squared magnitude drops the square root but introduces no extraneous solutions because both sides are non-negative.
Always check that the difference of squares factorisation is applied correctly to avoid sign errors.

Techniques used

form displacement vectors AB and BCsquare both magnitudes and equatesolve the resulting linear equation

(b)

$A$ , $B$ , $C$ and $D$ are the vertices of a rhombus.

Find the position vector of $D$ .

2M Similar

DifficultyMedium-Easy

Worked solution

Approach

In a rhombus with vertices labelled $A$ , $B$ , $C$ , $D$ in order, the opposite sides are equal as vectors: $\vec{AD} = \vec{BC}$ . Therefore $\vec{OD} = \vec{OA} + \vec{BC}$ .

Working

\vec{BC} = -(b+4)\mathbf{i} + 7\mathbf{j} - 5\mathbf{k}

Substituting $b = -\frac{1}{3}$ :

\vec{BC} = -\left(-\frac{1}{3}+4\right)\mathbf{i} + 7\mathbf{j} - 5\mathbf{k} = -\frac{11}{3}\mathbf{i} + 7\mathbf{j} - 5\mathbf{k}

Then:

\vec{OD} = \vec{OA} + \vec{BC} = \left(2 - \frac{11}{3}\right)\mathbf{i} + (-1 + 7)\mathbf{j} + (-6 - 5)\mathbf{k}

\vec{OD} = -\frac{5}{3}\mathbf{i} + 6\mathbf{j} - 11\mathbf{k}

Answer

\vec{OD} = -\frac{5}{3}\mathbf{i} + 6\mathbf{j} - 11\mathbf{k}

Final answer

OD = -5/3 i + 6j - 11k

Detailed explanation

Walkthrough

In any quadrilateral $ABCD$ (vertices taken in order), the side $\vec{AD}$ goes from $A$ to $D$ , while $\vec{BC}$ goes from $B$ to $C$ . When the figure is a rhombus, opposite sides are parallel and equal in length, and as vectors they are equal: $\vec{AD} = \vec{BC}$ .

To reach $D$ from the origin, we travel first from $O$ to $A$ (giving $\vec{OA}$ ) and then from $A$ to $D$ (giving $\vec{BC}$ ). So $\vec{OD} = \vec{OA} + \vec{AD} = \vec{OA} + \vec{BC}$ .

We already formed $\vec{BC} = -(b+4)\mathbf{i} + 7\mathbf{j} - 5\mathbf{k}$ in part (a). With $b = -1/3$ , the $i$ -component becomes $-(4 - 1/3) = -11/3$ . Adding component-wise to $\vec{OA} = 2\mathbf{i} - \mathbf{j} - 6\mathbf{k}$ gives the position vector of $D$ .

Key Takeaways

A rhombus has opposite sides equal both in length and direction (as vectors).
Adding $\vec{OA}$ and $\vec{AD}$ produces the position vector of $D$ .
The order in which vertices are named in a rhombus determines the direction of each side.

Common Mistakes

Using $\vec{AB}$ instead of $\vec{BC}$ to find $D$ — this would not give a rhombus.
Substituting $b = 0$ (forgetting the result of part (a)) and so losing the answer.
Reversing the direction of $\vec{BC}$ when adding, which would give the wrong vertex.

Things to Be Careful About

The rhombus property used is $\vec{AD} = \vec{BC}$ , not $\vec{AD} = -\vec{BC}$ . The order of vertices in the name $ABCD$ matters.
Per the mark scheme, the answer must be given in $\mathbf{i}$ , $\mathbf{j}$ , $\mathbf{k}$ notation (or as a column vector with bare numbers), not as a column of $\mathbf{i}$ -, $\mathbf{j}$ -, $\mathbf{k}$ -unit vectors.

Techniques used

use the rhombus property that opposite sides are equal vectorsadd the position vector of A to the displacement vector BCsimplify component-wise

(c)

Calculate angle $ABC$ .

3M Similar

DifficultyMedium

Worked solution

Approach

Use the scalar product formula $\vec{BA} \cdot \vec{BC} = |\vec{BA}|\,|\vec{BC}|\cos(\angle ABC)$ , solve for the cosine, and then take the inverse cosine.

