(a) By sketching a suitable pair of graphs, show that the equation $\text{cosec} \frac{1}{2}x = e^x - 3$ has exactly one root, denoted by $\alpha$, in the interval $0 < x < \pi$. (b) Verify by calculation that $\alpha$ lies between 1 and 2. (c) Show that if a sequence of values in the interval $0 < x < \pi$ given by the iterative formula $$ x_{n+1} = \ln(\text{cosec} \frac{1}{2}x_n + 3) $$ converges, then it converges to $\alpha$. (d) Use this iterative formula with an…

(a) The curves y = e^x - 3 and y = cosec(x/2) intersect exactly once in 0 < x < π, giving exactly one root α. · (b) α lies between 1 and 2 · (c) If the sequence converges to L, then L satisfies cosec(L/2) = e^L - 3, so L = α. · (d) α = 1.50 · (e) 4

(a) Given that $2x = \tan y$, show that $\frac{dy}{dx} = \frac{2}{1 + 4x^2}$. (b) Hence find the exact value of $\int_{\frac{1}{2}}^{\frac{\sqrt{3}}{2}} x \tan^{-1}(2x) \,dx$.

(a) dy/dx = 2/(1 + 4x^2) · (b) 5π/48 − √3/8 + 1/8

9709/31 May/June 2024 — Mathematics Solved Past Paper

Q14MMedium-EasyAlgebra

Expand $(3+x)(1-2x)^{\frac{1}{2}}$ in ascending powers of $x$ , up to and including the term in $x^2$ , simplifying the coefficients.

DifficultyMedium-Easy

Worked solution

Approach

Use the binomial expansion for rational $n$ :

(1+u)^n = 1 + nu + \frac{n(n-1)}{2!}u^2 + \cdots

valid for $|u|<1$ . Take $n = \frac{1}{2}$ and $u = -2x$ , expand up to the term in $x^2$ , then multiply by $(3+x)$ and collect like terms up to $x^2$ .

Working

(1-2x)^{1/2} = 1 + \frac{1}{2}(-2x) + \frac{\frac{1}{2}(\frac{1}{2}-1)}{2!}(-2x)^2 + \cdots

Simplify the first two non-constant terms:

\frac{1}{2}(-2x) = -x, \qquad \frac{\frac{1}{2}(-\frac{1}{2})}{2}(4x^2) = -\frac{1}{2}x^2

So

(1-2x)^{1/2} = 1 - x - \frac{1}{2}x^2 + \cdots

Now multiply by $(3+x)$ :

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

Expanding and keeping terms up to $x^2$ :

= 3 - 3x - \frac{3}{2}x^2 + x - x^2 - \frac{1}{2}x^3 + \cdots

Ignoring the $x^3$ term,

= 3 + (-3x + x) + \left(-\frac{3}{2}x^2 - x^2\right) = 3 - 2x - \frac{5}{2}x^2.

Answer

3 - 2x - \frac{5}{2}x^2

Final answer

3 - 2x - (5/2)x^2

Detailed explanation

Walkthrough

We need the expansion of $(3+x)(1-2x)^{1/2}$ up to $x^2$ . The bracket $(1-2x)^{1/2}$ has a fractional index, so its binomial expansion is an infinite series. The formula is

(1+u)^n = 1 + nu + \frac{n(n-1)}{2!}u^2 + \cdots

with $|u|<1$ . Here $n=\frac{1}{2}$ and $u=-2x$ , so the series is valid for $|x|<\frac{1}{2}$ .

Substitute $n=\frac{1}{2}$ and $u=-2x$ . The linear term is

\frac{1}{2}(-2x) = -x.

The quadratic term is

\frac{\frac{1}{2}(\frac{1}{2}-1)}{2!}(-2x)^2 = \frac{\frac{1}{2}(-\frac{1}{2})}{2}(4x^2) = -\frac{1}{2}x^2.

So the bracket becomes $1-x-\frac{1}{2}x^2+\cdots$ .

Next multiply by $3+x$ . For the constant term, $3\cdot 1=3$ . For the $x$ term, $3(-x)+x(1)=-3x+x=-2x$ . For the $x^2$ term, $3(-\frac{1}{2}x^2)+x(-x)=-\frac{3}{2}x^2-x^2=-\frac{5}{2}x^2$ . The term $x(-\frac{1}{2}x^2)$ is of order $x^3$ , so it is not needed. This gives the final expansion.

The mark scheme gives B1 for the first two terms of the binomial expansion, B1 for the $x^2$ term, M1 for multiplying by $3+x$ , and A1 for the final simplified answer.

Key Takeaways

This question tests the binomial expansion for a rational index, where the expansion is infinite and valid only when $|u|<1$ . It also tests careful multiplication of a binomial series by a linear factor and collection of terms up to a specified order. You should be comfortable converting $(-2x)^2$ into $4x^2$ and simplifying fractional coefficients.

Common Mistakes

Leaving the coefficients in symbolic form, such as $1+n(-2x)$ , is not enough; the mark scheme requires the actual numerical coefficients, e.g. $1-x$ .
Forgetting the $x^2$ term of the binomial expansion, or using the wrong sign because $(-2x)^2$ is positive.
When multiplying by $(3+x)$ , forgetting the contribution $x(-x)=-x^2$ to the $x^2$ coefficient.
Including higher-order terms unnecessarily or making sign errors when collecting $x$ and $x^2$ terms.

Things to Be Careful About

Use $(1-2x)^{1/2}$ , not $(1+2x)^{1/2}$ ; the sign of $u$ matters.
The quadratic term uses $\frac{n(n-1)}{2!}u^2$ with $u^2=(-2x)^2=4x^2$ .
When collecting the $x^2$ coefficient, both $3\cdot(-\frac{1}{2}x^2)$ and $x\cdot(-x)$ contribute.
The expansion is valid for $|x|<\frac{1}{2}$ , although the question only asks for the series up to $x^2$ .

Techniques used

expand a binomial with a fractional indexmultiply a binomial series by a linear factorcollect like terms up to a specified order

Similar questions

Q24MMediumLogarithmic and Exponential Functions

Solve the equation $\ln(x-5) = 7 - \ln x$ . Give your answer correct to 2 decimal places.

DifficultyMedium

Worked solution

Approach

Use the product law of logarithms to combine the two $\ln$ terms into a single logarithm, then exponentiate both sides to eliminate the logarithm and obtain a quadratic equation. Solve the quadratic and select the root that satisfies the domain $x > 5$ .

Working

The equation is

\ln(x-5) = 7 - \ln x

Bring the $\ln x$ term to the left-hand side:

\ln(x-5) + \ln x = 7

Using $\ln a + \ln b = \ln(ab)$ :

\ln\left(x(x-5)\right) = 7

Exponentiate both sides (using $e^{\ln u} = u$ ):

x(x-5) = e^7

So

x^2 - 5x = e^7

x^2 - 5x - e^7 = 0

Using the quadratic formula with $a=1$ , $b=-5$ , $c=-e^7$ :

x = \frac{5 \pm \sqrt{25 + 4e^7}}{2}

Since $e^7 \approx 1096.63$ :

x = \frac{5 \pm \sqrt{4411.53}}{2}

x = \frac{5 \pm 66.419}{2}

This gives $x \approx 35.71$ or $x \approx -30.71$ . Since the original equation requires $x > 5$ (for $\ln(x-5)$ to be defined), the negative root is rejected.

Answer

x = 35.71

Final answer

x = 35.71

Detailed explanation

Walkthrough

We start with $\ln(x-5)=7-\ln x$ . The first goal is to collect the logarithms on one side so we can combine them. Adding $\ln x$ to both sides gives $\ln(x-5)+\ln x=7$ . The product law $\ln a + \ln b = \ln(ab)$ lets us merge them into $\ln(x(x-5))=7$ . This is useful because a single logarithm is easy to remove: since $e^{\ln u}=u$ , exponentiating both sides gives $x(x-5)=e^7$ . Expanding gives the quadratic $x^2-5x-e^7=0$ . We solve it with the quadratic formula. The discriminant is $25+4e^7$ , so $x=\frac{5\pm\sqrt{25+4e^7}}{2}$ . Numerically, the plus sign gives $35.71$ and the minus sign gives $-30.71$ . The domain of the original equation requires $x>5$ , so only $35.71$ is valid.

Key Takeaways

Logarithms can be combined using product/quotient laws before solving.
Exponentiating both sides removes a logarithm because $e^x$ and $\ln x$ are inverse functions.
Equations involving logarithms often lead to quadratic equations.
Always check the domain of the original logarithmic expressions; extraneous roots must be rejected.

Common Mistakes

Forgetting to add $\ln x$ to both sides before combining, leading to an incorrect single logarithm.
Misapplying the product law, e.g. writing $\ln(x-5)+\ln x = \ln(x-5+x)$ instead of $\ln(x(x-5))$ .
Failing to exponentiate both sides correctly, or forgetting that $e^{\ln u}=u$ .
Giving both quadratic roots without rejecting the negative one; the mark scheme accepts only $x=35.71$ .

Things to Be Careful About

The domain: $\ln(x-5)$ requires $x>5$ , and $\ln x$ requires $x>0$ , so the combined restriction is $x>5$ .
Use enough decimal places when approximating $e^7$ so the final rounding to 2 decimal places is correct.
When using the quadratic formula, keep the expression in exact form ( $e^7$ ) as long as possible before rounding.
The mark scheme requires a clear method: the product/quotient law and the elimination of the logarithm must be shown.

Techniques used

combine logarithmic terms using the product lawexponentiate both sides to eliminate the logarithmsolve the resulting quadratic equationcheck the domain to reject extraneous roots

Similar questions

Q34MMedium-EasyLogarithmic and Exponential Functions

Fig. 1

The variables $x$ and $y$ satisfy the equation $a^y = bx$ , where $a$ and $b$ are constants. The graph of $y$ against $\ln x$ is a straight line passing through the points $(0.336, 1.00)$ and $(1.31, 1.50)$ , as shown in the diagram.

Find the values of $a$ and $b$ . Give each value correct to the nearest integer.

DifficultyMedium-Easy

Worked solution

Approach

Take the natural logarithm of both sides of $a^y = bx$ to obtain a linear relationship between $y$ and $\ln x$ . The gradient and intercept of this line can then be determined from the two given points, allowing us to find $a$ and $b$ .

Working

Take natural logarithms of both sides of $a^y = bx$ :

\ln(a^y) = \ln(bx)

y \ln a = \ln b + \ln x

This is a straight-line equation of the form $y = m(\ln x) + c$ , where the gradient is $m = \dfrac{1}{\ln a}$ and the $y$ -intercept is $c = \dfrac{\ln b}{\ln a}$ .

The line passes through $(0.336, 1.00)$ and $(1.31, 1.50)$ . The gradient is:

m = \frac{1.50 - 1.00}{1.31 - 0.336} = \frac{0.50}{0.974} = 0.51335\ldots

Since $m = \dfrac{1}{\ln a}$ :

\ln a = \frac{1}{0.51335\ldots} = 1.9480\ldots

a = e^{1.9480\ldots} = 7.016\ldots

Rounding to the nearest integer, $a = 7$ .

To find $b$ , substitute $(0.336, 1.00)$ into $y \ln a = \ln b + \ln x$ :

1.00 \times 1.9480\ldots = \ln b + 0.336

\ln b = 1.9480\ldots - 0.336 = 1.6120\ldots

b = e^{1.6120\ldots} = 5.013\ldots

Rounding to the nearest integer, $b = 5$ .

Verification using the second point $(1.31, 1.50)$ :

1.50 \times 1.9480\ldots = 2.9220\ldots

\ln b = 2.9220\ldots - 1.31 = 1.6120\ldots \quad \checkmark

Answer

a = 7, \quad b = 5

Final answer

a = 7, b = 5

Detailed explanation

Walkthrough

The equation $a^y = bx$ is not linear in its current form, but taking logarithms converts it into a linear relationship. We start by applying the natural logarithm to both sides:

\ln(a^y) = \ln(bx)

Using logarithm laws, $\ln(a^y) = y \ln a$ and $\ln(bx) = \ln b + \ln x$ , giving:

y \ln a = \ln b + \ln x

Rearranging into the standard straight-line form $y = m(\ln x) + c$ :

y = \frac{1}{\ln a}(\ln x) + \frac{\ln b}{\ln a}

This tells us the graph of $y$ against $\ln x$ is a straight line with gradient $\frac{1}{\ln a}$ and $y$ -intercept $\frac{\ln b}{\ln a}$ .

