The variables $x$ and $y$ satisfy the equation $Ay = b^x$, where $A$ and $b$ are constants. (a) Show that the graph of $\ln y$ against $x$ is a straight line. (b) When $x = 3.4$, $\ln y = 0.86$ and when $x = 5.7$, $\ln y = 2.56$. Find the value of $A$ and the value of $b$. Give your answers correct to 2 significant figures.

(a) ln y = (ln b)x - ln A, so the graph is a straight line. · (b) A = 5.2, b = 2.1

The constant $a$ is such that $\int_0^a x e^{\frac{1}{2}x} \,dx = 6$. (a) Show that $a = 2 + e^{-\frac{1}{2}a}$. (b) Verify by calculation that $a$ lies between 2.2 and 2.4. (c) Use an iterative formula based on the equation in part (a) to determine $a$ correct to 2 decimal places. Give the result of each iteration to 4 decimal places.

(a) a = 2 + e^(-a/2) · (b) a lies between 2.2 and 2.4 · (c) 2.31

9709/35 Oct/Nov 2025 — Mathematics Solved Past Paper

Q1Medium-EasyAlgebra

(a)

Sketch the graph of $y = |3x - 6|$ .

1M Similar

DifficultyEasy

Worked solution

Approach

The graph of $y = |3x - 6|$ is V-shaped. The vertex is where $3x - 6 = 0$ , i.e., $x = 2$ . The y-intercept is found by setting $x = 0$ .

Working

The expression $3x - 6 = 0$ when $x = 2$ , so the vertex is at $(2, 0)$ .

When $x = 0$ , $y = |3(0) - 6| = |-6| = 6$ , so the y-intercept is $(0, 6)$ .

For $x \geq 2$ , $y = 3x - 6$ (a straight line with gradient 3).

For $x < 2$ , $y = -(3x - 6) = 6 - 3x$ (a straight line with gradient $-3$ ).

The graph is V-shaped with vertex at $(2, 0)$ , crossing the y-axis at $(0, 6)$ , and the left arm extends into the second quadrant.

Answer

The graph is a V-shape with vertex at $(2, 0)$ and y-intercept at $(0, 6)$ .

Final answer

V-shaped graph with vertex at (2, 0) and y-intercept at (0, 6)

Detailed explanation

Walkthrough

The modulus function $y = |3x - 6|$ can be written as a piecewise function:

y = \begin{cases} 3x - 6 & \text{if } x \geq 2 \\ 6 - 3x & \text{if } x < 2 \end{cases}

The vertex (minimum point) occurs where the expression inside the modulus is zero: $3x - 6 = 0 \Rightarrow x = 2$ . At this point $y = 0$ , giving the vertex $(2, 0)$ .

To find the y-intercept, substitute $x = 0$ : $y = |3(0) - 6| = |-6| = 6$ , giving the point $(0, 6)$ .

For $x \geq 2$ , the graph is the straight line $y = 3x - 6$ with gradient 3. For $x < 2$ , the graph is $y = 6 - 3x$ with gradient $-3$ . The two arms meet at the vertex, forming a V-shape. The left arm (gradient $-3$ ) extends into the second quadrant since it passes through $(0, 6)$ and continues upward to the left.

Key Takeaways

The graph of $y = |ax + b|$ is always V-shaped with its vertex on the x-axis at $x = -b/a$ .
The vertex is found by setting the expression inside the modulus equal to zero.
The y-intercept is found by evaluating $|b|$ when $x = 0$ .
The two arms have gradients $+a$ and $-a$ .

Common Mistakes

Forgetting that the vertex is on the x-axis (y-coordinate is 0).
Plotting the wrong y-intercept (e.g., $-6$ instead of $6$ ).
Drawing a smooth curve instead of straight lines.
Not extending the left arm into the second quadrant.

Things to Be Careful About

The graph must consist of straight lines, not a curve.
The vertex must be exactly at $(2, 0)$ , not at some other point.
The y-intercept is $+6$ , not $-6$ .
The left arm must extend into the second quadrant (up and to the left from the y-axis).

Techniques used

identify vertex of modulus graphplot y-interceptsketch V-shaped graph

(b)

Solve the inequality $5x - 3 < |3x - 6|$ .

3M Similar

DifficultyMedium-Easy

Worked solution

Approach

The inequality $5x - 3 < |3x - 6|$ can be solved by considering two cases based on the sign of $3x - 6$ . Alternatively, one can solve the boundary equation $5x - 3 = |3x - 6|$ and then determine which side of the critical value satisfies the inequality.

Working

Case 1: $3x - 6 \geq 0$ , i.e., $x \geq 2$

Here $|3x - 6| = 3x - 6$ , so the inequality becomes:

5x - 3 < 3x - 6

2x < -3

x < -\frac{3}{2}

This contradicts the condition $x \geq 2$ , so there is no solution in this case.

Case 2: $3x - 6 < 0$ , i.e., $x < 2$

Here $|3x - 6| = 6 - 3x$ , so the inequality becomes:

5x - 3 < 6 - 3x

8x < 9

x < \frac{9}{8}

Since $\frac{9}{8} < 2$ , this is consistent with the case condition $x < 2$ .

Alternative method (squaring both sides):

Since both sides must be considered, note that $5x - 3 < |3x - 6|$ is satisfied when $5x - 3 < 0$ (i.e., $x < \frac{3}{5}$ ) or when $(5x - 3)^2 < (3x - 6)^2$ .

Solving $(5x - 3)^2 = (3x - 6)^2$ :

25x^2 - 30x + 9 = 9x^2 - 36x + 36

16x^2 + 6x - 27 = 0

(8x - 9)(2x + 3) = 0

Giving $x = \frac{9}{8}$ or $x = -\frac{3}{2}$ . The relevant critical value is $x = \frac{9}{8}$ , and testing shows the inequality holds for $x < \frac{9}{8}$ .

Answer

x < \frac{9}{8}

Final answer

x < 9/8

Detailed explanation

Walkthrough

We need to solve $5x - 3 < |3x - 6|$ . The modulus $|3x - 6|$ changes behavior at $x = 2$ , so we split into two cases.

Case 1: $x \geq 2$

Here $3x - 6 \geq 0$ , so $|3x - 6| = 3x - 6$ . The inequality becomes:

5x - 3 < 3x - 6

Subtracting $3x$ from both sides: $2x - 3 < -6$ . Adding 3: $2x < -3$ . Dividing by 2: $x < -\frac{3}{2}$ .

But we assumed $x \geq 2$ , and $x < -\frac{3}{2}$ contradicts this. So no solution from this case.

Case 2: $x < 2$

Here $3x - 6 < 0$ , so $|3x - 6| = -(3x - 6) = 6 - 3x$ . The inequality becomes:

5x - 3 < 6 - 3x

Adding $3x$ to both sides: $8x - 3 < 6$ . Adding 3: $8x < 9$ . Dividing by 8: $x < \frac{9}{8}$ .

Since $\frac{9}{8} = 1.125 < 2$ , this is fully consistent with our case condition $x < 2$ . So the solution from this case is $x < \frac{9}{8}$ .

Combining: The overall solution is $x < \frac{9}{8}$ .

Key Takeaways

When solving modulus inequalities, split into cases based on where the expression inside the modulus changes sign.
Always check that solutions are consistent with the case condition.
The critical value from the relevant case gives the boundary of the solution.
Solutions from one case may be empty if they contradict the case condition.

Common Mistakes

Forgetting to check that solutions satisfy the case condition (e.g., accepting $x < -\frac{3}{2}$ from Case 1 without noting it contradicts $x \geq 2$ ).
Incorrectly removing the modulus (e.g., writing $5x - 3 < 3x - 6$ in Case 2 instead of $5x - 3 < 6 - 3x$ ).
Sign errors when distributing the negative sign in $-(3x - 6)$ .
Not combining cases correctly at the end.

Things to Be Careful About

The solution must be checked against the case conditions.
Only one case typically gives a valid solution in simple modulus inequalities.
The final answer is just $x < \frac{9}{8}$ — no other answer is valid.
If using the squaring method, both roots $x = \frac{9}{8}$ and $x = -\frac{3}{2}$ appear, but only $x = \frac{9}{8}$ is the relevant critical value for the inequality direction.

Techniques used

split modulus inequality into casessolve resulting linear inequalitycheck against case condition

Q24MMedium-EasyComplex Numbers

On a sketch of an Argand diagram, shade the region which represents complex numbers $z$ satisfying both the inequalities $|z - 1 - 3i| \leq 2$ and $|z| \geq 4$ .

DifficultyMedium-Easy

Worked solution

Approach

The inequality $|z - (1 + 3\mathrm{i})| \leq 2$ represents all points $z$ whose distance from the point $1 + 3\mathrm{i}$ is at most 2. This is a closed disk (the interior and boundary of a circle) with centre $(1, 3)$ and radius 2.

The inequality $|z| \geq 4$ represents all points $z$ whose distance from the origin is at least 4. This is the region outside (and on the boundary of) a circle with centre $(0, 0)$ and radius 4.

We sketch both circles on an Argand diagram and shade the region that satisfies both conditions simultaneously.

Working

|z - (1 + 3\mathrm{i})| \leq 2

This is a circle with centre $(1, 3)$ and radius 2. The circle passes through $(1, 1)$ , $(1, 5)$ , $(-1, 3)$ , and $(3, 3)$ .

|z| \geq 4

This is the region outside a circle with centre $(0, 0)$ and radius 4. This circle intersects the axes at $4$ , $4\mathrm{i}$ , $-4$ , and $-4\mathrm{i}$ .

Answer

The shaded region is the part of the disk $|z - 1 - 3\mathrm{i}| \leq 2$ that lies outside or on the circle $|z| = 4$ .

Final answer

The region inside the circle |z - 1 - 3i| = 2 and outside the circle |z| = 4.

Detailed explanation

Walkthrough

First, we interpret the algebraic inequalities geometrically on the Argand diagram.