Working

\vec{BA} = \vec{OA} - \vec{OB} = \left(2+\frac{1}{3}\right)\mathbf{i} + (-1+2)\mathbf{j} + (-6-3)\mathbf{k} = \frac{7}{3}\mathbf{i} + \mathbf{j} - 9\mathbf{k}

\vec{BC} = -\frac{11}{3}\mathbf{i} + 7\mathbf{j} - 5\mathbf{k}

Scalar product:

\vec{BA} \cdot \vec{BC} = \left(\frac{7}{3}\right)\left(-\frac{11}{3}\right) + (1)(7) + (-9)(-5)

= -\frac{77}{9} + 7 + 45 = -\frac{77}{9} + \frac{52 \cdot 9}{9} = \frac{-77 + 468}{9} = \frac{391}{9}

Magnitudes:

|\vec{BA}|^2 = \left(\frac{7}{3}\right)^2 + 1^2 + 9^2 = \frac{49}{9} + \frac{9}{9} + \frac{729}{9} = \frac{787}{9}

|\vec{BC}|^2 = \left(\frac{11}{3}\right)^2 + 7^2 + 5^2 = \frac{121}{9} + 49 + 25 = \frac{121}{9} + \frac{441}{9} + \frac{225}{9} = \frac{787}{9}

So $|\vec{BA}|\,|\vec{BC}| = \dfrac{787}{9}$ . Therefore:

\cos(\angle ABC) = \frac{\vec{BA} \cdot \vec{BC}}{|\vec{BA}|\,|\vec{BC}|} = \frac{391/9}{787/9} = \frac{391}{787}

Finally:

\angle ABC = \cos^{-1}\left(\frac{391}{787}\right) \approx 60.2^\circ \approx 1.05 \text{ rad}

Answer

\angle ABC \approx 60.2^\circ \;\text{ (or } 1.05 \text{ radians)}

Final answer

60.2° (or 1.05 rad)

Detailed explanation

Walkthrough

The angle at vertex $B$ in triangle $ABC$ is formed by the two sides $\vec{BA}$ and $\vec{BC}$ . To find it, we use the scalar product identity $\vec{BA} \cdot \vec{BC} = |\vec{BA}|\,|\vec{BC}|\cos\theta$ , where $\theta$ is the angle between the two vectors. Rearranging gives $\cos\theta$ as a ratio.

We first build $\vec{BA}$ by subtracting the position vectors: $\vec{BA} = \vec{OA} - \vec{OB}$ . With $b = -1/3$ , each component simplifies: the $i$ -component becomes $2 - (-1/3) = 7/3$ , the $j$ -component is $1$ , and the $k$ -component is $-9$ . The vector $\vec{BC}$ was already found in part (b).

The scalar product is computed term by term, multiplying corresponding components and adding. We then compute $|\vec{BA}|^2$ and $|\vec{BC}|^2$ separately, which conveniently come out equal (both equal to $787/9$ ) — this confirms the rhombus property used in part (b).

The product of the magnitudes is $787/9$ , and the scalar product is $391/9$ , so the ratio is $391/787 \approx 0.4968$ . Taking the inverse cosine gives an angle of about $60.2°$ or $1.05$ radians.

Key Takeaways

The scalar product gives $\cos\theta$ directly when both magnitudes are known.
A clean simplification occurs when both sides of a triangle have equal length — their squared magnitudes are equal, so the product of the magnitudes is just the squared magnitude.
Either degrees or radians are accepted; the mark scheme accepts $60.2$ or $1.05$ .

Common Mistakes

Using $\vec{AB}$ and $\vec{BC}$ (or some other pair) instead of $\vec{BA}$ and $\vec{BC}$ ; the angle at $B$ must be formed by the two vectors emerging from $B$ .
Forgetting to take the square root after squaring magnitudes.
Using degrees versus radians inconsistently in the final answer.

Things to Be Careful About

The scalar product formula $\vec{u} \cdot \vec{v} = |\vec{u}||\vec{v}|\cos\theta$ requires the two vectors to share a common initial point — here that is $B$ , so $\vec{BA}$ and $\vec{BC}$ both start from $B$ .
Be careful with negative components when computing the scalar product, as they can produce a negative term that partially cancels with positive terms.

Techniques used

form BA and BC from position vectorscompute the scalar productcompute the product of the two magnitudestake the inverse cosine

Q118MMedium-HardDifferential Equations

The variables $x$ and $y$ satisfy the differential equation

(x^2 + 3)\frac{\text{d}y}{\text{d}x} = \text{e}^{3y}(x - 2).

It is given that $y = 0$ when $x = 0$ .

Solve the differential equation, and find the value of $y$ when $x = 2$ .

DifficultyMedium-Hard

Worked solution

Approach

Separate the variables so that all $y$ -terms are on one side and all $x$ -terms are on the other, then integrate both sides. Use the initial condition $y=0$ when $x=0$ to find the constant, and finally substitute $x=2$ .

