Mathematics 9709 / 33 — October/November 2024

DifficultyMedium-Easy

Worked solution

Approach

The condition $|z| = 2$ represents a circle of radius 2 centred at the origin. The condition $0 \leq \arg z \leq \frac{1}{4}\pi$ restricts this locus to a sector between the positive real axis and the line at angle $\frac{\pi}{4}$ in the first quadrant.

Working

$|z| = 2$ is an arc of a circle centred at the origin with radius 2.
$0 \leq \arg z \leq \frac{\pi}{4}$ restricts the locus to the sector from the positive real axis to the line $y = x$ in the first quadrant.
Sketch this arc on the Argand diagram, marking $2$ on the real axis and indicating the angle $\frac{\pi}{4}$ .

Answer

See diagram for part (a).

Final answer

Arc of circle centred at origin with radius 2, between arg 0 and arg pi/4

Detailed explanation

Walkthrough

First, we interpret the modulus condition. The equation $|z| = 2$ means the distance from the origin to the point $z$ is always 2. On an Argand diagram, this is a circle of radius 2 centred at the origin.

Next, we interpret the argument condition. The inequality $0 \leq \arg z \leq \frac{\pi}{4}$ means the angle that the line from the origin to $z$ makes with the positive real axis is between 0 and $\frac{\pi}{4}$ radians (0° to 45°). This restricts our circle to a specific sector in the first quadrant.

We combine these two conditions to draw an arc of radius 2 starting from the point $2$ on the real axis and ending at the point $\sqrt{2} + i\sqrt{2}$ on the line $\arg z = \frac{\pi}{4}$ .

Key Takeaways

The modulus $|z| = r$ represents a circle of radius $r$ centred at the origin.
The argument $\alpha \leq \arg z \leq \beta$ represents a sector between two rays from the origin.
Combining these gives an arc or sector on the Argand diagram.

Common Mistakes

Forgetting to restrict the circle to the correct sector based on the argument condition.
Drawing the arc with the wrong radius.
Not marking the key points or angles clearly on the diagram.

Things to Be Careful About

Ensure the scales on the real and imaginary axes are approximately equal.
The angle $\frac{1}{4}\pi$ is 45°, so the upper boundary of the sector should be at a 45° angle to the real axis.
Dashes can replace numbers on the axes, and arcs don't have to be perfectly circular as long as the intention is clear.

Techniques used

interpret modulus condition as circleinterpret argument condition as sectorsketch locus on Argand diagram

(b)

On the same diagram, sketch the locus of the points representing $z^2$ .

DifficultyMedium-Easy

Worked solution

Approach

If $z = r(\cos \theta + i \sin \theta)$ , then by De Moivre's theorem, $z^2 = r^2(\cos 2\theta + i \sin 2\theta)$ . We apply this to the locus of $z$ found in part (a) to find the locus of $z^2$ .

Working

For $z$ , we have $|z| = 2$ and $0 \leq \arg z \leq \frac{\pi}{4}$ .
For $z^2$ , the modulus is $|z^2| = |z|^2 = 2^2 = 4$ . This is an arc of a circle centred at the origin with radius 4.
The argument is $\arg(z^2) = 2 \arg z$ . Since $0 \leq \arg z \leq \frac{\pi}{4}$ , we have $0 \leq \arg(z^2) \leq 2 \times \frac{\pi}{4} = \frac{\pi}{2}$ .
Sketch this larger arc on the same Argand diagram, marking $4$ on the real axis and $4i$ on the imaginary axis.

Answer

See diagram for part (b).

Final answer

Arc of circle centred at origin with radius 4, between arg 0 and arg pi/2

Detailed explanation

Walkthrough

We use the polar form of complex numbers to understand the geometric effect of squaring. If $z$ has modulus $r$ and argument $\theta$ , then $z^2$ has modulus $r^2$ and argument $2\theta$ .

From part (a), the locus of $z$ is an arc of a circle with radius $r = 2$ and angles $\theta$ between $0$ and $\frac{\pi}{4}$ .

When we square $z$ :

The new modulus is $r^2 = 2^2 = 4$ . So the locus of $z^2$ lies on a circle of radius 4.
The new argument is $2\theta$ . Since $0 \leq \theta \leq \frac{\pi}{4}$ , multiplying by 2 gives $0 \leq 2\theta \leq \frac{\pi}{2}$ . This means the locus spans from the positive real axis to the positive imaginary axis (the first quadrant).

We draw an arc of radius 4 from the point $(4, 0)$ to the point $(0, 4)$ on the imaginary axis.

Key Takeaways

Squaring a complex number squares its modulus and doubles its argument.
This transforms a sector of angle $\alpha$ into a sector of angle $2\alpha$ and radius $r^2$ .
De Moivre's theorem provides a quick way to find the modulus and argument of powers of complex numbers.

Common Mistakes

Forgetting to square the modulus (using 2 instead of 4).
Forgetting to double the argument (using $\frac{\pi}{4}$ instead of $\frac{\pi}{2}$ ).
Drawing the arc in the wrong quadrant.

Things to Be Careful About

Ensure the scales on the axes are consistent with part (a) so the diagram is accurate.
Mark the key points $4$ and $4i$ to clearly show the radius of the new arc.
The upper boundary of the sector for $z^2$ is the positive imaginary axis, corresponding to $\arg = \frac{\pi}{2}$ .

Techniques used

multiply modulusdouble argumentsketch locus of z^2

Q2Medium-EasyNumerical Solution of Equations

Let $f(x) = 2x^3 - 5x^2 + 4$ .

(a)

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

x_{n+1} = \sqrt{\frac{4}{5 - 2x_n}}

converges, then it converges to a root of the equation $f(x) = 0$ .

DifficultyMedium-Easy

Worked solution

Approach

Suppose the sequence converges to a limit $L$ . Then both $x_n$ and $x_{n+1}$ tend to $L$ , so we can replace them by $L$ in the iterative formula and rearrange to obtain $f(L)=0$ .

Working

Let $x_n \to L$ as $n \to \infty$ . Then $x_{n+1} \to L$ also, so

L = \sqrt{\frac{4}{5-2L}}

Squaring both sides:

L^2 = \frac{4}{5-2L}

Multiply by $5-2L$ :

L^2(5-2L) = 4

Expanding:

5L^2 - 2L^3 = 4

Rearranging:

2L^3 - 5L^2 + 4 = 0

Since $f(x)=2x^3-5x^2+4$ , this is $f(L)=0$ . Therefore the limit of the sequence is a root of $f(x)=0$ .

Answer

If the sequence converges to $L$ , then $f(L)=0$ , so the sequence converges to a root of $f(x)=0$ .

Final answer

If the sequence converges to L, then f(L) = 0.

Detailed explanation

Walkthrough

When an iterative sequence $x_{n+1}=F(x_n)$ converges to a limit $L$ , the values $x_n$ get closer and closer to $L$ . Since $x_{n+1}$ is just the next term in the same sequence, it also tends to $L$ . Therefore, in the limit, the formula becomes $L=F(L)$ . Here $F(x)=\sqrt{\frac{4}{5-2x}}$ , so we set $L$ equal to that expression. Squaring removes the square root; multiplying by the denominator and rearranging gives the cubic $2L^3-5L^2+4=0$ , which is exactly $f(L)=0$ . This proves that any convergent limit of the iteration must be a root of the original equation.

Key Takeaways

A convergent iteration $x_{n+1}=F(x_n)$ must converge to a fixed point $x=F(x)$ .
Rearranging the fixed-point equation can recover the original equation whose root is being approximated.
This is why the iterative formula is a valid way to find roots of $f(x)=0$ .

Common Mistakes

Forgetting to square both sides when removing the square root.
Stopping at $L^2=\frac{4}{5-2L}$ without rearranging to the required cubic form.
Assuming convergence without noting that the argument only applies if the sequence converges.

Things to Be Careful About

The denominator $5-2L$ must be nonzero; the squaring step is valid for real $L$ where the square root is defined.
The question says "if ... converges", so no proof of convergence is required; only the limiting rearrangement is needed.
Use the same variable throughout, or consistently use $x_n$ and $x_{n+1}$ .

Techniques used

take limits of convergent iterative sequencessquare both sides of an equationrearrange to polynomial form

(b)

The equation has a root close to 1.2.

Use the iterative formula from part (a) and an initial value of 1.2 to determine the root correct to 2 decimal places. Give the result of each iteration to 4 decimal places.