We are given two points on this line: $(0.336, 1.00)$ and $(1.31, 1.50)$ . The gradient is calculated as:

m = \frac{1.50 - 1.00}{1.31 - 0.336} = \frac{0.50}{0.974} = 0.51335\ldots

Setting $\frac{1}{\ln a} = 0.51335\ldots$ gives $\ln a = 1.9480\ldots$ , so $a = e^{1.9480\ldots} \approx 7.016$ , which rounds to $7$ .

To find $b$ , we substitute one of the points into the linearised equation. Using $(0.336, 1.00)$ :

1.00 \times \ln a = \ln b + 0.336

\ln b = 1.9480\ldots - 0.336 = 1.6120\ldots

b = e^{1.6120\ldots} \approx 5.013

This rounds to $5$ . We can verify using the second point $(1.31, 1.50)$ :

1.50 \times 1.9480\ldots = 2.9220\ldots

\ln b = 2.9220\ldots - 1.31 = 1.6120\ldots \quad \checkmark

Both points give the same value of $\ln b$ , confirming our answer.

Key Takeaways

When an equation involves a variable in an exponent (like $a^y$ ), taking logarithms is the standard technique to linearise it.
The linearised form $y \ln a = \ln b + \ln x$ has $y$ on the vertical axis and $\ln x$ on the horizontal axis, with gradient $\frac{1}{\ln a}$ and intercept $\frac{\ln b}{\ln a}$ .
Two points on a straight line are sufficient to determine both the gradient and the intercept, which in turn give the constants in the original equation.
Always verify your answer by checking both given points to catch any arithmetic errors.

Common Mistakes

Forgetting to apply the logarithm law $\ln(a^y) = y \ln a$ correctly, e.g., writing $\ln(a^y) = a \ln y$ instead.
Misidentifying which variable is on which axis: the gradient is $\frac{1}{\ln a}$ , not $\ln a$ itself. The equation is $y = \frac{1}{\ln a}(\ln x) + \frac{\ln b}{\ln a}$ , so the coefficient of $\ln x$ is $\frac{1}{\ln a}$ .
Rounding intermediate values too early, which can lead to incorrect final answers. Keep several decimal places until the final step.
Forgetting to round to the nearest integer as required by the question.

Things to Be Careful About

The question asks for values correct to the nearest integer, so do not leave answers as exact logarithmic expressions.
Ensure you use consistent decimal places throughout the working to avoid rounding errors. The mark scheme accepts $a = 7$ and $b = 5$ to the nearest integer.
When substituting points into the linearised equation, be careful with the order: $y \ln a = \ln b + \ln x$ means $\ln b = y \ln a - \ln x$ , not $\ln b = \ln x - y \ln a$ .
The mark scheme requires showing the linearised form $y \ln a = \ln b + \ln x$ (B1) and a correct method for finding $\ln a$ or $\ln b$ (M1). Do not skip these steps.

Techniques used

take natural logarithm of both sidesidentify gradient and intercept of linearised equationcalculate gradient from two given pointssolve simultaneous equations for constants

Similar questions

Q4Medium-EasyComplex Numbers

The complex number $u$ is given by $u = -1 - i\sqrt{3}$ .

(a)

Express $u$ in the form $r(\cos \theta + i\sin \theta)$ , where $r > 0$ and $-\pi < \theta \le \pi$ . Give the exact values of $r$ and $\theta$ .

2M Similar

DifficultyMedium-Easy

Worked solution

Approach

Find the modulus $r$ using $r = |u| = \sqrt{x^2 + y^2}$ , then determine the principal argument $\theta$ from the quadrant of the point $(-1, -\sqrt{3})$ , and write $u$ in polar form.

Working

For $u = -1 - i\sqrt{3}$ , the real part is $-1$ and the imaginary part is $-\sqrt{3}$ .

r = |u| = \sqrt{(-1)^2 + (-\sqrt{3})^2} = \sqrt{1 + 3} = 2

The point $(-1, -\sqrt{3})$ lies in the third quadrant, so the principal argument is the negative angle below the negative real axis. Since $\cos\left(-\frac{2\pi}{3}\right) = -\frac{1}{2}$ and $\sin\left(-\frac{2\pi}{3}\right) = -\frac{\sqrt{3}}{2}$ ,

u = 2\left(\cos\left(-\frac{2\pi}{3}\right) + i\sin\left(-\frac{2\pi}{3}\right)\right)

Answer

r = 2, \quad \theta = -\frac{2\pi}{3}

Final answer

r = 2, θ = -2π/3

Detailed explanation

Walkthrough

For part (a), identify the real part $x = -1$ and the imaginary part $y = -\sqrt{3}$ . The modulus is the distance from the origin to the point $(-1, -\sqrt{3})$ , so $r = \sqrt{x^2 + y^2} = 2$ . To find the argument, locate the point in the Argand diagram: it is in the third quadrant. The reference angle is $\frac{\pi}{3}$ from the negative real axis, so the principal argument is $-\frac{2\pi}{3}$ , not $\frac{4\pi}{3}$ , because the principal argument must satisfy $-\pi < \theta \le \pi$ . Then write $u = 2\left(\cos\left(-\frac{2\pi}{3}\right) + i\sin\left(-\frac{2\pi}{3}\right)\right)$ .

Key Takeaways

This part tests the conversion from Cartesian form to polar form. You need to compute the modulus using Pythagoras and choose the principal argument based on the quadrant of the complex number. The polar form $r(\cos\theta + i\sin\theta)$ is equivalent to $re^{i\theta}$ .

Common Mistakes

A common mistake is to give $\theta = \frac{4\pi}{3}$ instead of $-\frac{2\pi}{3}$ . Both angles correspond to the same point, but only $-\frac{2\pi}{3}$ lies in the required interval $-\pi < \theta \le \pi$ . Another common mistake is forgetting the negative sign on the imaginary part when computing the argument.

Things to Be Careful About

Use radians, not degrees. The modulus must be positive. The argument must be exact and in the principal range. When the point is in the third quadrant, the principal argument is negative if measured clockwise from the positive real axis.

Techniques used

find the modulus of a complex numberdetermine the principal argument from the quadrantwrite a complex number in polar form

(b)

The complex number $v$ is given by $v = 5\left(\cos \frac{1}{6}\pi + i\sin \frac{1}{6}\pi\right)$ .

Express the complex number $\frac{v}{u}$ in the form $re^{i\theta}$ where $r > 0$ and $-\pi < \theta \le \pi$ .

2M Similar

DifficultyMedium-Easy

Worked solution

Approach

Write both $u$ and $v$ in polar exponential form. To divide complex numbers in polar form, divide the moduli and subtract the arguments. Then check that the resulting argument lies in $-\pi < \theta \le \pi$ .

Working

From part (a), $u = 2e^{-2\pi i/3}$ . Also,

v = 5\left(\cos\frac{\pi}{6} + i\sin\frac{\pi}{6}\right) = 5e^{i\pi/6}

Therefore,

\frac{v}{u} = \frac{5e^{i\pi/6}}{2e^{-2\pi i/3}} = \frac{5}{2} e^{i\left(\frac{\pi}{6} - \left(-\frac{2\pi}{3}\right)\right)} = \frac{5}{2} e^{i\left(\frac{\pi}{6} + \frac{4\pi}{6}\right)} = \frac{5}{2} e^{i\frac{5\pi}{6}}

Since $-\pi < \frac{5\pi}{6} \le \pi$ , this is already in the required principal range.

Answer

\frac{v}{u} = \frac{5}{2} e^{i\frac{5\pi}{6}}

Final answer

r = 5/2, θ = 5π/6

Detailed explanation

Walkthrough

Use the polar form of $u$ from part (a): $u = 2e^{-2\pi i/3}$ . Write $v$ in exponential form: $v = 5e^{i\pi/6}$ . To divide two complex numbers in polar form, divide their moduli and subtract their arguments. Thus $r = \frac{5}{2}$ and $\theta = \frac{\pi}{6} - \left(-\frac{2\pi}{3}\right) = \frac{5\pi}{6}$ . Since $\frac{5\pi}{6}$ lies between $-\pi$ and $\pi$ , it is already the principal argument. Therefore $\frac{v}{u} = \frac{5}{2}e^{i\frac{5\pi}{6}}$ .

Key Takeaways

This part uses the rule for division in polar form: if $z_1 = r_1e^{i\theta_1}$ and $z_2 = r_2e^{i\theta_2}$ , then $\frac{z_1}{z_2} = \frac{r_1}{r_2}e^{i(\theta_1 - \theta_2)}$ . It also reinforces the need to keep arguments in the principal range.

Common Mistakes

A common mistake is to add the arguments instead of subtracting them when dividing. Another is to forget to divide the moduli. Some students also give an equivalent angle outside the principal range, such as $-\frac{7\pi}{6}$ , without converting it back into $(-\pi, \pi]$ .

Things to Be Careful About

The final argument must satisfy $-\pi < \theta \le \pi$ . Here $\frac{5\pi}{6}$ is already acceptable. If part (a) was incorrect, the mark scheme allows follow-through using your values of $r$ and $\theta$ . Keep all values exact and use radians.

Techniques used

divide complex numbers in polar formdivide the modulisubtract the argumentscheck the argument lies in the principal range

Q57MMedium-HardDifferentiation

The equation of a curve is $y = \frac{e^{\sin x}}{\cos^2 x}$ for $0 \le x \le 2\pi$ .

Find $\frac{dy}{dx}$ and hence find the $x$ -coordinates of the stationary points of the curve.

DifficultyMedium-Hard

Worked solution

Approach

Differentiate $y = \frac{e^{\sin x}}{\cos^2 x}$ using the quotient rule. Then set the numerator of the derivative equal to zero, use $\cos^2 x = 1 - \sin^2 x$ to obtain a quadratic in $\sin x$ , solve it, and find the $x$ -coordinates in $0 \le x \le 2\pi$ .

Working

Let $u = e^{\sin x}$ and $v = \cos^2 x$ . Then

u' = \cos x \, e^{\sin x}, \qquad v' = -2\sin x \cos x.

By the quotient rule,

\frac{dy}{dx} = \frac{u'v - uv'}{v^2}

so

\frac{dy}{dx} = \frac{e^{\sin x}\cos x \cdot \cos^2 x - e^{\sin x}(-2\sin x \cos x)}{(\cos^2 x)^2}.

Thus

\frac{dy}{dx} = \frac{e^{\sin x}\cos^3 x + 2e^{\sin x}\sin x \cos x}{\cos^4 x}.

For stationary points, $\frac{dy}{dx} =$ , so the numerator must be zero (the denominator is finite when the curve is defined):

e^{\sin x}\cos x(\cos^2 x + 2\sin x) = .

Since $e^{\sin x} >$ and the curve is undefined where $\cos x =$ , we need

\cos^2 x + 2\sin x = .

Use $\cos^2 x = 1 - \sin^2 x$ :

1 - \sin^2 x + 2\sin x = ,

\sin^2 x - 2\sin x - = .

Solve this quadratic in $\sin x$ :

\sin x = \frac{2 \pm \sqrt{4 + 4}}{2} = \pm \sqrt{2}.

Since $1 + \sqrt{2} >$ , the only possible value is

\sin x = 1 - \sqrt{2} \approx -0.4142.

Let $\alpha = \arcsin(1 - \sqrt{2}) \approx -0.427$ radians. In $0 \le x \le 2\pi$ , the solutions are

x = \pi - \alpha \approx 3.57, \qquad x = 2\pi + \alpha \approx 5.86.

Answer

The $x$ -coordinates of the stationary points are

x \approx 3.57 \quad \text{and} \quad x \approx 5.86 \; \text{radians}.

Final answer

x = 3.57 and x = ̈5.86 radians

Detailed explanation

Walkthrough

We begin by differentiating $y = \frac{e^{\sin x}}{\cos^2 x}$ . The function is a quotient, so the quotient rule is the natural tool: $\frac{d}{dx}\left(\frac{u}{v}\right) = \frac{u'v - uv'}{v^2}$ . Here $u = e^{\sin x}$ , whose derivative uses the chain rule, giving $u' = \cos x \, e^{\sin x}$ , and $v = \cos^2 x$ , whose derivative is $v' = -2\sin x \cos x$ . Substituting these into the quotient rule gives the derivative.

For a stationary point we need $\frac{dy}{dx} =$ . Because the derivative is a fraction, this means the numerator must be zero (provided the denominator is defined). We factor the numerator. The factor $e^{\sin x}$ is never zero. The factor $\cos x$ would correspond to points where the original curve is undefined, because $\cos^2 x =$ in the denominator. Therefore the only stationary points come from $\cos^2 x + 2\sin x =$ .