The expression $|z - z_0|$ represents the distance between the complex number $z$ and the fixed complex number $z_0$ . Therefore, $|z - (1 + 3\mathrm{i})| \leq 2$ describes all points that are at a distance of 2 or less from the point $(1, 3)$ . This defines a solid disk (the interior and boundary) of a circle centred at $(1, 3)$ with radius 2.

Next, $|z|$ is the distance from $z$ to the origin $(0, 0)$ . The inequality $|z| \geq 4$ describes all points that are at a distance of 4 or more from the origin. This defines the region outside (and including the boundary of) a circle centred at the origin with radius 4.

To find the region satisfying both inequalities, we draw both circles on the same Argand diagram. The smaller circle is mostly contained within the larger circle, but a small segment near the top (around $3\mathrm{i}$ ) extends outside the larger circle. We shade only the part of the smaller disk that lies outside the larger circle, as this is the intersection of the two regions.

Key Takeaways

The locus $|z - z_0| = r$ is a circle with centre $z_0$ and radius $r$ on the Argand diagram.
$|z - z_0| \leq r$ represents the interior and boundary of that circle, while $|z - z_0| \geq r$ represents the exterior.
$|z| = r$ is a circle centred at the origin with radius $r$ .
To solve simultaneous modulus inequalities, sketch the loci and shade the overlapping region.

Common Mistakes

Confusing $|z - z_0| \leq r$ with the exterior of the circle instead of the interior.
Forgetting to include the boundary of the circles when the inequality is non-strict ( $\leq$ or $\geq$ ).
Drawing the centre of the first circle at the origin instead of at $(1, 3)$ .
Shading the region inside both circles instead of the intersection of the interior of the small circle and the exterior of the large circle.

Things to Be Careful About

Ensure the Argand diagram has a clear indication of scale on both the real and imaginary axes, as the mark scheme requires some indication of scale to be seen or implied.
The centre of the first circle is $1 + 3\mathrm{i}$ , which corresponds to the point $(1, 3)$ , not $(3, 1)$ .
The shaded region is a small cap of the smaller circle that protrudes outside the larger circle. Verify that the shaded area is indeed outside $|z| = 4$ .

Techniques used

interpret modulus inequalities geometricallysketch circles on Argand diagramidentify intersection region

Approach

Take natural logarithms of both sides of $Ay = b^x$ , then use the laws of logarithms to rearrange into the form $Y = mX + c$ , where $Y = \ln y$ and $X = x$ .

Working

\ln(Ay) = \ln\left(b^x\right)

Using $\ln(uv) = \ln u + \ln v$ and $\ln\left(b^x\right) = x\ln b$ :

\ln A + \ln y = x\ln b

Rearrange to make $\ln y$ the subject:

\ln y = (\ln b)x - \ln A

This has the form $Y = mX + c$ with $Y = \ln y$ , $X = x$ , gradient $m = \ln b$ and intercept $c = -\ln A$ . Hence the graph of $\ln y$ against $x$ is a straight line.

Answer

\ln y = (\ln b)x - \ln A

so the graph is a straight line with gradient $\ln b$ and vertical intercept $-\ln A$ .

Final answer

ln y = (ln b)x - ln A, so the graph is a straight line.

Detailed explanation

Walkthrough

We start with $Ay = b^x$ . The aim is to see how $\ln y$ depends on $x$ . Taking $\ln$ of both sides is useful because it turns the power $b^x$ into a multiplication $x\ln b$ , and it turns the product $Ay$ into a sum $\ln A + \ln y$ . This is exactly what is needed to get a linear expression. After rearranging, $\ln y = (\ln b)x - \ln A$ , which matches $Y = mX + c$ . So the graph is a straight line.

Key Takeaways

Logarithms convert products into sums and powers into products.
A relationship of the form $y = Ab^x$ can be linearised by plotting $\ln y$ against $x$ .
The gradient of the line is $\ln b$ and the intercept is $-\ln A$ .

Common Mistakes

Forgetting to take $\ln$ of both sides correctly, e.g. writing $\ln(Ay) = \ln A \ln y$ .
Not applying $\ln\left(b^x\right) = x\ln b$ .
Misidentifying the intercept: the intercept is $-\ln A$ , not $\ln A$ .

Things to Be Careful About

The base of the logarithm is not specified, but natural logarithms are intended because part (b) uses $\ln y$ .
The statement "graph of $\ln y$ against $x$ " means $x$ is the horizontal axis and $\ln y$ is the vertical axis; the line has gradient $\ln b$ .

Techniques used

take logarithms of both sidesapply logarithm lawsidentify the linear form

(b)

When $x = 3.4$ , $\ln y = 0.86$ and when $x = 5.7$ , $\ln y = 2.56$ .

Find the value of $A$ and the value of $b$ . Give your answers correct to 2 significant figures.

3M Similar

DifficultyMedium-Easy

Worked solution

Approach

Use the two given points $(x, \ln y)$ in the linear equation from part (a). This gives two simultaneous equations in $\ln A$ and $\ln b$ . Solve them, then exponentiate to find $A$ and $b$ .

Working

From part (a), $\ln y = x\ln b - \ln A$ .

Substitute $x = 3.4$ , $\ln y = 0.86$ :

0.86 = 3.4\ln b - \ln A

Substitute $x = 5.7$ , $\ln y = 2.56$ :

2.56 = 5.7\ln b - \ln A

Subtract the first equation from the second:

2.56 - 0.86 = (5.7 - 3.4)\ln b

1.70 = 2.3\ln b

\ln b = \frac{1.70}{2.3} = 0.739130\ldots

b = e^{0.739130\ldots} = 2.0939\ldots \approx 2.1 \quad (2\text{ sf})

Now use $0.86 = 3.4\ln b - \ln A$ :

\ln A = 3.4\ln b - 0.86

\ln A = 3.4(0.739130\ldots) - 0.86 = 1.65304\ldots

A = e^{1.65304\ldots} = 5.222\ldots \approx 5.2 \quad (2\text{ sf})

Answer

A = 5.2, \quad b = 2.1

(both correct to 2 significant figures).

Final answer

A = 5.2, b = 2.1

Detailed explanation

Walkthrough

We have two points on the straight line: $(x,\ln y) = (3.4,0.86)$ and $(5.7,2.56)$ . Substitute into $\ln y = x\ln b - \ln A$ to get two equations. Subtract to eliminate $\ln A$ , giving $1.70 = 2.3\ln b$ . Solve for $\ln b$ , then exponentiate to get $b$ . Then substitute back to find $\ln A$ and exponentiate to get $A$ . The values are $b \approx 2.1$ and $A \approx 5.2$ .

Key Takeaways

Two points on a straight line give two simultaneous equations for the unknown constants.
The slope of the line is $\ln b$ , so $b$ is found by exponentiating the slope.
The intercept gives $-\ln A$ , so $A$ is found by exponentiating the negative intercept.

Common Mistakes

Forgetting to exponentiate at the end; leaving answers as $\ln b$ and $\ln A$ .
Subtracting equations in the wrong order, leading to sign errors.
Rounding too early; keep more decimal places until the final answer.

Things to Be Careful About

The answers must be given to 2 significant figures: $A = 5.2$ , $b = 2.1$ . $A = 5.3$ is also accepted.
When substituting back, use the unrounded value of $\ln b$ to avoid rounding error in $A$ .
The equations can also be written as $Ae^{0.86} = b^{3.4}$ and $Ae^{2.56} = b^{5.7}$ ; either approach is valid.

Techniques used

substitute data points into the linearised equationsolve simultaneous linear equationsexponentiate to recover the constants

Q45MMedium-HardDifferentiation

The equation of a curve is $x^2 \ln 2y - y \ln(2 + x^2) = \ln 6$ .

Find the exact value of the gradient of the curve at the point $(2, 3)$ . Give your answer in simplified form.

DifficultyMedium-Hard

Worked solution

Approach

Differentiate both sides of the equation implicitly with respect to $x$ , using the product rule for the two product terms $x^2\ln(2y)$ and $y\ln(2+x^2)$ . Then substitute $x=2$ , $y=3$ and solve for $\frac{\mathrm{d}y}{\mathrm{d}x}$ .

Working

Differentiate $x^2\ln(2y)$ using the product rule:

\frac{\mathrm{d}}{\mathrm{d}x}(x^2\ln(2y)) = 2x\ln(2y) + x^2\cdot \frac{1}{y}\frac{\mathrm{d}y}{\mathrm{d}x}

Differentiate $y\ln(2+x^2)$ using the product rule:

\frac{\mathrm{d}}{\mathrm{d}x}(y\ln(2+x^2)) = \frac{\mathrm{d}y}{\mathrm{d}x}\ln(2+x^2) + y\cdot \frac{2x}{2+x^2}

Since the right-hand side $\ln 6$ is constant, its derivative is zero. Therefore:

2x\ln(2y) + \frac{x^2}{y}\frac{\mathrm{d}y}{\mathrm{d}x} - \left(\ln(2+x^2)\frac{\mathrm{d}y}{\mathrm{d}x} + \frac{2xy}{2+x^2}\right) = 0

Substitute $x = 2$ , $y = 3$ :

4\ln 6 + \frac{4}{3}\frac{\mathrm{d}y}{\mathrm{d}x} - \ln 6 \frac{\mathrm{d}y}{\mathrm{d}x} - 2 = 0

Rearrange to collect the $\frac{\mathrm{d}y}{\mathrm{d}x}$ terms:

\left(\frac{4}{3} - \ln 6\right)\frac{\mathrm{d}y}{\mathrm{d}x} = 2 - 4\ln 6

Hence

\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{2 - 4\ln 6}{\frac{4}{3} - \ln 6} = \frac{6 - 12\ln 6}{4 - 3\ln 6}

Answer

\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{6 - 12\ln 6}{4 - 3\ln 6}

Final answer

(6 - 12 ln 6)/(4 - 3 ln 6)

Detailed explanation

Walkthrough

This question is about finding the gradient of a curve that is defined implicitly, so the derivative must be taken with respect to $x$ on both sides of the equation. The left-hand side contains two products, so each product needs the product rule.