DifficultyMedium-Easy

Worked solution

Approach

Use the given iterative formula with $x_0=1.2$ , record each value to 4 decimal places, and confirm the root by checking a sign change in the interval $(1.275,1.285)$ .

Working

With $x_0=1.2000$ :

x_1 = \sqrt{\frac{4}{5-2(1.2)}} = \sqrt{\frac{4}{2.6}} = 1.2403

Continuing the iteration:

$n$	$x_n$
0	1.2000
1	1.2403
2	1.2601
3	1.2700
4	1.2752
5	1.2778
6	1.2792

The successive values are increasing and approaching a limit. To confirm the root to 2 decimal places, check the sign of $f$ on the interval $(1.275,1.285)$ :

f(1.275) = 2(1.275)^3 - 5(1.275)^2 + 4 \approx 0.0172 > 0

f(1.285) = 2(1.285)^3 - 5(1.285)^2 + 4 \approx -0.0125 < 0

There is a sign change, so the root lies between $1.275$ and $1.285$ . Hence, correct to 2 decimal places, the root is $1.28$ .

Answer

x = 1.28

Final answer

x = 1.28

Detailed explanation

Walkthrough

We start with $x_0=1.2$ . Substitute into $x_{n+1}=\sqrt{\frac{4}{5-2x_n}}$ to get $x_1$ . Then use $x_1$ to get $x_2$ , and so on. Each value is rounded to 4 decimal places as requested. The values increase: 1.2000, 1.2403, 1.2601, 1.2700, 1.2752, 1.2778, 1.2792. They appear to be settling near 1.28. To be certain the root rounds to 1.28, we check $f(1.275)>0$ and $f(1.285)<0$ ; the sign change shows the root is in $(1.275,1.285)$ , so every number in this interval rounds to 1.28 to 2 decimal places.

Key Takeaways

Iterative formulas produce a sequence of improving approximations.
Rounding intermediate results to the requested number of decimal places is part of the method.
A sign change over an interval whose endpoints both round to the same 2-dp value confirms the root's rounded value.

Common Mistakes

Using the wrong initial value or substituting into $f$ instead of the iterative formula.
Not showing enough iterations to justify the final rounded answer.
Rounding every intermediate value to 2 decimal places instead of 4.
Forgetting to check a sign change; an unsupported final answer may not earn full marks.

Things to Be Careful About

The formula requires careful use of brackets: $5-2x_n$ is the whole denominator.
Keep at least 4 decimal places during the iteration; premature rounding can shift the final digit.
The root is just above 1.28, so values below it such as 1.2792 still round to 1.28; the sign change justifies the rounding.

Techniques used

apply an iterative formula repeatedlyround intermediate values to 4 decimal placeslocate a root by sign change

Q3Medium-EasyLogarithmic and Exponential Functions

Fig. 3

The number of bacteria in a population, $P$ , at time $t$ hours is modelled by the equation $P = ae^{kt}$ , where $a$ and $k$ are constants. The graph of $\ln P$ against $t$ , shown in the diagram, has gradient $\frac{1}{20}$ and intersects the vertical axis at $(0, 3)$ .

(a)

State the value of $k$ and find the value of $a$ correct to 2 significant figures.

DifficultyMedium-Easy

Worked solution

Approach

Take the natural logarithm of both sides of $P = ae^{kt}$ to linearise the equation, then read off the gradient and y-intercept from the given graph of $\ln P$ against $t$ .

Working

Taking the natural logarithm of both sides:

\ln P = \ln\left(ae^{kt}\right)

Using the logarithm law $\ln(xy) = \ln x + \ln y$ :

\ln P = \ln a + \ln\left(e^{kt}\right)

Since $\ln\left(e^{kt}\right) = kt$ :

\ln P = \ln a + kt

This is a straight line of the form $y = c + mt$ with $y = \ln P$ and independent variable $t$ . The gradient is $k$ and the y-intercept is $\ln a$ .

From the graph, the gradient is $\frac{1}{20}$ , so:

k = \frac{1}{20}

The line intersects the vertical axis at $(0, 3)$ , so the y-intercept is $3$ :

\ln a = 3

Solving for $a$ :

a = e^3

Evaluating:

a \approx 20.0855...

Rounding to 2 significant figures:

a = 20

Answer

$k = \frac{1}{20}$ and $a = 20$

Final answer

k = 1/20, a = 20

Detailed explanation

Walkthrough

The model is $P = ae^{kt}$ , which is exponential. To extract the constants $a$ and $k$ from the graph of $\ln P$ against $t$ , we linearise by taking the natural logarithm of both sides.

Step 1: Linearise the equation.
Using the logarithm law $\ln(xy) = \ln x + \ln y$ , we write:

\ln P = \ln(ae^{kt}) = \ln a + \ln(e^{kt})

Since $\ln(e^x) = x$ , this simplifies to:

\ln P = \ln a + kt

Step 2: Match with the straight-line equation.
This has the form $y = c + mt$ where $y = \ln P$ , the gradient $m = k$ , and the y-intercept $c = \ln a$ .

Step 3: Read off $k$ from the gradient.
The graph has gradient $\frac{1}{20}$ , so $k = \frac{1}{20} = 0.05$ .

Step 4: Read off $a$ from the y-intercept.
The graph crosses the vertical axis at $\ln P = 3$ when $t = 0$ , so $\ln a = 3$ . Therefore $a = e^3 \approx 20.0855$ , which rounds to $20$ to 2 significant figures.

Key Takeaways

Taking the natural logarithm of an exponential model $P = ae^{kt}$ yields a linear equation $\ln P = \ln a + kt$ , allowing parameters to be read from a graph.
The gradient of the $\ln P$ vs $t$ graph gives $k$ , and the y-intercept gives $\ln a$ .
Always check the required number of significant figures when reporting numerical answers.

Common Mistakes

Using $\frac{dP}{dt} = kP$ to find $k$ instead of reading it directly from the graph. The mark scheme explicitly states $k = \frac{1}{20}$ must not come from differentiation.
Forgetting to round $a$ to 2 significant figures. $e^3 \approx 20.0855$ must be reported as $20$ , not $20.1$ or $20.09$ .
Writing $\ln P = kt + \ln a$ without correctly identifying which constant is the gradient and which is the intercept.

Things to Be Careful About

The value $k = \frac{1}{20}$ must be stated directly from the graph gradient, not derived via $\frac{dP}{dt} = kP$ .
The answer for $a$ must be to exactly 2 significant figures. $e^3 = 20.0855...$ rounds to $20$ (2 sf), not $20.1$ (3 sf).
The mark scheme notes that $\ln P = \frac{1}{20}t + 3$ alone does not earn marks until it is associated with the values of $a$ and $k$ . Always state the final values of $a$ and $k$ explicitly.

Techniques used

linearise exponential model by taking natural logarithmread gradient and y-intercept from linear graphevaluate e^3 to required significant figures

(b)

Find the time taken for $P$ to double. Give your answer correct to the nearest hour.

DifficultyMedium-Easy

Worked solution

Approach

At $t = 0$ , the population is $P = a$ . When the population has doubled, $P = 2a$ . Substitute this into the model and solve for $t$ .

Working

At $t = 0$ :

P = ae^{k \cdot 0} = a

When $P$ has doubled, $P = 2a$ . Substituting into the model $P = ae^{kt}$ :

2a = ae^{kt}

Dividing both sides by $a$ (since $a \neq 0$ ):

2 = e^{kt}

Taking the natural logarithm of both sides:

\ln 2 = kt

Substituting $k = \frac{1}{20}$ :

\ln 2 = \frac{1}{20}t

Solving for $t$ :

t = 20\ln 2

Evaluating:

t \approx 20 \times 0.693147... = 13.8629...

Rounding to the nearest hour:

t = 14

Answer

14 hours

Final answer

14 hours

Detailed explanation

Walkthrough

Step 1: Identify the doubling condition.
At $t = 0$ , $P = ae^0 = a$ . Doubling means $P = 2a$ . This is the key equation to solve.

Step 2: Substitute into the model.
Setting $P = 2a$ in $P = ae^{kt}$ :

2a = ae^{kt}

Step 3: Simplify.
Since $a \neq 0$ , divide both sides by $a$ :

2 = e^{kt}

Note that $a$ cancels out — this is why the exact value of $a$ is not needed for this part.