We now have an equation involving both $\cos^2 x$ and $\sin x$ . To solve it, use $\cos^2 x = 1 - \sin^2 x$ ,which turns it into a quadratic in $\sin x$ : $\sin^2 x - 2\sin x - =$ . Solving this quadratic gives $\sin x = 1 \pm \sqrt{2}$ . Since $1 + \sqrt{2} >$ , it cannot be a sine value; only $\sin x = - \sqrt{2}$ is possible.

Finally, we find all angles $x$ in $0 \le x \le 2\pi$ with that sine value. The principal value of $\arcsin(1 - \sqrt{2})$ is negative, about $-0.427$ radians. The two angles in the required interval are in the third and fourth quadrants: $x = \pi - \alpha$ and $x = 2\pi + \alpha$ . These give approximately $3.57$ and $5.86$ radians.

Key Takeaways

The quotient rule is used when differentiating a ratio of two functions; remember to square the denominator.
The chain rule is needed for $e^{\sin x}$ and for $\cos^2 x$ .
Stationary points occur where $\frac{dy}{dx} =$ , but for a quotient we only set the numerator to zero, not the denominator.
Trigonometric equations involving both $\sin x$ and $\cos^2 x$ can often been reduced to a quadratic in one trigonometric function using $\sin^2 x + \cos^2 x = 1$ .
When solving $\sin x = k$ in a given interval, there may be two solutions; use the quadrant diagram or the general solution formulae.

Common Mistakes

Forgetting the chain rule when differentiating $e^{\sin x}$ or $\cos^2 x$ .
Writing the quotient-rule numerator with the wrong sign: the second term must be $-uv'$ , so since $v'$ is negative, it becomes a plus term.
Setting $\cos x =$ as a stationary point: those $x$ -values are not in the domain of the original curve because $\cos^2 x =$ makes the denominator zero.
Taking only the principal value of $\arcsin$ ; this gives a negative angle, not the required $x$ -coordinates in $[0, 2\pi]$ .
Forgetting that $1 + \sqrt{2}$ is outside the range of $\sin x$ ,and so must be rejected.
The mark scheme requires solutions to at least 3 significant figures; give $3.57$ and $5.86$ , not just the principal value.

Things to Be Careful About

The curve is undefined where $\cos x =$ , i.e. at $x = \frac{\pi}{2}$ and $x = \frac{3\pi}{2}$ ; these are not stationary points.
The derivative can also be written as $\frac{e^{\sin x}(\cos^2 x + 2\sin x)}{\cos^3 x}$ , but the quotient-rule form is equally acceptable.
All $x$ -coordinates are in radians unless the question states otherwise.
When using the quadratic formula, check both roots against the range $-1 \le \sin x \le 1$ .
The final answers should be given to at least 3 significant figures.

Techniques used

apply the quotient rule to differentiate the given functionset the numerator of the derivative to zerouse the Pythagorean identity to rewrite the equation in terms of sin xsolve the quadratic in sin x and select solutions in the required interval

Similar questions

Q6MediumTrigonometryLogarithmic and Exponential FunctionsNumerical Solution of Equations

(a)

By sketching a suitable pair of graphs, show that the equation $\text{cosec} \frac{1}{2}x = e^x - 3$ has exactly one root, denoted by $\alpha$ , in the interval $0 < x < \pi$ .

2M Similar

DifficultyMedium-Easy

Worked solution

Approach

To show that $\text{cosec}\frac{1}{2}x = e^x - 3$ has exactly one root in $0 < x < \pi$ , we sketch the two curves $y = e^x - 3$ and $y = \text{cosec}\frac{1}{2}x$ on the same axes and observe their intersection.

Working

Curve 1: $y = e^x - 3$

This is the standard exponential $y = e^x$ shifted down by 3 units.

At $x = 0$ : $y = e^0 - 3 = 1 - 3 = -2$ , so the curve passes through $(0, -2)$ .
As $x$ increases, $y$ increases exponentially.
At $x = \pi$ : $y = e^\pi - 3 \approx 20.14$ .

Curve 2: $y = \text{cosec}\frac{1}{2}x$

For $0 < x < \pi$ , we have $0 < \frac{x}{2} < \frac{\pi}{2}$ , so $\sin\frac{x}{2}$ is positive and increasing from $0$ to $1$ .

As $x \to 0^+$ : $\sin\frac{x}{2} \to 0^+$ , so $\text{cosec}\frac{x}{2} \to +\infty$ .
At $x = \pi$ : $\text{cosec}\frac{\pi}{2} = \frac{1}{\sin\frac{\pi}{2}} = 1$ .
The curve decreases monotonically from $+\infty$ to $1$ on this interval, with a minimum value of $1$ at $x = \pi$ .

Intersection:

At $x = 0^+$ , $\text{cosec}\frac{x}{2} \to +\infty$ while $e^x - 3 \to -2$ , so $\text{cosec}\frac{x}{2} > e^x - 3$ .
At $x = \pi$ , $\text{cosec}\frac{\pi}{2} = 1$ while $e^\pi - 3 \approx 20.14$ , so $\text{cosec}\frac{\pi}{2} < e^\pi - 3$ .

Since $\text{cosec}\frac{x}{2}$ is decreasing and $e^x - 3$ is increasing on $(0, \pi)$ , they cross exactly once. This single intersection point is the root $\alpha$ .

Answer

The two curves intersect at exactly one point in $0 < x < \pi$ , confirming exactly one root $\alpha$ .

Final answer

The curves y = e^x - 3 and y = cosec(x/2) intersect exactly once in 0 < x < π, giving exactly one root α.

Detailed explanation

Walkthrough

We are asked to show that the equation $\text{cosec}\frac{1}{2}x = e^x - 3$ has exactly one root in the interval $(0, \pi)$ . The most direct approach is to treat each side as a separate function and sketch both graphs on the same axes.

Step 1: Sketch $y = e^x - 3$ . This is the standard exponential function $e^x$ shifted downward by 3. At $x = 0$ , $y = 1 - 3 = -2$ . The curve rises steeply as $x$ increases. It crosses the x-axis where $e^x = 3$ , i.e., at $x = \ln 3 \approx 1.099$ .

Step 2: Sketch $y = \text{cosec}\frac{x}{2}$ . Recall that $\text{cosec}\theta = \frac{1}{\sin\theta}$ . For $0 < x < \pi$ , the argument $\frac{x}{2}$ lies in $(0, \frac{\pi}{2})$ , where sine is positive and increasing. Therefore cosecant is positive and decreasing. As $x \to 0^+$ , $\sin\frac{x}{2} \to 0^+$ so $\text{cosec}\frac{x}{2} \to +\infty$ . At $x = \pi$ , $\text{cosec}\frac{\pi}{2} = 1$ . The curve is a smooth decreasing curve from $+\infty$ down to $1$ .

Step 3: Identify the intersection. At the left end ( $x \to 0^+$ ), the cosecant curve is very large while the exponential curve is near $-2$ . At the right end ( $x = \pi$ ), the cosecant curve is at $1$ while the exponential curve is at about $20$ . Since one curve is decreasing and the other is increasing, they must cross exactly once. That crossing point is the root $\alpha$ .

Key Takeaways

Sketching two curves and finding their intersection is a powerful method for showing the number of roots of an equation.
Understanding the behaviour of $\text{cosec}\frac{x}{2}$ on $(0, \pi)$ is essential: it decreases from $+\infty$ to $1$ .
The monotonicity of both curves (one increasing, one decreasing) guarantees exactly one intersection.

Common Mistakes

Forgetting that $\text{cosec}\frac{x}{2} \to +\infty$ as $x \to 0^+$ and drawing the cosecant curve starting from a finite value.
Not justifying why there is exactly one intersection — merely sketching is insufficient; the argument about one curve being increasing and the other decreasing is needed.
Using degrees instead of radians when evaluating cosecant values in later parts.

Things to Be Careful About

The interval is $0 < x < \pi$ (open interval), so $x = 0$ is excluded and the cosecant curve has a vertical asymptote there.
The minimum of $\text{cosec}\frac{x}{2}$ on this interval is $1$ at $x = \pi$ , which is above the x-axis.
The mark scheme awards a mark for justifying the statement about the single root, not just for the sketch.

Techniques used

sketch exponential curvesketch cosecant curveidentify intersection point

(b)

Verify by calculation that $\alpha$ lies between 1 and 2.

2M Similar

DifficultyMedium-Easy

Worked solution

Approach

Let $f(x) = \text{cosec}\frac{1}{2}x - (e^x - 3) = \text{cosec}\frac{1}{2}x - e^x + 3$ . We evaluate $f(1)$ and $f(2)$ and show they have opposite signs, which by the Intermediate Value Theorem implies a root lies between 1 and 2.

Working

At $x = 1$ :

\text{cosec}\frac{1}{2} = \frac{1}{\sin 0.5} \approx 2.0858

e^1 - 3 = e - 3 \approx -0.2817

Since $2.0858 > -0.2817$ , we have $f(1) = 2.0858 - (-0.2817) = 2.3675 > 0$ .

At $x = 2$ :

\text{cosec} 1 = \frac{1}{\sin 1} \approx 1.1884

e^2 - 3 \approx 7.3891 - 3 = 4.3891

Since $1.1884 < 4.3891$ , we have $f(2) = 1.1884 - 4.3891 = -3.2007 < 0$ .

Since $f(1) > 0$ and $f(2) < 0$ , and $f(x)$ is continuous on $(1, 2)$ , by the Intermediate Value Theorem there is a root $\alpha$ in $(1, 2)$ .

Answer

$f(1) \approx 2.368 > 0$ and $f(2) \approx -3.201 < 0$ , so $\alpha$ lies between 1 and 2.

Final answer

α lies between 1 and 2

Detailed explanation

Walkthrough

We need to verify that the root $\alpha$ (from part (a)) lies between $x = 1$ and $x = 2$ . The standard method is to define a function $f(x) = \text{cosec}\frac{1}{2}x - e^x + 3$ and show that $f(1)$ and $f(2)$ have opposite signs.

Step 1: Evaluate at $x = 1$ .

$\text{cosec}(0.5) = \frac{1}{\sin(0.5)}$ . Using a calculator in radian mode: $\sin(0.5) \approx 0.4794$ , so $\text{cosec}(0.5) \approx 2.0858$ .
$e^1 - 3 = e - 3 \approx 2.7183 - 3 = -0.2817$ .
So $f(1) = 2.0858 - (-0.2817) = 2.3675 > 0$ .

Step 2: Evaluate at $x = 2$ .

$\text{cosec}(1) = \frac{1}{\sin(1)}$ . Using a calculator: $\sin(1) \approx 0.8415$ , so $\text{cosec}(1) \approx 1.1884$ .
$e^2 - 3 \approx 7.3891 - 3 = 4.3891$ .
So $f(2) = 1.1884 - 4.3891 = -3.2007 < 0$ .

Step 3: Conclude. Since $f(1) > 0$ and $f(2) < 0$ , and $f$ is continuous on $[1, 2]$ , there must be a root in $(1, 2)$ .

Key Takeaways

To locate a root between two values, evaluate the function at both endpoints and check for a sign change.
Always ensure your calculator is in radian mode when dealing with trigonometric functions in calculus contexts.
At least 2 significant figures are acceptable for the values.

Common Mistakes

Using degrees instead of radians: $\text{cosec}(0.5°) \neq \text{cosec}(0.5 \text{ rad})$ . This would give completely wrong values and lose the method mark.
Forgetting to state the sign change argument — merely computing values is not enough; you must explicitly state that the signs are opposite and conclude a root lies between them.
Rounding too early: use at least 4 decimal places in intermediate calculations.

Things to Be Careful About

The mark scheme explicitly states: "Use of degrees is M0" — this is a common pitfall.
Values need at least 2 significant figures. The mark scheme accepts: $-0.282 < 2.086$ and $4.389 > 1.188$ , or equivalently $f(1) \approx 2.36 > 0$ and $f(2) \approx -3.20 < 0$ .
Truncation is condoned, but the argument must be complete.

Techniques used

evaluate functions at boundary pointscheck for sign changeapply intermediate value theorem

(c)

Show that if a sequence of values in the interval $0 < x < \pi$ given by the iterative formula

x_{n+1} = \ln(\text{cosec} \frac{1}{2}x_n + 3)

converges, then it converges to $\alpha$ .

1M Similar

DifficultyMedium-Easy

Worked solution

Approach

If the sequence $x_{n+1} = \ln\left(\text{cosec}\frac{1}{2}x_n + 3\right)$ converges, let the limit be $L$ . Then as $n \to \infty$ , both $x_n \to L$ and $x_{n+1} \to L$ . Substitute $L$ into the iterative formula and show it satisfies the original equation.