For the first term $x^2\ln(2y)$ , the product rule says to differentiate $x^2$ and $\ln(2y)$ separately. The derivative of $x^2$ is $2x$ , and the derivative of $\ln(2y)$ with respect to $x$ is $\frac{1}{y}\frac{\mathrm{d}y}{\mathrm{d}x}$ , because the chain rule introduces the factor $\frac{\mathrm{d}y}{\mathrm{d}x}$ . This gives $2x\ln(2y) + x^2\cdot \frac{1}{y}\frac{\mathrm{d}y}{\mathrm{d}x}$ .

For the second term $y\ln(2+x^2)$ , the product rule gives $\frac{\mathrm{d}y}{\mathrm{d}x}\ln(2+x^2) + y\cdot \frac{2x}{2+x^2}$ . The derivative of $\ln(2+x^2)$ is $\frac{2x}{2+x^2}$ by the chain rule.

The right-hand side is the constant $\ln 6$ , whose derivative is $0$ . Substituting $x=2$ and $y=3$ turns $\ln(2y)$ into $\ln 6$ and $\ln(2+x^2)$ into $\ln 6$ . This gives a linear equation in $\frac{\mathrm{d}y}{\mathrm{d}x}$ , which is then solved. Finally, the fraction is simplified so that there is no fraction inside a fraction.

Key Takeaways

Implicit differentiation: differentiate every term with respect to $x$ , multiplying by $\frac{\mathrm{d}y}{\mathrm{d}x}$ whenever an expression involving $y$ is differentiated.
The product rule is essential when a term is a product of two functions, such as $x^2\ln(2y)$ or $y\ln(2+x^2)$ .
Substitute the given point only after differentiating, before solving for $\frac{\mathrm{d}y}{\mathrm{d}x}$ .
The final answer should be simplified so that it does not contain a fraction within a fraction.

Common Mistakes

Forgetting the factor $\frac{\mathrm{d}y}{\mathrm{d}x}$ when differentiating $\ln(2y)$ or the $y$ in the second product.
Applying the product rule to only one of the two terms.
Substituting $x=2$ , $y=3$ before differentiating, which is invalid.
Leaving the gradient as $\frac{2-4\ln 6}{\frac{4}{3}-\ln 6}$ , which contains a fraction within a fraction and may lose the final mark.
Making a sign error when subtracting the derivative of $y\ln(2+x^2)$ .

Things to Be Careful About

The derivative of $\ln(2y)$ with respect to $x$ is $\frac{1}{y}\frac{\mathrm{d}y}{\mathrm{d}x}$ , not just $\frac{1}{2y}$ .
The derivative of $\ln(2+x^2)$ is $\frac{2x}{2+x^2}$ , using the chain rule.
The mark scheme requires evidence of substitution, so the substituted equation must be shown explicitly.
After solving, simplify to $\frac{6-12\ln 6}{4-3\ln 6}$ or equivalently $4-\frac{10}{4-3\ln 6}$ ; do not leave a fraction within a fraction.

Techniques used

differentiate the implicit equation with respect to xapply the product rule to two product termssubstitute the given coordinates into the derivativesolve the resulting linear equation for dy/dx

Approach

Let $z = x + \mathrm{i}y$ where $x, y \in \mathbb{R}$ . Then the conjugate is $z^* = x - \mathrm{i}y$ . Substitute both into the equation, expand using $\mathrm{i}^2 = -1$ , then equate the real parts and imaginary parts to zero separately. This gives a linear equation for $x$ and a quadratic equation for $y$ . Solve and combine to get the two complex solutions.

Working

Let $z = x + \mathrm{i}y$ , so $z^* = x - \mathrm{i}y$ .

Compute $zz^*$ :

zz^* = (x + \mathrm{i}y)(x - \mathrm{i}y) = x^2 - \mathrm{i}^2 y^2 = x^2 + y^2

Compute $5\mathrm{i}z$ :

5\mathrm{i}z = 5\mathrm{i}(x + \mathrm{i}y) = 5\mathrm{i}x + 5\mathrm{i}^2 y = 5\mathrm{i}x - 5y

Substitute into the equation:

x^2 + y^2 + 5\mathrm{i}x - 5y + 2 - 10\mathrm{i} = 0

Equate real parts:

x^2 + y^2 - 5y + 2 = 0

Equate imaginary parts:

5x - 10 = 0 \Rightarrow x = 2

Substitute $x = 2$ into the real part:

4 + y^2 - 5y + 2 = 0

y^2 - 5y + 6 = 0

(y - 2)(y - 3) = 0

So $y = 2$ or $y = 3$ .

Answer

z = 2 + 2\mathrm{i} \quad \text{or} \quad z = 2 + 3\mathrm{i}

Final answer

z = 2 + 2i or z = 2 + 3i

Detailed explanation

Walkthrough

The equation contains $z$ and its conjugate $z^*$ , so the first step is to write $z$ in terms of real and imaginary parts: $z = x + \mathrm{i}y$ , with $x$ and $y$ real. Then $z^* = x - \mathrm{i}y$ . Substituting these turns the complex equation into an equation we can separate.

First compute $zz^*$ : multiplying $(x + \mathrm{i}y)(x - \mathrm{i}y)$ uses the difference of squares pattern; because $\mathrm{i}^2 = -1$ , the $\mathrm{i}y$ terms cancel and we get $x^2 + y^2$ , which is a real number (it is the square of the modulus $|z|^2$ ).

Next compute $5\mathrm{i}z$ : multiplying $5\mathrm{i}$ by $(x + \mathrm{i}y)$ gives $5\mathrm{i}x + 5\mathrm{i}^2 y$ , and since $\mathrm{i}^2 = -1$ , this becomes $5\mathrm{i}x - 5y$ .

Now the whole equation becomes $(x^2 + y^2) + (5\mathrm{i}x - 5y) + 2 - 10\mathrm{i} = 0$ . Group the real parts $(x^2 + y^2 - 5y + 2)$ and the imaginary parts $(5x - 10)\mathrm{i}$ . A complex number is zero only if both its real and imaginary parts are zero, so we get two separate equations.

The imaginary part gives $5x - 10 = 0$ , so $x = 2$ immediately. The real part gives $x^2 + y^2 - 5y + 2 = 0$ . Substituting $x = 2$ gives $4 + y^2 - 5y + 2 = 0$ , i.e. $y^2 - 5y + 6 = 0$ , which factorises to $(y - 2)(y - 3) = 0$ . Hence $y = 2$ or $y = 3$ .

Combining with $x = 2$ gives the two solutions $z = 2 + 2\mathrm{i}$ and $z = 2 + 3\mathrm{i}$ .

Key Takeaways

Writing a complex number as $z = x + \mathrm{i}y$ lets you handle equations involving $z$ and $z^*$ by separating real and imaginary parts.
$zz^* = x^2 + y^2 = |z|^2$ , a real number — a very useful identity.
A complex equation is equivalent to two real equations (real part = 0 and imaginary part = 0).
$\mathrm{i}^2 = -1$ must be applied carefully when expanding products like $5\mathrm{i}z$ .

Common Mistakes

The mark scheme requires showing the substitution of $z = x + \mathrm{i}y$ and $z^*$ and the expansion — an unsupported final answer would not earn the method marks.
Forgetting that $\mathrm{i}^2 = -1$ when expanding $(x + \mathrm{i}y)(x - \mathrm{i}y)$ or $5\mathrm{i}(x + \mathrm{i}y)$ — this leads to wrong signs.
Mixing up real and imaginary parts when grouping terms.
Not equating both parts to zero — a complex number is zero only when both parts are zero.
Sign errors when substituting $x = 2$ into the real-part equation.

Things to Be Careful About

The term $5\mathrm{i}z$ contributes both a real part ( $-5y$ ) and an imaginary part ( $5\mathrm{i}x$ ) after expansion, since $\mathrm{i}^2 = -1$ .
$zz^*$ is always real, so it contributes only to the real part.
When equating imaginary parts, drop the $\mathrm{i}$ factor and set the coefficient $5x - 10$ equal to zero.
Both solutions must be presented in the form $x + \mathrm{i}y$ with $x$ and $y$ real.

Techniques used

substitute z = x + iy and its conjugateexpand using i^2 = -1equate real and imaginary parts to zerosolve the resulting linear and quadratic equations

Approach

Apply the quotient rule to $y = \dfrac{\tan x}{5 + 2\sin x}$ , then convert every trigonometric function to $\sin x$ and $\cos x$ using $\tan x = \dfrac{\sin x}{\cos x}$ and $\sec^2 x = \dfrac{1}{\cos^2 x}$ . Finally use $\cos^2 x = 1 - \sin^2 x$ to eliminate $\cos x$ and simplify the numerator to a polynomial in $\sin x$ .

Working

Let $u = \tan x$ and $v = 5 + 2\sin x$ . Then $u' = \sec^2 x$ and $v' = 2\cos x$ .

\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{(5 + 2\sin x)\sec^2 x - \tan x \cdot 2\cos x}{(5 + 2\sin x)^2}

Replace $\sec^2 x = \dfrac{1}{\cos^2 x}$ and $\tan x = \dfrac{\sin x}{\cos x}$ :

\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{\dfrac{5 + 2\sin x}{\cos^2 x} - \dfrac{\sin x}{\cos x}\cdot 2\cos x}{(5 + 2\sin x)^2}

The second term simplifies since $\dfrac{\sin x}{\cos x} \cdot 2\cos x = 2\sin x$ :

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

Multiply the numerator and the denominator by $\cos^2 x$ :

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

Substitute $\cos^2 x = 1 - \sin^2 x$ into the numerator:

5 + 2\sin x - 2\sin x(1 - \sin^2 x) = 5 + 2\sin x - 2\sin x + 2\sin^3 x = 5 + 2\sin^3 x

Hence:

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

Answer

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

Final answer

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

Detailed explanation

Walkthrough

We start with $y = \tan x / (5 + 2\sin x)$ , a quotient of two differentiable functions. The quotient rule says that if $y = u/v$ then $y' = (u'v - uv')/v^2$ . Here $u = \tan x$ and $v = 5 + 2\sin x$ , giving $u' = \sec^2 x$ (since the derivative of $\tan x$ is $\sec^2 x$ ) and $v' = 2\cos x$ (the derivative of $\sin x$ is $\cos x$ , and the 5 is a constant so disappears). Substituting these into the quotient rule produces the first line of the working.