Step 4: Solve for $t$ .
Take the natural logarithm:

\ln 2 = kt

Substitute $k = \frac{1}{20}$ :

\ln 2 = \frac{t}{20}

t = 20\ln 2 \approx 13.863

Step 5: Round appropriately.
The question asks for the answer to the nearest hour, so $t = 14$ hours.

Key Takeaways

When asked for the time to double (or halve), set $P = 2P_0$ (or $P = \frac{1}{2}P_0$ ) and solve.
The initial value $a$ often cancels out in doubling/halving problems, so you don't need its exact value.
Always use logarithms to solve equations where the unknown is in the exponent.

Common Mistakes

Forgetting to divide by $a$ and trying to solve $2a = ae^{kt}$ without simplifying.
Using $\log_{10}$ instead of $\ln$ (natural logarithm) when solving $2 = e^{kt}$ . While $\log_{10}$ works if you use $\log_{10} e$ correctly, it's error-prone.
Rounding too early: $20\ln 2 = 13.8629...$ , which rounds to $14$ , not $13$ .
Not stating the units (hours) in the final answer.

Things to Be Careful About

The question asks for the answer to the nearest hour, not to 2 significant figures or 1 decimal place. $13.863$ rounds to $14$ , not $13.9$ .
The mark scheme allows $13.75$ hours (13 hours 45 minutes) as an acceptable answer leading to $14$ hours, but $13.863$ is the correct value from $20\ln 2$ .
Always show the equation $2a = ae^{kt}$ or equivalent to earn the method mark, even if $a$ cancels.

Techniques used

set up doubling condition equationcancel common factorsolve exponential equation using logarithmsround to nearest hour

Q45MMediumComplex Numbers

Find the complex number $z$ satisfying the equation

\frac{z - 3i}{z + 3i} = \frac{2 - 9i}{5}

Give your answer in the form $x + iy$ , where $x$ and $y$ are real.

DifficultyMedium

Worked solution

Approach

Substitute $z = x + iy$ into the equation, cross-multiply, expand using $i^2 = -1$ , then equate real and imaginary parts to get two simultaneous equations in $x$ and $y$ .

Working

Let $z = x + iy$ , where $x,y \in \mathbb{R}$ . Then

\frac{x + (y - 3)i}{x + (y + 3)i} = \frac{2 - 9i}{5}

Cross-multiply:

5\left(x + (y - 3)i\right) = (2 - 9i)\left(x + (y + 3)i\right)

Expand the left-hand side:

5x + 5(y - 3)i

Expand the right-hand side using $i^2 = -1$ :

(2 - 9i)(x + (y + 3)i) = 2x + 2(y+3)i - 9xi - 9(y+3)i^2

= 2x + 2(y+3)i - 9xi + 9(y+3)

= (2x + 9y + 27) + (2y + 6 - 9x)i

So the equation becomes:

5x + 5(y - 3)i = (2x + 9y + 27) + (2y + 6 - 9x)i

Equate real parts:

5x = 2x + 9y + 27 \quad \Rightarrow \quad 3x - 9y = 27

Equate imaginary parts:

5(y - 3) = 2y + 6 - 9x \quad \Rightarrow \quad 9x + 3y = 21

Solve the simultaneous equations:

x - 3y = 9, \qquad 3x + y = 7

From $x - 3y = 9$ , $x = 9 + 3y$ . Substitute into $3x + y = 7$ :

3(9 + 3y) + y = 7

27 + 9y + y = 7

10y = -20

y = -2

Then $x = 9 + 3(-2) = 3$ .

Answer

z = 3 - 2i

Final answer

z = 3 - 2i

Detailed explanation

Walkthrough

The key idea is to turn the equation involving the unknown complex number $z$ into equations involving only real numbers. We write $z = x + iy$ , where $x$ and $y$ are real, so the left-hand side becomes a quotient of two complex numbers. Because $z + 3i \neq 0$ for the solution, we may multiply both sides by the denominator $z + 3i$ and by 5 to clear the fractions. This gives a complex equation whose two sides are products of linear expressions in $x$ and $y$ .

Next, expand both sides. The only special rule needed is $i^2 = -1$ ; this is what converts terms such as $-9i \cdot (y+3)i$ into the real number $9(y+3)$ . After expansion, each side is written in the form real + (imaginary)i.

Two complex numbers are equal if and only if their real parts are equal and their imaginary parts are equal. Therefore we equate the real parts to obtain one linear equation in $x$ and $y$ , and equate the imaginary parts to obtain another. Solving this pair of simultaneous equations gives $x = 3$ and $y = -2$ , so $z = 3 - 2i$ .

Key Takeaways

A complex equation can be solved by writing the unknown as $x + iy$ and separating real and imaginary parts.
The identity $i^2 = -1$ must be applied whenever multiplying complex numbers.
Equality of complex numbers means equality of real parts and equality of imaginary parts.
The final answer should be given in the requested Cartesian form $x + iy$ .

Common Mistakes

Forgetting to use $i^2 = -1$ , which leaves $i^2$ terms and gives a wrong equation.
Expanding the product incorrectly, especially the signs of terms involving $i^2$ .
Equating only real parts or only imaginary parts, instead of both.
Making sign errors when rearranging the simultaneous equations.
Giving an answer without showing the method; the mark scheme requires the working leading to $z = 3 - 2i$ .

Things to Be Careful About

The denominator $z + 3i$ must not be zero; the solution $z = 3 - 2i$ is valid because $z + 3i = 3 + i \neq 0$ .
When cross-multiplying, multiply the whole numerator by 5, not just part of it.
The simultaneous equations can be written in different equivalent forms; simplify them before solving to reduce sign errors.
If using the alternative method of solving for $z$ first and then dividing by a complex number, you must show the multiplication by the conjugate; simply stating the final value is not enough for full marks.

Techniques used

substitute z = x + iycross-multiply complex fractionsexpand products using i^2 = -1equate real and imaginary partssolve simultaneous linear equations

Approach

Start with the left-hand side. Factor $\cos^4\theta - \sin^4\theta$ as a difference of squares, and rewrite $4\sin^2\theta\cos^2\theta$ using $\sin 2\theta = 2\sin\theta\cos\theta$ . Then apply the Pythagorean identity $\sin^2 2\theta = 1 - \cos^2 2\theta$ to obtain the right-hand side.

Working

\cos^4\theta - \sin^4\theta = (\cos^2\theta + \sin^2\theta)(\cos^2\theta - \sin^2\theta) = \cos 2\theta

4\sin^2\theta\cos^2\theta = (2\sin\theta\cos\theta)^2 = \sin^2 2\theta

Therefore,

\cos^4\theta - \sin^4\theta - 4\sin^2\theta\cos^2\theta = \cos 2\theta - \sin^2 2\theta

Using $\sin^2 2\theta = 1 - \cos^2 2\theta$ ,

\cos 2\theta - \sin^2 2\theta = \cos 2\theta - (1 - \cos^2 2\theta) = \cos^2 2\theta + \cos 2\theta - 1

Answer

\cos^4\theta - \sin^4\theta - 4\sin^2\theta\cos^2\theta \equiv \cos^2 2\theta + \cos 2\theta - 1

Final answer

cos^4 theta - sin^4 theta - 4 sin^2 theta cos^2 theta = cos^2 2theta + cos 2theta - 1

Detailed explanation

Walkthrough

We need to prove an identity. Start with the left-hand side because it is more complicated. First factor $\cos^4\theta - \sin^4\theta$ : this is a difference of two squares, so it becomes $(\cos^2\theta + \sin^2\theta)(\cos^2\theta - \sin^2\theta)$ . Since $\cos^2\theta + \sin^2\theta = 1$ and $\cos^2\theta - \sin^2\theta = \cos 2\theta$ , this simplifies to $\cos 2\theta$ . Next, $4\sin^2\theta\cos^2\theta = (2\sin\theta\cos\theta)^2 = \sin^2 2\theta$ . The whole left-hand side becomes $\cos 2\theta - \sin^2 2\theta$ . Finally, use $\sin^2 2\theta = 1 - \cos^2 2\theta$ to replace it, giving $\cos 2\theta - (1 - \cos^2 2\theta) = \cos^2 2\theta + \cos 2\theta - 1$ , which is the right-hand side.

Key Takeaways

This question tests the double-angle formulae and the Pythagorean identity. Recognising $\cos^4\theta - \sin^4\theta$ as a difference of squares lets us simplify quickly, and rewriting $4\sin^2\theta\cos^2\theta$ as $\sin^2 2\theta$ makes the connection to $\cos 2\theta$ clear.