Working

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

L = \ln\left(\text{cosec}\frac{1}{2}L + 3\right)

Exponentiate both sides:

e^L = \text{cosec}\frac{1}{2}L + 3

Rearrange:

\text{cosec}\frac{1}{2}L = e^L - 3

This is exactly the original equation $\text{cosec}\frac{1}{2}x = e^x - 3$ , so $L = \alpha$ .

Answer

If the sequence converges, it converges to $\alpha$ .

Final answer

If the sequence converges to L, then L satisfies cosec(L/2) = e^L - 3, so L = α.

Detailed explanation

Walkthrough

This part asks us to show that if the iterative sequence converges, its limit must be the root $\alpha$ of the original equation.

Step 1: Assume convergence. Suppose the sequence $x_1, x_2, x_3, \ldots$ converges to some limit $L$ . This means $\lim_{n \to \infty} x_n = L$ and $\lim_{n \to \infty} x_{n+1} = L$ .

Step 2: Substitute into the iterative formula. The iterative formula is $x_{n+1} = \ln\left(\text{cosec}\frac{1}{2}x_n + 3\right)$ . Taking the limit of both sides as $n \to \infty$ :

L = \ln\left(\text{cosec}\frac{1}{2}L + 3\right)

Step 3: Rearrange to recover the original equation. Exponentiate both sides:

e^L = \text{cosec}\frac{1}{2}L + 3

Subtract 3:

\text{cosec}\frac{1}{2}L = e^L - 3

This is precisely the equation from part (a), whose unique root in $(0, \pi)$ is $\alpha$ . Therefore $L = \alpha$ .

Key Takeaways

When an iterative formula converges, the limit satisfies the fixed-point equation obtained by setting $x_{n+1} = x_n = L$ .
Rearranging the fixed-point equation back to the original form confirms the limit is the desired root.
This is a standard technique for justifying iterative methods.

Common Mistakes

Not explicitly stating the assumption that the sequence converges before substituting $L$ .
Forgetting to exponentiate to recover the original equation form.
Not concluding that $L = \alpha$ (the unique root).

Things to Be Careful About

The mark scheme awards a B1 for stating $x = \ln(\text{cosec}\frac{1}{2}x + 3)$ and rearranging to the given equation. The argument is straightforward but must be complete.
This part does not ask to prove convergence — only that IF it converges, then it converges to $\alpha$ .

Techniques used

assume convergence to limit Lsubstitute into iterative formularearrange to original equation

(d)

Use this iterative formula with an initial value of 1.4 to determine $\alpha$ correct to 2 decimal places. Give the result of each iteration to 4 decimal places.

3M Similar

DifficultyMedium

Worked solution

Approach

Use the iterative formula $x_{n+1} = \ln\left(\text{cosec}\frac{1}{2}x_n + 3\right)$ with $x_1 = 1.4$ , computing each value to 4 decimal places, until the value stabilises to 2 decimal places.

Working

Iteration 1: $x_1 = 1.4000$

Iteration 2:

x_2 = \ln\left(\text{cosec}\frac{1.4}{2} + 3\right) = \ln\left(\text{cosec}(0.7) + 3\right)

\sin(0.7) \approx 0.6442 \implies \text{cosec}(0.7) \approx 1.5523

x_2 = \ln(1.5523 + 3) = \ln(4.5523) \approx 1.5156

Iteration 3:

x_3 = \ln\left(\text{cosec}\frac{1.5156}{2} + 3\right) = \ln\left(\text{cosec}(0.7578) + 3\right)

\sin(0.7578) \approx 0.6874 \implies \text{cosec}(0.7578) \approx 1.4548

x_3 = \ln(1.4548 + 3) = \ln(4.4548) \approx 1.4940

Iteration 4:

x_4 = \ln\left(\text{cosec}\frac{1.4940}{2} + 3\right) = \ln\left(\text{cosec}(0.7470) + 3\right)

\sin(0.7470) \approx 0.6794 \implies \text{cosec}(0.7470) \approx 1.4719

x_4 = \ln(1.4719 + 3) = \ln(4.4719) \approx 1.4978

Iteration 5:

x_5 = \ln\left(\text{cosec}\frac{1.4978}{2} + 3\right) = \ln\left(\text{cosec}(0.7489) + 3\right)

\sin(0.7489) \approx 0.6808 \implies \text{cosec}(0.7489) \approx 1.4689

x_5 = \ln(1.4689 + 3) = \ln(4.4689) \approx 1.4971

Iteration 6:

x_6 = \ln\left(\text{cosec}\frac{1.4971}{2} + 3\right) = \ln\left(\text{cosec}(0.74855) + 3\right)

\sin(0.74855) \approx 0.6805 \implies \text{cosec}(0.74855) \approx 1.4696

x_6 = \ln(1.4696 + 3) = \ln(4.4696) \approx 1.4973

Iteration 7:

x_7 = \ln\left(\text{cosec}\frac{1.4973}{2} + 3\right) \approx \ln(4.4693) \approx 1.4972

The values are converging: $x_5 \approx 1.4971$ , $x_6 \approx 1.4973$ , $x_7 \approx 1.4972$ .

To confirm $\alpha = 1.50$ to 2 decimal places, we check that the root lies in $(1.495, 1.505)$ :

$f(1.495) = \text{cosec}(0.7475) - e^{1.495} + 3 \approx 1.4704 - 4.4594 + 3 = 0.0110 > 0$
$f(1.505) = \text{cosec}(0.7525) - e^{1.505} + 3 \approx 1.4634 - 4.5042 + 3 = -0.0408 < 0$

Since $f(1.495) > 0$ and $f(1.505) < 0$ , the root lies in $(1.495, 1.505)$ , confirming $\alpha = 1.50$ to 2 decimal places.

Answer

$\alpha = 1.50$ correct to 2 decimal places.

Final answer

α = 1.50

Detailed explanation

Walkthrough

We apply the iterative formula $x_{n+1} = \ln\left(\text{cosec}\frac{1}{2}x_n + 3\right)$ starting with $x_1 = 1.4$ , computing each result to 4 decimal places.

Iteration 2: $x_2 = \ln(\text{cosec}(0.7) + 3)$ . Since $\sin(0.7) \approx 0.6442$ , $\text{cosec}(0.7) \approx 1.5523$ . So $x_2 = \ln(4.5523) \approx 1.5156$ .

Iteration 3: $x_3 = \ln(\text{cosec}(0.7578) + 3)$ . Since $\sin(0.7578) \approx 0.6874$ , $\text{cosec}(0.7578) \approx 1.4548$ . So $x_3 = \ln(4.4548) \approx 1.4940$ .

Iteration 4: $x_4 = \ln(\text{cosec}(0.7470) + 3)$ . Since $\sin(0.7470) \approx 0.6794$ , $\text{cosec}(0.7470) \approx 1.4719$ . So $x_4 = \ln(4.4719) \approx 1.4978$ .

Iteration 5: $x_5 = \ln(\text{cosec}(0.7489) + 3)$ . Since $\sin(0.7489) \approx 0.6808$ , $\text{cosec}(0.7489) \approx 1.4689$ . So $x_5 = \ln(4.4689) \approx 1.4971$ .

Iteration 6: $x_6 = \ln(\text{cosec}(0.74855) + 3)$ . Since $\sin(0.74855) \approx 0.6805$ , $\text{cosec}(0.74855) \approx 1.4696$ . So $x_6 = \ln(4.4696) \approx 1.4973$ .

The sequence is oscillating and converging: 1.5156, 1.4940, 1.4978, 1.4971, 1.4973, ... All values round to 1.50 to 2 decimal places.

To formally confirm, we check that the function $f(x) = \text{cosec}\frac{1}{2}x - e^x + 3$ changes sign in $(1.495, 1.505)$ , which proves the root is 1.50 to 2 d.p.

Key Takeaways

When using iterative formulas, compute to at least 4 decimal places to ensure accuracy for 2 d.p. answers.
The sequence may oscillate around the root before converging — this is normal for this type of iteration.
To confirm the answer to 2 d.p., show that the value is in the correct rounding interval $(1.495, 1.505)$ .

Common Mistakes

Using degrees instead of radians in the cosecant function — this gives completely wrong iteration values.
Not computing enough iterations to show convergence.
Not verifying the final answer by checking the sign change in the rounding interval.
Rounding intermediate values too early, which can accumulate errors.

Things to Be Careful About

The mark scheme accepts values like 1.5156, 1.4940, 1.4978, 1.4971 as sufficient iterations.
The final answer must be 1.50 (not 1.5, which is only 1 d.p.).
You must show sufficient iterations to 4 d.p. to justify the answer, or show a sign change in $(1.495, 1.505)$ .
The iterative formula uses $\frac{1}{2}x_n$ , not $\frac{1}{2}x_n$ in degrees.

Techniques used

apply iterative formula repeatedlycompute to 4 decimal placescheck convergence to 2 decimal places

(e)

State the minimum number of calculated iterations needed with this initial value to determine $\alpha$ correct to 2 decimal places.

1M Similar

DifficultyEasy

Worked solution

Approach

Count the number of iterations needed from part (d) until the value is confirmed correct to 2 decimal places.

Working

From part (d), the iterations are:

$x_1 = 1.4000$
$x_2 = 1.5156$
$x_3 = 1.4940$
$x_4 = 1.4978$
$x_5 = 1.4971$
$x_6 = 1.4973$

After $x_4 = 1.4978$ and $x_5 = 1.4971$ , both round to $1.50$ to 2 decimal places. We then verified that the root lies in $(1.495, 1.505)$ using $f(1.495) > 0$ and $f(1.505) < 0$ .

The sign change in $(1.495, 1.505)$ was established using $x_4 = 1.4978$ (which is $> 1.495$ ) and $x_5 = 1.4971$ (which is $< 1.505$ ), but more precisely we evaluated $f$ at the boundary points. The iterations $x_4$ through $x_6$ all give values in $(1.495, 1.505)$ , confirming $\alpha = 1.50$ to 2 d.p.

The minimum number of calculated iterations needed is 4 (producing $x_2, x_3, x_4, x_5$ ), after which we can confirm convergence to 2 d.p.

Actually, counting from the initial value $x_1 = 1.4$ :

After 1 iteration: $x_2 = 1.5156$ (not yet stable to 2 d.p.)
After 2 iterations: $x_3 = 1.4940$ (rounds to 1.49)
After 3 iterations: $x_4 = 1.4978$ (rounds to 1.50)
After 4 iterations: $x_5 = 1.4971$ (rounds to 1.50)

With 4 calculated iterations ( $x_2$ through $x_5$ ), we have two consecutive values that both round to 1.50, and we can verify the root is in $(1.495, 1.505)$ .

Answer

4

Final answer

4

Detailed explanation

Walkthrough

In part (d), we performed iterations and obtained:

$x_1 = 1.4000$ (given)
$x_2 = 1.5156$ (1st iteration)
$x_3 = 1.4940$ (2nd iteration)
$x_4 = 1.4978$ (3rd iteration)
$x_5 = 1.4971$ (4th iteration)

After the 4th iteration ( $x_5 = 1.4971$ ), we have two consecutive values ( $x_4 = 1.4978$ and $x_5 = 1.4971$ ) that both round to $1.50$ to 2 decimal places. We then confirmed by checking $f(1.495) > 0$ and $f(1.505) < 0$ that the root indeed lies in the interval $(1.495, 1.505)$ , so $\alpha = 1.50$ to 2 d.p.

The minimum number of calculated iterations (not counting the initial value) is 4.

Key Takeaways

To determine the minimum number of iterations, count how many calculations are needed until the answer is confirmed to the required accuracy.
Consecutive iterations giving the same value to the required decimal places, combined with a sign change check, confirm the answer.

Common Mistakes

Counting the initial value $x_1$ as an iteration — it is given, not calculated.
Not performing enough iterations to confirm stability to 2 d.p.

Things to Be Careful About

The mark scheme answer is 4, meaning 4 calculated iterations ( $x_2, x_3, x_4, x_5$ ) are needed.
Some students might say 5 if they include $x_6$ , but 4 is sufficient when combined with the sign change verification.

Techniques used

count iterations to achieve required accuracy

Q7MediumComplex Numbers

(a)

On a single Argand diagram sketch the loci given by the equations $|z - 3 + 2i| = 2$ and $|w - 3 + 2i| = |w + 3 - 4i|$ where $z$ and $w$ are complex numbers.

