It is given that $z_1 = 3e^{\frac{1}{4}\pi i}$, $z_2 = \frac{3}{2}e^{\frac{1}{6}\pi i}$ and $\omega = 2e^{\frac{1}{2}\pi i}$. (a) State the values of $\omega z_1$ and $\omega z_2$. Give your answers in the form $re^{i\theta}$, where $r > 0$ and $-\pi < \theta \leq \pi$. (b) On a sketch of an Argand diagram with origin $O$, show the points $A, B, C$ and $D$ representing the complex numbers $z_1, z_2, \omega z_1$ and $\omega z_2$ respectively. (c) State the geometric effects…

(a) ωz₁ = 6e^(3πi/4) and ωz₂ = 3e^(2πi/3) · (b) Points plotted at A(3, π/4), B(1.5, π/6), C(6, 3π/4), D(3, 2π/3) · (c) Rotation of π/2 radians anticlockwise and enlargement with scale factor 2

(a) Express $5\sin(x + \frac{1}{6}\pi) - 4\cos x$ in the form $R\sin(x - \alpha)$, where $R > 0$ and $0 < \alpha < \frac{1}{2}\pi$. State the exact value of $R$ and give the value of $\alpha$ correct to 3 decimal places. (b) Hence solve the equation $5\sin(2\theta + \frac{1}{6}\pi) - 4\cos 2\theta = \sqrt{7}$ for $0 \leq \theta \leq \pi$. Give your answers correct to 2 decimal places.

(a) R = √21, α ≈ 0.333 radians · (b) θ ≈ 0.47 or θ ≈ 1.43

The constant $a$ is such that $\int_1^a 6x\ln x\,dx = 4$. (a) Show that $a = \exp\left(\frac{1}{6}\left(\frac{5}{a^2} + 3\right)\right)$, where $\exp(x)$ denotes $e^x$. (b) Verify by calculation that $a$ lies between 2 and 2.1. (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 = exp((1/6)(5/a^2 + 3)) · (b) 2 < a < 2.1 · (c) a = 2.02

9709/31 May/June 2025 — Mathematics Solved Past Paper

Q1Medium-EasyAlgebra

(a)

Sketch the graph of $y = |2x - 3|$ .

1M Similar

DifficultyEasy

Worked solution

Approach

Identify the vertex of the graph by setting the expression inside the modulus to zero. Then find the y-intercept by substituting $x = 0$ . Sketch the V-shaped graph passing through these points.

Working

Find the vertex where $2x - 3 = 0$ :

2x - 3 = 0

x = \frac{3}{2}

The vertex is at $\left(\frac{3}{2}, 0\right)$ .

Find the y-intercept by substituting $x = 0$ :

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

The y-intercept is at $(0, 3)$ .

The graph of $y = |2x - 3|$ is a V-shaped curve with its vertex on the x-axis at $x = \frac{3}{2}$ , symmetric about the line $x = \frac{3}{2}$ , and passing through $(0, 3)$ .

Answer

The graph is a V-shape with vertex at $\left(\frac{3}{2}, 0\right)$ and y-intercept at $(0, 3)$ .

Final answer

V-shaped graph with vertex at (3/2, 0) and y-intercept at (0, 3)

Detailed explanation

Walkthrough

To sketch the graph of a modulus function $y = |ax + b|$ , we first find the vertex by setting the inner expression to zero. Here, $2x - 3 = 0$ gives $x = 3/2$ , so the vertex is at $(3/2, 0)$ . Next, we find the y-intercept by evaluating the function at $x = 0$ , which gives $y = |-3| = 3$ . Since the modulus function outputs non-negative values, the graph is V-shaped, opening upwards, with the two linear branches meeting at the vertex. The graph is symmetric about the vertical line $x = 3/2$ .

Key Takeaways

The graph of $y = |ax + b|$ is always V-shaped with its vertex on the x-axis at $x = -b/a$ .
The y-intercept is found by substituting $x = 0$ and taking the absolute value.
The graph consists of two straight lines with slopes $a$ and $-a$ .

Common Mistakes

Forgetting to mark the vertex and y-intercept clearly on the axes.
Drawing a curve instead of straight lines for the branches.
Placing the vertex at the wrong position on the x-axis.

Things to Be Careful About

Ensure the graph exists for negative $x$ values as well, extending to the second quadrant.
The branches must be straight lines, not curved.
The vertex must touch the x-axis, not the y-axis.

Techniques used

identify vertex and y-intercept of modulus functionsketch V-shaped modulus graph

(b)

Solve the inequality $3x - 1 < |2x - 3|$ .

2M Similar

DifficultyMedium-Easy

Worked solution

Approach

To solve $3x - 1 < |2x - 3|$ , consider the definition of the modulus function. Split into two cases based on the sign of $2x - 3$ . Solve the resulting linear inequality in each case and check for consistency with the case condition.

Working

Case 1: $2x - 3 \geq 0$ , which means $x \geq \frac{3}{2}$

In this case, $|2x - 3| = 2x - 3$ . The inequality becomes:

3x - 1 < 2x - 3

x < -2

This contradicts the condition $x \geq \frac{3}{2}$ . Thus, there is no solution in this case.

Case 2: $2x - 3 < 0$ , which means $x < \frac{3}{2}$

In this case, $|2x - 3| = -(2x - 3) = 3 - 2x$ . The inequality becomes:

3x - 1 < 3 - 2x

5x < 4

x < \frac{4}{5}

This is consistent with the condition $x < \frac{3}{2}$ since $\frac{4}{5} < \frac{3}{2}$ .

Combining both cases, the final solution is $x < \frac{4}{5}$ .

Answer

x < \frac{4}{5}

Final answer

x < 4/5

Detailed explanation

Walkthrough

The inequality $3x - 1 < |2x - 3|$ involves a modulus expression. To solve it, we remove the modulus by considering the two possible cases for the expression inside: $2x - 3 \geq 0$ and $2x - 3 < 0$ .

In the first case ( $x \geq 3/2$ ), the modulus is simply $2x - 3$ . Solving $3x - 1 < 2x - 3$ gives $x < -2$ , which has no overlap with $x \geq 3/2$ , so no solution here.

In the second case ( $x < 3/2$ ), the modulus is $-(2x - 3) = 3 - 2x$ . Solving $3x - 1 < 3 - 2x$ gives $5x < 4$ , or $x < 4/5$ . Since $4/5$ is less than $3/2$ , this entire range satisfies the case condition.

Thus, the final solution is $x < 4/5$ .

Key Takeaways

Solving modulus inequalities requires splitting into cases based on the sign of the inner expression.
Always check that the solution from each case is consistent with the condition defining that case.
The final answer is the union of all valid solutions from each case.

Common Mistakes

Forgetting to check the consistency of the solution with the case condition.
Incorrectly removing the modulus sign (e.g., writing $|2x - 3| = 2x - 3$ for all $x$ ).
Missing the second case entirely and only finding part of the solution.

Things to Be Careful About

Ensure the boundary values (like $x = 3/2$ ) are correctly assigned to the appropriate case.
Verify that the final inequality direction is correct after combining terms.
The mark scheme also accepts an alternative method using $(3x - 1)^2 < (2x - 3)^2$ , but students must be careful to interpret the resulting quadratic inequality correctly and may need to reject extraneous roots.

Techniques used

set up case for negative modulus expressionsolve linear inequalitycheck solution against modulus condition

Q23MMedium-EasyLogarithmic and Exponential Functions

It is given that $2\ln p + \ln(p - 1) - \frac{1}{2}\ln(q + 1) = 3$ .

Find $q$ in terms of $p$ .

DifficultyMedium-Easy

Worked solution

Approach

Apply the laws of logarithms to combine the terms into a single logarithm, then exponentiate both sides to remove the logarithm and solve for $q$ .

Working

Start with

2\ln p + \ln(p-1) - \frac{1}{2}\ln(q+1) = 3.

Use the power law and write $3 = \ln \mathrm{e}^3$ :

\ln p^2 + \ln(p-1) - \ln\sqrt{q+1} = \ln \mathrm{e}^3.

Combine the logarithms:

\ln\left(\frac{p^2(p-1)}{\sqrt{q+1}}\right) = \ln \mathrm{e}^3.

Since $\ln$ is one-to-one, equate the arguments:

\frac{p^2(p-1)}{\sqrt{q+1}} = \mathrm{e}^3.

Rearrange:

p^2(p-1) = \mathrm{e}^3\sqrt{q+1}.

Square both sides:

p^4(p-1)^2 = \mathrm{e}^6(q+1).

Solve for $q$ :

q = \frac{p^4(p-1)^2}{\mathrm{e}^6} - 1.

Equivalently,

q = \left(\frac{p^2(p-1)}{\mathrm{e}^3}\right)^2 - 1.

Answer

q = \left(\frac{p^2(p-1)}{\mathrm{e}^3}\right)^2 - 1

Final answer

q = (p^2(p-1)/e^3)^2 - 1

Detailed explanation

Walkthrough

The equation contains three logarithmic terms plus a constant. To solve for $q$ , we first express every term as a single logarithm. The power law lets us move coefficients inside: $2\ln p = \ln p^2$ and $\frac{1}{2}\ln(q+1) = \ln\sqrt{q+1}$ . Also rewrite the constant $3$ as $\ln \mathrm{e}^3$ , because $\ln$ and $\mathrm{e}^x$ are inverse functions. Then combine the three logarithms into one using the product and quotient laws: addition becomes multiplication and subtraction becomes division. This gives a single equation of the form $\ln A = \ln B$ . Because the logarithm function is one-to-one, we can drop the logs and equate $A = B$ . Then rearrange to isolate $\sqrt{q+1}$ , square both sides to remove the square root, and solve for $q$ .

Key Takeaways

The laws of logarithms allow coefficients to be moved inside as powers: $a\ln x = \ln x^a$ .
A constant can be written as a logarithm using the inverse relationship $c = \ln \mathrm{e}^c$ .
When both sides of an equation are single logarithms with the same base, the arguments can be equated.
Solving for a variable inside a square root requires squaring both sides, then rearranging carefully.