Common Mistakes

Expanding $\cos^4\theta$ and $\sin^4\theta$ directly can lead to long, error-prone algebra; factoring is cleaner.
Forgetting that $4\sin^2\theta\cos^2\theta = \sin^2 2\theta$ , not $\sin 2\theta$ .
Sign errors when subtracting $\sin^2 2\theta$ after replacing it with $1 - \cos^2 2\theta$ .

Things to Be Careful About

This is an 'AG' (answer given) question, so every step must be shown clearly. The identity $\sin^2 2\theta = 1 - \cos^2 2\theta$ is just the Pythagorean identity applied to the angle $2\theta$ .

Techniques used

factor a difference of squaresapply the double-angle identityapply the Pythagorean identity

(b)

Solve the equation $\cos^4 \alpha - \sin^4 \alpha = 4 \sin^2 \alpha \cos^2 \alpha$ for $0^\circ \leq \alpha \leq 180^\circ$ .

DifficultyMedium

Worked solution

Approach

Use the identity from part (a) to rewrite the equation as a quadratic in $\cos 2\alpha$ . Solve the quadratic, reject any value outside $[-1,1]$ , then find all angles $\alpha$ in the given range.

Working

\cos^4\alpha - \sin^4\alpha = 4\sin^2\alpha\cos^2\alpha

Bring all terms to one side:

\cos^4\alpha - \sin^4\alpha - 4\sin^2\alpha\cos^2\alpha = 0

Using the identity from part (a), this becomes:

\cos^2 2\alpha + \cos 2\alpha - 1 = 0

Let $c = \cos 2\alpha$ . Then:

c^2 + c - 1 = 0

c = \frac{-1 \pm \sqrt{1^2 - 4(1)(-1)}}{2(1)} = \frac{-1 \pm \sqrt{5}}{2}

So $c = \frac{-1 + \sqrt{5}}{2} \approx 0.618$ or $c = \frac{-1 - \sqrt{5}}{2} \approx -1.618$ . Since $-1 \leq \cos 2\alpha \leq 1$ , reject $c = -1.618$ .

Therefore:

\cos 2\alpha = \frac{\sqrt{5} - 1}{2} \approx 0.618

For $0^\circ \leq \alpha \leq 180^\circ$ , we have $0^\circ \leq 2\alpha \leq 360^\circ$ . Cosine is positive in the first and fourth quadrants, so:

2\alpha = \cos^{-1}(0.618) \approx 51.8^\circ

or

2\alpha = 360^\circ - 51.8^\circ \approx 308.2^\circ

Thus:

\alpha \approx 25.9^\circ \quad \text{or} \quad \alpha \approx 154.1^\circ

Final answer

alpha = 25.9 degrees or alpha = 154.1 degrees

Detailed explanation

Walkthrough

The equation is exactly the left-hand side of the identity from part (a) set equal to zero. So replace $\cos^4\alpha - \sin^4\alpha - 4\sin^2\alpha\cos^2\alpha$ with $\cos^2 2\alpha + \cos 2\alpha - 1$ . This gives a quadratic in $\cos 2\alpha$ . Let $c = \cos 2\alpha$ and solve $c^2 + c - 1 = 0$ using the quadratic formula. One solution is $0.618$ and the other is $-1.618$ ; the latter is impossible because cosine is always between $-1$ and $1$ . Now solve $\cos 2\alpha = 0.618$ . Since $0^\circ \leq \alpha \leq 180^\circ$ , the angle $2\alpha$ ranges from $0^\circ$ to $360^\circ$ . Cosine is positive in the first and fourth quadrants, so $2\alpha$ can be $51.8^\circ$ or $308.2^\circ$ . Dividing by $2$ gives $\alpha = 25.9^\circ$ and $\alpha = 154.1^\circ$ .

Key Takeaways

This part connects a trigonometric identity to solving an equation. Once the identity is used, the problem becomes a quadratic equation followed by a standard trigonometric equation. You must always consider the full range of $2\alpha$ , not just the principal value of the inverse cosine.

Common Mistakes

Forgetting to bring all terms to one side before applying the identity.
Keeping the root $c = -1.618$ , which is outside the range of cosine.
Finding only $\alpha = 25.9^\circ$ and missing $\alpha = 154.1^\circ$ .
Mixing degrees and radians; the question uses degrees.

Things to Be Careful About

The mark scheme allows $154$ for $154.1$ , but you must give both solutions and no extra solutions in the range. If you work in radians, the solutions would be $0.452$ and $2.69$ , which is treated as a misread. Always check the range for $2\alpha$ before dividing by $2$ .

Techniques used

apply the identity from part (a)solve a quadratic equationfind all angles in the given range

Q6MediumVectors

The lines $l$ and $m$ have vector equations

l: \mathbf{r} = 2\mathbf{i} + \mathbf{j} - 3\mathbf{k} + \lambda(-\mathbf{i} + 2\mathbf{k}) \text{ and } m: \mathbf{r} = 2\mathbf{i} + \mathbf{j} - 3\mathbf{k} + \mu(2\mathbf{i} - \mathbf{j} + 5\mathbf{k})

Lines $l$ and $m$ intersect at the point $P$ .

(a)

State the coordinates of $P$ .

1M Similar

(b)

Find the exact value of the cosine of the acute angle between $l$ and $m$ .

DifficultyMedium-Easy

Worked solution

Approach

The angle between two lines is the angle between their direction vectors. Use the scalar product formula $\cos\theta = \frac{\mathbf{a}\cdot\mathbf{b}}{|\mathbf{a}||\mathbf{b}|}$ and take the acute angle.

Working

Direction vector of $l$ :

\mathbf{d}_l = -\mathbf{i} + 2\mathbf{k} = (-1, 0, 2)

Direction vector of $m$ :

\mathbf{d}_m = 2\mathbf{i} - \mathbf{j} + 5\mathbf{k} = (2, -1, 5)

Scalar product:

\mathbf{d}_l \cdot \mathbf{d}_m = (-1)(2) + (0)(-1) + (2)(5) = -2 + 0 + 10 = 8

Magnitudes:

|\mathbf{d}_l| = \sqrt{(-1)^2 + 0^2 + 2^2} = \sqrt{5}

|\mathbf{d}_m| = \sqrt{2^2 + (-1)^2 + 5^2} = \sqrt{30}

Therefore

\cos\theta = \frac{8}{\sqrt{5}\sqrt{30}} = \frac{8}{\sqrt{150}} = \frac{8}{5\sqrt{6}}

Since the scalar product is positive, this is the acute angle.

Answer

\cos\theta = \frac{8}{5\sqrt{6}}

Final answer

8/(5√6)

Detailed explanation

Walkthrough

To find the angle between two lines, use their direction vectors. For $l$ , the direction vector is $(-1, 0, 2)$ ; for $m$ , it is $(2, -1, 5)$ . Compute their scalar product:

(-1)(2) + (0)(-1) + (2)(5) = 8

Then compute the magnitudes:

|\mathbf{d}_l| = \sqrt{5}, \quad |\mathbf{d}_m| = \sqrt{30}

The cosine formula gives

\cos\theta = \frac{8}{\sqrt{5}\sqrt{30}} = \frac{8}{5\sqrt{6}}

Because the scalar product is positive, the angle is acute, so no sign adjustment is needed.

Key Takeaways

The cosine of the angle between two lines is found from the scalar product of their direction vectors divided by the product of their magnitudes. This is one of the main applications of the scalar product.

Common Mistakes

Using position vectors instead of direction vectors. Forgetting to take the absolute value when the scalar product is negative. Giving only a decimal approximation, which is not accepted for an exact-value question.

Things to Be Careful About

The scalar product can be negative, meaning the angle between the direction vectors is obtuse; for the acute angle between the lines, use the positive value. Simplify $\sqrt{5}\sqrt{30}$ to $5\sqrt{6}$ if desired. Equivalent exact forms such as $\frac{8}{\sqrt{150}}$ or $\frac{4\sqrt{6}}{15}$ are accepted.

Techniques used

extract direction vectors from the line equationscompute the scalar product of the direction vectorsdivide by the product of the magnitudestake the acute angle

(c)

The point $A$ on line $l$ has coordinates $(0, 1, 1)$ . The point $B$ on line $m$ has coordinates $(0, 2, -8)$ .

Find the exact area of triangle $APB$ .