4M Similar

DifficultyMedium

Worked solution

Approach

Rewrite each modulus equation in a standard form to identify the geometric locus:

$|z - z_0| = r$ represents a circle with centre $z_0$ and radius $r$ .
$|w - a| = |w - b|$ represents the perpendicular bisector of the line segment joining $a$ and $b$ .

Working

Locus 1: $|z - 3 + 2i| = 2$

Rewrite as:

|z - (3 - 2i)| = 2

This is a circle with centre $(3, -2)$ and radius $2$ .

Locus 2: $|w - 3 + 2i| = |w + 3 - 4i|$

Rewrite as:

|w - (3 - 2i)| = |w - (-3 + 4i)|

This means $w$ is equidistant from the points $(3, -2)$ and $(-3, 4)$ . The locus is the perpendicular bisector of the line segment joining these two points.

Find the midpoint of the segment joining $(3, -2)$ and $(-3, 4)$ :

\left(\frac{3 + (-3)}{2}, \frac{-2 + 4}{2}\right) = (0, 1)

Find the slope of the segment joining $(3, -2)$ and $(-3, 4)$ :

m = \frac{4 - (-2)}{-3 - 3} = \frac{6}{-6} = -1

The perpendicular bisector has slope $1$ (negative reciprocal of $-1$ ) and passes through $(0, 1)$ :

y - 1 = 1(x - 0) \implies y = x + 1

Sketch both loci on an Argand diagram:

Answer

The loci are: a circle centred at $(3, -2)$ with radius $2$ , and the line $y = x + 1$ (the perpendicular bisector of the segment from $(-3, 4)$ to $(3, -2)$ ).

Final answer

Circle centre (3, -2) radius 2; perpendicular bisector y = x + 1

Detailed explanation

Walkthrough

Step 1: Identify Locus 1. The equation $|z - 3 + 2i| = 2$ can be rewritten as $|z - (3 - 2i)| = 2$ . In the Argand diagram, $|z - z_0| = r$ is the set of all points at distance $r$ from $z_0$ , which is a circle. Here $z_0 = 3 - 2i$ , corresponding to the point $(3, -2)$ , and $r = 2$ . So this is a circle centred at $(3, -2)$ with radius $2$ .

Step 2: Identify Locus 2. The equation $|w - 3 + 2i| = |w + 3 - 4i|$ can be rewritten as $|w - (3 - 2i)| = |w - (-3 + 4i)|$ . This says the distance from $w$ to $(3, -2)$ equals the distance from $w$ to $(-3, 4)$ . The set of points equidistant from two fixed points is the perpendicular bisector of the segment joining them.

Step 3: Find the perpendicular bisector. The midpoint of the segment from $(3, -2)$ to $(-3, 4)$ is $(0, 1)$ . The slope of the segment is $\frac{4 - (-2)}{-3 - 3} = -1$ , so the perpendicular bisector has slope $1$ . Using point-slope form through $(0, 1)$ : $y - 1 = 1(x - 0)$ , giving $y = x + 1$ .

Step 4: Sketch. Plot the circle and the line on the Argand diagram with Re and Im axes.

Key Takeaways

$|z - z_0| = r$ is always a circle with centre $z_0$ and radius $r$ on the Argand diagram.
$|z - a| = |z - b|$ is always the perpendicular bisector of the segment joining $a$ and $b$ .
To find the perpendicular bisector, compute the midpoint and the negative reciprocal of the segment's slope.

Common Mistakes

Forgetting to rewrite $|z - 3 + 2i|$ as $|z - (3 - 2i)|$ and misidentifying the centre as $(3, 2)$ instead of $(3, -2)$ .
Not showing the midpoint or the point $(-3, 4)$ on the diagram, which are needed for the B1 marks.
Drawing the perpendicular bisector with the wrong slope (e.g., using slope $-1$ instead of $1$ ).
Forgetting that the locus is a full line, not just a ray or segment.

Things to Be Careful About

The centre $(3, -2)$ is in the fourth quadrant (positive Re, negative Im). Do not confuse the sign of the imaginary part.
The mark scheme awards B1FT for the perpendicular bisector based on the correct positions of $(-3, 4)$ and the circle centre, so ensure both points are clearly marked.
The line $y = x + 1$ passes through $(0, 1)$ and has gradient $1$ ; verify this matches the diagram.

Techniques used

interpret modulus equation as circle locusinterpret equal-distance equation as perpendicular bisectorfind midpoint and perpendicular slopesketch loci on Argand diagram

(b)

Hence find the least value of $|z - w|$ for points on these loci. Give your answer in an exact form.

2M Similar

DifficultyMedium

Worked solution

Approach

The minimum value of $|z - w|$ is the shortest distance between a point on the circle and a point on the line. This equals the perpendicular distance from the centre of the circle to the line, minus the radius of the circle.

Working

The centre of the circle is $(3, -2)$ and the line is $y = x + 1$ , or equivalently $x - y + 1 = 0$ .

The perpendicular from $(3, -2)$ to the line $y = x + 1$ has slope $-1$ (negative reciprocal of $1$ ). The line through $(3, -2)$ with slope $-1$ is:

y + 2 = -(x - 3) \implies y = -x + 1

Find the intersection with $y = x + 1$ :

-x + 1 = x + 1 \implies 2x = 0 \implies x = 0, \; y = 1

So the foot of the perpendicular from $(3, -2)$ to the line is $(0, 1)$ , which is the midpoint of the segment joining $(3, -2)$ and $(-3, 4)$ .

The distance from $(3, -2)$ to $(0, 1)$ is:

\sqrt{(3 - 0)^2 + (-2 - 1)^2} = \sqrt{9 + 9} = \sqrt{18} = 3\sqrt{2}

The minimum value of $|z - w|$ is this distance minus the radius $2$ :

|z - w|_{\min} = 3\sqrt{2} - 2

Answer

3\sqrt{2} - 2

Final answer

3√2 - 2

Detailed explanation

Walkthrough

Step 1: Understand the geometry. We want the minimum distance between any point $z$ on the circle and any point $w$ on the line. The shortest distance from a circle to a line is along the perpendicular from the centre of the circle to the line, minus the radius.

Step 2: Find the perpendicular foot. The line is $y = x + 1$ with gradient $1$ . The perpendicular from $(3, -2)$ has gradient $-1$ . The line through $(3, -2)$ with gradient $-1$ is $y = -x + 1$ . Setting $-x + 1 = x + 1$ gives $x = 0, y = 1$ . So the perpendicular meets the line at $(0, 1)$ .

Step 3: Calculate the distance. The distance from $(3, -2)$ to $(0, 1)$ is $\sqrt{3^2 + 3^2} = \sqrt{18} = 3\sqrt{2}$ .

Step 4: Subtract the radius. The minimum distance from the circle to the line is $3\sqrt{2} - 2$ .

Key Takeaways

The minimum distance between a circle and a line is the perpendicular distance from the centre to the line, minus the radius.
When the line is a perpendicular bisector, the perpendicular from one endpoint to the bisector meets it at the midpoint of the segment.
Always express the final answer in exact form (surds, not decimals).

Common Mistakes

Adding the radius instead of subtracting it, giving $3\sqrt{2} + 2$ .
Using the wrong distance formula or making arithmetic errors in $\sqrt{9 + 9}$ .
Not showing the method of finding the perpendicular distance, which is required for the M1 mark.
Giving a decimal approximation instead of the exact form $3\sqrt{2} - 2$ .

Things to Be Careful About

The mark scheme accepts $\sqrt{18} - 2$ or $3\sqrt{2} - 2$ ; both are correct.
Ensure the answer is in exact form — no decimal approximations.
The method must clearly show the distance from the centre to the line being calculated before subtracting the radius.

Techniques used

find perpendicular distance from centre to linesubtract radius to get minimum distance between circle and lineuse distance formula between two points

Q87MMedium-HardIntegration

Use the substitution $u = 1 - \sin x$ to find the exact value of

\int_{\pi}^{\frac{3}{2}\pi} \frac{\sin 2x}{\sqrt{1 - \sin x}} \,dx

Give your answer in the form $a + b\sqrt{2}$ where $a$ and $b$ are rational numbers to be determined.

DifficultyMedium-Hard

Worked solution

Approach

Use the substitution $u = 1 - \sin x$ . Differentiate to find $du$ , rewrite $\sin 2x$ using the double-angle identity, transform the integral into powers of $u$ , integrate, then evaluate using the new limits.

Working

Let

u = 1 - \sin x

Then

\frac{du}{dx} = -\cos x \quad \Rightarrow \quad \cos x\,dx = -du

Also

\sin 2x = 2\sin x \cos x = 2(1-u)\cos x

Therefore

\int_{\pi}^{\frac{3}{2}\pi} \frac{\sin 2x}{\sqrt{1-\sin x}}\,dx = \int_{u=1}^{2} \frac{2(1-u)(-du)}{\sqrt{u}} = -2\int_{1}^{2}\frac{1-u}{\sqrt{u}}\,du

Equivalently,

= 2\int_{1}^{2}\frac{u-1}{\sqrt{u}}\,du = 2\int_{1}^{2}\left(u^{1/2} - u^{-1/2}\right)du

Integrate:

2\left[\frac{2}{3}u^{3/2} - 2u^{1/2}\right]_{1}^{2} = \left[\frac{4}{3}u^{3/2} - 4u^{1/2}\right]_{1}^{2}

Evaluate at $u=2$ :

\frac{4}{3}(2\sqrt{2}) - 4\sqrt{2} = \frac{8\sqrt{2}}{3} - \frac{12\sqrt{2}}{3} = -\frac{4\sqrt{2}}{3}

Evaluate at $u=1$ :

\frac{4}{3} - 4 = -\frac{8}{3}

Subtract:

\left(-\frac{4\sqrt{2}}{3}\right) - \left(-\frac{8}{3}\right) = \frac{8}{3} - \frac{4\sqrt{2}}{3}

Answer

\frac{8}{3} - \frac{4}{3}\sqrt{2}

So $a = \frac{8}{3}$ and $b = -\frac{4}{3}$ .

Final answer

8/3 - (4/3)√2, so a = 8/3 and b = -4/3

Detailed explanation

Walkthrough

We need to evaluate a definite integral where the integrand contains $\sin 2x$ and $\sqrt{1-\sin x}$ . The substitution $u = 1 - \sin x$ is suggested because the denominator becomes $\sqrt{u}$ , and the derivative of $1-\sin x$ is $-\cos x$ , which appears in the numerator after using $\sin 2x = 2\sin x\cos x$ .

First differentiate the substitution:

\frac{du}{dx} = -\cos x \quad \Rightarrow \quad \cos x\,dx = -du

This tells us how to replace the $\cos x\,dx$ part of the integrand. Next, rewrite $\sin 2x$ as $2\sin x\cos x$ . Since $\sin x = 1-u$ , we get $\sin 2x = 2(1-u)\cos x$ . Substituting into the integral:

\int \frac{\sin 2x}{\sqrt{1-\sin x}}\,dx = \int \frac{2(1-u)\cos x}{\sqrt{u}}\,dx = \int \frac{2(1-u)(-du)}{\sqrt{u}}

So the integral becomes

-2\int \frac{1-u}{\sqrt{u}}\,du = 2\int \frac{u-1}{\sqrt{u}}\,du = 2\int \left(u^{1/2} - u^{-1/2}\right)du

For a definite integral, we must also change the limits. When $x = \pi$ , $\sin \pi = 0$ , so $u = 1$ . When $x = \frac{3}{2}\pi$ , $\sin \frac{3\pi}{2} = -1$ , so $u = 2$ . Thus the new limits are $u=1$ and $u=2$ .

Now integrate term by term:

2\int \left(u^{1/2} - u^{-1/2}\right)du = 2\left(\frac{2}{3}u^{3/2} - 2u^{1/2}\right) = \frac{4}{3}u^{3/2} - 4u^{1/2}

Evaluate at $u=2$ and $u=1$ :

\left[\frac{4}{3}u^{3/2} - 4u^{1/2}\right]_{1}^{2} = \left(\frac{8\sqrt{2}}{3} - 4\sqrt{2}\right) - \left(\frac{4}{3} - 4\right) = -\frac{4\sqrt{2}}{3} + \frac{8}{3}

This simplifies to $\frac{8}{3} - \frac{4\sqrt{2}}{3}$ , so $a = \frac{8}{3}$ and $b = -\frac{4}{3}$ .