Working

Separate the variables:

\frac{1}{e^{3y}}\,\text{d}y = \frac{x-2}{x^2+3}\,\text{d}x

e^{-3y}\,\text{d}y = \frac{x-2}{x^2+3}\,\text{d}x

Integrate both sides. Left-hand side:

\int e^{-3y}\,\text{d}y = -\frac{1}{3}e^{-3y}

Right-hand side:

\int \frac{x-2}{x^2+3}\,\text{d}x = \frac{1}{2}\ln(x^2+3) - \frac{2}{\sqrt{3}}\tan^{-1}\left(\frac{x}{\sqrt{3}}\right)

So the general solution is

-\frac{1}{3}e^{-3y} = \frac{1}{2}\ln(x^2+3) - \frac{2}{\sqrt{3}}\tan^{-1}\left(\frac{x}{\sqrt{3}}\right) + C

Use $x=0$ , $y=0$ :

-\frac{1}{3} = \frac{1}{2}\ln 3 + C

C = -\frac{1}{3} - \frac{1}{2}\ln 3

Therefore

-\frac{1}{3}e^{-3y} = \frac{1}{2}\ln(x^2+3) - \frac{2}{\sqrt{3}}\tan^{-1}\left(\frac{x}{\sqrt{3}}\right) - \frac{1}{3} - \frac{1}{2}\ln 3

At $x=2$ :

-\frac{1}{3}e^{-3y} = \frac{1}{2}\ln 7 - \frac{2}{\sqrt{3}}\tan^{-1}\left(\frac{2}{\sqrt{3}}\right) - \frac{1}{3} - \frac{1}{2}\ln 3

-\frac{1}{3}e^{-3y} \approx -0.8993

e^{-3y} \approx 2.6979

-3y \approx 0.9924

y \approx -0.331

Answer

y \approx -0.331

Final answer

y = -0.331

Detailed explanation

Walkthrough

This is a separable differential equation, so the first step is to rearrange it so that every term involving $y$ is on one side with $\text{d}y$ , and every term involving $x$ is on the other side with $\text{d}x$ .

Starting from

(x^2+3)\frac{\text{d}y}{\text{d}x} = e^{3y}(x-2),

we divide both sides by $e^{3y}$ and by $x^2+3$ :

e^{-3y}\,\text{d}y = \frac{x-2}{x^2+3}\,\text{d}x.

Now integrate both sides. The left side is a standard exponential integral:

\int e^{-3y}\,\text{d}y = -\frac{1}{3}e^{-3y}.

The right side is split into two standard forms:

\frac{x-2}{x^2+3} = \frac{x}{x^2+3} - \frac{2}{x^2+3}.

The first term integrates to $\frac{1}{2}\ln(x^2+3)$ , because the numerator is (apart from a constant) the derivative of the denominator. The second term integrates to $-\frac{2}{\sqrt{3}}\tan^{-1}\left(\frac{x}{\sqrt{3}}\right)$ , using the standard form $\int \frac{1}{x^2+a^2}\,\text{d}x = \frac{1}{a}\tan^{-1}\frac{x}{a}$ .

After integrating, we have a constant $C$ . The initial condition $y=0$ when $x=0$ lets us evaluate $C$ :

-\frac{1}{3} = \frac{1}{2}\ln 3 + C \quad \Rightarrow \quad C = -\frac{1}{3} - \frac{1}{2}\ln 3.

Substituting this back gives the particular solution. Finally, put $x=2$ and evaluate the right-hand side numerically to obtain $y \approx -0.331$ .

Key Takeaways

This question combines separation of variables with three standard integrals: $\int e^{ax}\,dx$ , $\int \frac{f'(x)}{f(x)}\,dx$ , and $\int \frac{1}{x^2+a^2}\,dx$ . It also tests the use of an initial condition to find the arbitrary constant and the evaluation of a particular solution.

Common Mistakes

Forgetting to divide by $e^{3y}$ correctly, so the left side is written as $e^{3y}\,\text{d}y$ instead of $e^{-3y}\,\text{d}y$ .
Missing the factor $\frac{1}{2}$ when integrating $\frac{x}{x^2+3}$ .
Missing the factor $\frac{1}{\sqrt{3}}$ inside the $\tan^{-1}$ and the factor $\frac{2}{\sqrt{3}}$ outside it when integrating $\frac{2}{x^2+3}$ .
Forgetting to use the initial condition, or using it before combining the two sides correctly.
Giving an unsupported numerical answer for $y$ ; the mark scheme requires the constant to be evaluated from the initial condition.

Things to Be Careful About

The right-hand side can be written as $\frac{1}{2}\ln(x^2+3)-\frac{2}{\sqrt{3}}\tan^{-1}\frac{x}{\sqrt{3}}+C$ . Do not omit the constant of integration before applying the initial condition.
When $x=0$ , $\tan^{-1}0=0$ and $\ln(x^2+3)=\ln 3$ , so the constant is not simply $-\frac{1}{3}$ .
The final answer is negative; check the sign of the $\tan^{-1}$ term when evaluating at $x=2$ .
The mark scheme allows equivalent forms, but the constant must be consistent with the integrated form used.

Techniques used

separate variablesintegrate exponential and rational functionsapply initial condition to determine constantevaluate particular solution numerically

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