DifficultyMedium

Worked solution

Approach

Use the sides of triangle $APB$ and the sine formula for area: $\text{Area} = \frac{1}{2}|\vec{PA}||\vec{PB}|\sin\angle APB$ . Find the side lengths, then use $\sin^2\theta = 1 - \cos^2\theta$ with the cosine from part (b).

Working

With $P = (2,1,-3)$ , $A = (0,1,1)$ and $B = (0,2,-8)$ :

\vec{PA} = A - P = (-2, 0, 4), \quad |\vec{PA}| = \sqrt{4 + 16} = 2\sqrt{5}

\vec{PB} = B - P = (-2, 1, -5), \quad |\vec{PB}| = \sqrt{4 + 1 + 25} = \sqrt{30}

\vec{AB} = B - A = (0, 1, -9), \quad |\vec{AB}| = \sqrt{82}

From part (b),

\cos^2\angle APB = \left(\frac{8}{5\sqrt{6}}\right)^2 = \frac{64}{150}

so

\sin^2\angle APB = 1 - \frac{64}{150} = \frac{86}{150}

\sin\angle APB = \sqrt{\frac{86}{150}}

Therefore the area is

\frac{1}{2} \times 2\sqrt{5} \times \sqrt{30} \times \sqrt{\frac{86}{150}} = \sqrt{150} \times \sqrt{\frac{86}{150}} = \sqrt{86}

Answer

\sqrt{86}

Final answer

√86

Detailed explanation

Walkthrough

We need the area of triangle $APB$ . With $P = (2,1,-3)$ , compute the displacement vectors from $P$ to the other vertices:

\vec{PA} = (-2, 0, 4), \quad \vec{PB} = (-2, 1, -5)

Their lengths are $2\sqrt{5}$ and $\sqrt{30}$ . We also compute $\vec{AB} = (0,1,-9)$ with length $\sqrt{82}$ . The area of a triangle using two sides and the included angle is $\frac{1}{2}ab\sin C$ . From part (b), $\cos^2\angle APB = \frac{64}{150}$ , so $\sin^2\angle APB = \frac{86}{150}$ . Therefore

\text{Area} = \frac{1}{2} \times 2\sqrt{5} \times \sqrt{30} \times \sqrt{\frac{86}{150}} = \sqrt{86}

Key Takeaways

The area of a triangle can be found from two side vectors and the sine of the included angle. When only the cosine is known, use the identity $\sin^2\theta = 1 - \cos^2\theta$ . Exact surd arithmetic is essential.

Common Mistakes

Forgetting the factor of $\frac{1}{2}$ in the area formula. Using $\cos$ instead of $\sin$ . Giving a decimal answer instead of the exact value $\sqrt{86}$ . Using $P - A$ instead of $A - P$ when forming $\vec{PA}$ .

Things to Be Careful About

The actual angle $\angle APB$ may be obtuse, but $\sin$ of an obtuse angle is still positive, so the area calculation is unaffected. Ensure the vectors are taken from the correct vertex. The final answer must be exact; the mark scheme accepts simplified exact equivalents.

Techniques used

find displacement vectors between pointscompute side lengths of the triangleuse the sine formula for triangle areaapply the Pythagorean identity to obtain sine from cosine

Q7Medium-HardDifferentiationTrigonometry

The parametric equations of a curve are

x = 3 \sin 2t, \quad y = \tan t + \cot t

for $0 < t < \frac{1}{2}\pi$ .

(a)

Show that $\frac{dy}{dx} = \frac{-2}{3 \sin^2 2t}$ .

DifficultyMedium-Hard

Worked solution

Approach

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

\frac{dy}{dx} = \frac{dy}{dt} \div \frac{dx}{dt}

Working

Differentiate $x = 3\sin 2t$ :

\frac{dx}{dt} = 6\cos 2t

Differentiate $y = \tan t + \cot t$ :

\frac{dy}{dt} = \sec^2 t - \csc^2 t

Therefore

\frac{dy}{dx} = \frac{\sec^2 t - \csc^2 t}{6\cos 2t}

Write the numerator as a single fraction:

\sec^2 t - \csc^2 t = \frac{1}{\cos^2 t} - \frac{1}{\sin^2 t} = \frac{\sin^2 t - \cos^2 t}{\sin^2 t \cos^2 t}

Since $\sin^2 t - \cos^2 t = -\cos 2t$ and $\sin t \cos t = \frac{1}{2}\sin 2t$ ,

\frac{dy}{dx} = \frac{-\cos 2t}{\frac{1}{4}\sin^2 2t} \cdot \frac{1}{6\cos 2t} = \frac{-4\cos 2t}{6\cos 2t \sin^2 2t}

Cancel $\cos 2t$ :

\frac{dy}{dx} = \frac{-4}{6\sin^2 2t} = \frac{-2}{3\sin^2 2t}

Answer

\frac{dy}{dx} = \frac{-2}{3\sin^2 2t}

Final answer

dy/dx = -2/(3 sin^2 2t)

Detailed explanation

Walkthrough

We are given $x$ and $y$ as functions of $t$ , so to find $\frac{dy}{dx}$ we cannot differentiate $y$ directly with respect to $x$ . Instead we differentiate both $x$ and $y$ with respect to the parameter $t$ , then use

\frac{dy}{dx} = \frac{dy}{dt} \div \frac{dx}{dt}

First, $x = 3\sin 2t$ gives $\frac{dx}{dt} = 6\cos 2t$ . Then $y = \tan t + \cot t$ gives $\frac{dy}{dt} = \sec^2 t - \csc^2 t$ . Dividing these gives

\frac{dy}{dx} = \frac{\sec^2 t - \csc^2 t}{6\cos 2t}

To simplify, write $\sec^2 t - \csc^2 t$ as a single fraction:

\frac{1}{\cos^2 t} - \frac{1}{\sin^2 t} = \frac{\sin^2 t - \cos^2 t}{\sin^2 t \cos^2 t}

Then use $\sin^2 t - \cos^2 t = -\cos 2t$ and $\sin t \cos t = \frac{1}{2}\sin 2t$ , so $\sin^2 t \cos^2 t = \frac{1}{4}\sin^2 2t$ . Substituting and cancelling the common $\cos 2t$ factor gives the required result.

Key Takeaways

Parametric differentiation uses the chain rule in the form

\frac{dy}{dx} = \frac{dy}{dt} \div \frac{dx}{dt}

Trigonometric identities such as the double-angle formulae are often needed to simplify derivatives. Cancellation is only valid when the factors in the numerator and denominator are identical.

Common Mistakes

Forgetting to divide by $\frac{dx}{dt}$ .
Writing $\frac{dy}{dx}$ as $\frac{dy}{dt}$ or $\frac{dx}{dt}$ .
Failing to combine $\sec^2 t - \csc^2 t$ into a single fraction.
Cancelling $\cos 2t$ before writing the numerator and denominator with identical factors.
Sign errors: $\sin^2 t - \cos^2 t = -\cos 2t$ , not $+\cos 2t$ .

Things to Be Careful About

The mark scheme requires full working; unsupported answers are not enough. At $t = \frac{\pi}{4}$ , both $\frac{dx}{dt}$ and $\frac{dy}{dt}$ are zero, so the quotient form is indeterminate at that single point; the simplified expression is obtained by cancellation and is valid by continuity. Alternatively, use $y = \frac{6}{x}$ to avoid this. Use the correct variable; slips between $t$ , $\theta$ and $x$ can lose the final mark if not corrected.

Techniques used

differentiate x and y with respect to the parameter tuse the parametric derivative formulacombine secant and cosecant terms into a single fractionapply double-angle identitiescancel common factors

(b)

Find the equation of the normal to the curve at the point where $t = \frac{1}{4}\pi$ . Give your answer in the form $py + qx + r = 0$ , where $p, q$ and $r$ are integers.

DifficultyMedium-Easy

Worked solution

Approach

Use the derivative from part (a) at $t = \frac{\pi}{4}$ to find the gradient of the tangent. The normal has gradient equal to the negative reciprocal, then use point-slope form.