The result is still in mixed trigonometric form, but the question asks for an answer purely in $\sin x$ . We therefore replace $\sec^2 x$ with $1/\cos^2 x$ and $\tan x$ with $\sin x / \cos x$ . A useful simplification happens immediately: $\tan x \cdot 2\cos x = (\sin x / \cos x) \cdot 2\cos x = 2\sin x$ , because the $\cos x$ factors cancel.

We now have $\dfrac{(5 + 2\sin x)/\cos^2 x - 2\sin x}{(5 + 2\sin x)^2}$ . To clear the $\cos^2 x$ from the denominator inside the numerator, we multiply both the top and bottom of the entire fraction by $\cos^2 x$ . The numerator becomes $(5 + 2\sin x) - 2\sin x\cos^2 x$ , and the denominator becomes $\cos^2 x (5 + 2\sin x)^2$ .

The final step uses the Pythagorean identity $\cos^2 x = 1 - \sin^2 x$ . Expanding $2\sin x(1 - \sin^2 x)$ gives $2\sin x - 2\sin^3 x$ , and this combines with the $2\sin x$ in the original numerator: $5 + 2\sin x - 2\sin x + 2\sin^3 x = 5 + 2\sin^3 x$ . Replacing $\cos^2 x$ in the denominator with $1 - \sin^2 x$ gives the fully factorised form required by the mark scheme.

Key Takeaways

The quotient rule is the natural tool when one function is divided by another.
Converting $\tan$ and $\sec$ into $\sin$ and $\cos$ is the standard way to express a trig derivative purely in $\sin x$ .
The Pythagorean identity $\cos^2 x = 1 - \sin^2 x$ is the bridge that turns a $\cos x$ expression into a polynomial in $\sin x$ .
Watch the denominator — the mark scheme requires it in factorised form, not expanded.

Common Mistakes

Forgetting to subtract the $uv'$ term in the quotient rule (i.e. writing $u'v/v^2$ alone).
Differentiating $\tan x$ as $\sec x \tan x$ instead of $\sec^2 x$ .
Leaving the answer with a fraction inside a fraction rather than multiplying through.
Expanding the denominator $(5 + 2\sin x)^2$ instead of leaving it in factorised form.

Things to Be Careful About

The sign in the Pythagorean identity: $\cos^2 x = 1 - \sin^2 x$ , not $\sin^2 x - 1$ .
The two $\sin x$ terms ( $+2\sin x$ and $-2\sin x$ ) cancel — this is what makes the final numerator so clean.
Domain restrictions: the original $y$ is undefined where $\cos x = 0$ (because of $\tan x$ ), and these are exactly the points where $1 - \sin^2 x = 0$ in the denominator of our derivative — the two viewpoints agree.

Techniques used

apply the quotient ruleconvert sec and tan to sin and cosexpand and collect terms in sin xuse the Pythagorean identity cos^2 x = 1 - sin^2 xfactorise the denominator in terms of sin x

(b)

Show that the curve has no stationary points.

2M Similar

DifficultyMedium-Easy

Worked solution

Approach

Stationary points occur when the derivative equals zero. With the expression from part (a), the denominator is non-zero wherever $y$ itself is defined, so we only need to check whether the numerator $5 + 2\sin^3 x$ can be zero for some valid value of $\sin x$ .

Working

Set the numerator of $\dfrac{\mathrm{d}y}{\mathrm{d}x}$ equal to zero:

5 + 2\sin^3 x = 0

Solve for $\sin x$ :

\sin^3 x = -\frac{5}{2} \quad\Rightarrow\quad \sin x = -\sqrt[3]{\tfrac{5}{2}}

The real cube root of $\dfrac{5}{2}$ is approximately $1.357$ , so the equation requires

\sin x \approx -1.357

But $\sin x$ is bounded: for every real $x$ , $-1 \le \sin x \le 1$ . Since $-1.357 < -1$ , no real value of $x$ can satisfy the equation.

Therefore the numerator is never zero on the domain of the curve, and the curve has no stationary points.

Answer

The equation $5 + 2\sin^3 x = 0$ has no solution because it would require $\sin x = -\sqrt[3]{5/2} \approx -1.357$ , which lies outside $[-1, 1]$ . Hence the curve has no stationary points.

Final answer

No stationary points — 5 + 2 sin^3 x = 0 requires sin x = -∛(5/2) ≈ -1.357, which is outside [-1, 1].

Detailed explanation

Walkthrough

A stationary point on a curve is a point where the gradient $\mathrm{d}y/\mathrm{d}x$ equals zero. From part (a) we have

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

The denominator is zero exactly when $\cos x = 0$ (so $1 - \sin^2 x = 0$ ) or when $5 + 2\sin x = 0$ (i.e. $\sin x = -5/2$ , which is itself impossible). Wherever the original curve $y = \tan x / (5 + 2\sin x)$ is defined, the denominator of $\mathrm{d}y/\mathrm{d}x$ is non-zero. So $\mathrm{d}y/\mathrm{d}x = 0$ if and only if the numerator $5 + 2\sin^3 x = 0$ .

That is a cubic equation in $\sin x$ . Taking cube roots gives the unique real candidate $\sin x = -\sqrt[3]{5/2} \approx -1.357$ . But $\sin x$ is trapped inside the closed interval $[-1, 1]$ for every real $x$ — a value of $-1.357$ is impossible. Therefore the numerator is never zero on the domain, and the curve has no stationary points anywhere it is defined.

Key Takeaways

A stationary point of $y = f(x)$ requires $f'(x) = 0$ at a point where $f$ itself is defined.
The mark scheme specifically requires explicit reference to the bound $-1 \le \sin x \le 1$ when concluding no solutions exist.
Recognising that a polynomial in $\sin x$ has no real solution because it would force $\sin x$ outside its natural range is a common A-level argument.

Common Mistakes

Setting the denominator of $\mathrm{d}y/\mathrm{d}x$ equal to zero instead of the numerator.
Solving $5 + 2\sin^3 x = 0$ correctly but then failing to state the bound $-1 \le \sin x \le 1$ explicitly — the mark scheme (AG) insists on this reference.
Concluding "no stationary points" without justifying why the absence of solutions to the cubic means no stationary points (i.e. without checking that the denominator is non-zero wherever the original curve is defined).

Things to Be Careful About

The mark for part (b) is awarded on AG (hence "must be from correct working only using a correct numerator from part (a)"). A numerically wrong numerator in (a) makes the proof in (b) invalid.
An alternative valid route is to bound $5 + 2\sin^3 x$ directly: since $-1 \le \sin x \le 1$ we have $-1 \le \sin^3 x \le 1$ , so $3 \le 5 + 2\sin^3 x \le 7$ , and in particular $5 + 2\sin^3 x \ne 0$ .
The denominator $(1 - \sin^2 x)(5 + 2\sin x)^2$ is always non-negative on the domain, so the sign of the gradient never changes — a further visual confirmation that no turning points exist.

Techniques used

set the numerator of dy/dx to zerorecognise the cubic equation in sin xapply the bound -1 ≤ sin x ≤ 1conclude no stationary points exist

Q7Medium-EasyIntegration

(a)

Use the substitution $u = x^2 - 3$ to show that

\int_{\sqrt{7}}^{\sqrt{12}} \frac{4x^3}{\sqrt{x^2 - 3}} \,dx = \int_a^b \frac{2(u + 3)}{\sqrt{u}} \,du

where $a$ and $b$ are values to be found.

4M Similar

DifficultyMedium-Easy

Worked solution

Approach

Use the substitution $u = x^2 - 3$ . Differentiate to obtain $\mathrm{d}u = 2x\,\mathrm{d}x$ , change the limits using the given values of $x$ , and rewrite $4x^3\,\mathrm{d}x$ in terms of $u$ and $\mathrm{d}u$ .

Working

Let $u = x^2 - 3$ . Then

\frac{\mathrm{d}u}{\mathrm{d}x} = 2x

so

\mathrm{d}u = 2x\,\mathrm{d}x.

Change the limits:

x = \sqrt{7} \Rightarrow u = 7 - 3 = 4,

x = \sqrt{12} \Rightarrow u = 12 - 3 = 9.

Also, since $x^2 = u + 3$ ,

4x^3\,\mathrm{d}x = 2x^2 \cdot 2x\,\mathrm{d}x = 2(u + 3)\,\mathrm{d}u.

Therefore

\int_{\sqrt{7}}^{\sqrt{12}} \frac{4x^3}{\sqrt{x^2 - 3}}\,\mathrm{d}x = \int_4^9 \frac{2(u + 3)}{\sqrt{u}}\,\mathrm{d}u.

Hence $a = 4$ and $b = 9$ .

Answer

a = 4, \quad b = 9

Final answer

a = 4, b = 9

Detailed explanation

Walkthrough

We are asked to use the substitution $u = x^2 - 3$ . The first step is to differentiate this substitution: since $u = x^2 - 3$ , we have $\frac{\mathrm{d}u}{\mathrm{d}x} = 2x$ , so $\mathrm{d}u = 2x\,\mathrm{d}x$ . This tells us how to replace $\mathrm{d}x$ in the integral.

Next, we change the limits. When $x = \sqrt{7}$ , we get $u = (\sqrt{7})^2 - 3 = 4$ . When $x = \sqrt{12}$ , we get $u = (\sqrt{12})^2 - 3 = 9$ . So the new limits are $4$ and $9$ .

Then we rewrite the numerator. We have $4x^3 = 2x^2 \cdot 2x$ . Since $2x\,\mathrm{d}x = \mathrm{d}u$ and $x^2 = u + 3$ , this becomes $2(u + 3)\,\mathrm{d}u$ . The denominator $\sqrt{x^2 - 3}$ becomes $\sqrt{u}$ . This gives exactly the required integral with $a = 4$ and $b = 9$ .