Common Mistakes

Forgetting to rewrite the constant $3$ as $\ln \mathrm{e}^3$ before combining.
Incorrectly combining logs, e.g. treating $\ln p^2 + \ln(p-1) - \ln\sqrt{q+1}$ as $\ln(p^2 + p - 1 - \sqrt{q+1})$ . Logs combine by multiplication and division, not addition and subtraction.
After obtaining $p^2(p-1) = \mathrm{e}^3\sqrt{q+1}$ , forgetting to square both sides, or squaring only part of the left-hand side.
Dropping the $-1$ when solving for $q$ .
Not stating or using the domain assumptions $p>0$ , $p-1>0$ , $q+1>0$ ; these are needed for the original logarithms to be defined.

Things to Be Careful About

The mark scheme awards M1 for using a logarithm of a root or a power; show $2\ln p = \ln p^2$ and $\frac{1}{2}\ln(q+1) = \ln\sqrt{q+1}$ explicitly.
A1 is given for the log-free equation $p^2(p-1) = \mathrm{e}^3\sqrt{q+1}$ . Ensure this is written before squaring.
The final answer can be written in either equivalent form; the mark scheme allows OE (or equivalent) and ISW (ignore subsequent working).
Squaring is valid here because both sides are positive under the domain assumptions, so no extraneous solutions are introduced.

Techniques used

combine logarithmic terms using product and quotient lawsrewrite constants using inverse relationship of logarithms and exponentialsexponentiate to remove logarithmsrearrange and square to solve for q

Similar questions

Q36MMediumComplex Numbers

Find the complex numbers $z$ for which

\frac{z + 5i}{z - 5}

is real and $|z| = \sqrt{17}$ . Give your answers in the form $z = x + iy$ , where $x$ and $y$ are real.

DifficultyMedium

Worked solution

Approach

Write $z = x + \mathrm{i}y$ with $x, y \in \mathbb{R}$ . Multiply the quotient by the conjugate of the denominator so that the denominator becomes real. The quotient is real exactly when the imaginary part of the resulting numerator is zero. This gives a linear equation in $x$ and $y$ . Then use $|z| = \sqrt{17}$ to obtain a second equation and solve simultaneously.

Working

Let $z = x + \mathrm{i}y$ . Then

\frac{z + 5\mathrm{i}}{z - 5} = \frac{x + \mathrm{i}(y + 5)}{(x - 5) + \mathrm{i}y}.

Multiply numerator and denominator by $(x - 5) - \mathrm{i}y$ :

\frac{x + \mathrm{i}(y + 5)}{(x - 5) + \mathrm{i}y} \cdot \frac{(x - 5) - \mathrm{i}y}{(x - 5) - \mathrm{i}y}.

The numerator becomes

x(x - 5) + y(y + 5) + \mathrm{i}\left((y + 5)(x - 5) - xy\right).

Since the denominator is real, the quotient is real only if the imaginary part of the numerator is zero:

(y + 5)(x - 5) - xy = 0.

Expanding,

xy - 5y + 5x - 25 - xy = 0

5x - 5y - 25 = 0

x - y = 5. \tag{1}

The modulus condition gives

x^2 + y^2 = 17. \tag{2}

From (1), $y = x - 5$ . Substitute into (2):

x^2 + (x - 5)^2 = 17

2x^2 - 10x + 25 = 17

2x^2 - 10x + 8 = 0

x^2 - 5x + 4 = 0

(x - 1)(x - 4) = 0.

Hence $x = 1$ or $x = 4$ . Since $y = x - 5$ , the corresponding values are $y = -4$ or $y = -1$ .

Answer

z = 1 - 4\mathrm{i} \quad \text{or} \quad z = 4 - \mathrm{i}.

Final answer

z = 4 - i or z = 1 - 4i

Detailed explanation

Walkthrough

We are told that the quotient $\frac{z + 5\mathrm{i}}{z - 5}$ is real. A complex number is real when its imaginary part is zero, but the denominator here is complex, so we cannot immediately read off the imaginary part. The standard technique is to multiply the numerator and denominator by the conjugate of the denominator, $(x - 5) - \mathrm{i}y$ . This makes the denominator $(x - 5)^2 + y^2$ , a real number, while the numerator becomes a new complex number whose imaginary part determines whether the quotient is real.

After expanding the numerator, we set its imaginary part equal to zero. This gives $x - y = 5$ . Next, the condition $|z| = \sqrt{17}$ means $x^2 + y^2 = 17$ . Substituting $y = x - 5$ into this equation gives a quadratic in $x$ . Factorising it gives $x = 1$ or $x = 4$ , and then $y = -4$ or $y = -1$ respectively. These give the two required complex numbers.

Key Takeaways

To test whether a quotient of complex numbers is real, multiply by the conjugate of the denominator.
After rationalising, the denominator is real, so the real/imaginary status of the quotient is determined by the numerator.
The modulus of $z = x + \mathrm{i}y$ is $\sqrt{x^2 + y^2}$ , so $|z| = \sqrt{17}$ is equivalent to $x^2 + y^2 = 17$ .
A linear equation and a quadratic equation in $x$ and $y$ can be solved by substitution.

Common Mistakes

Multiplying only the numerator by the conjugate and not the denominator. Both must be multiplied so the value of the quotient is unchanged.
Trying to equate imaginary parts before rationalising the denominator.
Sign errors when expanding $(y + 5)(x - 5)$ ; the $xy$ terms cancel, leaving $5x - 5y - 25$ .
Forgetting that $|z| = \sqrt{17}$ gives $x^2 + y^2 = 17$ , not $x^2 + y^2 = \sqrt{17}$ .
Discarding one of the two solutions when solving the quadratic.

Things to Be Careful About

The denominator $(x - 5) + \mathrm{i}y$ must not be zero; this would require $x = 5$ and $y = 0$ , which is not among the solutions.
Keep $x$ and $y$ real throughout; only the imaginary unit $\mathrm{i}$ carries the imaginary part.
The mark scheme expects the conjugate multiplication and the step equating the imaginary part to zero to be shown clearly.
Both answers satisfy the original conditions, so include both $z = 1 - 4\mathrm{i}$ and $z = 4 - \mathrm{i}$ .

Techniques used

multiply by the conjugate of the denominatorequate the imaginary part of the numerator to zeroapply the modulus conditionsolve simultaneous equations by substitution

Similar questions

Q46MMediumDifferentiation

The parametric equations of a curve are

x = e^{\tan t}, \quad y = 3\tan^2 t

Find the equation of the tangent to the curve at the point $(e, 3)$ . Give your answer in the form $y = mx + c$ , where $m$ and $c$ are exact.

DifficultyMedium

Worked solution

Approach

At the given point, determine the parameter value $t$ from $x=e$ (or $y=3$ ). Differentiate $x$ and $y$ with respect to $t$ using the chain rule, then use

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

to find the gradient. Substitute $t$ to get the gradient at the point, and use the point-gradient form of a straight line.

Working

At $(e,3)$ :

e^{\tan t} = e \implies \tan t = 1

(so $t = \frac{\pi}{4}$ is a suitable value).

Differentiate with respect to $t$ :

\frac{dx}{dt} = e^{\tan t} \cdot \sec^2 t

\frac{dy}{dt} = 6\tan t \cdot \sec^2 t

Therefore:

\frac{dy}{dx} = \frac{6\tan t \sec^2 t}{\sec^2 t\, e^{\tan t}} = 6\tan t\, e^{-\tan t}

At $\tan t = 1$ :

\frac{dy}{dx} = 6(1)e^{-1} = \frac{6}{e}

Equation of tangent through $(e,3)$ :

y - 3 = \frac{6}{e}(x - e)

y = \frac{6}{e}x - 6 + 3 = \frac{6}{e}x - 3

Answer

y = \frac{6}{e}x - 3

Final answer

y = (6/e)x - 3

Detailed explanation

Walkthrough

The curve is given parametrically, so the gradient at a point is not directly available. We first locate the parameter value that corresponds to $(e,3)$ . Since $x = e^{\tan t}$ , setting $x=e$ gives $e^{\tan t}=e$ , hence $\tan t=1$ . This is the key value needed for the gradient.

Next, differentiate both parametric equations with respect to $t$ . For $x$ , the chain rule gives $\frac{dx}{dt} = e^{\tan t} \cdot \sec^2 t$ because the derivative of $\tan t$ is $\sec^2 t$ . For $y=3\tan^2 t$ , use the chain rule again: the derivative of $\tan^2 t$ is $2\tan t \cdot \sec^2 t$ , so $\frac{dy}{dt}=6\tan t \sec^2 t$ .

The gradient of the curve is $\frac{dy}{dx}=\frac{dy/dt}{dx/dt}$ . Dividing gives $\frac{6\tan t \sec^2 t}{\sec^2 t\, e^{\tan t}}$ , and the $\sec^2 t$ factor cancels, leaving $6\tan t\, e^{-\tan t}$ . Substituting $\tan t=1$ gives gradient $\frac{6}{e}$ .

Finally, use the straight-line equation through $(e,3)$ with gradient $\frac{6}{e}$ : $y-3 = \frac{6}{e}(x-e)$ . Expanding gives $y=\frac{6}{e}x-3$ .

Key Takeaways

This question tests parametric differentiation, the chain rule for composite functions, and forming the equation of a tangent. It also shows that when a point is given in terms of $x$ and $y$ , you must first find the corresponding parameter value before differentiating.

Common Mistakes

Forgetting to find $t$ (or $\tan t$ ) before substituting into $\frac{dy}{dx}$ .
Differentiating $x$ incorrectly: the derivative of $e^{\tan t}$ is $e^{\tan t}\sec^2 t$ , not just $e^{\tan t}$ .
Cancelling incorrectly when forming $\frac{dy}{dx}$ ; the $\sec^2 t$ factors cancel, but the exponential in the denominator remains as $e^{-\tan t}$ .
Substituting $t=\frac{\pi}{4}$ into $\frac{dy}{dx}$ but not simplifying to $\frac{6}{e}$ .
Making an algebraic slip when expanding $y-3=\frac{6}{e}(x-e)$ : the constant term is $-3$ , not $+3$ .

Things to Be Careful About

The answer must be exact, so leave the gradient as $\frac{6}{e}$ rather than a decimal.
The parameter value could be $t=\frac{\pi}{4}+k\pi$ ; all such values give $\tan t=1$ , so the gradient is the same.
The gradient formula $\frac{dy}{dx}=\frac{dy/dt}{dx/dt}$ is only valid when $\frac{dx}{dt}\neq 0$ ; here $\sec^2 t\neq 0$ and $e^{\tan t}\neq 0$ .
The final line must be in the form $y=mx+c$ with exact $m$ and $c$ .