Working

At $t = \frac{\pi}{4}$ :

x = 3\sin\frac{\pi}{2} = 3

y = \tan\frac{\pi}{4} + \cot\frac{\pi}{4} = 1 + 1 = 2

From part (a),

\frac{dy}{dx} = \frac{-2}{3\sin^2 2t}

so at $t = \frac{\pi}{4}$ :

\frac{dy}{dx} = \frac{-2}{3\sin^2 \frac{\pi}{2}} = -\frac{2}{3}

Gradient of normal:

m_n = \frac{3}{2}

Equation of normal through $(3,2)$ :

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

Multiply by 2:

2y - 4 = 3x - 9

Rearrange:

2y - 3x + 5 = 0

Answer

2y - 3x + 5 = 0

Final answer

2y - 3x + 5 = 0

Detailed explanation

Walkthrough

At $t = \frac{\pi}{4}$ , compute the coordinates. Since $x = 3\sin 2t$ ,

x = 3\sin\frac{\pi}{2} = 3

and since $y = \tan t + \cot t$ ,

y = \tan\frac{\pi}{4} + \cot\frac{\pi}{4} = 1 + 1 = 2

The point is $(3,2)$ . From part (a), the tangent gradient is

\frac{dy}{dx} = -\frac{2}{3\sin^2 \frac{\pi}{2}} = -\frac{2}{3}

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

m_n = \frac{3}{2}

Using point-slope form $y - y_1 = m(x - x_1)$ with $(3,2)$ :

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

Multiply by 2 and rearrange to integer form:

2y - 4 = 3x - 9 \implies 2y - 3x + 5 = 0

Key Takeaways

To find a normal, take the negative reciprocal of the tangent gradient. Use the point-slope equation of a straight line. The final answer should be in the requested integer form $py + qx + r = 0$ .

Common Mistakes

Using the tangent gradient $-\frac{2}{3}$ instead of the normal gradient $\frac{3}{2}$ .
Using the wrong point, for example substituting $t$ values incorrectly.
Not multiplying through to clear the fraction.
Rearranging with sign errors.

Things to Be Careful About

The mark scheme allows a wrong $x$ or $y$ but not both when using the normal gradient. The final equation must have integer coefficients; any non-zero integer multiple is accepted. Check the point lies on the line: $2(2) - 3(3) + 5 = 4 - 9 + 5 = 0$ .

Techniques used

evaluate the parametric coordinates at the given parameterevaluate the tangent gradient from the derivativetake the negative reciprocal to find the normal gradientform the equation of a straight line

Q8MediumAlgebra

Let $f(x) = \frac{7a^2}{(a - 2x)(3a + x)}$ , where $a$ is a positive constant.

(a)

Express $f(x)$ in partial fractions.

DifficultyMedium-Easy

Worked solution

Approach

Express the given rational function as a sum of partial fractions with denominators $a-2x$ and $3a+x$ . Clear denominators and solve for the constants $A$ and $B$ .

Working

Let

\frac{7a^2}{(a-2x)(3a+x)} = \frac{A}{a-2x} + \frac{B}{3a+x}

Multiplying by $(a-2x)(3a+x)$ :

7a^2 = A(3a+x) + B(a-2x)

Substitute $x = \frac{a}{2}$ so that $a - 2x = 0$ :

7a^2 = A\left(3a + \frac{a}{2}\right) = \frac{7aA}{2}

Hence $A = 2a$ .

Substitute $x = -3a$ so that $3a + x = 0$ :

7a^2 = B(a + 6a) = 7aB

Hence $B = a$ .

Therefore

f(x) = \frac{2a}{a-2x} + \frac{a}{3a+x}

Answer

f(x) = \frac{2a}{a-2x} + \frac{a}{3a+x}

Final answer

f(x) = 2a/(a-2x) + a/(3a+x)

Detailed explanation

Walkthrough

We need to decompose $\frac{7a^2}{(a-2x)(3a+x)}$ into partial fractions. Because the denominator has two distinct linear factors, we write the expression as $\frac{A}{a-2x} + \frac{B}{3a+x}$ . Multiplying through by the denominator gives $7a^2 = A(3a+x) + B(a-2x)$ . To find $A$ , choose $x = \frac{a}{2}$ , which makes the $B$ term vanish; this gives $7a^2 = \frac{7aA}{2}$ , so $A = 2a$ . To find $B$ , choose $x = -3a$ , which makes the $A$ term vanish; this gives $7a^2 = 7aB$ , so $B = a$ . This method is efficient because each substitution isolates one unknown.

Key Takeaways

This question tests the standard partial fraction decomposition for distinct linear factors. It also shows how substituting values that make one factor zero can quickly determine constants without solving simultaneous equations.

Common Mistakes

Forgetting to include both partial fraction terms.
Substituting the wrong value for $x$ when eliminating a term.
Making sign errors when substituting $x = -3a$ .
Not simplifying $A$ and $B$ in terms of $a$ .

Things to Be Careful About

The denominators are $a-2x$ and $3a+x$ , not $x-a$ or $x+3a$ , so the signs in the substitutions must be handled carefully. Since $a$ is a positive constant, no absolute value issues arise, but the constants still contain $a$ .

Techniques used

express a rational function in partial fractionsclear denominatorssubstitute values to find constants

(b)

Hence obtain the expansion of $f(x)$ 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 term in the form $k(1+u)^{-1}$ and expand up to $x^2$ using the binomial series, then combine like terms.

Working

From part (a),

f(x) = \frac{2a}{a-2x} + \frac{a}{3a+x}

Rewrite the first term:

\frac{2a}{a-2x} = 2\left(1 - \frac{2x}{a}\right)^{-1}

Expanding:

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

Rewrite the second term:

\frac{a}{3a+x} = \frac{1}{3}\left(1 + \frac{x}{3a}\right)^{-1}

Expanding:

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

Adding the two expansions:

Constant term:

2 + \frac{1}{3} = \frac{7}{3}

Coefficient of $x$ :

\frac{4}{a} - \frac{1}{9a} = \frac{35}{9a}

Coefficient of $x^2$ :

\frac{8}{a^2} + \frac{1}{27a^2} = \frac{217}{27a^2}

Therefore

f(x) = \frac{7}{3} + \frac{35x}{9a} + \frac{217x^2}{27a^2} + \cdots

Answer

f(x) = \frac{7}{3} + \frac{35x}{9a} + \frac{217x^2}{27a^2}

Final answer

7/3 + 35x/(9a) + 217x^2/(27a^2)

Detailed explanation

Walkthrough

From part (a), $f(x) = \frac{2a}{a-2x} + \frac{a}{3a+x}$ . To expand in ascending powers of $x$ , rewrite each term so that the binomial has first term $1$ . For the first term, $\frac{2a}{a-2x} = 2(1 - \frac{2x}{a})^{-1}$ . Using $(1+u)^{-1} = 1 - u + u^2 - \cdots$ with $u = -\frac{2x}{a}$ , this becomes $2(1 + \frac{2x}{a} + \frac{4x^2}{a^2} + \cdots)$ . For the second term, $\frac{a}{3a+x} = \frac{1}{3}(1 + \frac{x}{3a})^{-1}$ , which expands to $\frac{1}{3}(1 - \frac{x}{3a} + \frac{x^2}{9a^2} - \cdots)$ . Adding the constant terms gives $\frac{7}{3}$ ; adding the $x$ coefficients gives $\frac{35}{9a}$ ; adding the $x^2$ coefficients gives $\frac{217}{27a^2}$ . Thus the expansion is $\frac{7}{3} + \frac{35x}{9a} + \frac{217x^2}{27a^2}$ .

Key Takeaways

This question tests the binomial expansion for negative powers and how to combine two expansions. It also reinforces the importance of rewriting expressions as $k(1+u)^n$ before expanding.

Common Mistakes

Forgetting the factor $2$ or $\frac{1}{3}$ when expanding.
Using the wrong sign in the expansion of $(1 + \frac{x}{3a})^{-1}$ .
Incorrectly adding the $x^2$ coefficients: $\frac{8}{a^2} + \frac{1}{27a^2} = \frac{217}{27a^2}$ , not $\frac{9}{27a^2}$ .
Including higher powers of $x$ in the final answer.

Things to Be Careful About

The expansion of $(1 - \frac{2x}{a})^{-1}$ has all positive signs, while $(1 + \frac{x}{3a})^{-1}$ alternates signs. When adding, align terms of the same power of $x$ . The final answer should contain no terms beyond $x^2$ .

Techniques used

rewrite partial fractions in binomial formexpand binomial series up to x^2collect like terms

(c)

State the set of values of $x$ for which the expansion in part (b) is valid.

1M Similar

DifficultyEasy

Worked solution

Approach

A binomial expansion $(1+u)^n$ is valid for $|u|<1$ . Apply this condition to both factors in the partial fraction expansion and take the stricter restriction.