Key Takeaways

When using substitution, always replace $\mathrm{d}x$ correctly using the derivative of the substitution.
Change the limits of integration when the variable changes from $x$ to $u$ .
Rewriting the integrand as a product can help match the substitution cleanly.

Common Mistakes

Forgetting to change the limits when moving from $x$ to $u$ .
Incorrectly writing $4x^3\,\mathrm{d}x$ ; it must be split as $2x^2 \cdot 2x\,\mathrm{d}x$ so that the $2x\,\mathrm{d}x$ becomes $\mathrm{d}u$ .
Losing the factor of $2$ when substituting.

Things to Be Careful About

The question asks to show a given result, so full working must be shown; an unsupported answer would not earn the final accuracy mark.
The limits must be stated correctly as $4$ and $9$ .
Keep the substitution consistent throughout: every $x$ must be replaced by the corresponding expression in $u$ .

Techniques used

differentiate the substitution u = x^2 - 3change the integration limitsrewrite the integrand in terms of u and du

(b)

Hence, find the exact value of $\int_{\sqrt{7}}^{\sqrt{12}} \frac{4x^3}{\sqrt{x^2 - 3}} \,dx$ .

4M Similar

DifficultyMedium-Easy

Worked solution

Approach

Use the result from part (a). Write the integrand as a sum of powers of $u$ , integrate term by term, then evaluate between $u = 4$ and $u = 9$ .

Working

From part (a),

I = \int_4^9 \frac{2(u + 3)}{\sqrt{u}}\,\mathrm{d}u.

Rewrite the integrand:

\frac{2(u + 3)}{\sqrt{u}} = 2u^{\frac{1}{2}} + 6u^{-\frac{1}{2}}.

Integrate:

\int \left(2u^{\frac{1}{2}} + 6u^{-\frac{1}{2}}\right)\,\mathrm{d}u = \frac{4}{3}u^{\frac{3}{2}} + 12u^{\frac{1}{2}}.

So

I = \left[\frac{4}{3}u^{\frac{3}{2}} + 12u^{\frac{1}{2}}\right]_4^9.

At $u = 9$ :

\frac{4}{3}(27) + 12(3) = 36 + 36 = 72.

At $u = 4$ :

\frac{4}{3}(8) + 12(2) = \frac{32}{3} + 24 = \frac{104}{3}.

Therefore

I = 72 - \frac{104}{3} = \frac{216}{3} - \frac{104}{3} = \frac{112}{3}.

Answer

\frac{112}{3}

or equivalently $37\frac{1}{3}$ .

Final answer

112/3

Detailed explanation

Walkthrough

Since part (a) has already converted the integral into $\int_4^9 \frac{2(u + 3)}{\sqrt{u}}\,\mathrm{d}u$ , we can use that result directly.

First, split the fraction:

\frac{2(u + 3)}{\sqrt{u}} = \frac{2u}{\sqrt{u}} + \frac{6}{\sqrt{u}} = 2u^{\frac{1}{2}} + 6u^{-\frac{1}{2}}.

Now integrate each term using the power rule. The integral of $2u^{\frac{1}{2}}$ is $\frac{4}{3}u^{\frac{3}{2}}$ , and the integral of $6u^{-\frac{1}{2}}$ is $12u^{\frac{1}{2}}$ .

Then substitute the limits. At $u = 9$ , the expression equals $72$ . At $u = 4$ , it equals $\frac{104}{3}$ . Subtracting gives $\frac{112}{3}$ .

Key Takeaways

A substitution often leaves an integral that can be handled by simple power integration.
Splitting a fraction into separate power terms makes integration straightforward.
The word “Hence” means the result from part (a) should be used directly.

Common Mistakes

Integrating in terms of $x$ instead of $u$ ; the question says “Hence”, so the integration must be done in terms of $u$ .
Forgetting to subtract the lower-limit value from the upper-limit value.
Making arithmetic errors with fractions such as $\frac{32}{3} + 24$ .

Things to Be Careful About

The final answer must be exact; a decimal answer would not be accepted.
If the answer is given without working, it earns no marks: the mark scheme requires the integration and limit substitution to be shown.
Be careful with the fractional powers: $u^{\frac{3}{2}}$ and $u^{\frac{1}{2}}$ must be evaluated correctly at $u = 9$ and $u = 4$ .

Techniques used

apply the result from part (a)split the integrand into power termsintegrate term by termevaluate the definite integral using the limits

Q8MediumTrigonometry

(a)

Express $3\sqrt{3} \sin(\theta + \frac{1}{6}\pi) - 2 \sin \theta$ in the form $R \sin(\theta + \alpha)$ , where $R > 0$ and $0 < \alpha < \frac{1}{2}\pi$ . Give the exact value of $R$ and state the value of $\alpha$ correct to 3 decimal places.

5M Similar

DifficultyMedium

Worked solution

Approach

Expand $\sin(\theta+\frac{1}{6}\pi)$ using the compound-angle identity, combine like terms, then write the result as $R\sin(\theta+\alpha)$ by equating coefficients.

Working

Expand the expression:

3\sqrt{3}\sin\left(\theta+\frac{\pi}{6}\right)-2\sin\theta = 3\sqrt{3}\left(\frac{\sqrt{3}}{2}\sin\theta+\frac{1}{2}\cos\theta\right)-2\sin\theta.

Using $\sin\left(\frac{\pi}{6}\right)=\frac{1}{2}$ and $\cos\left(\frac{\pi}{6}\right)=\frac{\sqrt{3}}{2}$ :

\frac{9}{2}\sin\theta+\frac{3\sqrt{3}}{2}\cos\theta-2\sin\theta = \frac{5}{2}\sin\theta+\frac{3\sqrt{3}}{2}\cos\theta.

Now compare with:

R\sin(\theta+\alpha) = R\sin\theta\cos\alpha+R\cos\theta\sin\alpha.

Equating coefficients of $\sin\theta$ and $\cos\theta$ :

R\cos\alpha=\frac{5}{2}, \qquad R\sin\alpha=\frac{3\sqrt{3}}{2}.

Therefore:

R=\sqrt{\left(\frac{5}{2}\right)^2+\left(\frac{3\sqrt{3}}{2}\right)^2} =\frac{1}{2}\sqrt{25+27} =\sqrt{13}.

For $\alpha$ , use the ratio:

\tan\alpha = \frac{R\sin\alpha}{R\cos\alpha} = \frac{\frac{3\sqrt{3}}{2}}{\frac{5}{2}} = \frac{3\sqrt{3}}{5}.

So:

\alpha = \tan^{-1}\left(\frac{3\sqrt{3}}{5}\right) = 0.805.

Thus $\alpha=0.805$ radians (to 3 d.p.).

Answer

3\sqrt{3}\sin\left(\theta+\frac{\pi}{6}\right)-2\sin\theta = \sqrt{13}\sin(\theta+0.805),

with $R=\sqrt{13}$ and $\alpha=0.805$ radians.

Final answer

R = sqrt(13), alpha = 0.805 radians

Detailed explanation

Walkthrough

Start with the expression $3\sqrt{3}\sin\left(\theta+\frac{\pi}{6}\right)-2\sin\theta$ . The first step is to expand $\sin\left(\theta+\frac{\pi}{6}\right)$ using the compound-angle identity:

\sin\left(\theta+\frac{\pi}{6}\right)=\sin\theta\cos\frac{\pi}{6}+\cos\theta\sin\frac{\pi}{6}.

Then substitute the exact values $\cos\frac{\pi}{6}=\frac{\sqrt{3}}{2}$ and $\sin\frac{\pi}{6}=\frac{1}{2}$ . Multiplying by $3\sqrt{3}$ gives $\frac{9}{2}\sin\theta+\frac{3\sqrt{3}}{2}\cos\theta$ . Subtracting $2\sin\theta$ combines the sine terms to give $\frac{5}{2}\sin\theta+\frac{3\sqrt{3}}{2}\cos\theta$ .

Next, write $R\sin(\theta+\alpha)$ in expanded form: $R\sin\theta\cos\alpha+R\cos\theta\sin\alpha$ . Comparing this with $\frac{5}{2}\sin\theta+\frac{3\sqrt{3}}{2}\cos\theta$ tells us that $R\cos\alpha=\frac{5}{2}$ and $R\sin\alpha=\frac{3\sqrt{3}}{2}$ . Squaring and adding removes $\alpha$ and gives $R^2=\left(\frac{5}{2}\right)^2+\left(\frac{3\sqrt{3}}{2}\right)^2=13$ , so $R=\sqrt{13}$ . To find $\alpha$ , divide the two equations: $\tan\alpha=\frac{R\sin\alpha}{R\cos\alpha}=\frac{3\sqrt{3}}{5}$ . Because both $\cos\alpha$ and $\sin\alpha$ are positive, $\alpha$ is acute, so $\alpha=\tan^{-1}\left(\frac{3\sqrt{3}}{5}\right)=0.805$ radians.

Key Takeaways

The question tests two connected ideas: expanding a compound angle and then converting a linear combination of $\sin$ and $\cos$ into the $R\sin(\theta+\alpha)$ form. The coefficient-matching step is central: $R\cos\alpha$ equals the coefficient of $\sin\theta$ and $R\sin\alpha$ equals the coefficient of $\cos\theta$ . Once this is set up, $R$ and $\alpha$ follow from Pythagoras and a tangent ratio.

Common Mistakes

Forgetting to subtract $2\sin\theta$ , which leaves $\frac{9}{2}\sin\theta+\frac{3\sqrt{3}}{2}\cos\theta$ and is marked as no method credit.
Writing $\sin\alpha=\frac{3\sqrt{3}}{2}$ or $\cos\alpha=\frac{5}{2}$ without dividing by $R$ ; the mark scheme explicitly does not allow this. The correct equations are $\sin\alpha=\frac{3\sqrt{3}}{2R}$ and $\cos\alpha=\frac{5}{2R}$ .
Giving $R$ as a decimal instead of the required exact value $\sqrt{13}$ .
Working in degrees instead of radians, since the question asks for $\alpha$ correct to 3 decimal places in radians.