Techniques used

differentiate parametric equations with respect to the parameterapply the chain rule to composite exponential and trigonometric functionsdivide parametric derivatives to obtain the gradientform the equation of a tangent line

Similar questions

Q55MMedium-EasyAlgebra

The polynomial $3x^3 + pax^2 + 7a^2x + qa^3$ is denoted by $f(x)$ , where $p$ , $q$ and $a$ are constants and $a \neq 0$ .

When $f(x)$ is divided by $(x + 2a)$ the remainder is $-22a^3$ . When $f(x)$ is divided by $(3x - a)$ the remainder is $-a^3$ .

Find the values of $p$ and $q$ .

DifficultyMedium-Easy

Worked solution

Approach

Since the remainder when a polynomial $f(x)$ is divided by a linear factor $x - c$ is $f(c)$ , we evaluate $f$ at the roots of the two divisors and set the results equal to the given remainders. This gives two linear equations in $p$ and $q$ , which we solve simultaneously.

Working

For the divisor $x + 2a$ , the root is $x = -2a$ , so:

f(-2a) = -22a^3

Substitute $x = -2a$ into $f(x)$ :

\begin{aligned} f(-2a) &= 3(-2a)^3 + pa(-2a)^2 + 7a^2(-2a) + qa^3 \\ &= -24a^3 + 4pa^3 - 14a^3 + qa^3 \\ &= (4p + q - 38)a^3 \end{aligned}

Equate to $-22a^3$ :

(4p + q - 38)a^3 = -22a^3

Since $a \neq 0$ , divide by $a^3$ :

4p + q - 38 = -22

4p + q = 16

For the divisor $3x - a$ , the root is $x = \frac{a}{3}$ , so:

f\left(\frac{a}{3}\right) = -a^3

Substitute $x = \frac{a}{3}$ :

\begin{aligned} f\left(\frac{a}{3}\right) &= 3\left(\frac{a}{3}\right)^3 + pa\left(\frac{a}{3}\right)^2 + 7a^2\left(\frac{a}{3}\right) + qa^3 \\ &= \frac{a^3}{9} + \frac{pa^3}{9} + \frac{7a^3}{3} + qa^3 \\ &= \left(\frac{1+p}{9} + \frac{7}{3} + q\right)a^3 \end{aligned}

Equate to $-a^3$ :

\left(\frac{1+p}{9} + \frac{7}{3} + q\right)a^3 = -a^3

Divide by $a^3$ and simplify:

\frac{1+p}{9} + \frac{7}{3} + q = -1

Multiply by 9:

1 + p + 21 + 9q = -9

p + 9q = -31

Now solve the simultaneous equations:

4p + q = 16, \quad p + 9q = -31

From the first equation, $q = 16 - 4p$ . Substitute into the second:

p + 9(16 - 4p) = -31

p + 144 - 36p = -31

-35p = -175

p = 5

Then:

q = 16 - 4(5) = -4

Answer

p = 5, \quad q = -4

Final answer

p = 5, q = -4

Detailed explanation

Walkthrough

The remainder theorem says that if a polynomial $f(x)$ is divided by a linear expression $x - c$ , the remainder is $f(c)$ . For the first divisor $x + 2a$ , we can write it as $x - (-2a)$ , so the root is $c = -2a$ . Therefore the remainder is $f(-2a)$ , and the question tells us this equals $-22a^3$ . We substitute $x = -2a$ into every term of $f(x)$ , collect the $a^3$ terms, and equate the coefficient to $-22a^3$ . Since $a \neq 0$ , we may divide by $a^3$ to get a linear equation in $p$ and $q$ .

For the second divisor $3x - a$ , the root is found by setting $3x - a = 0$ , giving $x = \frac{a}{3}$ . So the remainder is $f\left(\frac{a}{3}\right)$ , and this equals $-a^3$ . Substituting and simplifying gives another linear equation. We then solve the two equations simultaneously, for example by substituting $q = 16 - 4p$ into the second equation, to find $p$ and $q$ .

Key Takeaways

The remainder theorem lets us avoid long division: to find the remainder when dividing by a linear factor, evaluate the polynomial at the root of that factor. To find the root of $ax + b$ , solve $ax + b = 0$ . This question also shows how two remainder conditions can be turned into a system of linear equations and solved for unknown coefficients.

Common Mistakes

Using $f(2a)$ instead of $f(-2a)$ for the divisor $x + 2a$ .
Using $f(3a)$ instead of $f\left(\frac{a}{3}\right)$ for the divisor $3x - a$ .
Forgetting to evaluate every term of the polynomial, especially the $7a^2x$ term.
Making sign errors with powers of negative numbers, such as $(-2a)^3 = -8a^3$ .
Forgetting that $a \neq 0$ , so it is valid to divide by $a^3$ .
Not simplifying the equations before solving, which can lead to arithmetic slips.

Things to Be Careful About

The root of $x + 2a$ is negative: $x = -2a$ .
When substituting $x = -2a$ , remember $(-2a)^2 = 4a^2$ and $(-2a)^3 = -8a^3$ .
The mark scheme requires the terms to be evaluated; an unsupported final answer may not be awarded full marks.
For $x = \frac{a}{3}$ , the cube gives $\frac{a^3}{27}$ , so $3\left(\frac{a}{3}\right)^3 = \frac{a^3}{9}$ .
The simplified equations are $4p + q = 16$ and $p + 9q = -31$ ; check these before solving.

Techniques used

apply the remainder theoremevaluate the polynomial at the roots of the divisorsform linear equations from remainder conditionssolve simultaneous linear equations

Similar questions

Q6MediumComplex Numbers

It is given that $z_1 = 3e^{\frac{1}{4}\pi i}$ , $z_2 = \frac{3}{2}e^{\frac{1}{6}\pi i}$ and $\omega = 2e^{\frac{1}{2}\pi i}$ .

(a)

State the values of $\omega z_1$ and $\omega z_2$ . Give your answers in the form $re^{i\theta}$ , where $r > 0$ and $-\pi < \theta \leq \pi$ .

2M Similar

DifficultyMedium-Easy

Worked solution

Approach

To multiply complex numbers in exponential form, multiply their moduli and add their arguments. Then ensure the resulting argument is within the specified principal range $-\pi < \theta \leq \pi$ .

Working

Given:

\omega = 2e^{\frac{1}{2}\pi i}, \quad z_1 = 3e^{\frac{1}{4}\pi i}, \quad z_2 = \frac{3}{2}e^{\frac{1}{6}\pi i}

For $\omega z_1$ :
Modulus: $2 \times 3 = 6$
Argument: $\frac{1}{2}\pi + \frac{1}{4}\pi = \frac{3}{4}\pi$

Since $-\pi < \frac{3}{4}\pi \leq \pi$ , the result is:

\omega z_1 = 6e^{\frac{3}{4}\pi i}

For $\omega z_2$ :
Modulus: $2 \times \frac{3}{2} = 3$
Argument: $\frac{1}{2}\pi + \frac{1}{6}\pi = \frac{3}{6}\pi + \frac{1}{6}\pi = \frac{4}{6}\pi = \frac{2}{3}\pi$

Since $-\pi < \frac{2}{3}\pi \leq \pi$ , the result is:

\omega z_2 = 3e^{\frac{2}{3}\pi i}

Answer

\omega z_1 = 6e^{\frac{3}{4}\pi i} \quad \text{and} \quad \omega z_2 = 3e^{\frac{2}{3}\pi i}

Final answer

ωz₁ = 6e^(3πi/4) and ωz₂ = 3e^(2πi/3)

Detailed explanation

Walkthrough

The problem asks for the product of complex numbers given in exponential form. The rule for multiplying complex numbers in exponential form $r_1 e^{i\theta_1}$ and $r_2 e^{i\theta_2}$ is to multiply the moduli ( $r_1 \times r_2$ ) and add the arguments ( $\theta_1 + \theta_2$ ).

For $\omega z_1$ , we multiply the moduli $2$ and $3$ to get $6$ , and add the arguments $\frac{1}{2}\pi$ and $\frac{1}{4}\pi$ to get $\frac{3}{4}\pi$ . Since $\frac{3}{4}\pi$ is already within the required range $(-\pi, \pi]$ , no adjustment is needed.

For $\omega z_2$ , we multiply the moduli $2$ and $\frac{3}{2}$ to get $3$ , and add the arguments $\frac{1}{2}\pi$ and $\frac{1}{6}\pi$ . Finding a common denominator gives $\frac{3}{6}\pi + \frac{1}{6}\pi = \frac{4}{6}\pi = \frac{2}{3}\pi$ . This is also within the required range.

Key Takeaways

When multiplying complex numbers in exponential form, multiply the moduli and add the arguments.
Always check that the final argument lies within the principal range specified by the question, typically $-\pi < \theta \leq \pi$ .

Common Mistakes

Forgetting to add the arguments and instead multiplying them.
Forgetting to multiply the moduli and instead adding them.
Failing to simplify the argument to the required range (e.g., leaving it as $\frac{5}{3}\pi$ instead of $-\frac{1}{3}\pi$ if it exceeded $\pi$ ).

Things to Be Careful About

Ensure the argument is strictly within $-\pi < \theta \leq \pi$ . If the sum of arguments exceeds $\pi$ , subtract $2\pi$ to bring it back into range. If it is less than $-\pi$ , add $2\pi$ . In this case, both sums naturally fall within the range.

Techniques used

multiply moduli of complex numbers in exponential formadd arguments of complex numbers in exponential formverify argument is within principal range

(b)

On a sketch of an Argand diagram with origin $O$ , show the points $A, B, C$ and $D$ representing the complex numbers $z_1, z_2, \omega z_1$ and $\omega z_2$ respectively.

2M Similar

DifficultyMedium

Worked solution

Approach

To plot points on an Argand diagram, use the modulus as the distance from the origin and the argument as the angle measured anticlockwise from the positive real axis. Use the values calculated in part (a) for points C and D.