Working

For the factor $(1 - \frac{2x}{a})^{-1}$ :

\left|\frac{2x}{a}\right| < 1 \Rightarrow |x| < \frac{a}{2}

For the factor $(1 + \frac{x}{3a})^{-1}$ :

\left|\frac{x}{3a}\right| < 1 \Rightarrow |x| < 3a

Since $a > 0$ , the condition $|x| < \frac{a}{2}$ is stricter than $|x| < 3a$ .

Answer

-\frac{a}{2} < x < \frac{a}{2}

or equivalently $|x| < \frac{a}{2}$ .

Final answer

|x| < a/2

Detailed explanation

Walkthrough

For a binomial expansion $(1+u)^n$ , the series is valid when $|u|<1$ . The expansion in part (b) contains two binomial factors: $(1 - \frac{2x}{a})^{-1}$ and $(1 + \frac{x}{3a})^{-1}$ . The first requires $\left|\frac{2x}{a}\right|<1$ , i.e. $|x|<\frac{a}{2}$ . The second requires $\left|\frac{x}{3a}\right|<1$ , i.e. $|x|<3a$ . Since $a>0$ , $\frac{a}{2}<3a$ , so the stricter condition is $|x|<\frac{a}{2}$ . Therefore the expansion is valid for $-\frac{a}{2}<x<\frac{a}{2}$ .

Key Takeaways

This question tests the validity condition for binomial expansions with rational or negative powers. When several expansions are multiplied or added, the overall validity interval is the intersection of the individual intervals.

Common Mistakes

Only applying the condition to one factor and forgetting the other.
Choosing $|x|<3a$ instead of the stricter $|x|<\frac{a}{2}$ .
Omitting the absolute value or writing a one-sided inequality.

Things to Be Careful About

The condition must be stated clearly as a set of values. Because $a$ is positive, $\frac{a}{2}<3a$ ; if $a$ were negative the comparison would differ. The final answer can be given as $|x|<\frac{a}{2}$ or $-\frac{a}{2}<x<\frac{a}{2}$ .

Techniques used

apply binomial validity conditiontake intersection of intervalsstate interval of convergence

Q9MediumAlgebraIntegration

(a)

Find the quotient and remainder when $x^4 + 16$ is divided by $x^2 + 4$ .

DifficultyMedium-Easy

Worked solution

Approach

Divide the leading term of $x^4+16$ by the leading term of $x^2+4$ , multiply back, subtract, and repeat until the remainder has degree less than 2.

Working

x^4+16 - x^2(x^2+4) = -4x^2+16

-4x^2+16 - (-4)(x^2+4) = 32

Since the remainder $32$ has degree less than the divisor $x^2+4$ , the division stops. Therefore

x^4+16 = (x^2+4)(x^2-4) + 32

So the quotient is $x^2-4$ and the remainder is $32$ .

Answer

Quotient: $x^2-4$ ; remainder: $32$ .

Final answer

Quotient x^2 - 4; remainder 32

Detailed explanation

Walkthrough

We are dividing $x^4+16$ by $x^2+4$ . The first term of the quotient is $x^2$ , because $x^4 \div x^2 = x^2$ . Multiplying back gives $x^4+4x^2$ , and subtracting this from $x^4+16$ leaves $-4x^2+16$ . The next term is $-4$ , because $-4x^2 \div x^2 = -4$ . Multiplying back gives $-4x^2-16$ , and subtracting leaves $32$ . Since $32$ has degree $0$ , which is less than the degree $2$ of the divisor, we stop. Hence $x^4+16 = (x^2+4)(x^2-4)+32$ , so the quotient is $x^2-4$ and the remainder is $32$ .

Key Takeaways

Polynomial division lets us rewrite a rational expression as a polynomial plus a proper remainder. This is especially useful before integrating a rational function. The quotient and remainder can be checked by expanding $(x^2+4)(x^2-4)+32$ .

Common Mistakes

Students often forget the missing $0x^3$ , $0x^2$ and $0x$ terms when setting up long division. Sign errors are common when subtracting, especially with $-4x^2+16 - (-4x^2-16)$ . The mark scheme also penalises labelling a correct quotient and remainder incorrectly: if the quotient and remainder are stated wrongly, at most 2 of the 3 marks can be awarded.

Things to Be Careful About

The divisor is $x^2+4$ , not $x^2-4$ . The remainder must have degree less than the divisor, so $32$ is the final remainder. If using the alternative equating-coefficients method, expand $(x^2+4)(x^2+Bx+C)+D$ and compare coefficients carefully.

Techniques used

perform polynomial long divisionexpress dividend as divisor times quotient plus remainder

(b)

Hence show that

\int_2^{2\sqrt{3}} \frac{x^4 + 16}{x^2 + 4} \,dx = \frac{4}{3}(\pi + 4)

DifficultyMedium

Worked solution

Approach

Use the result of part (a) to rewrite the integrand as $x^2-4+\frac{32}{x^2+4}$ . Integrate term by term, using the standard integral $\int \frac{dx}{x^2+a^2} = \frac{1}{a}\tan^{-1}\left(\frac{x}{a}\right)$ . Then evaluate the definite integral between $2$ and $2\sqrt{3}$ .

Working

\frac{x^4+16}{x^2+4} = x^2 - 4 + \frac{32}{x^2+4}

\int \frac{x^4+16}{x^2+4}\,dx = \int \left(x^2 - 4 + \frac{32}{x^2+4}\right)dx

= \frac{x^3}{3} - 4x + 32\cdot \frac{1}{2}\tan^{-1}\frac{x}{2} + C

= \frac{x^3}{3} - 4x + 16\tan^{-1}\frac{x}{2} + C

At $x=2\sqrt{3}$ :

\frac{(2\sqrt{3})^3}{3} - 4(2\sqrt{3}) + 16\tan^{-1}\sqrt{3} = 8\sqrt{3} - 8\sqrt{3} + \frac{16\pi}{3} = \frac{16\pi}{3}

At $x=2$ :

\frac{2^3}{3} - 4(2) + 16\tan^{-1}1 = \frac{8}{3} - 8 + 4\pi = -\frac{16}{3} + 4\pi

Subtracting the lower value from the upper value:

\int_2^{2\sqrt{3}} \frac{x^4+16}{x^2+4}\,dx = \frac{16\pi}{3} - \left(-\frac{16}{3} + 4\pi\right) = \frac{16\pi}{3} - 4\pi + \frac{16}{3} = \frac{4\pi + 16}{3} = \frac{4}{3}(\pi + 4)

Answer

\frac{4}{3}(\pi + 4)

Final answer

4/3(π + 4)

Detailed explanation

Walkthrough

Part (a) gives $x^4+16 = (x^2+4)(x^2-4)+32$ , so dividing by $x^2+4$ gives $x^2-4+\frac{32}{x^2+4}$ . This is the key step: it turns a quartic-over-quadratic into a simple polynomial plus a term we can integrate. The polynomial part integrates to $\frac{x^3}{3}-4x$ . For $\frac{32}{x^2+4}$ , use $\int \frac{dx}{x^2+a^2} = \frac{1}{a}\tan^{-1}\frac{x}{a}$ with $a=2$ , so the integral is $32 \cdot \frac{1}{2}\tan^{-1}\frac{x}{2} = 16\tan^{-1}\frac{x}{2}$ . Evaluating at the upper limit $2\sqrt{3}$ , the $8\sqrt{3}$ and $-8\sqrt{3}$ cancel, leaving $16\tan^{-1}\sqrt{3} = \frac{16\pi}{3}$ . At the lower limit $2$ , the value is $\frac{8}{3}-8+16\tan^{-1}1 = -\frac{16}{3}+4\pi$ . Subtracting the lower value from the upper value gives $\frac{16\pi}{3} - (-\frac{16}{3}+4\pi) = \frac{4\pi+16}{3} = \frac{4}{3}(\pi+4)$ .

Key Takeaways

A rational function can often be simplified by polynomial division before integration. The integral $\int \frac{dx}{x^2+a^2}$ produces an inverse tangent, and the factor $\frac{1}{a}$ must not be forgotten. Exact values of $\tan^{-1}\sqrt{3}$ and $\tan^{-1}1$ are needed to obtain the final exact answer.