Things to Be Careful About

Ensure every trigonometric value used is exact: $\sin\frac{\pi}{6}=\frac{1}{2}$ and $\cos\frac{\pi}{6}=\frac{\sqrt{3}}{2}$ . Keep the algebra exact until the final decimal for $\alpha$ . Since both resulting coefficients are positive, the angle $\alpha$ is in the first quadrant, so the inverse tangent gives the correct value without adjustment. Set your calculator to radians before computing $\tan^{-1}$ .

Techniques used

expand a compound anglecombine sine and cosine termsequate coefficients with the R sin(theta+alpha) formuse Pythagoras and inverse tangent to find R and alpha

(b)

Hence, solve the equation $3\sqrt{3} \sin(2x + \frac{1}{6}\pi) - 2 \sin 2x = \sqrt{6}$ for $0 < x < \pi$ .

4M Similar

DifficultyMedium

Worked solution

Approach

Use the result from part (a) with $\theta=2x$ , so the left-hand side becomes $\sqrt{13}\sin(2x+\alpha)$ . Solve $\sin(2x+\alpha)=\frac{\sqrt{6}}{\sqrt{13}}$ , generate all general solutions, and keep only those corresponding to $0<x<\pi$ .

Working

From part (a), with $\theta=2x$ :

3\sqrt{3}\sin\left(2x+\frac{\pi}{6}\right)-2\sin 2x = \sqrt{13}\sin(2x+\alpha), \qquad \alpha=0.805.

The equation becomes:

\sqrt{13}\sin(2x+\alpha)=\sqrt{6} \quad\Rightarrow\quad \sin(2x+\alpha)=\frac{\sqrt{6}}{\sqrt{13}}.

The principal value is:

2x+\alpha = \sin^{-1}\left(\frac{\sqrt{6}}{\sqrt{13}}\right) = 0.747.

Hence the general solutions are:

2x+\alpha=0.747+2n\pi \quad\text{or}\quad 2x+\alpha=(\pi-0.747)+2n\pi=2.395+2n\pi.

Since $0<x<\pi$ , we have $0<2x<2\pi$ , so:

0.805<2x+\alpha<2\pi+0.805=7.088.

The values in this interval are:

2x+\alpha=2.395,\quad 7.030.

(The first candidate $0.747$ is below $0.805$ and so is invalid.)

Solving for $x$ :

x=\frac{2.395-0.805}{2}=0.795, \qquad x=\frac{7.030-0.805}{2}=3.11.

Answer

x=0.795 \quad \text{or} \quad x=3.11.

Both solutions are in $0<x<\pi$ .

Final answer

x = 0.795 or x = 3.11 (3 s.f.)

Detailed explanation

Walkthrough

Part (a) gives $3\sqrt{3}\sin\left(\theta+\frac{\pi}{6}\right)-2\sin\theta=\sqrt{13}\sin(\theta+\alpha)$ , valid for every angle $\theta$ . Replace $\theta$ by $2x$ so the left-hand side of the required equation becomes $\sqrt{13}\sin(2x+\alpha)$ . Therefore solve $\sin(2x+\alpha)=\frac{\sqrt{6}}{\sqrt{13}}$ .

Let $u=2x+\alpha$ . The principal solution is $u=\sin^{-1}\left(\frac{\sqrt{6}}{\sqrt{13}}\right)=0.747$ . But sine is also positive in the second quadrant, so the other base solution is $u=\pi-0.747=2.395$ . Adding multiples of $2\pi$ to either base gives further solutions, because sine has period $2\pi$ .

The interval for $x$ is $0<x<\pi$ , so multiplying by 2 gives $0<2x<2\pi$ . Adding $\alpha=0.805$ gives $0.805<u<2\pi+0.805=7.088$ . The candidates in this interval are $u=0.747$ (too small, below $0.805$ and therefore discarded), $u=2.395$ , and $u=7.030$ (from $0.747+2\pi$ ). The next candidate $8.677$ is above the upper limit.

Now convert each valid $u$ back to $x$ using $x=\frac{u-\alpha}{2}$ :

x=\frac{2.395-0.805}{2}=0.795,\qquad x=\frac{7.030-0.805}{2}=3.11.

Both lie in $0<x<\pi$ , so they are the required solutions.

Key Takeaways

This part shows why the $R\sin(\theta+\alpha)$ form is useful: it turns a complicated combination of sines and cosines into a single sine equation. Solving $\sin u=k$ requires remembering both the principal solution and the second-quadrant solution, and then adding periods. The interval for $u$ must be derived from the interval for $x$ before discarding or accepting candidates.

Common Mistakes

Stopping at $2x+\alpha=0.747$ and then getting a negative $x$ ; this solution is outside the given range.
Omitting the second-quadrant solution $\pi-0.747=2.395$ .
Omitting the periodic solution $0.747+2\pi=7.030$ , which is the one that gives $x=3.11$ .
Using the wrong interval $0<2x+\alpha<2\pi$ instead of $0.805<2x+\alpha<7.088$ .
Dividing incorrectly when solving for $x$ : the operation is $x=(u-\alpha)/2$ , not $u/2-\alpha$ .
Giving only one of the two valid answers; the mark scheme requires both and no others in the range.

Things to Be Careful About

Use radians throughout. Follow through from part (a): if a different $R$ or $\alpha$ was found, use that value consistently. Check every candidate by ensuring $0<x<\pi$ . The approximate values are accepted: $\alpha=0.805$ , principal value $0.747$ , and answers $0.795$ and $3.11$ . Since $3.11<\pi$ , both final answers are valid.

Techniques used

apply the R-form result by substitutionsolve a sine equation using inverse sinegenerate general solutions using periodicityrestrict solutions to the given interval

Q9MediumVectors

The equations of two lines are given by

\begin{aligned} l_1: \mathbf{r} &= (2\mathbf{i} + \mathbf{j} + 4\mathbf{k}) + \lambda(\mathbf{i} + 2\mathbf{j} - 3\mathbf{k}), \\ l_2: \mathbf{r} &= (3\mathbf{i} - \mathbf{j} + 5\mathbf{k}) + \mu(2\mathbf{i} + 3\mathbf{j} + a\mathbf{k}). \end{aligned}

(a)

Find the value of $a$ for which $l_1$ is perpendicular to $l_2$ .

2M Similar

(b)

Find the value of $a$ for which $l_1$ and $l_2$ intersect.

4M Similar

DifficultyMedium-Easy

Worked solution

Approach

For intersection of two lines, write each general point in component form, equate the $\mathbf{i}$ and $\mathbf{j}$ components to find $\lambda$ and $\mu$ , then use the $\mathbf{k}$ components to find $a$ .

Working

General point on $l_1$ :

\begin{pmatrix} 2 + \lambda \\ 1 + 2\lambda \\ 4 - 3\lambda \end{pmatrix}

General point on $l_2$ :

\begin{pmatrix} 3 + 2\mu \\ -1 + 3\mu \\ 5 + a\mu \end{pmatrix}

Equating $\mathbf{i}$ components:

2 + \lambda = 3 + 2\mu

\lambda - 2\mu = 1

Equating $\mathbf{j}$ components:

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

2\lambda - 3\mu = -2

From $\lambda - 2\mu = 1$ , $\lambda = 1 + 2\mu$ . Substitute into $2\lambda - 3\mu = -2$ :

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

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

\mu = -4

So

\lambda = 1 + 2(-4) = -7

Equating $\mathbf{k}$ components:

4 - 3\lambda = 5 + a\mu

Substitute $\lambda = -7$ and $\mu = -4$ :

4 - 3(-7) = 5 + a(-4)

25 = 5 - 4a

20 = -4a

a = -5

Answer

a = -5

Final answer

a = -5

Detailed explanation

Walkthrough

For the two lines to intersect, there must be a common point on both lines. Write the general position of each line using its own parameter. Equate the $\mathbf{i}$ and $\mathbf{j}$ components because they do not contain $a$ ; this gives two equations in $\lambda$ and $\mu$ . Solve these simultaneous equations to get $\lambda = -7$ and $\mu = -4$ . Then substitute these values into the $\mathbf{k}$ components to solve for $a$ .

Key Takeaways

Intersection of two lines is found by matching their general position vectors.
Use different parameters for different lines.
The components not involving the unknown parameter are used first to find the parameters.

Common Mistakes

Using the same parameter for both lines.
Making a sign error when solving $2(1+2\mu)-3\mu = -2$ .
Forgetting to substitute both parameters into the $\mathbf{k}$ component.

Things to Be Careful About

Show the component form clearly; the mark scheme awards a mark for equating the $\mathbf{i}$ and $\mathbf{j}$ components. Once $\lambda$ and $\mu$ are found, the $\mathbf{k}$ equation must be used to obtain $a = -5$ .

Techniques used

write general point in component formequate corresponding componentssolve simultaneous linear equationssubstitute to find remaining parameter

(c)

Find the values of $a$ for which the acute angle between $l_1$ and $l_2$ is equal to $\cos^{-1}(\frac{5}{14})$ .

4M Similar

DifficultyMedium

Worked solution

Approach

For the acute angle between two lines, use the scalar product of their direction vectors and the absolute value of the scalar product in the cosine formula. Set this equal to $\frac{5}{14}$ , square both sides, and solve the resulting quadratic.

Working

The direction vectors are

\mathbf{d}_1 = \mathbf{i} + 2\mathbf{j} - 3\mathbf{k}, \qquad \mathbf{d}_2 = 2\mathbf{i} + 3\mathbf{j} + a\mathbf{k}.