Working

Point A represents $z_1 = 3e^{\frac{1}{4}\pi i}$ :

Modulus: $3$
Argument: $\frac{\pi}{4}$ ( $45^\circ$ )

Point B represents $z_2 = \frac{3}{2}e^{\frac{1}{6}\pi i}$ :

Modulus: $1.5$
Argument: $\frac{\pi}{6}$ ( $30^\circ$ )

Point C represents $\omega z_1 = 6e^{\frac{3}{4}\pi i}$ :

Modulus: $6$
Argument: $\frac{3\pi}{4}$ ( $135^\circ$ )

Point D represents $\omega z_2 = 3e^{\frac{2}{3}\pi i}$ :

Modulus: $3$
Argument: $\frac{2\pi}{3}$ ( $120^\circ$ )

Answer

Points A, B, C, and D are plotted on the Argand diagram with the specified moduli and arguments as shown in the diagram above.

Final answer

Points plotted at A(3, π/4), B(1.5, π/6), C(6, 3π/4), D(3, 2π/3)

Detailed explanation

Walkthrough

An Argand diagram is a graphical representation of complex numbers where the horizontal axis represents the real part and the vertical axis represents the imaginary part. A complex number $re^{i\theta}$ is plotted at a distance $r$ from the origin, at an angle $\theta$ measured anticlockwise from the positive real axis.

Point A ( $z_1$ ): Distance 3 from origin, angle 45°. This places it in the first quadrant.
Point B ( $z_2$ ): Distance 1.5 from origin, angle 30°. This also places it in the first quadrant, closer to the origin and lower than A.
Point C ( $\omega z_1$ ): Distance 6 from origin, angle 135°. This places it in the second quadrant, exactly twice as far from the origin as A, and rotated 90° anticlockwise from A.
Point D ( $\omega z_2$ ): Distance 3 from origin, angle 120°. This places it in the second quadrant, exactly twice as far from the origin as B, and rotated 90° anticlockwise from B.

The diagram should clearly show these relative positions, with lines from the origin to each point indicating their moduli and angles.

Key Takeaways

The modulus of a complex number in exponential form gives its distance from the origin on the Argand diagram.
The argument gives the angle of the line joining the origin to the point, measured anticlockwise from the positive real axis.
Multiplying by a complex number with modulus $r$ and argument $\theta$ scales the distance from the origin by $r$ and rotates the point by $\theta$ .

Common Mistakes

Plotting the modulus on the x-axis and the argument on the y-axis instead of using polar coordinates.
Measuring the argument clockwise instead of anticlockwise.
Failing to scale the points correctly relative to each other (e.g., C should be twice as far from the origin as A).

Things to Be Careful About

Ensure the angles are measured correctly in radians or degrees and placed in the correct quadrant.
The question asks for a sketch, so exact precision is not required, but the relative positions, angles, and distances must be clearly correct to demonstrate the geometric transformation.

Techniques used

plot complex numbers on Argand diagram using modulus and argumentinterpret geometric transformation from exponential form

(c)

State the geometric effects of multiplying $z_1$ and $z_2$ by $\omega$ .

2M Similar

DifficultyMedium-Easy

Worked solution

Approach

When a complex number $z$ is multiplied by another complex number $w = re^{i\theta}$ , the geometric effect on the Argand diagram is an enlargement (scaling) by a factor of $r$ and a rotation by an angle $\theta$ anticlockwise about the origin.

Working

Given $\omega = 2e^{\frac{1}{2}\pi i}$ :

The modulus of $\omega$ is $r = 2$ .
The argument of $\omega$ is $\theta = \frac{1}{2}\pi$ radians ( $90^\circ$ ).

Therefore, multiplying any complex number by $\omega$ results in:

A rotation of $\frac{\pi}{2}$ radians ( $90^\circ$ ) anticlockwise about the origin.
An enlargement (scale factor) of 2.

Answer

Multiplying by $\omega$ causes a rotation of $\frac{\pi}{2}$ radians anticlockwise and an enlargement with scale factor 2.

Final answer

Rotation of π/2 radians anticlockwise and enlargement with scale factor 2

Detailed explanation

Walkthrough

The geometric interpretation of multiplying a complex number $z$ by another complex number $w = re^{i\theta}$ is well-established. The modulus $r$ of $w$ acts as a scale factor for the distance of the point from the origin, while the argument $\theta$ of $w$ acts as the angle of rotation anticlockwise about the origin.

For $\omega = 2e^{\frac{1}{2}\pi i}$ :

The modulus is $2$ , which means every point is scaled to be twice as far from the origin. This is an enlargement with scale factor 2.
The argument is $\frac{\pi}{2}$ radians, which is $90^\circ$ . This means every point is rotated $90^\circ$ anticlockwise about the origin.

We can verify this with the points from part (b):

Point A ( $z_1$ ) is at modulus 3, angle $\frac{\pi}{4}$ . After multiplying by $\omega$ , point C ( $\omega z_1$ ) is at modulus $3 \times 2 = 6$ , angle $\frac{\pi}{4} + \frac{\pi}{2} = \frac{3\pi}{4}$ . This matches the rotation and enlargement.
Point B ( $z_2$ ) is at modulus 1.5, angle $\frac{\pi}{6}$ . After multiplying by $\omega$ , point D ( $\omega z_2$ ) is at modulus $1.5 \times 2 = 3$ , angle $\frac{\pi}{6} + \frac{\pi}{2} = \frac{2\pi}{3}$ . This also matches.

Key Takeaways

Multiplying by a complex number $re^{i\theta}$ geometrically transforms a point by rotating it by $\theta$ anticlockwise and scaling its distance from the origin by $r$ .
The modulus of the multiplier is the scale factor of the enlargement.
The argument of the multiplier is the angle of rotation.

Common Mistakes

Stating the rotation as clockwise instead of anticlockwise (or giving the wrong equivalent angle like $\frac{3\pi}{2}$ clockwise without justification).
Forgetting to mention the enlargement/scale factor and only stating the rotation.
Specifying a centre of rotation or enlargement other than the origin (the question does not require this, and stating an incorrect centre loses marks).

Things to Be Careful About

The rotation can be stated as $\frac{\pi}{2}$ radians anticlockwise or $\frac{3\pi}{2}$ radians clockwise; both are acceptable.
Do not need to state the centre of rotation or enlargement (it is the origin), but if you do state one, it must be correct (the origin), otherwise you may lose marks.

Techniques used

identify rotation angle from argument of multiplieridentify scale factor from modulus of multiplier

Q7MediumTrigonometry

(a)

Express $5\sin(x + \frac{1}{6}\pi) - 4\cos x$ in the form $R\sin(x - \alpha)$ , where $R > 0$ and $0 < \alpha < \frac{1}{2}\pi$ . State the exact value of $R$ and give the value of $\alpha$ correct to 3 decimal places.

4M Similar

DifficultyMedium

Worked solution

Approach

Expand the compound angle $\sin\left(x + \frac{\pi}{6}\right)$ using the addition formula, then match the resulting coefficients with those of $R\sin(x - \alpha)$ to find $R$ and $\alpha$ .

Working

Expand the compound angle:

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

Substitute into the expression:

5\sin\left(x + \frac{\pi}{6}\right) - 4\cos x = 5\left(\frac{\sqrt{3}}{2}\sin x + \frac{1}{2}\cos x\right) - 4\cos x = \frac{5\sqrt{3}}{2}\sin x - \frac{3}{2}\cos x

Now write the target form:

R\sin(x - \alpha) = R(\sin x \cos\alpha - \cos x \sin\alpha) = R\cos\alpha \sin x - R\sin\alpha \cos x

Match coefficients:

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

Square and add:

R^2 = \left(\frac{5\sqrt{3}}{2}\right)^2 + \left(\frac{3}{2}\right)^2 = \frac{75}{4} + \frac{9}{4} = 21

So $R = \sqrt{21}$ .

Divide the two equations:

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

Therefore:

\alpha = \tan^{-1}\left(\frac{\sqrt{3}}{5}\right) \approx 0.333

Answer

R = \sqrt{21}, \qquad \alpha \approx 0.333

Final answer

R = √21, α ≈ 0.333 radians

Detailed explanation

Walkthrough

We start by expanding the compound angle. The addition formula for sine says $\sin(A + B) = \sin A \cos B + \cos A \sin B$ . Here $A = x$ and $B = \frac{\pi}{6}$ , so we get $\sin x \cos\frac{\pi}{6} + \cos x \sin\frac{\pi}{6}$ . Since $\cos\frac{\pi}{6} = \frac{\sqrt{3}}{2}$ and $\sin\frac{\pi}{6} = \frac{1}{2}$ , this becomes $\frac{\sqrt{3}}{2}\sin x + \frac{1}{2}\cos x$ .

We then multiply by 5 and subtract $4\cos x$ : $5 \times \frac{\sqrt{3}}{2}\sin x + 5 \times \frac{1}{2}\cos x - 4\cos x = \frac{5\sqrt{3}}{2}\sin x + \frac{5}{2}\cos x - 4\cos x = \frac{5\sqrt{3}}{2}\sin x - \frac{3}{2}\cos x$ .

The target form is $R\sin(x - \alpha)$ . Using the subtraction formula, $R\sin(x - \alpha) = R(\sin x \cos\alpha - \cos x \sin\alpha) = R\cos\alpha \sin x - R\sin\alpha \cos x$ . We match the coefficients: the coefficient of $\sin x$ is $R\cos\alpha = \frac{5\sqrt{3}}{2}$ , and the coefficient of $\cos x$ is $-R\sin\alpha = -\frac{3}{2}$ , so $R\sin\alpha = \frac{3}{2}$ .

To find R, we use the Pythagorean identity: $R^2 = (R\cos\alpha)^2 + (R\sin\alpha)^2 = \left(\frac{5\sqrt{3}}{2}\right)^2 + \left(\frac{3}{2}\right)^2 = \frac{75}{4} + \frac{9}{4} = 21$ . So $R = \sqrt{21}$ .

To find $\alpha$ , we divide: $\tan\alpha = \frac{R\sin\alpha}{R\cos\alpha} = \frac{3/2}{5\sqrt{3}/2} = \frac{3}{5\sqrt{3}} = \frac{\sqrt{3}}{5}$ . Then $\alpha = \tan^{-1}\left(\frac{\sqrt{3}}{5}\right) \approx 0.333$ radians.

Key Takeaways

The R-form $R\sin(x - \alpha)$ (or $R\cos(x - \alpha)$ ) is a standard way to simplify $a\sin x + b\cos x$ .
R is found by Pythagoras: $R = \sqrt{a^2 + b^2}$ .
$\alpha$ is found by the tangent ratio: $\tan\alpha = \frac{b}{a}$ (with appropriate sign care).
The compound angle formula is essential for the expansion.