Common Mistakes

The most common error is forgetting the $\frac{1}{2}$ factor when integrating $\frac{32}{x^2+4}$ : the antiderivative is $16\tan^{-1}\frac{x}{2}$ , not $32\tan^{-1}\frac{x}{2}$ . Another common error is subtracting the lower limit incorrectly: the whole lower value, including $4\pi$ , must be subtracted. The mark scheme states this is an AG (answer given) question, so the final result must be obtained from full and correct working; an unsupported answer is not enough.

Things to Be Careful About

Use the exact values $\tan^{-1}\sqrt{3} = \frac{\pi}{3}$ and $\tan^{-1}1 = \frac{\pi}{4}$ . Note that $\frac{8}{3}-8 = -\frac{16}{3}$ . Keep the limits in the correct order: evaluate at $2\sqrt{3}$ first and subtract the value at $2$ . Finally, combine $\frac{16\pi}{3}-4\pi$ as $\frac{4\pi}{3}$ and add $\frac{16}{3}$ to get $\frac{4(\pi+4)}{3}$ .

Techniques used

rewrite integrand using quotient and remainderintegrate polynomial termsuse standard inverse tangent integralevaluate definite integral at limitssimplify exact values

Q10MediumDifferential Equations

A water tank is in the shape of a cuboid with base area $40\,000\text{ cm}^2$ . At time $t$ minutes the depth of water in the tank is $h\text{ cm}$ . Water is pumped into the tank at a rate of $50\,000\text{ cm}^3$ per minute. Water is leaking out of the tank through a hole in the bottom at a rate of $600h\text{ cm}^3$ per minute.

(a)

Show that $200\frac{dh}{dt} = 250 - 3h$ .

DifficultyMedium-Easy

Worked solution

Approach

Let $V$ be the volume of water in the tank. The net rate of change of $V$ is the inflow minus the leakage.

Working

\frac{dV}{dt} = 50000 - 600h.

Since the tank is a cuboid with base area $40\,000\text{ cm}^2$ ,

V = 40000h, \quad \frac{dV}{dh} = 40000.

By the chain rule,

\frac{dV}{dt} = \frac{dV}{dh}\frac{dh}{dt} = 40000\frac{dh}{dt}.

Equating the two expressions for $\frac{dV}{dt}$ ,

40000\frac{dh}{dt} = 50000 - 600h.

Divide by $200$ :

200\frac{dh}{dt} =250 - 3h.

Answer

\boxed{200\frac{dh}{dt} =250 - 3h}

Final answer

200 dh/dt = 250 - 3h

Detailed explanation

Walkthrough

We need to connect the rate at which the depth $h$ changes to the rates given in the problem. Start by letting $V$ be the volume of water. The tank gains $50\,000\text{ cm}^3$ per minute from the pump and loses $600h\text{ cm}^3$ per minute through the hole, so the net rate is $\frac{dV}{dt} = 50000 - 600h$ .

Because the tank is a cuboid with base area $40\,000\text{ cm}^2$ , the volume is $V = 40000h$ . Differentiating with respect to $h$ gives $\frac{dV}{dh} = 40000$ . The chain rule links the rates:

\frac{dV}{dt} = \frac{dV}{dh}\frac{dh}{dt}.

Substituting $\frac{dV}{dh} = 40000$ and $\frac{dV}{dt} = 50000 - 600h$ gives

40000\frac{dh}{dt} = 50000 - 600h.

Finally divide every term by $200$ : $40000/200 = 200$ , $50000/200 = 250$ , $600/200 = 3$ , producing the required equation.

Key Takeaways

A rate-of-change statement can be translated directly into a differential equation by identifying the net rate of change.
For a cuboid, $V = \text{base area} \times h$ , so $\frac{dV}{dh}$ is constant.
The chain rule is essential when relating $\frac{dV}{dt}$ to $\frac{dh}{dt}$ .

Common Mistakes

Forgetting that the leak rate depends on $h$ , so it must be $600h$ , not a constant.
Using $\frac{dh}{dt} = \frac{dV}{dh} \div \frac{dV}{dt}$ instead of the correct chain-rule division.
Failing to divide by $200$ at the end.

Things to Be Careful About

The units are consistent: volumes in $\text{cm}^3$ , time in minutes, depth in cm.
The net rate $\frac{dV}{dt}$ can be positive or negative depending on $h$ ; here at $h=50$ it is positive, so depth initially increases.
The mark scheme requires the full chain-rule step to earn the method mark; do not jump straight to the final equation.

Techniques used

formulate net rate of change of volumeapply chain rule with volume-depth relationshipsimplify differential equation

(b)

It is given that when $t = 0$ , $h = 50$ .

Find the time taken for the depth of water in the tank to reach $80\text{ cm}$ . Give your answer correct to 2 significant figures.

DifficultyMedium

Worked solution

Approach

Separate the variables so that $h$ appears with $dh$ and $t$ appears with $dt$ , then integrate both sides. Use the initial condition $t=0, h=50$ to find the constant, then substitute $h=80$ and solve for $t$ .

Working

Separating variables gives

\int \frac{1}{250-3h}\,dh = \int \frac{1}{200}\,dt.

Integrating,

-\frac{1}{3}\ln|250-3h| = \frac{t}{200} + C.

At $t=0$ , $h=50$ :

-\frac{1}{3}\ln|250-150| = C \quad \Rightarrow \quad C = -\frac{1}{3}\ln 100.

So

-\frac{1}{3}\ln|250-3h| = \frac{t}{200} - \frac{1}{3}\ln 100.

When $h=80$ , $250-3h=10$ , so

-\frac{1}{3}\ln 10 = \frac{t}{200} - \frac{1}{3}\ln 100.

Rearrange:

\frac{t}{200} = \frac{1}{3}\left(\ln 100 - \ln 10\right) = \frac{1}{3}\ln 10.

Therefore

t = \frac{200}{3}\ln 10 \approx 153.5\ldots \approx 150\text{ minutes}.

Answer

\boxed{t = 150\text{ minutes}}

Final answer

t = 150 minutes (2 s.f.)

Detailed explanation

Walkthrough

The differential equation from part (a) is

200\frac{dh}{dt} =250 - 3h.

To solve it, separate the variables: put all $h$ -terms on one side with $dh$ , and all $t$ -terms on the other side with $dt$ .

\int \frac{1}{250-3h}\,dh = \int \frac{1}{200}\,dt.

Integrate the left side. Since the derivative of $250-3h$ is $-3$ , dividing by $-3$ gives $-\frac{1}{3}\ln|250-3h|$ . The right side integrates to $\frac{t}{200}$ .

Use $t=0, h=50$ :

-\frac{1}{3}\ln|250-150| = C,

so $C = -\frac{1}{3}\ln 100$ . Then substitute the target depth $h=80$ . Since $250-3(80)=10$ , we get

-\frac{1}{3}\ln 10 = \frac{t}{200} - \frac{1}{3}\ln 100.

Rearrange using $\ln 100 - \ln 10 = \ln(100/10) = \ln 10$ :

\frac{t}{200} = \frac{1}{3}\ln 10.

So $t = \frac{200}{3}\ln 10 \approx153.5$ minutes, which is $150$ minutes correct to 2 significant figures.

Key Takeaways

Separable first-order differential equations are solved by collecting each variable with its own differential and integrating.
The constant of integration is found from an initial condition.
Logarithm laws allow the final expression to be simplified before evaluating.

Common Mistakes

Forgetting the factor $-\frac{1}{3}$ when integrating $\frac{1}{250-3h}$ .
Omitting the constant of integration before using the initial condition.
Rounding $153.5$ to $150$ incorrectly;to 2 significant figures the first significant figure is 1, so the number is $1.5\times 10^2$ and becomes $150$ .
Using the limits incorrectly if attempting a definite-integral method.

Things to Be Careful About

The modulus signs in $\ln|250-3h|$ are mathematically important, though the mark scheme condones their omission.
Here $250-3h>0$ for the values used ( $h=50$ and $h=80$ ), so no sign issue arises.
The answer must be given to 2 significant figures; do not give $153.5$ or $154$ without rounding to the required accuracy.
The mark scheme accepts equivalent logarithmic forms, but the final numerical time must be $150$ .

Techniques used

separate variables and integrateuse initial condition to find constantsolve for time using logarithms

Q11Medium-HardDifferentiationTrigonometryIntegration

Fig. 11

The diagram shows the curve $y = 2 \sin x \sqrt{2 + \cos x}$ , for $0 \leq x \leq 2\pi$ , and its minimum point $M$ , where $x = a$ .

(a)

Find the value of $a$ correct to 2 decimal places.