Key Takeaways

In a substitution, always replace $dx$ using the derivative of the substitution, including any minus sign.
Use trigonometric identities such as $\sin 2x = 2\sin x\cos x$ to expose the derivative of the substituted expression.
For definite integrals, change the limits from $x$ -values to corresponding $u$ -values before evaluating.
Integrals involving $\sqrt{u}$ are handled as powers: $\sqrt{u} = u^{1/2}$ and $1/\sqrt{u} = u^{-1/2}$ .

Common Mistakes

Forgetting the minus sign from $du = -\cos x\,dx$ . This changes the sign of the whole integral.
Not changing the limits when using a substitution in a definite integral.
Using the wrong value of $\sin \pi$ : $\sin \pi = 0$ , so the lower limit is $u = 1$ , not $u = 0$ .
Incorrectly expanding $2(1-u)(-du)$ ; be careful that the factor $(1-u)$ becomes $u-1$ after taking out the minus sign.
Not simplifying the final surd expression to the required form $a + b\sqrt{2}$ .

Things to Be Careful About

At $x = \frac{3}{2}\pi$ , $\sin x = -1$ , so $u = 1 - (-1) = 2$ .
The antiderivative can also be written as $-4u^{1/2} + \frac{4}{3}u^{3/2}$ ; either form gives the same final answer when evaluated with the correct limits.
The final answer must be exact, with rational $a$ and $b$ : $a = \frac{8}{3}$ , $b = -\frac{4}{3}$ .
Be careful with the order of subtraction: evaluate the antiderivative at the upper limit minus the lower limit.

Techniques used

substitute u = 1 - sin xdifferentiate the substitution to express cos x dxuse the double-angle identity sin 2x = 2 sin x cos xintegrate powers of uevaluate the definite integral with new limits

Similar questions

Q9MediumVectors

The equations of two straight lines $l_1$ and $l_2$ are

l_1: \mathbf{r} = \mathbf{i} - 2\mathbf{j} + 3\mathbf{k} + \lambda(2\mathbf{i} - \mathbf{j} + a\mathbf{k}) \quad \text{and} \quad l_2: \mathbf{r} = -\mathbf{i} - \mathbf{j} - \mathbf{k} + \mu(3\mathbf{i} - 2\mathbf{j} - 2\mathbf{k})

where $a$ is a constant.

The lines $l_1$ and $l_2$ are perpendicular.

(a)

Show that $a = 4$ .

1M Similar

DifficultyMedium-Easy

Worked solution

Approach

The direction vectors of two perpendicular lines have a scalar product of zero. We extract the direction vectors from the vector forms of $l_1$ and $l_2$ , set their scalar product equal to zero, and solve for $a$ .

Working

The direction vector of $l_1$ is $\mathbf{d}_1 = 2\mathbf{i} - \mathbf{j} + a\mathbf{k}$ and the direction vector of $l_2$ is $\mathbf{d}_2 = 3\mathbf{i} - 2\mathbf{j} - 2\mathbf{k}$ .

Since $l_1 \perp l_2$ , the scalar product of the direction vectors is zero:

\mathbf{d}_1 \cdot \mathbf{d}_2 = 0

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

6 + 2 - 2a = 0

8 = 2a

a = 4

Answer

a = 4

Final answer

a = 4

Detailed explanation

Walkthrough

The direction vector of a line given in vector form $\mathbf{r} = \mathbf{a} + t\mathbf{b}$ is the vector $\mathbf{b}$ that multiplies the parameter. So for $l_1$ the direction vector is $2\mathbf{i} - \mathbf{j} + a\mathbf{k}$ , and for $l_2$ it is $3\mathbf{i} - 2\mathbf{j} - 2\mathbf{k}$ .

Two lines are perpendicular if and only if the scalar (dot) product of their direction vectors is zero. The scalar product is computed by multiplying corresponding components and summing them: $(2)(3) + (-1)(-2) + (a)(-2)$ .

Setting this equal to zero gives the linear equation $8 - 2a = 0$ , which we solve to obtain $a = 4$ .

Key Takeaways

The direction vector of a line $\mathbf{r} = \mathbf{a} + t\mathbf{b}$ is the multiplier $\mathbf{b}$ of the parameter $t$ .
Two lines are perpendicular if and only if the scalar product of their direction vectors is zero: $\mathbf{d}_1 \cdot \mathbf{d}_2 = 0$ .
The scalar product of $(p, q, r)$ and $(x, y, z)$ is $px + qy + rz$ .

Common Mistakes

Using the position vectors of the lines instead of their direction vectors.
Forgetting to set the scalar product equal to zero, or mistakenly equating it to some other value.
Sign errors when expanding the products (for example, writing $(-1)(-2) = -2$ instead of $+2$ ).

Things to Be Careful About

The perpendicularity condition is the scalar product equals zero, not the magnitudes being equal or the vectors being equal.
The unknown $a$ only appears in the direction of $l_1$ , so it appears in just one term of the scalar product.

Techniques used

extract direction vectors from the vector form of a lineapply the scalar product equals zero condition for perpendicular linessolve a linear equation for the unknown parameter

(b)

The lines $l_1$ and $l_2$ also intersect.

Find the position vector of the point of intersection.

4M Similar

DifficultyMedium

Worked solution

Approach

With $a = 4$ , write both lines in component form. Equate the components to obtain a system of three equations in the two unknowns $\lambda$ and $\mu$ , then solve using any two equations (the third acts as a check). Substitute back into either line to find the position vector of the intersection.

Working

With $a = 4$ , a general point on $l_1$ is:

\mathbf{r}_1 = (1 + 2\lambda)\mathbf{i} + (-2 - \lambda)\mathbf{j} + (3 + 4\lambda)\mathbf{k}

A general point on $l_2$ is:

\mathbf{r}_2 = (-1 + 3\mu)\mathbf{i} + (-1 - 2\mu)\mathbf{j} + (-1 - 2\mu)\mathbf{k}

For the lines to intersect, we require:

1 + 2\lambda = -1 + 3\mu \quad \quad (1)

-2 - \lambda = -1 - 2\mu \quad \quad (2)

3 + 4\lambda = -1 - 2\mu \quad \quad (3)

From equation (2):

\lambda = 2\mu - 1

Substituting into equation (1):

1 + 2(2\mu - 1) = -1 + 3\mu

1 + 4\mu - 2 = -1 + 3\mu

4\mu - 1 = -1 + 3\mu

\mu = 0

Hence $\lambda = 2(0) - 1 = -1$ .

Substituting $\lambda = -1$ into the equation of $l_1$ :

\mathbf{r} = (1 + 2(-1))\mathbf{i} + (-2 - (-1))\mathbf{j} + (3 + 4(-1))\mathbf{k} = -\mathbf{i} - \mathbf{j} - \mathbf{k}

Check with the third component: with $\mu = 0$ , the $l_2$ point is $(-1, -1, -1)$ , which matches the result from $l_1$ .

Answer

-\mathbf{i} - \mathbf{j} - \mathbf{k}

Final answer

-i - j - k

Detailed explanation

Walkthrough

With $a = 4$ in hand from part (a), we can write down the components of a general point on each line. A point on $l_1$ has $x$ -coordinate $1 + 2\lambda$ , $y$ -coordinate $-2 - \lambda$ , and $z$ -coordinate $3 + 4\lambda$ . A point on $l_2$ has $x$ -coordinate $-1 + 3\mu$ , $y$ -coordinate $-1 - 2\mu$ , and $z$ -coordinate $-1 - 2\mu$ .

For the lines to share a point, all three components must match for some pair $(\lambda, \mu)$ . This gives three equations in two unknowns. Using the second equation $-2 - \lambda = -1 - 2\mu$ we get $\lambda = 2\mu - 1$ . Substituting this into the first equation yields $\mu = 0$ , hence $\lambda = -1$ .

Substituting $\lambda = -1$ into the equation of $l_1$ gives the position vector $-\mathbf{i} - \mathbf{j} - \mathbf{k}$ . As a sanity check, $\mu = 0$ in the equation of $l_2$ gives the same point.

Key Takeaways

A line in vector form $\mathbf{r} = \mathbf{a} + t\mathbf{b}$ can be written in component form as $(a_x + tb_x,\ a_y + tb_y,\ a_z + tb_z)$ .
The intersection of two lines is found by equating their components and solving the resulting system.
An overdetermined system (more equations than unknowns) is consistent precisely when the two lines actually intersect.

Common Mistakes

Sign errors when expanding the line equations (e.g., writing $-1 - 2\mu$ as $1 - 2\mu$ ).
Confusing $\lambda$ and $\mu$ — they are different parameters for the two lines.
Forgetting to use the third component as a check on the answer.

Things to Be Careful About

The two lines are given with different parameters ( $\lambda$ and $\mu$ ); they are generally not the same value.
The third component equation is consistent with the first two only because the lines are given to intersect; it serves as a check that the answer is correct.

Techniques used

write a line in vector form as component equationsequate corresponding components of two lines to form a system of equationssolve a linear system of three equations in two unknownssubstitute the parameter back into the line equation to find the position vector

(c)

The point $A$ has position vector $-5\mathbf{i} + \mathbf{j} - 9\mathbf{k}$ .

Show that $A$ lies on $l_1$ .

2M Similar

DifficultyMedium-Easy

Worked solution

Approach

A point lies on a line in vector form if and only if its position vector can be written as $\mathbf{a} + \lambda \mathbf{b}$ for some value of $\lambda$ . We substitute the position vector of $A$ into the equation of $l_1$ , find $\lambda$ from one component, and check that it gives the correct values for the other two components.

Working

A general point on $l_1$ is:

\mathbf{r} = (1 + 2\lambda)\mathbf{i} + (-2 - \lambda)\mathbf{j} + (3 + 4\lambda)\mathbf{k}

Setting the $x$ -component equal to that of $A$ :

1 + 2\lambda = -5

2\lambda = -6

\lambda = -3

Check the $y$ -component with $\lambda = -3$ :

-2 - (-3) = 1

This matches the $y$ -component of $A$ ( $1$ ). ✓

Check the $z$ -component with $\lambda = -3$ :

3 + 4(-3) = 3 - 12 = -9

This matches the $z$ -component of $A$ ( $-9$ ). ✓

All three components agree, so $A$ lies on $l_1$ at $\lambda = -3$ .

Answer

$A$ lies on $l_1$ (at $\lambda = -3$ ).

Final answer

A lies on l_1 (at λ = -3)

Detailed explanation

Walkthrough

A point $A$ lies on a line in vector form $\mathbf{r} = \mathbf{a} + t\mathbf{b}$ if and only if its position vector can be written as $\mathbf{a} + t\mathbf{b}$ for some value of $t$ . So we need to find a value of $\lambda$ such that the equation of $l_1$ produces $A$ 's position vector.

Equating the $x$ -components gives $1 + 2\lambda = -5$ , hence $\lambda = -3$ . We then check that this same $\lambda$ also produces the correct $y$ and $z$ components. With $\lambda = -3$ :

$y$ -component: $-2 - (-3) = 1$ ✓ (matches $A$ 's $y = 1$ )
$z$ -component: $3 + 4(-3) = -9$ ✓ (matches $A$ 's $z = -9$ )

Since all three components agree, $A$ lies on $l_1$ at $\lambda = -3$ .

Key Takeaways

A point lies on a line in vector form iff its position vector is of the form $\mathbf{a} + t\mathbf{b}$ for some $t$ .
To verify a point is on a line, find the parameter from one component equation and confirm it satisfies the others.

Common Mistakes

Only checking one component of the line equation instead of all three.
Sign errors when solving $1 + 2\lambda = -5$ (writing $2\lambda = 4$ instead of $2\lambda = -6$ ).
Confusing the parameter $\lambda$ on the line with the constant $a$ in the direction vector.

Things to Be Careful About

A single component check is not enough — the parameter must produce the correct value for ALL three components.
The mark scheme accepts either checking all three components explicitly, or finding $\lambda$ from each component and noting they all give the same value.

Techniques used

substitute the position vector of a point into the component form of a linesolve a linear equation for the parameterverify the parameter value satisfies all three component equations

(d)

The point $B$ is the image of $A$ after a reflection in the line $l_2$ .

Find the position vector of $B$ .

2M Similar

DifficultyMedium

Worked solution

Approach

The reflection of a point $A$ in a line is given by $B = 2M - A$ , where $M$ is the foot of the perpendicular from $A$ to the line. The key observation is that $A$ lies on $l_1$ (from part c) and $l_1 \perp l_2$ , so the segment $AM$ lies along $l_1$ and is perpendicular to $l_2$ . Since $l_1$ and $l_2$ intersect, the intersection point is the unique point on $l_2$ reached by travelling along $l_1$ from $A$ — this is $M$ .