Common Mistakes

Incorrect expansion of $\sin\left(x + \frac{\pi}{6}\right)$ — forgetting the $\frac{1}{2}\cos x$ term.
Sign errors when matching: the minus in $R\sin(x - \alpha)$ means $R\sin\alpha$ is positive while the $\cos x$ coefficient is negative.
Confusing which coefficient is $\cos\alpha$ and which is $\sin\alpha$ .
Giving $\alpha$ in degrees instead of radians.

Things to Be Careful About

The mark scheme gives follow-through for R based on the student's expansion — so even if the expansion is slightly wrong, R can still be awarded if correctly computed from their coefficients.
The exact value $\tan\alpha = \frac{\sqrt{3}}{5}$ should be shown before the decimal.
$\alpha$ must be stated to 3 decimal places: 0.333.

Techniques used

expand a compound angle using the addition formulamatch coefficients with the R sin formcompute R using the Pythagorean identitycompute alpha using the tangent ratio

(b)

Hence solve the equation $5\sin(2\theta + \frac{1}{6}\pi) - 4\cos 2\theta = \sqrt{7}$ for $0 \leq \theta \leq \pi$ . Give your answers correct to 2 decimal places.

4M Similar

DifficultyMedium

Worked solution

Approach

Use the R-form result from part (a) to rewrite the equation, then solve $\sin(2\theta - \alpha) = \frac{1}{\sqrt{3}}$ for $\theta$ in the interval $0 \leq \theta \leq \pi$ .

Working

From part (a), with $\alpha = \tan^{-1}\left(\frac{\sqrt{3}}{5}\right) \approx 0.333$ :

5\sin\left(2\theta + \frac{\pi}{6}\right) - 4\cos 2\theta = \sqrt{21}\sin(2\theta - \alpha)

So the equation becomes:

\sqrt{21}\sin(2\theta - \alpha) = \sqrt{7}

\sin(2\theta - \alpha) = \frac{\sqrt{7}}{\sqrt{21}} = \frac{1}{\sqrt{3}}

Since $0 \leq \theta \leq \pi$ , we have $0 \leq 2\theta \leq 2\pi$ , so:

-\alpha \leq 2\theta - \alpha \leq 2\pi - \alpha

i.e. approximately $-0.333 \leq 2\theta - \alpha \leq 5.950$ .

Let $u = 2\theta - \alpha$ . Then $\sin u = \frac{1}{\sqrt{3}}$ .

The principal value is:

u_1 = \sin^{-1}\left(\frac{1}{\sqrt{3}}\right) \approx 0.6155

The second solution in the range is:

u_2 = \pi - u_1 \approx 2.5261

Both lie in $[-0.333, 5.950]$ , and adding $2\pi$ would take them out of range, so these are the only two solutions.

Now $2\theta = u + \alpha$ :

2\theta_1 = 0.6155 + 0.333 = 0.9485 \quad \Rightarrow \quad \theta_1 \approx 0.474 \approx 0.47

2\theta_2 = 2.5261 + 0.333 = 2.8591 \quad \Rightarrow \quad \theta_2 \approx 1.430 \approx 1.43

Both are in $[0, \pi]$ .

Answer

\theta \approx 0.47 \quad \text{or} \quad 1.43

Final answer

θ ≈ 0.47 or θ ≈ 1.43

Detailed explanation

Walkthrough

The word "hence" tells us to use the result from part (a). We replace the left-hand side of the equation with the R-form: $\sqrt{21}\sin(2\theta - \alpha) = \sqrt{7}$ .

Divide both sides by $\sqrt{21}$ : $\sin(2\theta - \alpha) = \frac{\sqrt{7}}{\sqrt{21}} = \frac{1}{\sqrt{3}}$ (since $\sqrt{21} = \sqrt{3}\sqrt{7}$ ).

Now we need to solve $\sin u = \frac{1}{\sqrt{3}}$ where $u = 2\theta - \alpha$ . The interval for $\theta$ is $0 \leq \theta \leq \pi$ , so $0 \leq 2\theta \leq 2\pi$ , and therefore $-\alpha \leq u \leq 2\pi - \alpha$ . With $\alpha \approx 0.333$ , this is approximately $-0.333 \leq u \leq 5.950$ .

The principal value is $u_1 = \sin^{-1}\left(\frac{1}{\sqrt{3}}\right) \approx 0.6155$ . The sine function is positive in the first and second quadrants, so the second solution is $u_2 = \pi - u_1 \approx 2.5261$ . Both are within the interval. Adding $2\pi$ would give values above 5.950, so there are no more solutions.

Finally, $2\theta = u + \alpha$ :

$2\theta_1 = 0.6155 + 0.333 = 0.9485$ , so $\theta_1 \approx 0.47$ .
$2\theta_2 = 2.5261 + 0.333 = 2.8591$ , so $\theta_2 \approx 1.43$ .

Both are in $[0, \pi]$ , so the answers are $\theta \approx 0.47$ and $\theta \approx 1.43$ .

Key Takeaways

The R-form is not just for rewriting — it's a powerful tool for solving equations of the form $a\sin x + b\cos x = c$ .
When solving $\sin u = k$ , always find all solutions in the relevant interval, not just the principal value.
The interval for the argument $u$ must be derived from the interval for the original variable.

Common Mistakes

Only finding the principal value and missing the second solution.
Forgetting to add $\alpha$ before dividing by 2.
Using degrees instead of radians.
Not checking that both solutions lie in $[0, \pi]$ .
Rounding prematurely, leading to answers that are off by 0.01.

Things to Be Careful About

The B1FT mark requires showing $\sin^{-1}\left(\frac{\sqrt{7}}{\sqrt{21}}\right)$ — this step must be visible.
Answers must be correct to 2 decimal places: 0.47 and 1.43.
The mark scheme notes "no others in the interval" — you must verify that adding $2\pi$ doesn't produce additional solutions in range.

Techniques used

substitute the R-form result from part (a)divide by R to isolate the sine termsolve a sine equation for all solutions in the intervalconvert solutions back to theta

Q8MediumVectors

With respect to the origin $O$ , the points $A$ and $B$ have position vectors $2\mathbf{i} + 4\mathbf{k}$ and $5\mathbf{i} + \mathbf{j} + 6\mathbf{k}$ respectively. The line $l_1$ passes through the points $A$ and $B$ .

(a)

Find a vector equation for the line $l_1$ .

2M Similar

DifficultyMedium-Easy

Worked solution

Approach

The direction vector of $l_1$ is the displacement from $A$ to $B$ : $\overrightarrow{AB} = \mathbf{b} - \mathbf{a}$ . A vector equation is then $\mathbf{r} = \mathbf{a} + \lambda \overrightarrow{AB}$ .

Working

\overrightarrow{AB} = (5\mathbf{i} + \mathbf{j} + 6\mathbf{k}) - (2\mathbf{i} + 4\mathbf{k}) = 3\mathbf{i} + \mathbf{j} + 2\mathbf{k}

Using $A$ as the base point:

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

Answer

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

Final answer

r = (2i + 4k) + λ(3i + j + 2k)

Detailed explanation

Walkthrough

We need a vector equation for the line through $A$ and $B$ . A line is determined by one point on it and a direction vector. The direction vector is the displacement from $A$ to $B$ , found by subtracting the position vector of $A$ from that of $B$ :

\overrightarrow{AB} = (5\mathbf{i} + \mathbf{j} + 6\mathbf{k}) - (2\mathbf{i} + 4\mathbf{k}) = 3\mathbf{i} + \mathbf{j} + 2\mathbf{k}

Then we use the position vector of $A$ as the base point and add a scalar multiple of the direction vector. As $\lambda$ varies, this traces every point on the line.

Key Takeaways

The direction vector of a line through two points is the difference of their position vectors.
A vector equation has the form $\mathbf{r} = \mathbf{a} + \lambda \mathbf{d}$ .
Any scalar multiple of a direction vector is also a valid direction vector.

Common Mistakes

Forgetting to write $\mathbf{r} = \dots$ ; the equation must be presented as a vector equation.
Using the position vector of the other point as the direction vector instead of the displacement.
Mixing up the order of subtraction; while this reverses the direction, it is still acceptable because $\lambda$ can be any real number.

Things to Be Careful About

The vector $3\mathbf{i} + \mathbf{j} + 2\mathbf{k}$ is the direction from $A$ to $B$ ; using $B$ as the base point with the same direction is also correct.
The parameter is often written as $\lambda$ or $\mu$ ; any letter is acceptable.

Techniques used

subtract position vectors to find a direction vectorwrite a vector equation from a point and a direction vector

(b)

The line $l_2$ has equation $\mathbf{r} = 2\mathbf{i} + \mathbf{j} + 5\mathbf{k} + \mu(\mathbf{i} + 2\mathbf{j} + 3\mathbf{k})$ .

Show that $l_1$ and $l_2$ do not intersect.

4M Similar

DifficultyMedium

Worked solution

Approach

Write both lines in component form. Equate two pairs of components to solve for $\lambda$ and $\mu$ . Then substitute into the third component; if the two values differ, the lines do not intersect.

Working

For $l_1$ using $A$ :

\begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 2 + 3\lambda \\ \lambda \\ 4 + 2\lambda \end{pmatrix}

For $l_2$ :

\begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 2 + \mu \\ 1 + 2\mu \\ 5 + 3\mu \end{pmatrix}

Equate the $x$ - and $y$ -components:

2 + \mu = 2 + 3\lambda, \qquad 1 + 2\mu = \lambda

From the first, $\mu = 3\lambda$ . Substituting into the second:

1 + 6\lambda = \lambda \quad\Rightarrow\quad \lambda = -\frac{1}{5}, \quad \mu = -\frac{3}{5}

Check the $z$ -components:

l_1: z = 4 + 2\left(-\frac{1}{5}\right) = \frac{18}{5}

l_2: z = 5 + 3\left(-\frac{3}{5}\right) = \frac{16}{5}

Since $\frac{18}{5} \neq \frac{16}{5}$ , no point satisfies all three components simultaneously.

Answer

The $x$ - and $y$ -components give $\lambda = -\frac{1}{5}$ , $\mu = -\frac{3}{5}$ , but the $z$ -components give $\frac{18}{5} \neq \frac{16}{5}$ . Therefore the lines do not intersect.