Scalar product:

\mathbf{d}_1 \cdot \mathbf{d}_2 = (1)(2) + (2)(3) + (-3)(a) = 8 - 3a

Magnitudes:

|\mathbf{d}_1| = \sqrt{1^2 + 2^2 + (-3)^2} = \sqrt{14}

|\mathbf{d}_2| = \sqrt{2^2 + 3^2 + a^2} = \sqrt{13 + a^2}

For the acute angle:

\cos\theta = \frac{|8 - 3a|}{\sqrt{14}\sqrt{13 + a^2}} = \frac{5}{14}

Cross-multiplying:

14|8 - 3a| = 5\sqrt{14}\sqrt{13 + a^2}

Squaring both sides:

196(8 - 3a)^2 = 350(13 + a^2)

Divide by 14:

14(8 - 3a)^2 = 25(13 + a^2)

Expand:

14(64 - 48a + 9a^2) = 325 + 25a^2

896 - 672a + 126a^2 = 325 + 25a^2

101a^2 - 672a + 571 = 0

Using the quadratic formula:

a = \frac{672 \pm \sqrt{672^2 - 4(101)(571)}}{2(101)}

a = \frac{672 \pm 470}{202}

a = 1 \quad \text{or} \quad a = \frac{571}{101}

Answer

a = 1 \quad \text{or} \quad a = \frac{571}{101}

Final answer

a = 1 or a = 571/101

Detailed explanation

Walkthrough

For a line in vector form, the direction vector is the vector multiplying the parameter. Here $\mathbf{d}_1 = \mathbf{i}+2\mathbf{j}-3\mathbf{k}$ and $\mathbf{d}_2 = 2\mathbf{i}+3\mathbf{j}+a\mathbf{k}$ .

The scalar product is $8-3a$ , and the magnitudes are $\sqrt{14}$ and $\sqrt{13+a^2}$ . The cosine of the acute angle between lines is the absolute value of the scalar product divided by the product of the magnitudes. If the scalar product is negative, the angle between the direction vectors is obtuse, but the acute angle between the lines is still positive.

Set this equal to $\frac{5}{14}$ and square both sides. Squaring removes the absolute value and gives a quadratic in $a$ : $101a^2 - 672a + 571=0$ . Using the quadratic formula gives $a=1$ and $a=\frac{571}{101}$ . Both values are valid because the absolute value ensures the acute angle condition is satisfied.

Key Takeaways

The acute angle between lines uses $\cos\theta = \frac{|\mathbf{d}_1\cdot\mathbf{d}_2|}{|\mathbf{d}_1||\mathbf{d}_2|}$ .
Squaring both sides is a valid method here because both sides are non-negative.
A quadratic equation can give two valid values of the parameter.

Common Mistakes

Omitting the absolute value and therefore not obtaining both valid solutions.
Squaring without expanding carefully, causing sign errors in the quadratic.
Forgetting to square both sides before dividing, leading to an incorrect equation.

Things to Be Careful About

The mark scheme accepts $101a^2 - 672a + 571=0$ or an equivalent multiple. If a value such as $a=1$ is spotted without full method, only a special-case mark may be awarded, so show the full quadratic and solving steps. Give exact values; $\frac{571}{101}$ is preferred over a decimal.

Techniques used

compute scalar product of direction vectorsform acute angle cosine with absolute valuesquare both sides to remove square rootsolve quadratic equation

Q10MediumIntegrationNumerical Solution of Equations

The constant $a$ is such that $\int_0^a x e^{\frac{1}{2}x} \,dx = 6$ .

(a)

Show that $a = 2 + e^{-\frac{1}{2}a}$ .

5M Similar

DifficultyMedium

Worked solution

Approach

Use integration by parts with $u = x$ and $\frac{dv}{dx} = e^{\frac{1}{2}x}$ . Evaluate the integral between $0$ and $a$ , set it equal to $6$ , and rearrange the resulting equation.

Working

Let $u = x$ and $\frac{dv}{dx} = e^{\frac{1}{2}x}$ . Then $\frac{du}{dx} = 1$ and $v = 2e^{\frac{1}{2}x}$ .

\begin{aligned} \int x e^{\frac{1}{2}x}\,dx &= 2x e^{\frac{1}{2}x} - \int 2 e^{\frac{1}{2}x}\,dx \\ &= 2x e^{\frac{1}{2}x} - 4 e^{\frac{1}{2}x} + C. \end{aligned}

Evaluate from $0$ to $a$ :

\int_0^a x e^{\frac{1}{2}x}\,dx = \left[2x e^{\frac{1}{2}x} - 4 e^{\frac{1}{2}x}\right]_0^a = \left(2a e^{\frac{1}{2}a} - 4 e^{\frac{1}{2}a}\right) - (-4).

So

2a e^{\frac{1}{2}a} - 4 e^{\frac{1}{2}a} + 4 = 6.

Simplify:

2(a-2)e^{\frac{1}{2}a} = 2

(a-2)e^{\frac{1}{2}a} = 1.

Divide by $e^{\frac{1}{2}a}$ :

a-2 = e^{-\frac{1}{2}a}

a = 2 + e^{-\frac{1}{2}a}.

Answer

a = 2 + e^{-\frac{1}{2}a}.

Final answer

a = 2 + e^(-a/2)

Detailed explanation

Walkthrough

The integrand is a product $x$ and $e^{\frac{1}{2}x}$ , so integration by parts is the natural method. Choose $u=x$ because its derivative is $1$ , and choose $\frac{dv}{dx}=e^{\frac{1}{2}x}$ because it integrates easily to $2e^{\frac{1}{2}x}$ .

After applying integration by parts, the second integral is $\int 2e^{\frac{1}{2}x}\,dx = 4e^{\frac{1}{2}x}$ , so the antiderivative is $2x e^{\frac{1}{2}x} - 4e^{\frac{1}{2}x}$ . Substituting $x=a$ and $x=0$ gives $\left(2a e^{\frac{1}{2}a} - 4e^{\frac{1}{2}a}\right) - (0 - 4)$ . The constant $4$ comes from subtracting the lower-limit value $-4$ ; this must not be forgotten.

Setting the definite integral equal to $6$ gives $2(a-2)e^{\frac{1}{2}a}=2$ , so $(a-2)e^{\frac{1}{2}a}=1$ . Dividing by the positive exponential gives exactly $a-2 = e^{-\frac{1}{2}a}$ .

Key Takeaways

This question tests integration by parts, including the correct antiderivative of $e^{\frac{1}{2}x}$ , and the careful substitution of limits in a definite integral. It also requires algebraic rearrangement of an equation into a stated form.

Common Mistakes

One common error is forgetting the contribution from the lower limit: substituting $x=0$ gives $-4$ , so this contributes $+4$ to the result. Another is using the wrong antiderivative for $e^{\frac{1}{2}x}$ ; it is $2e^{\frac{1}{2}x}$ , not $e^{\frac{1}{2}x}$ . Since part (a) asks to show the result, full working must be shown.

Things to Be Careful About

The factor $\frac{1}{2}$ in the exponent must be handled correctly throughout. When integrating by parts, the second integral is $\int 2e^{\frac{1}{2}x}\,dx = 4e^{\frac{1}{2}x}$ . Finally, divide by the exponential term rather than trying to combine it with $a$ without accounting for the factors of $2$ .

Techniques used

apply integration by partsevaluate the definite integral at its limitsrearrange the resulting equation into the target form

(b)

Verify by calculation that $a$ lies between 2.2 and 2.4.

2M Similar

DifficultyMedium-Easy

Worked solution

Approach

Define $\mathrm{f}(a) = a - 2 - e^{-\frac{1}{2}a}$ . Since $\mathrm{f}$ is continuous and the sign changes between $2.2$ and $2.4$ , there is a root in that interval.

Working

\mathrm{f}(2.2) = 2.2 - 2 - e^{-1.1} = 0.2 - 0.3329 = -0.1329 \quad (4 \text{ dp})

\mathrm{f}(2.4) = 2.4 - 2 - e^{-1.2} = 0.4 - 0.3012 = 0.0988 \quad (4 \text{ dp})

Since $\mathrm{f}(2.2) < 0$ and $\mathrm{f}(2.4) > 0$ , and $\mathrm{f}$ is continuous, the equation has a root between $2.2$ and $2.4$ .

Answer

2.2 < a < 2.4

Final answer

a lies between 2.2 and 2.4

Detailed explanation

Walkthrough

A convenient way to verify the root interval is to use the equation from part (a) rearranged as $\mathrm{f}(a) = a - 2 - e^{-\frac{1}{2}a}$ . The equation $a=2+e^{-\frac{1}{2}a}$ is equivalent to $\mathrm{f}(a)=0$ .

Substitute $a=2.2$ and $a=2.4$ . Because $\mathrm{f}(2.2)$ is negative and $\mathrm{f}(2.4)$ is positive, and $\mathrm{f}$ is continuous as a sum of a linear term and an exponential term, the intermediate value theorem shows that $\mathrm{f}(a)=0$ somewhere strictly between $2.2$ and $2.4$ .

Key Takeaways

This is a standard root-location argument: evaluate the function at the endpoints of the interval and look for a sign change. A sign change, together with continuity, guarantees at least one root in the interval.

Common Mistakes

The most common mistakes are to evaluate only one endpoint, to use an expression that is not equivalent to the original equation, or to forget to state the sign change. The comparison with $0$ must be clear.

Things to Be Careful About

The values used must be consistent with the expression being tested. It is enough to round to a few decimal places, but the signs of the two values must be opposite. If testing a pair that involves a comparison with $6$ , use the same function throughout.

Techniques used

form a sign-change functionevaluate the function at two endpointsapply the intermediate value argument

(c)

Use an iterative formula based on the equation in part (a) to determine $a$ correct to 2 decimal places. Give the result of each iteration to 4 decimal places.

3M Similar

DifficultyMedium-Easy

Worked solution

Approach

Use the rearranged equation $a = 2 + e^{-\frac{1}{2}a}$ as a fixed-point iteration. Start with $a_0 = 2.3$ inside the interval found in part (b) and compute

a_{n+1} = 2 + e^{-\frac{1}{2} a_n}.

Continue until consecutive values round consistently, recording each iteration to 4 decimal places.