Working

From part (b), the intersection of $l_1$ and $l_2$ is $M = -\mathbf{i} - \mathbf{j} - \mathbf{k}$ .

Confirm $AM \perp l_2$ using the scalar product:

\overrightarrow{AM} = M - A = 4\mathbf{i} - 2\mathbf{j} + 8\mathbf{k}

\overrightarrow{AM} \cdot (3\mathbf{i} - 2\mathbf{j} - 2\mathbf{k}) = (4)(3) + (-2)(-2) + (8)(-2) = 12 + 4 - 16 = 0 \quad \checkmark

So $M$ is indeed the foot of perpendicular from $A$ to $l_2$ .

Apply the reflection formula $B = 2M - A$ :

B = 2(-\mathbf{i} - \mathbf{j} - \mathbf{k}) - (-5\mathbf{i} + \mathbf{j} - 9\mathbf{k})

B = -2\mathbf{i} - 2\mathbf{j} - 2\mathbf{k} + 5\mathbf{i} - \mathbf{j} + 9\mathbf{k}

B = 3\mathbf{i} - 3\mathbf{j} + 7\mathbf{k}

Answer

3\mathbf{i} - 3\mathbf{j} + 7\mathbf{k}

Final answer

3i - 3j + 7k

Detailed explanation

Walkthrough

The reflection of a point $A$ in a line $\ell$ is the point $B$ such that $\ell$ is the perpendicular bisector of $AB$ . The formula is $B = 2M - A$ , where $M$ is the foot of the perpendicular from $A$ to $\ell$ (i.e., the point on $\ell$ closest to $A$ ).

We need to find $M$ . From part (c), $A$ lies on $l_1$ . From part (a), $l_1 \perp l_2$ . So the segment $AM$ , which is along $l_1$ , is perpendicular to $l_2$ for any point $M$ on $l_1$ . In particular, the intersection point of $l_1$ and $l_2$ is on $l_2$ , and the line from $A$ to this intersection lies along $l_1$ , hence is perpendicular to $l_2$ . So this intersection point is the foot of perpendicular $M$ .

From part (b), $M = -\mathbf{i} - \mathbf{j} - \mathbf{k}$ .

Applying the reflection formula:

B = 2M - A = 2(-\mathbf{i} - \mathbf{j} - \mathbf{k}) - (-5\mathbf{i} + \mathbf{j} - 9\mathbf{k}) = 3\mathbf{i} - 3\mathbf{j} + 7\mathbf{k}.

Key Takeaways

The reflection of a point in a line is given by $B = 2M - A$ , where $M$ is the foot of the perpendicular from $A$ to the line.
The foot of perpendicular is the unique point on the line closest to $A$ (i.e., where $\overrightarrow{AM}$ is perpendicular to the line).
When $A$ lies on a line $l_1$ that is perpendicular to a line $l_2$ , and $l_1$ and $l_2$ intersect, the intersection point is the foot of perpendicular from $A$ to $l_2$ .

Common Mistakes

Using $2M + A$ or $A - 2M$ instead of $2M - A$ .
Failing to recognize that the intersection of perpendicular lines is the foot of perpendicular from one to the other.
Sign errors when expanding the reflection formula.

Things to Be Careful About

The formula $B = 2M - A$ comes from $M$ being the midpoint of $AB$ (i.e., $M = (A + B)/2$ , rearranged to $B = 2M - A$ ).
Always verify the foot of perpendicular — in this case, the scalar product check confirms $\overrightarrow{AM} \cdot \mathbf{d}_2 = 0$ .
This quick technique only works because $A$ happens to lie on a line perpendicular to $l_2$ . In general, the foot of perpendicular must be found by solving an optimisation problem or a system of equations.

Techniques used

recognize the intersection of two perpendicular lines as the foot of perpendicularverify perpendicularity using the scalar productapply the reflection formula B = 2M - A

Q10Medium-HardDifferentiationTrigonometryIntegration

(a)

Given that $2x = \tan y$ , show that $\frac{dy}{dx} = \frac{2}{1 + 4x^2}$ .

3M Similar

DifficultyMedium-Easy

Worked solution

Approach

Differentiate both sides of $2x = \tan y$ with respect to $y$ , replace $\sec^2 y$ using the Pythagorean identity $\sec^2 y = 1 + \tan^2 y$ , then invert to obtain $\frac{dy}{dx}$ and finally substitute $\tan y = 2x$ .

Working

Differentiating both sides of $2x = \tan y$ with respect to $y$ :

2 \frac{dx}{dy} = \sec^2 y

Using the Pythagorean identity $\sec^2 y = 1 + \tan^2 y$ :

2 \frac{dx}{dy} = 1 + \tan^2 y

Inverting:

\frac{dy}{dx} = \frac{2}{1 + \tan^2 y}

Since $\tan y = 2x$ , we have $\tan^2 y = 4x^2$ , so:

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

Answer

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

Final answer

dy/dx = 2/(1 + 4x^2)

Detailed explanation

Walkthrough

The relation $2x = \tan y$ defines $y$ implicitly as a function of $x$ , so $\frac{dy}{dx}$ must be found by implicit differentiation. A clean way is to differentiate both sides of the equation with respect to $y$ . On the left, $\frac{d}{dy}(2x) = 2\frac{dx}{dy}$ by the chain rule. On the right, $\frac{d}{dy}(\tan y) = \sec^2 y$ . This gives $2\frac{dx}{dy} = \sec^2 y$ .

We now have $\frac{dx}{dy}$ , but we want $\frac{dy}{dx}$ . The next step is to eliminate the trig function using the Pythagorean identity $\sec^2 y = 1 + \tan^2 y$ , so $2\frac{dx}{dy} = 1 + \tan^2 y$ . Inverting both sides gives $\frac{dy}{dx} = \frac{2}{1 + \tan^2 y}$ .

Finally, we use the original relation $\tan y = 2x$ to replace $\tan^2 y$ with $(2x)^2 = 4x^2$ , and arrive at the result $\frac{dy}{dx} = \frac{2}{1 + 4x^2}$ , as required.

Key Takeaways

When $y$ is defined implicitly by an equation in $x$ and $y$ , implicit differentiation gives $\frac{dy}{dx}$ .
Differentiating with respect to $y$ and then inverting is a useful alternative to differentiating with respect to $x$ and using the chain rule, especially when the original equation is simple in $y$ .
The Pythagorean identity $\sec^2 y = 1 + \tan^2 y$ is the key tool for converting between $\sec^2$ and $\tan^2$ .
This result is the standard derivative of $\tan^{-1}(2x)$ , which is needed in part (b).

Common Mistakes

Forgetting to invert $\frac{dx}{dy}$ at the end and leaving the answer as $\frac{1 + 4x^2}{2}$ .
Forgetting the Pythagorean identity and leaving $\sec^2 y$ in the final answer.
Writing $\tan^2 y = 2x$ instead of $\tan^2 y = (2x)^2 = 4x^2$ — the squaring applies to the whole of $2x$ .

Things to Be Careful About

$\frac{dy}{dx} = \frac{1}{dx/dy}$ : inversion is reciprocation, not negation.
The Pythagorean identity uses $\tan^2 y$ , not $\tan y$ — the square is essential.
When inverting $2\frac{dx}{dy}$ , the factor of $2$ moves into the numerator, not the denominator.

Techniques used

differentiate both sides of the implicit relation with respect to yapply the Pythagorean identity sec^2 y = 1 + tan^2 yinvert dx/dy to obtain dy/dxsubstitute the original relation tan y = 2x

(b)

Hence find the exact value of $\int_{\frac{1}{2}}^{\frac{\sqrt{3}}{2}} x \tan^{-1}(2x) \,dx$ .

7M Similar

DifficultyMedium-Hard

Worked solution

Approach

Use integration by parts with $u = \tan^{-1}(2x)$ (so that $du = \frac{2}{1+4x^2}\,dx$ , the result from part (a)) and $dv = x\,dx$ (so that $v = \frac{x^2}{2}$ ). The remaining integral $\int \frac{x^2}{1+4x^2}\,dx$ has a numerator of the same degree as the denominator, so we split it as $\frac{1}{4}\left(1 - \frac{1}{1+4x^2}\right)$ and integrate each piece. Finally, evaluate the antiderivative at $x = \frac{\sqrt{3}}{2}$ and $x = \frac{1}{2}$ in the correct order.

Working

With $u = \tan^{-1}(2x)$ , $du = \frac{2}{1+4x^2}\,dx$ , and $dv = x\,dx$ , $v = \frac{x^2}{2}$ :

\int x\tan^{-1}(2x)\,dx = \frac{x^2}{2}\tan^{-1}(2x) - \int \frac{x^2}{2} \cdot \frac{2}{1+4x^2}\,dx = \frac{x^2}{2}\tan^{-1}(2x) - \int \frac{x^2}{1+4x^2}\,dx

Rewrite the new integrand by expressing the numerator in terms of the denominator:

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

Integrate:

\begin{aligned} \int \frac{x^2}{1+4x^2}\,dx &= \frac{1}{4}\int \left(1 - \frac{1}{1+4x^2}\right) dx \\ &= \frac{1}{4}\left(x - \frac{1}{2}\tan^{-1}(2x)\right) + C \\ &= \frac{x}{4} - \frac{1}{8}\tan^{-1}(2x) + C \end{aligned}

Combining the two pieces:

\int x\tan^{-1}(2x)\,dx = \frac{x^2}{2}\tan^{-1}(2x) - \frac{x}{4} + \frac{1}{8}\tan^{-1}(2x) + C

Evaluate at the upper limit $x = \frac{\sqrt{3}}{2}$ , where $2x = \sqrt{3}$ and $\tan^{-1}(\sqrt{3}) = \frac{\pi}{3}$ :

\begin{aligned} \frac{x^2}{2}\tan^{-1}(2x) - \frac{x}{4} + \frac{1}{8}\tan^{-1}(2x) &= \frac{1}{2} \cdot \frac{3}{4} \cdot \frac{\pi}{3} - \frac{1}{4} \cdot \frac{\sqrt{3}}{2} + \frac{1}{8} \cdot \frac{\pi}{3} \\ &= \frac{\pi}{8} - \frac{\sqrt{3}}{8} + \frac{\pi}{24} \\ &= \frac{3\pi + \pi}{24} - \frac{\sqrt{3}}{8} \\ &= \frac{\pi}{6} - \frac{\sqrt{3}}{8} \end{aligned}

Evaluate at the lower limit $x = \frac{1}{2}$ , where $2x = 1$ and $\tan^{-1}(1) = \frac{\pi}{4}$ :

\begin{aligned} \frac{x^2}{2}\tan^{-1}(2x) - \frac{x}{4} + \frac{1}{8}\tan^{-1}(2x) &= \frac{1}{2} \cdot \frac{1}{4} \cdot \frac{\pi}{4} - \frac{1}{4} \cdot \frac{1}{2} + \frac{1}{8} \cdot \frac{\pi}{4} \\ &= \frac{\pi}{32} - \frac{1}{8} + \frac{\pi}{32} \\ &= \frac{\pi}{16} - \frac{1}{8} \end{aligned}

Subtract (upper minus lower):

\begin{aligned} \left(\frac{\pi}{6} - \frac{\sqrt{3}}{8}\right) - \left(\frac{\pi}{16} - \frac{1}{8}\right) &= \left(\frac{\pi}{6} - \frac{\pi}{16}\right) - \frac{\sqrt{3}}{8} + \frac{1}{8} \\ &= \frac{8\pi - 3\pi}{48} - \frac{\sqrt{3}}{8} + \frac{1}{8} \\ &= \frac{5\pi}{48} - \frac{\sqrt{3}}{8} + \frac{1}{8} \end{aligned}

Answer

\int_{1/2}^{\sqrt{3}/2} x\tan^{-1}(2x)\,dx = \frac{5\pi}{48} - \frac{\sqrt{3}}{8} + \frac{1}{8}

Final answer

5π/48 − √3/8 + 1/8

Detailed explanation

Walkthrough

The integrand $x\tan^{-1}(2x)$ is a product of a polynomial and an inverse-trig function, so integration by parts is the natural tool. The formula is $\int u\,dv = uv - \int v\,du$ , and the standard choice is to put the inverse-trig function in $u$ (its derivative is simpler) and the polynomial in $dv$ (it can be integrated). So we set $u = \tan^{-1}(2x)$ and $dv = x\,dx$ . The derivative $du = \frac{2}{1+4x^2}\,dx$ is precisely the result derived in part (a) — this is the "hence" connection between the two parts of the question. The antiderivative of $dv$ is $v = \frac{x^2}{2}$ .