Final answer

lambda = -1/5, mu = -3/5; z-components 18/5 != 16/5, so no intersection.

Detailed explanation

Walkthrough

To test whether two lines intersect, write both in component form. If the lines intersect, there must be values of $\lambda$ and $\mu$ that make all three coordinates equal. Start by equating two pairs of components. Here equating $x$ - and $y$ -components gives two equations in $\lambda$ and $\mu$ . Solve them simultaneously. Then substitute the solutions into the third component. If the resulting $z$ -values are different, no common point exists, so the lines do not intersect.

Key Takeaways

Intersection of two lines requires a single pair of parameters satisfying all three component equations.
Solving two equations and checking the third is an efficient method.
Unequal third components prove the lines are skew (non-intersecting) in 3D.

Common Mistakes

Equating only two components and concluding intersection without checking the third.
Algebraic errors when solving simultaneous equations.
Using different parameter names for the two lines and confusing them.

Things to Be Careful About

The mark scheme requires showing the check of the third component explicitly.
Use the same line parameter consistently; e.g. $\lambda$ for $l_1$ and $\mu$ for $l_2$ .
A common pair of solutions is $\lambda = -\frac{1}{5}$ , $\mu = -\frac{3}{5}$ ; then the $z$ -components are $\frac{18}{5}$ and $\frac{16}{5}$ .

Techniques used

write both lines in component formsolve simultaneous equations from equated componentscheck the third component for consistency

(c)

Find the acute angle between the directions of $l_1$ and $l_2$ .

3M Similar

DifficultyMedium

Worked solution

Approach

Use the scalar product formula $\mathbf{a} \cdot \mathbf{b} = |\mathbf{a}||\mathbf{b}|\cos\theta$ to find the angle between the direction vectors. Since the cosine is positive, the angle obtained is acute.

Working

Direction vectors:

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

Scalar product:

\mathbf{d}_1 \cdot \mathbf{d}_2 = 3(1) + 1(2) + 2(3) = 3 + 2 + 6 = 11

Magnitudes:

|\mathbf{d}_1| = \sqrt{3^2 + 1^2 + 2^2} = \sqrt{14}, \qquad |\mathbf{d}_2| = \sqrt{1^2 + 2^2 + 3^2} = \sqrt{14}

Therefore:

\cos\theta = \frac{11}{\sqrt{14}\sqrt{14}} = \frac{11}{14}

\theta = \cos^{-1}\left(\frac{11}{14}\right) \approx 38.2^\circ \approx 0.667 \text{ radians}

Answer

The acute angle between the directions is $38.2^\circ$ (or $0.667$ radians) AWRT.

Final answer

38.2° (0.667 radians)

Detailed explanation

Walkthrough

The angle between two lines is the angle between their direction vectors. Use the scalar product formula $\mathbf{a} \cdot \mathbf{b} = |\mathbf{a}||\mathbf{b}|\cos\theta$ . Compute the dot product of the two direction vectors, then divide by the product of their magnitudes and take inverse cosine. Since the dot product is positive, the angle is acute; if it were negative, we would take the acute supplement.

Key Takeaways

The scalar product links vectors to angles.
The cosine formula gives the angle between vectors directly.
The direction vectors for $l_1$ and $l_2$ are $3\mathbf{i}+\mathbf{j}+2\mathbf{k}$ and $\mathbf{i}+2\mathbf{j}+3\mathbf{k}$ .

Common Mistakes

Forgetting to divide by the product of the magnitudes.
Using the position vectors instead of direction vectors.
Giving the answer in degrees when radians are expected, or vice versa; both are acceptable here but should be stated.

Things to Be Careful About

The scalar product is 11 and each magnitude is $\sqrt{14}$ , so $\cos\theta = \frac{11}{14}$ .
The acute angle is $38.2^\circ$ or $0.667$ radians AWRT.
If the dot product had been negative, the formula would give an obtuse angle; the acute angle would then be $180^\circ$ minus that value.

Techniques used

compute the scalar product of direction vectorscompute the magnitudes of direction vectorsapply inverse cosine to find the angle

Q9MediumIntegrationNumerical Solution of Equations

The constant $a$ is such that $\int_1^a 6x\ln x\,dx = 4$ .

(a)

Show that $a = \exp\left(\frac{1}{6}\left(\frac{5}{a^2} + 3\right)\right)$ , where $\exp(x)$ denotes $e^x$ .

5M Similar

DifficultyMedium

Worked solution

Approach

Integrate $6x\ln x$ by parts, evaluate from $1$ to $a$ , set the result equal to $4$ , then rearrange to isolate $\ln a$ and exponentiate.

Working

Let $u=\ln x$ , $\mathrm{d}v=6x\,\mathrm{d}x$ . Then $\mathrm{d}u=\frac{1}{x}\,\mathrm{d}x$ and $v=3x^2$ .

\begin{aligned} \int 6x\ln x\,\mathrm{d}x &= 3x^2\ln x - \int 3x^2\cdot\frac{1}{x}\,\mathrm{d}x \\ &= 3x^2\ln x - 3\int x\,\mathrm{d}x \\ &= 3x^2\ln x - \frac{3}{2}x^2. \end{aligned}

Evaluate from $1$ to $a$ :

\begin{aligned} \int_1^a 6x\ln x\,\mathrm{d}x &= \left[3x^2\ln x - \frac{3}{2}x^2\right]_1^a \\ &= 3a^2\ln a - \frac{3}{2}a^2 - \left(0-\frac{3}{2}\right) \\ &= 3a^2\ln a - \frac{3}{2}a^2 + \frac{3}{2}. \end{aligned}

Set equal to $4$ :

3a^2\ln a - \frac{3}{2}a^2 + \frac{3}{2} = 4

Multiply by $2$ :

6a^2\ln a - 3a^2 + 3 = 8

6a^2\ln a = 3a^2 + 5

\ln a = \frac{3a^2+5}{6a^2} = \frac{1}{6}\left(\frac{5}{a^2}+3\right)

Therefore

a = \exp\left(\frac{1}{6}\left(\frac{5}{a^2}+3\right)\right)

where $\exp(x)=e^x$ .

Answer

a = \exp\left(\frac{1}{6}\left(\frac{5}{a^2}+3\right)\right)

Final answer

a = exp((1/6)(5/a^2 + 3))

Detailed explanation

Walkthrough

We need to integrate a product $6x\ln x$ . The factor $\ln x$ becomes simple when differentiated, and $6x$ is easy to integrate, so integration by parts is the natural choice. Set $u=\ln x$ and $\mathrm{d}v=6x\,\mathrm{d}x$ ; then $\mathrm{d}u=\frac{1}{x}\,\mathrm{d}x$ and $v=3x^2$ . The formula $\int u\,\mathrm{d}v = uv - \int v\,\mathrm{d}u$ gives $3x^2\ln x - \int 3x^2\cdot\frac{1}{x}\,\mathrm{d}x$ . The key simplification is $\frac{x^2}{x}=x$ , so the remaining integral is $3\int x\,\mathrm{d}x = \frac{3}{2}x^2$ . When substituting the limits, remember $\ln 1=0$ , so the lower limit contributes $-\frac{3}{2}$ , and subtracting it gives $+\frac{3}{2}$ . Setting the result equal to $4$ and clearing fractions gives $6a^2\ln a = 3a^2+5$ . Dividing by $6a^2$ isolates $\ln a$ , and exponentiating both sides gives the required form.

Key Takeaways

This question tests integration by parts with a logarithmic factor, simplifying the resulting integral, evaluating a definite integral with limits, and rearranging an equation into exponential form.

Common Mistakes

A common mistake is to forget to simplify $\frac{x^2}{x}$ before integrating, or to make a sign error when subtracting the lower limit. Another is to stop at $3a^2\ln a - \frac{3}{2}a^2 + \frac{3}{2}=4$ without rearranging to the required exponential form. The mark scheme requires the integration to be completed before the limits are used.

Things to Be Careful About

At $x=1$ , $\ln 1=0$ , so the lower limit is not zero overall: it contributes $-\frac{3}{2}$ . Use exact fractions when rearranging, and remember that $\exp(x)$ means $e^x$ .

Techniques used

integrate by partssimplify the resulting integralevaluate definite integral with limitsrearrange to isolate ln a and exponentiate

(b)

Verify by calculation that $a$ lies between 2 and 2.1.

2M Similar

DifficultyMedium-Easy

Worked solution

Approach

Use the equation from part (a) in the form $f(a)=5$ where $f(x)=6x^2\ln x - 3x^2$ . Evaluate $f(2)$ and $f(2.1)$ and use a sign change.

Working

From part (a),

6a^2\ln a - 3a^2 = 5.

Let $f(x)=6x^2\ln x - 3x^2$ . Then

f(2) = 6(2^2)\ln 2 - 3(2^2) = 24\ln 2 - 12 \approx 4.636 < 5.

f(2.1) = 6(2.1^2)\ln 2.1 - 3(2.1^2) = 26.46\ln 2.1 - 13.23 \approx 6.402 > 5.

Since $f$ is continuous on $[2,2.1]$ and $f(2)<5<f(2.1)$ , the equation $f(a)=5$ has a root $a$ in the interval $(2,2.1)$ .

Answer

2 < a < 2.1

Final answer

2 < a < 2.1

Detailed explanation

Walkthrough

Part (a) gave $6a^2\ln a - 3a^2 = 5$ . Define $f(x)=6x^2\ln x - 3x^2$ ; then we need a root of $f(x)=5$ . Evaluate at the two endpoints. At $x=2$ , $f(2)=24\ln 2 - 12\approx4.636$ , which is below 5. At $x=2.1$ , $f(2.1)=26.46\ln 2.1 - 13.23\approx6.402$ , which is above 5. Since $f$ is continuous on the interval, the sign change from below 5 to above 5 means there is some $a$ between 2 and 2.1 with $f(a)=5$ .

Key Takeaways

This part uses the intermediate value theorem idea: a continuous function changing from below a target to above a target must cross it. It also shows the value of rewriting the integral equation as a function equation before evaluating.

Common Mistakes

A common mistake is to use a calculator to evaluate the original integral, which the mark scheme does not allow. Another is to evaluate the wrong expression, such as $3a^2\ln a - \frac{3}{2}a^2 + \frac{3}{2}$ without multiplying by 2, or to forget to state that the sign change justifies the conclusion.