Working

\begin{array}{c|c} n & a_n \text{ to 4 dp} \\ \hline 0 & 2.3000 \\ 1 & 2.3166 \\ 2 & 2.3140 \\ 3 & 2.3144 \\ 4 & 2.3144 \end{array}

For example,

a_1 = 2 + e^{-1.15} = 2.3166

a_2 = 2 + e^{-1.1583} = 2.3140

a_3 = 2 + e^{-1.1570} = 2.3144.

The iterates settle at $2.3144$ to 4 decimal places, so correct to 2 decimal places:

a \approx 2.31.

Answer

a \approx 2.31

Final answer

2.31

Detailed explanation

Walkthrough

Part (c) asks for an iterative method. The equation $a=2+e^{-\frac{1}{2}a}$ can be used directly as a fixed-point iteration: $a_{n+1}=2+e^{-\frac{1}{2}a_n}$ . Starting inside the interval found in part (b), say $a_0=2.3$ , the successive substitutions quickly approach the root.

Each iteration should be recorded to 4 decimal places. After substituting, the successive values are $2.3166$ , $2.3140$ , $2.3144$ , $2.3144$ to 4 dp. Both of the last two iterates round to $2.31$ to 2 dp.

Key Takeaways

This demonstrates how an equation rearranged to the form $x=F(x)$ can be used iteratively, and how a sequence of approximations can estimate a root to a specified accuracy. Recording each iterate to one or two extra decimal places is essential for a reliable final rounding.

Common Mistakes

Failing to give each iteration to 4 decimal places, stopping after one iteration, or using an iteration formula that is not algebraically equivalent to the equation in part (a) are the main errors. Stopping too early may not justify the final 2-decimal-place value.

Things to Be Careful About

Start with a value reasonably close to the root, such as $2.3$ . Keep enough decimal places throughout so that the final rounding is valid. Consecutive iterates to 4 dp are $2.3166, 2.3140, 2.3144, 2.3144$ , so the root rounds to $2.31$ to 2 dp.

Techniques used

rearrange the equation into an iterative formapply repeated substitutionround each iterate to the requested precision

Q11Medium-HardDifferential EquationsIntegration

A fungal disease is affecting some of the trees in a forest. The fraction of the trees affected after $t$ years is denoted by $x$ . The rate of increase of $x$ is proportional to the product of the fraction of the trees affected and the fraction of the trees not affected.

(a)

Explain why, after $t$ years, $\frac{dx}{dt} = kx(1 - x)$ , where $k$ is a constant.

1M Similar

DifficultyEasy

Worked solution

Approach

Translate the verbal rate statement into a differential equation. The rate of increase is $\frac{dx}{dt}$ , and it is proportional to the product of the affected fraction $x$ and the unaffected fraction $1-x$ .

Working

The fraction of trees affected is $x$ , so the fraction not affected is $1-x$ . Since the rate of increase is proportional to the product of these two fractions,

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

where $k$ is the constant of proportionality.

Answer

$\frac{dx}{dt} = kx(1-x)$ , with $1-x$ representing the fraction of trees not affected.

Final answer

dx/dt = kx(1 - x), with 1 - x the fraction not affected

Detailed explanation

Walkthrough

The problem states that the rate of increase of $x$ is proportional to the product of the fraction affected and the fraction not affected. The fraction affected is $x$ . Since $x$ is a fraction of the whole forest, the fraction not affected is $1-x$ . The rate of increase is the derivative $\frac{dx}{dt}$ , so we write $\frac{dx}{dt} = kx(1-x)$ , where $k$ is the constant of proportionality.

Key Takeaways

This part tests translating a real-life proportionality statement into a differential equation. The important skill is identifying the two quantities being multiplied: the affected fraction and the unaffected fraction.

Common Mistakes

Forgetting the factor $1-x$ and writing $\frac{dx}{dt}=kx$ .
Not explaining that $1-x$ is the fraction not affected, which is required because the question says 'explain why'.
Confusing the constant $k$ with a rate or percentage.

Things to Be Careful About

$x$ is a fraction, so $0 \leq x \leq 1$ ; this makes $1-x$ non-negative.
The mark scheme awards the mark only if the factor $1-x$ is explained, so state clearly where it comes from.

Techniques used

translate a proportionality statement into a differential equationidentify the fraction not affected as 1 - x

(b)

When the disease is first detected, one quarter of the trees are affected.
Two years later, one third of the trees are affected.

Solve the differential equation to find the number of years from the time when the disease is first detected until the time when three quarters of the trees are affected. Give your answer correct to the nearest year.

8M Similar

DifficultyMedium-Hard

Worked solution

Approach

Separate the variables so that all terms involving $x$ are on one side and all terms involving $t$ are on the other. Integrate both sides, using partial fractions for $\frac{1}{x(1-x)}$ . Then use the two given data points to find the constant of integration and the value of $k$ . Finally substitute $x=\frac34$ and solve for $t$ .

Working

Separating variables:

\int \frac{1}{x(1-x)}\,\mathrm{d}x = \int k\,\mathrm{d}t

Write the integrand in partial fractions:

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

Multiplying by $x(1-x)$ :

A(1-x) + Bx = 1

Comparing coefficients gives $A=1$ and $B=1$ , so

\frac{1}{x(1-x)} = \frac{1}{x} + \frac{1}{1-x}

Integrating:

\ln x - \ln(1-x) = kt + C

or equivalently

\ln\left(\frac{x}{1-x}\right) = kt + C

Use $t=0$ , $x=\frac14$ :

C = \ln\left(\frac{\frac14}{1-\frac14}\right) = \ln\left(\frac13\right)

Use $t=2$ , $x=\frac13$ :

\ln\left(\frac{\frac13}{\frac23}\right) = 2k + \ln\left(\frac13\right)

\ln\left(\frac12\right) = 2k + \ln\left(\frac13\right)

2k = \ln\left(\frac12\right) - \ln\left(\frac13\right) = \ln\left(\frac32\right)

k = \frac12 \ln\left(\frac32\right)

Therefore

\ln\left(\frac{x}{1-x}\right) = \frac12 \ln\left(\frac32\right)t + \ln\left(\frac13\right)

Now set $x=\frac34$ . Since $\frac{\frac34}{\frac14}=3$ ,

\ln 3 = \frac12 \ln\left(\frac32\right)t + \ln\left(\frac13\right)

Using $\ln\left(\frac13\right)=-\ln 3$ :

2\ln 3 = \frac12 \ln\left(\frac32\right)t

t = \frac{2\ln 3}{\frac12 \ln\left(\frac32\right)} = \frac{4\ln 3}{\ln\left(\frac32\right)} \approx 10.838

Answer

The number of years from first detection until three quarters of the trees are affected is $11$ years, to the nearest year.

Final answer

11 years

Detailed explanation

Walkthrough

We begin with the differential equation $\frac{dx}{dt}=kx(1-x)$ . To solve it, separate variables: bring all $x$ terms to the left with $dx$ and all $t$ terms to the right with $dt$ . This gives $\int \frac{1}{x(1-x)}\,dx = \int k\,dt$ .

The integrand on the left is not a standard integral, so we decompose it into partial fractions. Write $\frac{1}{x(1-x)} = \frac{A}{x}+\frac{B}{1-x}$ . Multiplying through by $x(1-x)$ gives $A(1-x)+Bx=1$ , so $A=1$ and $B=1$ . Thus the integral becomes $\int \left(\frac{1}{x}+\frac{1}{1-x}\right)dx$ . The integral of $\frac{1}{x}$ is $\ln x$ ; the integral of $\frac{1}{1-x}$ is $-\ln(1-x)$ because the derivative of $\ln(1-x)$ is $-\frac{1}{1-x}$ . So the left side is $\ln x - \ln(1-x)$ , which can be combined as $\ln\left(\frac{x}{1-x}\right)$ .

Now use the data. At first detection, $t=0$ and $x=\frac14$ . Substitute to find the constant $C$ : $C = \ln\left(\frac{\frac14}{\frac34}\right)=\ln\frac13$ . Then use $t=2$ , $x=\frac13$ to find $k$ : $\ln\frac12 = 2k + \ln\frac13$ , so $k=\frac12\ln\frac32$ .

Finally, set $x=\frac34$ . Then $\frac{x}{1-x}=3$ , so $\ln 3 = \frac12\ln\frac32\, t + \ln\frac13$ . Since $\ln\frac13=-\ln 3$ , this gives $2\ln 3 = \frac12\ln\frac32\, t$ , and hence $t=\frac{4\ln 3}{\ln(3/2)}\approx 10.838$ . To the nearest year, this is 11 years.

Key Takeaways

This question combines separable differential equations, partial fractions, and logarithms. It also shows how initial conditions determine the arbitrary constant and the parameter $k$ in a real-life model.

Common Mistakes

Forgetting the constant of integration when integrating.
Making a sign error in the partial fractions or in integrating $\frac{1}{1-x}$ ; the mark scheme condones a sign error between the fractions but it must be corrected later.
Using $t=1$ and $t=3$ instead of $t=0$ and $t=2$ ; this leads to $t=12$ years rather than $11$ years.
Not showing the final equation $\ln\frac{x}{1-x}=\frac12\ln\frac32\,t+\ln\frac13$ ; the mark scheme requires it to be seen.
Rounding intermediate values too early, which can change the final nearest-year answer.

Things to Be Careful About

The time origin is first detection, so $t=0$ corresponds to $x=\frac14$ , and $t=2$ corresponds to $x=\frac13$ .
Since $x$ is a fraction, $0<x<1$ throughout, so $\ln x$ and $\ln(1-x)$ are defined.
The mark scheme allows equivalent forms of the general solution, e.g. $\ln\frac{x}{1-x}=(0.2027...)t-1.0986...$ , but the final equation must be seen.
The answer must be given to the nearest year: $10.838$ rounds to $11$ , not $10$ .

Techniques used

separate variables and integrate both sidesdecompose a rational integrand into partial fractionsintegrate using natural logarithmsapply initial conditions to determine constantssolve a logarithmic equation for time

Mathematics 9709/35 — October/November 2025

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