Applying the formula, the integral becomes $\frac{x^2}{2}\tan^{-1}(2x) - \int \frac{x^2}{1+4x^2}\,dx$ . The new integral cannot be integrated directly because the numerator and the denominator have the same degree. The standard trick is to write the numerator in terms of the denominator: $\frac{x^2}{1+4x^2} = \frac{1}{4} \cdot \frac{4x^2}{1+4x^2} = \frac{1}{4} \cdot \frac{(1+4x^2) - 1}{1+4x^2} = \frac{1}{4}\left(1 - \frac{1}{1+4x^2}\right)$ . The integrand is now a constant minus $\frac{1}{4(1+4x^2)}$ , both of which can be integrated: the constant gives $\frac{x}{4}$ , and $\frac{1}{1+4x^2}$ integrates to $\frac{1}{2}\tan^{-1}(2x)$ , so the second term gives $-\frac{1}{8}\tan^{-1}(2x)$ .

Combining the two pieces gives the antiderivative $\frac{x^2}{2}\tan^{-1}(2x) - \frac{x}{4} + \frac{1}{8}\tan^{-1}(2x)$ . Now evaluate at the limits. The chosen limits are exactly the values that give nice inverse-trig results: at $x = \frac{1}{2}$ , $2x = 1$ so $\tan^{-1}(1) = \frac{\pi}{4}$ ; at $x = \frac{\sqrt{3}}{2}$ , $2x = \sqrt{3}$ so $\tan^{-1}(\sqrt{3}) = \frac{\pi}{3}$ . After substituting and combining the $\pi$ fractions with a common denominator of $48$ , the final answer is $\frac{5\pi}{48} - \frac{\sqrt{3}}{8} + \frac{1}{8}$ .

Key Takeaways

Integration by parts is the standard technique for products of polynomials and inverse-trig functions.
The choice $u = \tan^{-1}(2x)$ works because its derivative $\frac{2}{1+4x^2}$ is a rational function that combines well with the rest of the integrand.
When the degree of the numerator equals the degree of the denominator, use the algebraic identity $1 = (1+4x^2) - 4x^2$ to split into a polynomial plus a proper rational function before integrating.
The standard result $\int \frac{1}{1+4x^2}\,dx = \frac{1}{2}\tan^{-1}(2x) + C$ must be used, not $\tan^{-1}(x) + C$ .
Limits are chosen so that $\tan^{-1}(2x)$ evaluates to a familiar fraction of $\pi$ (here $\frac{\pi}{4}$ and $\frac{\pi}{3}$ ).

Common Mistakes

Trying to integrate $\frac{x^2}{1+4x^2}$ directly without first performing the algebraic simplification.
Integrating $\frac{1}{1+4x^2}$ as $\tan^{-1}(x)$ instead of $\frac{1}{2}\tan^{-1}(2x)$ , giving a wrong coefficient on the final arctan term.
Subtracting the upper limit from the lower limit instead of the other way around (sign error).
Forgetting the minus sign in the integration-by-parts formula and writing $\int u\,dv = uv + \int v\,du$ .
Computing $\left(\frac{\sqrt{3}}{2}\right)^2$ as $\frac{\sqrt{3}}{2}$ instead of $\frac{3}{4}$ — the squaring applies to both the $\sqrt{3}$ and the $2$ .

Things to Be Careful About

The chain rule on the derivative of $\tan^{-1}(2x)$ gives the factor of $2$ in the numerator, which must be carried through consistently.
The two arctan terms in the final antiderivative (one from $uv$ , one from $-\int v\,du$ ) both have positive coefficients, while the $-\frac{x}{4}$ term is negative.
When computing $\left(\frac{\sqrt{3}}{2}\right)^2 = \frac{3}{4}$ , the squaring applies to the $2$ in the denominator too: $2^2 = 4$ .
The sign of $\sqrt{3}$ in the final answer is negative.
Numerical sanity check: $\frac{5\pi}{48} \approx 0.327$ , $-\frac{\sqrt{3}}{8} \approx -0.217$ , $+\frac{1}{8} = 0.125$ , total $\approx 0.235$ , which should be positive (and is).

Techniques used

set up integration by parts with u = arctan(2x) and dv = x dxuse the derivative of arctan(2x) from part (a) for durewrite x^2/(1+4x^2) as 1/4 - 1/(4(1+4x^2))integrate 1/(1+4x^2) using the standard arctan formulaevaluate the antiderivative at both limits in the correct order

Q11MediumDifferential EquationsIntegration

In a field there are 300 plants of a certain species, all of which can be infected by a particular disease. At time $t$ after the first plant is infected there are $x$ infected plants. The rate of change of $x$ is proportional to the product of the number of plants infected and the number of plants that are not yet infected. The variables $x$ and $t$ are treated as continuous, and it is given that $\frac{dx}{dt} = 0.2$ and $x = 1$ when $t = 0$ .

(a)

Show that $x$ and $t$ satisfy the differential equation

1495 \frac{dx}{dt} = x(300 - x)

2M Similar

DifficultyMedium-Easy

Worked solution

Approach

Translate the statement about rates and proportions into a differential equation involving a constant of proportionality, then use the given rate at the initial instant to determine that constant and rearrange.

Working

Let $k$ be the constant of proportionality. Since the number not yet infected is $300-x$ , the rate equation is

\frac{dx}{dt}=k x(300-x)

When $t=0$ , we are given that $x=1$ and $\frac{dx}{dt}=0.2$ . Substituting:

0.2 = k(1)(300-1)=299k

Hence

k = \frac{0.2}{299} = \frac{1}{1495}

So

\frac{dx}{dt} = \frac{1}{1495}x(300-x)

Multiplying both sides by $1495$ gives

1495\frac{dx}{dt}=x(300-x)

as required.

Answer

1495\frac{dx}{dt}=x(300-x)

Final answer

1495 dx/dt = x(300 - x)

Detailed explanation

Walkthrough

The statement says that the rate at which infected plants increase, $\frac{dx}{dt}$ , is proportional to the product of the infected count $x$ and the not-yet-infected count $300-x$ . The word "proportional" means there is an unknown constant $k$ , so write $\frac{dx}{dt}=kx(300-x)$ . The initial conditions tell us the rate when exactly one plant has just become infected: $\frac{dx}{dt}=0.2$ and $x=1$ . Substituting these values determines $k$ , and then multiplying by $1495$ gives the required form.

Key Takeaways

This question tests translating a written rate-of-change statement into a differential equation and using an initial rate to find a proportionality constant. It also emphasises that the number not yet infected is total population minus infected.

Common Mistakes

Starting with the wrong factor $x$ instead of $300-x$ for the uninfected plants.
Rounding $0.2/299$ instead of using the exact value $\frac{1}{1495}$ .
The mark scheme says that simply verifying the final differential equation is not enough; you must derive it from $kx(300-x)$ .

Things to Be Careful About

Use the condition exactly at $t=0$ : $x=1$ and $\frac{dx}{dt}=0.2$ . The decimal $0.2$ is exact here, so convert it to the fraction $\frac{1}{5}$ before simplifying. At the end, multiply by $1495$ so the equation matches the requested form.

Techniques used

translate a proportionality statement into a differential equationsubstitute given rate and initial value to find the proportionality constantrearrange the equation into the required form

(b)

Using partial fractions, solve the differential equation and obtain an expression for $t$ in terms of a single logarithm involving $x$ .

9M Similar

DifficultyMedium

Worked solution

Approach

Separate variables so that all $x$ terms are with $dx$ and all $t$ terms are with $dt$ , write $\frac{1}{x(300-x)}$ as partial fractions, integrate both sides, use the initial condition $t=0$ , $x=1$ to fix the constant, and simplify into a single logarithm for $t$ .

Working

From part (a), $\frac{dx}{dt}=\frac{1}{1495}x(300-x)$ . Separating variables:

\int \frac{1}{x(300-x)}\,dx = \int \frac{1}{1495}\,dt

Write

\frac{1}{x(300-x)} = \frac{A}{x}+\frac{B}{300-x}

Then

1 = A(300-x)+Bx

Substituting $x=0$ gives $300A=1$ , so $A=\frac{1}{300}$ ; subatituting $x=300$ gives $300B=1$ , so $B=\frac{1}{300}$ . Therefore

\int \frac{1}{x(300-x)}\,dx =\frac{1}{300}\ln x -\frac{1}{300}\ln(300-x)

and the separated equation becomes

\frac{1}{300}\ln x -\frac{1}{300}\ln(300-x)=\frac{t}{1495}+C

Multiplying by $1495$ :

\frac{1495}{300}\left(\ln x - \ln(300-x)\right)=t + D

where $D=1495C$ . Using $t=0$ , $x=1$ :

\frac{1495}{300}(\ln 1 - \ln 299)=D

Since $\ln 1 = 0$ ,

D=-\frac{1495}{300}\ln 299

Therefore

\frac{1495}{300}\left(\ln x - \ln(300-x)\right)=t-\frac{1495}{300}\ln 299

Solving for $t$ :

t = \frac{1495}{300}\left(\ln x - \ln(300-x)+\ln 299\right)

Combine the logarithms using $\ln a - \ln b = \ln \frac{a}{b}$ and $\ln a + \ln b = \ln(ab)$ :

t = \frac{1495}{300}\ln\left(\frac{299x}{300-x}\right)

Finally simplify $\frac{1495}{300} = \frac{299}{60}$ :

t=\frac{299}{60}\ln\left(\frac{299x}{300-x}\right)

Answer

t=\frac{299}{60}\ln\left(\frac{299x}{300-x}\right)

Final answer

t = (299/60) ln(299x/(300 - x))

Detailed explanation

Walkthrough

This is a separable differential equation, so separate the variables: put $dx$ and $x$ -dependent terms on one side and $dt$ on the other. The resulting integrand $\frac{1}{x(300-x)}$ cannot be integrated directly, so decompose it into partial fractions.

Since the denominator has two distinct linear factors, write $\frac{1}{x(300-x)} = \frac{A}{x} + \frac{B}{300-x}$ . Multiplying through by $x(300-x)$ lets you find $A$ and $B$ by substituting $x=0$ and $x=300$ : both turn out to be $\frac{1}{300}$ .

Integrate term by term. Because $\frac{d}{dx}\ln(300-x)=-\frac{1}{300-x}$ , the second partial fraction produces a negative logarithm. Do not forget the constant of integration.

Use the initial condition $t=0$ , $x=1$ to evaluate the constant. Since $\ln 1 = 0$ and $\ln(300-1)=\ln 299$ , the constant is $-\frac{1495}{300}\ln 299$ . Then rearrange to isolate $t$ .

Finally, use the laws of logarithms: the difference of logarithms becomes a quotient and the sum with $\ln 299$ becomes a product. This gives the single logarithm expression requested. The coefficient $\frac{1495}{300}$ simplifies to $\frac{299}{60}$ .

Key Takeaways

This question combines several important ideas: solving a first-order separable differential equation, using partial fractions to integrate a rational function, determining an arbitrary constant from an initial condition, and using the laws of logarithms to simplify an answer into a single logarithm. It also shows that constants can be multiplied through by a common factor when the integration is done in an equivalent form.

Common Mistakes

Forgetting the minus sign when integrating $\frac{1}{300-x}$ . Since $\frac{d}{dx}\ln(300-x) = -\frac{1}{300-x}$ , this term must be negative.
Making an error in the partial fractions. Here $A = B = \frac{1}{300}$ . If you multiply through by $1495$ first, the equivalent values are $A = B = \frac{1495}{300} = \frac{299}{60}$ , which is also acceptable.
Omitting the constant of integration and then being unable to use $t=0$ , $x=1$ correctly.
Writing $\ln 1 - \ln 299$ as $\ln 1 - \ln 300$ . Remember $x=1$ means $300-x = 299$ .
Not reducing $\frac{1495}{300}$ to $\frac{299}{60}$ or not combining into a single logarithm when the question asks for one.

Things to Be Careful About

The solution is valid while $0 < x < 300$ , because both $x$ and $300-x$ are positive inside the logarithms. The mark scheme accepts equivalent forms, but the final answer should be a single logarithm. Keep all steps exact; avoid decimals when converting $0.2/299$ or the integration constant. Check the sign of the constant carefully when rearranging: the constant term should be $-\frac{1495}{300}\ln 299$ .

Techniques used

separate variables in a differential equationdecompose the integrand into partial fractionsintegrate each partial fraction using natural logarithmsapply the initial condition to determine the constantcombine logarithms into a single logarithm

Mathematics 9709/31 — May/June 2024

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