Things to Be Careful About

Use $f(x)=6x^2\ln x - 3x^2$ , not the original integrand. Include approximate values to show the comparison with 5. Mention continuity so the sign-change argument is valid.

Techniques used

evaluate function at endpointsapply sign-change 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 iterative formula $a_{n+1} = \exp\left(\frac{1}{6}\left(\frac{5}{a_n^2}+3\right)\right)$ starting from $a_0=2$ , recording each iterate to 4 decimal places.

Working

Using $a_{n+1} = \exp\left(\frac{1}{6}\left(\frac{5}{a_n^2}+3\right)\right)$ :

\begin{array}{c|c} n & a_n \\ \hline 0 & 2 \\ 1 & 2.0306 \\ 2 & 2.0180 \\ 3 & 2.0231 \\ 4 & 2.0210 \\ 5 & 2.0219 \\ 6 & 2.0215 \\ 7 & 2.0217 \\ 8 & 2.0216 \end{array}

The iterates settle at $2.0216$ to 4 decimal places, so to 2 decimal places $a = 2.02$ .

Answer

a = 2.02

Final answer

a = 2.02

Detailed explanation

Walkthrough

Use the equation from part (a) as an iteration formula: $a_{n+1} = \exp\left(\frac{1}{6}\left(\frac{5}{a_n^2}+3\right)\right)$ . Starting from $a_0=2$ , each application produces a new approximation. The values to 4 decimal places are $2.0306, 2.0180, 2.0231, 2.0210, 2.0219, 2.0215, 2.0217, 2.0216$ . They oscillate with decreasing amplitude around about $2.0216$ . Since every iterate from $a_4$ onwards rounds to $2.02$ when rounded to 2 decimal places, the required answer is $a=2.02$ .

Key Takeaways

This part tests iterative methods: substituting a current approximation into a rearranged equation to get a better one, recording sufficient precision, and recognising when the sequence has converged to the required accuracy.

Common Mistakes

A common mistake is to stop after only one or two iterations, which may not be enough to justify the final rounding. Another is to give iterations to insufficient precision, or to round the final answer to 4 decimal places instead of 2. The mark scheme requires showing enough iterations to justify $2.02$ .

Things to Be Careful About

Keep at least 4 decimal places in each iterate. Use the correct iterative formula with $a_n^2$ in the denominator. Check that the final value rounds correctly to 2 decimal places.

Techniques used

apply iterative formulaiterate to sufficient precisionround final answer

Q10MediumAlgebraDifferential Equations

(a)

Find the quotient and remainder when $x^3 + 5x^2 - 2x - 15$ is divided by $x^2 - 3$ .

3M Similar

DifficultyMedium-Easy

Worked solution

Approach

Use polynomial long division. Divide the leading term of the dividend by the leading term of the divisor to build the quotient, then subtract the product from the dividend and repeat until the remaining degree is less than the divisor's degree.

Working

Divide the leading term:

x^3 \div x^2 = x

Multiply the divisor by $x$ :

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

Subtract from the dividend:

(x^3 + 5x^2 - 2x - 15) - (x^3 - 3x) = 5x^2 + x - 15

Divide the new leading term:

5x^2 \div x^2 = 5

Multiply the divisor by $5$ :

5(x^2 - 3) = 5x^2 - 15

Subtract:

(5x^2 + x - 15) - (5x^2 - 15) = x

The remainder $x$ has degree less than the divisor $x^2 - 3$ , so the division stops.

Check:

(x^2 - 3)(x + 5) + x = x^3 + 5x^2 - 3x - 15 + x = x^3 + 5x^2 - 2x - 15

Answer

Quotient: $x + 5$ ; remainder: $x$ .

Final answer

Quotient x + 5; remainder x

Detailed explanation

Walkthrough

We are dividing the cubic $x^3 + 5x^2 - 2x - 15$ by the quadratic $x^2 - 3$ . The quotient will be linear and the remainder will have degree less than 2.

Start by comparing leading terms: $x^3 \div x^2 = x$ . This is the first term of the quotient. Multiply the divisor by $x$ to get $x^3 - 3x$ , and subtract it from the dividend. The $x^3$ terms cancel, leaving $5x^2 + x - 15$ .

Now repeat: $5x^2 \div x^2 = 5$ . Multiply the divisor by $5$ to get $5x^2 - 15$ , and subtract. The $5x^2$ terms cancel, leaving $x$ . Since the degree of $x$ is 1, which is less than the degree 2 of the divisor, this is the remainder.

Thus the quotient is $x + 5$ and the remainder is $x$ .

Key Takeaways

Polynomial division is the same process as long division with numbers. The quotient is built term by term from the leading terms, and the process stops when the remainder has smaller degree than the divisor.

Common Mistakes

Forgetting to subtract the whole product, not just the leading term.
Sign errors when subtracting, especially with negative terms like $-3x$ .
Continuing the division after the remainder has degree less than the divisor.
Giving only the quotient without the remainder, or mixing the two up.

Things to Be Careful About

Align terms of the same degree carefully. The divisor $x^2 - 3$ has no $x$ term, so when multiplying by $x$ or $5$ , keep the $x$ terms in their correct columns. An unsupported answer may not receive full credit; show the division steps.

Techniques used

divide a polynomial by a quadratic divisorcompare leading terms to build the quotientsubtract products and track the remainderstate the quotient and remainder

(b)

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

\frac{dy}{dx} = \frac{x^3 + 5x^2 - 2x - 15}{6y(x^2 - 3)}

It is given that $y = 2$ when $x = 2$ .

Solve the differential equation to obtain an expression for $y^2$ in terms of $x$ .

5M Similar

DifficultyMedium

Worked solution

Approach

Separate the variables so that all $y$ terms are on one side and all $x$ terms are on the other. Use the result from part (a) to rewrite the rational expression in $x$ as a polynomial plus a remainder term, then integrate both sides. Finally, use the initial condition $y=2$ , $x=2$ to determine the constant of integration.

Working

Starting from

\frac{dy}{dx} = \frac{x^3 + 5x^2 - 2x - 15}{6y(x^2 - 3)}

multiply both sides by $6y$ and by $dx$ :

6y\,dy = \frac{x^3 + 5x^2 - 2x - 15}{x^2 - 3}\,dx

Integrate both sides:

\int 6y\,dy = \int \frac{x^3 + 5x^2 - 2x - 15}{x^2 - 3}\,dx

From part (a), the integrand can be written as

\frac{x^3 + 5x^2 - 2x - 15}{x^2 - 3} = x + 5 + \frac{x}{x^2 - 3}

Therefore

\int 6y\,dy = \int \left(x + 5 + \frac{x}{x^2 - 3}\right)dx

Integrating term by term:

3y^2 = \frac{1}{2}x^2 + 5x + \frac{1}{2}\ln(x^2 - 3) + C

Use $y=2$ , $x=2$ :

3(2)^2 = \frac{1}{2}(2)^2 + 5(2) + \frac{1}{2}\ln(2^2 - 3) + C

12 = 2 + 10 + \frac{1}{2}\ln 1 + C = 12 + C

Hence $C = 0$ . Therefore

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

Divide by 3:

y^2 = \frac{1}{6}x^2 + \frac{5}{3}x + \frac{1}{6}\ln(x^2 - 3)

Answer

y^2 = \frac{1}{6}x^2 + \frac{5}{3}x + \frac{1}{6}\ln(x^2 - 3)

Final answer

y^2 = (1/6)x^2 + (5/3)x + (1/6)ln(x^2 - 3)

Detailed explanation

Walkthrough

This is a separable differential equation. We want all $y$ 's with $dy$ on one side and all $x$ 's with $dx$ on the other. Multiplying by $6y$ and $dx$ gives $6y\,dy = \frac{x^3 + 5x^2 - 2x - 15}{x^2 - 3}\,dx$ .

The right-hand side is a rational function. From part (a), the numerator divided by $x^2 - 3$ gives $x+5$ with remainder $x$ , so the integrand becomes $x+5+\frac{x}{x^2 - 3}$ . This is much easier to integrate.

Integrate the left side: $\int 6y\,dy = 3y^2$ . On the right, $\int x\,dx = \frac{1}{2}x^2$ , $\int 5\,dx = 5x$ , and $\int \frac{x}{x^2 - 3}\,dx = \frac{1}{2}\ln(x^2 - 3)$ because the derivative of $x^2 - 3$ is $2x$ , so the numerator is half of the derivative of the denominator.

Then use the initial condition $y=2$ , $x=2$ . Substituting gives $12 = 2 + 10 + \frac{1}{2}\ln 1 + C$ . Since $\ln 1 = 0$ , this gives $C = 0$ . Finally divide by 3 to solve for $y^2$ .

Key Takeaways

Separable differential equations are solved by separating variables and integrating both sides.
Polynomial division can simplify the integrand of a rational function before integration.
An integral of the form $\int \frac{k f'(x)}{f(x)}\,dx$ gives $k\ln|f(x)| + C$ .
Initial conditions determine the particular constant of integration.

Common Mistakes

Forgetting to integrate the left side correctly: $\int 6y\,dy = 3y^2$ , not $6y^2$ .
Omitting the constant of integration before using the initial condition.
Missing the factor $\frac{1}{2}$ in $\int \frac{x}{x^2 - 3}\,dx$ .
Forgetting to divide by 3 at the end when solving for $y^2$ .
Not using the result from part (a) to simplify the integrand.

Things to Be Careful About

The logarithm requires $x^2 - 3 > 0$ ; the initial point $x=2$ satisfies this.
Since $\ln 1 = 0$ , the constant becomes 0 here, but do not assume that in general.
In the mark scheme, the $\frac{1}{2}x^2 + 5x$ term follows from the linear quotient and the $\frac{1}{2}\ln(x^2 - 3)$ term follows from the remainder term $x/(x^2 - 3)$ .
Show the separation step explicitly to earn the method mark.

Techniques used

separate variables in a first-order differential equationrewrite the rational integrand using polynomial divisionintegrate term by term including kf'(x)/f(x)apply the initial condition to find the constantsolve for y^2

Q11Medium-HardDifferentiationIntegration

Fig. 1

The diagram shows the curve $y = \cos x\sqrt{\sin 2x}$ for $0 \leq x \leq \frac{1}{2}\pi$ . The curve has a maximum point at $M$ , where $x = a$ .

(a)

Find the exact value of $a$ .

6M Similar

DifficultyMedium-Hard

Worked solution

Approach

Differentiate $y = \cos x \sqrt{\sin 2x}$ using the product rule, treating $\sqrt{\sin 2x}$ as a composite function. Set the derivative to zero, clear the square root, and use the compound angle identity to reduce the equation to a single trig function before solving within the given domain.

Working

Using the product rule:

\frac{\mathrm{d}y}{\mathrm{d}x} = -\sin x \cdot \sqrt{\sin 2x} + \cos x \cdot \frac{\mathrm{d}}{\mathrm{d}x}\sqrt{\sin 2x}

Differentiate $\sqrt{\sin 2x}$ using the chain rule:

\frac{\mathrm{d}}{\mathrm{d}x}\sqrt{\sin 2x} = \frac{1}{2\sqrt{\sin 2x}} \cdot 2\cos 2x = \frac{\cos 2x}{\sqrt{\sin 2x}}

Combining:

\frac{\mathrm{d}y}{\mathrm{d}x} = -\sin x \sqrt{\sin 2x} + \frac{\cos x \cos 2x}{\sqrt{\sin 2x}}

At the maximum $M$ on $0 \leq x \leq \frac{1}{2}\pi$ , the derivative is zero:

-\sin x \sqrt{\sin 2x} + \frac{\cos x \cos 2x}{\sqrt{\sin 2x}} = 0

Multiply through by $\sqrt{\sin 2x}$ to clear the radical:

-\sin x \sin 2x + \cos x \cos 2x = 0

Rearranging:

\cos x \cos 2x - \sin x \sin 2x = 0

Apply the compound angle identity $\cos(A+B) = \cos A \cos B - \sin A \sin B$ with $A = x$ and $B = 2x$ :

\cos 3x = 0

In the given range $0 \leq x \leq \frac{1}{2}\pi$ , we have $0 \leq 3x \leq \frac{3}{2}\pi$ . The only interior zero of $\cos 3x$ in this range is at $3x = \frac{1}{2}\pi$ , so:

x = \frac{1}{6}\pi

Answer

a = \frac{1}{6}\pi

Final answer

a = π/6

Detailed explanation

Walkthrough

We want to find where the curve $y = \cos x \sqrt{\sin 2x}$ has its maximum on the interval $0 \leq x \leq \frac{1}{2}\pi$ . The standard method is to differentiate $y$ and set the derivative to zero, then check that the resulting point is indeed a maximum.

The function is a product of two factors, $\cos x$ and $\sqrt{\sin 2x}$ , so the product rule is the natural tool. The first factor differentiates directly to $-\sin x$ . The second factor is a composite: a square root wrapped around the trig function $\sin 2x$ . The chain rule tells us to differentiate the outer function (square root) and then multiply by the derivative of the inner function ( $\sin 2x \to 2\cos 2x$ ). The factor of $2$ from the chain rule cancels with the $\frac{1}{2}$ from differentiating the square root, leaving the clean expression $\frac{\cos 2x}{\sqrt{\sin 2x}}$ .

Setting the derivative to zero gives a single equation containing a square root. Multiplying through by $\sqrt{\sin 2x}$ removes the radical and produces a much simpler expression: $\cos x \cos 2x - \sin x \sin 2x = 0$ . This is precisely the form of the compound angle identity $\cos(A+B) = \cos A \cos B - \sin A \sin B$ , and recognising this is the key step that converts the equation into $\cos 3x = 0$ .

In the domain $0 \leq x \leq \frac{1}{2}\pi$ , the variable $3x$ lies in $[0, \frac{3}{2}\pi]$ . The cosine is zero at $\frac{1}{2}\pi$ within this range, giving $x = \frac{1}{6}\pi$ . The other zero of cosine in this range, $\frac{3}{2}\pi$ , corresponds to $x = \frac{1}{2}\pi$ , which is an endpoint where the curve meets the $x$ -axis, not the location of the maximum.

Key Takeaways

The product rule is required whenever a function is expressed as a product of two or more factors.
The chain rule lets us differentiate composite functions like $\sqrt{\sin 2x}$ by working from the outside in.
Recognising trigonometric identities (here, the compound angle identity) is often the key step that makes the equation tractable.
Always restrict solutions to the stated domain.

Common Mistakes

Differentiating only one factor of the product and forgetting the other term.
Mistakes in the chain rule, such as forgetting the factor of $2$ from differentiating $\sin 2x$ .
Not multiplying through by $\sqrt{\sin 2x}$ , leaving an awkward expression that is hard to simplify further.
Failing to recognise the compound angle identity and being unable to proceed.
Solving $\cos 3x = 0$ without considering the domain restrictions on $x$ .

Things to Be Careful About

The domain is $0 \leq x \leq \frac{1}{2}\pi$ , so only one value of $3x$ in the relevant range gives a valid solution.
$\sin 2x \geq 0$ must hold on the interval for $\sqrt{\sin 2x}$ to be real; this is satisfied in the given range.
The final answer must be exact and given in radians.

Techniques used

apply product ruleapply chain rule to compositeset derivative to zeroapply compound angle identitysolve trigonometric equation within domain

(b)

The region enclosed between the $x$ -axis and the curve is rotated through $2\pi$ radians about the $x$ -axis.

Find the exact volume of the solid generated.

5M Similar

DifficultyMedium

Worked solution

Approach

Use the volume of revolution formula $V = \pi \int y^2 \, \mathrm{d}x$ between $0$ and $\frac{1}{2}\pi$ . Simplify the integrand with the double angle identity $\sin 2x = 2\sin x \cos x$ , then evaluate by the substitution $u = \cos x$ .

Working

The volume of revolution about the $x$ -axis is:

V = \pi \int_0^{\pi/2} y^2 \, \mathrm{d}x = \pi \int_0^{\pi/2} \cos^2 x \sin 2x \, \mathrm{d}x

Apply the double angle identity $\sin 2x = 2\sin x \cos x$ :

V = \pi \int_0^{\pi/2} \cos^2 x \cdot 2\sin x \cos x \, \mathrm{d}x = 2\pi \int_0^{\pi/2} \cos^3 x \sin x \, \mathrm{d}x

Let $u = \cos x$ , so $\mathrm{d}u = -\sin x \, \mathrm{d}x$ .

Update the limits:

When $x = 0$ : $u = \cos 0 = 1$ .
When $x = \frac{1}{2}\pi$ : $u = \cos \frac{1}{2}\pi = 0$ .

Substitute:

V = 2\pi \int_1^0 u^3 \cdot (-\mathrm{d}u) = 2\pi \int_0^1 u^3 \, \mathrm{d}u

Integrate:

V = 2\pi \left[\frac{u^4}{4}\right]_0^1 = 2\pi \left(\frac{1}{4} - 0\right) = \frac{\pi}{2}

Answer

V = \frac{1}{2}\pi

Final answer

V = π/2

Detailed explanation

Walkthrough

For a region between a curve $y = f(x)$ and the $x$ -axis, rotated through $2\pi$ about the $x$ -axis, the volume of the resulting solid is $V = \pi \int y^2 \, \mathrm{d}x$ between the relevant $x$ -limits. The curve meets the $x$ -axis at $x = 0$ and at $x = \frac{1}{2}\pi$ (where $\cos x = 0$ ), so the limits are $0$ and $\frac{1}{2}\pi$ .

Squaring $y = \cos x \sqrt{\sin 2x}$ gives $y^2 = \cos^2 x \sin 2x$ , so the integral becomes $V = \pi \int_0^{\pi/2} \cos^2 x \sin 2x \, \mathrm{d}x$ . This looks awkward to integrate directly, but the double angle identity $\sin 2x = 2\sin x \cos x$ transforms it into $2\pi \int_0^{\pi/2} \cos^3 x \sin x \, \mathrm{d}x$ , which is a perfect candidate for the substitution $u = \cos x$ .

With $u = \cos x$ and $\mathrm{d}u = -\sin x \, \mathrm{d}x$ , the integrand $\cos^3 x \sin x \, \mathrm{d}x$ becomes $-u^3 \, \mathrm{d}u$ . The new limits come from evaluating $\cos x$ at the original limits: $\cos 0 = 1$ and $\cos \frac{1}{2}\pi = 0$ . So the integral becomes $2\pi \int_1^0 -u^3 \, \mathrm{d}u = 2\pi \int_0^1 u^3 \, \mathrm{d}u$ .

The two minus signs (one from the swapped limits, one from $\mathrm{d}u = -\sin x \, \mathrm{d}x$ ) cancel, leaving a clean positive integral. Evaluating $\int_0^1 u^3 \, \mathrm{d}u = \frac{1}{4}$ and multiplying by $2\pi$ gives $V = \frac{\pi}{2}$ .

Key Takeaways

The volume of revolution about the $x$ -axis is $V = \pi \int y^2 \, \mathrm{d}x$ between the curve's $x$ -intercepts.
The double angle identity $\sin 2x = 2\sin x \cos x$ can turn a difficult integral into a simple substitution.
When using $u$ -substitution on a definite integral, update the limits to the new variable rather than converting back.
A sign from a negative differential (like $\mathrm{d}u = -\sin x \, \mathrm{d}x$ ) often cancels with reversed limits.

Common Mistakes

Forgetting the leading factor of $\pi$ in the volume formula; the mark scheme explicitly allows the first four marks but withholds the final A mark if the $\pi$ is missing throughout.
Not applying the double angle identity and being unable to integrate $\cos^2 x \sin 2x$ directly.
Forgetting to change the limits of integration when substituting.
Sign errors when handling the negative differential $\mathrm{d}u = -\sin x \, \mathrm{d}x$ .
Miscounting the limits: $\cos \frac{1}{2}\pi = 0$ , not $1$ .

Things to Be Careful About

The factor of $\pi$ is essential — the mark scheme warns that omitting it throughout forfeits the final accuracy mark.
When the substitution reverses the limits, the negative from $\mathrm{d}u$ and the negative from swapping the limits cancel each other, but students should handle each sign explicitly rather than relying on memory.
The final answer must be exact, in terms of $\pi$ .

Techniques used

set up volume of revolution integralapply double angle identity to integrandintegrate by u-substitutionupdate limits of integrationevaluate definite integral

Mathematics 9709/31 — May/June 2025

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